Differential forms provide a modern, coordinate-free framework for calculus on manifolds, unifying classical vector analysis with the language of algebraic topology. Rather than working with gradient, curl, and divergence as separate operations on vector fields, this course develops a single geometric object—the differential form—whose behavior under integration, differentiation, and pullback captures all of multivariable calculus in a unified system. The course explores how forms encode information about smooth spaces and reveals deep connections between the local analytic properties of a manifold (captured by the [exterior derivative](/theorems/1525)) and its global topological structure. The chapters progress systematically from foundations to applications. We begin with exterior algebra, the algebraic machinery underlying forms, then introduce differential forms concretely on open subsets of ℝⁿ where calculus intuition is strongest. The [exterior derivative](/theorems/1525) emerges as the natural generalization of classical differentiation, and the pullback operation shows how forms behave under smooth maps. Once these tools are mastered, we extend the theory to smooth manifolds—abstract curved spaces where these ideas become essential—and develop integration via orientation and the generalised Stokes theorem. The Poincaré lemma establishes when closed forms are exact, leading naturally to de Rham cohomology: a way of extracting topological invariants from the algebra of forms. De Rham's theorem, the course's central result, reveals that the cohomology computed from differential forms coincides exactly with singular cohomology, a fundamental topological invariant. This identification shows that smooth and purely topological information are intimately related, and the final chapters explore how this perspective illuminates problems across differential geometry, physics, and topology. Throughout, the framework develops the conceptual tools that professional mathematicians and physicists use daily to understand smooth manifolds. # Introduction This opening chapter sets the perspective for the course. Differential forms give a single language for line integrals, surface flux, volume integration, change of variables, and the boundary terms that appear in Stokes-type theorems. The first aim is to replace several coordinate-dependent vector calculus constructions by one functorial object. The second aim is to understand why the equation $d\omega = 0$ carries topological information about the space on which $\omega$ lives. The course starts from linear algebra on finite-dimensional real vector spaces and then moves to smooth manifolds. Smooth manifolds will be used as spaces that locally look like open subsets of $\mathbb R^n$, so the early chapters develop every construction first on open sets. Later chapters globalise these constructions by checking that they behave correctly under change of coordinates. ## Why Forms Replace Vector Calculus Notation What goes wrong if line integrals, surface integrals, and volume integrals are treated as unrelated operations? The formulas of vector calculus depend heavily on the coordinates chosen to write them down. A line integral such as $P\,dx + Q\,dy$ does not transform like a vector field; it transforms by pullback along parametrisations. Differential forms are designed so that the integrand already contains the transformation rule needed for integration. [definition: Differential Form On An Open Set] Let $U \subset \mathbb R^n$ be open. A differential $k$-form on $U$ is a smooth assignment $\omega$ sending each point $x \in U$ to an alternating $k$-[linear map](/page/Linear%20Map) \begin{align*} \omega_x : (\mathbb R^n)^k \to \mathbb R. \end{align*} The [vector space](/page/Vector%20Space) of smooth $k$-forms on $U$ is denoted $\Omega^k(U)$. For $k=0$, set $\Omega^0(U)=C^\infty(U)$. [/definition] The word alternating means that the value changes sign when two input vectors are exchanged and is zero when two input vectors are equal. Thus a $1$-form eats one tangent vector, a $2$-form eats an ordered pair of tangent vectors, and a top-degree $n$-form eats an ordered basis of $\mathbb R^n$. [example: Line Integral As A One Form] Let $U\subset \mathbb R^2$ be open, and let \begin{align*} \omega=P\,dx+Q\,dy\in\Omega^1(U), \end{align*} where $P,Q\in C^\infty(U)$. Let $\gamma:[a,b]\to U$ be a smooth curve, and write \begin{align*} \gamma(t)=(\gamma_1(t),\gamma_2(t)),\qquad \dot\gamma(t)=(\dot\gamma_1(t),\dot\gamma_2(t)). \end{align*} For $p\in U$ and $v=(v_1,v_2)\in\mathbb R^2$, the coordinate covectors satisfy \begin{align*} dx_p(v)=v_1,\qquad dy_p(v)=v_2. \end{align*} The integral of $\omega$ along $\gamma$ is \begin{align*} \int_\gamma\omega =\int_a^b\left(P(\gamma(t))\dot\gamma_1(t)+Q(\gamma(t))\dot\gamma_2(t)\right)\,dt. \end{align*} By the definition of integrating a $1$-form over a parametrised curve, \begin{align*} \int_\gamma\omega =\int_a^b \omega_{\gamma(t)}(\dot\gamma(t))\,dt. \end{align*} For each $t\in[a,b]$, evaluate the covector $\omega_{\gamma(t)}$ on the velocity vector: \begin{align*} \omega_{\gamma(t)}(\dot\gamma(t)) &=\left(P(\gamma(t))\,dx_{\gamma(t)}+Q(\gamma(t))\,dy_{\gamma(t)}\right)(\dot\gamma(t)) \\ &=P(\gamma(t))\,dx_{\gamma(t)}(\dot\gamma(t)) +Q(\gamma(t))\,dy_{\gamma(t)}(\dot\gamma(t)) \\ &=P(\gamma(t))\,dx_{\gamma(t)}(\dot\gamma_1(t),\dot\gamma_2(t)) +Q(\gamma(t))\,dy_{\gamma(t)}(\dot\gamma_1(t),\dot\gamma_2(t)) \\ &=P(\gamma(t))\dot\gamma_1(t)+Q(\gamma(t))\dot\gamma_2(t). \end{align*} Substituting this pointwise identity into the defining integral gives \begin{align*} \int_\gamma\omega &=\int_a^b \omega_{\gamma(t)}(\dot\gamma(t))\,dt \\ &=\int_a^b\left(P(\gamma(t))\dot\gamma_1(t)+Q(\gamma(t))\dot\gamma_2(t)\right)\,dt. \end{align*} The parametrised curve contributes the velocity vector $\dot\gamma(t)$, and the $1$-form contributes the covector $P(\gamma(t))\,dx_{\gamma(t)}+Q(\gamma(t))\,dy_{\gamma(t)}$. Feeding the velocity into this covector produces exactly the coordinate integrand $P(\gamma(t))\dot\gamma_1(t)+Q(\gamma(t))\dot\gamma_2(t)$, so the ordinary planar line integral is naturally the integral of a $1$-form. [/example] Coordinate notation is useful because it gives a basis for all forms on an open subset of Euclidean space. If $x=(x_1,\dots,x_n)$ are the standard coordinates, then $dx_i$ denotes the $i$-th coordinate covector. [example: Coordinate Basis For Forms] Let $U\subset \mathbb R^n$ be open, and let \begin{align*} \omega\in\Omega^k(U). \end{align*} Write $e_1,\dots,e_n$ for the standard basis of $\mathbb R^n$, and write $dx_1,\dots,dx_n$ for the dual coordinate covectors, so \begin{align*} dx_i(e_j)= \begin{cases} 1,& i=j,\\ 0,& i\ne j. \end{cases} \end{align*} For a strictly increasing multi-index $I=(i_1,\dots,i_k)$, set \begin{align*} dx_I:=dx_{i_1}\wedge\cdots\wedge dx_{i_k}. \end{align*} There are unique smooth functions $a_I\in C^\infty(U)$ such that \begin{align*} \omega=\sum_{1\le i_1<\cdots<i_k\le n} a_{i_1\cdots i_k}\,dx_{i_1}\wedge\cdots\wedge dx_{i_k}. \end{align*} For $n=3$ and $k=2$, this becomes \begin{align*} \omega=A\,dy\wedge dz+B\,dz\wedge dx+C\,dx\wedge dy. \end{align*} Fix $x\in U$. Since $\omega_x$ is an alternating $k$-[linear map](/page/Linear%20Map), its values on arbitrary vectors are determined by its values on ordered $k$-tuples of basis vectors. Define \begin{align*} a_{i_1\cdots i_k}(x):=\omega_x(e_{i_1},\dots,e_{i_k}) \end{align*} for every $1\le i_1<\cdots<i_k\le n$. Now evaluate the coordinate wedge $dx_I$ on an ordered basis tuple $(e_{j_1},\dots,e_{j_k})$. By the defining determinant formula for the wedge of coordinate covectors, \begin{align*} dx_I(e_{j_1},\dots,e_{j_k}) &=(dx_{i_1}\wedge\cdots\wedge dx_{i_k})(e_{j_1},\dots,e_{j_k})\\ &=\det\left(dx_{i_p}(e_{j_q})\right)_{1\le p,q\le k}. \end{align*} If $(j_1,\dots,j_k)=(i_1,\dots,i_k)$, the matrix is the identity matrix, so \begin{align*} dx_I(e_{i_1},\dots,e_{i_k})=\det(I_k)=1. \end{align*} If some $j_q$ is not one of $i_1,\dots,i_k$, then the $q$-th column of the matrix has all entries $0$, so \begin{align*} dx_I(e_{j_1},\dots,e_{j_k})=0. \end{align*} If $(j_1,\dots,j_k)$ is a permutation of $(i_1,\dots,i_k)$, then the matrix is the corresponding permutation matrix, so \begin{align*} dx_I(e_{j_1},\dots,e_{j_k})=\operatorname{sgn}(j_1,\dots,j_k). \end{align*} Set \begin{align*} \widetilde\omega_x =\sum_{1\le i_1<\cdots<i_k\le n} a_{i_1\cdots i_k}(x)\,dx_{i_1}\wedge\cdots\wedge dx_{i_k}. \end{align*} For every strictly increasing tuple $J=(j_1,\dots,j_k)$, \begin{align*} \widetilde\omega_x(e_{j_1},\dots,e_{j_k}) &=\sum_I a_I(x)\,dx_I(e_{j_1},\dots,e_{j_k})\\ &=a_J(x)\\ &=\omega_x(e_{j_1},\dots,e_{j_k}). \end{align*} Both $\widetilde\omega_x$ and $\omega_x$ are alternating and $k$-linear, so equality on the ordered basis tuples gives equality on all $k$ input vectors. Hence \begin{align*} \omega_x=\sum_{1\le i_1<\cdots<i_k\le n} a_{i_1\cdots i_k}(x)\,dx_{i_1}\wedge\cdots\wedge dx_{i_k}. \end{align*} Since this holds for every $x\in U$, \begin{align*} \omega=\sum_{1\le i_1<\cdots<i_k\le n} a_{i_1\cdots i_k}\,dx_{i_1}\wedge\cdots\wedge dx_{i_k}. \end{align*} The functions $a_I$ are smooth because they are the coordinate coefficient functions of the smooth form $\omega$. For uniqueness, suppose also that \begin{align*} \omega=\sum_I b_I\,dx_I. \end{align*} Evaluating both sides at $x$ on $(e_{i_1},\dots,e_{i_k})$ gives \begin{align*} \omega_x(e_{i_1},\dots,e_{i_k}) &=\sum_J b_J(x)\,dx_J(e_{i_1},\dots,e_{i_k})\\ &=b_{i_1\cdots i_k}(x). \end{align*} But by the definition of $a_{i_1\cdots i_k}$, \begin{align*} \omega_x(e_{i_1},\dots,e_{i_k})=a_{i_1\cdots i_k}(x). \end{align*} Therefore \begin{align*} a_{i_1\cdots i_k}(x)=b_{i_1\cdots i_k}(x) \end{align*} for every $x\in U$ and every strictly increasing multi-index, so the coefficients are unique. For $n=3$ and $k=2$, the strictly increasing pairs are \begin{align*} (1,2),\qquad (1,3),\qquad (2,3). \end{align*} Thus every $2$-form on $U\subset\mathbb R^3$ has the form \begin{align*} \omega=a_{12}\,dx\wedge dy+a_{13}\,dx\wedge dz+a_{23}\,dy\wedge dz. \end{align*} Using $dx\wedge dz=-dz\wedge dx$, this can be rewritten as \begin{align*} \omega &=a_{23}\,dy\wedge dz-a_{13}\,dz\wedge dx+a_{12}\,dx\wedge dy. \end{align*} If we set \begin{align*} A=a_{23},\qquad B=-a_{13},\qquad C=a_{12}, \end{align*} then \begin{align*} \omega=A\,dy\wedge dz+B\,dz\wedge dx+C\,dx\wedge dy. \end{align*} Coordinate wedge products form the natural coordinate basis for differential forms. The coefficients are not arbitrary decorations: each coefficient is recovered by feeding $\omega_x$ the corresponding ordered coordinate vectors. In dimension $3$, the three basic $2$-forms $dy\wedge dz$, $dz\wedge dx$, and $dx\wedge dy$ record the three oriented coordinate planes. [/example] ## The Operations This Course Builds Which operations on integrands survive change of coordinates and still remember orientation, dimension, and boundary? The course develops three fundamental constructions: the wedge product, the [exterior derivative](/theorems/1525), and pullback. Together they replace dot products, cross products, gradients, curls, divergences, and Jacobian determinants by coordinate-free operations. [definition: Wedge Product] Let $U \subset \mathbb R^n$ be open. The wedge product is a bilinear operation \begin{align*} \wedge : \Omega^p(U)\times \Omega^q(U)\to \Omega^{p+q}(U) \end{align*} whose value at each point is the exterior product of alternating covectors. On coordinate $1$-forms it is determined by multilinearity and the relations \begin{align*} dx_i\wedge dx_j=-dx_j\wedge dx_i,\qquad dx_i\wedge dx_i=0. \end{align*} [/definition] The wedge product encodes signed area, signed volume, and higher-dimensional oriented content. The relation $dx_i\wedge dx_i=0$ is the algebraic expression of the fact that a parallelogram with two equal directions has zero area. To do calculus with forms, one also needs an operation that differentiates their coefficient functions while raising degree by one. The exterior derivative is designed to generalize gradient, curl, and divergence in a single coordinate-independent formalism. [definition: Exterior Derivative] Let $U \subset \mathbb R^n$ be open. The [exterior derivative](/theorems/1525) is the family of linear maps \begin{align*} d: \Omega^k(U)\to \Omega^{k+1}(U) \end{align*} defined in coordinates by \begin{align*} d\left(\sum_I a_I\,dx_{i_1}\wedge\cdots\wedge dx_{i_k}\right) =\sum_I\sum_{j=1}^n \frac{\partial a_I}{\partial x_j}\,dx_j\wedge dx_{i_1}\wedge\cdots\wedge dx_{i_k}, \end{align*} where $I=(i_1,\dots,i_k)$ ranges over strictly increasing multi-indices. [/definition] For a function $f\in C^\infty(U)$, this gives $df=\sum_i (\partial f/\partial x_i)\,dx_i$. For a $1$-form on $\mathbb R^3$, it packages the curl; for a $2$-form on $\mathbb R^3$, it packages the divergence. The coordinate formula would be much less useful if it were only a way to compute examples. To support calculus with forms, it must interact predictably with wedge products, and applying it twice must create no new obstruction beyond equality of mixed partial derivatives. These are the identities that make exterior calculus behave like a coherent differential algebra. [quotetheorem:1525] [citeproof:1525] The identity $d^2=0$ is the algebraic seed of de Rham cohomology. It says that every form produced by an [exterior derivative](/theorems/1525) automatically satisfies a compatibility equation. [example: Curl And Divergence Inside Exterior Derivative] Work on $\mathbb R^3$ with standard coordinates $(x,y,z)$ and coordinate $1$-forms $dx,dy,dz$. Let $P,Q,R,A,B,C\in C^\infty(\mathbb R^3)$, and set \begin{align*} \alpha=P\,dx+Q\,dy+R\,dz \end{align*} and \begin{align*} \beta=A\,dy\wedge dz+B\,dz\wedge dx+C\,dx\wedge dy. \end{align*} We use the wedge relations \begin{align*} dx\wedge dx=dy\wedge dy=dz\wedge dz=0 \end{align*} and \begin{align*} dy\wedge dx=-dx\wedge dy,\qquad dz\wedge dy=-dy\wedge dz,\qquad dx\wedge dz=-dz\wedge dx. \end{align*} The [exterior derivative](/theorems/1525) of the $1$-form $\alpha$ records the curl components: \begin{align*} d\alpha &=\left(\frac{\partial R}{\partial y}-\frac{\partial Q}{\partial z}\right)dy\wedge dz +\left(\frac{\partial P}{\partial z}-\frac{\partial R}{\partial x}\right)dz\wedge dx +\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dx\wedge dy. \end{align*} The [exterior derivative](/theorems/1525) of the $2$-form $\beta$ records the divergence component: \begin{align*} d\beta=\left(\frac{\partial A}{\partial x}+\frac{\partial B}{\partial y}+\frac{\partial C}{\partial z}\right)dx\wedge dy\wedge dz. \end{align*} By linearity of $d$ and the coordinate definition of the [exterior derivative](/theorems/1525), \begin{align*} d\alpha &=d(P\,dx)+d(Q\,dy)+d(R\,dz)\\ &=\left(\frac{\partial P}{\partial x}dx+\frac{\partial P}{\partial y}dy+\frac{\partial P}{\partial z}dz\right)\wedge dx\\ &\quad+\left(\frac{\partial Q}{\partial x}dx+\frac{\partial Q}{\partial y}dy+\frac{\partial Q}{\partial z}dz\right)\wedge dy\\ &\quad+\left(\frac{\partial R}{\partial x}dx+\frac{\partial R}{\partial y}dy+\frac{\partial R}{\partial z}dz\right)\wedge dz. \end{align*} Expand each wedge product term by term: \begin{align*} d\alpha &=\frac{\partial P}{\partial x}dx\wedge dx +\frac{\partial P}{\partial y}dy\wedge dx +\frac{\partial P}{\partial z}dz\wedge dx\\ &\quad+\frac{\partial Q}{\partial x}dx\wedge dy +\frac{\partial Q}{\partial y}dy\wedge dy +\frac{\partial Q}{\partial z}dz\wedge dy\\ &\quad+\frac{\partial R}{\partial x}dx\wedge dz +\frac{\partial R}{\partial y}dy\wedge dz +\frac{\partial R}{\partial z}dz\wedge dz. \end{align*} The repeated-coordinate terms vanish, so \begin{align*} d\alpha &=\frac{\partial P}{\partial y}dy\wedge dx +\frac{\partial P}{\partial z}dz\wedge dx +\frac{\partial Q}{\partial x}dx\wedge dy +\frac{\partial Q}{\partial z}dz\wedge dy\\ &\quad+\frac{\partial R}{\partial x}dx\wedge dz +\frac{\partial R}{\partial y}dy\wedge dz. \end{align*} Rewrite every term in the ordered $2$-form basis $dy\wedge dz$, $dz\wedge dx$, $dx\wedge dy$: \begin{align*} d\alpha &=-\frac{\partial P}{\partial y}dx\wedge dy +\frac{\partial P}{\partial z}dz\wedge dx +\frac{\partial Q}{\partial x}dx\wedge dy -\frac{\partial Q}{\partial z}dy\wedge dz\\ &\quad-\frac{\partial R}{\partial x}dz\wedge dx +\frac{\partial R}{\partial y}dy\wedge dz. \end{align*} Collect coefficients of each basis wedge: \begin{align*} d\alpha &=\left(\frac{\partial R}{\partial y}-\frac{\partial Q}{\partial z}\right)dy\wedge dz +\left(\frac{\partial P}{\partial z}-\frac{\partial R}{\partial x}\right)dz\wedge dx +\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dx\wedge dy. \end{align*} Now compute $d\beta$. By linearity and the coordinate definition of $d$, \begin{align*} d\beta &=d(A\,dy\wedge dz)+d(B\,dz\wedge dx)+d(C\,dx\wedge dy)\\ &=\left(\frac{\partial A}{\partial x}dx+\frac{\partial A}{\partial y}dy+\frac{\partial A}{\partial z}dz\right)\wedge dy\wedge dz\\ &\quad+\left(\frac{\partial B}{\partial x}dx+\frac{\partial B}{\partial y}dy+\frac{\partial B}{\partial z}dz\right)\wedge dz\wedge dx\\ &\quad+\left(\frac{\partial C}{\partial x}dx+\frac{\partial C}{\partial y}dy+\frac{\partial C}{\partial z}dz\right)\wedge dx\wedge dy. \end{align*} Expand term by term: \begin{align*} d\beta &=\frac{\partial A}{\partial x}dx\wedge dy\wedge dz +\frac{\partial A}{\partial y}dy\wedge dy\wedge dz +\frac{\partial A}{\partial z}dz\wedge dy\wedge dz\\ &\quad+\frac{\partial B}{\partial x}dx\wedge dz\wedge dx +\frac{\partial B}{\partial y}dy\wedge dz\wedge dx +\frac{\partial B}{\partial z}dz\wedge dz\wedge dx\\ &\quad+\frac{\partial C}{\partial x}dx\wedge dx\wedge dy +\frac{\partial C}{\partial y}dy\wedge dx\wedge dy +\frac{\partial C}{\partial z}dz\wedge dx\wedge dy. \end{align*} Every term with a repeated coordinate factor vanishes: \begin{align*} d\beta &=\frac{\partial A}{\partial x}dx\wedge dy\wedge dz +\frac{\partial B}{\partial y}dy\wedge dz\wedge dx +\frac{\partial C}{\partial z}dz\wedge dx\wedge dy. \end{align*} The remaining two cyclic permutations have positive sign: \begin{align*} dy\wedge dz\wedge dx &=-dy\wedge dx\wedge dz\\ &=dx\wedge dy\wedge dz, \end{align*} and \begin{align*} dz\wedge dx\wedge dy &=-dx\wedge dz\wedge dy\\ &=dx\wedge dy\wedge dz. \end{align*} Therefore \begin{align*} d\beta &=\frac{\partial A}{\partial x}dx\wedge dy\wedge dz +\frac{\partial B}{\partial y}dx\wedge dy\wedge dz +\frac{\partial C}{\partial z}dx\wedge dy\wedge dz\\ &=\left(\frac{\partial A}{\partial x}+\frac{\partial B}{\partial y}+\frac{\partial C}{\partial z}\right)dx\wedge dy\wedge dz. \end{align*} Applying $d$ to a $1$-form on $\mathbb R^3$ produces the three signed curl coefficients in the $2$-form basis $dy\wedge dz$, $dz\wedge dx$, $dx\wedge dy$. Applying $d$ to a $2$-form produces the single divergence coefficient multiplying the volume form $dx\wedge dy\wedge dz$. Thus curl and divergence are the same operation, the [exterior derivative](/theorems/1525), applied in adjacent degrees. [/example] ## Integration And Orientation Why does a change of parametrisation sometimes preserve an integral and sometimes change its sign? Integration of forms depends on orientation. A top-degree form measures signed volume, so reversing the order of a basis reverses the sign of the measured volume. [definition: Orientation Of A Vector Space] Let $V$ be an $n$-dimensional real [vector space](/page/Vector%20Space). An orientation of $V$ is a choice of one of the two classes of ordered bases of $V$, where two ordered bases are in the same class when the change-of-basis matrix between them has positive determinant. [/definition] On a manifold, an orientation is a smoothly varying choice of orientation on every tangent space. This condition is exactly what lets local integrals in coordinate charts combine into a global integral. Before global integration can be defined, the coordinate-domain case must be fixed precisely. A top-degree form on an oriented open subset of $\mathbb R^n$ has one scalar coefficient against the standard volume form. The problem is to turn that coefficient into a signed number without losing the orientation information carried by the ordered coordinates. [definition: Integral Of A Top Degree Form On A Coordinate Domain] Let $U\subset\mathbb R^n$ be open and let $\omega=f\,dx_1\wedge\cdots\wedge dx_n\in\Omega^n(U)$, where $f\in C_c^\infty(U)$. The integral of $\omega$ over $U$ with the standard orientation is \begin{align*} \int_U \omega := \int_U f\,d\mathcal L^n. \end{align*} [/definition] The determinant appears because it is the top-degree alternating form applied to the columns of a Jacobian matrix. This is why differential forms absorb the Jacobian factor in the change of variables formula. The next issue is whether this definition depends on the chosen coordinates. A valid theory of integration for forms needs the integral to stay unchanged under orientation-preserving coordinate changes and to change sign exactly when the orientation is reversed. [quotetheorem:3554] [citeproof:3554] This theorem is the bridge from multivariable calculus to integration on manifolds. Once the integral is independent of the oriented coordinate chart, it can be defined by choosing charts, integrating locally, and summing with a [partition of unity](/page/Partition%20of%20Unity). [example: Polar Coordinates And The Area Form] Let \begin{align*} F:(0,\infty)\times(0,2\pi)\to\mathbb R^2\setminus\{(x,0):x\ge0\} \end{align*} be the polar coordinate map \begin{align*} F(r,\theta)=(r\cos\theta,r\sin\theta). \end{align*} Write $x,y$ for the standard coordinates on $\mathbb R^2$, and write $r,\theta$ for the standard coordinates on $(0,\infty)\times(0,2\pi)$. The standard area form on $\mathbb R^2$ is \begin{align*} dx\wedge dy. \end{align*} The pullback of the standard area form is \begin{align*} F^*(dx\wedge dy)=r\,dr\wedge d\theta. \end{align*} Thus the polar Jacobian factor $r$ is already contained in the pullback of the area form. By definition of pullback on coordinate functions, \begin{align*} x\circ F(r,\theta)&=r\cos\theta,\\ y\circ F(r,\theta)&=r\sin\theta. \end{align*} Therefore \begin{align*} F^*(dx)&=d(x\circ F)=d(r\cos\theta),\\ F^*(dy)&=d(y\circ F)=d(r\sin\theta). \end{align*} Using the product rule for differentials, \begin{align*} d(r\cos\theta) &=\cos\theta\,dr+r\,d(\cos\theta)\\ &=\cos\theta\,dr-r\sin\theta\,d\theta, \end{align*} and \begin{align*} d(r\sin\theta) &=\sin\theta\,dr+r\,d(\sin\theta)\\ &=\sin\theta\,dr+r\cos\theta\,d\theta. \end{align*} By compatibility of pullback with wedge products, \begin{align*} F^*(dx\wedge dy) &=F^*(dx)\wedge F^*(dy)\\ &=d(r\cos\theta)\wedge d(r\sin\theta)\\ &=(\cos\theta\,dr-r\sin\theta\,d\theta) \wedge(\sin\theta\,dr+r\cos\theta\,d\theta). \end{align*} Expand by bilinearity of the wedge product: \begin{align*} F^*(dx\wedge dy) &=\cos\theta\sin\theta\,dr\wedge dr +r\cos^2\theta\,dr\wedge d\theta\\ &\quad-r\sin^2\theta\,d\theta\wedge dr -r^2\sin\theta\cos\theta\,d\theta\wedge d\theta. \end{align*} The repeated wedge products vanish: \begin{align*} dr\wedge dr=0,\qquad d\theta\wedge d\theta=0. \end{align*} Also \begin{align*} d\theta\wedge dr=-dr\wedge d\theta. \end{align*} Substituting these wedge relations gives \begin{align*} F^*(dx\wedge dy) &=r\cos^2\theta\,dr\wedge d\theta -r\sin^2\theta(-dr\wedge d\theta)\\ &=r\cos^2\theta\,dr\wedge d\theta +r\sin^2\theta\,dr\wedge d\theta\\ &=r(\cos^2\theta+\sin^2\theta)\,dr\wedge d\theta\\ &=r\,dr\wedge d\theta. \end{align*} Pulling back the area form means pulling back both coordinate covectors and then wedging them. In polar coordinates, this produces the factor $r$ from the expansion of $d(r\cos\theta)\wedge d(r\sin\theta)$. The usual polar area element $r\,dr\,d\theta$ is therefore not an added correction term; it is the coordinate expression of the pulled-back form $F^*(dx\wedge dy)$. [/example] ## Stokes As The Unifying Principle What single theorem contains the [fundamental theorem of calculus](/theorems/632), Green's theorem, the [divergence theorem](/theorems/2754), and the Kelvin-Stokes theorem? The answer is the generalised Stokes theorem. Its strength is that the same formula holds in every dimension and for every degree of form. [quotetheorem:3555] [citeproof:3555] The theorem says that integration of a derivative over a region is the same as integration over the oriented boundary. The many classical vector calculus theorems differ only in how forms are translated into vector-field notation. [example: Fundamental Theorem As Stokes] Let $M=[a,b]$ with its standard orientation, so the positive coordinate vector on the interior is $\partial/\partial x$. Let $f\in C^\infty([a,b])$, and regard $f$ as a $0$-form on $M$. For an oriented point $+p$, integration of a $0$-form is evaluation: \begin{align*} \int_{+p} f=f(p). \end{align*} For the same point with the opposite orientation, denoted $-p$, integration changes sign: \begin{align*} \int_{-p} f=-f(p). \end{align*} The generalised Stokes formula on the oriented interval $[a,b]$ is exactly \begin{align*} \int_a^b f'(x)\,dx=f(b)-f(a). \end{align*} The signs come from the induced boundary orientation \begin{align*} \partial[a,b]=\{b\}-\{a\}. \end{align*} Since $f$ is a $0$-form, the coordinate definition of the [exterior derivative](/theorems/1525) gives \begin{align*} df=\frac{\partial f}{\partial x}\,dx=f'(x)\,dx. \end{align*} Therefore, by the definition of integration of a top-degree form on the coordinate interval, \begin{align*} \int_M df &=\int_{[a,b]} f'(x)\,dx\\ &=\int_a^b f'(x)\,dx. \end{align*} Now determine the boundary orientation. At the right endpoint $b$, the outward-pointing vector is $+\partial/\partial x$, which agrees with the standard orientation of $[a,b]$. Thus $b$ receives positive orientation: \begin{align*} b=+b. \end{align*} At the left endpoint $a$, the outward-pointing vector is $-\partial/\partial x$, which is the negative of the standard positive vector. Thus $a$ receives negative orientation: \begin{align*} a=-a. \end{align*} Hence the oriented boundary is \begin{align*} \partial M=+b+(-a)=\{b\}-\{a\}. \end{align*} Using the definition of integration over oriented points, \begin{align*} \int_{\partial M} f &=\int_{\{b\}-\{a\}} f\\ &=\int_{+b} f+\int_{-a} f\\ &=f(b)-f(a). \end{align*} By *Generalised Stokes Theorem*, applied to the $0$-form $f$ on the oriented $1$-manifold $M=[a,b]$, \begin{align*} \int_M df=\int_{\partial M} f. \end{align*} Substituting the two computed sides gives \begin{align*} \int_a^b f'(x)\,dx=f(b)-f(a). \end{align*} The ordinary [fundamental theorem of calculus](/theorems/632) is the $1$-dimensional case of Stokes theorem. The right endpoint appears with a plus sign because its outward direction agrees with the orientation of the interval, and the left endpoint appears with a minus sign because its outward direction is opposite to the orientation. [/example] The next case raises the dimension by one. Instead of endpoints, the boundary is now an oriented curve, and the same sign convention becomes the positive orientation around a planar region. [example: Green Theorem As Stokes] Let $D\subset\mathbb R^2$ be a compact oriented region with positively oriented boundary $\partial D$. Let $P,Q$ be smooth functions on an [open set](/page/Open%20Set) containing $D$, and set \begin{align*} \omega=P\,dx+Q\,dy. \end{align*} Orient $D$ by the standard area form $dx\wedge dy$. The generalised Stokes theorem applied to $\omega$ gives Green's theorem: \begin{align*} \int_{\partial D} P\,dx+Q\,dy =\int_D\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)d\mathcal L^2. \end{align*} By linearity of the [exterior derivative](/theorems/1525) and the coordinate definition of $d$, \begin{align*} d\omega &=d(P\,dx+Q\,dy)\\ &=d(P\,dx)+d(Q\,dy)\\ &=\left(\frac{\partial P}{\partial x}dx+\frac{\partial P}{\partial y}dy\right)\wedge dx +\left(\frac{\partial Q}{\partial x}dx+\frac{\partial Q}{\partial y}dy\right)\wedge dy. \end{align*} Expand the wedge products term by term: \begin{align*} d\omega &=\frac{\partial P}{\partial x}dx\wedge dx +\frac{\partial P}{\partial y}dy\wedge dx +\frac{\partial Q}{\partial x}dx\wedge dy +\frac{\partial Q}{\partial y}dy\wedge dy. \end{align*} Using \begin{align*} dx\wedge dx=0,\qquad dy\wedge dy=0,\qquad dy\wedge dx=-dx\wedge dy, \end{align*} this becomes \begin{align*} d\omega &=0-\frac{\partial P}{\partial y}dx\wedge dy +\frac{\partial Q}{\partial x}dx\wedge dy+0\\ &=\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dx\wedge dy. \end{align*} By *Generalised Stokes Theorem*, applied to the oriented $2$-manifold with boundary $D$ and the $1$-form $\omega$, \begin{align*} \int_D d\omega=\int_{\partial D}\omega. \end{align*} Substitute the expression for $d\omega$: \begin{align*} \int_{\partial D}\omega &=\int_D d\omega\\ &=\int_D\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dx\wedge dy. \end{align*} By the definition of integration of a top-degree form on a coordinate domain with the standard orientation, \begin{align*} \int_D\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dx\wedge dy =\int_D\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)d\mathcal L^2. \end{align*} Since $\omega=P\,dx+Q\,dy$, the boundary integral is \begin{align*} \int_{\partial D}\omega=\int_{\partial D}P\,dx+Q\,dy. \end{align*} Combining these equalities gives \begin{align*} \int_{\partial D}P\,dx+Q\,dy =\int_D\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)d\mathcal L^2. \end{align*} Green's theorem is the case of Stokes theorem with $M=D\subset\mathbb R^2$ and $\omega=P\,dx+Q\,dy$. The [exterior derivative](/theorems/1525) turns the line-integrand coefficients $P$ and $Q$ into the signed area coefficient $\partial Q/\partial x-\partial P/\partial y$, while the induced positive boundary orientation supplies the orientation used in the boundary line integral. [/example] ## Cohomology From Closed And Exact Forms If $d^2=0$, what information is lost when a closed form is not itself a derivative? This question leads from analysis to topology. The failure of a closed form to be exact can detect holes in the underlying space. [definition: Closed And Exact Forms] Let $M$ be a smooth manifold. A form $\omega\in\Omega^k(M)$ is closed if $d\omega=0$. It is exact if there exists $\eta\in\Omega^{k-1}(M)$ such that $\omega=d\eta$. [/definition] Since $d^2=0$, every exact form is closed. The quotient by exact forms measures the closed forms that remain after derivatives have been ignored. [definition: De Rham Cohomology] Let $M$ be a smooth manifold. The $k$-th de Rham cohomology group of $M$ is \begin{align*} H_{\mathrm{dR}}^k(M):=\frac{\ker(d:\Omega^k(M)\to\Omega^{k+1}(M))}{\operatorname{im}(d:\Omega^{k-1}(M)\to\Omega^k(M))}. \end{align*} [/definition] A class in $H_{\mathrm{dR}}^k(M)$ is represented by a closed $k$-form, and changing the representative by an exact form does not change the class. Stokes theorem explains this quotient: exact corrections integrate to zero over closed cycles. To separate local behavior from global topology, one needs to know what closed forms look like on domains with no holes. The Poincare lemma supplies this local model: on a sufficiently simple open set, closed positive-degree forms are already exact. [quotetheorem:832] [citeproof:832] The Poincare lemma says that local cohomology in positive degree vanishes on sufficiently simple coordinate domains. De Rham cohomology is therefore a global invariant: it appears when local primitives cannot be chosen consistently across the whole space. [example: A Closed One Form On The Punctured Plane] Let $M=\mathbb R^2\setminus\{(0,0)\}$, with standard coordinates $x,y$. Define \begin{align*} \omega=\frac{-y}{x^2+y^2}\,dx+\frac{x}{x^2+y^2}\,dy. \end{align*} Since $x^2+y^2>0$ on $M$, the coefficient functions are smooth on $M$. Write \begin{align*} P(x,y)=\frac{-y}{x^2+y^2},\qquad Q(x,y)=\frac{x}{x^2+y^2}, \end{align*} so that $\omega=P\,dx+Q\,dy$. Let \begin{align*} \gamma:[0,2\pi]\to M,\qquad \gamma(t)=(\cos t,\sin t) \end{align*} be the counterclockwise parametrisation of the unit circle. The $1$-form $\omega$ is closed but not exact on $M$. More precisely, \begin{align*} d\omega=0 \end{align*} and \begin{align*} \int_\gamma\omega=2\pi. \end{align*} The nonzero integral around the closed curve $\gamma$ rules out the existence of a global smooth function $f$ on $M$ with $\omega=df$. By the [coordinate formula for the exterior derivative](/theorems/3564) of a $1$-form on an open subset of $\mathbb R^2$, \begin{align*} d\omega &=d(P\,dx+Q\,dy)\\ &=d(P\,dx)+d(Q\,dy)\\ &=\left(\frac{\partial P}{\partial x}dx+\frac{\partial P}{\partial y}dy\right)\wedge dx +\left(\frac{\partial Q}{\partial x}dx+\frac{\partial Q}{\partial y}dy\right)\wedge dy\\ &=\frac{\partial P}{\partial x}dx\wedge dx +\frac{\partial P}{\partial y}dy\wedge dx +\frac{\partial Q}{\partial x}dx\wedge dy +\frac{\partial Q}{\partial y}dy\wedge dy. \end{align*} Using \begin{align*} dx\wedge dx=0,\qquad dy\wedge dy=0,\qquad dy\wedge dx=-dx\wedge dy, \end{align*} this becomes \begin{align*} d\omega &=-\frac{\partial P}{\partial y}dx\wedge dy +\frac{\partial Q}{\partial x}dx\wedge dy\\ &=\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dx\wedge dy. \end{align*} Now compute the two partial derivatives: \begin{align*} \frac{\partial Q}{\partial x} &=\frac{\partial}{\partial x}\left(x(x^2+y^2)^{-1}\right)\\ &=(x^2+y^2)^{-1}+x\left(-1\right)(x^2+y^2)^{-2}(2x)\\ &=\frac{1}{x^2+y^2}-\frac{2x^2}{(x^2+y^2)^2}\\ &=\frac{x^2+y^2-2x^2}{(x^2+y^2)^2}\\ &=\frac{y^2-x^2}{(x^2+y^2)^2}, \end{align*} and \begin{align*} \frac{\partial P}{\partial y} &=\frac{\partial}{\partial y}\left(-y(x^2+y^2)^{-1}\right)\\ &=-(x^2+y^2)^{-1}+(-y)\left(-1\right)(x^2+y^2)^{-2}(2y)\\ &=-\frac{1}{x^2+y^2}+\frac{2y^2}{(x^2+y^2)^2}\\ &=\frac{-(x^2+y^2)+2y^2}{(x^2+y^2)^2}\\ &=\frac{y^2-x^2}{(x^2+y^2)^2}. \end{align*} Therefore \begin{align*} \frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y} &=\frac{y^2-x^2}{(x^2+y^2)^2} -\frac{y^2-x^2}{(x^2+y^2)^2}\\ &=0, \end{align*} so \begin{align*} d\omega=0. \end{align*} Thus $\omega$ is closed. Next compute the pullback along $\gamma$. Since \begin{align*} x\circ\gamma(t)=\cos t,\qquad y\circ\gamma(t)=\sin t, \end{align*} we have \begin{align*} \gamma^*(dx)=d(\cos t)=-\sin t\,dt,\qquad \gamma^*(dy)=d(\sin t)=\cos t\,dt. \end{align*} Also \begin{align*} (P\circ\gamma)(t) &=\frac{-\sin t}{\cos^2 t+\sin^2 t}\\ &=-\sin t, \end{align*} and \begin{align*} (Q\circ\gamma)(t) &=\frac{\cos t}{\cos^2 t+\sin^2 t}\\ &=\cos t. \end{align*} By the definition of pullback of a $1$-form, \begin{align*} \gamma^*\omega &=(P\circ\gamma)\,\gamma^*(dx)+(Q\circ\gamma)\,\gamma^*(dy)\\ &=(-\sin t)(-\sin t\,dt)+(\cos t)(\cos t\,dt)\\ &=\sin^2 t\,dt+\cos^2 t\,dt\\ &=(\sin^2 t+\cos^2 t)\,dt\\ &=dt. \end{align*} Therefore \begin{align*} \int_\gamma\omega &=\int_0^{2\pi}\gamma^*\omega\\ &=\int_0^{2\pi}dt\\ &=\int_0^{2\pi}1\,dt\\ &=2\pi-0\\ &=2\pi. \end{align*} It remains to show that $\omega$ is not exact. Suppose, for contradiction, that there exists $f\in C^\infty(M)$ such that \begin{align*} \omega=df. \end{align*} Let $C=S^1\subset M$ be the unit circle with the orientation induced by $\gamma$. Since $C$ is a closed oriented $1$-manifold, \begin{align*} \partial C=\varnothing. \end{align*} By *Generalised Stokes Theorem*, applied to the $0$-form $f|_C$ on $C$, \begin{align*} \int_C df=\int_{\partial C} f. \end{align*} Because $\partial C=\varnothing$, \begin{align*} \int_{\partial C} f=0. \end{align*} Thus \begin{align*} \int_C df=0. \end{align*} But $\omega=df$, and $\gamma$ parametrises $C$ once with its chosen orientation, so \begin{align*} 0 &=\int_C df\\ &=\int_C \omega\\ &=\int_\gamma\omega\\ &=2\pi. \end{align*} This contradiction shows that no such global smooth function $f$ exists. Hence $\omega$ is not exact on $M$. The form \begin{align*} \omega=\frac{-y}{x^2+y^2}\,dx+\frac{x}{x^2+y^2}\,dy \end{align*} is closed because the two mixed coefficient derivatives agree exactly. It is not exact because its integral around the unit circle is $2\pi$, while an exact $1$-form has integral $0$ over every closed oriented $1$-manifold by Stokes theorem. Therefore $\omega$ represents a nonzero de Rham cohomology class in $H_{\mathrm{dR}}^1(\mathbb R^2\setminus\{0\})$, detecting the hole at the origin. [/example] ## How The Course Progresses What background should a reader keep active while moving through the course? The main prerequisites are multivariable calculus, linear algebra of dual spaces and determinants, and basic topology of open sets, compactness, and continuity. Smooth manifolds enter gradually: first as spaces covered by coordinate charts, then as objects on which forms, pullbacks, orientations, and integrals are intrinsic. The first part of the course is algebraic. It constructs alternating tensors, wedge products, and bases for $\Lambda^k(V^*)$. This explains why $k$-forms have the coordinate expressions used throughout the notes. The second part is analytic and geometric. It defines differential forms on open sets and manifolds, proves the coordinate invariance of [exterior derivative](/theorems/1525) and pullback, develops oriented integration, and proves the generalised Stokes theorem. The final part is topological. It introduces closed and exact forms, proves the Poincare lemma, computes de Rham cohomology for standard examples, and states de Rham's comparison theorem relating differential forms to singular cohomology. The guiding principle is that a form is an integrand with its transformation law built in. Once this viewpoint is in place, the major theorems of vector calculus become shadows of one theorem, and the obstruction to finding primitives becomes a computable invariant of the underlying space. The exterior algebra provides the necessary algebraic foundation for differential forms. We begin by studying alternating tensors on a single [vector space](/page/Vector%20Space), which will become, at each point of a manifold, the domain on which differential k-forms act. # 1. Exterior Algebra This opening chapter builds the algebraic language that later becomes the language of differential forms on manifolds. At a single point of a manifold, a differential $k$-form is an alternating $k$-linear function on tangent vectors, so the first task is to understand alternating tensors on an ordinary finite-dimensional real [vector space](/page/Vector%20Space). The chapter moves from multilinearity, to antisymmetrisation, to the wedge product, and ends with the coordinate basis and the determinant as the model top-degree form. ## Multilinear Maps and Alternating Tensors What algebraic object should receive $k$ vectors and produce a number in a way compatible with linear algebra in each input? For integration and orientation, the answer must also change sign when two directions are swapped, because reversing an oriented parallelepiped should reverse its signed volume. Let $V$ be a finite-dimensional real [vector space](/page/Vector%20Space). A map of $k$ vector inputs records how $k$ directions interact, but multilinearity is the condition that lets the map be determined by its values on a basis. [definition: Multilinear Map] Let $k \in \mathbb N$. A $k$-[linear map](/page/Linear%20Map) on $V$ is a function $T: V^k \to \mathbb R$ such that for each $j \in \{1, \dots, k\}$, each fixed choice of the other $k-1$ inputs, and all $a,b \in \mathbb R$ and $v,w \in V$, \begin{align*} T(v_1, \dots, av + bw, \dots, v_k) &= aT(v_1, \dots, v, \dots, v_k) + bT(v_1, \dots, w, \dots, v_k). \end{align*} The [vector space](/page/Vector%20Space) of all $k$-linear maps $V^k \to \mathbb R$ is denoted by $\operatorname{Mult}^k(V)$. [/definition] For $k=1$, this is just the [dual space](/page/Dual%20Space) $V^*$. For larger $k$, the new feature is the ability to compare what happens under a rearrangement of inputs. [definition: Alternating Multilinear Map] Let $k \in \mathbb N$. A $k$-[linear map](/page/Linear%20Map) $\alpha: V^k \to \mathbb R$ is alternating if \begin{align*} \alpha(v_1, \dots, v_k) = 0 \end{align*} whenever $v_i = v_j$ for some distinct $i,j \in \{1, \dots, k\}$. The [vector space](/page/Vector%20Space) of alternating $k$-linear maps $V^k \to \mathbb R$ is denoted by $\Lambda^k(V^*)$. We set $\Lambda^0(V^*) = \mathbb R$. [/definition] The notation $\Lambda^k(V^*)$ anticipates the construction of exterior powers. Its elements are called alternating $k$-tensors, exterior $k$-forms on $V$, or simply $k$-forms when the [vector space](/page/Vector%20Space) is fixed. The definition uses vanishing on repeated inputs, but computations usually require knowing what happens under an arbitrary permutation of the inputs. The key structural fact is that alternation forces every swap, and therefore every permutation, to contribute exactly its sign. [quotetheorem:3556] [citeproof:3556] This theorem is the algebraic source of orientation signs. Alternating tensors do not merely vanish on degenerate $k$-tuples; they remember the order of independent directions up to the sign of the corresponding permutation. [example: A Two Form on the Plane] Let $V=\mathbb R^2$, and let $dx,dy \in V^*$ be the coordinate covectors. Thus for a vector $(x,y)\in \mathbb R^2$, \begin{align*} dx(x,y)=x, \qquad dy(x,y)=y. \end{align*} Define the two-input map $\omega:V^2\to \mathbb R$ by \begin{align*} \omega(v,w)=dx(v)dy(w)-dx(w)dy(v). \end{align*} Take \begin{align*} v=(a,b), \qquad w=(c,d). \end{align*} The value of $\omega$ on $(v,w)$ is \begin{align*} \omega(v,w)=ad-bc. \end{align*} Equivalently, $\omega(v,w)$ is the determinant of the $2\times 2$ matrix whose columns are $v$ and $w$. By the defining property of the coordinate covectors, \begin{align*} dx(v)&=dx(a,b)=a, & dy(v)&=dy(a,b)=b,\\ dx(w)&=dx(c,d)=c, & dy(w)&=dy(c,d)=d. \end{align*} Substituting these four values into the definition of $\omega$ gives \begin{align*} \omega(v,w) &=dx(v)dy(w)-dx(w)dy(v)\\ &=a\cdot d-c\cdot b\\ &=ad-cb\\ &=ad-bc. \end{align*} The matrix with first column $v=(a,b)$ and second column $w=(c,d)$ is \begin{align*} \begin{pmatrix} a & c\\ b & d \end{pmatrix}. \end{align*} By the defining formula for the determinant of a $2\times 2$ matrix, \begin{align*} \det\begin{pmatrix} a & c\\ b & d \end{pmatrix} =ad-cb =ad-bc. \end{align*} Therefore \begin{align*} \omega(v,w) = \det\begin{pmatrix} a & c\\ b & d \end{pmatrix}. \end{align*} The two-form $\omega$ assigns to the ordered pair $(v,w)$ the signed area determinant of the parallelogram spanned by $v$ and $w$. Reversing the order of the vectors changes the sign, because \begin{align*} \omega(w,v)=dx(w)dy(v)-dx(v)dy(w)=cb-ad=-(ad-bc)=-\omega(v,w). \end{align*} [/example] The example already contains the pattern of the wedge product: take products of covectors and antisymmetrise them so that repeated directions disappear and interchanges of directions introduce signs. ## Antisymmetrisation and the Wedge Product How can an arbitrary multilinear expression be turned into an alternating one without changing the alternating part it already contains? The operation is to average over all permutations with the signs attached, producing a projection onto $\Lambda^k(V^*)$. [definition: Antisymmetrisation] Let $T \in \operatorname{Mult}^k(V)$. The antisymmetrisation of $T$ is the $k$-[linear map](/page/Linear%20Map) $\operatorname{Alt}(T): V^k \to \mathbb R$ defined by \begin{align*} \operatorname{Alt}(T)(v_1, \dots, v_k) = \frac{1}{k!}\sum_{\sigma \in S_k} \operatorname{sgn}(\sigma) T(v_{\sigma(1)}, \dots, v_{\sigma(k)}). \end{align*} [/definition] Antisymmetrisation is the formal version of keeping only the oriented part of a multilinear expression. The factor $1/k!$ makes it a projection rather than only a signed sum. The obstruction is that the defining sum ranges over all permutations, so it is not automatic from the formula alone that the result has exactly the alternating behaviour desired, or that already alternating tensors are not accidentally rescaled. To use antisymmetrisation as a construction of forms, we need it to be the canonical projection onto the alternating subspace. [quotetheorem:3557] [citeproof:3557] The [projection theorem](/theorems/1985) explains why antisymmetrisation is the right operation rather than just a convenient formula. Linearity means it can be applied term-by-term after expanding a tensor, the identity property means already alternating tensors are left unchanged, and idempotence means applying it a second time does not alter the result. The factor $1/k!$ is responsible for that idempotence; without it, alternating tensors would be multiplied by $k!$ rather than fixed. This theorem does not say that $\operatorname{Alt}$ respects tensor products. Even if two factors are alternating, their ordinary [tensor product](/page/Tensor%20Product) need not be alternating in all variables, so the [tensor product](/page/Tensor%20Product) must be antisymmetrised before it becomes a form of higher degree. [example: Tensor Product Before Antisymmetrisation] Let $V=\mathbb R^2$, and let $dx,dy\in V^*$ be the coordinate covectors: \begin{align*} dx(x,y)=x, \qquad dy(x,y)=y. \end{align*} Let \begin{align*} e_1=(1,0), \qquad e_2=(0,1). \end{align*} Set \begin{align*} T=dx\otimes dy, \end{align*} so for $v,w\in V$, \begin{align*} T(v,w)=(dx\otimes dy)(v,w)=dx(v)dy(w). \end{align*} For comparison, the tensor $dy\otimes dx$ is defined by \begin{align*} (dy\otimes dx)(v,w)=dy(v)dx(w). \end{align*} The tensor $dx\otimes dy$ is not alternating. Its antisymmetrisation is \begin{align*} \operatorname{Alt}(dx\otimes dy)=\frac{1}{2}(dx\otimes dy-dy\otimes dx), \end{align*} and the wedge-product normalisation gives \begin{align*} dx\wedge dy=dx\otimes dy-dy\otimes dx. \end{align*} The coordinate covectors give \begin{align*} dx(e_1)&=dx(1,0)=1, & dy(e_1)&=dy(1,0)=0,\\ dx(e_2)&=dx(0,1)=0, & dy(e_2)&=dy(0,1)=1. \end{align*} Therefore \begin{align*} (dx\otimes dy)(e_1,e_2) &=dx(e_1)dy(e_2)\\ &=1\cdot 1\\ &=1, \end{align*} while \begin{align*} (dx\otimes dy)(e_2,e_1) &=dx(e_2)dy(e_1)\\ &=0\cdot 0\\ &=0. \end{align*} If a bilinear map $B:V^2\to\mathbb R$ is alternating, then \begin{align*} B(e_1,e_1)=0,\qquad B(e_2,e_2)=0,\qquad B(e_1+e_2,e_1+e_2)=0. \end{align*} Expanding the last equality by bilinearity gives \begin{align*} 0 &=B(e_1+e_2,e_1+e_2)\\ &=B(e_1,e_1)+B(e_1,e_2)+B(e_2,e_1)+B(e_2,e_2)\\ &=0+B(e_1,e_2)+B(e_2,e_1)+0\\ &=B(e_1,e_2)+B(e_2,e_1). \end{align*} Hence every alternating bilinear map satisfies \begin{align*} B(e_2,e_1)=-B(e_1,e_2). \end{align*} For $T=dx\otimes dy$, this would require \begin{align*} T(e_2,e_1)=-T(e_1,e_2)=-1. \end{align*} But the computed value is \begin{align*} T(e_2,e_1)=0. \end{align*} Since $0\neq -1$, the tensor $dx\otimes dy$ is not alternating. Now compute its antisymmetrisation. For $k=2$, the two permutations are the identity, with sign $+1$, and the transposition, with sign $-1$. Thus for arbitrary $v,w\in V$, \begin{align*} \operatorname{Alt}(T)(v,w) &=\frac{1}{2}\bigl(T(v,w)-T(w,v)\bigr)\\ &=\frac{1}{2}\bigl(dx(v)dy(w)-dx(w)dy(v)\bigr). \end{align*} On the other hand, \begin{align*} (dx\otimes dy-dy\otimes dx)(v,w) &=(dx\otimes dy)(v,w)-(dy\otimes dx)(v,w)\\ &=dx(v)dy(w)-dy(v)dx(w)\\ &=dx(v)dy(w)-dx(w)dy(v), \end{align*} using commutativity of multiplication in $\mathbb R$. Therefore, for every $v,w\in V$, \begin{align*} \operatorname{Alt}(dx\otimes dy)(v,w) =\frac{1}{2}(dx\otimes dy-dy\otimes dx)(v,w), \end{align*} so \begin{align*} \operatorname{Alt}(dx\otimes dy) =\frac{1}{2}(dx\otimes dy-dy\otimes dx). \end{align*} For one-forms, the wedge product has the normalising factor \begin{align*} \frac{(1+1)!}{1!1!}=\frac{2!}{1}=2. \end{align*} Hence \begin{align*} dx\wedge dy &=2\,\operatorname{Alt}(dx\otimes dy)\\ &=2\cdot \frac{1}{2}(dx\otimes dy-dy\otimes dx)\\ &=dx\otimes dy-dy\otimes dx. \end{align*} The ordinary [tensor product](/page/Tensor%20Product) $dx\otimes dy$ records the first coordinate of the first input and the second coordinate of the second input, but it does not enforce alternating sign behavior under an interchange of inputs. Antisymmetrisation removes the non-alternating part by subtracting the reversed tensor. The wedge product then rescales this alternating part so that \begin{align*} (dx\wedge dy)(e_1,e_2)=1,\qquad (dx\wedge dy)(e_2,e_1)=-1. \end{align*} [/example] The wedge product is therefore defined by multiplying two alternating tensors and then applying antisymmetrisation. The normalising constant is chosen so that wedge products of basis covectors have the expected determinant formula without extra factorials. The example suggests the general construction: first form the ordinary tensor product, then keep only its alternating part with the correct normalization. This gives an intrinsic multiplication that raises degree and turns exterior forms into an algebra. [definition: Wedge Product] Let $\alpha \in \Lambda^p(V^*)$ and $\beta \in \Lambda^q(V^*)$. Their wedge product is the element $\alpha \wedge \beta \in \Lambda^{p+q}(V^*)$ defined by \begin{align*} \alpha \wedge \beta = \frac{(p+q)!}{p!q!}\operatorname{Alt}(\alpha \otimes \beta), \end{align*} where \begin{align*} (\alpha \otimes \beta)(v_1, \dots, v_{p+q}) = \alpha(v_1, \dots, v_p)\beta(v_{p+1}, \dots, v_{p+q}). \end{align*} [/definition] The formula says that $\alpha$ is evaluated on $p$ of the vectors, $\beta$ on the remaining $q$, and then every way of choosing and ordering those inputs is combined with the correct orientation sign. For computation, the full antisymmetrisation formula is too redundant: many terms differ only by permutations inside the $p$ inputs of $\alpha$ or inside the $q$ inputs of $\beta$. Since $\alpha$ and $\beta$ are already alternating, those internal permutations should collapse without changing the value. The needed result is a shuffle formula that keeps only the genuinely different ways to split the inputs. [quotetheorem:3558] [citeproof:3558] The shuffle formula is often the most efficient way to compute wedges by hand. In small degrees it recovers familiar determinant expressions. [example: Wedge of Two One Forms in Three Dimensions] Let $V=\mathbb R^3$ with coordinate covectors $dx,dy,dz\in V^*$. Thus, for $(x,y,z)\in\mathbb R^3$, \begin{align*} dx(x,y,z)=x,\qquad dy(x,y,z)=y,\qquad dz(x,y,z)=z. \end{align*} Let $a_1,a_2,a_3,b_1,b_2,b_3\in\mathbb R$, and define one-forms \begin{align*} \alpha = a_1dx+a_2dy+a_3dz, \qquad \beta = b_1dx+b_2dy+b_3dz. \end{align*} The wedge product of $\alpha$ and $\beta$ is \begin{align*} \alpha \wedge \beta = (a_1b_2-a_2b_1)dx\wedge dy + (a_1b_3-a_3b_1)dx\wedge dz + (a_2b_3-a_3b_2)dy\wedge dz. \end{align*} First record the two identities for wedges of one-forms that will be used in the expansion. Let $\eta,\theta\in V^*$ and let $v,w\in V$. By the definition of the wedge product with $p=q=1$, \begin{align*} \eta\wedge\theta &=\frac{(1+1)!}{1!1!}\operatorname{Alt}(\eta\otimes\theta)\\ &=2\operatorname{Alt}(\eta\otimes\theta). \end{align*} By the definition of antisymmetrisation for two inputs, \begin{align*} \operatorname{Alt}(\eta\otimes\theta)(v,w) &=\frac{1}{2}\bigl((\eta\otimes\theta)(v,w)-(\eta\otimes\theta)(w,v)\bigr)\\ &=\frac{1}{2}\bigl(\eta(v)\theta(w)-\eta(w)\theta(v)\bigr). \end{align*} Therefore \begin{align*} (\eta\wedge\theta)(v,w) &=2\cdot \frac{1}{2}\bigl(\eta(v)\theta(w)-\eta(w)\theta(v)\bigr)\\ &=\eta(v)\theta(w)-\eta(w)\theta(v). \end{align*} Taking $\theta=\eta$ gives \begin{align*} (\eta\wedge\eta)(v,w) &=\eta(v)\eta(w)-\eta(w)\eta(v)\\ &=0, \end{align*} so $\eta\wedge\eta=0$. Also, \begin{align*} (\theta\wedge\eta)(v,w) &=\theta(v)\eta(w)-\theta(w)\eta(v)\\ &=\eta(w)\theta(v)-\eta(v)\theta(w)\\ &=-\bigl(\eta(v)\theta(w)-\eta(w)\theta(v)\bigr)\\ &=-(\eta\wedge\theta)(v,w), \end{align*} so $\theta\wedge\eta=-\eta\wedge\theta$. Applying these identities to $dx,dy,dz$ gives \begin{align*} dx\wedge dx=dy\wedge dy=dz\wedge dz=0, \end{align*} and \begin{align*} dy\wedge dx=-dx\wedge dy,\qquad dz\wedge dx=-dx\wedge dz,\qquad dz\wedge dy=-dy\wedge dz. \end{align*} Now expand $\alpha\wedge\beta$. The wedge product is bilinear because it is defined by the [tensor product](/page/Tensor%20Product) followed by the [linear map](/page/Linear%20Map) $\operatorname{Alt}$. Hence \begin{align*} \alpha\wedge\beta &=(a_1dx+a_2dy+a_3dz)\wedge(b_1dx+b_2dy+b_3dz)\\ &=a_1b_1\,dx\wedge dx +a_1b_2\,dx\wedge dy +a_1b_3\,dx\wedge dz\\ &\quad +a_2b_1\,dy\wedge dx +a_2b_2\,dy\wedge dy +a_2b_3\,dy\wedge dz\\ &\quad +a_3b_1\,dz\wedge dx +a_3b_2\,dz\wedge dy +a_3b_3\,dz\wedge dz. \end{align*} Substituting the zero and sign identities gives \begin{align*} \alpha\wedge\beta &=a_1b_1\cdot 0 +a_1b_2\,dx\wedge dy +a_1b_3\,dx\wedge dz\\ &\quad +a_2b_1(-dx\wedge dy) +a_2b_2\cdot 0 +a_2b_3\,dy\wedge dz\\ &\quad +a_3b_1(-dx\wedge dz) +a_3b_2(-dy\wedge dz) +a_3b_3\cdot 0\\ &=a_1b_2\,dx\wedge dy -a_2b_1\,dx\wedge dy +a_1b_3\,dx\wedge dz -a_3b_1\,dx\wedge dz\\ &\quad +a_2b_3\,dy\wedge dz -a_3b_2\,dy\wedge dz\\ &=(a_1b_2-a_2b_1)dx\wedge dy +(a_1b_3-a_3b_1)dx\wedge dz +(a_2b_3-a_3b_2)dy\wedge dz. \end{align*} The wedge product keeps only the antisymmetric pairwise coordinate combinations. The coefficient of $dx\wedge dy$ is the signed $dx,dy$ minor $a_1b_2-a_2b_1$, the coefficient of $dx\wedge dz$ is $a_1b_3-a_3b_1$, and the coefficient of $dy\wedge dz$ is $a_2b_3-a_3b_2$. Thus $\alpha\wedge\beta$ measures the oriented two-dimensional coordinate areas determined by the two one-forms. [/example] The wedge product is the multiplication law of exterior algebra. Its two structural features are associativity and a sign rule recording the degrees of the factors. A multiplication on forms would be awkward if products of three or more factors depended on parenthesisation. The first structural question is therefore whether the normalization in the definition makes iterated wedge products unambiguous. [quotetheorem:3559] [citeproof:3559] Associativity is the point at which the normalisation in the wedge product pays for itself. The two parenthesisations both reduce to the same signed sum over $(p,q,r)$-shuffles, with no extra binomial coefficients left over from choosing intermediate blocks. Thus exterior forms assemble into an algebra $\Lambda^\ast(V^*)=\bigoplus_{k=0}^n \Lambda^k(V^*)$ whose multiplication is the wedge product. Because of associativity, expressions such as $dx\wedge dy\wedge dz$ need no parentheses. The remaining question is what happens when two homogeneous factors are interchanged. Forms cannot commute in the ordinary sense, because one-forms already satisfy $dx\wedge dy=-dy\wedge dx$, but the sign should depend only on the degrees of the factors. [quotetheorem:3560] [citeproof:3560] For one-forms this gives anticommutativity, while for a two-form and another two-form it gives commutativity. In particular, if $\alpha$ has odd degree, then $\alpha\wedge\alpha=-\alpha\wedge\alpha$, so over $\mathbb R$ we get $\alpha\wedge\alpha=0$. This conclusion uses the sign and the ability to divide by $2$; in characteristic $2$ the sign disappears, so alternating behaviour has to be imposed separately rather than recovered from graded commutativity. The theorem also belongs to exterior forms, not to arbitrary multilinear tensors: without alternation of both factors, the shuffle-sign argument has no reason to apply. ## Coordinate Bases and Dimension Once a basis of $V$ is chosen, which alternating tensors are needed to describe every $k$-form? Since alternating tensors vanish on repeated vectors, only ordered collections of distinct coordinate covectors can survive. Let $\dim V=n$, and let $e_1,\dots,e_n$ be a basis of $V$ with [dual basis](/theorems/414) $dx_1,\dots,dx_n \in V^*$. The notation $dx_i$ is chosen to match the later differential-geometric setting, where these covectors are differentials of coordinate functions. [definition: Coordinate Wedge] For indices $1 \le i_1 < \dots < i_k \le n$, the coordinate $k$-form \begin{align*} dx_{i_1}\wedge \dots \wedge dx_{i_k} \in \Lambda^k(V^*) \end{align*} is the wedge product of the corresponding [dual basis](/theorems/414) covectors. [/definition] The strict inequality in the indices is not cosmetic. If an index is repeated, the wedge is zero; if the same indices are written in a different order, the result changes by the sign of the ordering permutation. To compute with $k$-forms, one needs to know that these increasing coordinate wedges are not merely convenient examples. They should form a basis, so that every alternating tensor has unique coefficients indexed by increasing coordinate subsets. [quotetheorem:3561] [citeproof:3561] This theorem is the computational backbone of exterior algebra. It turns the abstract [vector space](/page/Vector%20Space) $\Lambda^k(V^*)$ into a concrete coordinate space whose coordinates are indexed by increasing $k$-element subsets of $\{1,\dots,n\}$. [example: Two Forms in Three Dimensions] Let $V=\mathbb R^3$ with standard basis \begin{align*} e_1=(1,0,0),\qquad e_2=(0,1,0),\qquad e_3=(0,0,1), \end{align*} and [dual basis](/theorems/414) $dx,dy,dz\in V^*$. Thus \begin{align*} dx(e_1)&=1, & dy(e_1)&=0, & dz(e_1)&=0,\\ dx(e_2)&=0, & dy(e_2)&=1, & dz(e_2)&=0,\\ dx(e_3)&=0, & dy(e_3)&=0, & dz(e_3)&=1. \end{align*} Let $\omega\in \Lambda^2(V^*)$. By *Basis of Exterior Powers*, there are unique scalars $A,B,C\in\mathbb R$ such that \begin{align*} \omega = A\,dx\wedge dy + B\,dx\wedge dz + C\,dy\wedge dz. \end{align*} The coefficients are recovered by evaluating $\omega$ on the increasing pairs of standard basis vectors: \begin{align*} A = \omega(e_1,e_2), \qquad B = \omega(e_1,e_3), \qquad C = \omega(e_2,e_3). \end{align*} First compute the relevant coordinate wedge values. For one-forms $\eta,\theta\in V^*$ and vectors $r,s\in V$, the definition of the wedge product with $p=q=1$ gives \begin{align*} \eta\wedge\theta &=\frac{(1+1)!}{1!1!}\operatorname{Alt}(\eta\otimes\theta)\\ &=2\operatorname{Alt}(\eta\otimes\theta). \end{align*} The definition of antisymmetrisation gives \begin{align*} \operatorname{Alt}(\eta\otimes\theta)(r,s) &=\frac{1}{2}\bigl((\eta\otimes\theta)(r,s)-(\eta\otimes\theta)(s,r)\bigr)\\ &=\frac{1}{2}\bigl(\eta(r)\theta(s)-\eta(s)\theta(r)\bigr). \end{align*} Therefore \begin{align*} (\eta\wedge\theta)(r,s) &=2\cdot \frac{1}{2}\bigl(\eta(r)\theta(s)-\eta(s)\theta(r)\bigr)\\ &=\eta(r)\theta(s)-\eta(s)\theta(r). \end{align*} Now evaluate the three coordinate wedges on $(e_1,e_2)$: \begin{align*} (dx\wedge dy)(e_1,e_2) &=dx(e_1)dy(e_2)-dx(e_2)dy(e_1)\\ &=1\cdot 1-0\cdot 0\\ &=1,\\ (dx\wedge dz)(e_1,e_2) &=dx(e_1)dz(e_2)-dx(e_2)dz(e_1)\\ &=1\cdot 0-0\cdot 0\\ &=0,\\ (dy\wedge dz)(e_1,e_2) &=dy(e_1)dz(e_2)-dy(e_2)dz(e_1)\\ &=0\cdot 0-1\cdot 0\\ &=0. \end{align*} Using linearity of evaluation, \begin{align*} \omega(e_1,e_2) &=A(dx\wedge dy)(e_1,e_2) +B(dx\wedge dz)(e_1,e_2) +C(dy\wedge dz)(e_1,e_2)\\ &=A\cdot 1+B\cdot 0+C\cdot 0\\ &=A. \end{align*} On $(e_1,e_3)$, \begin{align*} (dx\wedge dy)(e_1,e_3) &=dx(e_1)dy(e_3)-dx(e_3)dy(e_1)\\ &=1\cdot 0-0\cdot 0\\ &=0,\\ (dx\wedge dz)(e_1,e_3) &=dx(e_1)dz(e_3)-dx(e_3)dz(e_1)\\ &=1\cdot 1-0\cdot 0\\ &=1,\\ (dy\wedge dz)(e_1,e_3) &=dy(e_1)dz(e_3)-dy(e_3)dz(e_1)\\ &=0\cdot 1-0\cdot 0\\ &=0. \end{align*} Hence \begin{align*} \omega(e_1,e_3) &=A\cdot 0+B\cdot 1+C\cdot 0\\ &=B. \end{align*} On $(e_2,e_3)$, \begin{align*} (dx\wedge dy)(e_2,e_3) &=dx(e_2)dy(e_3)-dx(e_3)dy(e_2)\\ &=0\cdot 0-0\cdot 1\\ &=0,\\ (dx\wedge dz)(e_2,e_3) &=dx(e_2)dz(e_3)-dx(e_3)dz(e_2)\\ &=0\cdot 1-0\cdot 0\\ &=0,\\ (dy\wedge dz)(e_2,e_3) &=dy(e_2)dz(e_3)-dy(e_3)dz(e_2)\\ &=1\cdot 1-0\cdot 0\\ &=1. \end{align*} Therefore \begin{align*} \omega(e_2,e_3) &=A\cdot 0+B\cdot 0+C\cdot 1\\ &=C. \end{align*} It remains to see why these three values determine all values on basis pairs. Since $\omega$ is alternating, \begin{align*} \omega(e_i,e_i)=0 \end{align*} for $i=1,2,3$. For distinct $i,j$, alternation and bilinearity give \begin{align*} 0 &=\omega(e_i+e_j,e_i+e_j)\\ &=\omega(e_i,e_i)+\omega(e_i,e_j)+\omega(e_j,e_i)+\omega(e_j,e_j)\\ &=0+\omega(e_i,e_j)+\omega(e_j,e_i)+0, \end{align*} so \begin{align*} \omega(e_j,e_i)=-\omega(e_i,e_j). \end{align*} Thus the reversed basis-pair values are \begin{align*} \omega(e_2,e_1)=-A,\qquad \omega(e_3,e_1)=-B,\qquad \omega(e_3,e_2)=-C. \end{align*} For arbitrary vectors \begin{align*} r=x_1e_1+x_2e_2+x_3e_3,\qquad s=y_1e_1+y_2e_2+y_3e_3, \end{align*} bilinearity gives \begin{align*} \omega(r,s) &=\sum_{i=1}^3\sum_{j=1}^3 x_i y_j\,\omega(e_i,e_j)\\ &=x_1y_1\cdot 0+x_1y_2A+x_1y_3B +x_2y_1(-A)+x_2y_2\cdot 0+x_2y_3C\\ &\quad +x_3y_1(-B)+x_3y_2(-C)+x_3y_3\cdot 0\\ &=A(x_1y_2-x_2y_1)+B(x_1y_3-x_3y_1)+C(x_2y_3-x_3y_2). \end{align*} So knowing $A$, $B$, and $C$ is the same as knowing the value of $\omega$ on every pair of vectors. A two-form on $\mathbb R^3$ is determined by its three oriented coordinate-plane measurements. The coefficient of $dx\wedge dy$ is the value on the oriented $e_1,e_2$ plane, the coefficient of $dx\wedge dz$ is the value on the oriented $e_1,e_3$ plane, and the coefficient of $dy\wedge dz$ is the value on the oriented $e_2,e_3$ plane. Reversing an ordered basis pair changes the sign, and repeated basis directions contribute zero. [/example] The binomial coefficient reflects the fact that a $k$-form chooses $k$ independent coordinate directions at a time. This is why exterior algebra is finite in dimension even though wedge products can be iterated formally. ## Top-Degree Forms and the Determinant What remains in degree $n$ when $V$ itself has dimension $n$? There is only one increasing choice of all coordinate directions, so the space of top-degree forms is one-dimensional. [quotetheorem:3561] [citeproof:3561] The finite-dimensional hypothesis is essential here. In an infinite-dimensional [vector space](/page/Vector%20Space) there is no largest finite degree, and exterior powers do not collapse to a single top-dimensional line. For finite-dimensional $V$, an orientation can be described as a choice of one of the two possible positive rays in the one-dimensional space $\Lambda^n(V^*)$; choosing a nonzero top form fixes which ordered bases are positively oriented. Dually, $\Lambda^n(V)$ is also one-dimensional and pairs naturally with $\Lambda^n(V^*)$ by evaluation. To connect this one-dimensional space with concrete volume, we need to know what the standard coordinate top form does to a list of vectors. The determinant is the only alternating $n$-linear measurement normalized to give $1$ on the standard basis, so it is the transformation factor that top-degree forms must recover. [quotetheorem:393] [citeproof:393] This theorem connects exterior algebra with the Jacobian determinants from multivariable calculus. Later, when forms are pulled back along smooth maps, this determinant factor will appear automatically from the algebra of top-degree forms. [example: The Standard Volume Form in Three Dimensions] Let $V=\mathbb R^3$ with standard coordinate covectors $dx,dy,dz\in V^*$. Thus, for $r=(r_1,r_2,r_3)\in \mathbb R^3$, \begin{align*} dx(r)=r_1,\qquad dy(r)=r_2,\qquad dz(r)=r_3. \end{align*} Set \begin{align*} \Omega = dx\wedge dy\wedge dz. \end{align*} Let \begin{align*} u=(u_1,u_2,u_3),\qquad v=(v_1,v_2,v_3),\qquad w=(w_1,w_2,w_3). \end{align*} The matrix with columns $u,v,w$ is \begin{align*} A= \begin{pmatrix} u_1 & v_1 & w_1\\ u_2 & v_2 & w_2\\ u_3 & v_3 & w_3 \end{pmatrix}. \end{align*} The value of the standard volume form on $(u,v,w)$ is \begin{align*} \Omega(u,v,w)=\det A. \end{align*} First compute the two-form $dx\wedge dy$. For arbitrary $r,s\in \mathbb R^3$, the wedge product of one-forms gives \begin{align*} (dx\wedge dy)(r,s) &=dx(r)dy(s)-dx(s)dy(r). \end{align*} Let \begin{align*} \varphi=dx\wedge dy. \end{align*} Then $\Omega=\varphi\wedge dz$. For $T=\varphi\otimes dz$, \begin{align*} T(r,s,t)=\varphi(r,s)dz(t). \end{align*} Using the wedge product definition with degrees $2$ and $1$, \begin{align*} \Omega=3\,\operatorname{Alt}(T). \end{align*} For three inputs, the six signed permutation terms give \begin{align*} \Omega(u,v,w) &=\frac{1}{2}\bigl( T(u,v,w)-T(v,u,w)-T(u,w,v)\\ &\quad +T(w,u,v)+T(v,w,u)-T(w,v,u) \bigr). \end{align*} Since $\varphi$ is alternating, \begin{align*} T(v,u,w)&=\varphi(v,u)dz(w)=-\varphi(u,v)dz(w)=-T(u,v,w),\\ T(w,u,v)&=\varphi(w,u)dz(v)=-\varphi(u,w)dz(v)=-T(u,w,v),\\ T(w,v,u)&=\varphi(w,v)dz(u)=-\varphi(v,w)dz(u)=-T(v,w,u). \end{align*} Substituting these three identities, \begin{align*} \Omega(u,v,w) &=\frac{1}{2}\bigl( T(u,v,w)+T(u,v,w)-T(u,w,v)\\ &\quad -T(u,w,v)+T(v,w,u)+T(v,w,u) \bigr)\\ &=T(u,v,w)-T(u,w,v)+T(v,w,u). \end{align*} Now expand each term: \begin{align*} T(u,v,w) &=\varphi(u,v)dz(w)\\ &=\bigl(dx(u)dy(v)-dx(v)dy(u)\bigr)dz(w)\\ &=(u_1v_2-v_1u_2)w_3\\ &=u_1v_2w_3-v_1u_2w_3,\\ T(u,w,v) &=\varphi(u,w)dz(v)\\ &=\bigl(dx(u)dy(w)-dx(w)dy(u)\bigr)dz(v)\\ &=(u_1w_2-w_1u_2)v_3\\ &=u_1w_2v_3-w_1u_2v_3,\\ T(v,w,u) &=\varphi(v,w)dz(u)\\ &=\bigl(dx(v)dy(w)-dx(w)dy(v)\bigr)dz(u)\\ &=(v_1w_2-w_1v_2)u_3\\ &=v_1w_2u_3-w_1v_2u_3. \end{align*} Therefore \begin{align*} \Omega(u,v,w) &=u_1v_2w_3-v_1u_2w_3-u_1w_2v_3+w_1u_2v_3\\ &\quad+v_1w_2u_3-w_1v_2u_3. \end{align*} On the other hand, expanding $\det A$ along the first row gives \begin{align*} \det A &=u_1 \begin{vmatrix} v_2 & w_2\\ v_3 & w_3 \end{vmatrix} -v_1 \begin{vmatrix} u_2 & w_2\\ u_3 & w_3 \end{vmatrix} +w_1 \begin{vmatrix} u_2 & v_2\\ u_3 & v_3 \end{vmatrix}\\ &=u_1(v_2w_3-w_2v_3)-v_1(u_2w_3-w_2u_3)+w_1(u_2v_3-v_2u_3)\\ &=u_1v_2w_3-u_1w_2v_3-v_1u_2w_3+v_1w_2u_3+w_1u_2v_3-w_1v_2u_3. \end{align*} This is the same scalar expression as the one obtained for $\Omega(u,v,w)$, with terms reordered using commutativity of multiplication in $\mathbb R$. Hence \begin{align*} (dx\wedge dy\wedge dz)(u,v,w)=\det A. \end{align*} The three-form $dx\wedge dy\wedge dz$ evaluates an ordered triple of vectors by the determinant of the matrix having those vectors as columns. Thus it measures the signed volume of the parallelepiped spanned by $u,v,w$: positive for the standard orientation, negative when the orientation is reversed, and zero when the three vectors are linearly dependent. [/example] The determinant interpretation also explains why volume is sensitive to linear dependence rather than merely to the lengths of the input vectors. The next example refines the repeated-direction case by perturbing one vector and watching only the independent component survive. [example: A Nearly Repeated Direction] Let $V=\mathbb R^3$ with coordinate covectors $dx,dy,dz\in V^*$. Set \begin{align*} \Omega=dx\wedge dy\wedge dz. \end{align*} Fix vectors $u,v,w\in \mathbb R^3$ and a scalar $\varepsilon\in\mathbb R$. Since $\Omega\in \Lambda^3(V^*)$, it is a trilinear alternating map $V^3\to\mathbb R$. The value of $\Omega$ on the nearly repeated triple $(u,u+\varepsilon v,w)$ is \begin{align*} \Omega(u,u+\varepsilon v,w) = \varepsilon \Omega(u,v,w). \end{align*} Equivalently, \begin{align*} (dx\wedge dy\wedge dz)(u,u+\varepsilon v,w) = \varepsilon (dx\wedge dy\wedge dz)(u,v,w). \end{align*} Apply trilinearity of $\Omega$ in the second input while keeping the first and third inputs fixed: \begin{align*} \Omega(u,u+\varepsilon v,w) &=\Omega(u,1\cdot u+\varepsilon v,w)\\ &=1\cdot \Omega(u,u,w)+\varepsilon \Omega(u,v,w)\\ &=\Omega(u,u,w)+\varepsilon \Omega(u,v,w). \end{align*} Because $\Omega$ is alternating, it vanishes whenever two inputs are equal. In the triple $(u,u,w)$, the first and second inputs are both $u$, so \begin{align*} \Omega(u,u,w)=0. \end{align*} Substituting this into the previous expression gives \begin{align*} \Omega(u,u+\varepsilon v,w) &=0+\varepsilon \Omega(u,v,w)\\ &=\varepsilon \Omega(u,v,w). \end{align*} Replacing $\Omega$ by $dx\wedge dy\wedge dz$ yields \begin{align*} (dx\wedge dy\wedge dz)(u,u+\varepsilon v,w) = \varepsilon (dx\wedge dy\wedge dz)(u,v,w). \end{align*} The repeated part of the second vector contributes no signed volume, because alternating three-forms vanish on repeated directions. Only the perturbation term $\varepsilon v$ contributes, and it contributes linearly with the scalar factor $\varepsilon$. Thus the top-degree form measures precisely the component of the second input that creates new oriented volume together with $u$ and $w$. [/example] Exterior algebra therefore packages three facts into a single structure: multilinearity, antisymmetry, and determinant-like volume. The next chapter globalises this pointwise algebra by assigning to every point of an [open set](/page/Open%20Set) or manifold an alternating tensor on the tangent space, producing differential forms as fields of exterior-algebra elements. With the algebraic structure of exterior algebra in place at a single point, we now globalize it by letting these constructions vary smoothly across an [open set](/page/Open%20Set) of ℝⁿ. This produces differential forms as smooth fields of exterior algebra elements. # 2. Differential k-Forms on Open Subsets of Rⁿ The first chapter built exterior algebra at a single [vector space](/page/Vector%20Space): alternating covectors, their wedge product, and the determinant as the top-degree case. We now let those algebraic objects vary smoothly from point to point on an [open set](/page/Open%20Set) $U\subset \mathbb R^n$. The resulting objects are differential forms, and they are the basic quantities that will later be differentiated, pulled back, and integrated. The guiding shift is from a single alternating map to a field of alternating maps. A $k$-form assigns to each point $x\in U$ an element of $\Lambda^k(T_x^*U)$, with coordinate coefficients depending smoothly on $x$. On open subsets of Euclidean space, this can be written without bundle machinery because each tangent space $T_xU$ is naturally identified with $\mathbb R^n$. ## Forms as Smooth Covector Fields How can the alternating covectors from exterior algebra become objects on an [open set](/page/Open%20Set) rather than at a single point? The coordinate functions $x_1,\dots,x_n$ on $U\subset\mathbb R^n$ provide a standard covector basis at every point, so a form is described by smooth coefficient functions multiplying the exterior algebra basis. [definition: Coordinate One-Forms] Let $U\subset \mathbb R^n$ be open. For $1\le i\le n$, the coordinate one-form $dx_i$ assigns to $x\in U$ the covector $dx_i|_x:T_xU\to\mathbb R$ defined by \begin{align*} dx_i|_x(v)=v_i \end{align*} for $v=(v_1,\dots,v_n)\in T_xU$. [/definition] These coordinate one-forms are the moving version of the [dual basis](/theorems/414) from exterior algebra. Since $U$ is open in $\mathbb R^n$, the notation suppresses the identification $T_xU\cong \mathbb R^n$. Having a covector basis at each point is not yet a differential form; the coefficients must vary smoothly from point to point. The next definition packages this requirement so that a form is both an alternating object in each tangent space and a smooth field over the open set. [definition: Smooth k-Form On An Open Set] Let $U\subset\mathbb R^n$ be open and let $0\le k\le n$. A smooth $k$-form $\omega$ on $U$ is an assignment \begin{align*} x\mapsto \omega_x\in \Lambda^k(T_x^*U) \end{align*} for which there exist functions $a_I\in C^\infty(U)$, indexed by increasing $k$-tuples $I=(i_1<\dots<i_k)$, satisfying \begin{align*} \omega_x=\sum_I a_I(x)\,dx_{i_1}|_x\wedge\dots\wedge dx_{i_k}|_x \end{align*} for every $x\in U$. The set of all smooth $k$-forms on $U$ is denoted $\Omega^k(U)$. [/definition] The standard shorthand omits the point $x$ and writes \begin{align*} \omega=\sum_I a_I\,dx_{i_1}\wedge\dots\wedge dx_{i_k}. \end{align*} The smoothness requirement is not cosmetic. If the coefficients were merely assigned pointwise with no regularity, the exterior algebra operations would still make sense at individual tangent spaces, but the next operations in the course would fail to be stable: differentiating the coefficients would not produce smooth forms, and integration over smooth parametrised pieces would lose the regularity needed for change-of-variables arguments. Thus $\Omega^k(U)$ is the class where algebra, differentiation, and integration can be developed together. When $k=0$, the space $\Lambda^0(T_x^*U)$ is $\mathbb R$, so a smooth $0$-form is a smooth function $f\in C^\infty(U)$. When $k=n$, there is only one basis form, $dx_1\wedge\dots\wedge dx_n$, so every $n$-form is $a\,dx_1\wedge\dots\wedge dx_n$ for a unique $a\in C^\infty(U)$. [quotetheorem:3562] The theorem says that differential forms on an open subset of $\mathbb R^n$ can be manipulated by their coefficient functions, provided the exterior basis is kept in increasing order. The increasing-index convention prevents multiple names for the same term: for instance $dx_j\wedge dx_i=-dx_i\wedge dx_j$ and $dx_i\wedge dx_i=0$, so an unrestricted sum would contain redundant or vanishing pieces. Writing every term in increasing order makes the coefficient of each basis element well-defined and gives the uniqueness statement real content. The free $C^\infty(U)$-module structure is the computational reason for using coordinates at this stage. It lets later operators be defined by formulas on basis elements and smooth coefficient functions, then extended by linearity and product rules. Without smooth coefficients, this coordinate method would produce expressions that no longer lie in the same class of objects. [example: Polynomial One-Form On The Plane] Let $U=\mathbb R^2$ with coordinate functions $x,y$. Define \begin{align*} \omega=x\,dy-y\,dx. \end{align*} At a point $p=(x,y)\in\mathbb R^2$, let $v=(v_1,v_2)\in T_p\mathbb R^2\cong\mathbb R^2$. The form $\omega$ is a smooth $1$-form on $\mathbb R^2$, and \begin{align*} \omega_p(v)=xv_2-yv_1. \end{align*} Write $\omega$ in the coordinate basis $dx,dy$: \begin{align*} \omega=x\,dy-y\,dx=(-y)\,dx+x\,dy. \end{align*} The coefficient of $dx$ is the polynomial function $-y$, and the coefficient of $dy$ is the polynomial function $x$. Both functions lie in $C^\infty(\mathbb R^2)$, so by the definition of a smooth $1$-form on an [open set](/page/Open%20Set), $\omega\in\Omega^1(\mathbb R^2)$. By the definition of the coordinate one-forms, \begin{align*} dx|_p(v)&=v_1,\\ dy|_p(v)&=v_2. \end{align*} Therefore \begin{align*} \omega_p(v) &=(x\,dy-y\,dx)_p(v)\\ &=x(p)\,dy|_p(v)-y(p)\,dx|_p(v)\\ &=x\,v_2-y\,v_1. \end{align*} The polynomial expression $x\,dy-y\,dx$ defines a smooth covector field on the plane. At each point $(x,y)$, it sends the tangent vector $(v_1,v_2)$ to the scalar $xv_2-yv_1$. [/example] ## The Wedge Product and the Graded Algebra of Forms How do forms of different degrees multiply while remembering orientation and sign? The answer is to apply the exterior algebra wedge product pointwise, then track the degree of each form. [definition: Total Space Of Differential Forms] Let $U\subset\mathbb R^n$ be open. The graded space of smooth differential forms on $U$ is \begin{align*} \Omega^*(U)=\bigoplus_{k=0}^n\Omega^k(U). \end{align*} An element of $\Omega^k(U)$ has degree $k$. [/definition] The direct sum keeps the degree visible. Functions live in degree $0$, ordinary work-type integrands live in degree $1$, and top-degree forms live in degree $n$. Putting all degrees in one space creates a new operation to define: forms should multiply in a way that raises degree, vanishes on repeated directions, and changes sign when inputs are interchanged. The obstruction is that this multiplication must be both pointwise algebraic and smoothly dependent on the base point. [definition: Wedge Product Of Differential Forms] Let $\alpha\in\Omega^p(U)$ and $\beta\in\Omega^q(U)$. The wedge product $\alpha\wedge\beta\in\Omega^{p+q}(U)$ is defined by \begin{align*} (\alpha\wedge\beta)_x=\alpha_x\wedge\beta_x \end{align*} for every $x\in U$, with $\Omega^m(U)=0$ for $m>n$. [/definition] In coordinates, if $\alpha=\sum_I a_I dx_I$ and $\beta=\sum_J b_J dx_J$, then \begin{align*} \alpha\wedge\beta=\sum_{I,J}a_Ib_J\,dx_I\wedge dx_J, \end{align*} with repeated coordinate covectors giving zero and the remaining terms reordered into increasing order with the sign of the required permutation. A reliable computation has three steps: expand bilinearly, delete any term containing a repeated coordinate covector, and reorder each surviving wedge into increasing index order while multiplying by the sign of the permutation. To use this multiplication as algebra rather than just notation, we also need the pointwise construction to satisfy the expected associativity, unit, and graded sign laws across all degrees. [quotetheorem:3563] This result is the global form of the exterior algebra identities from the previous chapter. The new point is that coefficient functions multiply in $C^\infty(U)$ while the signs are still controlled by degrees. Ordinary commutativity would forget the orientation reversal caused by swapping the inputs of alternating forms; graded-commutativity records exactly that sign. When $\alpha$ has odd degree, the identity gives $\alpha\wedge\alpha=-\alpha\wedge\alpha$, hence $\alpha\wedge\alpha=0$ over $\mathbb R$. This is why $1$-forms behave like oriented infinitesimal directions rather than scalar factors. The same sign rule is what later makes the [exterior derivative](/theorems/1525) satisfy a signed product rule: \begin{align*} d(\alpha\wedge\beta)=d\alpha\wedge\beta+(-1)^p\alpha\wedge d\beta \end{align*} for $\alpha\in\Omega^p(U)$. The algebraic signs in this theorem are therefore not bookkeeping only; they are the mechanism that makes calculus of forms consistent across degrees. [example: Wedge Product Computation In Three Variables] On $\mathbb R^3$ with coordinates $x,y,z$, let \begin{align*} \alpha=x\,dx+y\,dy,\qquad \beta=z\,dy+dx. \end{align*} Their wedge product is \begin{align*} \alpha\wedge\beta=(xz-y)\,dx\wedge dy. \end{align*} Use bilinearity of the wedge product over smooth coefficient functions: \begin{align*} \alpha\wedge\beta &=(x\,dx+y\,dy)\wedge(z\,dy+dx)\\ &=(x\,dx)\wedge(z\,dy)+(x\,dx)\wedge dx+(y\,dy)\wedge(z\,dy)+(y\,dy)\wedge dx\\ &=xz\,(dx\wedge dy)+x\,(dx\wedge dx)+yz\,(dy\wedge dy)+y\,(dy\wedge dx). \end{align*} The wedge product is alternating on one-forms, so \begin{align*} dx\wedge dx&=0,\\ dy\wedge dy&=0. \end{align*} Also, swapping two one-forms changes the sign: \begin{align*} dy\wedge dx=-dx\wedge dy. \end{align*} Substitute these three identities into the expanded expression: \begin{align*} \alpha\wedge\beta &=xz\,(dx\wedge dy)+x\cdot 0+yz\cdot 0+y\,(-dx\wedge dy)\\ &=xz\,dx\wedge dy-y\,dx\wedge dy\\ &=(xz-y)\,dx\wedge dy. \end{align*} The only surviving basis term is $dx\wedge dy$. The $dx\wedge dx$ and $dy\wedge dy$ terms vanish because repeated one-forms wedge to zero, and the $dy\wedge dx$ term contributes $-y\,dx\wedge dy$ after reordering into increasing coordinate order. [/example] ## The Vector Calculus Dictionary in Three Dimensions How does the notation of gradients, line integrals, fluxes, and volume integrals fit into the language of forms? In $\mathbb R^3$, the usual vector calculus integrands appear as forms of degree $0,1,2,3$, but the translation uses the Euclidean metric and the chosen orientation. [explanation: Vector Calculus Dictionary In R Three] Let $U\subset\mathbb R^3$ be open with coordinates $x,y,z$. The basic dictionary is \begin{align*} \text{scalar function } f&\longleftrightarrow f\in\Omega^0(U),\\ \text{work integrand for }F=(P,Q,R)&\longleftrightarrow \alpha_F=P\,dx+Q\,dy+R\,dz\in\Omega^1(U),\\ \text{flux integrand for }F=(P,Q,R)&\longleftrightarrow \beta_F=P\,dy\wedge dz+Q\,dz\wedge dx+R\,dx\wedge dy\in\Omega^2(U),\\ \text{volume integrand } g&\longleftrightarrow \gamma_g=g\,dx\wedge dy\wedge dz\in\Omega^3(U). \end{align*} A $1$-form evaluates on one tangent vector, matching the way a line integral samples a velocity vector along a curve. A $2$-form evaluates on an ordered pair of tangent vectors, matching the oriented parallelogram element used in flux. A $3$-form evaluates on an ordered triple of tangent vectors, matching signed volume. [/explanation] This dictionary explains why two different forms can be associated to the same vector field $F$: the $1$-form $\alpha_F$ is the work integrand, while the $2$-form $\beta_F$ is the flux integrand. They are related by Euclidean metric and orientation data, not by exterior algebra alone. Later, the [exterior derivative](/theorems/1525) will turn this dictionary into the familiar operations grad, curl, and div. [remark: Metric Dependence Of The Dictionary] The spaces $\Omega^k(U)$ and their wedge product are defined without an inner product. The identifications between vector fields and $1$-forms or $2$-forms in $\mathbb R^3$ use the dot product and the standard orientation. On a general manifold, these identifications require extra geometric structure. [/remark] ## Guiding Examples What examples should be kept in mind before exterior differentiation and integration are introduced? Two models are especially useful: the angular form on the punctured plane, which detects rotation around a missing point, and coordinate area forms in $\mathbb R^3$, which encode oriented projected area. [illustration:forms-angular-one-form] [example: Angle One-Form On The Punctured Plane] Let $U=\mathbb R^2\setminus\{0\}$ with coordinate functions $x,y$. Define \begin{align*} d\theta=\frac{-y\,dx+x\,dy}{x^2+y^2} =\frac{-y}{x^2+y^2}\,dx+\frac{x}{x^2+y^2}\,dy. \end{align*} Let $p=(a,b)\in U$. The radial vector at $p$ is $r_p=(a,b)$, and the angular vector at $p$ is $t_p=(-b,a)$. The expression $d\theta$ defines a smooth $1$-form on $U$, and it satisfies \begin{align*} d\theta_p(r_p)&=0,\\ d\theta_p(t_p)&=1. \end{align*} Since $p=(a,b)\in U$, we have $(a,b)\ne(0,0)$, so \begin{align*} a^2+b^2>0. \end{align*} Thus the function $x^2+y^2$ is nonzero at every point of $U$. The coordinate functions $x,y$ are smooth, $x^2+y^2$ is smooth, and the reciprocal of a nonvanishing smooth function is smooth. Hence \begin{align*} \frac{-y}{x^2+y^2},\qquad \frac{x}{x^2+y^2} \end{align*} are smooth functions on $U$. Therefore $d\theta$ is a smooth $1$-form on $U$ by the definition of a smooth $1$-form in coordinates. Now let $v=(v_1,v_2)\in T_pU\cong\mathbb R^2$. By the definition of the coordinate one-forms, \begin{align*} dx|_p(v)&=v_1,\\ dy|_p(v)&=v_2. \end{align*} Therefore \begin{align*} d\theta_p(v) &=\left(\frac{-y}{x^2+y^2}\,dx+\frac{x}{x^2+y^2}\,dy\right)_p(v)\\ &=\frac{-b}{a^2+b^2}\,dx|_p(v)+\frac{a}{a^2+b^2}\,dy|_p(v)\\ &=\frac{-b}{a^2+b^2}\,v_1+\frac{a}{a^2+b^2}\,v_2\\ &=\frac{-bv_1+av_2}{a^2+b^2}. \end{align*} For the radial vector $r_p=(a,b)$, substitute $v_1=a$ and $v_2=b$: \begin{align*} d\theta_p(r_p) &=\frac{-ba+ab}{a^2+b^2}\\ &=\frac{ab-ab}{a^2+b^2}\\ &=\frac{0}{a^2+b^2}\\ &=0. \end{align*} For the angular vector $t_p=(-b,a)$, substitute $v_1=-b$ and $v_2=a$: \begin{align*} d\theta_p(t_p) &=\frac{-b(-b)+a(a)}{a^2+b^2}\\ &=\frac{b^2+a^2}{a^2+b^2}\\ &=\frac{a^2+b^2}{a^2+b^2}\\ &=1. \end{align*} The form $d\theta$ is a smooth covector field on the punctured plane. At each point $p=(a,b)$, it gives value $0$ on motion directly away from the origin and value $1$ on the counterclockwise angular direction $(-b,a)$, so it records infinitesimal angular motion around the missing origin. [/example] The notation $d\theta$ is local notation from polar coordinates. There is no globally defined smooth angle function $\theta:U\to\mathbb R$ on the whole punctured plane, a fact that will later reappear as the first basic example of de Rham cohomology. This is a useful warning about forms: a formula can make perfect sense as a smooth $1$-form even when it is not the differential of any globally defined smooth function. The missing origin creates a topological obstruction to choosing a single-valued angle, so differential forms can record global failure modes that scalar potentials cannot see. [illustration:forms-area-two-form-projection] [example: Coordinate Area Two-Form In R Three] Let $U=\mathbb R^3$ with coordinates $x,y,z$, and consider the $2$-form \begin{align*} dx\wedge dy\in\Omega^2(\mathbb R^3). \end{align*} Fix a point $p\in\mathbb R^3$ and tangent vectors \begin{align*} v=(v_1,v_2,v_3),\qquad w=(w_1,w_2,w_3) \end{align*} in $T_p\mathbb R^3\cong\mathbb R^3$. The value of $dx\wedge dy$ on $v,w$ is \begin{align*} (dx\wedge dy)|_p(v,w)=v_1w_2-v_2w_1. \end{align*} This is the signed area of the parallelogram obtained by projecting $v$ and $w$ to the $xy$-plane. By the definition of the coordinate one-forms, \begin{align*} dx|_p(v)&=v_1,& dy|_p(v)&=v_2,\\ dx|_p(w)&=w_1,& dy|_p(w)&=w_2. \end{align*} For one-forms, the wedge product is the alternating product \begin{align*} (\alpha\wedge\beta)(v,w)=\alpha(v)\beta(w)-\alpha(w)\beta(v). \end{align*} Using $\alpha=dx|_p$ and $\beta=dy|_p$ gives \begin{align*} (dx\wedge dy)|_p(v,w) &=dx|_p(v)\,dy|_p(w)-dx|_p(w)\,dy|_p(v)\\ &=v_1w_2-w_1v_2\\ &=v_1w_2-v_2w_1. \end{align*} Now project the two tangent vectors to the $xy$-plane: \begin{align*} \pi_{xy}(v)&=(v_1,v_2),\\ \pi_{xy}(w)&=(w_1,w_2). \end{align*} The signed area of the parallelogram spanned by two vectors $(a,b)$ and $(c,d)$ in the oriented $xy$-plane is the determinant \begin{align*} \det\begin{pmatrix} a & c\\ b & d \end{pmatrix} =ad-bc. \end{align*} Substituting $(a,b)=(v_1,v_2)$ and $(c,d)=(w_1,w_2)$ gives \begin{align*} \det\begin{pmatrix} v_1 & w_1\\ v_2 & w_2 \end{pmatrix} &=v_1w_2-v_2w_1. \end{align*} Therefore \begin{align*} (dx\wedge dy)|_p(v,w) = \det\begin{pmatrix} v_1 & w_1\\ v_2 & w_2 \end{pmatrix}. \end{align*} The same calculation for the companion coordinate area forms gives \begin{align*} (dy\wedge dz)|_p(v,w) &=dy|_p(v)\,dz|_p(w)-dy|_p(w)\,dz|_p(v)\\ &=v_2w_3-w_2v_3\\ &=v_2w_3-v_3w_2, \end{align*} and \begin{align*} (dz\wedge dx)|_p(v,w) &=dz|_p(v)\,dx|_p(w)-dz|_p(w)\,dx|_p(v)\\ &=v_3w_1-w_3v_1\\ &=v_3w_1-v_1w_3. \end{align*} Thus $dy\wedge dz$ measures signed projected area in the $yz$-plane, and $dz\wedge dx$ measures signed projected area in the $zx$-plane. The $2$-form $dx\wedge dy$ ignores the $z$-components of its input vectors and records exactly the oriented area of their projection to the $xy$-plane. Its companion forms $dy\wedge dz$ and $dz\wedge dx$ record the corresponding oriented projected areas in the other coordinate planes. [/example] The chapter has converted the fixed exterior algebra of a [vector space](/page/Vector%20Space) into a smooth graded algebra over every [open set](/page/Open%20Set) $U\subset\mathbb R^n$. The next operation to add is the [exterior derivative](/theorems/1525), which differentiates coefficient functions and raises degree by one while respecting the wedge product through a signed product rule. With differential forms defined on open sets of ℝⁿ, we now add the derivative operator that respects their algebraic structure. The [exterior derivative](/theorems/1525) raises degree by one, respects the wedge product through a signed rule, and squares to zero. # 3. The Exterior Derivative The [exterior derivative](/theorems/1525) is the operation that turns the algebra of differential forms into a differential complex. In the previous chapters, forms were built from alternating covectors and the wedge product; now we add the derivative compatible with that alternating algebra. The guiding point is that antisymmetry removes the symmetric second-derivative terms, so the same operator explains gradients, curls, divergences, and the obstruction theory behind de Rham cohomology. ## Axioms for Differentiating Forms What should a derivative of forms do that ordinary partial differentiation does not already do? It must raise degree by one, obey a signed product rule for the wedge product, commute with change of variables, and reduce to the usual differential on functions. If we merely differentiated coefficients without the wedge signs, the result would depend on the chosen coordinate expression and the cancellation behind $d^2=0$ would no longer be forced by antisymmetry. [definition: Degree Plus One Antiderivation] Let $U \subset \mathbb R^n$ be open. A family of $\mathbb R$-linear maps $(D_k:\Omega^k(U) \to \Omega^{k+1}(U))_{k\geq 0}$ is a degree plus one antiderivation if, for every $\omega \in \Omega^k(U)$ and every $\eta \in \Omega^\ell(U)$, \begin{align*} D(\omega \wedge \eta)=D\omega \wedge \eta+(-1)^k\omega \wedge D\eta. \end{align*} [/definition] The sign records the degree of the form past which the derivative moves. It is the same sign convention that makes the wedge product graded-commutative. The definition describes the product-rule shape any candidate derivative should have, but it does not guarantee that such an operator exists or is uniquely determined by natural requirements. The key question is whether these axioms force a single coordinate-independent differential on forms. [quotetheorem:1525] [citeproof:1525] This theorem is the reason the notation $d$ is used without choosing coordinates. The coordinate formula is a way to compute the operator, while the axioms explain why the result is invariant under smooth changes of variables. Naturality is essential on manifolds: without compatibility with pullback, formulas written in two overlapping charts might define different forms. Uniqueness matters because it says these requirements leave no room for a second [exterior derivative](/theorems/1525) with different signs or extra terms. The condition $d^2=0$ is what turns the spaces of forms into a complex, so dropping it would lose the later connection with closed forms, exact forms, and cohomology. ## The Coordinate Formula How do we compute $d\omega$ once a form is written in coordinates? The answer is to differentiate only the coefficient functions and wedge the resulting $1$-forms onto the existing basis wedge. For an increasing multi-index $I=(i_1<\cdots<i_k)$, write \begin{align*} dx_I=dx_{i_1}\wedge\cdots\wedge dx_{i_k}. \end{align*} Every $k$-form on $U\subset\mathbb R^n$ has a unique expression \begin{align*} \omega=\sum_{|I|=k} a_I dx_I, \end{align*} with $a_I\in C^\infty(U)$. [quotetheorem:3564] [citeproof:3564] The formula is local, but it is compatible on overlaps because of naturality. On a manifold, this lets us compute $d$ in any chart $(U,\varphi)$ and obtain a globally defined form. The assumption that the coefficients are smooth ensures that all derivatives appearing in the formula exist and that the mixed second derivatives needed for $d^2=0$ agree. The increasing multi-index convention gives each basis wedge a preferred order, so signs are tracked by reordering rather than by duplicating terms. The same ordering convention reappears when top-degree forms are integrated with an orientation. [example: Exterior Derivative Of A One Form] On $\mathbb R^3$ with coordinates $(x,y,z)$, let \begin{align*} \omega=P\,dx+Q\,dy+R\,dz, \end{align*} where $P,Q,R\in C^\infty(\mathbb R^3)$. The [exterior derivative](/theorems/1525) is \begin{align*} d\omega &=\left(\frac{\partial R}{\partial y}-\frac{\partial Q}{\partial z}\right)dy\wedge dz +\left(\frac{\partial P}{\partial z}-\frac{\partial R}{\partial x}\right)dz\wedge dx +\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dx\wedge dy. \end{align*} By $\mathbb R$-linearity of $d$, \begin{align*} d\omega=d(P\,dx)+d(Q\,dy)+d(R\,dz). \end{align*} For the first term, the signed product rule gives \begin{align*} d(P\,dx)=dP\wedge dx+P\,d(dx). \end{align*} Since $dx=d x$, we have $d(dx)=d(d x)=0$, so \begin{align*} d(P\,dx)=dP\wedge dx. \end{align*} Also, \begin{align*} dP=\frac{\partial P}{\partial x}\,dx+\frac{\partial P}{\partial y}\,dy+\frac{\partial P}{\partial z}\,dz, \end{align*} and therefore \begin{align*} d(P\,dx) &=\left(\frac{\partial P}{\partial x}\,dx+\frac{\partial P}{\partial y}\,dy+\frac{\partial P}{\partial z}\,dz\right)\wedge dx\\ &=\frac{\partial P}{\partial x}\,dx\wedge dx +\frac{\partial P}{\partial y}\,dy\wedge dx +\frac{\partial P}{\partial z}\,dz\wedge dx\\ &=0-\frac{\partial P}{\partial y}\,dx\wedge dy +\frac{\partial P}{\partial z}\,dz\wedge dx\\ &=\frac{\partial P}{\partial z}\,dz\wedge dx -\frac{\partial P}{\partial y}\,dx\wedge dy. \end{align*} Similarly, \begin{align*} d(Q\,dy)=dQ\wedge dy \end{align*} because $d(dy)=d(d y)=0$. Since \begin{align*} dQ=\frac{\partial Q}{\partial x}\,dx+\frac{\partial Q}{\partial y}\,dy+\frac{\partial Q}{\partial z}\,dz, \end{align*} we get \begin{align*} d(Q\,dy) &=\left(\frac{\partial Q}{\partial x}\,dx+\frac{\partial Q}{\partial y}\,dy+\frac{\partial Q}{\partial z}\,dz\right)\wedge dy\\ &=\frac{\partial Q}{\partial x}\,dx\wedge dy +\frac{\partial Q}{\partial y}\,dy\wedge dy +\frac{\partial Q}{\partial z}\,dz\wedge dy\\ &=\frac{\partial Q}{\partial x}\,dx\wedge dy +0-\frac{\partial Q}{\partial z}\,dy\wedge dz\\ &=-\frac{\partial Q}{\partial z}\,dy\wedge dz +\frac{\partial Q}{\partial x}\,dx\wedge dy. \end{align*} For the last term, \begin{align*} d(R\,dz)=dR\wedge dz \end{align*} because $d(dz)=d(d z)=0$. Since \begin{align*} dR=\frac{\partial R}{\partial x}\,dx+\frac{\partial R}{\partial y}\,dy+\frac{\partial R}{\partial z}\,dz, \end{align*} we have \begin{align*} d(R\,dz) &=\left(\frac{\partial R}{\partial x}\,dx+\frac{\partial R}{\partial y}\,dy+\frac{\partial R}{\partial z}\,dz\right)\wedge dz\\ &=\frac{\partial R}{\partial x}\,dx\wedge dz +\frac{\partial R}{\partial y}\,dy\wedge dz +\frac{\partial R}{\partial z}\,dz\wedge dz\\ &=-\frac{\partial R}{\partial x}\,dz\wedge dx +\frac{\partial R}{\partial y}\,dy\wedge dz +0\\ &=\frac{\partial R}{\partial y}\,dy\wedge dz -\frac{\partial R}{\partial x}\,dz\wedge dx. \end{align*} Adding the three displayed formulas gives \begin{align*} d\omega &=\left(\frac{\partial P}{\partial z}\,dz\wedge dx -\frac{\partial P}{\partial y}\,dx\wedge dy\right) +\left(-\frac{\partial Q}{\partial z}\,dy\wedge dz +\frac{\partial Q}{\partial x}\,dx\wedge dy\right)\\ &\qquad +\left(\frac{\partial R}{\partial y}\,dy\wedge dz -\frac{\partial R}{\partial x}\,dz\wedge dx\right)\\ &=\left(\frac{\partial R}{\partial y}-\frac{\partial Q}{\partial z}\right)dy\wedge dz +\left(\frac{\partial P}{\partial z}-\frac{\partial R}{\partial x}\right)dz\wedge dx +\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dx\wedge dy. \end{align*} The coefficient of each basis $2$-form is the difference of the two cross-partial terms forced by the antisymmetry of the wedge product. In the ordered basis $dy\wedge dz$, $dz\wedge dx$, $dx\wedge dy$, these are exactly the three curl-type combinations of $P,Q,R$. [/example] This example is the first sign that the [exterior derivative](/theorems/1525) contains curl. The signs are not an added convention; they are forced by alternating multiplication. [example: Exterior Derivative Of A Two Form] On $\mathbb R^3$ with coordinates $(x,y,z)$, let \begin{align*} \beta=P\,dy\wedge dz+Q\,dz\wedge dx+R\,dx\wedge dy, \end{align*} where $P,Q,R\in C^\infty(\mathbb R^3)$. The [exterior derivative](/theorems/1525) is \begin{align*} d\beta=\left(\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z}\right)dx\wedge dy\wedge dz. \end{align*} By $\mathbb R$-linearity of $d$, \begin{align*} d\beta=d(P\,dy\wedge dz)+d(Q\,dz\wedge dx)+d(R\,dx\wedge dy). \end{align*} The $\mathbb R$-linearity, signed product rule, and identity $d^2=0$ are the axioms in *[Exterior Derivative](/theorems/1525) Characterisation*. For the first summand, the signed product rule gives \begin{align*} d(P\,dy\wedge dz)=dP\wedge dy\wedge dz+P\,d(dy\wedge dz). \end{align*} Since $dy=d y$ and $dz=d z$, \begin{align*} d(dy\wedge dz) &=d(dy)\wedge dz-dy\wedge d(dz)\\ &=d(d y)\wedge dz-dy\wedge d(d z)\\ &=0\wedge dz-dy\wedge 0\\ &=0. \end{align*} Thus \begin{align*} d(P\,dy\wedge dz)=dP\wedge dy\wedge dz. \end{align*} Since \begin{align*} dP=\frac{\partial P}{\partial x}\,dx+\frac{\partial P}{\partial y}\,dy+\frac{\partial P}{\partial z}\,dz, \end{align*} we get \begin{align*} d(P\,dy\wedge dz) &=\left(\frac{\partial P}{\partial x}\,dx+\frac{\partial P}{\partial y}\,dy+\frac{\partial P}{\partial z}\,dz\right)\wedge dy\wedge dz\\ &=\frac{\partial P}{\partial x}\,dx\wedge dy\wedge dz +\frac{\partial P}{\partial y}\,dy\wedge dy\wedge dz +\frac{\partial P}{\partial z}\,dz\wedge dy\wedge dz\\ &=\frac{\partial P}{\partial x}\,dx\wedge dy\wedge dz+0+0\\ &=\frac{\partial P}{\partial x}\,dx\wedge dy\wedge dz. \end{align*} For the second summand, \begin{align*} d(Q\,dz\wedge dx)=dQ\wedge dz\wedge dx+Q\,d(dz\wedge dx). \end{align*} Since $dz=d z$ and $dx=d x$, \begin{align*} d(dz\wedge dx) &=d(dz)\wedge dx-dz\wedge d(dx)\\ &=d(d z)\wedge dx-dz\wedge d(d x)\\ &=0\wedge dx-dz\wedge 0\\ &=0. \end{align*} Thus \begin{align*} d(Q\,dz\wedge dx)=dQ\wedge dz\wedge dx. \end{align*} Since \begin{align*} dQ=\frac{\partial Q}{\partial x}\,dx+\frac{\partial Q}{\partial y}\,dy+\frac{\partial Q}{\partial z}\,dz, \end{align*} we get \begin{align*} d(Q\,dz\wedge dx) &=\left(\frac{\partial Q}{\partial x}\,dx+\frac{\partial Q}{\partial y}\,dy+\frac{\partial Q}{\partial z}\,dz\right)\wedge dz\wedge dx\\ &=\frac{\partial Q}{\partial x}\,dx\wedge dz\wedge dx +\frac{\partial Q}{\partial y}\,dy\wedge dz\wedge dx +\frac{\partial Q}{\partial z}\,dz\wedge dz\wedge dx\\ &=0+\frac{\partial Q}{\partial y}\,dy\wedge dz\wedge dx+0\\ &=\frac{\partial Q}{\partial y}\,(-dy\wedge dx\wedge dz)\\ &=\frac{\partial Q}{\partial y}\,dx\wedge dy\wedge dz. \end{align*} For the third summand, \begin{align*} d(R\,dx\wedge dy)=dR\wedge dx\wedge dy+R\,d(dx\wedge dy). \end{align*} Since $dx=d x$ and $dy=d y$, \begin{align*} d(dx\wedge dy) &=d(dx)\wedge dy-dx\wedge d(dy)\\ &=d(d x)\wedge dy-dx\wedge d(d y)\\ &=0\wedge dy-dx\wedge 0\\ &=0. \end{align*} Thus \begin{align*} d(R\,dx\wedge dy)=dR\wedge dx\wedge dy. \end{align*} Since \begin{align*} dR=\frac{\partial R}{\partial x}\,dx+\frac{\partial R}{\partial y}\,dy+\frac{\partial R}{\partial z}\,dz, \end{align*} we get \begin{align*} d(R\,dx\wedge dy) &=\left(\frac{\partial R}{\partial x}\,dx+\frac{\partial R}{\partial y}\,dy+\frac{\partial R}{\partial z}\,dz\right)\wedge dx\wedge dy\\ &=\frac{\partial R}{\partial x}\,dx\wedge dx\wedge dy +\frac{\partial R}{\partial y}\,dy\wedge dx\wedge dy +\frac{\partial R}{\partial z}\,dz\wedge dx\wedge dy\\ &=0+0+\frac{\partial R}{\partial z}\,dz\wedge dx\wedge dy\\ &=\frac{\partial R}{\partial z}\,(-dx\wedge dz\wedge dy)\\ &=\frac{\partial R}{\partial z}\,dx\wedge dy\wedge dz. \end{align*} Adding the three summands gives \begin{align*} d\beta &=\frac{\partial P}{\partial x}\,dx\wedge dy\wedge dz +\frac{\partial Q}{\partial y}\,dx\wedge dy\wedge dz +\frac{\partial R}{\partial z}\,dx\wedge dy\wedge dz\\ &=\left(\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z}\right)dx\wedge dy\wedge dz. \end{align*} The [exterior derivative](/theorems/1525) of this $2$-form is a scalar multiple of the standard oriented volume form $dx\wedge dy\wedge dz$. Under the encoding $(P,Q,R)\leftrightarrow P\,dy\wedge dz+Q\,dz\wedge dx+R\,dx\wedge dy$, that scalar is the divergence-type sum $\partial P/\partial x+\partial Q/\partial y+\partial R/\partial z$. [/example] The second example shows the divergence pattern. The coordinate formula therefore packages several vector calculus operations into one construction. ## Closed And Exact Forms When does the differential equation $d\eta=\omega$ have a solution, and what condition must $\omega$ satisfy before such a solution can exist? The identity $d^2=0$ supplies the universal compatibility condition. [definition: Closed Differential Form] Let $U\subset\mathbb R^n$ be open, and let $\omega\in\Omega^k(U)$. The form $\omega$ is closed if \begin{align*} d\omega=0. \end{align*} [/definition] Closure is a differential condition on the coefficients of $\omega$. For a $1$-form it says that the mixed first derivatives line up in the pattern detected by curl. The compatibility condition is only the first half of the equation $d\eta=\omega$. To discuss whether a closed form actually comes from a potential, we need a separate name for forms that are produced as exterior derivatives. [definition: Exact Differential Form] Let $U\subset\mathbb R^n$ be open, and let $\omega\in\Omega^k(U)$. The form $\omega$ is exact if there exists $\eta\in\Omega^{k-1}(U)$ such that \begin{align*} \omega=d\eta. \end{align*} [/definition] Exactness is a solvability condition: $\eta$ is a potential for $\omega$. The central question of de Rham cohomology is how far closed forms are from being exact. [quotetheorem:3565] [citeproof:3565] The converse is both a local question and a global question. Closedness is checked locally from the coefficients of $d\omega$, while exactness asks for a single potential defined on the whole domain. Locally, the Poincare lemma later proves that closed forms have potentials on star-shaped regions; globally, holes in the domain can obstruct the existence of a potential. De Rham cohomology records exactly this obstruction by measuring closed forms modulo exact forms. [example: Closed One Form On The Punctured Plane] Let $U=\mathbb R^2\setminus\{(0,0)\}$ with coordinates $(x,y)$, and set \begin{align*} \omega=\frac{-y}{x^2+y^2}\,dx+\frac{x}{x^2+y^2}\,dy. \end{align*} Write \begin{align*} P(x,y)=\frac{-y}{x^2+y^2},\qquad Q(x,y)=\frac{x}{x^2+y^2}, \end{align*} so that $\omega=P\,dx+Q\,dy$. Since $x^2+y^2\neq 0$ on $U$, both $P$ and $Q$ are smooth on $U$. The form $\omega$ is closed on $U$, but it is not exact on $U$. First compute $d\omega$. By $\mathbb R$-linearity and the product rule for $d$, \begin{align*} d\omega &=d(P\,dx)+d(Q\,dy)\\ &=dP\wedge dx+P\,d(dx)+dQ\wedge dy+Q\,d(dy). \end{align*} Since $dx=d x$ and $dy=d y$, the identity $d^2=0$ gives \begin{align*} d(dx)=d(d x)=0,\qquad d(dy)=d(d y)=0. \end{align*} Thus \begin{align*} d\omega=dP\wedge dx+dQ\wedge dy. \end{align*} Now \begin{align*} dP=\frac{\partial P}{\partial x}\,dx+\frac{\partial P}{\partial y}\,dy, \qquad dQ=\frac{\partial Q}{\partial x}\,dx+\frac{\partial Q}{\partial y}\,dy, \end{align*} so \begin{align*} dP\wedge dx &=\left(\frac{\partial P}{\partial x}\,dx+\frac{\partial P}{\partial y}\,dy\right)\wedge dx\\ &=\frac{\partial P}{\partial x}\,dx\wedge dx+\frac{\partial P}{\partial y}\,dy\wedge dx\\ &=0-\frac{\partial P}{\partial y}\,dx\wedge dy, \end{align*} and \begin{align*} dQ\wedge dy &=\left(\frac{\partial Q}{\partial x}\,dx+\frac{\partial Q}{\partial y}\,dy\right)\wedge dy\\ &=\frac{\partial Q}{\partial x}\,dx\wedge dy+\frac{\partial Q}{\partial y}\,dy\wedge dy\\ &=\frac{\partial Q}{\partial x}\,dx\wedge dy+0. \end{align*} Therefore \begin{align*} d\omega &=\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dx\wedge dy. \end{align*} It remains to compute the two partial derivatives. For $Q=x(x^2+y^2)^{-1}$, \begin{align*} \frac{\partial Q}{\partial x} &=(x^2+y^2)^{-1}+x\left(-1\right)(x^2+y^2)^{-2}(2x)\\ &=\frac{1}{x^2+y^2}-\frac{2x^2}{(x^2+y^2)^2}\\ &=\frac{x^2+y^2}{(x^2+y^2)^2}-\frac{2x^2}{(x^2+y^2)^2}\\ &=\frac{y^2-x^2}{(x^2+y^2)^2}. \end{align*} For $P=-y(x^2+y^2)^{-1}$, \begin{align*} \frac{\partial P}{\partial y} &=-(x^2+y^2)^{-1}+(-y)\left(-1\right)(x^2+y^2)^{-2}(2y)\\ &=-\frac{1}{x^2+y^2}+\frac{2y^2}{(x^2+y^2)^2}\\ &=-\frac{x^2+y^2}{(x^2+y^2)^2}+\frac{2y^2}{(x^2+y^2)^2}\\ &=\frac{y^2-x^2}{(x^2+y^2)^2}. \end{align*} Hence \begin{align*} \frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y} &=\frac{y^2-x^2}{(x^2+y^2)^2}-\frac{y^2-x^2}{(x^2+y^2)^2}\\ &=0, \end{align*} and so \begin{align*} d\omega=0. \end{align*} Thus $\omega$ is closed on $U$. To show that $\omega$ is not exact on $U$, consider the closed parametrised curve \begin{align*} \gamma:[0,2\pi]\to U,\qquad \gamma(t)=(\cos t,\sin t). \end{align*} Along this curve, \begin{align*} x=\cos t,\qquad y=\sin t,\qquad dx=-\sin t\,dt,\qquad dy=\cos t\,dt, \end{align*} and \begin{align*} x^2+y^2=\cos^2 t+\sin^2 t=1. \end{align*} Therefore \begin{align*} \gamma^*\omega &=\frac{-\sin t}{1}(-\sin t\,dt)+\frac{\cos t}{1}(\cos t\,dt)\\ &=\sin^2 t\,dt+\cos^2 t\,dt\\ &=(\sin^2 t+\cos^2 t)\,dt\\ &=dt. \end{align*} Thus \begin{align*} \int_\gamma \omega &=\int_0^{2\pi}\gamma^*\omega\\ &=\int_0^{2\pi}dt\\ &=2\pi. \end{align*} Now suppose, for contradiction, that $\omega$ were exact on $U$. Then there would be a smooth function $f:U\to\mathbb R$ such that $\omega=df$. Along $\gamma$, \begin{align*} \int_\gamma \omega &=\int_\gamma df\\ &=\int_0^{2\pi}\frac{d}{dt}\bigl(f(\gamma(t))\bigr)\,dt\\ &=f(\gamma(2\pi))-f(\gamma(0))\\ &=f(1,0)-f(1,0)\\ &=0. \end{align*} This contradicts the computation $\int_\gamma\omega=2\pi$. Hence no such global potential $f$ exists, and $\omega$ is not exact on $U$. The equality $d\omega=0$ is a local differential condition, verified by the equality $\partial Q/\partial x=\partial P/\partial y$ on $U$. Exactness is stronger: it would force every closed-curve integral of $\omega$ to vanish. The unit circle integral equals $2\pi$, so this closed form is not exact on the punctured plane. [/example] This example motivates the quotient of closed forms by exact forms. That quotient is the de Rham cohomology group, introduced after integration and homotopy invariance are available. ## Gradient Curl And Divergence Why do the identities $\nabla\times\nabla f=0$ and $\nabla\cdot(\nabla\times F)=0$ have the same proof? After choosing the Euclidean metric and the standard orientation on $\mathbb R^3$, they are both the single identity $d^2=0$ written in different degrees. A smooth function $f\in C^\infty(\mathbb R^3)$ is a $0$-form. A vector field $F=(P,Q,R)$ may be encoded either as the $1$-form \begin{align*} \alpha_F=P\,dx+Q\,dy+R\,dz \end{align*} or as the $2$-form \begin{align*} \beta_F=P\,dy\wedge dz+Q\,dz\wedge dx+R\,dx\wedge dy. \end{align*} The first encoding is used for curl, and the second encoding is used for divergence. [quotetheorem:3566] [citeproof:3566] The theorem also explains why the classical operators occur in the order gradient, curl, divergence. They are the degree-by-degree pieces of the de Rham complex on $\mathbb R^3$: \begin{align*} 0\longrightarrow \Omega^0(\mathbb R^3)\xrightarrow{d}\Omega^1(\mathbb R^3)\xrightarrow{d}\Omega^2(\mathbb R^3)\xrightarrow{d}\Omega^3(\mathbb R^3)\longrightarrow 0. \end{align*} [example: A Curl Computation] On $\mathbb R^3$ with coordinates $(x,y,z)$, let \begin{align*} F=(yz,xz,xy). \end{align*} Encode $F$ as the $1$-form \begin{align*} \alpha_F=yz\,dx+xz\,dy+xy\,dz. \end{align*} Write \begin{align*} P=yz,\qquad Q=xz,\qquad R=xy, \end{align*} so that $\alpha_F=P\,dx+Q\,dy+R\,dz$. We have \begin{align*} d\alpha_F=0. \end{align*} Hence, under the standard identification in *Vector Calculus From [Exterior Derivative](/theorems/1525)*, $\nabla\times F=0$. Moreover, $F=\nabla(xyz)$. By *Coordinate Formula For [Exterior Derivative](/theorems/1525)*, applied to the $1$-form $P\,dx+Q\,dy+R\,dz$, \begin{align*} d\alpha_F &=\left(\frac{\partial R}{\partial y}-\frac{\partial Q}{\partial z}\right)dy\wedge dz +\left(\frac{\partial P}{\partial z}-\frac{\partial R}{\partial x}\right)dz\wedge dx +\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dx\wedge dy. \end{align*} Compute each partial derivative: \begin{align*} \frac{\partial R}{\partial y} &=\frac{\partial (xy)}{\partial y} =x\frac{\partial y}{\partial y}+y\frac{\partial x}{\partial y} =x\cdot 1+y\cdot 0 =x,\\ \frac{\partial Q}{\partial z} &=\frac{\partial (xz)}{\partial z} =x\frac{\partial z}{\partial z}+z\frac{\partial x}{\partial z} =x\cdot 1+z\cdot 0 =x,\\ \frac{\partial P}{\partial z} &=\frac{\partial (yz)}{\partial z} =y\frac{\partial z}{\partial z}+z\frac{\partial y}{\partial z} =y\cdot 1+z\cdot 0 =y,\\ \frac{\partial R}{\partial x} &=\frac{\partial (xy)}{\partial x} =x\frac{\partial y}{\partial x}+y\frac{\partial x}{\partial x} =x\cdot 0+y\cdot 1 =y,\\ \frac{\partial Q}{\partial x} &=\frac{\partial (xz)}{\partial x} =x\frac{\partial z}{\partial x}+z\frac{\partial x}{\partial x} =x\cdot 0+z\cdot 1 =z,\\ \frac{\partial P}{\partial y} &=\frac{\partial (yz)}{\partial y} =y\frac{\partial z}{\partial y}+z\frac{\partial y}{\partial y} =y\cdot 0+z\cdot 1 =z. \end{align*} Substituting these values into the formula for $d\alpha_F$ gives \begin{align*} d\alpha_F &=(x-x)\,dy\wedge dz+(y-y)\,dz\wedge dx+(z-z)\,dx\wedge dy\\ &=0\,dy\wedge dz+0\,dz\wedge dx+0\,dx\wedge dy\\ &=0. \end{align*} By *Vector Calculus From [Exterior Derivative](/theorems/1525)*, the $2$-form $d\alpha_F$ corresponds to $\nabla\times F$. Therefore \begin{align*} \nabla\times F=(0,0,0). \end{align*} Finally, let $f=xyz$. Then \begin{align*} \nabla f &=\left(\frac{\partial (xyz)}{\partial x},\frac{\partial (xyz)}{\partial y},\frac{\partial (xyz)}{\partial z}\right)\\ &=(yz,xz,xy)\\ &=F. \end{align*} Thus $F$ is the gradient of the globally defined smooth function $xyz$. The vanishing of $d\alpha_F$ comes from the three cancellations $x-x=0$, $y-y=0$, and $z-z=0$ in the curl-type coefficients. The vector field is not merely locally curl-free: it is globally a gradient field on $\mathbb R^3$, with potential $xyz$. [/example] The curl example illustrates the degree-$1$ part of the complex: a $1$-form is differentiated into a $2$-form, and vanishing means that a local potential may exist. The divergence example moves one step further, from $2$-forms to top-degree $3$-forms, where the output is a scalar multiple of the oriented volume form. Reading the two examples together shows how $d$ links the vector calculus operators in sequence rather than treating them as unrelated formulas. [example: A Divergence Computation] On $\mathbb R^3$ with coordinates $(x,y,z)$, let \begin{align*} F=(x^2,y^2,z^2). \end{align*} Encode $F$ as the $2$-form \begin{align*} \beta_F=x^2\,dy\wedge dz+y^2\,dz\wedge dx+z^2\,dx\wedge dy. \end{align*} Write \begin{align*} P=x^2,\qquad Q=y^2,\qquad R=z^2, \end{align*} so that \begin{align*} \beta_F=P\,dy\wedge dz+Q\,dz\wedge dx+R\,dx\wedge dy. \end{align*} We have \begin{align*} d\beta_F=2(x+y+z)\,dx\wedge dy\wedge dz. \end{align*} Under the standard identification in *Vector Calculus From [Exterior Derivative](/theorems/1525)*, this says \begin{align*} \nabla\cdot F=2x+2y+2z. \end{align*} By the $2$-form computation in *Coordinate Formula For [Exterior Derivative](/theorems/1525)*, \begin{align*} d\beta_F =\left(\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z}\right)dx\wedge dy\wedge dz. \end{align*} For the three coefficient functions, \begin{align*} \frac{\partial P}{\partial x} &=\frac{\partial}{\partial x}(x^2) =2x,\\ \frac{\partial Q}{\partial y} &=\frac{\partial}{\partial y}(y^2) =2y,\\ \frac{\partial R}{\partial z} &=\frac{\partial}{\partial z}(z^2) =2z. \end{align*} Substituting these three derivatives gives \begin{align*} d\beta_F &=(2x+2y+2z)\,dx\wedge dy\wedge dz\\ &=2(x+y+z)\,dx\wedge dy\wedge dz. \end{align*} By *Vector Calculus From [Exterior Derivative](/theorems/1525)*, the coefficient of $dx\wedge dy\wedge dz$ in $d\beta_F$ corresponds to the divergence of $F$. Therefore \begin{align*} \nabla\cdot F &=2x+2y+2z. \end{align*} The [exterior derivative](/theorems/1525) of the encoded $2$-form records the sum of the three coordinate derivatives of the vector field components. For $F=(x^2,y^2,z^2)$, those derivatives are $2x$, $2y$, and $2z$, so the divergence is $2x+2y+2z$. [/example] The [exterior derivative](/theorems/1525) is therefore not a new vector calculus operator beside grad, curl, and div. It is the coordinate-free operator whose shadows in dimension three are precisely those familiar operations, and whose square-zero property is the algebraic source of the compatibility conditions used throughout the rest of the course. Beyond the [exterior derivative](/theorems/1525), we must understand how forms behave under smooth maps between spaces. Pullback is the operation that transports a differential form from a target space back to a source space, and it will be compatible with the [exterior derivative](/theorems/1525) in a way that underpins [Stokes' theorem](/theorems/1530). # 4. Pullback of Forms Pullback answers the question: how does a differential form on a target space become a form on a source space when we have a smooth map into that target? The construction is forced by the interpretation of a $k$-form as something that takes $k$ tangent vectors as input. A map pushes tangent vectors forward by its total derivative, so a form pulls back by first pushing the input vectors forward and then evaluating the original form. This chapter develops that operation, proves its compatibility with wedge products and exterior derivatives, and explains why the determinant in multivariable change of variables is a top-degree pullback coefficient. ## Pulling Forms Back Along Smooth Maps What data are needed to measure a $k$-dimensional infinitesimal parallelepiped in the source using a form defined on the target? Suppose $U$ and $V$ are open subsets of Euclidean spaces and $f: U \to V$ is smooth. A tangent vector at $x \in U$ is sent to a tangent vector at $f(x)$ by $Df_x$, so the natural way to evaluate a target form on source vectors is to feed their images into the target form. [definition: Pullback Of A Differential Form] Let $U \subseteq \mathbb{R}^m$ and $V \subseteq \mathbb{R}^n$ be open sets, let $f: U \to V$ be smooth, and let $\omega \in \Omega^k(V)$. The pullback of $\omega$ by $f$ is the $k$-form $f^*\omega \in \Omega^k(U)$ defined by \begin{align*} (f^*\omega)_x(v_1,\dots,v_k) = \omega_{f(x)}(Df_x(v_1),\dots,Df_x(v_k)) \end{align*} for $x \in U$ and $v_1,\dots,v_k \in \mathbb{R}^m$. [/definition] The definition is pointwise, but smoothness of $f$ ensures that the resulting coefficients vary smoothly in $x$. For $0$-forms, which are smooth functions, this recovers composition: if $g \in C^\infty(V)$, then $f^*g = g \circ f$. The same formula works for smooth maps $F:M\to N$ between manifolds, replacing $Df_x$ by the differential $dF_p:T_pM\to T_{F(p)}N$. [example: Pullback Of A One-Form On The Plane] Let $(u,v)$ be the source coordinates on $\mathbb{R}^2$, and let $(x,y)$ be the target coordinates on $\mathbb{R}^2$. Define \begin{align*} f(u,v)=(x(u,v),y(u,v))=(u^2-v,uv), \end{align*} and let \begin{align*} \omega=y\,dx+x\,dy. \end{align*} The pullback of $\omega$ by $f$ is \begin{align*} f^*\omega &=(uv)(2u\,du-dv)+(u^2-v)(v\,du+u\,dv) \\ &=(3u^2v-v^2)\,du+(u^3-2uv)\,dv. \end{align*} Pullback of coefficient functions is composition with $f$, so \begin{align*} f^*x&=x\circ f=u^2-v,\\ f^*y&=y\circ f=uv. \end{align*} For coordinate differentials, pullback replaces the target differential by the differential of the pulled-back coordinate function: \begin{align*} f^*(dx)&=d(f^*x)=d(u^2-v),\\ f^*(dy)&=d(f^*y)=d(uv). \end{align*} Therefore \begin{align*} f^*\omega &=f^*(y\,dx+x\,dy)\\ &=f^*y\,f^*(dx)+f^*x\,f^*(dy)\\ &=(uv)\,d(u^2-v)+(u^2-v)\,d(uv). \end{align*} Using linearity of $d$ and the ordinary product rule for functions, \begin{align*} d(u^2-v) &=d(u^2)-d(v)\\ &=2u\,du-dv, \end{align*} and \begin{align*} d(uv) &=u\,dv+v\,du\\ &=v\,du+u\,dv. \end{align*} Substituting these two differentials gives \begin{align*} f^*\omega &=(uv)(2u\,du-dv)+(u^2-v)(v\,du+u\,dv). \end{align*} Expanding the first product, \begin{align*} (uv)(2u\,du-dv) &=(uv)(2u\,du)-(uv)\,dv\\ &=2u^2v\,du-uv\,dv. \end{align*} Expanding the second product, \begin{align*} (u^2-v)(v\,du+u\,dv) &=(u^2-v)v\,du+(u^2-v)u\,dv\\ &=(u^2v-v^2)\,du+(u^3-uv)\,dv. \end{align*} Adding the $du$-terms and the $dv$-terms separately, \begin{align*} f^*\omega &=\bigl(2u^2v\,du-uv\,dv\bigr)+\bigl((u^2v-v^2)\,du+(u^3-uv)\,dv\bigr)\\ &=\bigl(2u^2v+u^2v-v^2\bigr)\,du+\bigl(-uv+u^3-uv\bigr)\,dv\\ &=(3u^2v-v^2)\,du+(u^3-2uv)\,dv. \end{align*} Pullback does two things at once: it substitutes $x=u^2-v$ and $y=uv$ into the coefficient functions, and it replaces $dx$ and $dy$ by the differentials of the pulled-back coordinate functions. Thus the target form $y\,dx+x\,dy$ becomes a source form expressed entirely in $du$ and $dv$. [/example] The calculation gives a candidate formula for pullback, but an operator on forms needs more than a successful example. If $f$ has a derivative whose rank changes from point to point, the expression obtained by inserting $Df_x$ into $\omega_{f(x)}$ must still vary smoothly with $x$. The obstruction is therefore a smoothness question: pointwise alternating multilinearity is inherited from $\omega$, while smooth dependence on the base point has to be checked. Without this check, pullback would be only a pointwise recipe rather than an operation on smooth differential forms. The needed result is that smoothness of $f$ is exactly enough to promote the coordinate formula to a genuine map $\Omega^k(V)\to\Omega^k(U)$ for every smooth map of coordinate domains. [quotetheorem:3567] [citeproof:3567] This result lets us use $f^*$ as an operator $\Omega^k(V) \to \Omega^k(U)$, not merely as a pointwise formula. Smoothness of $f$ is essential here: a merely continuous map has no derivative with which to push tangent vectors forward, and even a differentiable map with poorly behaved derivative need not produce smooth coefficients. Degenerate points of $Df_x$ cause no problem; they simply make some pulled-back forms vanish on directions that are collapsed by the map. The next issue is whether pullback respects the algebraic structure built in the first chapters. ## Algebraic Behaviour Of Pullback If differential forms are multiplied by wedge product, what kind of map should pullback be on the exterior algebra? Since pullback applies the same derivative $Df_x$ to every input vector before evaluation, it should preserve all alternating algebraic operations. It also reverses the direction of maps: a map $f: U \to V$ gives a map on forms from $V$ back to $U$. [quotetheorem:3568] [citeproof:3568] Functoriality is the formal reason pullback is the right operation for changing coordinates. It says that changing from one coordinate system to another and then to a third gives the same result as changing directly. The direction reversal is important: maps of spaces compose as $U \to V \to W$, while the induced maps on forms go $\Omega^k(W) \to \Omega^k(V) \to \Omega^k(U)$. This contravariance is also the pattern that later reappears when smooth maps induce maps on de Rham cohomology. Functoriality alone does not say how pullback interacts with the wedge product. Since forms are multiplied pointwise using alternating algebra, we still need to know that pulling back a product gives the same result as multiplying the pulled-back factors. [quotetheorem:3569] [citeproof:3569] This theorem says that every algebraic identity among forms survives pullback. For instance, if $\omega\wedge\omega=0$ because $\omega$ has odd degree, then $f^*\omega\wedge f^*\omega=0$ as well. [example: Pullback Under The Circle Inclusion] Let $\iota:S^1\to\mathbb{R}^2\setminus\{0\}$ be the inclusion. On the angular chart $t$ for $S^1\setminus\{(1,0)\}$, write \begin{align*} \iota(t)=(x(t),y(t))=(\cos t,\sin t). \end{align*} On $\mathbb{R}^2\setminus\{0\}$ consider the $1$-form \begin{align*} d\theta=\frac{-y\,dx+x\,dy}{x^2+y^2}. \end{align*} On this angular chart, \begin{align*} \iota^*(d\theta)=dt. \end{align*} Pulling back the target coordinate functions means composing them with $\iota$, so \begin{align*} \iota^*x&=x\circ\iota=\cos t,\\ \iota^*y&=y\circ\iota=\sin t. \end{align*} For the coordinate differentials, \begin{align*} \iota^*(dx)&=d(\iota^*x)=d(\cos t)=-\sin t\,dt,\\ \iota^*(dy)&=d(\iota^*y)=d(\sin t)=\cos t\,dt. \end{align*} Now pull back the numerator: \begin{align*} \iota^*(-y\,dx+x\,dy) &=-(\iota^*y)\,\iota^*(dx)+(\iota^*x)\,\iota^*(dy)\\ &=-(\sin t)(-\sin t\,dt)+(\cos t)(\cos t\,dt)\\ &=\sin^2 t\,dt+\cos^2 t\,dt\\ &=(\sin^2 t+\cos^2 t)\,dt\\ &=dt. \end{align*} The denominator pulls back to \begin{align*} \iota^*(x^2+y^2) &=(\iota^*x)^2+(\iota^*y)^2\\ &=(\cos t)^2+(\sin t)^2\\ &=\cos^2 t+\sin^2 t\\ &=1. \end{align*} Therefore \begin{align*} \iota^*(d\theta) &=\iota^*\left(\frac{-y\,dx+x\,dy}{x^2+y^2}\right)\\ &=\frac{\iota^*(-y\,dx+x\,dy)}{\iota^*(x^2+y^2)}\\ &=\frac{dt}{1}\\ &=dt. \end{align*} Along the unit circle, the form $d\theta$ pulls back to the ordinary angular differential $dt$. Thus $d\theta$ measures infinitesimal angular displacement when restricted to the circle. [/example] The form $d\theta$ is locally the differential of an angle function, but it is not globally the differential of a single-valued smooth function on the punctured plane. Pullback to the circle records the winding behaviour that later becomes visible in de Rham cohomology. ## Pullback And The Exterior Derivative Can differentiation of forms commute with changing variables? It must do so if the [exterior derivative](/theorems/1525) is to be independent of the coordinates used to compute it. The central compatibility result is naturality of $d$: pull back first and then differentiate, or differentiate first and then pull back, and the result is the same. [quotetheorem:1525] [citeproof:1525] This theorem is the reason closed and exact forms behave well under smooth maps. If $d\omega=0$, then $d(f^*\omega)=0$; if $\omega=d\eta$, then $f^*\omega=d(f^*\eta)$. [remark: Pullback Preserves Closed And Exact Forms] For a smooth map $f: U \to V$, pullback sends closed forms on $V$ to closed forms on $U$ and exact forms on $V$ to exact forms on $U$. Therefore $f^*$ descends to a map on de Rham cohomology groups, once those groups have been defined. [/remark] The remark previews the role of pullback in cohomology. Smooth maps between spaces will induce linear maps between cohomology groups in the opposite direction. ## Coordinate Formulae How do we compute pullbacks without returning to the multilinear definition each time? The answer is to express a form in coordinate differentials and replace each target coordinate function by its component under $f$. This produces a compact formula that is the practical workhorse for examples. [quotetheorem:3570] [citeproof:3570] This formula is often the shortest path through a computation. It also records dimension effects automatically: if $k>m$, then every pulled-back $k$-form on $U\subseteq\mathbb{R}^m$ is zero because there are not enough independent coordinate differentials on the source. The map $f$ need not be injective for pullback to make sense; non-injectivity affects global integration, but the pulled-back form is still defined pointwise from $Df_x$. The same coordinate formula also shows why the Jacobian appears when pulling back top-degree forms. [quotetheorem:3571] [citeproof:3571] The determinant is therefore not an additional ingredient imposed on integration. It is the coefficient by which a smooth map rescales the oriented top-degree form. [illustration:forms-polar-coordinate-sector] [example: Polar Coordinates And Area] Let $(r,\theta)$ be the source coordinates on $(0,\infty)\times(0,2\pi)$, and let $(x,y)$ be the target coordinates on $\mathbb{R}^2$. Define \begin{align*} f:(0,\infty)\times(0,2\pi)&\to \mathbb{R}^2\setminus\{(x,0):x\ge 0\},\\ f(r,\theta)&=(x(r,\theta),y(r,\theta))=(r\cos\theta,r\sin\theta). \end{align*} We pull back the standard area form $dx\wedge dy$. The pullback of $dx\wedge dy$ under polar coordinates is \begin{align*} f^*(dx\wedge dy)=r\,dr\wedge d\theta. \end{align*} Pulling back the target coordinate functions means composing them with $f$: \begin{align*} f^*x&=x\circ f=r\cos\theta,\\ f^*y&=y\circ f=r\sin\theta. \end{align*} For coordinate differentials, \begin{align*} f^*(dx)&=d(f^*x)=d(r\cos\theta),\\ f^*(dy)&=d(f^*y)=d(r\sin\theta). \end{align*} Using the product rule for functions and the one-variable identities $d(\cos\theta)=-\sin\theta\,d\theta$ and $d(\sin\theta)=\cos\theta\,d\theta$, \begin{align*} d(r\cos\theta) &=r\,d(\cos\theta)+\cos\theta\,dr\\ &=r(-\sin\theta\,d\theta)+\cos\theta\,dr\\ &=\cos\theta\,dr-r\sin\theta\,d\theta, \end{align*} and \begin{align*} d(r\sin\theta) &=r\,d(\sin\theta)+\sin\theta\,dr\\ &=r(\cos\theta\,d\theta)+\sin\theta\,dr\\ &=\sin\theta\,dr+r\cos\theta\,d\theta. \end{align*} By compatibility of pullback with wedge products, \begin{align*} f^*(dx\wedge dy) &=f^*(dx)\wedge f^*(dy)\\ &=(\cos\theta\,dr-r\sin\theta\,d\theta)\wedge(\sin\theta\,dr+r\cos\theta\,d\theta). \end{align*} Expanding by bilinearity of the wedge product, \begin{align*} f^*(dx\wedge dy) &=(\cos\theta\,dr)\wedge(\sin\theta\,dr) +(\cos\theta\,dr)\wedge(r\cos\theta\,d\theta)\\ &\quad+(-r\sin\theta\,d\theta)\wedge(\sin\theta\,dr) +(-r\sin\theta\,d\theta)\wedge(r\cos\theta\,d\theta)\\ &=\cos\theta\sin\theta\,dr\wedge dr +r\cos^2\theta\,dr\wedge d\theta\\ &\quad-r\sin^2\theta\,d\theta\wedge dr -r^2\sin\theta\cos\theta\,d\theta\wedge d\theta. \end{align*} By alternation, $dr\wedge dr=0$ and $d\theta\wedge d\theta=0$. Also $d\theta\wedge dr=-dr\wedge d\theta$. Therefore \begin{align*} f^*(dx\wedge dy) &=0+r\cos^2\theta\,dr\wedge d\theta -r\sin^2\theta(-dr\wedge d\theta)-0\\ &=r\cos^2\theta\,dr\wedge d\theta+r\sin^2\theta\,dr\wedge d\theta\\ &=r(\cos^2\theta+\sin^2\theta)\,dr\wedge d\theta\\ &=r\,dr\wedge d\theta. \end{align*} The same coefficient is the determinant of the Jacobian matrix: \begin{align*} Jf_{(r,\theta)} &= \begin{pmatrix} \partial x/\partial r & \partial x/\partial \theta\\ \partial y/\partial r & \partial y/\partial \theta \end{pmatrix}\\ &= \begin{pmatrix} \cos\theta & -r\sin\theta\\ \sin\theta & r\cos\theta \end{pmatrix}, \end{align*} so \begin{align*} \det(Jf_{(r,\theta)}) &=(\cos\theta)(r\cos\theta)-(-r\sin\theta)(\sin\theta)\\ &=r\cos^2\theta+r\sin^2\theta\\ &=r(\cos^2\theta+\sin^2\theta)\\ &=r. \end{align*} In polar coordinates, the Euclidean area form pulls back to $r\,dr\wedge d\theta$. Thus the familiar polar area factor $r$ is exactly the Jacobian determinant appearing as the coefficient of the pulled-back top-degree form. [/example] Coordinate formulae also make sense for maps that are not between spaces of the same dimension. Pulling a top-degree form from a higher-dimensional target to a lower-dimensional source produces the form that measures the target quantity along the parametrised source. [illustration:forms-inverse-stereographic-area] [example: Stereographic Projection And The Area Form] Let $N=(0,0,1)$, and let $\sigma:\mathbb{R}^2\to S^2\setminus\{N\}\subset\mathbb{R}^3$ be inverse stereographic projection from the north pole: \begin{align*} \sigma(u,v) &=\left(\frac{2u}{1+u^2+v^2},\frac{2v}{1+u^2+v^2},\frac{u^2+v^2-1}{1+u^2+v^2}\right). \end{align*} Set \begin{align*} s=1+u^2+v^2, \end{align*} so the component functions of $\sigma$ are \begin{align*} X=\frac{2u}{s},\qquad Y=\frac{2v}{s},\qquad Z=\frac{u^2+v^2-1}{s}. \end{align*} Orient $S^2\setminus\{N\}$ by the inverse stereographic coordinates $(u,v)$, and let $\omega_{S^2}$ be the corresponding Riemannian area form. The pullback of the spherical area form is \begin{align*} \sigma^*\omega_{S^2} =\frac{4}{(1+u^2+v^2)^2}\,du\wedge dv. \end{align*} Since $s=1+u^2+v^2$, its partial derivatives are \begin{align*} \frac{\partial s}{\partial u}=2u,\qquad \frac{\partial s}{\partial v}=2v. \end{align*} Using the quotient rule, \begin{align*} X_u &=\frac{2s-(2u)(2u)}{s^2}\\ &=\frac{2s-4u^2}{s^2}\\ &=\frac{2(1+u^2+v^2)-4u^2}{s^2}\\ &=\frac{2(1-u^2+v^2)}{s^2}, \end{align*} and \begin{align*} X_v &=\frac{0\cdot s-(2u)(2v)}{s^2}\\ &=-\frac{4uv}{s^2}. \end{align*} Similarly, \begin{align*} Y_u &=\frac{0\cdot s-(2v)(2u)}{s^2}\\ &=-\frac{4uv}{s^2}, \end{align*} and \begin{align*} Y_v &=\frac{2s-(2v)(2v)}{s^2}\\ &=\frac{2s-4v^2}{s^2}\\ &=\frac{2(1+u^2-v^2)}{s^2}. \end{align*} For $Z=(u^2+v^2-1)/s$, \begin{align*} Z_u &=\frac{(2u)s-(u^2+v^2-1)(2u)}{s^2}\\ &=\frac{2u\bigl(s-(u^2+v^2-1)\bigr)}{s^2}\\ &=\frac{2u\bigl((1+u^2+v^2)-(u^2+v^2-1)\bigr)}{s^2}\\ &=\frac{4u}{s^2}, \end{align*} and \begin{align*} Z_v &=\frac{(2v)s-(u^2+v^2-1)(2v)}{s^2}\\ &=\frac{2v\bigl(s-(u^2+v^2-1)\bigr)}{s^2}\\ &=\frac{4v}{s^2}. \end{align*} Therefore the coordinate tangent vectors are \begin{align*} \sigma_u &=\frac{1}{s^2}\bigl(2(1-u^2+v^2),-4uv,4u\bigr),\\ \sigma_v &=\frac{1}{s^2}\bigl(-4uv,2(1+u^2-v^2),4v\bigr). \end{align*} Compute the first inner product: \begin{align*} \sigma_u\cdot\sigma_u &=\frac{4(1-u^2+v^2)^2+16u^2v^2+16u^2}{s^4}. \end{align*} Now \begin{align*} (1-u^2+v^2)^2 &=1+u^4+v^4-2u^2+2v^2-2u^2v^2, \end{align*} so \begin{align*} 4(1-u^2+v^2)^2+16u^2v^2+16u^2 &=4+4u^4+4v^4-8u^2+8v^2-8u^2v^2\\ &\quad+16u^2v^2+16u^2\\ &=4+4u^4+4v^4+8u^2+8v^2+8u^2v^2\\ &=4(1+u^2+v^2)^2\\ &=4s^2. \end{align*} Hence \begin{align*} \sigma_u\cdot\sigma_u=\frac{4s^2}{s^4}=\frac{4}{s^2}. \end{align*} For the second inner product, \begin{align*} \sigma_v\cdot\sigma_v &=\frac{16u^2v^2+4(1+u^2-v^2)^2+16v^2}{s^4}. \end{align*} Since \begin{align*} (1+u^2-v^2)^2 &=1+u^4+v^4+2u^2-2v^2-2u^2v^2, \end{align*} we get \begin{align*} 16u^2v^2+4(1+u^2-v^2)^2+16v^2 &=16u^2v^2+4+4u^4+4v^4+8u^2\\ &\quad-8v^2-8u^2v^2+16v^2\\ &=4+4u^4+4v^4+8u^2+8v^2+8u^2v^2\\ &=4s^2. \end{align*} Thus \begin{align*} \sigma_v\cdot\sigma_v=\frac{4}{s^2}. \end{align*} The mixed inner product is \begin{align*} \sigma_u\cdot\sigma_v &=\frac{2(1-u^2+v^2)(-4uv)+(-4uv)2(1+u^2-v^2)+(4u)(4v)}{s^4}\\ &=\frac{-8uv(1-u^2+v^2)-8uv(1+u^2-v^2)+16uv}{s^4}\\ &=\frac{-8uv\bigl((1-u^2+v^2)+(1+u^2-v^2)\bigr)+16uv}{s^4}\\ &=\frac{-8uv(2)+16uv}{s^4}\\ &=0. \end{align*} For a positively oriented pair of tangent vectors $A,B$ on an oriented Riemannian surface, the area form satisfies \begin{align*} \omega(A,B)=\sqrt{(A\cdot A)(B\cdot B)-(A\cdot B)^2}. \end{align*} Applying this to $A=\sigma_u$ and $B=\sigma_v$ gives \begin{align*} \omega_{S^2}(\sigma_u,\sigma_v) &=\sqrt{\left(\frac{4}{s^2}\right)\left(\frac{4}{s^2}\right)-0^2}\\ &=\sqrt{\frac{16}{s^4}}\\ &=\frac{4}{s^2}, \end{align*} because $s=1+u^2+v^2>0$. By the definition of pullback, \begin{align*} (\sigma^*\omega_{S^2})_{(u,v)}(\partial_u,\partial_v) &=(\omega_{S^2})_{\sigma(u,v)}(D\sigma_{(u,v)}\partial_u,D\sigma_{(u,v)}\partial_v)\\ &=(\omega_{S^2})_{\sigma(u,v)}(\sigma_u,\sigma_v)\\ &=\frac{4}{s^2}. \end{align*} Since $du\wedge dv(\partial_u,\partial_v)=1$, the coefficient of $du\wedge dv$ is $\frac{4}{s^2}$. Therefore \begin{align*} \sigma^*\omega_{S^2} &=\frac{4}{s^2}\,du\wedge dv\\ &=\frac{4}{(1+u^2+v^2)^2}\,du\wedge dv. \end{align*} Pulling the spherical area form back along inverse stereographic projection turns the geometric area form on $S^2$ into the concrete planar form $\frac{4}{(1+u^2+v^2)^2}\,du\wedge dv$. The factor $\frac{4}{(1+u^2+v^2)^2}$ is the local area distortion: a small coordinate parallelogram in the $(u,v)$-plane has its area scaled by this amount on the sphere. [/example] The stereographic formula is a model for later manifold calculations: even when the form lives geometrically on a curved space, pullback to a coordinate chart turns it into an ordinary differential form on an open subset of Euclidean space. ## Change Of Variables From Pullback Why does the change-of-variables theorem place an absolute value around the Jacobian determinant, while pullback of the volume form gives an oriented determinant? Differential forms integrate over oriented domains, so orientation-preserving maps contribute $\det(Jf_x)$ and orientation-reversing maps contribute a negative sign. The classical theorem for non-oriented volume uses the positive density $|\det(Jf_x)|$. [quotetheorem:3554] [citeproof:3554] This theorem recovers the familiar formula for ordinary multiple integrals by replacing the oriented form with its associated density. [example: Orientation Reversal On The Line] Let $x$ be the source coordinate on $(-1,1)$ and let $y$ be the target coordinate on $(-1,1)$. Define \begin{align*} f:(-1,1)&\to(-1,1),\\ f(x)&=-x. \end{align*} Let \begin{align*} \omega=a(y)\,dy \end{align*} be a compactly supported $1$-form on the target interval. The pullback is \begin{align*} f^*\omega=-a(-x)\,dx. \end{align*} Consequently, oriented integration records the orientation reversal by a minus sign, while the ordinary Lebesgue change-of-variables formula uses $|f'(x)|=1$ and records only unsigned length. Pulling back the coefficient function means composing it with $f$: \begin{align*} f^*(a(y)) &=a(y)\circ f\\ &=a(f(x))\\ &=a(-x). \end{align*} The target coordinate function $y$ pulls back to \begin{align*} f^*y &=y\circ f\\ &=f(x)\\ &=-x. \end{align*} For the coordinate differential, \begin{align*} f^*(dy) &=d(f^*y)\\ &=d(-x)\\ &=-d(x)\\ &=-dx, \end{align*} where the third line uses linearity of $d$. Now use multiplication of the pulled-back coefficient with the pulled-back $1$-form: \begin{align*} f^*\omega &=f^*(a(y)\,dy)\\ &=f^*(a(y))\,f^*(dy)\\ &=a(-x)(-dx)\\ &=-a(-x)\,dx. \end{align*} With the standard orientations on the source and target intervals, integrate this pulled-back $1$-form over $(-1,1)$: \begin{align*} \int_{-1}^{1} f^*\omega &=\int_{-1}^{1} -a(-x)\,dx. \end{align*} Set $y=-x$. Then $dy=-dx$, so $dx=-dy$. The endpoints change as follows: \begin{align*} x=-1&\implies y=1,\\ x=1&\implies y=-1. \end{align*} Therefore \begin{align*} \int_{-1}^{1} -a(-x)\,dx &=\int_{1}^{-1} -a(y)(-dy)\\ &=\int_{1}^{-1} a(y)\,dy\\ &=-\int_{-1}^{1} a(y)\,dy. \end{align*} Thus \begin{align*} \int_{-1}^{1} f^*\omega=-\int_{-1}^{1}\omega. \end{align*} For the ordinary unsigned substitution formula, compute \begin{align*} f'(x)=-1, \end{align*} so \begin{align*} |f'(x)|=|-1|=1. \end{align*} Hence the Lebesgue-measure change of variables uses the factor $1$: \begin{align*} \int_{-1}^{1} a(y)\,d\mathcal L^1(y) &=\int_{-1}^{1} a(f(x))\,|f'(x)|\,d\mathcal L^1(x)\\ &=\int_{-1}^{1} a(-x)\cdot 1\,d\mathcal L^1(x)\\ &=\int_{-1}^{1} a(-x)\,d\mathcal L^1(x). \end{align*} The pullback keeps the sign of the derivative: $dy$ becomes $-dx$, so oriented integration sees that $f(x)=-x$ reverses orientation. Ordinary Lebesgue integration uses $|f'(x)|=1$, so it keeps the size of the stretch but discards the orientation sign. [/example] The line example is the simplest case of the general phenomenon. Pullback keeps track of whether a parametrisation preserves or reverses the chosen orientation, whereas densities record only unsigned size. The obstruction in comparing the two integrals is the sign of the Jacobian determinant. Oriented integration must keep that sign because it records whether ordered volume elements are reversed, while Lebesgue integration measures only unsigned volume. A precise comparison therefore has to identify how the signed determinant in pullback becomes the absolute value in the ordinary change-of-variables formula. [quotetheorem:22] [citeproof:22] The absolute value is therefore the price of forgetting orientation. Differential forms keep the sign because they are designed to integrate over oriented parametrised objects. [remark: What Pullback Prepares] Pullback is the operation that makes integration on manifolds local. To integrate a form over a parametrised curve, surface, or coordinate chart, we pull the form back to a [Euclidean domain](/page/Euclidean%20Domain) and integrate there. The compatibility $f^*(d\omega)=d(f^*\omega)$ will later be the mechanism behind [Stokes' theorem](/theorems/1530) and the induced maps on de Rham cohomology. [/remark] So far, all constructions—exterior algebra, differential forms, pullback, and the [exterior derivative](/theorems/1525)—have lived in coordinates on open subsets of ℝⁿ. To fully unlock the geometric meaning of these tools, we must reformulate them on abstract manifolds, where forms become genuinely coordinate-free geometric objects. # 5. Smooth Manifolds and Forms on Manifolds Smooth manifolds are the setting in which differential forms stop being coordinate-dependent formulae and become geometric objects. The preceding chapters built exterior algebra, forms on open subsets of Euclidean space, pullback, wedge product, and [exterior derivative](/theorems/1525) in coordinates. This chapter explains how those constructions survive a change of chart, how they assemble on an abstract manifold, and why the same algebraic and differential operations are available globally. The guiding question is: if a manifold is only locally an open subset of $\mathbb R^n$, what data must be required so that a $k$-form written in one coordinate system is the same geometric object as the corresponding expression in another? The answer is compatibility under pullback by smooth transition maps. Once that compatibility is in place, $\Omega^*(M)$ becomes a graded algebra and $d$ becomes a global operator, giving the language needed for integration and [Stokes' theorem](/theorems/1530) in the next part of the course. ## Smooth Atlases and the Cotangent Bundle How much structure is needed on a topological space before calculus makes sense on it? A chart identifies a small part of the space with an open subset of $\mathbb R^n$, but calculus also requires that overlapping charts use compatible coordinate changes. [definition: Smooth Chart] Let $M$ be a topological space. An $n$-dimensional chart on $M$ is a pair $(U,\varphi)$ such that $U \subset M$ is open, $\varphi: U \to \varphi(U) \subseteq \mathbb R^n$ is a homeomorphism, and $\varphi(U)$ is open in $\mathbb R^n$. [/definition] A chart gives local coordinates $x_1,\dots,x_n$ on $U$ by writing $\varphi(p)=(x_1(p),\dots,x_n(p))$. The point is not that $M$ is a subset of Euclidean space, but that near each point it admits a coordinate description in which limits, derivatives, and forms can be computed. The danger is that two coordinate systems can describe the same points while giving incompatible calculus; the next condition rules out such fake coordinate changes. [definition: Smoothly Compatible Charts] Two $n$-dimensional charts $(U,\varphi)$ and $(V,\psi)$ on $M$ are smoothly compatible if either $U \cap V=\varnothing$, or the transition map \begin{align*} \psi \circ \varphi^{-1}:\varphi(U\cap V)\longrightarrow \psi(U\cap V) \end{align*} is a diffeomorphism between open subsets of $\mathbb R^n$. [/definition] The transition map is the change of coordinates from the $\varphi$-coordinates to the $\psi$-coordinates. Smooth compatibility is what permits a derivative computed in one chart to be transported to another chart without changing the underlying geometric statement. Compatible charts alone still do not determine a satisfactory global space for calculus. The underlying topology must separate points well enough for limits to be meaningful, and it must be countable enough for local constructions to be assembled without pathological bookkeeping. To turn compatible coordinate patches into a usable space for global calculus, the topology also has to rule out pathological identifications and support countable local arguments. The object we need must remember both the coordinate-change rules and these topological safeguards. That package is what allows forms, integration, and partitions of unity to be defined without depending on a particular chart list. [definition: Smooth Manifold] A smooth $n$-manifold is a second-countable Hausdorff topological space $M$ equipped with a maximal atlas $\mathcal A$ of smoothly compatible $n$-dimensional charts whose domains cover $M$. [/definition] The maximality condition means that every chart smoothly compatible with the atlas is already included. In practice, one usually specifies a smaller atlas and understands the smooth structure to be the maximal atlas it generates. [example: Standard Smooth Structure on the Sphere] Let \begin{align*} S^n=\{(x_1,\dots,x_n,x_{n+1})\in\mathbb R^{n+1}:x_1^2+\cdots+x_n^2+x_{n+1}^2=1\}. \end{align*} Write $x'=(x_1,\dots,x_n)$, let $N=(0,\dots,0,1)$ be the north pole, and let $S=(0,\dots,0,-1)$ be the south pole. Define \begin{align*} U_N=S^n\setminus\{N\}, \qquad U_S=S^n\setminus\{S\}. \end{align*} Then $U_N\cup U_S=S^n$. On these open sets define stereographic coordinate maps \begin{align*} \sigma_N(x',x_{n+1})=\frac{x'}{1-x_{n+1}}, \qquad \sigma_S(x',x_{n+1})=\frac{x'}{1+x_{n+1}}. \end{align*} The denominators are nonzero on the indicated domains: if $x\in U_N$, then $x_{n+1}\ne 1$, so $1-x_{n+1}\ne 0$; if $x\in U_S$, then $x_{n+1}\ne -1$, so $1+x_{n+1}\ne 0$. The two charts $(U_N,\sigma_N)$ and $(U_S,\sigma_S)$ are smoothly compatible, and the maximal atlas generated by them gives $S^n$ its standard smooth $n$-manifold structure. For $u=(u_1,\dots,u_n)\in\mathbb R^n$, write $|u|^2=u_1^2+\cdots+u_n^2$. Define \begin{align*} \tau_N(u)=\left(\frac{2u}{|u|^2+1},\frac{|u|^2-1}{|u|^2+1}\right). \end{align*} First check that $\tau_N(u)\in S^n$: \begin{align*} \left|\frac{2u}{|u|^2+1}\right|^2+\left(\frac{|u|^2-1}{|u|^2+1}\right)^2 &=\frac{4|u|^2}{(|u|^2+1)^2}+\frac{(|u|^2-1)^2}{(|u|^2+1)^2}\\ &=\frac{4|u|^2+|u|^4-2|u|^2+1}{(|u|^2+1)^2}\\ &=\frac{|u|^4+2|u|^2+1}{(|u|^2+1)^2}\\ &=\frac{(|u|^2+1)^2}{(|u|^2+1)^2}\\ &=1. \end{align*} Also the last coordinate of $\tau_N(u)$ is never $1$, since \begin{align*} \frac{|u|^2-1}{|u|^2+1}=1 \end{align*} would imply $|u|^2-1=|u|^2+1$, hence $-1=1$. Thus $\tau_N(u)\in U_N$. Now compute $\sigma_N(\tau_N(u))$: \begin{align*} \sigma_N(\tau_N(u)) &=\frac{\frac{2u}{|u|^2+1}}{1-\frac{|u|^2-1}{|u|^2+1}}\\ &=\frac{\frac{2u}{|u|^2+1}}{\frac{|u|^2+1-(|u|^2-1)}{|u|^2+1}}\\ &=\frac{\frac{2u}{|u|^2+1}}{\frac{2}{|u|^2+1}}\\ &=u. \end{align*} Conversely, if $x=(x',x_{n+1})\in U_N$ and $u=\sigma_N(x)=x'/(1-x_{n+1})$, then \begin{align*} |u|^2 &=\frac{|x'|^2}{(1-x_{n+1})^2}\\ &=\frac{1-x_{n+1}^2}{(1-x_{n+1})^2}\\ &=\frac{(1-x_{n+1})(1+x_{n+1})}{(1-x_{n+1})^2}\\ &=\frac{1+x_{n+1}}{1-x_{n+1}}. \end{align*} Therefore \begin{align*} |u|^2+1 &=\frac{1+x_{n+1}}{1-x_{n+1}}+1\\ &=\frac{1+x_{n+1}+1-x_{n+1}}{1-x_{n+1}}\\ &=\frac{2}{1-x_{n+1}}. \end{align*} The first $n$ coordinates of $\tau_N(u)$ are \begin{align*} \frac{2u}{|u|^2+1} &=\frac{2\frac{x'}{1-x_{n+1}}}{\frac{2}{1-x_{n+1}}}\\ &=x', \end{align*} and the last coordinate is \begin{align*} \frac{|u|^2-1}{|u|^2+1} &=\frac{\frac{1+x_{n+1}}{1-x_{n+1}}-1}{\frac{2}{1-x_{n+1}}}\\ &=\frac{\frac{1+x_{n+1}-1+x_{n+1}}{1-x_{n+1}}}{\frac{2}{1-x_{n+1}}}\\ &=x_{n+1}. \end{align*} So $\tau_N(\sigma_N(x))=x$. Hence $\sigma_N:U_N\to\mathbb R^n$ is a homeomorphism with inverse $\tau_N$. Similarly, \begin{align*} \tau_S(v)=\left(\frac{2v}{|v|^2+1},\frac{1-|v|^2}{|v|^2+1}\right) \end{align*} is the inverse of $\sigma_S:U_S\to\mathbb R^n$. Thus $(U_N,\sigma_N)$ and $(U_S,\sigma_S)$ are charts. Their overlap is \begin{align*} U_N\cap U_S=S^n\setminus\{N,S\}. \end{align*} In $\sigma_N$-coordinates, the removed south pole corresponds to $u=0$, because \begin{align*} \sigma_N(S)=\sigma_N(0,\dots,0,-1)=\frac{0}{1-(-1)}=0. \end{align*} Thus $\sigma_N(U_N\cap U_S)=\mathbb R^n\setminus\{0\}$. For $u\ne 0$, the transition map from north coordinates to south coordinates is \begin{align*} (\sigma_S\circ\tau_N)(u) &=\frac{\frac{2u}{|u|^2+1}}{1+\frac{|u|^2-1}{|u|^2+1}}\\ &=\frac{\frac{2u}{|u|^2+1}}{\frac{|u|^2+1+|u|^2-1}{|u|^2+1}}\\ &=\frac{\frac{2u}{|u|^2+1}}{\frac{2|u|^2}{|u|^2+1}}\\ &=\frac{u}{|u|^2}. \end{align*} The coordinate functions are \begin{align*} u_i\longmapsto \frac{u_i}{u_1^2+\cdots+u_n^2}, \end{align*} which are smooth on $\mathbb R^n\setminus\{0\}$ because the denominator is nonzero there. This transition map is its own inverse. Indeed, for $u\ne 0$, \begin{align*} \left|\frac{u}{|u|^2}\right|^2 &=\frac{|u|^2}{|u|^4}\\ &=\frac{1}{|u|^2}, \end{align*} and hence \begin{align*} \frac{\frac{u}{|u|^2}}{\left|\frac{u}{|u|^2}\right|^2} &=\frac{\frac{u}{|u|^2}}{\frac{1}{|u|^2}}\\ &=u. \end{align*} Therefore the transition map is a diffeomorphism of $\mathbb R^n\setminus\{0\}$ with smooth inverse. The two stereographic charts cover $S^n$ and have a smooth transition map on their overlap. They therefore form a smooth atlas, and the maximal atlas generated by this atlas is the standard smooth structure on the sphere. [/example] Once a smooth structure is fixed, each point has its own tangent space and hence its own space of covectors. A global $1$-form cannot be described by a single vector space, because its value at $p$ lives in $T_p^*M$ and its value at $q$ lives in $T_q^*M$. The next object solves this bookkeeping problem by collecting all covector spaces into one space lying over $M$. With that bundle available, covector fields and differential forms can be described as smoothly varying choices of elements from the appropriate fiber. [definition: Cotangent Bundle] Let $M$ be a smooth $n$-manifold. The cotangent bundle of $M$ is \begin{align*} T^*M=\bigsqcup_{p\in M} T_p^*M, \end{align*} where $T_p^*M$ is the dual [vector space](/page/Vector%20Space) of the tangent space $T_pM$. [/definition] In a chart $(U,\varphi)$ with coordinates $x_1,\dots,x_n$, the coordinate covectors $dx_1|_p,\dots,dx_n|_p$ form a basis of $T_p^*M$ for each $p\in U$. Thus a local covector field can be written as $\sum_i a_i\,dx_i$, with coefficient functions $a_i:U\to\mathbb R$. [example: Cotangent Coordinates on the Circle] Let $U,V\subset S^1$ be two angular coordinate neighbourhoods with coordinates $\theta:U\to I\subset\mathbb R$ and $\tilde\theta:V\to \tilde I\subset\mathbb R$. On $U\cap V$, define the coordinate change \begin{align*} h=\tilde\theta\circ\theta^{-1}: \theta(U\cap V)\longrightarrow \tilde\theta(U\cap V). \end{align*} Let a covector field be written in the two coordinates as \begin{align*} \omega_\theta=f(\theta)\,d\theta, \qquad \omega_{\tilde\theta}=\tilde f(\tilde\theta)\,d\tilde\theta. \end{align*} On the overlap, the two coefficient functions represent the same covector field exactly when \begin{align*} f(\theta)=\tilde f(h(\theta))\,h'(\theta). \end{align*} Equivalently, if $\tilde\theta=h(\theta)$, then \begin{align*} \tilde f(\tilde\theta)=\frac{f(\theta)}{h'(\theta)}. \end{align*} The compatibility law says that the $\tilde\theta$-coordinate expression must pull back to the $\theta$-coordinate expression: \begin{align*} h^*(\omega_{\tilde\theta})=\omega_\theta. \end{align*} Now compute the left-hand side. Since $\omega_{\tilde\theta}=\tilde f(\tilde\theta)\,d\tilde\theta$, pullback gives \begin{align*} h^*(\omega_{\tilde\theta}) &=h^*(\tilde f(\tilde\theta)\,d\tilde\theta)\\ &=h^*(\tilde f(\tilde\theta))\,h^*(d\tilde\theta)\\ &=\tilde f(h(\theta))\,d(h(\theta)). \end{align*} Because $h$ is a one-variable smooth function, \begin{align*} d(h(\theta))=h'(\theta)\,d\theta. \end{align*} Therefore \begin{align*} h^*(\omega_{\tilde\theta}) &=\tilde f(h(\theta))\,h'(\theta)\,d\theta. \end{align*} For this to equal $\omega_\theta=f(\theta)\,d\theta$, the coefficients of $d\theta$ must agree: \begin{align*} f(\theta)=\tilde f(h(\theta))\,h'(\theta). \end{align*} If $\tilde\theta=h(\theta)$, then solving for the coefficient in the second chart gives \begin{align*} \tilde f(\tilde\theta) &=\tilde f(h(\theta))\\ &=\frac{f(\theta)}{h'(\theta)}. \end{align*} A covector field on $S^1$ is not just a coefficient function. When the angular coordinate changes from $\theta$ to $\tilde\theta=h(\theta)$, the basis covector changes by $d\tilde\theta=h'(\theta)\,d\theta$, so the coefficient changes by the reciprocal factor. This is the one-dimensional form of the pullback compatibility rule for differential forms on manifolds. [/example] ## Differential Forms Defined by Charts A formula such as $f(x,y)\,dx\wedge dy$ is meaningful on a coordinate patch, but it should not depend on the chosen coordinates. The obstruction is that the same geometric object may have different coefficient functions in overlapping charts, because both the coordinates and the coordinate covectors change. A global form is therefore not a single formula in one chart, but a family of local formulae that agree after pulling back along every transition map. [definition: Differential Form on a Manifold] Let $M$ be a smooth $n$-manifold with smooth atlas $\mathcal A$. A smooth $k$-form on $M$ is an assignment which, for each chart $(U,\varphi)\in\mathcal A$, gives a smooth $k$-form $\omega_\varphi\in\Omega^k(\varphi(U))$ such that for any two charts $(U,\varphi)$ and $(V,\psi)$, the compatibility condition \begin{align*} (\psi\circ\varphi^{-1})^*(\omega_\psi|_{\psi(U\cap V)})=\omega_\varphi|_{\varphi(U\cap V)} \end{align*} holds on $\varphi(U\cap V)$. [/definition] The compatibility condition says that if a form is written in $\psi$-coordinates and then pulled back through the coordinate change to $\varphi$-coordinates, the result is the form already written in $\varphi$-coordinates. This is the same rule that appeared for pullbacks on open subsets of Euclidean space, now used as the glue holding the local pieces together. Without it, one could assign $dx$ in one coordinate and $2\,dx$ in another coordinate on the same interval; the two formulae would not describe a single covector field on the overlap. [definition: Space of Smooth Forms] For a smooth manifold $M$, the [vector space](/page/Vector%20Space) of smooth $k$-forms on $M$ is denoted by $\Omega^k(M)$. The full space of smooth differential forms is \begin{align*} \Omega^*(M)=\bigoplus_{k=0}^{n}\Omega^k(M). \end{align*} [/definition] Here $\Omega^0(M)$ is the algebra $C^\infty(M)$ of smooth functions on $M$. For $k>n$, the space $\Omega^k(M)$ is zero because each tangent space has dimension $n$. [example: Forms on the Sphere by Restriction] Let $i:S^n\hookrightarrow\mathbb R^{n+1}$ be the inclusion map. For every $\alpha\in\Omega^k(\mathbb R^{n+1})$, the pullback $i^*\alpha$ is a $k$-form on $S^n$. Now specialize to $S^2\subset\mathbb R^3$. Write a point of $S^2$ as \begin{align*} x=(x_1,x_2,x_3), \qquad x_1^2+x_2^2+x_3^2=1. \end{align*} Let \begin{align*} \sigma=x_1\,dx_2\wedge dx_3+x_2\,dx_3\wedge dx_1+x_3\,dx_1\wedge dx_2 \end{align*} be a Euclidean $2$-form on $\mathbb R^3$. The pulled-back form $i^*\sigma$ is the standard oriented area form on the unit sphere $S^2$. Let $v,w\in T_xS^2$, and write \begin{align*} v=(v_1,v_2,v_3), \qquad w=(w_1,w_2,w_3). \end{align*} Since $i$ is the inclusion, its differential sends each tangent vector to the same vector regarded in $\mathbb R^3$: \begin{align*} di_x(v)=v, \qquad di_x(w)=w. \end{align*} By the definition of pullback, \begin{align*} (i^*\sigma)_x(v,w)=\sigma_x(v,w). \end{align*} Evaluate each wedge term using \begin{align*} (dx_a\wedge dx_b)_x(v,w)=dx_a(v)\,dx_b(w)-dx_a(w)\,dx_b(v) =v_a w_b-w_a v_b. \end{align*} Thus \begin{align*} \sigma_x(v,w) &=x_1(dx_2\wedge dx_3)_x(v,w) +x_2(dx_3\wedge dx_1)_x(v,w) +x_3(dx_1\wedge dx_2)_x(v,w)\\ &=x_1(v_2w_3-w_2v_3) +x_2(v_3w_1-w_3v_1) +x_3(v_1w_2-w_1v_2). \end{align*} Since scalar multiplication is commutative, \begin{align*} w_2v_3=v_3w_2, \qquad w_3v_1=v_1w_3, \qquad w_1v_2=v_2w_1. \end{align*} Therefore \begin{align*} \sigma_x(v,w) &=x_1(v_2w_3-v_3w_2) +x_2(v_3w_1-v_1w_3) +x_3(v_1w_2-v_2w_1). \end{align*} The cross product is \begin{align*} v\times w =\bigl(v_2w_3-v_3w_2,\; v_3w_1-v_1w_3,\; v_1w_2-v_2w_1\bigr). \end{align*} Taking the Euclidean dot product with $x$ gives \begin{align*} x\cdot(v\times w) &=x_1(v_2w_3-v_3w_2) +x_2(v_3w_1-v_1w_3) +x_3(v_1w_2-v_2w_1). \end{align*} Comparing the two displayed formulae, \begin{align*} (i^*\sigma)_x(v,w)=x\cdot(v\times w). \end{align*} Because $x\in S^2$, the vector $x$ has length $1$. Also, $x$ is normal to $T_xS^2$: differentiating the constraint $x\cdot x=1$ along a tangent vector $v$ gives \begin{align*} 0=d(x\cdot x)_x(v)=2x\cdot v, \end{align*} so $x\cdot v=0$, and similarly $x\cdot w=0$. Thus $x$ is the outward unit normal to the tangent plane. If $(e_1,e_2)$ is an oriented orthonormal basis of $T_xS^2$, then $e_1\times e_2=x$. Hence \begin{align*} (i^*\sigma)_x(e_1,e_2) &=x\cdot(e_1\times e_2)\\ &=x\cdot x\\ &=1. \end{align*} This is exactly the defining normalization of the standard oriented area form on $S^2$. Pulling back the ambient form $\sigma$ by the inclusion restricts it to tangent vectors on the sphere. At each $x\in S^2$, the value on tangent vectors $v,w$ is \begin{align*} (i^*\sigma)_x(v,w)=x\cdot(v\times w), \end{align*} the signed Euclidean area of the parallelogram spanned by $v$ and $w$ with outward normal $x$. Therefore $i^*\sigma$ is the standard area form on the unit sphere. [/example] This example illustrates an important source of forms on submanifolds: forms from the ambient Euclidean space restrict by pullback. Not every form on a manifold has to be presented in this way, but many geometric examples arise from ambient formulae. There remains a structural issue: the chart-based definition of a form should agree with the coordinate-free picture of a smooth section of an exterior power of the cotangent bundle. Establishing this equivalence lets later arguments switch between computations in coordinates and intrinsic bundle language. [quotetheorem:3572] [citeproof:3572] This theorem justifies switching between two viewpoints. In calculations, forms are local coordinate expressions; conceptually, they are coordinate-independent sections of exterior powers of the cotangent bundle. The Hausdorff condition keeps the local pieces of the space separated enough for tangent and cotangent spaces to assemble into bundles, while second-countability supplies the countability hypotheses used later for partitions of unity. The section viewpoint is what makes restriction, pullback, support, and integration statements independent of a chosen atlas; the chart viewpoint remains the practical way to compute coefficients. ## Wedge Product as a Global Operation The wedge product was defined algebraically on each cotangent space and analytically on open subsets of $\mathbb R^n$. The issue on a manifold is whether multiplying two compatible families of local forms gives another compatible family. [definition: Wedge Product on a Manifold] Let $\alpha\in\Omega^p(M)$ and $\beta\in\Omega^q(M)$. Their wedge product $\alpha\wedge\beta\in\Omega^{p+q}(M)$ is the form whose expression in every chart is \begin{align*} (\alpha\wedge\beta)_\varphi=\alpha_\varphi\wedge\beta_\varphi. \end{align*} [/definition] The definition is local, but it is not arbitrary: pullback commutes with the wedge product, so the local expressions continue to agree under transition maps. The next result records that no algebraic law is lost when we pass from Euclidean coordinate patches to an abstract manifold. This matters because integration and [Stokes' theorem](/theorems/1530) will multiply forms of different degrees, so the signs and degrees must be globally consistent rather than chartwise conventions. [quotetheorem:3573] [citeproof:3573] The theorem is the first place where the coordinate-free formalism pays off. A global algebraic identity between forms is proved by reducing to the Euclidean coordinate calculation and using compatibility to glue the results. The sign $(-1)^{pq}$ is not optional: if $\alpha$ has odd degree, then $\alpha\wedge\alpha=-\alpha\wedge\alpha$, so over $\mathbb R$ this forces $\alpha\wedge\alpha=0$. On a space with non-smooth transition functions, the coefficient transformations needed for this algebra may fail to be smooth; on badly non-Hausdorff spaces, the local pieces need not assemble into well-behaved global sections. [example: Wedge Products on a Surface] Let $M$ be a smooth surface and let $(U,\varphi)$ be a chart with coordinates $x_1,x_2$. On $U$, let \begin{align*} \alpha=a_1\,dx_1+a_2\,dx_2, \qquad \beta=b_1\,dx_1+b_2\,dx_2, \end{align*} where $a_1,a_2,b_1,b_2$ are smooth functions on $U$. In these coordinates, \begin{align*} \alpha\wedge\beta=(a_1b_2-a_2b_1)\,dx_1\wedge dx_2. \end{align*} The coefficient $a_1b_2-a_2b_1$ is the determinant of the two coefficient rows. Using bilinearity of the wedge product, \begin{align*} \alpha\wedge\beta &=(a_1\,dx_1+a_2\,dx_2)\wedge(b_1\,dx_1+b_2\,dx_2)\\ &=a_1b_1\,dx_1\wedge dx_1 +a_1b_2\,dx_1\wedge dx_2 +a_2b_1\,dx_2\wedge dx_1 +a_2b_2\,dx_2\wedge dx_2. \end{align*} Since the wedge product is alternating, \begin{align*} dx_1\wedge dx_1=0, \qquad dx_2\wedge dx_2=0, \qquad dx_2\wedge dx_1=-dx_1\wedge dx_2. \end{align*} Substituting these identities gives \begin{align*} \alpha\wedge\beta &=a_1b_1\cdot 0 +a_1b_2\,dx_1\wedge dx_2 +a_2b_1(-dx_1\wedge dx_2) +a_2b_2\cdot 0\\ &=a_1b_2\,dx_1\wedge dx_2-a_2b_1\,dx_1\wedge dx_2\\ &=(a_1b_2-a_2b_1)\,dx_1\wedge dx_2. \end{align*} The coefficient is the determinant \begin{align*} \det \begin{pmatrix} a_1 & a_2\\ b_1 & b_2 \end{pmatrix} =a_1b_2-a_2b_1. \end{align*} Now let $y_1,y_2$ be another coordinate system on an overlap. Write \begin{align*} dx_1=\frac{\partial x_1}{\partial y_1}\,dy_1+\frac{\partial x_1}{\partial y_2}\,dy_2, \qquad dx_2=\frac{\partial x_2}{\partial y_1}\,dy_1+\frac{\partial x_2}{\partial y_2}\,dy_2. \end{align*} Then \begin{align*} dx_1\wedge dx_2 &=\left(\frac{\partial x_1}{\partial y_1}\,dy_1+\frac{\partial x_1}{\partial y_2}\,dy_2\right) \wedge \left(\frac{\partial x_2}{\partial y_1}\,dy_1+\frac{\partial x_2}{\partial y_2}\,dy_2\right)\\ &=\frac{\partial x_1}{\partial y_1}\frac{\partial x_2}{\partial y_1}\,dy_1\wedge dy_1 +\frac{\partial x_1}{\partial y_1}\frac{\partial x_2}{\partial y_2}\,dy_1\wedge dy_2\\ &\quad +\frac{\partial x_1}{\partial y_2}\frac{\partial x_2}{\partial y_1}\,dy_2\wedge dy_1 +\frac{\partial x_1}{\partial y_2}\frac{\partial x_2}{\partial y_2}\,dy_2\wedge dy_2\\ &=\left( \frac{\partial x_1}{\partial y_1}\frac{\partial x_2}{\partial y_2} -\frac{\partial x_1}{\partial y_2}\frac{\partial x_2}{\partial y_1} \right)dy_1\wedge dy_2\\ &=\det\left(\frac{\partial(x_1,x_2)}{\partial(y_1,y_2)}\right)dy_1\wedge dy_2. \end{align*} Thus the local coefficient of a $2$-form changes by the Jacobian determinant of the coordinate change. On a surface, wedging two $1$-forms produces a $2$-form whose coordinate coefficient is the determinant of their coefficient rows. Under a change of coordinates, the basis form $dx_1\wedge dx_2$ transforms by the Jacobian determinant, so the compatibility law for forms ensures that this $2$-form is a geometric object, not an artifact of the chosen chart. [/example] ## Exterior Derivative on Manifolds The [exterior derivative](/theorems/1525) should differentiate a form without requiring a preferred coordinate system. Since the Euclidean [exterior derivative](/theorems/1525) commutes with pullback by smooth maps, it can be applied chart by chart and then glued. [definition: Exterior Derivative on a Manifold] Let $\omega\in\Omega^k(M)$. The [exterior derivative](/theorems/1525) $d\omega\in\Omega^{k+1}(M)$ is the form whose expression in every chart is \begin{align*} (d\omega)_\varphi=d(\omega_\varphi). \end{align*} [/definition] This gives the same operation as ordinary differentiation on functions: if $f\in\Omega^0(M)=C^\infty(M)$, then in local coordinates \begin{align*} df=\sum_{i=1}^n \frac{\partial f}{\partial x_i}\,dx_i. \end{align*} For higher-degree forms, $d$ differentiates the coefficient functions and wedges the corresponding coordinate covectors in front. Because the definition was made chart by chart, the remaining question is whether it is independent of the chart and still satisfies the algebraic rules proved in Euclidean space. Those identities must survive coordinate changes before $d$ can be treated as a global operator on manifolds. [quotetheorem:1525] [citeproof:1525] This is the mechanism behind much of the theory of forms on manifolds: prove a natural identity in Euclidean coordinates, check that it commutes with pullback, and then regard it as a statement on every smooth manifold. The identity $d^2=0$ is the algebraic reason closed forms and exact forms form a cochain complex, which is the starting point of de Rham cohomology. Locality is just as important: if the chart representatives did not satisfy the overlap compatibility condition, differentiating them would produce unrelated local $(k+1)$-forms rather than a single global object. [example: Differentiating a One-Form in Local Coordinates] Let $(U,x,y)$ be a coordinate patch on a smooth surface, and let \begin{align*} \omega=P\,dx+Q\,dy \end{align*} be a smooth $1$-form on $U$, where $P,Q:U\to\mathbb R$ are smooth functions. In the coordinates $x,y$, \begin{align*} d\omega=\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dx\wedge dy. \end{align*} By linearity of the [exterior derivative](/theorems/1525), \begin{align*} d\omega &=d(P\,dx+Q\,dy)\\ &=d(P\,dx)+d(Q\,dy). \end{align*} For a coefficient function multiplied by a coordinate $1$-form, the coordinate formula for $d$ gives \begin{align*} d(P\,dx)=dP\wedge dx, \qquad d(Q\,dy)=dQ\wedge dy. \end{align*} Therefore \begin{align*} d\omega=dP\wedge dx+dQ\wedge dy. \end{align*} The differentials of the coefficient functions are \begin{align*} dP=\frac{\partial P}{\partial x}\,dx+\frac{\partial P}{\partial y}\,dy, \qquad dQ=\frac{\partial Q}{\partial x}\,dx+\frac{\partial Q}{\partial y}\,dy. \end{align*} Substitute these into the expression for $d\omega$: \begin{align*} d\omega &=\left(\frac{\partial P}{\partial x}\,dx+\frac{\partial P}{\partial y}\,dy\right)\wedge dx +\left(\frac{\partial Q}{\partial x}\,dx+\frac{\partial Q}{\partial y}\,dy\right)\wedge dy. \end{align*} Using bilinearity of the wedge product, \begin{align*} d\omega &=\frac{\partial P}{\partial x}\,dx\wedge dx +\frac{\partial P}{\partial y}\,dy\wedge dx +\frac{\partial Q}{\partial x}\,dx\wedge dy +\frac{\partial Q}{\partial y}\,dy\wedge dy. \end{align*} Since the wedge product is alternating, \begin{align*} dx\wedge dx=0, \qquad dy\wedge dy=0, \qquad dy\wedge dx=-dx\wedge dy. \end{align*} Thus \begin{align*} d\omega &=\frac{\partial P}{\partial x}\cdot 0 +\frac{\partial P}{\partial y}(-dx\wedge dy) +\frac{\partial Q}{\partial x}\,dx\wedge dy +\frac{\partial Q}{\partial y}\cdot 0\\ &=-\frac{\partial P}{\partial y}\,dx\wedge dy +\frac{\partial Q}{\partial x}\,dx\wedge dy\\ &=\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dx\wedge dy. \end{align*} The [exterior derivative](/theorems/1525) of the local $1$-form $P\,dx+Q\,dy$ is the $2$-form whose coefficient is $\partial Q/\partial x-\partial P/\partial y$, the scalar curl expression from vector calculus. On a manifold, another chart gives the pulled-back coordinate expression of this same global $2$-form. [/example] This computation also shows why $d$ is more rigid than ordinary partial differentiation of coefficients. The alternating combination is exactly what survives a change of coordinates. [remark: Locality of the Exterior Derivative] If two forms agree on an open subset $U\subset M$, then their exterior derivatives agree on $U$. This locality is inherited from the coordinate definition and will be used when partitions of unity and integration are introduced. [/remark] ## Pullback of Forms Between Manifolds How should a form on one manifold be transported to another when the two manifolds need not sit inside a common Euclidean space? A smooth map $F:M\to N$ compares tangent vectors by its differential, so a form on $N$ can be pulled back to a form on $M$ by applying $dF_p$ to tangent vectors before evaluating the original form. [definition: Pullback on Manifolds] Let $F:M\to N$ be a smooth map between smooth manifolds, and let $\omega\in\Omega^k(N)$. The pullback $F^*\omega\in\Omega^k(M)$ is defined pointwise by \begin{align*} (F^*\omega)_p(v_1,\dots,v_k)=\omega_{F(p)}(dF_p(v_1),\dots,dF_p(v_k)) \end{align*} for $p\in M$ and $v_1,\dots,v_k\in T_pM$. [/definition] In local coordinates, this is the same substitution rule already used for forms on open subsets of Euclidean space. The coordinate-free pointwise formula explains why the construction is independent of charts. The remaining issue is structural: a pointwise definition of $F^*\omega$ would be too weak if it failed to respect the algebra and calculus of forms. To use pullback in later integration and cohomology arguments, it must preserve wedge products, commute with exterior differentiation, and compose in the contravariant order forced by maps of manifolds. [quotetheorem:3574] [citeproof:3574] Pullback is contravariant: forms move from the target of a map back to its source. Smoothness of $F$ is essential, because the formula uses $dF_p$ and the identity $d(F^*\alpha)=F^*(d\alpha)$ differentiates the pulled-back coefficients. If $F$ is only continuous, there is no derivative with which to pull back positive-degree forms; if $F$ has only finite differentiability, pulling back and then applying $d$ consumes derivatives. This theorem is the naturality statement that later makes de Rham cohomology functorial. ## Manifolds with Boundary and Collars Integration on manifolds requires a precise meaning of the boundary. The local model changes from all of $\mathbb R^n$ to a half-space, because near a boundary point only one side of the manifold is present. [definition: Half-Space] The closed upper half-space in $\mathbb R^n$ is \begin{align*} \mathbb H^n=\{(x_1,\dots,x_n)\in\mathbb R^n:x_n\ge 0\}. \end{align*} [/definition] A boundary chart maps an open neighbourhood in the manifold to an open subset of $\mathbb H^n$, with transition maps smooth in the sense that they extend smoothly to open subsets of $\mathbb R^n$. The obstruction is that ordinary manifold charts cannot distinguish a point that has neighbourhoods on all sides from a point that has only inward directions available. A manifold with boundary must allow half-space charts while still enforcing smooth compatibility on overlaps, so calculus remains well-defined up to the boundary hyperplane. [definition: Smooth Manifold with Boundary] A smooth $n$-manifold with boundary is a second-countable Hausdorff topological space $M$ equipped with a maximal atlas of charts $(U,\varphi)$ where $\varphi:U\to\varphi(U)\subseteq\mathbb H^n$ is a homeomorphism onto a relatively open subset of $\mathbb H^n$, and the transition maps are smooth up to the boundary. [/definition] The boundary should consist of the points that are sent to the hyperplane $x_n=0$ in a boundary chart. The obstruction is chart-dependence: the same point may be described by different half-space coordinates, so the definition must rely on the fact that smooth transition maps preserve the distinction between interior points and boundary points. Having allowed half-space charts, we need a coordinate-independent way to name the actual boundary set. This is the set that later carries boundary orientation and appears in Stokes' theorem. [definition: Boundary of a Smooth Manifold with Boundary] Let $M$ be a smooth $n$-manifold with boundary. Its boundary is \begin{align*} \partial M=\{p\in M: \varphi_n(p)=0 \text{ in some boundary chart }(U,\varphi)\text{ around }p\}. \end{align*} [/definition] The set $M\setminus\partial M$ is the interior of $M$, and $\partial M$ itself carries a natural smooth structure of dimension $n-1$. Boundary orientation will be addressed when integration is introduced. Stokes' theorem is a global statement, so local boundary charts alone are not enough: the later integrals must be reducible to finitely controlled coordinate calculations. Compactness supplies that finiteness by preventing mass from escaping through infinitely many coordinate pieces and by making finite covers and partitions of unity available. [definition: Compact Manifold with Boundary] A compact smooth manifold with boundary is a smooth manifold with boundary whose underlying topological space is compact. [/definition] Near the boundary, compactness alone does not tell us that the manifold separates into a tangential boundary direction and a normal inward direction. Without such a product model, it is hard to compare forms on the boundary with forms in a neighbourhood of the boundary, or to formulate the boundary term in Stokes' theorem cleanly. [quotetheorem:3575] This theorem is quoted without proof in this course. It says that near its boundary, a compact smooth manifold with boundary looks like a product of the boundary with a half-open interval; this product structure is essential for stating and proving [Stokes' theorem](/theorems/1530) cleanly. [illustration:forms-collar-neighbourhood] [example: The Closed Ball] Let $M=\overline{B}(0,1)\subset\mathbb R^n$, so \begin{align*} M=\{x\in\mathbb R^n:|x|\le 1\}. \end{align*} Its boundary is \begin{align*} \partial M=S^{n-1}=\{p\in\mathbb R^n:|p|=1\}. \end{align*} Fix $0<\varepsilon<1$, and define \begin{align*} \Phi:S^{n-1}\times[0,\varepsilon)\longrightarrow M, \qquad \Phi(p,t)=(1-t)p. \end{align*} The map $\Phi$ is a collar of the boundary of the closed ball. Its image is the open shell \begin{align*} U=\{x\in M:1-\varepsilon<|x|\le 1\}, \end{align*} and $\Phi(p,0)=p$ for every $p\in S^{n-1}$. First, if $p\in S^{n-1}$ and $0\le t<\varepsilon<1$, then $1-t>0$. Since $|p|=1$, \begin{align*} |\Phi(p,t)| &=|(1-t)p|\\ &=(1-t)|p|\\ &=1-t. \end{align*} Because $0\le t<\varepsilon$, we have \begin{align*} 1-\varepsilon<1-t\le 1. \end{align*} Thus \begin{align*} \Phi(p,t)\in \{x\in M:1-\varepsilon<|x|\le 1\}=U. \end{align*} Next, if $x\in U$, then $1-\varepsilon<|x|\le 1$, so $|x|>0$. Define \begin{align*} p=\frac{x}{|x|}, \qquad t=1-|x|. \end{align*} Then \begin{align*} |p| &=\left|\frac{x}{|x|}\right|\\ &=\frac{|x|}{|x|}\\ &=1, \end{align*} so $p\in S^{n-1}$. Also, since $1-\varepsilon<|x|\le 1$, \begin{align*} 0\le 1-|x|<\varepsilon, \end{align*} so $t\in[0,\varepsilon)$. Now \begin{align*} \Phi(p,t) &=(1-t)p\\ &=(1-(1-|x|))\frac{x}{|x|}\\ &=|x|\frac{x}{|x|}\\ &=x. \end{align*} Therefore every point of $U$ lies in the image of $\Phi$. If $\Phi(p,t)=\Phi(q,s)$, then \begin{align*} (1-t)p=(1-s)q. \end{align*} Taking norms gives \begin{align*} |(1-t)p|&=|(1-s)q|,\\ (1-t)|p|&=(1-s)|q|,\\ 1-t&=1-s, \end{align*} because $|p|=|q|=1$ and $1-t,1-s>0$. Hence $t=s$. Substituting back, \begin{align*} (1-t)p=(1-t)q. \end{align*} Since $1-t>0$, dividing by $1-t$ gives \begin{align*} p=q. \end{align*} Thus $\Phi$ is injective. The inverse map on $U$ is therefore \begin{align*} \Phi^{-1}(x)=\left(\frac{x}{|x|},\,1-|x|\right). \end{align*} Both coordinate expressions are smooth for $|x|>0$, and every point of $U$ satisfies $|x|>1-\varepsilon>0$. Hence $\Phi^{-1}$ is smooth on $U$. The map $\Phi(p,t)=(1-t)p$ is smooth in the variables $(p,t)$ because each coordinate is a product of a coordinate function of $p$ with the smooth function $1-t$. Therefore $\Phi$ is a diffeomorphism from $S^{n-1}\times[0,\varepsilon)$ onto $U$. Finally, for $p\in S^{n-1}$, \begin{align*} \Phi(p,0) &=(1-0)p\\ &=p. \end{align*} So the boundary is fixed at parameter value $t=0$. The collar identifies a thin neighbourhood of the boundary of the closed ball with \begin{align*} S^{n-1}\times[0,\varepsilon). \end{align*} The parameter $t$ records inward radial distance from the boundary, since \begin{align*} |\Phi(p,t)|=1-t. \end{align*} Thus $t=0$ is the boundary sphere, and increasing $t$ moves inward along the radius through $p$. [/example] ## Orientability and the Möbius Band Top-degree forms are local volume elements. The question for integration is whether these local volume elements can be chosen consistently across the whole manifold. [definition: Orientation by an Atlas] An orientation on a smooth $n$-manifold $M$ is a smooth atlas such that every transition map has positive Jacobian determinant, considered up to refinement by atlases with the same positivity property. [/definition] This definition matches the intuition that a change of coordinates should preserve the chosen sign of a local volume form. The remaining question is whether this atlas-based sign convention is the same as choosing a globally nonzero top-degree form. That equivalence matters because integration is expressed using forms, while orientability is often checked using charts. [quotetheorem:3576] [citeproof:3576] The course will use orientability to define integration of top-degree forms. Non-orientable manifolds show why the hypothesis is not cosmetic: there may be no globally consistent choice of sign for volume. Connectedness is not required for the theorem: each connected component may be oriented independently, and a nowhere-vanishing top-degree form records all those choices at once. Passing to an orientable double cover can restore a consistent local sign upstairs, but the sign may reverse under the covering symmetry; the Möbius band below is the basic example where this obstruction appears. [illustration:forms-mobius-orientation-reversal] [example: Möbius Band] Let $M$ be the quotient of $[0,1]\times[-1,1]$ by the relation \begin{align*} (0,t)\sim(1,-t). \end{align*} Equivalently, let \begin{align*} \rho:\mathbb R\times[-1,1]\longrightarrow \mathbb R\times[-1,1], \qquad \rho(s,t)=(s+1,-t), \end{align*} and write \begin{align*} M=(\mathbb R\times[-1,1])/\langle \rho\rangle. \end{align*} Let $\pi:\mathbb R\times[-1,1]\to M$ be the quotient map. The central circle is the image of the set $\{(s,0):s\in\mathbb R\}$. The space $M$ is a smooth surface with boundary, but it is not orientable. Consequently, $M$ admits no smooth nowhere-vanishing global $2$-form. Away from $t=\pm 1$, small rectangles in $\mathbb R\times(-1,1)$ give coordinate charts modelled on $\mathbb R^2$. Near $t=1$, use half-rectangles \begin{align*} (a-\delta,a+\delta)\times(1-\eta,1], \end{align*} with boundary coordinate $r=1-t\ge 0$. Near $t=-1$, use half-rectangles \begin{align*} (a-\delta,a+\delta)\times[-1,-1+\eta), \end{align*} with boundary coordinate $r=t+1\ge 0$. Thus boundary points are modelled on $\mathbb H^2$. The only nontrivial gluing is generated by $\rho$. Its derivative matrix is \begin{align*} D\rho= \begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix}, \end{align*} so \begin{align*} \det(D\rho)=1\cdot(-1)-0\cdot 0=-1. \end{align*} Thus $\rho$ is smooth with smooth inverse \begin{align*} \rho^{-1}(s,t)=(s-1,-t), \end{align*} and the quotient has a smooth surface-with-boundary structure. Now compute what the gluing does to the local area form $ds\wedge dt$: \begin{align*} \rho^*(ds\wedge dt) &=d(s+1)\wedge d(-t)\\ &=ds\wedge(-dt)\\ &=-\,ds\wedge dt. \end{align*} So one trip around the central circle reverses the sign of the local area element. To see that no nowhere-vanishing global $2$-form can exist, suppose that $\omega$ is such a form on $M$. Since $ds\wedge dt$ spans the $2$-forms in the covering coordinates, write \begin{align*} \pi^*\omega=f(s,t)\,ds\wedge dt \end{align*} for a smooth function $f$. Because $\omega$ is nowhere zero and $\pi$ is locally a diffeomorphism, $f(s,t)\ne 0$ everywhere. Since $\pi\circ\rho=\pi$, the pullbacks agree: \begin{align*} \rho^*(\pi^*\omega)=\pi^*\omega. \end{align*} Compute the left-hand side: \begin{align*} \rho^*(\pi^*\omega) &=\rho^*\bigl(f(s,t)\,ds\wedge dt\bigr)\\ &=f(s+1,-t)\,\rho^*(ds\wedge dt)\\ &=f(s+1,-t)(-\,ds\wedge dt)\\ &=-f(s+1,-t)\,ds\wedge dt. \end{align*} Comparing with $\pi^*\omega=f(s,t)\,ds\wedge dt$ gives \begin{align*} f(s,t)=-f(s+1,-t). \end{align*} Set $t=0$ and define $g(s)=f(s,0)$. Then \begin{align*} g(s)=-g(s+1), \end{align*} so in particular \begin{align*} g(1)=-g(0). \end{align*} Since $g$ is continuous and nowhere zero on $[0,1]$, it cannot change sign on that interval. But the equality $g(1)=-g(0)$ forces the endpoint signs to be opposite. This contradiction shows that no smooth nowhere-vanishing global $2$-form exists. By *Orientability and Top-Degree Forms*, a smooth surface is orientable exactly when it admits a nowhere-vanishing smooth top-degree form. Therefore $M$ is not orientable. The Möbius band is locally a smooth surface with boundary, but its gluing map reverses the transverse coordinate. Algebraically, this is the sign change \begin{align*} \rho^*(ds\wedge dt)=-ds\wedge dt. \end{align*} Thus a local area form returns with the opposite sign after one circuit around the central circle, preventing a globally consistent orientation. [/example] The Möbius band is the standard warning that local data are not enough. Differential forms can always be defined locally, but global statements about integration and orientation require compatibility around loops as well as across individual overlaps. With differential forms defined on manifolds, the next task is to integrate them over domains. This requires equipping manifolds with an orientation, a consistent choice of sign for the top-degree form that lets us assign [real numbers](/page/Real%20Numbers) to integrals. # 6. Orientation and Integration of Forms This chapter turns differential forms from algebraic objects into quantities that can be integrated. The central difficulty is that integration is signed: a coordinate system tells us whether $dx_1 \wedge \cdots \wedge dx_n$ is positive or negative, and different coordinate systems need to make compatible choices. Once this compatibility is encoded by an orientation, compactly supported top-degree forms have a well-defined integral over a manifold, and the familiar line, surface, and volume integrals become instances of the same construction. The chapter also prepares the boundary convention needed for Stokes theorem. The orientation on $\partial M$ is not an extra arbitrary choice once $M$ is oriented; it is fixed by the outward normal first convention. This is the sign convention that makes the general Stokes theorem match Green theorem, the [divergence theorem](/theorems/2754), and the classical orientation rules for curves and surfaces. ## Orientability and Top-Degree Forms What data lets a manifold decide whether a local coordinate volume element is positive? On an $n$-manifold, the space of top-degree covectors at each point is one-dimensional, so choosing a positive side of that line at every point is the geometric content of orientation. [definition: Orientation of a Tangent Space] Let $V$ be a real [vector space](/page/Vector%20Space) of dimension $n$. An orientation of $V$ is a choice of one of the two equivalence classes of ordered bases of $V$, where two ordered bases $(v_1,\dots,v_n)$ and $(w_1,\dots,w_n)$ are equivalent when the change-of-basis matrix from the first basis to the second has positive determinant. [/definition] Thus an ordered basis is either positively oriented or negatively oriented relative to the chosen class. For a one-dimensional [vector space](/page/Vector%20Space), this amounts to choosing which nonzero vectors point in the positive direction. On a manifold, this choice has to be made at every tangent space, but arbitrary pointwise choices would be useless for integration: the sign could jump from point to point or disagree with local coordinates. The definition therefore requires local coordinate bases that are positive throughout a neighbourhood. [definition: Orientation of a Manifold] Let $M$ be a smooth $n$-manifold. An orientation of $M$ is a choice of orientation of each tangent space $T_pM$, for $p \in M$, such that around every point there is a coordinate chart $(U,\varphi)$ whose coordinate basis \begin{align*} \left(\frac{\partial}{\partial x_1}\bigg|_p,\dots,\frac{\partial}{\partial x_n}\bigg|_p\right) \end{align*} is positively oriented for every $p \in U$. [/definition] The local compatibility condition prevents the orientation from changing sign discontinuously from point to point. A manifold admitting such a choice is called orientable. For integration, an orientation should also be detectable by the forms being integrated. A bare choice of positive bases does not by itself produce an integrand, while a top-degree form can both be integrated and test the sign of an ordered basis. A nowhere-vanishing top-degree form gives exactly such a detector. Because it never evaluates to zero on a basis, its sign can consistently separate positive bases from negative ones. The next notion is therefore the form-level representative of an orientation, and it is the object that later appears under an integral sign. [definition: Volume Form] Let $M$ be a smooth $n$-manifold. A volume form on $M$ is a smooth differential form $\omega \in \Omega^n(M)$ such that $\omega_p \ne 0$ in $\Lambda^n(T_p^*M)$ for every $p \in M$. [/definition] A volume form chooses positive bases by evaluation: an ordered basis $(v_1,\dots,v_n)$ of $T_pM$ is positive when $\omega_p(v_1,\dots,v_n)>0$. The point that needs proof is that this pointwise sign rule is locally compatible, and conversely that every locally compatible orientation can be represented by some nowhere-vanishing top-degree form. [quotetheorem:3576] [citeproof:3576] This theorem is useful because a nowhere-vanishing top-degree form is often easier to construct than an orientation atlas. On a submanifold of Euclidean space, a normal vector field can also encode an orientation, as surface examples below show. [example: Standard Orientation of Euclidean Space] On $\mathbb R^n$ with global coordinates $(x_1,\dots,x_n)$, let \begin{align*} \omega=dx_1\wedge\cdots\wedge dx_n. \end{align*} For each $p\in\mathbb R^n$, write \begin{align*} e_i\big|_p=\frac{\partial}{\partial x_i}\bigg|_p. \end{align*} The form $\omega$ determines the standard orientation of $\mathbb R^n$: an ordered basis $(v_1,\dots,v_n)$ of $T_p\mathbb R^n$ is positive exactly when its change-of-basis determinant relative to $(e_1|_p,\dots,e_n|_p)$ is positive. First evaluate $\omega_p$ on the standard ordered basis. Since $dx_i(e_j)=\delta_{ij}$, the evaluation rule for the wedge product gives \begin{align*} \omega_p(e_1,\dots,e_n) &= \det \begin{pmatrix} dx_1(e_1) & dx_1(e_2) & \cdots & dx_1(e_n)\\ dx_2(e_1) & dx_2(e_2) & \cdots & dx_2(e_n)\\ \vdots & \vdots & \ddots & \vdots\\ dx_n(e_1) & dx_n(e_2) & \cdots & dx_n(e_n) \end{pmatrix}\\ &= \det \begin{pmatrix} 1 & 0 & \cdots & 0\\ 0 & 1 & \cdots & 0\\ \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & \cdots & 1 \end{pmatrix}\\ &=1. \end{align*} Thus $\omega_p\ne 0$ for every $p$, so $\omega$ is a nowhere-vanishing top-degree form. Now let $(v_1,\dots,v_n)$ be any ordered basis of $T_p\mathbb R^n$. Write each vector in the standard basis: \begin{align*} v_j=\sum_{i=1}^n a_{ij}e_i. \end{align*} The matrix \begin{align*} A=(a_{ij}) \end{align*} is the change-of-basis matrix from $(e_1,\dots,e_n)$ to $(v_1,\dots,v_n)$, with the $j$th column containing the coordinates of $v_j$. For each pair $(i,j)$, linearity of $dx_i$ gives \begin{align*} dx_i(v_j) &=dx_i\left(\sum_{k=1}^n a_{kj}e_k\right)\\ &=\sum_{k=1}^n a_{kj}dx_i(e_k)\\ &=\sum_{k=1}^n a_{kj}\delta_{ik}\\ &=a_{ij}. \end{align*} Therefore \begin{align*} \omega_p(v_1,\dots,v_n) &= \det \begin{pmatrix} dx_1(v_1) & dx_1(v_2) & \cdots & dx_1(v_n)\\ dx_2(v_1) & dx_2(v_2) & \cdots & dx_2(v_n)\\ \vdots & \vdots & \ddots & \vdots\\ dx_n(v_1) & dx_n(v_2) & \cdots & dx_n(v_n) \end{pmatrix}\\ &= \det \begin{pmatrix} a_{11} & a_{12} & \cdots & a_{1n}\\ a_{21} & a_{22} & \cdots & a_{2n}\\ \vdots & \vdots & \ddots & \vdots\\ a_{n1} & a_{n2} & \cdots & a_{nn} \end{pmatrix}\\ &=\det A. \end{align*} The form $dx_1\wedge\cdots\wedge dx_n$ evaluates positively exactly on those ordered bases whose change-of-basis determinant relative to $(e_1,\dots,e_n)$ is positive. Hence it selects the usual standard orientation of $\mathbb R^n$. [/example] The Euclidean example is the reference model: orientation is detected by a nowhere-zero top-degree form whose sign is globally consistent. The next example shows exactly what fails when that local sign cannot be transported consistently around the manifold. [example: Non-Orientability of the Mobius Strip] Let $M$ be the Mobius strip obtained from $[0,1]\times(-1,1)$ by the identification \begin{align*} (0,t)\sim(1,-t). \end{align*} Equivalently, use the covering strip \begin{align*} \widetilde M=\mathbb R\times(-1,1) \end{align*} with deck transformation \begin{align*} T(s,t)=(s+1,-t). \end{align*} The quotient map \begin{align*} \pi:\widetilde M\to M \end{align*} satisfies $\pi\circ T=\pi$. Write $(s,t)$ for the standard coordinates on $\widetilde M$. The Mobius strip admits no volume form. Hence, by *Orientability by Volume Form*, it is not orientable. Suppose, for contradiction, that $M$ admits a volume form $\Omega$. Since $\pi$ is a local diffeomorphism, the pullback \begin{align*} \beta=\pi^*\Omega \end{align*} is a nowhere-vanishing $2$-form on $\widetilde M$. In the coordinates $(s,t)$, every $2$-form on $\widetilde M$ has the form \begin{align*} \beta=f(s,t)\,ds\wedge dt \end{align*} for a smooth function $f$. Since $\beta$ is nowhere zero and $ds\wedge dt$ is a basis of $\Lambda^2(T_{(s,t)}^*\widetilde M)$, we have \begin{align*} f(s,t)\ne 0 \end{align*} for every $(s,t)\in\widetilde M$. Now use the fact that $\pi\circ T=\pi$. Pulling back $\Omega$ gives \begin{align*} T^*\beta &=T^*(\pi^*\Omega)\\ &=(\pi\circ T)^*\Omega\\ &=\pi^*\Omega\\ &=\beta. \end{align*} Compute the pullback of the coordinate differentials under $T(s,t)=(s+1,-t)$: \begin{align*} T^*ds&=d(s+1)=ds,\\ T^*dt&=d(-t)=-dt. \end{align*} Therefore \begin{align*} T^*\beta &=T^*\bigl(f(s,t)\,ds\wedge dt\bigr)\\ &=f(T(s,t))\,T^*ds\wedge T^*dt\\ &=f(s+1,-t)\,ds\wedge(-dt)\\ &=-f(s+1,-t)\,ds\wedge dt. \end{align*} Since $T^*\beta=\beta$, comparison with $\beta=f(s,t)\,ds\wedge dt$ gives \begin{align*} f(s,t)\,ds\wedge dt=-f(s+1,-t)\,ds\wedge dt. \end{align*} Because $ds\wedge dt$ is nonzero, the coefficients are equal: \begin{align*} f(s,t)=-f(s+1,-t). \end{align*} Set $t=0$. Then \begin{align*} f(s,0)=-f(s+1,0). \end{align*} In particular, at $s=0$, \begin{align*} f(0,0)=-f(1,0), \end{align*} so \begin{align*} f(1,0)=-f(0,0). \end{align*} Define \begin{align*} h:[0,1]\to\mathbb R,\qquad h(s)=f(s,0). \end{align*} The function $h$ is continuous and satisfies \begin{align*} h(1)=-h(0). \end{align*} Since $f$ is nowhere zero, $h(0)\ne 0$. If $h(0)>0$, then $h(1)<0$. If $h(0)<0$, then $h(1)>0$. Thus $h$ takes values of opposite signs on $[0,1]$. Since the image of a connected interval under a continuous function is an interval, there exists $c\in[0,1]$ such that \begin{align*} h(c)=0. \end{align*} Equivalently, \begin{align*} f(c,0)=0, \end{align*} contradicting the fact that $f$ is nowhere zero. The same sign change is visible in the Jacobian matrix of the gluing map: \begin{align*} DT = \begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix}, \qquad \det(DT)=(1)(-1)-(0)(0)=-1. \end{align*} The transverse coordinate reversal forces a top-degree form to pick up a minus sign after one trip around the strip. A volume form would have to be both nowhere zero and compatible with the identification $(s,t)\sim(s+1,-t)$. Compatibility forces its coefficient to change sign along the central circle, while nowhere-vanishing continuity forbids such a sign change. Hence the Mobius strip has no volume form and is not orientable. [/example] ## Oriented Atlases and Boundary Orientation How can an orientation be recorded in coordinates, and what sign should a boundary inherit? Coordinate changes compare local choices by their Jacobian determinants, while the boundary convention is fixed by placing the outward normal before the boundary basis. [definition: Oriented Atlas] Let $M$ be a smooth $n$-manifold. An oriented atlas on $M$ is an atlas $\{(U_i,\varphi_i)\}_{i\in I}$ such that for every overlap $U_i\cap U_j$, the transition map \begin{align*} \varphi_j\circ \varphi_i^{-1}:\varphi_i(U_i\cap U_j)\to \varphi_j(U_i\cap U_j) \end{align*} has positive Jacobian determinant at every point of its domain. [/definition] An oriented atlas is a coordinate-level expression of the same orientation from the previous section. Its advantage is computational: it tells us which coordinate formula for an $n$-form should be integrated with a positive sign. The remaining question is whether this coordinate description is merely convenient or actually equivalent to the tangent-space definition of orientation. The comparison matters because integration is defined in charts, while the orientation of a manifold was defined intrinsically on tangent spaces. [quotetheorem:3577] [citeproof:3577] This equivalence lets us move freely between the two languages. In calculations, positive charts determine the sign of coordinate integrals; in geometric statements, tangent-space orientations make the construction independent of a chosen atlas. The positivity of transition Jacobians is exactly what keeps these two descriptions from assigning opposite signs on overlaps. The boundary now needs its own orientation, but it cannot be chosen independently of the ambient manifold. Stokes' theorem compares an integral over $M$ with an integral over $\partial M$, so the boundary sign must be fixed by a convention involving the outward normal direction. [definition: Boundary Orientation] Let $M$ be an oriented smooth $n$-manifold with boundary. The induced orientation on $\partial M$ is the orientation for which an ordered basis $(v_1,\dots,v_{n-1})$ of $T_p(\partial M)$ is positive exactly when $(\nu,v_1,\dots,v_{n-1})$ is a positive ordered basis of $T_pM$, where $\nu\in T_pM$ is any outward-pointing vector transverse to $T_p(\partial M)$. [/definition] The outward vector is placed first. For a region in the plane with the standard orientation, this produces counterclockwise orientation on the outer boundary and clockwise orientation on holes. [example: Boundary of the Unit Interval] Give the interval $[0,1]$ the standard orientation determined by the $1$-form $dt$. Thus, at each interior point, the ordered basis \begin{align*} \left(\frac{\partial}{\partial t}\right) \end{align*} is positive because \begin{align*} dt\left(\frac{\partial}{\partial t}\right)=1>0. \end{align*} The boundary is the $0$-manifold \begin{align*} \partial[0,1]=\{0,1\}. \end{align*} At a point of a $0$-manifold, the tangent space is the zero [vector space](/page/Vector%20Space), whose ordered basis is the empty basis $()$. The induced boundary orientation assigns sign $+1$ to the endpoint $1$ and sign $-1$ to the endpoint $0$. By the boundary orientation convention, the empty basis $()$ of $T_p(\partial[0,1])$ is positive exactly when \begin{align*} (\nu) \end{align*} is a positive ordered basis of $T_p[0,1]$, where $\nu$ is an outward-pointing vector at $p$. At the right endpoint $p=1$, the outward direction points toward increasing $t$, so we may take \begin{align*} \nu_1=\frac{\partial}{\partial t}\bigg|_1. \end{align*} Its sign relative to the standard orientation is determined by evaluation against $dt$: \begin{align*} dt_1(\nu_1) &= dt_1\left(\frac{\partial}{\partial t}\bigg|_1\right)\\ &=1\\ &>0. \end{align*} Therefore $(\nu_1)$ is a positive basis of $T_1[0,1]$. Hence the empty basis $()$ is positive in $T_1(\partial[0,1])$, so the point $1$ receives sign $+1$. At the left endpoint $p=0$, the outward direction points toward decreasing $t$, so we may take \begin{align*} \nu_0=-\frac{\partial}{\partial t}\bigg|_0. \end{align*} Again evaluate against $dt$: \begin{align*} dt_0(\nu_0) &= dt_0\left(-\frac{\partial}{\partial t}\bigg|_0\right)\\ &= -\,dt_0\left(\frac{\partial}{\partial t}\bigg|_0\right)\\ &=-1\\ &<0. \end{align*} Therefore $(\nu_0)$ is a negative basis of $T_0[0,1]$. Hence the empty basis $()$ has the negative induced orientation at $0$, so the point $0$ receives sign $-1$. With the outward-normal-first boundary convention, the oriented boundary of the standard oriented interval is \begin{align*} \partial[0,1]=\{1\}-\{0\}. \end{align*} Thus the right endpoint contributes positively and the left endpoint contributes negatively. [/example] The interval is the one-dimensional version of the same outward-normal-first rule. In dimension two, the rule becomes the familiar direction of travel around a boundary curve. [example: Boundary Orientation of the Unit Square] Let \begin{align*} Q=[0,1]^2\subset \mathbb R^2 \end{align*} with coordinates $(x,y)$. Give $Q$ the standard orientation determined by the $2$-form \begin{align*} dx\wedge dy. \end{align*} Thus an ordered basis $(u,v)$ of a tangent plane is positive exactly when \begin{align*} (dx\wedge dy)(u,v)>0. \end{align*} For tangent vectors $u,v$, the wedge evaluation is \begin{align*} (dx\wedge dy)(u,v)=dx(u)dy(v)-dx(v)dy(u). \end{align*} The induced boundary orientation on $\partial Q$ runs counterclockwise: rightward along the bottom edge, upward along the right edge, leftward along the top edge, and downward along the left edge. By the boundary orientation convention, a tangent vector $v$ along $\partial Q$ is positive exactly when $(\nu,v)$ is a positive ordered basis of $T_pQ$, where $\nu$ is outward-pointing and transverse to the boundary. On the bottom edge $y=0$, the outward direction is \begin{align*} \nu=-\frac{\partial}{\partial y}. \end{align*} Test $v=\frac{\partial}{\partial x}$. Since \begin{align*} dx\left(-\frac{\partial}{\partial y}\right)&=0,& dy\left(\frac{\partial}{\partial x}\right)&=0,\\ dx\left(\frac{\partial}{\partial x}\right)&=1,& dy\left(-\frac{\partial}{\partial y}\right)&=-1, \end{align*} we get \begin{align*} (dx\wedge dy)\left(-\frac{\partial}{\partial y},\frac{\partial}{\partial x}\right) &= dx\left(-\frac{\partial}{\partial y}\right) dy\left(\frac{\partial}{\partial x}\right) - dx\left(\frac{\partial}{\partial x}\right) dy\left(-\frac{\partial}{\partial y}\right)\\ &=(0)(0)-(1)(-1)\\ &=1. \end{align*} Therefore the positive boundary direction on the bottom edge is $\frac{\partial}{\partial x}$. On the right edge $x=1$, the outward direction is \begin{align*} \nu=\frac{\partial}{\partial x}. \end{align*} Test $v=\frac{\partial}{\partial y}$. Since \begin{align*} dx\left(\frac{\partial}{\partial x}\right)&=1,& dy\left(\frac{\partial}{\partial y}\right)&=1,\\ dx\left(\frac{\partial}{\partial y}\right)&=0,& dy\left(\frac{\partial}{\partial x}\right)&=0, \end{align*} we get \begin{align*} (dx\wedge dy)\left(\frac{\partial}{\partial x},\frac{\partial}{\partial y}\right) &= dx\left(\frac{\partial}{\partial x}\right) dy\left(\frac{\partial}{\partial y}\right) - dx\left(\frac{\partial}{\partial y}\right) dy\left(\frac{\partial}{\partial x}\right)\\ &=(1)(1)-(0)(0)\\ &=1. \end{align*} Therefore the positive boundary direction on the right edge is $\frac{\partial}{\partial y}$. On the top edge $y=1$, the outward direction is \begin{align*} \nu=\frac{\partial}{\partial y}. \end{align*} Test $v=-\frac{\partial}{\partial x}$. Since \begin{align*} dx\left(\frac{\partial}{\partial y}\right)&=0,& dy\left(-\frac{\partial}{\partial x}\right)&=0,\\ dx\left(-\frac{\partial}{\partial x}\right)&=-1,& dy\left(\frac{\partial}{\partial y}\right)&=1, \end{align*} we get \begin{align*} (dx\wedge dy)\left(\frac{\partial}{\partial y},-\frac{\partial}{\partial x}\right) &= dx\left(\frac{\partial}{\partial y}\right) dy\left(-\frac{\partial}{\partial x}\right) - dx\left(-\frac{\partial}{\partial x}\right) dy\left(\frac{\partial}{\partial y}\right)\\ &=(0)(0)-(-1)(1)\\ &=1. \end{align*} Therefore the positive boundary direction on the top edge is $-\frac{\partial}{\partial x}$. On the left edge $x=0$, the outward direction is \begin{align*} \nu=-\frac{\partial}{\partial x}. \end{align*} Test $v=-\frac{\partial}{\partial y}$. Since \begin{align*} dx\left(-\frac{\partial}{\partial x}\right)&=-1,& dy\left(-\frac{\partial}{\partial y}\right)&=-1,\\ dx\left(-\frac{\partial}{\partial y}\right)&=0,& dy\left(-\frac{\partial}{\partial x}\right)&=0, \end{align*} we get \begin{align*} (dx\wedge dy)\left(-\frac{\partial}{\partial x},-\frac{\partial}{\partial y}\right) &= dx\left(-\frac{\partial}{\partial x}\right) dy\left(-\frac{\partial}{\partial y}\right) - dx\left(-\frac{\partial}{\partial y}\right) dy\left(-\frac{\partial}{\partial x}\right)\\ &=(-1)(-1)-(0)(0)\\ &=1. \end{align*} Therefore the positive boundary direction on the left edge is $-\frac{\partial}{\partial y}$. Starting on the bottom edge, the induced orientation moves right, then up the right edge, then left along the top edge, then down the left edge. This is exactly counterclockwise orientation of the boundary of the unit square. [/example] ## Integrating Compactly Supported Top-Degree Forms How can we integrate an $n$-form on a manifold that may not have one global coordinate system? The construction is local: break the form using a [partition of unity](/page/Partition%20of%20Unity), integrate each piece in a positive coordinate chart, and prove that the sum is independent of every auxiliary choice. [definition: Local Integral of a Top-Degree Form] Let $(U,\varphi)$ be a positive coordinate chart on an oriented smooth $n$-manifold $M$, with coordinates $(x_1,\dots,x_n)$. If $\omega\in\Omega_c^n(U)$ has coordinate expression \begin{align*} \omega = f\,dx_1\wedge\cdots\wedge dx_n, \end{align*} then its local integral over $U$ is \begin{align*} \int_U \omega := \int_{\varphi(U)} f\circ\varphi^{-1}\,d\mathcal L^n. \end{align*} [/definition] This definition uses a positive chart. If a negative chart were used instead, the sign would change, since the determinant in the change-of-variables formula would have the opposite sign. Before local integrals can be used as geometric quantities, they must be independent of the particular positive coordinates used on the same region. The possible obstruction is the Jacobian factor in a coordinate change; orientation should make that factor enter with the correct sign and recover the ordinary change-of-variables formula. [quotetheorem:3578] [citeproof:3578] The theorem says that a local integral belongs to the oriented region and the form, not to the chart used to compute it. This is the essential local well-definedness step: without it, a partition-of-unity construction would depend on arbitrary coordinate choices. Positive charts are still required, because switching to a negative chart would reverse the sign of a top-degree form integral. Now the global integral can be assembled by cutting a compactly supported form into chart-supported pieces. Compact support ensures that only finitely many partition-of-unity terms contribute after choosing a locally finite cover. [definition: Integral of a Compactly Supported Top-Degree Form] Let $M$ be an oriented smooth $n$-manifold and let $\omega\in\Omega_c^n(M)$. Choose a locally finite cover by positive coordinate charts $(U_i,\varphi_i)$ and a smooth [partition of unity](/page/Partition%20of%20Unity) $(\rho_i)_{i\in I}$ subordinate to this cover. The integral of $\omega$ over $M$ is \begin{align*} \int_M \omega := \sum_{i\in I}\int_{U_i} \rho_i\omega. \end{align*} [/definition] The definition still contains auxiliary choices: the positive chart cover and the subordinate partition of unity. The next well-definedness result says that these choices do not affect the answer, so the notation $\int_M\omega$ refers to an invariant of the oriented manifold and the compactly supported top-degree form. [quotetheorem:3579] [citeproof:3579] This result is the analytic foundation for the rest of the course. From now on, $\int_M\omega$ is meaningful whenever $M$ is oriented and $\omega$ is compactly supported of degree $\dim M$. The compact-support hypothesis is what keeps the partition sum under control on noncompact manifolds, while orientation is what fixes the sign of each local contribution. [quotetheorem:3580] [citeproof:3580] This theorem shows that integration of forms depends on the orientation, not just on the underlying smooth manifold. This differs from measure-theoretic integration, where changing an orientation has no effect on the integral of a nonnegative density. [example: Integrating the Standard Area Form on the Unit Square] Let \begin{align*} Q=[0,1]^2\subset\mathbb R^2 \end{align*} with coordinates $(x,y)$. Give $Q$ the standard orientation determined by the top-degree form \begin{align*} dx\wedge dy. \end{align*} Let \begin{align*} \omega=dx\wedge dy. \end{align*} The integral of $\omega$ over $Q$ is \begin{align*} \int_Q \omega=1. \end{align*} If $Q$ is given the opposite orientation, then the same form integrates to $-1$. Use the global coordinate chart \begin{align*} \varphi:Q\to[0,1]^2,\qquad \varphi(x,y)=(x,y). \end{align*} This chart is positive for the standard orientation because the coordinate basis \begin{align*} \left(\frac{\partial}{\partial x},\frac{\partial}{\partial y}\right) \end{align*} satisfies \begin{align*} (dx\wedge dy)\left(\frac{\partial}{\partial x},\frac{\partial}{\partial y}\right) &= dx\left(\frac{\partial}{\partial x}\right) dy\left(\frac{\partial}{\partial y}\right) - dx\left(\frac{\partial}{\partial y}\right) dy\left(\frac{\partial}{\partial x}\right)\\ &=(1)(1)-(0)(0)\\ &=1\\ &>0. \end{align*} In this positive chart, \begin{align*} \omega=dx\wedge dy=1\,dx\wedge dy. \end{align*} By the definition of the local integral of a top-degree form, \begin{align*} \int_Q \omega &=\int_{[0,1]^2}1\,d\mathcal L^2(x,y). \end{align*} Using iterated integration over the rectangle $[0,1]\times[0,1]$, \begin{align*} \int_{[0,1]^2}1\,d\mathcal L^2(x,y) &=\int_0^1\int_0^1 1\,dx\,dy\\ &=\int_0^1 \left(\int_0^1 1\,dx\right)dy\\ &=\int_0^1 \left(x\bigg|_{x=0}^{x=1}\right)dy\\ &=\int_0^1 (1-0)\,dy\\ &=\int_0^1 1\,dy\\ &=y\bigg|_{y=0}^{y=1}\\ &=1-0\\ &=1. \end{align*} Therefore \begin{align*} \int_Q dx\wedge dy=1. \end{align*} Now give the same square the opposite orientation. A positive chart for the opposite orientation is obtained by reflecting one coordinate, for instance \begin{align*} u=-x,\qquad v=y. \end{align*} Then \begin{align*} du=d(-x)=-dx,\qquad dv=dy, \end{align*} so \begin{align*} du\wedge dv &=(-dx)\wedge dy\\ &=-(dx\wedge dy). \end{align*} Hence \begin{align*} dx\wedge dy=-du\wedge dv. \end{align*} The reflected coordinate image of $Q$ is \begin{align*} [-1,0]\times[0,1]. \end{align*} Thus, in the positive coordinates $(u,v)$ for the opposite orientation, \begin{align*} \int_{-Q} dx\wedge dy &=\int_{[-1,0]\times[0,1]} (-1)\,d\mathcal L^2(u,v)\\ &=\int_0^1\int_{-1}^0 (-1)\,du\,dv\\ &=\int_0^1 \left(-u\bigg|_{u=-1}^{u=0}\right)dv\\ &=\int_0^1 (0-1)\,dv\\ &=\int_0^1 (-1)\,dv\\ &=-v\bigg|_{v=0}^{v=1}\\ &=-1. \end{align*} With the standard orientation, $dx\wedge dy$ integrates over the unit square to its ordinary Euclidean area, namely $1$. Reversing the orientation changes the sign of the top-degree form in positive coordinates, so the same form integrates to $-1$ on the oppositely oriented square. [/example] The square illustrates orientation on a compact coordinate domain. The next example records the noncompact case where compact support, rather than compactness of the manifold, makes the integral finite and well-defined. [example: A Compactly Supported Form on the Plane] Let $\eta\in C_c^\infty(\mathbb R^2)$, and give $\mathbb R^2$ the standard orientation determined by \begin{align*} dx\wedge dy. \end{align*} Define the compactly supported $2$-form \begin{align*} \omega=\eta(x,y)\,dx\wedge dy. \end{align*} Since $\eta$ has compact support, the coefficient of $\omega$ is zero outside the compact set $\operatorname{supp}\eta$, so $\omega\in\Omega_c^2(\mathbb R^2)$. The manifold integral of $\omega$ is the ordinary double integral of $\eta$: \begin{align*} \int_{\mathbb R^2}\omega = \int_{\mathbb R^2}\eta(x,y)\,d\mathcal L^2(x,y). \end{align*} Use the global coordinate chart \begin{align*} \varphi:\mathbb R^2\to\mathbb R^2,\qquad \varphi(x,y)=(x,y). \end{align*} This chart is positive for the standard orientation because \begin{align*} (dx\wedge dy)\left(\frac{\partial}{\partial x},\frac{\partial}{\partial y}\right) &= dx\left(\frac{\partial}{\partial x}\right) dy\left(\frac{\partial}{\partial y}\right) - dx\left(\frac{\partial}{\partial y}\right) dy\left(\frac{\partial}{\partial x}\right)\\ &=(1)(1)-(0)(0)\\ &=1\\ &>0. \end{align*} The one-chart cover $\{\mathbb R^2\}$ has subordinate [partition of unity](/page/Partition%20of%20Unity) consisting of the single function \begin{align*} \rho(x,y)=1. \end{align*} Therefore \begin{align*} \rho\omega &=1\cdot \eta(x,y)\,dx\wedge dy\\ &=\eta(x,y)\,dx\wedge dy. \end{align*} In the coordinate expression \begin{align*} \rho\omega=f\,dx\wedge dy, \end{align*} the coefficient is \begin{align*} f(x,y)=\eta(x,y). \end{align*} Since $\varphi$ is the identity map, its inverse is also the identity map: \begin{align*} \varphi^{-1}(x,y)=(x,y). \end{align*} Hence \begin{align*} (f\circ\varphi^{-1})(x,y) &=f(x,y)\\ &=\eta(x,y). \end{align*} By the definition of the local integral of a top-degree form in a positive chart, \begin{align*} \int_{\mathbb R^2}\omega &=\int_{\varphi(\mathbb R^2)} f\circ\varphi^{-1}\,d\mathcal L^2\\ &=\int_{\mathbb R^2} \eta(x,y)\,d\mathcal L^2(x,y). \end{align*} On the standard oriented plane, the manifold integral of the compactly supported form $\eta(x,y)\,dx\wedge dy$ is exactly the usual Lebesgue double integral of its coefficient function $\eta$. [/example] ## Line and Surface Integrals as Form Integrals Why do line integrals, flux integrals, and surface integrals have different formulas in vector calculus? They look different because they are written using coordinates and Euclidean vector identifications, but differential forms express them by the same rule: integrate a top-degree form over an oriented manifold of the matching dimension. For a curve, the top degree is $1$. If $\gamma:[a,b]\to M$ is a smooth parametrised curve compatible with the chosen orientation, integrating a $1$-form $\alpha$ over the oriented curve means integrating the pullback $\gamma^*\alpha$ over $[a,b]$. [definition: Integral of a One-Form Along a Parametrised Curve] Let $M$ be a smooth manifold, let $\alpha\in\Omega^1(M)$, and let $\gamma:[a,b]\to M$ be a smooth curve. The line integral of $\alpha$ along $\gamma$ is \begin{align*} \int_\gamma \alpha := \int_{[a,b]}\gamma^*\alpha. \end{align*} [/definition] If $\alpha=P\,dx+Q\,dy+R\,dz$ on an open subset of $\mathbb R^3$ and $\gamma(t)=(x(t),y(t),z(t))$, then \begin{align*} \gamma^*\alpha=(P(\gamma(t))x^{\prime}(t)+Q(\gamma(t))y^{\prime}(t)+R(\gamma(t))z^{\prime}(t))\,dt. \end{align*} Thus the familiar dot-product line integral is the pullback formula for $1$-forms. [example: Work Integral as a One-Form Integral] Let $\alpha=-y\,dx+x\,dy$ be the $1$-form on $\mathbb R^2$, and let \begin{align*} \gamma:[0,2\pi]\to\mathbb R^2,\qquad \gamma(t)=(\cos t,\sin t). \end{align*} Write $x,y$ for the standard coordinate functions on $\mathbb R^2$. The curve $\gamma$ traces the unit circle once counterclockwise, so it parametrises the positively oriented unit circle. The integral of $\alpha$ around the positively oriented unit circle is \begin{align*} \int_\gamma \alpha=2\pi. \end{align*} By the definition of the line integral of a $1$-form along a parametrised curve, \begin{align*} \int_\gamma \alpha=\int_{[0,2\pi]}\gamma^*\alpha. \end{align*} First compute the coordinate pullbacks. Since \begin{align*} x\circ\gamma(t)&=\cos t,\\ y\circ\gamma(t)&=\sin t, \end{align*} we have \begin{align*} \gamma^*dx &=d(x\circ\gamma)\\ &=d(\cos t)\\ &=-\sin t\,dt, \end{align*} and \begin{align*} \gamma^*dy &=d(y\circ\gamma)\\ &=d(\sin t)\\ &=\cos t\,dt. \end{align*} Using linearity of pullback and multiplication by pulled-back coefficient functions, \begin{align*} \gamma^*\alpha &=\gamma^*(-y\,dx+x\,dy)\\ &=-(y\circ\gamma)\,\gamma^*dx+(x\circ\gamma)\,\gamma^*dy\\ &=-(\sin t)(-\sin t\,dt)+(\cos t)(\cos t\,dt)\\ &=\sin^2 t\,dt+\cos^2 t\,dt\\ &=(\sin^2 t+\cos^2 t)\,dt\\ &=1\,dt. \end{align*} Therefore \begin{align*} \int_\gamma \alpha &=\int_{[0,2\pi]}1\,dt\\ &=\int_0^{2\pi}1\,dt\\ &=t\bigg|_{t=0}^{t=2\pi}\\ &=2\pi-0\\ &=2\pi. \end{align*} The form $-y\,dx+x\,dy$ pulls back along the positively oriented unit circle to the standard length-one form $dt$ on the parameter interval. Since the parameter runs from $0$ to $2\pi$, the integral is $2\pi$. [/example] For an oriented surface in $\mathbb R^3$, the top degree is $2$, so surface flux should be expressible as the integral of a $2$-form. The coordinate problem is to encode the three components of a vector field so that pulling the form back along a parametrised surface produces the classical dot product with the oriented normal. This is a sign-sensitive encoding problem: each component of the vector field must pair with the oriented coordinate area perpendicular to that component. The wedge terms below are arranged so that a parametrisation $r(u,v)$ produces the scalar triple product $F(r(u,v))\cdot(r_u\times r_v)$, which is the usual flux density. [definition: Flux Form] Let $U\subset\mathbb R^3$ be open and let $F:U\to\mathbb R^3$ be a smooth vector field, written $F=(F_1,F_2,F_3)$. The flux form associated to $F$ is the $2$-form \begin{align*} \omega_F := F_1\,dy\wedge dz+F_2\,dz\wedge dx+F_3\,dx\wedge dy. \end{align*} [/definition] For a parametrised oriented surface $r:D\subset\mathbb R^2\to\mathbb R^3$, the pullback $r^*\omega_F$ is a multiple of $du\wedge dv$. The coefficient is the classical scalar triple product $F(r(u,v))\cdot(r_u\times r_v)$ when $(u,v)$ is a positive parameter orientation. What remains is to justify that this local coefficient identity really recovers the full surface integral, rather than only matching one parametrisation. The result below is the bridge between the abstract integral of the flux form and the classical vector-calculus flux integral with the chosen orientation. [quotetheorem:3581] [citeproof:3581] The formula identifies the abstract integral of a $2$-form with the vector-calculus flux integral whenever the parametrisation and orientation agree. The following example uses the simplest closed oriented surface to check the sign convention. [example: Outward Flux Through the Unit Sphere] Let \begin{align*} S^2=\{(x,y,z)\in\mathbb R^3:x^2+y^2+z^2=1\} \end{align*} with the outward orientation, and let \begin{align*} F(x,y,z)=(x,y,z). \end{align*} The associated flux form is \begin{align*} \omega_F=x\,dy\wedge dz+y\,dz\wedge dx+z\,dx\wedge dy. \end{align*} Use the spherical parametrisation \begin{align*} r:[0,\pi]\times[0,2\pi]\to S^2,\qquad r(\theta,\phi)=(\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta). \end{align*} The integral of $\omega_F$ over $S^2$ is \begin{align*} \int_{S^2}\omega_F=4\pi. \end{align*} On the interior of the parameter rectangle, \begin{align*} r_\theta&=(\cos\theta\cos\phi,\cos\theta\sin\phi,-\sin\theta),\\ r_\phi&=(-\sin\theta\sin\phi,\sin\theta\cos\phi,0). \end{align*} Their cross product is \begin{align*} r_\theta\times r_\phi &= \begin{pmatrix} \cos\theta\sin\phi\cdot 0-(-\sin\theta)(\sin\theta\cos\phi)\\ (-\sin\theta)(-\sin\theta\sin\phi)-\cos\theta\cos\phi\cdot 0\\ \cos\theta\cos\phi\cdot\sin\theta\cos\phi-\cos\theta\sin\phi\cdot(-\sin\theta\sin\phi) \end{pmatrix}\\ &= \begin{pmatrix} \sin^2\theta\cos\phi\\ \sin^2\theta\sin\phi\\ \sin\theta\cos\theta(\cos^2\phi+\sin^2\phi) \end{pmatrix}\\ &= \sin\theta(\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta)\\ &=\sin\theta\,r(\theta,\phi). \end{align*} Since $\sin\theta>0$ for $0<\theta<\pi$, the normal $r_\theta\times r_\phi$ points in the same direction as $r(\theta,\phi)$, which is the outward radial direction. Thus $(\theta,\phi)$ is compatible with the outward orientation away from the boundary of the parameter rectangle. The boundary has $\mathcal L^2$-measure zero, so it does not affect the ordinary integral of the pullback coefficient. Now compute the pullback of $\omega_F$. The coordinate functions along $r$ are \begin{align*} x\circ r&=\sin\theta\cos\phi,\\ y\circ r&=\sin\theta\sin\phi,\\ z\circ r&=\cos\theta. \end{align*} Therefore \begin{align*} r^*dx&=d(\sin\theta\cos\phi) =\cos\theta\cos\phi\,d\theta-\sin\theta\sin\phi\,d\phi,\\ r^*dy&=d(\sin\theta\sin\phi) =\cos\theta\sin\phi\,d\theta+\sin\theta\cos\phi\,d\phi,\\ r^*dz&=d(\cos\theta) =-\sin\theta\,d\theta. \end{align*} First, \begin{align*} r^*(dy\wedge dz) &=r^*dy\wedge r^*dz\\ &=(\cos\theta\sin\phi\,d\theta+\sin\theta\cos\phi\,d\phi)\wedge(-\sin\theta\,d\theta)\\ &=-(\cos\theta\sin\phi)(\sin\theta)\,d\theta\wedge d\theta -(\sin\theta\cos\phi)(\sin\theta)\,d\phi\wedge d\theta\\ &=0-\sin^2\theta\cos\phi\,d\phi\wedge d\theta\\ &=\sin^2\theta\cos\phi\,d\theta\wedge d\phi. \end{align*} Second, \begin{align*} r^*(dz\wedge dx) &=r^*dz\wedge r^*dx\\ &=(-\sin\theta\,d\theta)\wedge(\cos\theta\cos\phi\,d\theta-\sin\theta\sin\phi\,d\phi)\\ &=-(\sin\theta)(\cos\theta\cos\phi)\,d\theta\wedge d\theta +(\sin\theta)(\sin\theta\sin\phi)\,d\theta\wedge d\phi\\ &=0+\sin^2\theta\sin\phi\,d\theta\wedge d\phi. \end{align*} Third, \begin{align*} r^*(dx\wedge dy) &=r^*dx\wedge r^*dy\\ &=(\cos\theta\cos\phi\,d\theta-\sin\theta\sin\phi\,d\phi) \wedge (\cos\theta\sin\phi\,d\theta+\sin\theta\cos\phi\,d\phi)\\ &=(\cos\theta\cos\phi)(\sin\theta\cos\phi)\,d\theta\wedge d\phi\\ &\quad -(\sin\theta\sin\phi)(\cos\theta\sin\phi)\,d\phi\wedge d\theta\\ &=\sin\theta\cos\theta\cos^2\phi\,d\theta\wedge d\phi +\sin\theta\cos\theta\sin^2\phi\,d\theta\wedge d\phi\\ &=\sin\theta\cos\theta(\cos^2\phi+\sin^2\phi)\,d\theta\wedge d\phi\\ &=\sin\theta\cos\theta\,d\theta\wedge d\phi. \end{align*} Hence \begin{align*} r^*\omega_F &=(\sin\theta\cos\phi)(\sin^2\theta\cos\phi\,d\theta\wedge d\phi)\\ &\quad +(\sin\theta\sin\phi)(\sin^2\theta\sin\phi\,d\theta\wedge d\phi)\\ &\quad +(\cos\theta)(\sin\theta\cos\theta\,d\theta\wedge d\phi)\\ &=(\sin^3\theta\cos^2\phi+\sin^3\theta\sin^2\phi+\sin\theta\cos^2\theta)\,d\theta\wedge d\phi\\ &=(\sin^3\theta(\cos^2\phi+\sin^2\phi)+\sin\theta\cos^2\theta)\,d\theta\wedge d\phi\\ &=(\sin^3\theta+\sin\theta\cos^2\theta)\,d\theta\wedge d\phi\\ &=\sin\theta(\sin^2\theta+\cos^2\theta)\,d\theta\wedge d\phi\\ &=\sin\theta\,d\theta\wedge d\phi. \end{align*} Therefore \begin{align*} \int_{S^2}\omega_F &=\int_0^{2\pi}\int_0^\pi \sin\theta\,d\theta\,d\phi\\ &=\int_0^{2\pi}\left(-\cos\theta\bigg|_{\theta=0}^{\theta=\pi}\right)d\phi\\ &=\int_0^{2\pi}\bigl(-\cos\pi+\cos 0\bigr)d\phi\\ &=\int_0^{2\pi}(1+1)\,d\phi\\ &=\int_0^{2\pi}2\,d\phi\\ &=2\phi\bigg|_{\phi=0}^{\phi=2\pi}\\ &=4\pi. \end{align*} The outward-oriented flux form for $F(x,y,z)=(x,y,z)$ pulls back under spherical coordinates to $\sin\theta\,d\theta\wedge d\phi$. Its integral over the parameter rectangle is $4\pi$, so the outward flux through the unit sphere is $4\pi$. [/example] Flux forms also recover unsigned surface area once a unit normal has been chosen to convert area into an oriented $2$-form. This is the bridge between the metric notion of area and the orientation-sensitive integration of forms. [example: Surface Area Form from a Parametrisation] Let $r:D\to\mathbb R^3$ be an orientation-compatible parametrisation of a smooth oriented surface $S$, where $D\subset\mathbb R^2$ has coordinates $(u,v)$. Write \begin{align*} r(u,v)=(x(u,v),y(u,v),z(u,v)). \end{align*} Assume the chosen unit normal field along $S$ is $n$. Since $r$ is orientation-compatible, the vector $r_u\times r_v$ points in the same direction as $n$ wherever $r$ is regular. Thus \begin{align*} n(r(u,v))=\frac{r_u(u,v)\times r_v(u,v)}{|r_u(u,v)\times r_v(u,v)|}. \end{align*} The flux form associated to $n=(n_1,n_2,n_3)$ is \begin{align*} \omega_n=n_1\,dy\wedge dz+n_2\,dz\wedge dx+n_3\,dx\wedge dy. \end{align*} Integrating the flux form of the chosen unit normal over $S$ gives the surface area of $S$: \begin{align*} \int_S\omega_n=\int_D |r_u\times r_v|\,d\mathcal L^2. \end{align*} First compute the coordinate pullbacks: \begin{align*} r^*dx&=x_u\,du+x_v\,dv,\\ r^*dy&=y_u\,du+y_v\,dv,\\ r^*dz&=z_u\,du+z_v\,dv. \end{align*} Using bilinearity and antisymmetry of the wedge product, \begin{align*} r^*(dy\wedge dz) &=r^*dy\wedge r^*dz\\ &=(y_u\,du+y_v\,dv)\wedge(z_u\,du+z_v\,dv)\\ &=y_uz_u\,du\wedge du+y_uz_v\,du\wedge dv+y_vz_u\,dv\wedge du+y_vz_v\,dv\wedge dv\\ &=0+y_uz_v\,du\wedge dv-y_vz_u\,du\wedge dv+0\\ &=(y_uz_v-y_vz_u)\,du\wedge dv. \end{align*} Similarly, \begin{align*} r^*(dz\wedge dx) &=r^*dz\wedge r^*dx\\ &=(z_u\,du+z_v\,dv)\wedge(x_u\,du+x_v\,dv)\\ &=z_ux_u\,du\wedge du+z_ux_v\,du\wedge dv+z_vx_u\,dv\wedge du+z_vx_v\,dv\wedge dv\\ &=(z_ux_v-z_vx_u)\,du\wedge dv, \end{align*} and \begin{align*} r^*(dx\wedge dy) &=r^*dx\wedge r^*dy\\ &=(x_u\,du+x_v\,dv)\wedge(y_u\,du+y_v\,dv)\\ &=x_uy_u\,du\wedge du+x_uy_v\,du\wedge dv+x_vy_u\,dv\wedge du+x_vy_v\,dv\wedge dv\\ &=(x_uy_v-x_vy_u)\,du\wedge dv. \end{align*} Therefore \begin{align*} r^*\omega_n &=n_1(r)\,r^*(dy\wedge dz)+n_2(r)\,r^*(dz\wedge dx)+n_3(r)\,r^*(dx\wedge dy)\\ &=\Bigl(n_1(r)(y_uz_v-y_vz_u)+n_2(r)(z_ux_v-z_vx_u)+n_3(r)(x_uy_v-x_vy_u)\Bigr)\,du\wedge dv. \end{align*} Now \begin{align*} r_u&=(x_u,y_u,z_u),\\ r_v&=(x_v,y_v,z_v), \end{align*} so \begin{align*} r_u\times r_v &=(y_uz_v-z_uy_v,\ z_ux_v-x_uz_v,\ x_uy_v-y_ux_v)\\ &=(y_uz_v-y_vz_u,\ z_ux_v-z_vx_u,\ x_uy_v-x_vy_u). \end{align*} Hence the coefficient of $du\wedge dv$ in $r^*\omega_n$ is \begin{align*} n(r)\cdot(r_u\times r_v). \end{align*} Since $n(r)$ is the unit normal in the same direction as $r_u\times r_v$, \begin{align*} n(r)\cdot(r_u\times r_v) &=\frac{r_u\times r_v}{|r_u\times r_v|}\cdot(r_u\times r_v)\\ &=\frac{(r_u\times r_v)\cdot(r_u\times r_v)}{|r_u\times r_v|}\\ &=\frac{|r_u\times r_v|^2}{|r_u\times r_v|}\\ &=|r_u\times r_v|. \end{align*} Thus \begin{align*} r^*\omega_n=|r_u\times r_v|\,du\wedge dv. \end{align*} By the definition of integration over an oriented parametrised surface, \begin{align*} \int_S\omega_n &=\int_D r^*\omega_n\\ &=\int_D |r_u\times r_v|\,d\mathcal L^2. \end{align*} The flux form of the chosen unit normal turns the oriented surface integral into the usual scalar surface-area integral. The orientation condition is exactly what makes the coefficient positive: $n\cdot(r_u\times r_v)=|r_u\times r_v|$, not its negative. [/example] The point of these examples is not that vector calculus disappears, but that its sign conventions become part of the orientation of the domain. In the next chapter, exterior differentiation and boundary orientation combine in Stokes theorem, which turns the identity between an integral over a boundary and an integral over the interior into a single formula for all degrees. With orientations and integration in place, we can now state the theorem unifying all of vector calculus. The generalized Stokes theorem relates the integral of an [exterior derivative](/theorems/1525) over an oriented manifold to an integral of the original form over its boundary. # 7. The Generalised Stokes Theorem The previous chapters built the machinery needed to integrate differential forms: exterior products, the [exterior derivative](/theorems/1525), pullbacks, orientations, and integration over oriented manifolds. This chapter brings those constructions together in a single theorem. The generalised Stokes theorem says that integrating an [exterior derivative](/theorems/1525) over a manifold is the same as integrating the original form over the oriented boundary. This is the point at which differential forms repay the setup. The [fundamental theorem of calculus](/theorems/632), the theorem of Green, the classical Stokes theorem, and the [divergence theorem](/theorems/2754) become the same statement in different dimensions, with different choices of form. ## Why Compact Support and Boundary Orientation Are Needed What hypotheses make the two integrals in Stokes formula finite and give them compatible signs? The compact-support condition prevents contributions from escaping to infinity on a non-compact manifold. On a compact manifold it is automatic, but on an open manifold it is the condition that makes [integration by parts](/theorems/2098) have only the geometric boundary term. [definition: Compactly Supported Differential Form] Let $M$ be a smooth manifold. A differential form $\omega \in \Omega^k(M)$ is compactly supported if \begin{align*} \operatorname{supp}\omega := \overline{\{p \in M : \omega_p \ne 0\}} \end{align*} is compact in $M$. The subspace of compactly supported $k$-forms is \begin{align*} \Omega_c^k(M) := \{\omega \in \Omega^k(M) : \operatorname{supp}\omega \text{ is compact in } M\}\subset \Omega^k(M). \end{align*} [/definition] The sign on the boundary integral is not an extra convention attached to the theorem; it is forced by the orientation convention for manifolds with boundary. Before Stokes' theorem can even be stated, the boundary must inherit an orientation from the ambient manifold in a way that records which side is outward. The half-space model shows why the outward normal must be built into the sign rule. If the boundary orientation is chosen incorrectly, the one-dimensional [fundamental theorem of calculus](/theorems/632) appears with the wrong sign when Stokes' theorem is specialized to an interval. The convention below is the one that makes the local model and the global theorem agree. [definition: Boundary Orientation] Let $M$ be an oriented smooth $n$-manifold with boundary. The boundary orientation on $\partial M$ is the orientation for which a basis $(v_1,\dots,v_{n-1})$ of $T_p\partial M$ is positive precisely when $(\nu,v_1,\dots,v_{n-1})$ is a positive basis of $T_pM$, where $\nu \in T_pM$ is any outward-pointing vector transverse to $T_p\partial M$. [/definition] For the model half-space $H^n=\{x\in \mathbb R^n:x_n\ge 0\}$ with its standard orientation, the outward normal is $-\partial_{x_n}$ along $\partial H^n$. Hence the induced boundary orientation is represented by $(-1)^n dx_1\wedge\cdots\wedge dx_{n-1}$. [illustration:forms-halfspace-boundary-orientation] ## The Generalised Stokes Theorem How can the integral of a derivative over all of $M$ be detected only from values on $\partial M$? The interior contributions should cancel by an integration-by-parts principle, but a manifold has many coordinate patches and the signs on their artificial boundaries must cancel consistently. The only boundary that should remain is the genuine boundary $\partial M$, with the orientation fixed above. [quotetheorem:3555] The theorem is local in nature: away from the boundary, compact support makes all coordinate-direction boundary terms cancel; near the boundary, the outward normal direction is the only source of a remaining term. The theorem should be read as an integration-by-parts formula without coordinates. The [exterior derivative](/theorems/1525) supplies the differentiated quantity, the orientation supplies the sign, and the pullback $\iota^*\omega$ is the operation that restricts the form to the boundary. If the orientation of $M$ is reversed, both sides of the formula change sign because the boundary orientation is induced from the orientation of $M$. If the orientation of $M$ is fixed but $\partial M$ is given the opposite orientation, only the boundary integral changes sign, so the stated equality no longer has the correct sign. The compact-support hypothesis is also part of the theorem rather than a technical decoration. It removes boundary terms at infinity on non-compact manifolds. [example: Failure Without Compact Support On The Line] Let $M=\mathbb R$ with its standard orientation, and let $f(x)=\arctan x$, regarded as a smooth $0$-form on $\mathbb R$. The manifold $\mathbb R$ has no boundary, so $\partial\mathbb R=\varnothing$. The Stokes formula cannot be extended from compactly supported $0$-forms to all smooth $0$-forms on $\mathbb R$ without adding a term at infinity. Since $f'(x)=\frac{1}{1+x^2}$, the [exterior derivative](/theorems/1525) of the $0$-form $f$ is \begin{align*} df = f'(x)\,dx = \frac{1}{1+x^2}\,dx. \end{align*} The integral of $df$ over $\mathbb R$ is the improper integral \begin{align*} \int_{\mathbb R} df &= \int_{-\infty}^{\infty}\frac{1}{1+x^2}\,dx \\ &= \lim_{A\to-\infty,\ B\to\infty}\int_A^B \frac{1}{1+x^2}\,dx \\ &= \lim_{A\to-\infty,\ B\to\infty}\bigl(\arctan B-\arctan A\bigr) \\ &= \frac{\pi}{2}-\left(-\frac{\pi}{2}\right) \\ &= \pi. \end{align*} On the other hand, because $\partial\mathbb R=\varnothing$, integration over the boundary gives \begin{align*} \int_{\partial\mathbb R} f = \int_{\varnothing} f = 0. \end{align*} Thus the two sides one would expect from Stokes' formula are \begin{align*} \int_{\mathbb R} df=\pi, \qquad \int_{\partial\mathbb R} f=0. \end{align*} The missing quantity is \begin{align*} \lim_{x\to\infty} f(x)-\lim_{x\to-\infty} f(x) = \frac{\pi}{2}-\left(-\frac{\pi}{2}\right) = \pi. \end{align*} This is a boundary contribution at infinity. Compact support rules out this contribution because a compactly supported $0$-form on $\mathbb R$ has the same limiting value $0$ at both ends. [/example] This is the analytic reason compact support appears in the global statement. ## The Half-Space Calculation Which coordinate computation is hidden inside the proof of the theorem? The local model is $H^n=\{x\in\mathbb R^n:x_n\ge 0\}$ with standard coordinates. For a compactly supported $(n-1)$-form on $H^n$, write \begin{align*} \omega = \sum_{i=1}^n f_i\, dx_1\wedge\cdots\wedge dx_{i-1}\wedge dx_{i+1}\wedge\cdots\wedge dx_n. \end{align*} Then \begin{align*} d\omega = \sum_{i=1}^n (-1)^{i-1}\frac{\partial f_i}{\partial x_i}\, dx_1\wedge\cdots\wedge dx_n. \end{align*} For $i<n$, integration in the $x_i$ direction gives zero because $f_i$ has compact support. The normal term gives \begin{align*} \int_{H^n} (-1)^{n-1}\frac{\partial f_n}{\partial x_n}\,d\mathcal L^n = (-1)^n\int_{\mathbb R^{n-1}} f_n(x_1,\dots,x_{n-1},0)\,d\mathcal L^{n-1}. \end{align*} This is precisely the integral of $\omega$ over $\partial H^n$ with the boundary orientation described above. This computation is the [fundamental theorem of calculus](/theorems/632) applied in the coordinate normal to the boundary. [example: A Compactly Supported Form On The Half-Plane] Let $H^2=\{(x,y)\in\mathbb R^2:y\ge 0\}$ with standard orientation $dx\wedge dy$. Let \begin{align*} \omega=f(x,y)\,dx, \end{align*} where $f\in C_c^\infty(H^2)$. Choose $R>0$ such that \begin{align*} \operatorname{supp}f\subset (-R,R)\times [0,R). \end{align*} Then $f(x,y)=0$ whenever $|x|\ge R$ or $y\ge R$. Let $\iota:\partial H^2\hookrightarrow H^2$ be the inclusion. For this compactly supported $1$-form, \begin{align*} \int_{H^2} d\omega = \int_{\partial H^2}\iota^*\omega = \int_{\mathbb R} f(x,0)\,dx. \end{align*} First compute the exterior derivative. Since \begin{align*} df = \frac{\partial f}{\partial x}\,dx+\frac{\partial f}{\partial y}\,dy, \end{align*} and $d(dx)=0$, we have \begin{align*} d\omega &= d(f\,dx) \\ &= df\wedge dx+f\,d(dx) \\ &= \left(\frac{\partial f}{\partial x}\,dx+\frac{\partial f}{\partial y}\,dy\right)\wedge dx \\ &= \frac{\partial f}{\partial x}\,dx\wedge dx+\frac{\partial f}{\partial y}\,dy\wedge dx \\ &= 0-\frac{\partial f}{\partial y}\,dx\wedge dy \\ &= -\frac{\partial f}{\partial y}\,dx\wedge dy. \end{align*} The term $dx\wedge dx$ is zero by alternatingness, and $dy\wedge dx=-dx\wedge dy$ by antisymmetry of the wedge product. Because $d\omega$ is compactly supported, *[Fubini's Theorem](/theorems/2961)* applies. Using the positive area form $dx\wedge dy$, we get \begin{align*} \int_{H^2}d\omega &= \int_{-\infty}^{\infty}\int_0^\infty -\frac{\partial f}{\partial y}(x,y)\,dy\,dx \\ &= \int_{-R}^{R}\int_0^R -\frac{\partial f}{\partial y}(x,y)\,dy\,dx. \end{align*} For each fixed $x\in[-R,R]$, the one-variable *[Fundamental Theorem of Calculus](/theorems/632)* gives \begin{align*} \int_0^R -\frac{\partial f}{\partial y}(x,y)\,dy &= -\left(f(x,R)-f(x,0)\right) \\ &= f(x,0)-f(x,R). \end{align*} Since $f(x,R)=0$, this becomes \begin{align*} \int_0^R -\frac{\partial f}{\partial y}(x,y)\,dy = f(x,0). \end{align*} Therefore \begin{align*} \int_{H^2}d\omega &= \int_{-R}^{R} f(x,0)\,dx \\ &= \int_{\mathbb R} f(x,0)\,dx, \end{align*} because $f(x,0)=0$ for $|x|\ge R$. Now compute the boundary integral. Along $\partial H^2=\{(x,0):x\in\mathbb R\}$, the outward normal is $-\partial_y$. The tangent vector $\partial_x$ is positively oriented on the boundary because the ordered basis $(-\partial_y,\partial_x)$ has coordinate matrix \begin{align*} \begin{pmatrix} 0 & 1\\ -1 & 0 \end{pmatrix} \end{align*} relative to $(\partial_x,\partial_y)$, and \begin{align*} \det \begin{pmatrix} 0 & 1\\ -1 & 0 \end{pmatrix} =1. \end{align*} Thus the parametrization $\tau:\mathbb R\to\partial H^2$ given by $\tau(t)=(t,0)$ is orientation-preserving. Pulling back $\omega$ gives \begin{align*} \tau^*(\iota^*\omega) &= \tau^*(f(x,y)\,dx) \\ &= f(t,0)\,dt. \end{align*} Hence \begin{align*} \int_{\partial H^2}\iota^*\omega &= \int_{\mathbb R} f(t,0)\,dt \\ &= \int_{\mathbb R} f(x,0)\,dx. \end{align*} The normal derivative term in $d\omega$ contributes exactly the value of $f$ on the boundary line $y=0$. The outward-normal-first boundary orientation makes $\partial_x$ positive on $\partial H^2$, so the boundary integral has the same sign as the interior calculation: \begin{align*} \int_{H^2}d\omega = \int_{\partial H^2}\iota^*\omega = \int_{\mathbb R} f(x,0)\,dx. \end{align*} [/example] The example shows why the outward-normal-first convention is the right convention for this course. It makes the boundary contribution agree with the sign produced by integration in the inward coordinate direction. ## Boundaries of Boundaries What remains if the form being integrated is already an [exterior derivative](/theorems/1525)? The algebraic identity $d^2=0$ becomes a geometric statement after applying Stokes. It says that the oriented boundary of an oriented boundary contributes nothing to integration. [quotetheorem:3582] [citeproof:3582] This corollary is often the first hint of cohomology, but it still depends on the same support and integrability hypotheses as Stokes theorem. Without compact support, an exact form on a non-compact boundary can retain a contribution from infinity: for $M=H^2$ and $\eta(x,y)=\arctan x$, the restriction to $\partial H^2\cong\mathbb R$ satisfies \begin{align*} \int_{\partial H^2} d(\iota^*\eta)=\int_{\mathbb R}\frac{1}{1+x^2}\,dx=\pi. \end{align*} The result also does not say that every closed form is exact. It says that exact forms integrate to zero over oriented boundaries under the stated hypotheses; the converse question is local in the Poincare lemma and global in de Rham cohomology. ## Recovering The Classical Integral Theorems What familiar formulas appear when the dimension and the form are chosen in the standard Euclidean ways? Each classical theorem is obtained by choosing $M$ and $\omega$ so that $d\omega$ is the usual integrand in the interior and $\omega$ is the usual line, surface, or endpoint integrand on the boundary. [example: Fundamental Theorem Of Calculus] Let $M=[a,b]$ with its standard orientation, represented by the positive $1$-form $dx$. Let $f\in C^\infty([a,b])$ be regarded as a smooth $0$-form on $M$. The generalised Stokes theorem on the oriented interval $[a,b]$ gives \begin{align*} \int_a^b \frac{df}{dx}\,dx=f(b)-f(a). \end{align*} Since $f$ is a $0$-form, its [exterior derivative](/theorems/1525) is the $1$-form \begin{align*} df=\frac{df}{dx}\,dx. \end{align*} Therefore integration over $M=[a,b]$ with orientation $dx$ gives \begin{align*} \int_M df &=\int_{[a,b]} \frac{df}{dx}\,dx \\ &=\int_a^b \frac{df}{dx}\,dx. \end{align*} The boundary of the interval is \begin{align*} \partial [a,b]=\{a\}\cup\{b\}. \end{align*} At $a$, the outward-pointing vector is $-\partial_x$, so the boundary point $a$ has sign $-1$. At $b$, the outward-pointing vector is $\partial_x$, so the boundary point $b$ has sign $+1$. Hence the oriented boundary is \begin{align*} \partial[a,b]=\{b\}-\{a\}. \end{align*} The pullback of the $0$-form $f$ to a boundary point is just its value at that point, so \begin{align*} \int_{\partial M} f &=\int_{\{b\}-\{a\}} f \\ &=\int_{\{b\}} f-\int_{\{a\}} f \\ &=f(b)-f(a). \end{align*} Applying *Generalised Stokes Theorem* to the $0$-form $f$ gives \begin{align*} \int_M df=\int_{\partial M} f. \end{align*} Substituting the two computations above yields \begin{align*} \int_a^b \frac{df}{dx}\,dx=f(b)-f(a). \end{align*} In dimension $1$, the generalised Stokes theorem says that integrating the derivative of a function over an oriented interval records the value at the positively oriented endpoint minus the value at the negatively oriented endpoint. Thus Stokes theorem recovers the [fundamental theorem of calculus](/theorems/632): \begin{align*} \int_a^b f'(x)\,dx=f(b)-f(a). \end{align*} [/example] This is the one-dimensional case from which the theorem gets its sign convention. The higher-dimensional statements are the same principle applied to forms of higher degree. [example: Planar Green Formula From Differential Forms] Let $D\subset\mathbb R^2$ be a compact oriented region with smooth boundary, oriented by the standard area form $dx\wedge dy$. Let $P,Q\in C^\infty(D)$ and define the $1$-form \begin{align*} \omega=P\,dx+Q\,dy. \end{align*} Since $D$ is compact, every smooth form on $D$ has compact support in $D$, so $\omega$ satisfies the support hypothesis in [Stokes' theorem](/theorems/1530). The differential-form version of Stokes theorem gives Green's theorem in circulation form: \begin{align*} \int_{\partial D} P\,dx+Q\,dy =\int_D \left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)\,d\mathcal L^2. \end{align*} First compute $d\omega$. By linearity of the [exterior derivative](/theorems/1525), \begin{align*} d\omega &=d(P\,dx+Q\,dy)\\ &=d(P\,dx)+d(Q\,dy). \end{align*} For a smooth function $P$, \begin{align*} dP=\frac{\partial P}{\partial x}\,dx+\frac{\partial P}{\partial y}\,dy. \end{align*} Using the graded Leibniz rule and $d(dx)=0$, \begin{align*} d(P\,dx) &=dP\wedge dx+P\,d(dx)\\ &=\left(\frac{\partial P}{\partial x}\,dx+\frac{\partial P}{\partial y}\,dy\right)\wedge dx\\ &=\frac{\partial P}{\partial x}\,dx\wedge dx+\frac{\partial P}{\partial y}\,dy\wedge dx\\ &=0+\frac{\partial P}{\partial y}(-dx\wedge dy)\\ &=-\frac{\partial P}{\partial y}\,dx\wedge dy. \end{align*} Here $dx\wedge dx=0$ by alternatingness, and $dy\wedge dx=-dx\wedge dy$ by antisymmetry. Similarly, \begin{align*} dQ=\frac{\partial Q}{\partial x}\,dx+\frac{\partial Q}{\partial y}\,dy, \end{align*} so \begin{align*} d(Q\,dy) &=dQ\wedge dy+Q\,d(dy)\\ &=\left(\frac{\partial Q}{\partial x}\,dx+\frac{\partial Q}{\partial y}\,dy\right)\wedge dy\\ &=\frac{\partial Q}{\partial x}\,dx\wedge dy+\frac{\partial Q}{\partial y}\,dy\wedge dy\\ &=\frac{\partial Q}{\partial x}\,dx\wedge dy+0\\ &=\frac{\partial Q}{\partial x}\,dx\wedge dy. \end{align*} Combining the two terms gives \begin{align*} d\omega &=-\frac{\partial P}{\partial y}\,dx\wedge dy+\frac{\partial Q}{\partial x}\,dx\wedge dy\\ &=\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dx\wedge dy. \end{align*} Because $D$ is oriented by $dx\wedge dy$, integration of the top-degree form above is \begin{align*} \int_D d\omega &=\int_D \left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dx\wedge dy\\ &=\int_D \left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)\,d\mathcal L^2. \end{align*} Now apply [Stokes' Theorem](/theorems/1530) to the compact oriented surface-with-boundary $D$ and the $1$-form $\omega$: \begin{align*} \int_D d\omega=\int_{\partial D}\iota^*\omega, \end{align*} where $\iota:\partial D\hookrightarrow D$ is the inclusion. By linearity of pullback, \begin{align*} \iota^*\omega &=\iota^*(P\,dx+Q\,dy)\\ &=\iota^*(P\,dx)+\iota^*(Q\,dy). \end{align*} Thus the boundary integral is the usual line integral notation \begin{align*} \int_{\partial D}\iota^*\omega =\int_{\partial D}P\,dx+Q\,dy. \end{align*} Substituting the computed value of $\int_D d\omega$ into Stokes' formula gives \begin{align*} \int_{\partial D}P\,dx+Q\,dy =\int_D \left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)\,d\mathcal L^2. \end{align*} It remains to identify the boundary orientation. Let $\gamma(t)=(x(t),y(t))$ be a positively oriented parametrization of a boundary component, and let $T=\gamma'(t)=(x'(t),y'(t))$. If $D$ lies locally to the left of $\gamma$, then the outward normal direction is represented by \begin{align*} \nu_{\mathrm{out}}=\frac{1}{\sqrt{(x'(t))^2+(y'(t))^2}}\bigl(y'(t),-x'(t)\bigr). \end{align*} The coordinate matrix of $(\nu_{\mathrm{out}},T)$ relative to $(\partial_x,\partial_y)$ is \begin{align*} \begin{pmatrix} \dfrac{y'(t)}{\sqrt{(x'(t))^2+(y'(t))^2}} & x'(t)\\ -\dfrac{x'(t)}{\sqrt{(x'(t))^2+(y'(t))^2}} & y'(t) \end{pmatrix}. \end{align*} Its determinant is \begin{align*} \frac{(y'(t))^2}{\sqrt{(x'(t))^2+(y'(t))^2}} +\frac{(x'(t))^2}{\sqrt{(x'(t))^2+(y'(t))^2}} =\sqrt{(x'(t))^2+(y'(t))^2}>0. \end{align*} Therefore $(\nu_{\mathrm{out}},T)$ is positively oriented in $\mathbb R^2$, so the boundary-orientation convention agrees with the usual positive planar orientation: the boundary is traversed with $D$ on the left. For the outer boundary component this is counterclockwise; for inner boundary components it is clockwise. For the $1$-form $\omega=P\,dx+Q\,dy$, the [exterior derivative](/theorems/1525) records the scalar curl coefficient: \begin{align*} d\omega =\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dx\wedge dy. \end{align*} [Stokes' theorem](/theorems/1530) turns the integral of this area-density form over $D$ into the circulation of $P\,dx+Q\,dy$ around the induced oriented boundary: \begin{align*} \int_{\partial D} P\,dx+Q\,dy =\int_D \left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)\,d\mathcal L^2. \end{align*} This is exactly Green's theorem in circulation form. [/example] In this example the two-form $d\omega$ is the signed area-density form whose coefficient is the scalar curl of the planar vector field $(P,Q)$. [example: Classical Stokes Theorem For Curl] Let $\Sigma\subset\mathbb R^3$ be an oriented compact smooth surface with smooth boundary. Let $F=(P,Q,R)$ be a smooth vector field on an open neighbourhood of $\Sigma$, and define the ambient $1$-form \begin{align*} \omega=P\,dx+Q\,dy+R\,dz. \end{align*} Use the same symbol $\omega$ for its restriction to $\Sigma$. Since $\Sigma$ is compact, this restricted $1$-form has compact support on $\Sigma$. Let $\nu$ be the unit normal field compatible with the orientation of $\Sigma$, and give $\partial\Sigma$ the induced boundary orientation. The generalised Stokes theorem gives the classical curl formula \begin{align*} \int_{\partial\Sigma} F\cdot dr =\int_\Sigma (\operatorname{curl}F)\cdot \nu\,dS. \end{align*} Write subscripts for partial derivatives, for example $P_y=\frac{\partial P}{\partial y}$. By linearity of $d$, \begin{align*} d\omega &=d(P\,dx+Q\,dy+R\,dz)\\ &=d(P\,dx)+d(Q\,dy)+d(R\,dz). \end{align*} For the first term, \begin{align*} d(P\,dx) &=dP\wedge dx+P\,d(dx)\\ &=(P_x\,dx+P_y\,dy+P_z\,dz)\wedge dx\\ &=P_x\,dx\wedge dx+P_y\,dy\wedge dx+P_z\,dz\wedge dx\\ &=0-P_y\,dx\wedge dy+P_z\,dz\wedge dx. \end{align*} Here $dx\wedge dx=0$ by alternatingness, and $dy\wedge dx=-dx\wedge dy$ by antisymmetry. For the second term, \begin{align*} d(Q\,dy) &=dQ\wedge dy+Q\,d(dy)\\ &=(Q_x\,dx+Q_y\,dy+Q_z\,dz)\wedge dy\\ &=Q_x\,dx\wedge dy+Q_y\,dy\wedge dy+Q_z\,dz\wedge dy\\ &=Q_x\,dx\wedge dy+0-Q_z\,dy\wedge dz. \end{align*} For the third term, \begin{align*} d(R\,dz) &=dR\wedge dz+R\,d(dz)\\ &=(R_x\,dx+R_y\,dy+R_z\,dz)\wedge dz\\ &=R_x\,dx\wedge dz+R_y\,dy\wedge dz+R_z\,dz\wedge dz\\ &=-R_x\,dz\wedge dx+R_y\,dy\wedge dz+0. \end{align*} Adding the three expressions gives \begin{align*} d\omega &=(-P_y\,dx\wedge dy+P_z\,dz\wedge dx) +(Q_x\,dx\wedge dy-Q_z\,dy\wedge dz)\\ &\qquad+(-R_x\,dz\wedge dx+R_y\,dy\wedge dz)\\ &=(R_y-Q_z)\,dy\wedge dz+(P_z-R_x)\,dz\wedge dx+(Q_x-P_y)\,dx\wedge dy. \end{align*} Since \begin{align*} \operatorname{curl}F =\left(R_y-Q_z,\ P_z-R_x,\ Q_x-P_y\right), \end{align*} the form $d\omega$ is the flux $2$-form associated to $\operatorname{curl}F$. To see this identification in coordinates, let $\phi(u,v)=(x(u,v),y(u,v),z(u,v))$ be a positively oriented parametrization of a piece of $\Sigma$. Put \begin{align*} N=\phi_u\times \phi_v. \end{align*} Because $\phi$ is positively oriented, $N=\|N\|\nu$. The pullbacks of the basic $2$-forms are \begin{align*} \phi^*(dy\wedge dz) &=(y_u\,du+y_v\,dv)\wedge(z_u\,du+z_v\,dv)\\ &=(y_u z_v-y_v z_u)\,du\wedge dv,\\ \phi^*(dz\wedge dx) &=(z_u\,du+z_v\,dv)\wedge(x_u\,du+x_v\,dv)\\ &=(z_u x_v-z_v x_u)\,du\wedge dv,\\ \phi^*(dx\wedge dy) &=(x_u\,du+x_v\,dv)\wedge(y_u\,du+y_v\,dv)\\ &=(x_u y_v-x_v y_u)\,du\wedge dv. \end{align*} Thus \begin{align*} \phi^*(d\omega) &=\Bigl((R_y-Q_z)(y_u z_v-y_v z_u)\\ &\qquad +(P_z-R_x)(z_u x_v-z_v x_u)\\ &\qquad +(Q_x-P_y)(x_u y_v-x_v y_u)\Bigr)\,du\wedge dv\\ &=\bigl((\operatorname{curl}F)\circ\phi\bigr)\cdot(\phi_u\times\phi_v)\,du\wedge dv\\ &=\bigl((\operatorname{curl}F)\circ\phi\bigr)\cdot\nu\,\|N\|\,du\wedge dv. \end{align*} Since $dS=\|N\|\,du\,dv$, integration over $\Sigma$ gives \begin{align*} \int_\Sigma d\omega =\int_\Sigma (\operatorname{curl}F)\cdot\nu\,dS. \end{align*} Now compute the boundary integral. Let $\gamma(t)=(x(t),y(t),z(t))$ be an orientation-preserving parametrization of a boundary component. Then \begin{align*} \gamma^*dx=x'(t)\,dt,\qquad \gamma^*dy=y'(t)\,dt,\qquad \gamma^*dz=z'(t)\,dt. \end{align*} Therefore \begin{align*} \gamma^*\omega &=P(\gamma(t))x'(t)\,dt+Q(\gamma(t))y'(t)\,dt+R(\gamma(t))z'(t)\,dt\\ &=F(\gamma(t))\cdot\gamma'(t)\,dt. \end{align*} Summing over the oriented boundary components gives \begin{align*} \int_{\partial\Sigma}\omega =\int_{\partial\Sigma}F\cdot dr. \end{align*} Apply [Stokes' Theorem](/theorems/1530) to the compact oriented surface-with-boundary $\Sigma$ and the compactly supported $1$-form $\omega$: \begin{align*} \int_\Sigma d\omega=\int_{\partial\Sigma}\omega. \end{align*} Substituting the two identifications above yields \begin{align*} \int_\Sigma (\operatorname{curl}F)\cdot\nu\,dS =\int_{\partial\Sigma}F\cdot dr. \end{align*} For the $1$-form $\omega=P\,dx+Q\,dy+R\,dz$, the [exterior derivative](/theorems/1525) is the flux form of $\operatorname{curl}F$: \begin{align*} d\omega =(R_y-Q_z)\,dy\wedge dz+(P_z-R_x)\,dz\wedge dx+(Q_x-P_y)\,dx\wedge dy. \end{align*} The boundary integral of $\omega$ is the line integral of $F$ around the induced oriented boundary. Hence the generalised Stokes theorem is exactly the classical curl theorem: \begin{align*} \int_{\partial\Sigma} F\cdot dr =\int_\Sigma (\operatorname{curl}F)\cdot \nu\,dS. \end{align*} [/example] This is the version traditionally called Stokes theorem in vector calculus. The differential-form statement removes the need to remember a separate curl theorem. [example: Divergence Theorem From Differential Forms] Let $V\subset\mathbb R^3$ be a compact oriented solid region with smooth boundary, oriented by the volume form $dx\wedge dy\wedge dz$. Let $F=(P,Q,R)$ be smooth on an open neighbourhood of $V$, and define the $2$-form \begin{align*} \omega=P\,dy\wedge dz+Q\,dz\wedge dx+R\,dx\wedge dy. \end{align*} Since $V$ is compact, $\omega$ has compact support on $V$. Let $\iota:\partial V\hookrightarrow V$ be the inclusion, and let $\nu$ be the outward unit normal along $\partial V$. The generalised Stokes theorem gives the [divergence theorem](/theorems/2754): \begin{align*} \int_{\partial V} F\cdot \nu\,dS =\int_V \left(\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z}\right)\,d\mathcal L^3. \end{align*} Write \begin{align*} P_x=\frac{\partial P}{\partial x},\qquad Q_y=\frac{\partial Q}{\partial y},\qquad R_z=\frac{\partial R}{\partial z}. \end{align*} By linearity of the [exterior derivative](/theorems/1525), \begin{align*} d\omega &=d(P\,dy\wedge dz)+d(Q\,dz\wedge dx)+d(R\,dx\wedge dy). \end{align*} For the first term, the graded Leibniz rule and $d(dy)=d(dz)=0$ give \begin{align*} d(P\,dy\wedge dz) &=dP\wedge dy\wedge dz+P\,d(dy\wedge dz)\\ &=dP\wedge dy\wedge dz\\ &=(P_x\,dx+P_y\,dy+P_z\,dz)\wedge dy\wedge dz\\ &=P_x\,dx\wedge dy\wedge dz +P_y\,dy\wedge dy\wedge dz +P_z\,dz\wedge dy\wedge dz\\ &=P_x\,dx\wedge dy\wedge dz+0+0\\ &=P_x\,dx\wedge dy\wedge dz. \end{align*} The terms with repeated factors vanish by alternatingness of the wedge product. For the second term, \begin{align*} d(Q\,dz\wedge dx) &=dQ\wedge dz\wedge dx+Q\,d(dz\wedge dx)\\ &=dQ\wedge dz\wedge dx\\ &=(Q_x\,dx+Q_y\,dy+Q_z\,dz)\wedge dz\wedge dx\\ &=Q_x\,dx\wedge dz\wedge dx +Q_y\,dy\wedge dz\wedge dx +Q_z\,dz\wedge dz\wedge dx\\ &=0+Q_y\,dy\wedge dz\wedge dx+0. \end{align*} Since \begin{align*} dy\wedge dz\wedge dx &=-dy\wedge dx\wedge dz\\ &=dx\wedge dy\wedge dz, \end{align*} we get \begin{align*} d(Q\,dz\wedge dx) =Q_y\,dx\wedge dy\wedge dz. \end{align*} For the third term, \begin{align*} d(R\,dx\wedge dy) &=dR\wedge dx\wedge dy+R\,d(dx\wedge dy)\\ &=dR\wedge dx\wedge dy\\ &=(R_x\,dx+R_y\,dy+R_z\,dz)\wedge dx\wedge dy\\ &=R_x\,dx\wedge dx\wedge dy +R_y\,dy\wedge dx\wedge dy +R_z\,dz\wedge dx\wedge dy\\ &=0+0+R_z\,dz\wedge dx\wedge dy. \end{align*} Since \begin{align*} dz\wedge dx\wedge dy &=-dx\wedge dz\wedge dy\\ &=dx\wedge dy\wedge dz, \end{align*} we get \begin{align*} d(R\,dx\wedge dy) =R_z\,dx\wedge dy\wedge dz. \end{align*} Adding the three terms gives \begin{align*} d\omega &=P_x\,dx\wedge dy\wedge dz +Q_y\,dx\wedge dy\wedge dz +R_z\,dx\wedge dy\wedge dz\\ &=(P_x+Q_y+R_z)\,dx\wedge dy\wedge dz\\ &=\left(\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z}\right)dx\wedge dy\wedge dz. \end{align*} Because $V$ is oriented by $dx\wedge dy\wedge dz$, integration of this top-degree form is \begin{align*} \int_V d\omega =\int_V \left(\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z}\right)\,d\mathcal L^3. \end{align*} Now identify the boundary integral. Let \begin{align*} \psi(u,v)=(x(u,v),y(u,v),z(u,v)) \end{align*} be an orientation-preserving parametrization of a boundary patch. Put \begin{align*} \psi_u=(x_u,y_u,z_u),\qquad \psi_v=(x_v,y_v,z_v). \end{align*} The boundary orientation means that \begin{align*} (\nu,\psi_u,\psi_v) \end{align*} is a positively oriented ordered basis of $\mathbb R^3$. Hence \begin{align*} \nu\cdot(\psi_u\times\psi_v)>0, \end{align*} so \begin{align*} \psi_u\times\psi_v=\|\psi_u\times\psi_v\|\,\nu. \end{align*} Pull back the three basic $2$-forms: \begin{align*} \psi^*(dy\wedge dz) &=(y_u\,du+y_v\,dv)\wedge(z_u\,du+z_v\,dv)\\ &=y_u z_u\,du\wedge du+y_u z_v\,du\wedge dv +y_v z_u\,dv\wedge du+y_v z_v\,dv\wedge dv\\ &=(y_u z_v-y_v z_u)\,du\wedge dv, \end{align*} \begin{align*} \psi^*(dz\wedge dx) &=(z_u\,du+z_v\,dv)\wedge(x_u\,du+x_v\,dv)\\ &=(z_u x_v-z_v x_u)\,du\wedge dv, \end{align*} and \begin{align*} \psi^*(dx\wedge dy) &=(x_u\,du+x_v\,dv)\wedge(y_u\,du+y_v\,dv)\\ &=(x_u y_v-x_v y_u)\,du\wedge dv. \end{align*} Therefore \begin{align*} \psi^*(\iota^*\omega) &=\Bigl(P(\psi)(y_u z_v-y_v z_u) +Q(\psi)(z_u x_v-z_v x_u) +R(\psi)(x_u y_v-x_v y_u)\Bigr)\,du\wedge dv. \end{align*} But \begin{align*} \psi_u\times\psi_v =\bigl(y_u z_v-y_v z_u,\ z_u x_v-z_v x_u,\ x_u y_v-x_v y_u\bigr), \end{align*} so \begin{align*} \psi^*(\iota^*\omega) &=\bigl(F(\psi)\cdot(\psi_u\times\psi_v)\bigr)\,du\wedge dv\\ &=\bigl(F(\psi)\cdot\nu\bigr)\|\psi_u\times\psi_v\|\,du\wedge dv. \end{align*} Since $dS=\|\psi_u\times\psi_v\|\,du\,dv$, this says on each positively oriented boundary patch that \begin{align*} \iota^*\omega=F\cdot\nu\,dS. \end{align*} Summing over oriented boundary patches gives \begin{align*} \int_{\partial V}\iota^*\omega =\int_{\partial V}F\cdot\nu\,dS. \end{align*} Apply [Stokes' Theorem](/theorems/1530) to the compact oriented $3$-manifold-with-boundary $V$ and the compactly supported $2$-form $\omega$: \begin{align*} \int_V d\omega=\int_{\partial V}\iota^*\omega. \end{align*} Substituting the two computations above yields \begin{align*} \int_V \left(\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z}\right)\,d\mathcal L^3 =\int_{\partial V}F\cdot\nu\,dS. \end{align*} The $2$-form \begin{align*} \omega=P\,dy\wedge dz+Q\,dz\wedge dx+R\,dx\wedge dy \end{align*} is the flux form of the vector field $F$. Its [exterior derivative](/theorems/1525) is the divergence density: \begin{align*} d\omega=(\operatorname{div}F)\,dx\wedge dy\wedge dz. \end{align*} The boundary orientation is exactly the one for which the induced surface element uses the outward unit normal. Thus the generalised Stokes theorem becomes \begin{align*} \int_{\partial V} F\cdot \nu\,dS =\int_V \operatorname{div}F\,d\mathcal L^3, \end{align*} which is the classical [divergence theorem](/theorems/2754). [/example] The [divergence theorem](/theorems/2754) is therefore Stokes theorem for a two-form on a three-dimensional manifold. The boundary orientation is what enforces the outward normal in the usual statement. ## Applying The Theorem Reliably When using Stokes theorem in a calculation, which choices must be fixed before computing? First choose the oriented manifold $M$ and record whether it has boundary. Next choose the $(n-1)$-form $\omega$ whose [exterior derivative](/theorems/1525) is the desired interior integrand. Then check compact support, or compactness of $M$, so that the integrals are defined without an additional term at infinity. The last step is the orientation check. The boundary orientation determines whether the boundary is traversed counterclockwise, clockwise, outward-normal-first, or with another sign convention inherited from $M$. Most sign errors in applications of Stokes come from replacing this induced orientation with an informal geometric guess. This chapter completes the integration theory of differential forms. The next part of the course studies what happens when $d\omega=0$ and asks whether such a closed form must be of the form $d\eta$; the answer is de Rham cohomology. Having developed the full integration machinery, the fundamental question becomes: when can we solve dη = ω for a given closed form ω? The Poincaré lemma answers this locally on contractible domains, but the failure of exactness globally will measure the topology of the space. # 8. The Poincaré Lemma The previous chapters developed differential forms, pullback, exterior differentiation, and [Stokes' theorem](/theorems/1530) as a unified language for integration. This chapter asks when the equation $d\eta=\omega$ can be solved, given the necessary condition $d\omega=0$. The answer is local: on domains that can be contracted to a point in a controlled way, closed forms of positive degree are exact. The same statement fails on spaces with holes, and the failure becomes the first visible link between differential forms and topology. The central tool is a homotopy operator. It turns a geometric contraction into an algebraic identity $dK+Kd=\operatorname{id}$ on positive-degree forms, so the proof is not a search for primitives by guesswork. The angle form on the punctured plane then shows why the hypothesis matters: a closed form can have nonzero integral around a loop, and no exact form can do that. ## Closed Forms and the Local Exactness Problem When does the differential equation $d\eta=\omega$ have a solution for a given differential form $\omega$? Since $d^2=0$, every exact form is closed, so closedness is the first compatibility condition. The Poincare lemma says that, on a star-shaped region of Euclidean space, this compatibility condition is also sufficient in positive degrees. [definition: Closed and Exact Forms] Let $U \subseteq \mathbb R^n$ be open, and let $\omega \in \Omega^k(U)$. The form $\omega$ is closed if $d\omega=0$. For $k \ge 1$, the form $\omega$ is exact if there exists $\eta \in \Omega^{k-1}(U)$ such that $\omega=d\eta$. [/definition] This terminology packages the cohomological question into a quotient: closed forms are potential cohomology classes, and exact forms are the classes counted as zero. Thus \begin{align*} H^k_{\mathrm{dR}}(U) = \frac{\ker(d:\Omega^k(U)\to \Omega^{k+1}(U))}{\operatorname{im}(d:\Omega^{k-1}(U)\to \Omega^k(U))}. \end{align*} The lemma will prove that this quotient vanishes in positive degrees for star-shaped open sets. We now isolate the geometric condition under which the proof will build an actual primitive. The point is not only that the domain has no holes, but that it admits one fixed straight-line contraction usable at every point. Naming this condition lets the Poincare lemma state precisely where the radial homotopy argument applies. [definition: Star-Shaped Open Set] Let $U \subseteq \mathbb R^n$ be open and let $a \in U$. The set $U$ is star-shaped with centre $a$ if, for every $x \in U$ and every $t \in [0,1]$, the point $a+t(x-a)$ lies in $U$. [/definition] Star-shaped domains include balls, convex open sets, and all of $\mathbb R^n$. The hypothesis is stronger than contractibility, but it is exactly the condition needed for the explicit radial homotopy used in the proof. [illustration:forms-radial-contraction] With the radial contraction available, the local cohomology question becomes concrete: can every closed positive-degree form be written as the derivative of a form obtained by integrating along those radial segments? The Poincare lemma answers this by showing that the radial homotopy kills all positive-degree de Rham classes on such domains. [quotetheorem:832] The statement says that positive-degree de Rham cohomology has no local content on such domains. Any obstruction to exactness must therefore come from the global shape of the domain rather than from the differential equation alone. The proof reduces the lemma to the construction and verification of $K$. The next section gives the formula, since the same pattern will later prove homotopy invariance of cohomology. [example: Exact One-Forms on the Plane] Let \begin{align*} \omega=P(x,y)\,dx+Q(x,y)\,dy\in\Omega^1(\mathbb R^2), \end{align*} where $P,Q\in C^\infty(\mathbb R^2)$. Since \begin{align*} d\omega =dP\wedge dx+dQ\wedge dy =(\partial_xP\,dx+\partial_yP\,dy)\wedge dx +(\partial_xQ\,dx+\partial_yQ\,dy)\wedge dy =(\partial_xQ-\partial_yP)\,dx\wedge dy, \end{align*} the assumption $d\omega=0$ is exactly \begin{align*} \partial_xQ=\partial_yP. \end{align*} There is a smooth function $f:\mathbb R^2\to\mathbb R$ such that $df=\omega$. Define \begin{align*} f(x,y)=\int_0^1\bigl(xP(tx,ty)+yQ(tx,ty)\bigr)\,dt. \end{align*} The integrand is smooth in $(t,x,y)$, and the interval $[0,1]$ is compact, so differentiation under the integral sign is valid. Hence \begin{align*} \partial_x f(x,y) &=\int_0^1 \partial_x\bigl(xP(tx,ty)+yQ(tx,ty)\bigr)\,dt\\ &=\int_0^1\bigl(P(tx,ty)+xt\,\partial_xP(tx,ty)+yt\,\partial_xQ(tx,ty)\bigr)\,dt\\ &=\int_0^1\bigl(P(tx,ty)+xt\,\partial_xP(tx,ty)+yt\,\partial_yP(tx,ty)\bigr)\,dt\\ &=\int_0^1 \frac{d}{dt}\bigl(tP(tx,ty)\bigr)\,dt\\ &=tP(tx,ty)\big|_{t=0}^{t=1}\\ &=P(x,y). \end{align*} The third line uses $\partial_xQ=\partial_yP$, and the fourth line follows from \begin{align*} \frac{d}{dt}\bigl(tP(tx,ty)\bigr) =P(tx,ty)+t\bigl(x\,\partial_xP(tx,ty)+y\,\partial_yP(tx,ty)\bigr). \end{align*} Similarly, \begin{align*} \partial_y f(x,y) &=\int_0^1 \partial_y\bigl(xP(tx,ty)+yQ(tx,ty)\bigr)\,dt\\ &=\int_0^1\bigl(xt\,\partial_yP(tx,ty)+Q(tx,ty)+yt\,\partial_yQ(tx,ty)\bigr)\,dt\\ &=\int_0^1\bigl(xt\,\partial_xQ(tx,ty)+Q(tx,ty)+yt\,\partial_yQ(tx,ty)\bigr)\,dt\\ &=\int_0^1 \frac{d}{dt}\bigl(tQ(tx,ty)\bigr)\,dt\\ &=tQ(tx,ty)\big|_{t=0}^{t=1}\\ &=Q(x,y). \end{align*} Therefore \begin{align*} df=\partial_xf\,dx+\partial_yf\,dy=P\,dx+Q\,dy=\omega. \end{align*} The function $f$ is an explicit global potential for the closed one-form $\omega$ on $\mathbb R^2$. The formula integrates $\omega$ along the straight radial segment from $0$ to $(x,y)$, and the equality $\partial_xQ=\partial_yP$ is exactly what makes the two partial derivatives of this potential recover the original coefficients $P$ and $Q$. [/example] This example is the differential-form version of finding a potential function for a conservative vector field. The point is that the formula uses the whole straight segment from $0$ to $(x,y)$, so it depends on the star-shaped geometry of the domain. ## The Homotopy Operator How can a contraction of a domain produce a primitive for every closed form at once? The answer is to integrate the form along the contraction direction while keeping the remaining variables as tangent inputs. This produces an operator of degree $-1$, and its commutator with $d$ measures the difference between the two ends of the homotopy. [definition: Homotopy Operator for a Radial Contraction] Let $U \subseteq \mathbb R^n$ be star-shaped with centre $a$. For $t\in[0,1]$, set \begin{align*} H_t(x)=a+t(x-a). \end{align*} For each $k\ge 1$, the homotopy operator is the [linear map](/page/Linear%20Map) \begin{align*} K &: \Omega^k(U)\to\Omega^{k-1}(U) \end{align*} defined by \begin{align*} (K\omega)_x(v_1,\dots,v_{k-1}) = \int_0^1 t^{k-1}\,\omega_{a+t(x-a)}(x-a,v_1,\dots,v_{k-1})\,dt, \end{align*} where $\omega\in\Omega^k(U)$, $x\in U$, and $v_1,\dots,v_{k-1}\in\mathbb R^n$. For $k=0$, set $Kf=0$ for $f\in\Omega^0(U)$. [/definition] The factor $t^{k-1}$ records how the spatial tangent vectors are scaled by the radial map, while the inserted vector $x-a$ is the contraction direction. The star-shaped hypothesis ensures that every point $a+t(x-a)$ at which the integrand is evaluated remains inside $U$. [quotetheorem:3583] This is the chain-homotopy identity behind the Poincare lemma. It says that the identity pullback and the constant pullback induce the same map on cohomology, with $K$ recording the explicit coboundary between them. The hypotheses are doing real work: the homotopy must be smooth so that pullback, exterior differentiation, and integration in $t$ interact as stated, and the star-shaped condition keeps the entire segment $a+t(x-a)$ inside $U$. If the segment leaves $U$, then $H_t$ is not a homotopy through maps $U\to U$, the formula for $K$ is not defined on all of $U$, and global periods such as those on the punctured plane can survive. The formula also explains why functions behave differently. For $k=0$, closedness means $df=0$, and the identity becomes $Kd f=f-f(a)$, which is the usual [fundamental theorem of calculus](/theorems/632) along the line from $a$ to $x$. [example: Recovering the Fundamental Theorem Along Rays] Let $U\subseteq\mathbb R^n$ be star-shaped with centre $a\in U$, and let $f\in C^\infty(U)$. Fix $x\in U$, and define the radial path \begin{align*} \gamma_x:[0,1]\to U, \qquad \gamma_x(t)=a+t(x-a). \end{align*} The star-shaped condition ensures that $\gamma_x(t)\in U$ for every $t\in[0,1]$. The homotopy operator recovers the radial fundamental theorem formula \begin{align*} Kdf(x)=f(x)-f(a). \end{align*} By the definition of the homotopy operator for a $1$-form, the factor is $t^{1-1}=1$, so \begin{align*} Kdf(x) &=\int_0^1 t^{1-1}\,(df)_{a+t(x-a)}(x-a)\,dt\\ &=\int_0^1 (df)_{\gamma_x(t)}(x-a)\,dt. \end{align*} The derivative of the path is \begin{align*} \gamma_x'(t) &=\frac{d}{dt}\bigl(a+t(x-a)\bigr)\\ &=x-a. \end{align*} By the chain rule, \begin{align*} \frac{d}{dt}\bigl(f(\gamma_x(t))\bigr) &=(df)_{\gamma_x(t)}(\gamma_x'(t))\\ &=(df)_{\gamma_x(t)}(x-a). \end{align*} Therefore \begin{align*} Kdf(x) &=\int_0^1 \frac{d}{dt}\bigl(f(\gamma_x(t))\bigr)\,dt\\ &=f(\gamma_x(1))-f(\gamma_x(0)) \qquad\text{by the *one-variable fundamental theorem of calculus*.} \end{align*} The endpoint values are \begin{align*} \gamma_x(1)&=a+1(x-a)=a+x-a=x,\\ \gamma_x(0)&=a+0(x-a)=a. \end{align*} Substituting these endpoints gives \begin{align*} Kdf(x) &=f(x)-f(a). \end{align*} The same equality is the degree-zero instance of the *Homotopy Operator Identity*. For the $0$-form $f$, the constant map $c_a$ satisfies \begin{align*} (c_a^*f)(x)=f(c_a(x))=f(a), \end{align*} and $Kf=0$ by definition. Hence the identity gives \begin{align*} dKf+Kdf&=f-c_a^*f,\\ 0+Kdf&=f-c_a^*f,\\ Kdf(x)&=f(x)-f(a). \end{align*} For functions, the homotopy operator is exactly integration of the derivative along the radial segment from $a$ to $x$. Thus the homotopy identity specializes to the ordinary [fundamental theorem of calculus](/theorems/632) on the path $t\mapsto a+t(x-a)$. [/example] Thus $K$ is not an artificial device. It is the higher-degree analogue of integrating a derivative along a path, with the contraction direction inserted into the form. ## The Angle Form on the Punctured Plane What breaks when the domain has a hole? The punctured plane $\mathbb R^2\setminus\{0\}$ admits loops that cannot be contracted through the domain, and a closed form can record how many times such a loop winds around the missing point. On $\mathbb R^2\setminus\{0\}$, define the angle form \begin{align*} \omega_\theta=\frac{-y\,dx+x\,dy}{x^2+y^2}. \end{align*} On any angular coordinate chart, this form is the differential of the local angle coordinate $\theta$. There is no globally defined smooth function $\theta:\mathbb R^2\setminus\{0\}\to\mathbb R$ whose differential is $\omega_\theta$, and the integral around the unit circle detects this failure. [illustration:forms-angle-winding-puncture] [quotetheorem:3584] This gives a practical test for non-exactness. To disprove exactness of a closed $1$-form, it is enough to find one closed curve on which the integral is nonzero. The converse requires global information: on a simply connected domain, closed $1$-forms have path-independent integrals and are exact, but on a general domain a closed form may be locally exact while nonzero periods obstruct a global primitive. In higher degree, the analogous obstruction is the integral of a closed $k$-form over a closed $k$-cycle rather than over a loop. The angle form passes the closedness test but fails the period test. That is exactly the phenomenon excluded by the Poincare lemma on star-shaped domains. [example: The Angle Form Is Closed But Not Exact] Let $M=\mathbb R^2\setminus\{0\}$, with coordinate functions $x,y$, and define \begin{align*} \omega_\theta=\frac{-y\,dx+x\,dy}{x^2+y^2}. \end{align*} Since $x^2+y^2>0$ on $M$, this is a smooth $1$-form on $M$. Write \begin{align*} P(x,y)=\frac{-y}{x^2+y^2}, \qquad Q(x,y)=\frac{x}{x^2+y^2}, \end{align*} so that $\omega_\theta=P\,dx+Q\,dy$. The form $\omega_\theta$ is closed but not exact. Hence \begin{align*} H^1_{\mathrm{dR}}(\mathbb R^2\setminus\{0\})\ne0. \end{align*} Set \begin{align*} s=x^2+y^2. \end{align*} Then $P=-ys^{-1}$ and $Q=xs^{-1}$. We compute \begin{align*} \partial_yP &=\partial_y(-ys^{-1})\\ &=-s^{-1}+(-y)\partial_y(s^{-1})\\ &=-s^{-1}+(-y)(-s^{-2}\partial_y s)\\ &=-s^{-1}+(-y)(-2ys^{-2})\\ &=-\frac{1}{s}+\frac{2y^2}{s^2}\\ &=\frac{-s+2y^2}{s^2}\\ &=\frac{-(x^2+y^2)+2y^2}{s^2}\\ &=\frac{y^2-x^2}{s^2}. \end{align*} Similarly, \begin{align*} \partial_xQ &=\partial_x(xs^{-1})\\ &=s^{-1}+x\partial_x(s^{-1})\\ &=s^{-1}+x(-s^{-2}\partial_xs)\\ &=s^{-1}-2x^2s^{-2}\\ &=\frac{1}{s}-\frac{2x^2}{s^2}\\ &=\frac{s-2x^2}{s^2}\\ &=\frac{(x^2+y^2)-2x^2}{s^2}\\ &=\frac{y^2-x^2}{s^2}. \end{align*} Using $d(dx)=d(dy)=0$, $dx\wedge dx=0$, $dy\wedge dy=0$, and $dy\wedge dx=-dx\wedge dy$, we get \begin{align*} d\omega_\theta &=d(P\,dx+Q\,dy)\\ &=dP\wedge dx+dQ\wedge dy\\ &=(\partial_xP\,dx+\partial_yP\,dy)\wedge dx +(\partial_xQ\,dx+\partial_yQ\,dy)\wedge dy\\ &=\partial_xP\,dx\wedge dx+\partial_yP\,dy\wedge dx +\partial_xQ\,dx\wedge dy+\partial_yQ\,dy\wedge dy\\ &=0-\partial_yP\,dx\wedge dy+\partial_xQ\,dx\wedge dy+0\\ &=(\partial_xQ-\partial_yP)\,dx\wedge dy\\ &=\left(\frac{y^2-x^2}{s^2}-\frac{y^2-x^2}{s^2}\right)\,dx\wedge dy\\ &=0. \end{align*} Thus $\omega_\theta$ is closed. Now parametrize the unit circle by \begin{align*} \gamma:[0,2\pi]\to M, \qquad \gamma(t)=(\cos t,\sin t). \end{align*} This curve is closed because \begin{align*} \gamma(0)=(1,0)=\gamma(2\pi). \end{align*} Along $\gamma$, \begin{align*} \gamma^*x&=\cos t, & \gamma^*y&=\sin t,\\ \gamma^*dx&=d(\cos t)=-\sin t\,dt, & \gamma^*dy&=d(\sin t)=\cos t\,dt. \end{align*} Also, \begin{align*} \gamma^*(x^2+y^2) &=(\cos t)^2+(\sin t)^2\\ &=1. \end{align*} Therefore \begin{align*} \gamma^*\omega_\theta &=\frac{-(\gamma^*y)\gamma^*dx+(\gamma^*x)\gamma^*dy}{\gamma^*(x^2+y^2)}\\ &=\frac{-(\sin t)(-\sin t\,dt)+(\cos t)(\cos t\,dt)}{1}\\ &=(\sin^2t+\cos^2t)\,dt\\ &=dt. \end{align*} Hence \begin{align*} \int_\gamma\omega_\theta &=\int_0^{2\pi}\gamma^*\omega_\theta\\ &=\int_0^{2\pi}dt\\ &=t\big|_{0}^{2\pi}\\ &=2\pi. \end{align*} Suppose, for contradiction, that $\omega_\theta$ were exact. Then there would be a smooth function $f:M\to\mathbb R$ such that $\omega_\theta=df$. Pulling back along $\gamma$ gives \begin{align*} \gamma^*\omega_\theta &=\gamma^*df\\ &=d(f\circ\gamma). \end{align*} Thus \begin{align*} \int_\gamma\omega_\theta &=\int_0^{2\pi}d(f\circ\gamma)\\ &=\int_0^{2\pi}\frac{d}{dt}\bigl(f(\gamma(t))\bigr)\,dt\\ &=f(\gamma(2\pi))-f(\gamma(0)) \qquad\text{by the *one-variable fundamental theorem of calculus*}\\ &=f(1,0)-f(1,0)\\ &=0. \end{align*} This contradicts the earlier computation \begin{align*} \int_\gamma\omega_\theta=2\pi\ne0. \end{align*} Therefore $\omega_\theta$ is not exact. Since $\omega_\theta$ is closed but not exact, its class $[\omega_\theta]$ is a nonzero element of \begin{align*} H^1_{\mathrm{dR}}(M) =\frac{\ker(d:\Omega^1(M)\to\Omega^2(M))}{\operatorname{im}(d:\Omega^0(M)\to\Omega^1(M))}. \end{align*} Thus \begin{align*} H^1_{\mathrm{dR}}(\mathbb R^2\setminus\{0\})\ne0. \end{align*} The angle form has zero [exterior derivative](/theorems/1525), so it is locally compatible with being a differential $df$. Its integral around the unit circle is $2\pi$, so no global potential function can exist on the punctured plane. The nonzero period records the hole at the missing origin. [/example] The notation $d\theta$ is therefore local notation on the punctured plane, not the differential of a global real-valued angle function. Its nonzero period measures the obstruction to choosing a continuous single-valued angle around the origin. [remark: Topological Obstruction] The contrapositive of the Poincare lemma is useful: if an [open set](/page/Open%20Set) admits a closed positive-degree form that is not exact, then the [open set](/page/Open%20Set) is not star-shaped. In the punctured plane, the nonzero period of $\omega_\theta$ detects the missing origin. Later computations refine this observation by identifying $H^1_{\mathrm{dR}}(\mathbb R^2\setminus\{0\})$ with a one-dimensional [vector space](/page/Vector%20Space) generated by $[\omega_\theta]$. [/remark] The contrast with the plane is sharp. Removing a single point changes which loops can be collapsed, and the angle form records that change through its period. The next example returns to $\mathbb R^2$ itself to isolate the role of the missing point: once the radial contraction is available again, every closed one-form has a global potential. [example: Contrast with the Whole Plane] Let $\alpha\in\Omega^1(\mathbb R^2)$ be closed, so $d\alpha=0$. The set $\mathbb R^2$ is star-shaped with centre $0$, because for every $(x,y)\in\mathbb R^2$ and every $t\in[0,1]$, \begin{align*} 0+t((x,y)-0)=(tx,ty)\in\mathbb R^2. \end{align*} The cohomology class $[\alpha]\in H^1_{\mathrm{dR}}(\mathbb R^2)$ is zero. Since $\mathbb R^2$ is star-shaped and $\alpha$ is a closed $1$-form, the *Poincare Lemma for Star-Shaped Domains* gives a smooth function $f:\mathbb R^2\to\mathbb R$ such that \begin{align*} df=\alpha. \end{align*} By the definition of first de Rham cohomology, \begin{align*} H^1_{\mathrm{dR}}(\mathbb R^2) = \frac{\ker(d:\Omega^1(\mathbb R^2)\to\Omega^2(\mathbb R^2))} {\operatorname{im}(d:\Omega^0(\mathbb R^2)\to\Omega^1(\mathbb R^2))}. \end{align*} The equality $df=\alpha$ says exactly that \begin{align*} \alpha\in \operatorname{im}(d:\Omega^0(\mathbb R^2)\to\Omega^1(\mathbb R^2)). \end{align*} Therefore $\alpha$ represents the zero coset in the quotient: \begin{align*} [\alpha]=0\in H^1_{\mathrm{dR}}(\mathbb R^2). \end{align*} Now let $\gamma:[0,1]\to\mathbb R^2$ be any smooth closed curve. Since $\alpha=df$, \begin{align*} \int_\gamma \alpha &=\int_\gamma df\\ &=\int_0^1 \gamma^*(df). \end{align*} Pullback commutes with exterior differentiation on functions, so \begin{align*} \gamma^*(df)=d(f\circ\gamma). \end{align*} Thus \begin{align*} \int_\gamma \alpha &=\int_0^1 d(f\circ\gamma)\\ &=\int_0^1 \frac{d}{dt}\bigl(f(\gamma(t))\bigr)\,dt\\ &=f(\gamma(1))-f(\gamma(0)) \qquad\text{by the *one-variable fundamental theorem of calculus*.} \end{align*} Because $\gamma$ is closed, \begin{align*} \gamma(1)=\gamma(0), \end{align*} and hence \begin{align*} f(\gamma(1))-f(\gamma(0))=0. \end{align*} Therefore \begin{align*} \int_\gamma \alpha=0 \end{align*} for every smooth closed curve $\gamma$ in $\mathbb R^2$. On the whole plane, every closed $1$-form is exact, so every class in $H^1_{\mathrm{dR}}(\mathbb R^2)$ is zero. The punctured plane differs because the missing origin prevents a global radial contraction, and the angle form has a nonzero integral around loops enclosing that missing point. [/example] ## Homotopy Invariance of de Rham Cohomology Why should a deformation of a map preserve the induced map on cohomology? The homotopy operator proves that two pullbacks along homotopic maps differ by an exact term on closed forms. This turns de Rham cohomology into a topological invariant rather than a construction depending only on coordinates. [definition: Smooth Homotopy] Let $M$ and $N$ be smooth manifolds, and let $f,g:M\to N$ be smooth maps. A smooth homotopy from $f$ to $g$ is a smooth map $F:M\times[0,1]\to N$ such that $F(p,0)=f(p)$ and $F(p,1)=g(p)$ for every $p\in M$. [/definition] The interval direction in $M\times[0,1]$ plays the same role as the radial direction in the proof of the Poincare lemma. Integrating the contraction of $F^*\omega$ against $\partial_t$ gives a degree $-1$ operator between forms on $N$ and forms on $M$. The point of constructing this operator is to compare the two endpoint pullbacks $f^*$ and $g^*$. If their difference is a coboundary operator of the form $dK+Kd$, then the difference vanishes after passing to cohomology classes. [quotetheorem:3585] This theorem is the global version of the homotopy operator identity. It says that cohomology cannot distinguish maps that are connected by a smooth deformation. The proof uses no special feature of Euclidean domains beyond the existence of a smooth homotopy. This is why the Poincare lemma extends from star-shaped subsets of $\mathbb R^n$ to smoothly contractible manifolds. A smoothly contractible manifold has a homotopy from the identity map to a constant map. Homotopy invariance then forces its cohomology to agree with the cohomology of a point. The obstruction being removed is global rather than coordinate-level: once every point can be deformed to one base point through smooth maps, every positive-degree cohomology class must disappear. [quotetheorem:3586] This form of the lemma says that positive-degree de Rham cohomology is invariant under smooth contraction. It also explains why nonzero cohomology classes are obstructions to contractibility. The angle form on $\mathbb R^2\setminus\{0\}$ is therefore not just a counterexample to a naive exactness statement. It proves that the punctured plane is not smoothly contractible, and it anticipates the general principle that de Rham cohomology detects holes through integrals of closed forms over cycles. The Poincaré lemma showed that closed forms are not always exact. De Rham cohomology makes this obstruction precise by quotient—it measures the space of closed forms modulo exact forms—and the resulting invariant captures the global topology of the manifold. # 9. de Rham Cohomology This chapter turns the calculus of differential forms into a cohomology theory. The guiding question is: when does a closed form arise as an [exterior derivative](/theorems/1525)? The quotient by exact forms measures the obstruction, and the answer depends only on the global shape of the manifold rather than on a particular coordinate system. We use the Poincare lemma from the previous chapter as the local input and Mayer-Vietoris as the mechanism for assembling local information into global computations. ## Closed Forms Modulo Exact Forms The equation $d\alpha=0$ is a local compatibility condition, but solving $\alpha=d\beta$ asks for a single primitive defined on all of $M$. The obstruction appears when local primitives disagree as one moves around loops or across overlaps of coordinate charts. De Rham cohomology is designed to keep exactly this residual obstruction: two closed forms represent the same class when their difference is globally exact. [definition: Closed Differential Form] Let $M$ be a smooth manifold. A $k$-form $\alpha\in \Omega^k(M)$ is closed if $d\alpha = 0$. [/definition] Closedness is the equation imposed by the [exterior derivative](/theorems/1525). Exactness is stronger because it asks the form to have a global primitive. [definition: Exact Differential Form] Let $M$ be a smooth manifold. A $k$-form $\alpha\in \Omega^{k}(M)$ is exact if there exists $\beta\in \Omega^{k-1}(M)$ such that $\alpha = d\beta$. [/definition] Since $d^2=0$, every exact form is closed. The failure of the converse is the central object of this chapter. To measure it, we keep all closed forms but identify two of them when they differ by an exact form, since exact differences represent a change of global primitive rather than a new obstruction. The quotient below turns that comparison into an abelian group in each degree. [definition: De Rham Cohomology] Let $M$ be a smooth manifold. The $k$-th de Rham cohomology group of $M$ is \begin{align*} H^k_{\mathrm{dR}}(M) := \frac{\ker(d:\Omega^k(M)\to \Omega^{k+1}(M))}{\operatorname{im}(d:\Omega^{k-1}(M)\to \Omega^k(M))}. \end{align*} An element of $H^k_{\mathrm{dR}}(M)$ is called a de Rham cohomology class. [/definition] We write $[\alpha]$ for the class of a closed form $\alpha$. Thus $[\alpha]=[\alpha']$ exactly when $\alpha-\alpha'=d\beta$ for some $\beta\in\Omega^{k-1}(M)$. [example: Closed Not Exact On The Circle] Let $\gamma:[0,2\pi]\to S^1$ be the positively oriented parametrisation \begin{align*} \gamma(t)=(\cos t,\sin t). \end{align*} Let $x,y$ be the standard coordinate functions on $\mathbb R^2$, and let \begin{align*} \alpha=d\theta:=(-y\,dx+x\,dy)|_{S^1}. \end{align*} This is globally defined on $S^1$ because it is the restriction of a globally defined $1$-form on $\mathbb R^2$. The form $d\theta$ is closed but not exact, and therefore defines a non-zero class in $H^1_{\mathrm{dR}}(S^1)$. Since $S^1$ is one-dimensional, every $2$-form on $S^1$ is zero: if $A$ is alternating bilinear on the one-dimensional space $T_pS^1$ and $u=av$, $w=bv$, then \begin{align*} A(u,w)=A(av,bv)=abA(v,v)=0. \end{align*} Thus $\Omega^2(S^1)=0$. Since $d\alpha\in\Omega^2(S^1)$, we get \begin{align*} d\alpha=0. \end{align*} So $d\theta$ is closed. Now compute its integral around the positively oriented circle. Along $\gamma$, \begin{align*} \gamma^*x&=\cos t, & \gamma^*y&=\sin t,\\ \gamma^*dx&=d(\cos t)=-\sin t\,dt, & \gamma^*dy&=d(\sin t)=\cos t\,dt. \end{align*} Therefore \begin{align*} \gamma^*\alpha &=\gamma^*(-y\,dx+x\,dy)\\ &=-(\gamma^*y)(\gamma^*dx)+(\gamma^*x)(\gamma^*dy)\\ &=-(\sin t)(-\sin t\,dt)+(\cos t)(\cos t\,dt)\\ &=(\sin^2 t+\cos^2 t)\,dt\\ &=dt. \end{align*} Hence \begin{align*} \oint_{S^1}d\theta &=\int_0^{2\pi}\gamma^*\alpha\\ &=\int_0^{2\pi}dt\\ &=\int_0^{2\pi}1\,dt\\ &=\bigl[t\bigr]_{0}^{2\pi}\\ &=2\pi. \end{align*} Suppose, for contradiction, that $d\theta$ were exact. Then there would be a smooth function $f:S^1\to\mathbb R$ such that \begin{align*} \alpha=df. \end{align*} Let $F=f\circ\gamma:[0,2\pi]\to\mathbb R$. Pulling back $\alpha=df$ along $\gamma$ gives \begin{align*} dt &=\gamma^*\alpha\\ &=\gamma^*(df)\\ &=d(f\circ\gamma)\\ &=dF\\ &=F'(t)\,dt. \end{align*} Thus $F'(t)=1$ for all $t\in[0,2\pi]$. By the [fundamental theorem of calculus](/theorems/632), \begin{align*} F(2\pi)-F(0) &=\int_0^{2\pi}F'(t)\,dt\\ &=\int_0^{2\pi}1\,dt\\ &=2\pi. \end{align*} But $\gamma(2\pi)=\gamma(0)=(1,0)$, so \begin{align*} F(2\pi) &=f(\gamma(2\pi))\\ &=f(\gamma(0))\\ &=F(0). \end{align*} Therefore \begin{align*} F(2\pi)-F(0)=0, \end{align*} contradicting $F(2\pi)-F(0)=2\pi$. Hence $d\theta$ is not exact. The form $d\theta$ satisfies the local closedness equation $d(d\theta)=0$, but it has no globally defined smooth primitive on $S^1$. In the quotient defining de Rham cohomology, exact closed forms represent the zero class. Since $d\theta$ is closed and not exact, \begin{align*} [d\theta]\ne 0\in H^1_{\mathrm{dR}}(S^1). \end{align*} [/example] This example is the prototype: the equation $d\alpha=0$ is local, while being $d\beta$ for a globally defined $\beta$ is a global condition. ## Low-Degree Cohomology Before computing new examples, it is useful to ask what the definition says in the lowest degrees. Degree zero is controlled by functions with zero derivative, while positive-degree cohomology of a point vanishes because there are no positive-degree forms on a point. [quotetheorem:3587] [citeproof:3587] The zeroth group therefore detects connected components, but the hypotheses are doing real work. The proof uses the fact that connected components of a smooth manifold are path connected, so a zero derivative forces constancy along paths. In a general topological space this implication has no differential meaning, and even in pathological smooth settings without the usual manifold assumptions the relation between components and paths can fail. If $M$ has infinitely many connected components, the notation $\mathbb R^{\pi_0(M)}$ means the [vector space](/page/Vector%20Space) of all real-valued functions on the set of components, not necessarily a finite-dimensional space. Higher groups detect higher-dimensional holes, as the computations below will show. The simplest calibration case is the one-point manifold: it has one component and no positive-dimensional directions. Computing its cohomology fixes the normalization for later contractibility arguments. [quotetheorem:3588] This follows from the description of $H^0$ and the fact that $\Omega^k(\{p\})=0$ for $k\ge 1$: the tangent space of a zero-dimensional manifold is zero, so there are no non-zero alternating $k$-linear forms once $k\ge 1$. The dimension is essential here; even a one-dimensional manifold can carry non-zero $1$-forms, and a circle carries a closed $1$-form which is not exact. Contractibility removes that global obstruction in the local Euclidean situation: the Poincare lemma says that every closed positive-degree form on a contractible open subset of $\mathbb R^n$ is exact. Without contractibility, local exactness need not assemble into a global primitive, which is precisely the failure measured by de Rham cohomology. [example: Disconnected Manifold] Let \begin{align*} M=S^1\sqcup S^2. \end{align*} A $0$-form on $M$ is a smooth function $f:M\to\mathbb R$. Write \begin{align*} f_1=f|_{S^1},\qquad f_2=f|_{S^2}. \end{align*} The zeroth de Rham cohomology of $M$ is \begin{align*} H^0_{\mathrm{dR}}(S^1\sqcup S^2)\cong \mathbb R^2. \end{align*} By the definition of de Rham cohomology, \begin{align*} H^0_{\mathrm{dR}}(M) &= \frac{\ker(d:\Omega^0(M)\to\Omega^1(M))} {\operatorname{im}(d:\Omega^{-1}(M)\to\Omega^0(M))}. \end{align*} There are no $(-1)$-forms, so \begin{align*} \Omega^{-1}(M)&=0,\\ \operatorname{im}(d:\Omega^{-1}(M)\to\Omega^0(M))&=0. \end{align*} Thus \begin{align*} H^0_{\mathrm{dR}}(M) &=\ker(d:\Omega^0(M)\to\Omega^1(M)). \end{align*} A smooth function on the disjoint union is the same as a pair of smooth functions on the two components: \begin{align*} C^\infty(M,\mathbb R) &\cong C^\infty(S^1,\mathbb R)\oplus C^\infty(S^2,\mathbb R),\\ f&\longmapsto (f_1,f_2). \end{align*} The inverse sends a pair $(g,h)$ to the function \begin{align*} p\longmapsto \begin{cases} g(p), & p\in S^1,\\ h(p), & p\in S^2. \end{cases} \end{align*} Now suppose $df=0$. Since exterior differentiation commutes with restriction to open components, \begin{align*} df_1=0\quad\text{on }S^1,\qquad df_2=0\quad\text{on }S^2. \end{align*} Let $X$ be either $S^1$ or $S^2$, and let $g:X\to\mathbb R$ be a smooth function with $dg=0$. Both $S^1$ and $S^2$ are path connected, so for any $p,q\in X$ there is a piecewise smooth path $\sigma:[0,1]\to X$ with $\sigma(0)=p$ and $\sigma(1)=q$. Set $G=g\circ\sigma$. On every smooth subinterval of $\sigma$, \begin{align*} G'(t) &=dg_{\sigma(t)}(\sigma'(t))\\ &=0. \end{align*} If $0=t_0<t_1<\cdots<t_m=1$ are the breakpoints of the piecewise smooth path, then by the [fundamental theorem of calculus](/theorems/632), \begin{align*} g(q)-g(p) &=G(1)-G(0)\\ &=\sum_{j=1}^{m}\bigl(G(t_j)-G(t_{j-1})\bigr)\\ &=\sum_{j=1}^{m}\int_{t_{j-1}}^{t_j}G'(t)\,dt\\ &=\sum_{j=1}^{m}\int_{t_{j-1}}^{t_j}0\,dt\\ &=0. \end{align*} So $g$ is constant on $X$. Applying this to $f_1$ and $f_2$, there exist [real numbers](/page/Real%20Numbers) $a,b\in\mathbb R$ such that \begin{align*} f(p)= \begin{cases} a, & p\in S^1,\\ b, & p\in S^2. \end{cases} \end{align*} Conversely, any function of this form has zero differential. Indeed, if $p\in S^1$ and $v\in T_pS^1$, then for any smooth curve $c$ with $c(0)=p$ and $c'(0)=v$, \begin{align*} df_p(v) &=\left.\frac{d}{dt}\right|_{t=0} f(c(t))\\ &=\left.\frac{d}{dt}\right|_{t=0} a\\ &=0. \end{align*} The same calculation on $S^2$ gives $df_p(v)=0$ there as well. Hence the closed $0$-forms on $M$ are exactly the functions that are constant on each component. Define \begin{align*} \Phi:H^0_{\mathrm{dR}}(M)&\to \mathbb R^2,\\ [f]&\mapsto (a,b), \end{align*} where $a$ is the value of $f$ on $S^1$ and $b$ is the value of $f$ on $S^2$. Since there are no exact $0$-forms, two closed $0$-forms represent the same class exactly when they are the same function. Thus $\Phi$ is well-defined. If $\Phi([f])=(0,0)$, then $f$ is zero on both components, so \begin{align*} [f]=0. \end{align*} Thus $\Phi$ is injective. Given any $(a,b)\in\mathbb R^2$, the function \begin{align*} f_{a,b}(p)= \begin{cases} a, & p\in S^1,\\ b, & p\in S^2 \end{cases} \end{align*} is smooth, closed, and satisfies \begin{align*} \Phi([f_{a,b}])=(a,b). \end{align*} Thus $\Phi$ is surjective. Therefore \begin{align*} H^0_{\mathrm{dR}}(S^1\sqcup S^2)\cong \mathbb R^2. \end{align*} The two coordinates of $H^0_{\mathrm{dR}}(S^1\sqcup S^2)$ record the two independent constant values of a closed $0$-form: one value on the circle component $S^1$ and one value on the sphere component $S^2$. [/example] ## Mayer-Vietoris For De Rham Cohomology Local computations are useful only if they can be glued. Suppose $M=U\cup V$ with $U$ and $V$ open. A form on $M$ restricts to forms on $U$ and $V$ which agree on $U\cap V$, and the Mayer-Vietoris sequence packages this gluing condition into a long exact sequence in cohomology. [quotetheorem:3589] [citeproof:3589] The openness of $U$ and $V$ is not a cosmetic assumption. Differential forms restrict naturally to open submanifolds, and smooth partitions of unity subordinate to open covers are what make the short exact sequence of complexes exact on the right. For arbitrary closed covers, extension by zero can fail to be smooth at the boundary, so the same proof does not apply without extra collar or relative hypotheses. Mayer-Vietoris also does not assert that $H^k_{\mathrm{dR}}(M)$ splits as a direct sum of the neighbouring groups; exactness controls kernels and images, while the connecting homomorphism records the obstruction to such naive splitting. The connecting homomorphism is the part of the sequence that carries the geometry. It turns a class on the overlap into the obstruction to choosing compatible primitives on the two pieces. [explanation: Connecting Homomorphism] Given a closed $k$-form $\eta\in\Omega^k(U\cap V)$, choose forms $\beta\in\Omega^k(U)$ and $\gamma\in\Omega^k(V)$ such that $\beta|_{U\cap V}-\gamma|_{U\cap V}=\eta$. The forms $d\beta$ and $d\gamma$ agree on $U\cap V$, so they glue to a closed $(k+1)$-form on $M$. The cohomology class of this glued form is $\delta[\eta]\in H^{k+1}_{\mathrm{dR}}(M)$. Different choices of $\beta$ and $\gamma$ change the glued form by an exact form, so the class is well-defined. [/explanation] Exactness means that the image of each map is the kernel of the next. In computations, this lets us recover unknown groups by comparing the known cohomology of $U$, $V$, and $U\cap V$. ## Spheres From Mayer-Vietoris The sphere is the first major test case: it is locally contractible, but its top-dimensional cohomology should remember the fundamental enclosed volume. Mayer-Vietoris proves this by covering the sphere by two contractible caps whose overlap deformation retracts onto an equatorial sphere. [quotetheorem:3590] [citeproof:3590] The restriction $n\ge 1$ avoids the exceptional zero-sphere: $S^0$ is two points, so its cohomology is concentrated in degree zero with $H^0_{\mathrm{dR}}(S^0)\cong\mathbb R^2$. For $n\ge 1$, the overlap in the two-cap cover must deformation retract onto $S^{n-1}$ because Mayer-Vietoris reads the cohomology of the whole sphere from the cohomology carried by this equatorial overlap. If the overlap were contractible, the sequence would lose the class that shifts into the top degree. The resulting top class is represented by a volume form, and its non-vanishing can also be detected by integration: an exact top-degree form has integral zero over a compact boundaryless oriented manifold, while a positive volume form has non-zero integral. This computation is also the template for later examples: choose a cover whose pieces are simple, identify the overlap, and use exactness to see which overlap classes become global classes one degree higher. [example: Full Mayer-Vietoris Computation For The Two-Sphere] Let $N=(0,0,1)$ and $S=(0,0,-1)$ be the north and south poles of $S^2\subset \mathbb R^3$. Take \begin{align*} U=S^2\setminus\{S\},\qquad V=S^2\setminus\{N\}. \end{align*} Then $U$ is an open neighbourhood of the northern hemisphere, $V$ is an open neighbourhood of the southern hemisphere, and \begin{align*} S^2=U\cup V. \end{align*} Write $H^k(X)$ for $H^k_{\mathrm{dR}}(X)$. The Mayer-Vietoris sequence for this cover gives \begin{align*} H^0(S^2)\cong \mathbb R,\qquad H^1(S^2)=0,\qquad H^2(S^2)\cong \mathbb R. \end{align*} Stereographic projection identifies both $U$ and $V$ with $\mathbb R^2$. Since $\mathbb R^2$ is star-shaped, *Poincare Lemma On Star Shaped Domains* gives \begin{align*} H^k(U)=H^k(V)=0\qquad(k\ge 1), \end{align*} and *Zeroth De Rham Cohomology* gives \begin{align*} H^0(U)\cong \mathbb R,\qquad H^0(V)\cong \mathbb R. \end{align*} The intersection is \begin{align*} U\cap V=S^2\setminus\{N,S\}. \end{align*} For $(x,y,z)\in U\cap V$, we have $x^2+y^2>0$. The map \begin{align*} \Phi:U\cap V&\to S^1\times(-1,1),\\ (x,y,z)&\mapsto \left(\frac{(x,y)}{\sqrt{x^2+y^2}},z\right) \end{align*} has inverse \begin{align*} \Psi:S^1\times(-1,1)&\to U\cap V,\\ ((a,b),t)&\mapsto \left(\sqrt{1-t^2}\,a,\sqrt{1-t^2}\,b,t\right). \end{align*} Thus $U\cap V$ is diffeomorphic to $S^1\times(-1,1)$, which deformation retracts onto $S^1\times\{0\}$ by \begin{align*} R_\tau(q,t)=(q,(1-\tau)t). \end{align*} By *[Homotopy Invariance of De Rham Cohomology](/theorems/3585)* and *[De Rham Cohomology Of Spheres](/theorems/3590)* in the case $n=1$, \begin{align*} H^0(U\cap V)&\cong \mathbb R,\\ H^1(U\cap V)&\cong \mathbb R,\\ H^k(U\cap V)&=0\qquad(k\ge 2). \end{align*} By *Mayer-Vietoris Long Exact Sequence*, the relevant part of the long exact sequence is \begin{align*} 0\to H^0(S^2) \xrightarrow{r_0}\mathbb R\oplus\mathbb R \xrightarrow{s_0}\mathbb R \xrightarrow{\delta_0}H^1(S^2) \xrightarrow{r_1}0\oplus 0 \xrightarrow{s_1}\mathbb R \xrightarrow{\delta_1}H^2(S^2) \xrightarrow{r_2}0\oplus 0. \end{align*} The map $s_0$ is induced by \begin{align*} s([\beta],[\gamma])=[\beta|_{U\cap V}-\gamma|_{U\cap V}]. \end{align*} Under the identifications $H^0(U)\cong\mathbb R$, $H^0(V)\cong\mathbb R$, and $H^0(U\cap V)\cong\mathbb R$, an element $(a,b)\in \mathbb R\oplus\mathbb R$ represents the constant value $a$ on $U$ and the constant value $b$ on $V$. Therefore \begin{align*} s_0(a,b) &=a-b. \end{align*} Its kernel is \begin{align*} \ker s_0 &=\{(a,b)\in\mathbb R^2:a-b=0\}\\ &=\{(a,b)\in\mathbb R^2:a=b\}\\ &=\{(a,a):a\in\mathbb R\}\\ &\cong \mathbb R. \end{align*} Its image is all of $\mathbb R$, because for every $c\in\mathbb R$, \begin{align*} s_0(c,0)=c-0=c. \end{align*} So \begin{align*} \operatorname{im}s_0=\mathbb R. \end{align*} Exactness at $\mathbb R\oplus\mathbb R$ gives \begin{align*} \operatorname{im}r_0=\ker s_0=\{(a,a):a\in\mathbb R\}. \end{align*} Exactness at $H^0(S^2)$ gives $\ker r_0=0$, so $r_0$ identifies $H^0(S^2)$ with the diagonal copy of $\mathbb R$. Hence \begin{align*} H^0(S^2)\cong \mathbb R. \end{align*} Next, exactness at $H^0(U\cap V)\cong\mathbb R$ gives \begin{align*} \ker\delta_0=\operatorname{im}s_0=\mathbb R. \end{align*} Since the domain of $\delta_0$ is $\mathbb R$, this says $\delta_0$ is the zero map, so \begin{align*} \operatorname{im}\delta_0=0. \end{align*} The next map is \begin{align*} r_1:H^1(S^2)\to 0\oplus 0. \end{align*} Since every element maps to $0$, its kernel is all of $H^1(S^2)$: \begin{align*} \ker r_1=H^1(S^2). \end{align*} Exactness at $H^1(S^2)$ gives \begin{align*} H^1(S^2) &=\ker r_1\\ &=\operatorname{im}\delta_0\\ &=0. \end{align*} Finally, the map \begin{align*} s_1:0\oplus 0\to \mathbb R \end{align*} has image $0$. Exactness at $H^1(U\cap V)\cong\mathbb R$ gives \begin{align*} \ker\delta_1=\operatorname{im}s_1=0. \end{align*} So $\delta_1$ is injective. Exactness at $H^2(S^2)$ gives \begin{align*} \operatorname{im}\delta_1=\ker r_2. \end{align*} But $r_2$ maps $H^2(S^2)$ to $0\oplus 0$, so \begin{align*} \ker r_2=H^2(S^2). \end{align*} Therefore \begin{align*} \operatorname{im}\delta_1=H^2(S^2). \end{align*} Thus $\delta_1:\mathbb R\to H^2(S^2)$ is both injective and surjective, and hence \begin{align*} H^2(S^2)\cong \mathbb R. \end{align*} The two contractible pieces $U$ and $V$ contribute no positive-degree cohomology. The overlap $U\cap V$ carries one degree-one class coming from the equatorial circle, and Mayer-Vietoris shifts that class into degree $2$ on $S^2$. Thus $S^2$ has one connected component, no degree-one cohomology, and one top-degree cohomology class: \begin{align*} H^0(S^2)\cong \mathbb R,\qquad H^1(S^2)=0,\qquad H^2(S^2)\cong \mathbb R. \end{align*} [/example] This computation is the model for higher spheres. The overlap carries the cohomology one dimension lower, and the connecting map shifts it into the top degree of the whole sphere. ## The Torus And Product Behaviour The torus has two independent circular directions, so we expect two independent degree-one classes and one degree-two class obtained by wedging them. This is formalised by the Kunneth theorem, which describes the cohomology of a product from the cohomology of its factors. [quotetheorem:3591] [citeproof:3591] The product formula uses the exterior product of forms and homological algebra; in this course we use it as a computational tool. The real coefficients matter: over a field, tensor products behave cleanly and there are no torsion correction terms. By contrast, integral singular cohomology has a Kunneth formula with possible $\operatorname{Tor}$ terms, so the displayed formula should not be copied unchanged into integer-coefficient topology. The finite-dimensional smooth-manifold setting also keeps the product differential forms under ordinary algebraic control; outside this setting, topological completions and infinite-dimensional phenomena require additional care. Applying the product formula to $T^2=S^1\times S^1$ should produce one degree-zero class, two degree-one classes, and one degree-two class. The calculation below records this expected pattern as a concrete benchmark for later examples. [quotetheorem:3592] [citeproof:3592] If $\theta_1$ and $\theta_2$ are angular coordinates on the two circle factors, then $d\theta_1$ and $d\theta_2$ represent the two degree-one classes, while $d\theta_1\wedge d\theta_2$ represents the degree-two class. [example: Betti Numbers Of The Torus] Let $T^2=S^1\times S^1$. For each $k\ge 0$, the $k$-th Betti number is \begin{align*} b_k(T^2)=\dim H^k_{\mathrm{dR}}(T^2). \end{align*} Write $H^k(T^2)$ for $H^k_{\mathrm{dR}}(T^2)$. The Betti numbers in degrees $0$, $1$, and $2$ are \begin{align*} b_0(T^2)=1,\qquad b_1(T^2)=2,\qquad b_2(T^2)=1. \end{align*} By *De Rham Cohomology Of The Two-Torus*, \begin{align*} H^0(T^2)&\cong \mathbb R,\\ H^1(T^2)&\cong \mathbb R^2,\\ H^2(T^2)&\cong \mathbb R,\\ H^k(T^2)&=0\qquad(k\ge 3). \end{align*} Isomorphic vector spaces have the same dimension, so \begin{align*} b_0(T^2) &=\dim H^0(T^2)\\ &=\dim \mathbb R\\ &=1, \end{align*} because $\{1\}$ is a basis of $\mathbb R$. Similarly, \begin{align*} b_1(T^2) &=\dim H^1(T^2)\\ &=\dim \mathbb R^2\\ &=2, \end{align*} because $\{(1,0),(0,1)\}$ is a basis of $\mathbb R^2$. Finally, \begin{align*} b_2(T^2) &=\dim H^2(T^2)\\ &=\dim \mathbb R\\ &=1. \end{align*} For comparison with the exterior algebra description, let $E$ be a two-dimensional real [vector space](/page/Vector%20Space) with basis $e_1,e_2$, corresponding to the two degree-one circle classes on the two factors of $S^1\times S^1$. Its exterior algebra has graded pieces \begin{align*} \Lambda^0E&=\mathbb R\cdot 1,\\ \Lambda^1E&=\mathbb R e_1\oplus \mathbb R e_2,\\ \Lambda^2E&=\mathbb R(e_1\wedge e_2). \end{align*} There are no non-zero degree-three wedge products, since any wedge of three elements of the two-element basis repeats one basis vector, and alternating products satisfy \begin{align*} e_i\wedge e_i=0,\qquad e_2\wedge e_1=-e_1\wedge e_2. \end{align*} Thus \begin{align*} \dim \Lambda^0E=1,\qquad \dim \Lambda^1E=2,\qquad \dim \Lambda^2E=1, \end{align*} which gives the same row \begin{align*} 1,\ 2,\ 1. \end{align*} The number $b_0(T^2)=1$ records that the torus has one connected component. The number $b_1(T^2)=2$ records the two independent circular directions, one from each $S^1$ factor. The number $b_2(T^2)=1$ records the fundamental area class represented by wedging the two degree-one classes. [/example] ## Betti Numbers And Compact Manifolds The examples above produced finite-dimensional cohomology groups, but this need not be taken for granted from the definition: the spaces of differential forms are infinite-dimensional. Compactness is the analytic hypothesis that brings finite-dimensionality into the theory. [quotetheorem:3593] In a full Hodge theory course, this theorem is proved by choosing a Riemannian metric, identifying each cohomology class with a unique harmonic representative, and using elliptic regularity to show that the space of harmonic $k$-forms is finite-dimensional. Here we use the result as a structural fact about compact manifolds. [definition: Betti Number] Let $M$ be a compact smooth manifold. The $k$-th Betti number of $M$ is \begin{align*} b_k(M):=\dim H^k_{\mathrm{dR}}(M). \end{align*} [/definition] Betti numbers turn cohomology groups into numerical invariants. They forget the preferred representatives of cohomology classes, but retain the number of independent classes in each degree. [example: Betti Numbers Of Spheres] Let $n\ge 1$, and let $S^n$ be the $n$-sphere. For each $k\ge 0$, write \begin{align*} H^k(S^n)=H^k_{\mathrm{dR}}(S^n), \end{align*} and define the $k$-th Betti number by \begin{align*} b_k(S^n)=\dim H^k(S^n). \end{align*} The Betti numbers of $S^n$ are \begin{align*} b_k(S^n)= \begin{cases} 1, & k=0,\\ 1, & k=n,\\ 0, & \text{otherwise}. \end{cases} \end{align*} By *[De Rham Cohomology Of Spheres](/theorems/3590)*, for $n\ge 1$, \begin{align*} H^k(S^n)\cong \begin{cases} \mathbb R, & k=0,\\ \mathbb R, & k=n,\\ 0, & \text{otherwise}. \end{cases} \end{align*} Isomorphic real vector spaces have the same dimension. For $k=0$, this gives \begin{align*} b_0(S^n) &=\dim H^0(S^n)\\ &=\dim \mathbb R\\ &=1, \end{align*} because $\{1\}$ is a basis of $\mathbb R$. For $k=n$, this gives \begin{align*} b_n(S^n) &=\dim H^n(S^n)\\ &=\dim \mathbb R\\ &=1. \end{align*} For $k\ne 0$ and $k\ne n$, the same theorem gives \begin{align*} H^k(S^n)=0. \end{align*} The zero [vector space](/page/Vector%20Space) has only the zero vector, so its basis is the empty set and its dimension is $0$. Hence \begin{align*} b_k(S^n) &=\dim H^k(S^n)\\ &=\dim 0\\ &=0. \end{align*} Combining the three cases, \begin{align*} b_k(S^n)= \begin{cases} 1, & k=0,\\ 1, & k=n,\\ 0, & \text{otherwise}. \end{cases} \end{align*} The equality $b_0(S^n)=1$ records that $S^n$ has one connected component. The equality $b_n(S^n)=1$ records that there is one independent top-degree de Rham class. All other Betti numbers vanish, so $S^n$ has no de Rham cohomology in degrees other than $0$ and $n$. [/example] De Rham cohomology is therefore a bridge between analysis and topology. It starts with smooth forms and the differential operator $d$, but its answers are stable under global deformation and reveal connected components, loops, enclosed voids, and their higher-dimensional analogues. De Rham cohomology encodes topological information through the [exterior derivative](/theorems/1525), but to make this precise, we must connect it to topology constructed from singular simplices. De Rham's theorem shows that these two cohomology theories coincide, revealing that differential forms compute topological invariants. # 10. de Rham's Theorem and Comparison with Singular Cohomology The earlier chapters built differential forms, exterior differentiation, pullback, integration on oriented manifolds, and the generalized Stokes theorem. This chapter compares that analytic complex with a topological complex built from singular simplices. The main point is that integrating closed forms over cycles loses no information: de Rham cohomology is naturally the same as singular cohomology with real coefficients. This comparison turns computations with forms into topological invariants and explains why periods, degrees, and Euler characteristics appear in the same theory. ## Singular Chains and Real Cochains How can a topological space be converted into algebra that remembers its holes? Singular homology answers this by mapping standard simplices into the space, allowing many parameterized pieces with formal real coefficients. Cohomology then takes linear functionals on those chains and studies which functionals vanish on boundaries. [definition: Standard Simplex] For $k \ge 0$, the standard $k$-simplex is \begin{align*} \Delta^k = \left\{(t_0,\dots,t_k) \in \mathbb R^{k+1} : t_i \ge 0, \sum_{i=0}^k t_i = 1\right\}. \end{align*} [/definition] The vertices of $\Delta^k$ are the coordinate vectors $e_0,\dots,e_k$. Faces are obtained by omitting one of these vertices. To use simplices as probes of an arbitrary space, the simplex itself should remain the standard parameter domain while the geometry is carried by a map into the space. This lets one record points, paths, triangles, and higher-dimensional pieces without requiring the space to come with a preferred triangulation. The resulting object is the basic cell from which singular chains are assembled. [definition: Singular Simplex] A singular $k$-simplex in a topological space $M$ is a continuous map $\sigma:\Delta^k \to M$. [/definition] For a smooth manifold, the underlying topological space is being used here. In the integration construction below, smooth singular simplices are used first; the smoothing theorem identifies their cohomology with ordinary singular cohomology with real coefficients. [definition: Singular Chain Group] For $k \ge 0$, the real singular chain group $C_k(M;\mathbb R)$ is the real [vector space](/page/Vector%20Space) of finite formal sums \begin{align*} c = \sum_{j=1}^N a_j \sigma_j, \end{align*} where $a_j \in \mathbb R$ and each $\sigma_j:\Delta^k \to M$ is a singular $k$-simplex. [/definition] A chain is a finite linear combination of parameterized simplices. The next step is to make these formal sums interact with integration and homology: each simplex must contribute an oriented edge-chain, and adjacent faces must cancel when chains are glued. The boundary operator supplies this signed rule, so it is the algebraic device that will later make Stokes' theorem match the topological boundary of a chain. [definition: Boundary Operator] Let $\iota_i:\Delta^{k-1}\to \Delta^k$ be the affine map whose image is the face opposite $e_i$, determined on vertices by \begin{align*} \iota_i(e_j)= \begin{cases} e_j, & j<i,\\ e_{j+1}, & j\ge i. \end{cases} \end{align*} The boundary of a singular $k$-simplex $\sigma:\Delta^k \to M$ is \begin{align*} \partial \sigma = \sum_{i=0}^k (-1)^i \sigma \circ \iota_i, \end{align*} and $\partial:C_k(M;\mathbb R)\to C_{k-1}(M;\mathbb R)$ is extended linearly. [/definition] The alternating signs encode the orientation of the faces. They are chosen so that taking the boundary twice cancels every codimension-two face. [illustration:forms-simplex-boundary-signs] The sign convention is justified by the basic consistency test for any boundary theory: an edge of an edge should not remain. Each codimension-two face appears twice, from two different orders of deleting vertices, so the algebra must force those two appearances to cancel. The next formal result records this cancellation as the chain-complex identity. [quotetheorem:2232] [citeproof:2232] The identity $\partial^2=0$ is what makes homology possible: boundaries automatically become cycles, so quotienting cycles by boundaries is consistent. Without this cancellation, a boundary could have a nonzero boundary of its own, and the phrase "cycles modulo boundaries" would not define the intended invariant. This is the chain-level analogue of $d^2=0$ for differential forms. Cochains reverse direction: they assign numbers to chains. This dual viewpoint is the one naturally reached by integrating differential forms, since integration turns each parameterized simplex into a real number. To compare integration with cohomology, the cochains need their own differential, obtained by testing a cochain on the boundary of a chain. [definition: Singular Cochain Complex] The degree $k$ singular cochain group with real coefficients is \begin{align*} C^k_{\mathrm{sing}}(M;\mathbb R)=\operatorname{Hom}_{\mathbb R}(C_k(M;\mathbb R),\mathbb R). \end{align*} The coboundary map $\delta:C^k_{\mathrm{sing}}(M;\mathbb R)\to C^{k+1}_{\mathrm{sing}}(M;\mathbb R)$ is defined by \begin{align*} (\delta \varphi)(c)=\varphi(\partial c) \end{align*} for every $c \in C_{k+1}(M;\mathbb R)$. [/definition] Since $\partial^2=0$, the coboundary also squares to zero. Thus singular cochains form a cochain complex. The cohomology groups now arise by the same kernel-modulo-image pattern as de Rham cohomology, with cocycles measuring compatible assignments to cycles and coboundaries measuring assignments that come from one degree lower. [definition: Singular Cohomology] The singular cocycles, coboundaries, and cohomology groups are \begin{align*} Z^k_{\mathrm{sing}}(M;\mathbb R) &= \ker\left(\delta:C^k_{\mathrm{sing}}(M;\mathbb R)\to C^{k+1}_{\mathrm{sing}}(M;\mathbb R)\right),\\ B^k_{\mathrm{sing}}(M;\mathbb R) &= \operatorname{im}\left(\delta:C^{k-1}_{\mathrm{sing}}(M;\mathbb R)\to C^k_{\mathrm{sing}}(M;\mathbb R)\right),\\ H^k_{\mathrm{sing}}(M;\mathbb R) &= Z^k_{\mathrm{sing}}(M;\mathbb R)/B^k_{\mathrm{sing}}(M;\mathbb R). \end{align*} [/definition] The analogy with de Rham cohomology is now visible. In de Rham theory, closed forms are killed by $d$, exact forms are images under $d$, and $d^2=0$. In singular cohomology, cocycles are killed by $\delta$, coboundaries are images under $\delta$, and $\delta^2=0$. Real coefficients are essential for this comparison because integration produces [real numbers](/page/Real%20Numbers). Integer cohomology contains torsion information that real-valued periods cannot see; for example, torsion classes vanish after tensoring with $\mathbb R$ and therefore have no de Rham representative. [example: Zero-Dimensional Cohomology] Let $M$ be a nonempty path-connected smooth manifold. A singular $0$-simplex is a map $\Delta^0\to M$, and since $\Delta^0=\{e_0\}$, it is determined by one point $p\in M$. Thus a $0$-cochain $\varphi\in C^0_{\mathrm{sing}}(M;\mathbb R)$ assigns a real number $\varphi(p)$ to each point $p\in M$. $H^0_{\mathrm{sing}}(M;\mathbb R)\cong \mathbb R$, and this matches $H^0_{\mathrm{dR}}(M)\cong \mathbb R$. Let $\sigma:\Delta^1\to M$ be a singular $1$-simplex. The two face maps $\iota_0,\iota_1:\Delta^0\to\Delta^1$ satisfy $\iota_0(e_0)=e_1$ and $\iota_1(e_0)=e_0$, so the boundary formula gives \begin{align*} \partial\sigma &= (-1)^0\sigma\circ\iota_0+(-1)^1\sigma\circ\iota_1\\ &= \sigma\circ\iota_0-\sigma\circ\iota_1. \end{align*} If $q=\sigma(e_1)$ and $p=\sigma(e_0)$, then this is \begin{align*} \partial\sigma=q-p. \end{align*} By the definition of the coboundary, \begin{align*} (\delta\varphi)(\sigma) &=\varphi(\partial\sigma)\\ &=\varphi(q-p)\\ &=\varphi(q)-\varphi(p), \end{align*} where the last equality uses linearity of $\varphi$. Thus $\varphi$ is a $0$-cocycle exactly when, for every singular path $\sigma$ from $p$ to $q$, \begin{align*} 0=(\delta\varphi)(\sigma)=\varphi(q)-\varphi(p), \end{align*} or equivalently $\varphi(q)=\varphi(p)$. Since $M$ is path-connected, for any two points $p,q\in M$ there is a continuous path from $p$ to $q$, hence a singular $1$-simplex with endpoints $p$ and $q$. Therefore every $0$-cocycle has the same value at every point of $M$. Conversely, if $\varphi(p)=a$ for all $p\in M$, then for every singular $1$-simplex with endpoints $p,q$, \begin{align*} (\delta\varphi)(\sigma)=\varphi(q)-\varphi(p)=a-a=0, \end{align*} so every constant $0$-cochain is a cocycle. There are no degree $-1$ singular cochains in this complex, so \begin{align*} B^0_{\mathrm{sing}}(M;\mathbb R)=0. \end{align*} Hence \begin{align*} H^0_{\mathrm{sing}}(M;\mathbb R) &=Z^0_{\mathrm{sing}}(M;\mathbb R)/B^0_{\mathrm{sing}}(M;\mathbb R)\\ &=Z^0_{\mathrm{sing}}(M;\mathbb R)/0\\ &\cong Z^0_{\mathrm{sing}}(M;\mathbb R)\\ &\cong \mathbb R, \end{align*} where the final isomorphism sends a constant cocycle $\varphi$ to its common value. On the de Rham side, a $0$-form is a smooth function $f:M\to\mathbb R$. The condition $df=0$ says that $f$ has zero derivative in every tangent direction. Along any smooth path $\gamma:[0,1]\to M$, \begin{align*} \frac{d}{dt}(f\circ\gamma)(t)=df_{\gamma(t)}(\gamma'(t))=0, \end{align*} so $f\circ\gamma$ is constant. Path-connectedness then forces $f$ to be constant on $M$. There are no degree $-1$ differential forms, so there are no exact $0$-forms. Therefore $H^0_{\mathrm{dR}}(M)\cong\mathbb R$ as well. Degree-zero cohomology records connected components. Since $M$ has exactly one path component, both singular cohomology and de Rham cohomology in degree $0$ consist of one real parameter: the constant value on that component. [/example] ## Integration as a Cochain Map What must be checked before integration over simplices can define a map on cohomology? A $k$-form can be pulled back to a smooth $k$-simplex and integrated over the standard simplex, giving a real number. The decisive point is compatibility with the two differentials: the [exterior derivative](/theorems/1525) $d$ on forms and the coboundary $\delta$ on singular cochains. [definition: Integration Cochain] Let $M$ be a smooth manifold and let $\omega \in \Omega^k(M)$. For a smooth singular $k$-simplex $\sigma:\Delta^k\to M$, define \begin{align*} I_k(\omega)(\sigma)=\int_{\Delta^k} \sigma^*\omega. \end{align*} Extend $I_k(\omega)$ linearly to smooth singular $k$-chains. [/definition] Smoothness of $\sigma$ is needed because $\sigma^*\omega$ uses derivatives of $\sigma$. For a merely continuous singular simplex, the pullback of a differential form is not defined in the usual smooth sense. The smoothing theorem is what permits this smooth-chain construction to recover ordinary singular cohomology. The orientation on $\Delta^k$ is the standard one induced by the ordered vertices $(e_0,\dots,e_k)$. With this convention, the boundary signs in singular homology match the boundary orientation in Stokes theorem. This compatibility is the critical local check for integration to descend from forms and chains to cohomology classes. If differentiating a form did not correspond to applying the coboundary after integration, closed forms would not reliably produce cocycles and exact forms would not reliably produce coboundaries. The comparison map therefore rests on the following Stokes-type identity for the integration cochain. [quotetheorem:3594] [citeproof:3594] This identity is the whole mechanism behind the comparison map. Closed forms give singular cocycles, and exact forms give singular coboundaries, so integration respects the equivalence relations on both sides. What remains is to promote the chain-level construction to a well-defined map between cohomology groups; the next statement records the induced homomorphism and the period formula that makes it computable. [quotetheorem:3595] [citeproof:3595] The formula says that a de Rham class is detected by its periods over cycles. In practice, to identify a class one chooses cycles representing a basis of homology, integrates the closed form over those cycles, and compares the resulting numbers with a known [dual basis](/theorems/414) in cohomology. On a compact connected oriented $n$-manifold, a normalized top-degree form represents the generator dual to the fundamental class when its integral over $M$ is $1$. The next example is the basic model for all period computations. [illustration:forms-s1-period-pairing] [example: The Period Pairing on the Circle] Let $S^1=\{(x,y)\in \mathbb R^2:x^2+y^2=1\}$ with its positive orientation, and let \begin{align*} \alpha=\frac{1}{2\pi}(-y\,dx+x\,dy)\big|_{S^1}\in\Omega^1(S^1). \end{align*} Identify the standard $1$-simplex $\Delta^1$ with the interval $[0,1]$ using the coordinate $t$. Let \begin{align*} \gamma:[0,1]\to S^1,\qquad \gamma(t)=(\cos 2\pi t,\sin 2\pi t). \end{align*} This path satisfies $\gamma(0)=(1,0)=\gamma(1)$, so it represents the positively oriented fundamental $1$-cycle of $S^1$. The period of $\alpha$ over the fundamental cycle is \begin{align*} I([\alpha])([\gamma])=1. \end{align*} Write the coordinate functions of $\gamma$ as \begin{align*} x(t)&=\cos 2\pi t,\\ y(t)&=\sin 2\pi t. \end{align*} Then \begin{align*} dx &= x'(t)\,dt = -2\pi\sin 2\pi t\,dt,\\ dy &= y'(t)\,dt = 2\pi\cos 2\pi t\,dt. \end{align*} Pulling back the numerator of $\alpha$ along $\gamma$ gives \begin{align*} \gamma^*(-y\,dx+x\,dy) &= -y(t)x'(t)\,dt+x(t)y'(t)\,dt\\ &= -\sin 2\pi t\left(-2\pi\sin 2\pi t\right)dt +\cos 2\pi t\left(2\pi\cos 2\pi t\right)dt\\ &= 2\pi\sin^2 2\pi t\,dt+2\pi\cos^2 2\pi t\,dt\\ &= 2\pi\left(\sin^2 2\pi t+\cos^2 2\pi t\right)dt\\ &= 2\pi\,dt. \end{align*} Therefore \begin{align*} \gamma^*\alpha &=\gamma^*\left(\frac{1}{2\pi}(-y\,dx+x\,dy)\big|_{S^1}\right)\\ &=\frac{1}{2\pi}\gamma^*(-y\,dx+x\,dy)\\ &=\frac{1}{2\pi}(2\pi\,dt)\\ &=dt. \end{align*} By the definition of the integration cochain, \begin{align*} I_1(\alpha)(\gamma) &=\int_{\Delta^1}\gamma^*\alpha\\ &=\int_0^1 dt\\ &=1-0\\ &=1. \end{align*} Since $\gamma$ represents the positively oriented fundamental homology class, this is exactly \begin{align*} I([\alpha])([\gamma])=1. \end{align*} The normalized angular form $\alpha$ has period $1$ around the positively oriented circle. Thus $[\alpha]$ is the de Rham cohomology class dual to the fundamental homology class of $S^1$. [/example] ## The De Rham Comparison Theorem Why should integration over cycles capture every singular cohomology class and identify only exact ambiguity? This is the central comparison question. Local contractibility makes both theories look the same on coordinate balls, while partitions of unity allow the local comparison to be assembled over the manifold. [quotetheorem:3596] [citeproof:3596] The theorem is stronger than equality of dimensions. It gives a canonical way to replace a differential form computation by a singular cochain computation, and conversely to represent real singular cohomology classes by closed differential forms. [remark: Ring Structure] The comparison is compatible with products: wedge product of forms corresponds to the cup product in singular cohomology. Thus [de Rham theorem](/theorems/3596) identifies $H^*_{\mathrm{dR}}(M)$ and $H^*_{\mathrm{sing}}(M;\mathbb R)$ as graded real algebras, after using the standard singular cup product conventions. [/remark] Naturality is especially useful when a map acts on top-degree cohomology. It turns the analytic operation of pulling back a volume form into the topological degree of a map. [example: Degree of a Smooth Self-Map of the Sphere] Let $n\ge 1$, let $S^n$ carry its positive orientation, and let $f:S^n\to S^n$ be smooth. Choose an orientation form $\omega\in\Omega^n(S^n)$ normalized by \begin{align*} \int_{S^n}\omega=1. \end{align*} Let $[S^n]\in H_n^{\mathrm{sing}}(S^n;\mathbb R)$ denote the positively oriented fundamental class. The topological degree of $f$ is the integer satisfying \begin{align*} f_*[S^n]=\deg(f)[S^n]. \end{align*} Let $I:H^n_{\mathrm{dR}}(S^n)\to H^n_{\mathrm{sing}}(S^n;\mathbb R)$ be the de Rham comparison map defined by integration. The degree is computed by the integral \begin{align*} \int_{S^n}f^*\omega=\deg(f). \end{align*} Since $S^n$ is an $n$-manifold, $\Omega^{n+1}(S^n)=0$, so \begin{align*} d\omega=0. \end{align*} Thus $\omega$ defines a de Rham cohomology class $[\omega]\in H^n_{\mathrm{dR}}(S^n)$. The class $[\omega]$ is nonzero. If $[\omega]=0$, then $\omega=d\eta$ for some $\eta\in\Omega^{n-1}(S^n)$, and *Generalized Stokes Theorem* gives \begin{align*} 1 &=\int_{S^n}\omega\\ &=\int_{S^n}d\eta\\ &=\int_{\partial S^n}\eta\\ &=\int_{\varnothing}\eta\\ &=0, \end{align*} a contradiction. By *[De Rham Cohomology Of Spheres](/theorems/3590)*, \begin{align*} H^n_{\mathrm{dR}}(S^n)\cong \mathbb R, \end{align*} so the nonzero class $[\omega]$ is a basis. Therefore there is a unique real number $\lambda$ such that \begin{align*} f^*[\omega]=\lambda[\omega]. \end{align*} This equality of cohomology classes means that for some $\eta\in\Omega^{n-1}(S^n)$, \begin{align*} f^*\omega-\lambda\omega=d\eta. \end{align*} Integrating both sides over $S^n$ gives \begin{align*} \int_{S^n}f^*\omega-\lambda\int_{S^n}\omega &=\int_{S^n}(f^*\omega-\lambda\omega)\\ &=\int_{S^n}d\eta\\ &=\int_{\partial S^n}\eta\\ &=0. \end{align*} Using $\int_{S^n}\omega=1$, this becomes \begin{align*} \int_{S^n}f^*\omega-\lambda=0, \end{align*} so \begin{align*} \int_{S^n}f^*\omega=\lambda. \end{align*} It remains to identify $\lambda$ with $\deg(f)$. By the naturality statement in *[De Rham Theorem](/theorems/3596)*, \begin{align*} I(f^*[\omega])=f_{\mathrm{sing}}^*I([\omega]). \end{align*} Evaluate this identity on the fundamental class: \begin{align*} \lambda &=\lambda\int_{S^n}\omega\\ &=I(\lambda[\omega])([S^n])\\ &=I(f^*[\omega])([S^n])\\ &=(f_{\mathrm{sing}}^*I([\omega]))([S^n])\\ &=I([\omega])(f_*[S^n])\\ &=I([\omega])(\deg(f)[S^n])\\ &=\deg(f)I([\omega])([S^n])\\ &=\deg(f)\int_{S^n}\omega\\ &=\deg(f). \end{align*} Thus \begin{align*} \int_{S^n}f^*\omega=\lambda=\deg(f). \end{align*} The normalized top-degree form $\omega$ represents the cohomology class dual to the fundamental class of the sphere. Pulling $\omega$ back by $f$ multiplies this top cohomology class by the same scalar by which $f$ multiplies the fundamental homology class. That scalar is the topological degree, and the normalization $\int_{S^n}\omega=1$ makes it equal to the single integral $\int_{S^n}f^*\omega$. [/example] ## Poincare Duality, Betti Numbers, and Euler Characteristic What topological information becomes available once de Rham cohomology has been identified with singular cohomology? The comparison theorem imports the major structural results of algebraic topology into differential forms. It also gives analytic formulas for invariants that were originally defined by triangulations or singular chains. [definition: De Rham Betti Number] If $H^k_{\mathrm{dR}}(M)$ is finite-dimensional, the degree $k$ Betti number of $M$ is \begin{align*} b_k(M)=\dim_{\mathbb R} H^k_{\mathrm{dR}}(M). \end{align*} [/definition] By [de Rham theorem](/theorems/3596), this agrees with $\dim_{\mathbb R}H^k_{\mathrm{sing}}(M;\mathbb R)$. This raises a useful invariance question: although $H^k_{\mathrm{dR}}(M)$ is built from smooth forms and derivatives, does its dimension depend only on the underlying topological space? The comparison with singular cohomology gives exactly that conclusion for homeomorphic smooth manifolds. [quotetheorem:3597] This consequence is conceptually important: the de Rham groups are defined using smooth forms, but their isomorphism type depends only on the topology. Smoothness is still needed to define the complex $\Omega^*(M)$ and to produce explicit pullback maps on forms; a general homeomorphism need not pull back smooth forms to smooth forms. [De Rham theorem](/theorems/3596) supplies the missing bridge by passing through singular cohomology, so homeomorphic smooth manifolds have isomorphic de Rham groups even when the homeomorphism is not a diffeomorphism. [quotetheorem:3598] [citeproof:3598] Poincare duality is usually obtained either from singular Poincare duality together with [de Rham theorem](/theorems/3596), or analytically from Hodge theory after choosing a Riemannian metric. The hypotheses matter: non-orientability removes the global fundamental class needed for this real pairing, and noncompactness can make the pairing degenerate unless compactly supported cohomology is used. For instance, $\mathbb R$ has $H^0_{\mathrm{dR}}(\mathbb R)\cong \mathbb R$ and $H^1_{\mathrm{dR}}(\mathbb R)=0$, so the symmetry $b_0=b_1$ fails without compactness. After recording individual Betti numbers and their duality symmetries, it is useful to compress the whole list into a single alternating invariant. The alternating signs make cancellations visible: paired even and odd cohomology groups offset each other, while unpaired groups contribute to a global count. This motivates defining Euler characteristic directly from the de Rham Betti numbers. [definition: Euler Characteristic] For a smooth manifold $M$ with finite-dimensional de Rham cohomology and only finitely many nonzero Betti numbers, its Euler characteristic is \begin{align*} \chi(M)=\sum_{k\ge 0}(-1)^k b_k(M). \end{align*} [/definition] This formula is often the most efficient way to compute $\chi(M)$ once de Rham cohomology is known. The equality with the topological Euler characteristic follows from [de Rham theorem](/theorems/3596) and the corresponding singular cohomology formula. [example: Euler Characteristic of the Circle] Let $S^1$ be the circle. Use the known de Rham cohomology groups \begin{align*} H^0_{\mathrm{dR}}(S^1)&\cong \mathbb R,\\ H^1_{\mathrm{dR}}(S^1)&\cong \mathbb R,\\ H^k_{\mathrm{dR}}(S^1)&=0 \quad \text{for every } k\ge 2. \end{align*} For each $k$, let \begin{align*} b_k(S^1)=\dim_{\mathbb R}H^k_{\mathrm{dR}}(S^1) \end{align*} be the degree $k$ de Rham Betti number. The Euler characteristic is \begin{align*} \chi(S^1)=\sum_{k\ge 0}(-1)^k b_k(S^1). \end{align*} The period computation above used \begin{align*} \alpha=\frac{1}{2\pi}(-y\,dx+x\,dy)\big|_{S^1} \end{align*} and the positively oriented loop \begin{align*} \gamma(t)=(\cos 2\pi t,\sin 2\pi t) \end{align*} to show \begin{align*} I([\alpha])([\gamma])=1. \end{align*} The Euler characteristic of the circle is \begin{align*} \chi(S^1)=0. \end{align*} Since vector-space isomorphisms preserve dimension, \begin{align*} b_0(S^1) &=\dim_{\mathbb R}H^0_{\mathrm{dR}}(S^1)\\ &=\dim_{\mathbb R}\mathbb R\\ &=1, \end{align*} and similarly \begin{align*} b_1(S^1) &=\dim_{\mathbb R}H^1_{\mathrm{dR}}(S^1)\\ &=\dim_{\mathbb R}\mathbb R\\ &=1. \end{align*} For every $k\ge 2$, \begin{align*} b_k(S^1) &=\dim_{\mathbb R}H^k_{\mathrm{dR}}(S^1)\\ &=\dim_{\mathbb R}0\\ &=0. \end{align*} Therefore the defining alternating sum becomes \begin{align*} \chi(S^1) &=\sum_{k\ge 0}(-1)^k b_k(S^1)\\ &=(-1)^0b_0(S^1)+(-1)^1b_1(S^1)+\sum_{k\ge 2}(-1)^k b_k(S^1)\\ &=1\cdot 1+(-1)\cdot 1+\sum_{k\ge 2}(-1)^k\cdot 0\\ &=1-1+0\\ &=0. \end{align*} The same data also identifies a concrete generator in degree one. Since \begin{align*} I([\alpha])([\gamma])=1\neq 0, \end{align*} the class $[\alpha]$ cannot be zero: if $[\alpha]=0$, then by linearity of the de Rham comparison map, \begin{align*} I([\alpha])([\gamma]) &=I(0)([\gamma])\\ &=0([\gamma])\\ &=0, \end{align*} contradicting the displayed period value. Thus $[\alpha]\neq 0$. Since \begin{align*} \dim_{\mathbb R}H^1_{\mathrm{dR}}(S^1)=1, \end{align*} every nonzero class in $H^1_{\mathrm{dR}}(S^1)$ spans it, so $[\alpha]$ is a generator of the degree-one de Rham cohomology group. The circle has one degree-zero cohomology parameter and one degree-one cohomology parameter. Their contributions to the Euler characteristic cancel: \begin{align*} \chi(S^1)=b_0(S^1)-b_1(S^1)=1-1=0. \end{align*} The normalized angular form $\alpha$ realizes the nonzero degree-one class whose presence is responsible for the negative term in this cancellation. [/example] The chapter closes the circle begun with exterior algebra. Alternating multilinear forms led to differential forms; the [exterior derivative](/theorems/1525) produced a cochain complex; Stokes theorem made integration compatible with boundaries; [de Rham theorem](/theorems/3596) identifies the resulting cohomology with singular cohomology. From this point onward, computations with differential forms can be read as computations of topological invariants. With de Rham's theorem establishing the link between differential forms and topological invariants, we now apply this machinery to concrete geometric problems. The final chapter demonstrates how questions about global manifold structure reduce to explicit calculations with differential forms. # 11. Selected Applications The earlier chapters built the basic machine of differential forms: exterior differentiation, pullback, integration on oriented manifolds, [Stokes' theorem](/theorems/1530), the Poincare lemma, Mayer-Vietoris, and de Rham cohomology. This final chapter shows how that machine turns global geometric questions into integrals of differential forms. The applications here are selective rather than exhaustive: degree detects how a sphere wraps around itself, Gauss-Bonnet turns curvature into topology, and Hodge theory explains why Riemannian geometry can choose canonical representatives of cohomology classes. The connecting theme is that integration over top-dimensional cycles pairs differential forms with topology. A closed form may carry more information than a local formula suggests, because its integral over a cycle can be unchanged under deformation. The degree, the Euler characteristic, and [harmonic representatives](/theorems/2747) are three ways this principle appears in geometry. ## Degree of Smooth Maps The first problem is to measure the global winding of a smooth map $f:S^n \to S^n$. Pointwise data such as the derivative can change under a homotopy, but the total signed number of times the domain covers the target should remain fixed. Differential forms give a compact definition: compare the integral of a volume form on the target with the integral of its pullback to the domain. [definition: Degree of a Smooth Map Between Spheres] Let $S^n$ carry its standard orientation. For a smooth map $f:S^n \to S^n$ and a volume form $\omega \in \Omega^n(S^n)$ with $\int_{S^n}\omega \ne 0$, the degree of $f$ is \begin{align*} \deg(f) = \frac{\int_{S^n} f^*\omega}{\int_{S^n} \omega}. \end{align*} [/definition] The definition uses a choice of $\omega$, but the answer is forced by the one-dimensionality of $H^n_{\mathrm{dR}}(S^n)$. The pullback $f^*$ acts on top cohomology by multiplication by a scalar, and the displayed ratio computes that scalar. For computations, the integral definition should agree with the geometric picture of counting preimages with signs. This requires choosing a regular value so that each preimage point has a nonsingular derivative and therefore a well-defined local orientation sign. The following result connects the cohomological scalar with that signed finite count. [quotetheorem:3599] [citeproof:3599] The regular-value formula is useful because it turns a global integral into local orientation data. The condition that $y$ is regular is essential: at a critical value the preimage need not be a finite collection of points, and the derivative may not decide an orientation sign. The sphere hypothesis keeps the target oriented and gives $H^n_{\mathrm{dR}}(S^n)\cong \mathbb R$, so pullback on top cohomology is multiplication by a single scalar. For a proper smooth map between compact connected oriented $n$-manifolds the same construction defines a degree; without compactness or properness, preimages can escape to infinity and the signed count need not be stable. [example: Antipodal Map on the Sphere] Let $A:S^n\to S^n$ be the antipodal map $A(x)=-x$. We orient $S^n\subset \mathbb R^{n+1}$ by the outward-normal convention: an ordered basis $(v_1,\ldots,v_n)$ of $T_pS^n$ is positive exactly when \begin{align*} (p,v_1,\ldots,v_n) \end{align*} is a positive ordered basis of $\mathbb R^{n+1}$. The ambient derivative of $A$ is the [linear map](/page/Linear%20Map) $-I_{n+1}$, so \begin{align*} \det(-I_{n+1})=(-1)(-1)\cdots(-1)=(-1)^{n+1}. \end{align*} The [degree of the antipodal map](/theorems/2247) is \begin{align*} \deg(A)=(-1)^{n+1}. \end{align*} Fix $y\in S^n$. The equation $A(x)=y$ means \begin{align*} -x&=y,\\ x&=-y, \end{align*} so \begin{align*} A^{-1}(y)=\{-y\}. \end{align*} The derivative at $-y$ is the restriction of $-I_{n+1}$ to the tangent space: \begin{align*} DA_{-y}:T_{-y}S^n&\to T_yS^n,\\ v&\mapsto -v. \end{align*} Since $-I_{n+1}$ is invertible, this restricted map is an isomorphism, so every $y\in S^n$ is a regular value. Choose a positive oriented basis $(v_1,\ldots,v_n)$ of $T_{-y}S^n$. By the outward-normal convention, \begin{align*} (-y,v_1,\ldots,v_n) \end{align*} is a positive basis of $\mathbb R^{n+1}$. Its image tangent basis at $y$ is \begin{align*} (DA_{-y}v_1,\ldots,DA_{-y}v_n)=(-v_1,\ldots,-v_n). \end{align*} To test the orientation of this basis in $T_yS^n$, prepend the outward normal $y$: \begin{align*} (y,-v_1,\ldots,-v_n) &=(-1)\cdot(-1)\cdots(-1)\,(-y,v_1,\ldots,v_n)\\ &=(-1)^{n+1}(-y,v_1,\ldots,v_n). \end{align*} Thus the ordered basis $(-v_1,\ldots,-v_n)$ of $T_yS^n$ is positive if $(-1)^{n+1}=1$ and negative if $(-1)^{n+1}=-1$. Therefore \begin{align*} \operatorname{sgn}(JA_{-y})=(-1)^{n+1}. \end{align*} By *Degree Theorem for Spheres*, the degree is the signed count over any regular value. Since $y$ has exactly one preimage, \begin{align*} \deg(A) &=\sum_{x\in A^{-1}(y)}\operatorname{sgn}(JA_x)\\ &=\operatorname{sgn}(JA_{-y})\\ &=(-1)^{n+1}. \end{align*} The antipodal map preserves the orientation of $S^n$ when $n$ is odd and reverses it when $n$ is even. Its degree is exactly the orientation sign forced by applying $-I_{n+1}$ to the outward normal together with an oriented tangent basis. [/example] The antipodal computation has a sharp topological consequence: in even dimensions $n=2k$, $\deg(A)=-1\ne 1=\deg(\operatorname{id})$, so the antipodal map is not homotopic to the identity on $S^{2k}$. This is the precise obstruction that powers the [hairy ball theorem](/theorems/2248) and the Borsuk-Ulam theorem in even dimensions. In odd dimensions, by contrast, $\deg(A)=+1$, consistent with the fact that $S^{2k-1}$ admits nowhere-vanishing tangent vector fields whose flows interpolate between the identity and the antipodal map. The degree is also a faithful invariant in lower dimensions. For $n=1$, the degree recovers the classical [winding number](/page/Winding%20Number) of a map $S^1\to S^1$: the map $z\mapsto z^m$ wraps the circle $m$ times around itself, so its degree is $m$ for $m\ge 1$ and $-|m|$ for $m\le -1$. Two circle maps are homotopic if and only if they have the same degree, so $[S^1,S^1]\cong\mathbb Z$ via the degree. The Hopf degree theorem extends this to $[S^n,S^n]\cong\mathbb Z$ for every $n\ge 1$. This turns the classical existence of complex roots into a test case for degree theory. A nonconstant polynomial has a boundary map on a large circle that winds nontrivially around the origin; if the polynomial had no zero inside, that winding would have to extend across the disk and become trivial. For degree theory, the essential problem is to turn this boundary winding obstruction into an existence statement about zeros. The [fundamental theorem of algebra](/theorems/347) supplies exactly the formal target: every nonconstant complex polynomial must have a zero, so the nonzero degree seen at infinity cannot be filled in by a zero-free disk map. [quotetheorem:347] [citeproof:347] The theorem is a model application of de Rham cohomology: the obstruction to extending a circle map over the disk is detected by integrating a closed $1$-form around the boundary. Three structural inputs are worth isolating. First, the closed disk is compact with boundary $S^1$, so [Stokes' theorem](/theorems/1530) forces the boundary integral of an exact form to vanish; on a non-compact replacement of the disk this conclusion would fail. Second, the disk is contractible, so the Poincare lemma converts every closed form into an exact form there; on a domain with a hole, the closed pullback could carry a nonzero period. Third, the boundary map $F_R$ would have to be smooth on the entire disk for the pullback to be defined, which is why the hypothetical absence of a zero matters: a zero of $p$ on the disk is precisely the singularity that would prevent the extension. The same circle of ideas generalises far beyond polynomials. Replacing $S^1\to S^1$ with $S^n\to S^n$ and the disk with $\overline{B}(0,1)\subset\mathbb R^{n+1}$ gives Brouwer's fixed-point theorem and the no-retraction principle, since a continuous retraction of the ball onto its boundary would contradict the degree of the identity map. The same degree machinery underlies the Hopf degree theorem identifying $[S^n,S^n]$ with $\mathbb Z$, the topological proof of the [hairy ball theorem](/theorems/2248) on $S^{2k}$, and the more general Poincare-Hopf theorem relating vector field indices to the Euler characteristic. The [fundamental theorem of algebra](/theorems/347) is the one-dimensional, polynomial-flavoured instance of this pattern. [illustration:forms-fta-winding-obstruction] [example: The Form Detecting Winding Number] Let \begin{align*} U=\mathbb R^2\setminus\{0\}. \end{align*} On $U$, define the $1$-form \begin{align*} \alpha=\frac{x\,dy-y\,dx}{x^2+y^2}. \end{align*} The denominator is nonzero on $U$, since $x^2+y^2=0$ holds only at $(0,0)$. Let \begin{align*} \gamma:[0,2\pi]&\to U, & \gamma(\theta)&=(\cos\theta,\sin\theta) \end{align*} be the positively oriented parametrization of the unit circle. The integral of $\alpha$ around $\gamma$ is \begin{align*} \int_\gamma \alpha=2\pi, \end{align*} and $\alpha$ is not exact on $\mathbb R^2\setminus\{0\}$. Consequently, $\alpha$ represents a nonzero class in $H^1_{\mathrm{dR}}(\mathbb R^2\setminus\{0\})$. First compute the pullback of the coordinate functions and their differentials: \begin{align*} \gamma^*x&=x\circ\gamma=\cos\theta, & \gamma^*y&=y\circ\gamma=\sin\theta,\\ \gamma^*dx&=d(\gamma^*x)=d(\cos\theta)=-\sin\theta\,d\theta, & \gamma^*dy&=d(\gamma^*y)=d(\sin\theta)=\cos\theta\,d\theta. \end{align*} Therefore \begin{align*} \gamma^*\alpha &=\gamma^*\left(\frac{x\,dy-y\,dx}{x^2+y^2}\right)\\ &=\frac{(\gamma^*x)(\gamma^*dy)-(\gamma^*y)(\gamma^*dx)}{(\gamma^*x)^2+(\gamma^*y)^2}\\ &=\frac{(\cos\theta)(\cos\theta\,d\theta)-(\sin\theta)(-\sin\theta\,d\theta)}{\cos^2\theta+\sin^2\theta}\\ &=\frac{\cos^2\theta\,d\theta+\sin^2\theta\,d\theta}{1}\\ &=(\cos^2\theta+\sin^2\theta)\,d\theta\\ &=d\theta. \end{align*} Hence \begin{align*} \int_\gamma\alpha &=\int_{0}^{2\pi}\gamma^*\alpha\\ &=\int_{0}^{2\pi}d\theta\\ &=\int_{0}^{2\pi}1\,d\theta\\ &=2\pi-0\\ &=2\pi. \end{align*} Next verify that $\alpha$ is closed. Write \begin{align*} \alpha=P\,dx+Q\,dy, \qquad P=-\frac{y}{x^2+y^2}, \qquad Q=\frac{x}{x^2+y^2}. \end{align*} Using $dx\wedge dx=0$, $dy\wedge dy=0$, and $dy\wedge dx=-dx\wedge dy$, \begin{align*} d\alpha &=dP\wedge dx+dQ\wedge dy\\ &=\left(\frac{\partial P}{\partial x}\,dx+\frac{\partial P}{\partial y}\,dy\right)\wedge dx +\left(\frac{\partial Q}{\partial x}\,dx+\frac{\partial Q}{\partial y}\,dy\right)\wedge dy\\ &=\frac{\partial P}{\partial y}\,dy\wedge dx+\frac{\partial Q}{\partial x}\,dx\wedge dy\\ &=\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dx\wedge dy. \end{align*} The two partial derivatives are \begin{align*} \frac{\partial Q}{\partial x} &=\frac{(x^2+y^2)\cdot 1-x\cdot 2x}{(x^2+y^2)^2}\\ &=\frac{x^2+y^2-2x^2}{(x^2+y^2)^2}\\ &=\frac{y^2-x^2}{(x^2+y^2)^2}, \end{align*} and \begin{align*} \frac{\partial P}{\partial y} &=\frac{(x^2+y^2)(-1)-(-y)(2y)}{(x^2+y^2)^2}\\ &=\frac{-x^2-y^2+2y^2}{(x^2+y^2)^2}\\ &=\frac{y^2-x^2}{(x^2+y^2)^2}. \end{align*} Thus \begin{align*} \frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y} &=\frac{y^2-x^2}{(x^2+y^2)^2}-\frac{y^2-x^2}{(x^2+y^2)^2}\\ &=0, \end{align*} so \begin{align*} d\alpha=0. \end{align*} It remains to rule out exactness. Suppose, for contradiction, that $\alpha=d\beta$ for some smooth function $\beta:U\to\mathbb R$. Since pullback commutes with exterior differentiation on functions, \begin{align*} \gamma^*\alpha &=\gamma^*(d\beta)\\ &=d(\beta\circ\gamma). \end{align*} Integrating over $[0,2\pi]$ gives \begin{align*} \int_\gamma\alpha &=\int_0^{2\pi}d(\beta\circ\gamma)\\ &=\int_0^{2\pi}\frac{d}{d\theta}\beta(\cos\theta,\sin\theta)\,d\theta\\ &=\beta(\cos(2\pi),\sin(2\pi))-\beta(\cos 0,\sin 0)\\ &=\beta(1,0)-\beta(1,0)\\ &=0. \end{align*} This contradicts the earlier computation \begin{align*} \int_\gamma\alpha=2\pi. \end{align*} Therefore $\alpha$ is not exact on $U$. The form $\alpha$ is closed but not exact on $\mathbb R^2\setminus\{0\}$. Its integral around the unit circle is $2\pi$, so it detects one counterclockwise winding around the missing origin. Since a nonzero integral over a closed curve is impossible for a globally exact $1$-form, the class $[\alpha]$ is nonzero in $H^1_{\mathrm{dR}}(\mathbb R^2\setminus\{0\})$. [/example] ## Gauss-Bonnet and Curvature as a Cohomology Class The next problem is more geometric: can the total curvature of a surface remember its topology? Gaussian curvature is defined from a Riemannian metric and may vary from point to point, while the Euler characteristic is a homotopy invariant. Gauss-Bonnet says that the integral of curvature is exactly the bridge between them. [definition: Euler Characteristic of a Closed Surface] Let $M$ be a compact connected oriented smooth surface. Its Euler characteristic is \begin{align*} \chi(M)=b_0(M)-b_1(M)+b_2(M), \qquad b_k(M)=\dim H^k_{\mathrm{dR}}(M). \end{align*} [/definition] For a connected closed oriented surface, $b_0=b_2=1$, so the topology is controlled by $b_1$. If $M$ has genus $g$, then $b_1=2g$ and $\chi(M)=2-2g$. To compare this topological number with geometry, curvature must be packaged as something integrable over the whole surface. The metric supplies a pointwise Gaussian curvature $K$ and an oriented area form $dA$; multiplying them produces the global $2$-form whose integral can be paired with the fundamental class. [definition: Curvature Two-Form of an Oriented Surface] Let $(M,g)$ be an oriented Riemannian surface. Let $K:M\to\mathbb R$ be its Gaussian curvature and let $dA$ be its Riemannian area form. The curvature two-form is \begin{align*} \Omega_K = K\,dA \in \Omega^2(M). \end{align*} [/definition] Since every $2$-form on a surface is closed, $\Omega_K$ defines a de Rham cohomology class. The natural question is whether this metric-dependent class has a metric-independent period. Gauss-Bonnet answers by saying that the total curvature is forced to equal a purely topological invariant: after division by $2\pi$, the curvature class is the Euler class of the tangent bundle. [quotetheorem:3600] [citeproof:3600] This proof places Gauss-Bonnet inside the same framework as the earlier chapters: local differential forms are integrated, [Stokes' theorem](/theorems/1530) moves the information to boundaries, and cancellation leaves a topological invariant. Each hypothesis pulls its weight. Orientability gives a global area form $dA$ and a consistent sign on each triangle, so the boundary integrals cancel between adjacent faces rather than doubling up; on a non-orientable surface such as $\mathbb RP^2$ or the Klein bottle, the global area form fails to exist and the statement must be reformulated through the Euler class of the orthonormal frame bundle, with $K\,dA$ promoted to a twisted-density. Closedness of $M$ is equally essential: any boundary $\partial M$ would contribute an extra geodesic-curvature term $\int_{\partial M}\kappa_g\,ds$, recovering the boundary version $\int_M K\,dA + \int_{\partial M}\kappa_g\,ds = 2\pi\chi(M)$. Smoothness lets the connection $1$-forms be differentiated, while finiteness of the triangulation lets the vertex contributions add to a finite combinatorial number. On a non-compact surface, the integral $\int_M K\,dA$ need not converge, and even when it does, no triangulation identity links it to a Betti-number sum. [illustration:forms-triangulated-surface-curvature] [explanation: Cohomological Form of Gauss-Bonnet] The curvature form gives a class \begin{align*} \left[\frac{\Omega_K}{2\pi}\right]\in H^2_{\mathrm{dR}}(M). \end{align*} Gauss-Bonnet says that pairing this class with the fundamental class $[M]$ gives the Euler characteristic: \begin{align*} \int_M \frac{\Omega_K}{2\pi}=\chi(M). \end{align*} In this form, the metric dependence has disappeared from the final number. Different Riemannian metrics change the representative $\Omega_K$, but not its cohomology class when paired with $[M]$. [/explanation] Two examples make the metric-independence concrete. The first inspects a single closed surface — the round two-sphere — under metrics of different sizes and verifies that the total curvature is unchanged. The second reads Gauss-Bonnet in the opposite direction: fixing the topology of a surface forces the total curvature to take a specific value, regardless of how the metric is deformed. [example: Round Two-Sphere] Let $S^2_R\subset\mathbb R^3$ be the round sphere of radius $R>0$, with outward orientation. Thus \begin{align*} S^2_R=\{(x,y,z)\in\mathbb R^3:x^2+y^2+z^2=R^2\}. \end{align*} Let $K$ be its Gaussian curvature and let $dA$ be its Riemannian area form. The curvature integral is \begin{align*} \int_{S^2_R}K\,dA=4\pi. \end{align*} Consequently, *Gauss-Bonnet Theorem for Closed Oriented Surfaces* gives \begin{align*} \chi(S^2)=2. \end{align*} Parametrize the sphere away from the poles by \begin{align*} \Phi:(0,\pi)\times(0,2\pi)&\to S^2_R,\\ \Phi(\varphi,\theta)&=(R\sin\varphi\cos\theta,\ R\sin\varphi\sin\theta,\ R\cos\varphi). \end{align*} The coordinate tangent vectors are \begin{align*} \Phi_\varphi &=(R\cos\varphi\cos\theta,\ R\cos\varphi\sin\theta,\ -R\sin\varphi),\\ \Phi_\theta &=(-R\sin\varphi\sin\theta,\ R\sin\varphi\cos\theta,\ 0). \end{align*} Their cross product is \begin{align*} \Phi_\varphi\times\Phi_\theta &=(R^2\sin^2\varphi\cos\theta,\ R^2\sin^2\varphi\sin\theta,\ R^2\sin\varphi\cos\varphi)\\ &=R^2\sin\varphi(\sin\varphi\cos\theta,\ \sin\varphi\sin\theta,\ \cos\varphi). \end{align*} For $0<\varphi<\pi$, $\sin\varphi>0$, so this vector points in the outward normal direction. Its norm is \begin{align*} |\Phi_\varphi\times\Phi_\theta| &=R^2\sin\varphi \sqrt{\sin^2\varphi\cos^2\theta+\sin^2\varphi\sin^2\theta+\cos^2\varphi}\\ &=R^2\sin\varphi \sqrt{\sin^2\varphi(\cos^2\theta+\sin^2\theta)+\cos^2\varphi}\\ &=R^2\sin\varphi \sqrt{\sin^2\varphi+\cos^2\varphi}\\ &=R^2\sin\varphi. \end{align*} Therefore the area is \begin{align*} \operatorname{Area}(S^2_R) &=\int_0^{2\pi}\int_0^\pi R^2\sin\varphi\,d\varphi\,d\theta\\ &=\int_0^{2\pi}R^2[-\cos\varphi]_{\varphi=0}^{\varphi=\pi}\,d\theta\\ &=\int_0^{2\pi}R^2(-\cos\pi+\cos0)\,d\theta\\ &=\int_0^{2\pi}R^2(1+1)\,d\theta\\ &=\int_0^{2\pi}2R^2\,d\theta\\ &=2R^2(2\pi-0)\\ &=4\pi R^2. \end{align*} The outward unit normal is \begin{align*} N(p)=\frac{p}{R}. \end{align*} For $v\in T_pS^2_R$, differentiating $N(p)=p/R$ gives \begin{align*} dN_p(v)=\frac{1}{R}v. \end{align*} Thus, in any orthonormal basis of $T_pS^2_R$, the matrix of $dN_p$ is \begin{align*} \begin{pmatrix} 1/R&0\\ 0&1/R \end{pmatrix}. \end{align*} Hence the Gaussian curvature is \begin{align*} K &=\det(dN_p)\\ &=\det \begin{pmatrix} 1/R&0\\ 0&1/R \end{pmatrix}\\ &=\frac{1}{R}\cdot\frac{1}{R}-0\cdot0\\ &=\frac{1}{R^2}. \end{align*} Since $K$ is constant, linearity of integration gives \begin{align*} \int_{S^2_R}K\,dA &=\int_{S^2_R}\frac{1}{R^2}\,dA\\ &=\frac{1}{R^2}\int_{S^2_R}dA\\ &=\frac{1}{R^2}\operatorname{Area}(S^2_R)\\ &=\frac{1}{R^2}(4\pi R^2)\\ &=4\pi\frac{R^2}{R^2}\\ &=4\pi. \end{align*} By *Gauss-Bonnet Theorem for Closed Oriented Surfaces*, \begin{align*} \int_{S^2_R}K\,dA=2\pi\chi(S^2_R). \end{align*} Substituting the computed integral gives \begin{align*} 4\pi&=2\pi\chi(S^2_R). \end{align*} Since $2\pi\ne0$, division by $2\pi$ gives \begin{align*} \chi(S^2_R) &=\frac{4\pi}{2\pi}\\ &=2. \end{align*} The sphere $S^2_R$ is diffeomorphic to $S^2$, so \begin{align*} \chi(S^2)=2. \end{align*} The pointwise curvature $K=1/R^2$ decreases as the radius increases, while the total area $4\pi R^2$ increases by the reciprocal factor. Their product is always $4\pi$. Thus the total curvature is independent of the radius and records the topological invariant $\chi(S^2)=2$. [/example] For surfaces of higher genus, the same equation runs in the opposite direction: the topology is fixed first, and Gauss-Bonnet then constrains the total curvature. This is the form in which the theorem becomes a rigidity statement, since no smooth metric on a genus-$g$ surface can violate the prescribed total. The torus case is especially striking, because it forces any embedded torus in $\mathbb R^3$ to balance positive and negative curvature exactly. [example: Surfaces of Genus g] Let $M_g$ be a compact connected oriented smooth surface of genus $g$, and let $g_R$ be any Riemannian metric on $M_g$. Let $K:M_g\to\mathbb R$ be the Gaussian curvature of this metric, and let $dA$ be its Riemannian area form. The Euler characteristic of a compact connected oriented surface of genus $g$ is \begin{align*} \chi(M_g)=2-2g. \end{align*} Every Riemannian metric on $M_g$ satisfies \begin{align*} \int_{M_g}K\,dA=2\pi(2-2g). \end{align*} Since $M_g$ is compact, connected, oriented, and two-dimensional, *Gauss-Bonnet Theorem for Closed Oriented Surfaces* applies. Therefore \begin{align*} \int_{M_g}K\,dA &=2\pi\chi(M_g). \end{align*} Substituting the genus formula for the Euler characteristic gives \begin{align*} \int_{M_g}K\,dA &=2\pi(2-2g). \end{align*} Expanding the right-hand side, \begin{align*} 2\pi(2-2g) &=(2\pi)\cdot 2-(2\pi)\cdot 2g\\ &=4\pi-4\pi g\\ &=4\pi(1-g). \end{align*} Thus the same identity may also be written as \begin{align*} \int_{M_g}K\,dA=4\pi(1-g). \end{align*} For $g=1$, the formula gives \begin{align*} \int_{M_1}K\,dA &=2\pi(2-2\cdot 1)\\ &=2\pi(2-2)\\ &=2\pi\cdot 0\\ &=0. \end{align*} Thus every Riemannian metric on a torus has total curvature zero. If a torus is embedded in $\mathbb R^3$ with its induced Riemannian metric, then the same calculation applies to that induced metric: \begin{align*} \int_{M_1}K\,dA=0. \end{align*} Writing the integral as a sum over the regions where $K$ is positive, negative, and zero, \begin{align*} 0 &=\int_{M_1}K\,dA\\ &=\int_{\{K>0\}}K\,dA+\int_{\{K<0\}}K\,dA+\int_{\{K=0\}}K\,dA\\ &=\int_{\{K>0\}}K\,dA+\int_{\{K<0\}}K\,dA+0. \end{align*} Hence \begin{align*} \int_{\{K>0\}}K\,dA &=-\int_{\{K<0\}}K\,dA. \end{align*} So the positive and negative curvature contributions cancel in total. For a flat quotient torus, the curvature satisfies \begin{align*} K(p)=0 \end{align*} for every $p\in M_1$. Therefore \begin{align*} \int_{M_1}K\,dA &=\int_{M_1}0\,dA\\ &=0, \end{align*} which agrees with the genus-one case of Gauss-Bonnet. The total curvature of a compact connected oriented genus-$g$ surface is fixed by topology: \begin{align*} \int_{M_g}K\,dA=2\pi(2-2g). \end{align*} Changing the Riemannian metric can move curvature around the surface, but it cannot change this total. For a torus, the forced total is zero; for a sphere it is $4\pi$; and for every genus $g\ge 2$, it is negative. [/example] ## Harmonic Representatives and the Hodge Decomposition The last problem asks for a preferred representative of a de Rham cohomology class. A class is an equivalence class of closed forms modulo exact forms, so it usually has many representatives. Once a Riemannian metric is chosen on a compact manifold, Hodge theory selects the representative satisfying an elliptic differential equation. [example: Oscillating Representatives on the Circle] Let $S^1=\mathbb R/(2\pi\mathbb Z)$ with angular coordinate $\theta$ and the standard metric \begin{align*} g=d\theta^2. \end{align*} For $N\in\mathbb N$, define \begin{align*} \alpha_0&=d\theta,\\ \alpha_N&=d\theta+d(\sin N\theta). \end{align*} The forms $\alpha_0$ and $\alpha_N$ represent the same class in $H^1_{\mathrm{dR}}(S^1)$, but the representative $\alpha_N$ has oscillation and pointwise size depending on $N$. Since $\sin(N(\theta+2\pi))=\sin(N\theta+2\pi N)=\sin(N\theta)$, the function \begin{align*} f_N:S^1&\to\mathbb R, & f_N(\theta)&=\sin(N\theta) \end{align*} is well-defined. Its differential is \begin{align*} df_N &=d(\sin N\theta)\\ &=N\cos(N\theta)\,d\theta. \end{align*} Therefore \begin{align*} \alpha_N &=d\theta+d(\sin N\theta)\\ &=d\theta+N\cos(N\theta)\,d\theta\\ &=(1+N\cos(N\theta))\,d\theta. \end{align*} Both forms are closed. Indeed, \begin{align*} d\alpha_0 &=d(d\theta)\\ &=0, \end{align*} and \begin{align*} d\alpha_N &=d\bigl((1+N\cos(N\theta))\,d\theta\bigr)\\ &=d(1+N\cos(N\theta))\wedge d\theta\\ &=(-N^2\sin(N\theta)\,d\theta)\wedge d\theta\\ &=-N^2\sin(N\theta)\,(d\theta\wedge d\theta)\\ &=0. \end{align*} Their difference is exact: \begin{align*} \alpha_N-\alpha_0 &=\bigl(d\theta+d(\sin N\theta)\bigr)-d\theta\\ &=d(\sin N\theta). \end{align*} Thus $\alpha_N$ and $\alpha_0$ differ by an exact $1$-form, so they define the same de Rham cohomology class: \begin{align*} [\alpha_N]=[\alpha_0]\in H^1_{\mathrm{dR}}(S^1). \end{align*} Now compute the pointwise size. For the standard metric $g=d\theta^2$, the covector $d\theta$ has norm \begin{align*} |d\theta|_g=1. \end{align*} Hence \begin{align*} |\alpha_N|_g &=|(1+N\cos(N\theta))\,d\theta|_g\\ &=|1+N\cos(N\theta)|\,|d\theta|_g\\ &=|1+N\cos(N\theta)|. \end{align*} At $\theta=0$, \begin{align*} |\alpha_N(0)|_g &=|1+N\cos 0|\\ &=1+N. \end{align*} If $N\ge 1$, then at $\theta=\pi/N$, \begin{align*} |\alpha_N(\pi/N)|_g &=|1+N\cos\pi|\\ &=|1-N|. \end{align*} The coefficient $1+N\cos(N\theta)$ oscillates between \begin{align*} 1-N \end{align*} and \begin{align*} 1+N. \end{align*} So the added exact term changes the pointwise representative by an amount whose size is controlled by $N$. The forms $d\theta$ and $d\theta+d(\sin N\theta)$ lie in the same de Rham cohomology class because their difference is exact. Nevertheless, the second representative has coefficient $1+N\cos(N\theta)$, so its oscillation and pointwise norm depend on $N$. This shows that a cohomology class does not by itself choose a geometrically preferred representative. [/example] This example shows why a quotient description does not by itself choose a canonical form. To select preferred representatives, de Rham theory must be supplemented by metric data that can measure the size and orthogonality of forms. The first problem is to integrate inner products of differential forms in a coordinate-free way. At each point $p\in M$, a Riemannian metric induces a pointwise inner product $(\cdot,\cdot)_g$ on $\Lambda^kT_p^*M$; the Hodge star is the operator that converts this pointwise measurement into wedge products against the Riemannian volume form. [definition: Hodge Star] Let $(M,g)$ be an oriented Riemannian $n$-manifold with Riemannian volume form $dV_g$, and let $(\cdot,\cdot)_g$ denote the fiberwise inner product induced by $g$ on each exterior power $\Lambda^kT_p^*M$. The Hodge star is the bundle map $*:\Lambda^kT^*M\to \Lambda^{n-k}T^*M$, extended to global sections as $*:\Omega^k(M)\to \Omega^{n-k}(M)$, determined pointwise by \begin{align*} \alpha_p\wedge (*\beta)_p = (\alpha_p,\beta_p)_g\,(dV_g)_p \end{align*} for all $\alpha_p,\beta_p\in \Lambda^kT_p^*M$ and all $p\in M$. [/definition] The Hodge star turns the metric into an $L^2$ inner product on forms. For compact $M$, define \begin{align*} (\alpha,\beta)_{L^2}=\int_M \alpha\wedge *\beta. \end{align*} Closed forms are insensitive to adding exact terms, so a metric choice still needs a way to reject the directions that come from exact noise. The right orthogonality condition is expressed by the adjoint of the exterior derivative with respect to the $L^2$ inner product. This adjoint is the codifferential, and it measures the complementary first-order condition to closedness. [definition: Codifferential] Let $(M,g)$ be a compact oriented Riemannian manifold. The codifferential $d^*:\Omega^k(M)\to\Omega^{k-1}(M)$ is the formal adjoint of $d$ with respect to the $L^2$ inner product: \begin{align*} (d\alpha,\beta)_{L^2}=(\alpha,d^*\beta)_{L^2} \end{align*} for all compatible degrees. [/definition] The codifferential supplies the missing orthogonality condition, but using it effectively requires more than keeping a second equation beside closedness. The next step is to combine $d\alpha=0$ and $d^*\alpha=0$ into one operator whose kernel can be studied by elliptic theory. This operator is the Hodge Laplacian, and it is the analytic mechanism that turns the two balance conditions into a finite-dimensional space of canonical candidates. [definition: Hodge Laplacian] Let $(M,g)$ be a compact oriented Riemannian manifold. The Hodge Laplacian on $k$-forms is \begin{align*} \Delta = dd^*+d^*d:\Omega^k(M)\to\Omega^k(M). \end{align*} [/definition] The Hodge Laplacian is a second-order elliptic operator that reduces, on functions in Euclidean space, to minus the ordinary Laplacian. Its symmetry under the $L^2$ inner product follows from the adjoint relation between $d$ and $d^*$, and this symmetry is the reason its kernel singles out a distinguished subspace of forms. The remaining obstruction is uniqueness inside a cohomology class: arbitrary representatives can be changed by exact terms, so there must be a condition that selects the balanced representative and excludes all nonzero exact perturbations. Forms in the kernel of the Hodge Laplacian satisfy precisely this balance condition; they are the candidates for canonical representatives of de Rham cohomology classes. [definition: Harmonic Form] Let $(M,g)$ be a compact oriented Riemannian manifold. A $k$-form $\alpha\in\Omega^k(M)$ is harmonic if \begin{align*} \Delta\alpha=0. \end{align*} The [vector space](/page/Vector%20Space) of harmonic $k$-forms is denoted $\mathcal H^k(M)$. [/definition] On a compact manifold without boundary, the identity \begin{align*} (\Delta\alpha,\alpha)_{L^2}=\|d\alpha\|_{L^2}^2+\|d^*\alpha\|_{L^2}^2 \end{align*} shows that a harmonic form is exactly a form satisfying $d\alpha=0$ and $d^*\alpha=0$. The central question is whether this analytic condition merely produces special examples or actually captures every de Rham cohomology class. If each class has one and only one harmonic representative, then the quotient space of closed forms modulo exact forms can be replaced by a finite-dimensional space of solutions to an elliptic equation. Hodge's theorem is the statement that this replacement is valid on compact oriented Riemannian manifolds. [quotetheorem:2745] [citeproof:2745] The theorem relies on elliptic regularity and Fredholm theory for the Hodge Laplacian, which go beyond the differential-form tools developed in this course. The hypotheses are doing real work. Compactness gives Fredholm behaviour and finite-dimensional harmonic spaces; on non-compact manifolds, harmonic forms can fail to represent de Rham cohomology in this clean way. The metric is also essential: without it there is no Hodge star, no codifferential, and no Laplacian selecting a preferred representative. Orientability is part of the ordinary-form presentation used here, since it supplies a global volume form for the Hodge star and the integral pairing; non-orientable manifolds require density or twisted-coefficient versions. In these notes, the theorem is used as a structural statement explaining how analysis refines de Rham cohomology. [example: Harmonic Forms on the Flat Torus] Let \begin{align*} T^2=S^1\times S^1=\mathbb R^2/(2\pi\mathbb Z)^2 \end{align*} with angular coordinates $(\theta,\phi)$ and flat product metric \begin{align*} g=d\theta^2+d\phi^2. \end{align*} Orient $T^2$ by the area form \begin{align*} dA=d\theta\wedge d\phi. \end{align*} The $1$-forms $d\theta$ and $d\phi$ are the translation-invariant forms descending from $\mathbb R^2$. The harmonic forms on the flat torus are \begin{align*} \mathcal H^0(T^2)&=\operatorname{span}\{1\},\\ \mathcal H^1(T^2)&=\operatorname{span}\{d\theta,d\phi\},\\ \mathcal H^2(T^2)&=\operatorname{span}\{d\theta\wedge d\phi\}. \end{align*} The coframe $(d\theta,d\phi)$ is orthonormal for the flat product metric, so \begin{align*} (d\theta,d\theta)_g&=1, & (d\phi,d\phi)_g&=1, & (d\theta,d\phi)_g&=0. \end{align*} By the defining rule for the Hodge star, \begin{align*} \alpha\wedge *\beta=(\alpha,\beta)_g\,dA. \end{align*} Thus \begin{align*} *1&=d\theta\wedge d\phi,\\ *d\theta&=d\phi, \end{align*} because \begin{align*} d\theta\wedge dphi=dA. \end{align*} Also \begin{align*} *d\phi&=-d\theta, \end{align*} because \begin{align*} d\phi\wedge(-d\theta) &=-d\phi\wedge d\theta\\ &=d\theta\wedge d\phi\\ &=dA. \end{align*} Finally, \begin{align*} *(d\theta\wedge d\phi)=1, \end{align*} because \begin{align*} (d\theta\wedge d\phi)\wedge 1=dA. \end{align*} Now compute exterior derivatives. Since $d^2=0$, \begin{align*} d(d\theta)&=0, & d(d\phi)&=0. \end{align*} Also \begin{align*} d(d\theta\wedge d\phi) &=d(d\theta)\wedge d\phi-d\theta\wedge d(d\phi)\\ &=0\wedge d\phi-d\theta\wedge0\\ &=0. \end{align*} Hence $1$, $d\theta$, $d\phi$, and $d\theta\wedge d\phi$ are closed in their respective degrees. For the codifferential on a $k$-form on this oriented $2$-manifold, use \begin{align*} d^*\eta=(-1)^{2(k+1)+1}*d*\eta=-*d*\eta. \end{align*} For $0$-forms, $d^*$ has target $\Omega^{-1}(T^2)=0$, so \begin{align*} d^*1=0. \end{align*} For $d\theta$, \begin{align*} d^*(d\theta) &=-*d*(d\theta)\\ &=-*d(d\phi)\\ &=-*0\\ &=0. \end{align*} For $d\phi$, \begin{align*} d^*(d\phi) &=-*d*(d\phi)\\ &=-*d(-d\theta)\\ &=-*(-d(d\theta))\\ &=-*0\\ &=0. \end{align*} For the area form, \begin{align*} d^*(d\theta\wedge d\phi) &=-*d*(d\theta\wedge d\phi)\\ &=-*d1\\ &=-*0\\ &=0. \end{align*} Therefore each listed form satisfies both $d\alpha=0$ and $d^*\alpha=0$. Since \begin{align*} \Delta=dd^*+d^*d, \end{align*} each listed form is harmonic: \begin{align*} \Delta\alpha &=d(d^*\alpha)+d^*(d\alpha)\\ &=d0+d^*0\\ &=0. \end{align*} Thus \begin{align*} \operatorname{span}\{1\}&\subseteq\mathcal H^0(T^2),\\ \operatorname{span}\{d\theta,d\phi\}&\subseteq\mathcal H^1(T^2),\\ \operatorname{span}\{d\theta\wedge d\phi\}&\subseteq\mathcal H^2(T^2). \end{align*} These spans have dimensions $1$, $2$, and $1$. For degree $1$, if \begin{align*} a\,d\theta+b\,d\phi=0, \end{align*} then evaluating on the coordinate vector fields gives \begin{align*} 0&=(a\,d\theta+b\,d\phi)(\partial_\theta)=a,\\ 0&=(a\,d\theta+b\,d\phi)(\partial_\phi)=b. \end{align*} For degree $2$, if \begin{align*} c\,d\theta\wedge d\phi=0, \end{align*} then evaluating on $(\partial_\theta,\partial_\phi)$ gives \begin{align*} 0 &=c(d\theta\wedge d\phi)(\partial_\theta,\partial_\phi)\\ &=c\bigl(d\theta(\partial_\theta)d\phi(\partial_\phi)-d\theta(\partial_\phi)d\phi(\partial_\theta)\bigr)\\ &=c(1\cdot1-0\cdot0)\\ &=c. \end{align*} So the displayed spans have dimensions \begin{align*} 1,\quad 2,\quad 1. \end{align*} The earlier Mayer-Vietoris computation for the torus gives \begin{align*} b_0(T^2)&=1, & b_1(T^2)&=2, & b_2(T^2)&=1. \end{align*} By *[Hodge Decomposition](/theorems/2745)*, the natural map \begin{align*} \mathcal H^k(T^2)&\to H^k_{\mathrm{dR}}(T^2), & \alpha&\mapsto[\alpha] \end{align*} is an isomorphism. Hence \begin{align*} \dim\mathcal H^0(T^2)&=1,\\ \dim\mathcal H^1(T^2)&=2,\\ \dim\mathcal H^2(T^2)&=1. \end{align*} Since each displayed span is contained in the corresponding harmonic space and has the same dimension, the inclusions are equalities: \begin{align*} \mathcal H^0(T^2)&=\operatorname{span}\{1\},\\ \mathcal H^1(T^2)&=\operatorname{span}\{d\theta,d\phi\},\\ \mathcal H^2(T^2)&=\operatorname{span}\{d\theta\wedge d\phi\}. \end{align*} On the flat torus, the [harmonic representatives](/theorems/2747) are exactly the constant-coefficient forms. The metric makes $d\theta$ and $d\phi$ orthonormal, the Hodge star sends them to constant-coefficient forms again, and both $d$ and $d^*$ vanish on the listed generators. Hodge theory then identifies these $1$, $2$, and $1$ dimensional harmonic spaces with the de Rham cohomology of $T^2$. [/example] The flat torus is the cleanest case because translation-invariance picks out the [harmonic representatives](/theorems/2747) without any analysis: the constant-coefficient forms automatically satisfy both $d\alpha=0$ and $d^*\alpha=0$. Once the geometry is no longer flat, however, the harmonic equation becomes a genuine elliptic problem and curvature enters the picture. The sphere is the simplest such case: its constant positive curvature changes the answer in degree $1$ and illustrates how Hodge theory interacts with Riemannian geometry rather than merely with topology. [example: Harmonic Representatives on the Sphere] Equip $S^2$ with the round metric of constant Gaussian curvature $K\equiv 1$, and orient it by its area form $dA$. Its de Rham cohomology is \begin{align*} H^0_{\mathrm{dR}}(S^2)&\cong\mathbb R, & H^1_{\mathrm{dR}}(S^2)&=0, & H^2_{\mathrm{dR}}(S^2)&\cong\mathbb R. \end{align*} All norms, covariant derivatives, Hodge stars, codifferentials, and Laplacians below are computed with the round metric. For a form $\eta$ on a compact manifold without boundary, the Hodge energy identity gives \begin{align*} (\Delta\eta,\eta)_{L^2} =\|d\eta\|_{L^2}^2+\|d^*\eta\|_{L^2}^2. \end{align*} Thus a harmonic form satisfies $d\eta=0$ and $d^*\eta=0$. The harmonic forms on the round sphere are \begin{align*} \mathcal H^0(S^2)&=\operatorname{span}\{1\},\\ \mathcal H^1(S^2)&=\{0\},\\ \mathcal H^2(S^2)&=\operatorname{span}\{dA\}. \end{align*} First consider degree $0$. Let $f\in\Omega^0(S^2)$ be harmonic. Then \begin{align*} 0 &=(\Delta f,f)_{L^2}\\ &=\|df\|_{L^2}^2+\|d^*f\|_{L^2}^2. \end{align*} Since $d^*f$ has degree $-1$, it is zero. Hence \begin{align*} \|df\|_{L^2}^2 &=\int_{S^2}|df|^2\,dA\\ &=0. \end{align*} The function $|df|^2$ is continuous and nonnegative, so $|df|^2=0$ everywhere. Therefore $df=0$. If $c:[0,1]\to S^2$ is any smooth path, then \begin{align*} \frac{d}{dt}(f\circ c)(t) &=df_{c(t)}(c'(t))\\ &=0. \end{align*} Since $S^2$ is connected, any two points can be joined by a path, so $f$ is constant. Conversely, if $f$ is constant, then \begin{align*} df&=0, & d^*f&=0, \end{align*} and therefore \begin{align*} \Delta f &=dd^*f+d^*df\\ &=d0+d^*0\\ &=0. \end{align*} Thus \begin{align*} \mathcal H^0(S^2)=\operatorname{span}\{1\}. \end{align*} Now consider degree $1$. Let $\alpha\in\Omega^1(S^2)$ be harmonic. Then \begin{align*} d\alpha&=0, & d^*\alpha&=0. \end{align*} By the *Bochner formula for one-forms on a closed Riemannian surface*, \begin{align*} \int_{S^2}|\nabla\alpha|^2\,dA +\int_{S^2}K|\alpha|^2\,dA = \int_{S^2}\bigl(|d\alpha|^2+|d^*\alpha|^2\bigr)\,dA. \end{align*} Substituting $K=1$, $d\alpha=0$, and $d^*\alpha=0$ gives \begin{align*} \int_{S^2}|\nabla\alpha|^2\,dA +\int_{S^2}|\alpha|^2\,dA &=\int_{S^2}(0+0)\,dA\\ &=0. \end{align*} Both integrands $|\nabla\alpha|^2$ and $|\alpha|^2$ are nonnegative. Hence \begin{align*} \int_{S^2}|\alpha|^2\,dA=0. \end{align*} Since $|\alpha|^2$ is continuous and nonnegative, this forces $|\alpha|^2=0$ everywhere, so $\alpha=0$. The zero form is harmonic because \begin{align*} \Delta 0=0. \end{align*} Therefore \begin{align*} \mathcal H^1(S^2)=\{0\}. \end{align*} Finally consider degree $2$. At any point choose an oriented orthonormal coframe $(e^1,e^2)$, so \begin{align*} dA=e^1\wedge e^2. \end{align*} The defining rule for the Hodge star gives \begin{align*} *1&=dA, & *(dA)&=1. \end{align*} Also $\Omega^3(S^2)=0$, so \begin{align*} d(dA)=0. \end{align*} For a $2$-form on a $2$-manifold, \begin{align*} d^*\eta=(-1)^{2(2+1)+1}*d*\eta=-*d*\eta. \end{align*} Thus \begin{align*} d^*(dA) &=-*d*(dA)\\ &=-*d1\\ &=-*0\\ &=0. \end{align*} Therefore $dA$ is harmonic, and every constant multiple $c\,dA$ is harmonic by linearity. Conversely, let $\beta\in\Omega^2(S^2)$ be harmonic. Since $\Lambda^2T_p^*S^2$ is one-dimensional with basis $dA_p$, there is a smooth function $h:S^2\to\mathbb R$ such that \begin{align*} \beta=h\,dA. \end{align*} The condition $d^*\beta=0$ gives \begin{align*} 0 &=d^*(h\,dA)\\ &=-*d*(h\,dA)\\ &=-*d(h*(dA))\\ &=-*d(h\cdot 1)\\ &=-*dh. \end{align*} The Hodge star is a pointwise isomorphism, so $dh=0$. As in degree $0$, connectedness of $S^2$ implies that $h$ is constant. Hence \begin{align*} \beta=c\,dA \end{align*} for some $c\in\mathbb R$. Therefore \begin{align*} \mathcal H^2(S^2)=\operatorname{span}\{dA\}. \end{align*} The round sphere has one-dimensional harmonic spaces in degrees $0$ and $2$, represented by constant functions and constant multiples of the area form. In degree $1$, positive curvature enters through the Bochner formula and forces every harmonic $1$-form to vanish. This is the geometric contrast with the flat torus, where $d\theta$ and $d\phi$ survive as nonzero harmonic $1$-forms. [/example] Hodge theory closes the circle of the course. De Rham cohomology began as a quotient of closed forms by exact forms; with a metric, the same cohomology can be represented by solutions of the elliptic equation $\Delta\alpha=0$. This is the point where topology, geometry, and analysis meet. ## References