Differential Forms II: Manifolds and Cohomology develops the language and tools of differential geometry on smooth manifolds, then uses them to build a bridge from calculus to topology. The course begins by defining smooth manifolds and the tangent and cotangent bundles that replace the familiar linear structure of Euclidean space. From there it introduces differential forms as the natural objects for differentiation and integration on manifolds, leading to orientations, integration, and the general Stokes’ theorem as the central analytic principle tying local computations to global results. The later chapters shift from analysis to topology through de Rham cohomology, which measures global obstructions to solving differential equations on manifolds. The Poincaré lemma provides the local exactness theory that makes cohomology meaningful, while the Mayer-Vietoris sequence and explicit computations show how to assemble local data into global invariants. The course culminates in de Rham’s theorem, which identifies de Rham cohomology with singular cohomology with real coefficients, and then explores applications and further directions that show how differential forms connect geometry, topology, and analysis in a unified framework. # Introduction This course continues the exterior calculus of Differential Forms I by moving from open subsets of $\mathbb R^n$ to smooth manifolds. The central question is how much of calculus depends on coordinates, and how much survives on spaces that only look Euclidean locally. We shall build the language needed to integrate forms on manifolds, prove the manifold version of Stokes theorem, and then use closed and exact forms to extract topological information. The course has two intertwined themes. The analytic theme is that differentiation, pullback, integration, and Stokes theorem can be formulated without a preferred coordinate system. The topological theme is that the failure of a closed form to be exact records global structure, leading to de Rham cohomology and ultimately to de Rham theorem. ## The Problem of Coordinate-Free Calculus In Differential Forms I, forms lived on open sets $U \subset \mathbb R^n$, where vectors, covectors, and coordinates are available from the start. Many geometric spaces are not single open subsets of Euclidean space: spheres, tori, projective spaces, and matrix groups require several coordinate systems. The first problem of the course is to formulate calculus so that changing coordinates changes the description but not the object being studied. [explanation: Local Descriptions and Global Objects] A smooth manifold is a space that can be studied through coordinate charts, but the charts are not part of the final geometric answer. A function, vector field, differential form, or integral must therefore have local coordinate formulae that agree on overlaps. This is the same discipline already present in exterior calculus on $\mathbb R^n$: the coordinate expression of a form may change, while the alternating multilinear object it represents remains the same. The guiding test for every definition in the first part of the course is compatibility under change of coordinates. If a construction is meaningful only after choosing a chart and does not transform correctly on chart overlaps, it is not an intrinsic construction on the manifold. [/explanation] The simplest place where this issue appears is already one-dimensional. A circle can be drawn in the plane, but its intrinsic calculus should not depend on choosing a preferred point at which to cut it open. [example: The Circle Requires More Than One Coordinate] Consider $S^1=\{(x,y)\in \mathbb R^2:x^2+y^2=1\}$. A single global real coordinate would be a homeomorphism from $S^1$ onto an open subset of $\mathbb R$, but $S^1$ is compact and no nonempty open subset of $\mathbb R$ is compact, so such a coordinate cannot exist. Let $N=(0,1)$ and $S=(0,-1)$. Stereographic coordinates give charts \begin{align*} \varphi_N(x,y)=\frac{x}{1-y}\quad\text{on }S^1\setminus\{N\}, \qquad \varphi_S(x,y)=\frac{x}{1+y}\quad\text{on }S^1\setminus\{S\}. \end{align*} Solving for the inverse of the first chart gives \begin{align*} \varphi_N^{-1}(t)=\left(\frac{2t}{t^2+1},\frac{t^2-1}{t^2+1}\right), \end{align*} and on the overlap $S^1\setminus\{N,S\}$ the transition map is \begin{align*} (\varphi_S\circ \varphi_N^{-1})(t)=\frac{1}{t}. \end{align*} The inverse transition has the same formula, so both transition maps are smooth on $\mathbb R\setminus\{0\}$. Calculus on $S^1$ is therefore built from local coordinates, and the smooth transition function $t\mapsto 1/t$ is what lets derivatives and differential forms agree on the overlap. [/example] This example foreshadows the general pattern: a manifold is locally Euclidean, but global questions come from how the local pieces fit together. ## Smooth Manifolds as the Setting What hypotheses on a space make it a suitable domain for calculus? Local Euclidean behaviour is necessary, but topology also matters: we need enough open sets to separate points, and enough countability to support partitions of unity and integration theory. The first chapter after this introduction therefore begins with topological and smooth manifolds. [definition: Topological Manifold] A topological manifold of dimension $n$ is a topological space $M$ such that: 1. $M$ is Hausdorff; 2. $M$ is second-countable; 3. every point $p \in M$ has an open neighbourhood $U \subset M$ and a homeomorphism $\varphi: U \to \varphi(U) \subset \mathbb R^n$, where $\varphi(U)$ is open. [/definition] A pair $(U,\varphi)$ as in the definition is called a coordinate chart. The Hausdorff and second-countability assumptions prevent pathological spaces from entering the theory and ensure that the usual tools of analysis behave as expected. [definition: Smooth Atlas] A smooth atlas on a topological manifold $M$ is a collection of coordinate charts $\{(U_i,\varphi_i)\}_{i \in I}$ such that: 1. $M = \bigcup_{i \in I} U_i$; 2. for every $i,j \in I$, 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*} is smooth whenever $U_i \cap U_j \ne \varnothing$. [/definition] The transition maps are the mechanism by which Euclidean calculus becomes manifold calculus. Once they are smooth, derivatives and differential forms transform in a controlled way from chart to chart. [example: The Standard Smooth Structure on the Sphere] Let $N=(0,\ldots,0,1)$ and $S=(0,\ldots,0,-1)$, and write a point of $\mathbb R^{n+1}$ as $(u,z)$ with $u\in\mathbb R^n$ and $z\in\mathbb R$. Define \begin{align*} \varphi_N(u,z)=\frac{u}{1-z} \end{align*} on $S^n\setminus\{N\}$ and \begin{align*} \varphi_S(u,z)=\frac{u}{1+z} \end{align*} on $S^n\setminus\{S\}$. If $a=\varphi_N(u,z)$, then $u=a(1-z)$ and $|u|^2+z^2=1$, so \begin{align*} |a|^2(1-z)^2+z^2&=1,\\ |a|^2(1-z)^2-(1-z)(1+z)&=0,\\ (1-z)\bigl(|a|^2(1-z)-(1+z)\bigr)&=0. \end{align*} Since $(u,z)\ne N$, we have $1-z\ne0$, hence \begin{align*} |a|^2(1-z)&=1+z,\\ |a|^2-1&=(|a|^2+1)z,\\ z&=\frac{|a|^2-1}{|a|^2+1}. \end{align*} Therefore \begin{align*} u &=a\left(1-\frac{|a|^2-1}{|a|^2+1}\right)\\ &=a\left(\frac{2}{|a|^2+1}\right)\\ &=\frac{2a}{|a|^2+1}, \end{align*} so \begin{align*} \varphi_N^{-1}(a)=\left(\frac{2a}{|a|^2+1},\frac{|a|^2-1}{|a|^2+1}\right). \end{align*} This point lies on $S^n$ because \begin{align*} \left|\frac{2a}{|a|^2+1}\right|^2+\left(\frac{|a|^2-1}{|a|^2+1}\right)^2 &=\frac{4|a|^2}{(|a|^2+1)^2}+\frac{(|a|^2-1)^2}{(|a|^2+1)^2}\\ &=\frac{4|a|^2+|a|^4-2|a|^2+1}{(|a|^2+1)^2}\\ &=\frac{|a|^4+2|a|^2+1}{(|a|^2+1)^2}\\ &=1. \end{align*} The same calculation with $1+z$ in place of $1-z$ gives \begin{align*} \varphi_S^{-1}(b)=\left(\frac{2b}{|b|^2+1},\frac{1-|b|^2}{|b|^2+1}\right). \end{align*} On the overlap $S^n\setminus\{N,S\}$, the coordinate $a=\varphi_N(u,z)$ is nonzero. Indeed, if $a=0$, then $u=0$, and $|u|^2+z^2=1$ gives $z=\pm1$, so the point is either $N$ or $S$. For $a\in\mathbb R^n\setminus\{0\}$, \begin{align*} (\varphi_S\circ\varphi_N^{-1})(a) &=\frac{\frac{2a}{|a|^2+1}}{1+\frac{|a|^2-1}{|a|^2+1}}\\ &=\frac{\frac{2a}{|a|^2+1}}{\frac{2|a|^2}{|a|^2+1}}\\ &=\frac{a}{|a|^2}. \end{align*} Similarly, for $b\in\mathbb R^n\setminus\{0\}$, \begin{align*} (\varphi_N\circ\varphi_S^{-1})(b) &=\frac{\frac{2b}{|b|^2+1}}{1-\frac{1-|b|^2}{|b|^2+1}}\\ &=\frac{\frac{2b}{|b|^2+1}}{\frac{2|b|^2}{|b|^2+1}}\\ &=\frac{b}{|b|^2}. \end{align*} Each component of $a\mapsto a/|a|^2$ is the rational function $a_i/(a_1^2+\cdots+a_n^2)$, whose denominator is nonzero on $\mathbb R^n\setminus\{0\}$. Thus both transition maps are smooth on the overlap. The two stereographic charts cover $S^n$, and their smooth transition maps make them a smooth atlas on the sphere. [/example] ## Tangent and Cotangent Bundles Once manifolds replace open sets in $\mathbb R^n$, the next question is what should replace a fixed vector space of directions. At each point of a manifold there is a tangent space, and these tangent spaces vary from point to point. Differential forms live naturally on the dual spaces, so the cotangent bundle becomes the main stage for exterior calculus. [definition: Tangent Space] Let $M$ be a smooth manifold and let $p \in M$. The tangent space $T_pM$ is the vector space of tangent vectors at $p$, defined using equivalence classes of smooth curves through $p$. [/definition] The course will compare the curve-based and derivation-based definitions and use whichever is more convenient. Curves make the geometric meaning visible, while derivations interact well with functions and coordinate computations. Many constructions in exterior calculus are not built directly from tangent vectors, but from linear measurements of tangent vectors. To make those measurements intrinsic at each point, we pass from a tangent space to its dual space. [definition: Cotangent Space] Let $M$ be a smooth manifold and let $p \in M$. The cotangent space at $p$ is the dual vector space \begin{align*} T_p^*M := (T_pM)^*. \end{align*} [/definition] Differential $k$-forms at $p$ are alternating $k$-linear maps on $T_pM$. As $p$ varies, these spaces assemble into vector bundles over $M$. [example: Tangent Spaces of a Sphere] Let $p \in S^n \subset \mathbb R^{n+1}$, so $p\cdot p=1$. The tangent vectors at $p$ are exactly the Euclidean vectors orthogonal to $p$: \begin{align*} T_pS^n=\{v\in\mathbb R^{n+1}:v\cdot p=0\}. \end{align*} Indeed, differentiating the constraint $\gamma(t)\cdot\gamma(t)=1$ along any curve in $S^n$ through $p$ shows that its velocity is orthogonal to $p$. Conversely, if $v\cdot p=0$, the normalized curve \begin{align*} \gamma(t)=\frac{p+tv}{|p+tv|} \end{align*} lies on $S^n$ for $t$ near $0$ and has initial velocity $v$. Geometrically, the tangent space is the hyperplane through the origin perpendicular to the radius vector $p$. [/example] ## Differential Forms and Stokes Theorem The next problem is how to integrate on a space without a single coordinate system. Differential forms are designed for this: pullbacks describe change of variables, top-degree forms are the correct objects to integrate, and the exterior derivative is the coordinate-free derivative compatible with integration over boundaries. [definition: Differential Form] Let $M$ be a smooth manifold. A differential $k$-form on $M$ is a smooth section of the vector bundle $\Lambda^k T^*M$. The space of smooth $k$-forms on $M$ is denoted $\Omega^k(M)$. [/definition] The exterior derivative is the operator that turns a $k$-form into a $(k+1)$-form and extends the differential from functions to all forms. Its defining properties are locality, compatibility with the Euclidean formula in charts, and the identity $d \circ d = 0$. The local coordinate formula for $d$ is only useful on a manifold if it gives the same form on overlaps of charts. The immediate question is therefore not yet an integration theorem, but a gluing problem: does the Euclidean exterior derivative commute with smooth changes of coordinates well enough to define a global operator on differential forms? [quotetheorem:3899] [citeproof:3899] This theorem is the main bridge from local exterior calculus to global exterior calculus. The smoothness of the transition maps is essential: without smooth transition maps, the coordinate formula for $d\omega$ need not transform by pullback in a way that agrees on overlaps. The theorem also does not say that every locally defined formula automatically globalises; it globalises precisely because the Euclidean exterior derivative is natural under smooth coordinate changes. Thus the construction is coordinate-free, but its verification is deliberately local. [quotetheorem:3900] [citeproof:3900] The hypotheses in Stokes theorem are not cosmetic. Orientation is needed to make the signs of the integrals coherent across charts, and the induced boundary orientation is what fixes the sign on $\partial M$. Compact support, or a suitable compactness assumption on $M$, ensures that the integral is controlled by finitely many local computations after passing to a locally finite partition of unity. Smoothness of the manifold, boundary, and form is also part of the theorem: the statement does not by itself cover singular domains, rough boundaries, or arbitrary measurable forms. [example: Green Theorem as Stokes Theorem] Let $M \subset \mathbb R^2$ be compact and oriented by $dx\wedge dy$, with smooth boundary, and let $P,Q$ be smooth functions on a neighbourhood of $M$. For \begin{align*} \omega=P\,dx+Q\,dy, \end{align*} we compute its exterior derivative term by term: \begin{align*} d\omega &=d(P\,dx)+d(Q\,dy)\\ &=dP\wedge dx+P\,d(dx)+dQ\wedge dy+Q\,d(dy)\\ &=dP\wedge dx+dQ\wedge dy, \end{align*} since $d(dx)=0$ and $d(dy)=0$. Writing \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*} gives \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*} because $dx\wedge dx=0$ and $dy\wedge dx=-dx\wedge dy$. Similarly, \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 &=-\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*} Applying *Stokes Theorem on Manifolds* to the $1$-form $\omega$ gives \begin{align*} \int_M d\omega=\int_{\partial M}\omega, \end{align*} with $\partial M$ carrying the boundary orientation induced by $dx\wedge dy$. Substituting the computed formula for $d\omega$ and the definition of $\omega$ yields \begin{align*} \int_M \left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)\,dx\wedge dy = \int_{\partial M} P\,dx+Q\,dy. \end{align*} Thus Green theorem is exactly the two-dimensional case of Stokes theorem for the form $P\,dx+Q\,dy$. [/example] ## Closed Forms, Exact Forms, and Cohomology Stokes theorem immediately raises a global question. Since $d^2=0$, every exact form is closed, but not every closed form has a global primitive. The obstruction to solving $\omega=d\eta$ is measured by de Rham cohomology. [definition: Closed and Exact Forms] Let $M$ be a smooth manifold and let $\omega \in \Omega^k(M)$. The form $\omega$ is closed if $d\omega=0$. The form $\omega$ is exact if there exists $\eta \in \Omega^{k-1}(M)$ such that $\omega=d\eta$. [/definition] The identity $d^2=0$ implies every exact form is closed. Cohomology records the quotient of closed forms by exact forms, turning the failure of global primitives into an algebraic invariant. [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*} [/definition] The first test case should show that the quotient is not empty formalism. A puncture in the plane creates a closed angular form whose failure to be exact can be detected by integration around the missing point. [example: A Closed Non-Exact Form on the Punctured Plane] On $\mathbb R^2\setminus\{0\}$, define \begin{align*} \omega=\frac{-y\,dx+x\,dy}{x^2+y^2}. \end{align*} We show that $\omega$ is closed but not exact. Write \begin{align*} \omega=f\,dx+g\,dy, \qquad f=\frac{-y}{x^2+y^2}, \qquad g=\frac{x}{x^2+y^2}. \end{align*} Since $d(dx)=0$ and $d(dy)=0$, \begin{align*} d\omega &=d(f\,dx)+d(g\,dy)\\ &=df\wedge dx+f\,d(dx)+dg\wedge dy+g\,d(dy)\\ &=df\wedge dx+dg\wedge dy. \end{align*} Now \begin{align*} \frac{\partial f}{\partial y} &=\frac{(-1)(x^2+y^2)-(-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*} and \begin{align*} \frac{\partial g}{\partial x} &=\frac{(1)(x^2+y^2)-x(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*} Therefore \begin{align*} df\wedge 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\\ &=-\frac{\partial f}{\partial y}\,dx\wedge dy, \end{align*} and \begin{align*} dg\wedge dy &=\left(\frac{\partial g}{\partial x}\,dx+\frac{\partial g}{\partial y}\,dy\right)\wedge dy\\ &=\frac{\partial g}{\partial x}\,dx\wedge dy+\frac{\partial g}{\partial y}\,dy\wedge dy\\ &=\frac{\partial g}{\partial x}\,dx\wedge dy. \end{align*} Thus \begin{align*} d\omega &=\left(\frac{\partial g}{\partial x}-\frac{\partial f}{\partial y}\right)dx\wedge dy\\ &=\left(\frac{y^2-x^2}{(x^2+y^2)^2}-\frac{y^2-x^2}{(x^2+y^2)^2}\right)dx\wedge dy\\ &=0. \end{align*} So $\omega$ is closed. To show that $\omega$ is not exact, parametrize the unit circle by \begin{align*} \gamma(t)=(\cos t,\sin t), \qquad 0\le t\le 2\pi. \end{align*} Along this curve, \begin{align*} \gamma^*x=\cos t,\qquad \gamma^*y=\sin t,\qquad \gamma^*dx=-\sin t\,dt,\qquad \gamma^*dy=\cos t\,dt. \end{align*} Hence \begin{align*} \gamma^*\omega &=\frac{-(\sin t)(-\sin t\,dt)+(\cos t)(\cos t\,dt)}{\cos^2 t+\sin^2 t}\\ &=\frac{\sin^2 t\,dt+\cos^2 t\,dt}{1}\\ &=dt. \end{align*} Therefore \begin{align*} \int_{\gamma}\omega &=\int_0^{2\pi}\gamma^*\omega\\ &=\int_0^{2\pi}dt\\ &=2\pi. \end{align*} If $\omega=df$ for some smooth function $f:\mathbb R^2\setminus\{0\}\to\mathbb R$, then \begin{align*} \gamma^*\omega &=\gamma^*(df)\\ &=d(f\circ\gamma), \end{align*} so \begin{align*} \int_{\gamma}\omega &=\int_0^{2\pi}d(f\circ\gamma)\\ &=\int_0^{2\pi}\frac{d}{dt}(f(\gamma(t)))\,dt\\ &=f(\gamma(2\pi))-f(\gamma(0))\\ &=f(1,0)-f(1,0)\\ &=0. \end{align*} This contradicts $\int_{\gamma}\omega=2\pi$, so $\omega$ is not exact. Thus $\omega$ is a closed non-exact $1$-form on the punctured plane, detecting the missing origin. [/example] This example is the prototype for de Rham cohomology: a local primitive exists on small arcs, but no single-valued global primitive exists on the punctured plane. The obstruction detects the hole at the origin. This is the point of the ordinary de Rham Poincare lemma: locally, closed positive-degree forms have primitives on sufficiently simple domains, so cohomology is not caused by local calculus. The punctured plane shows why the local hypothesis cannot simply be globalized. Although the angular form is locally exact on small arcs, the missing origin prevents those local primitives from assembling into one global primitive. Thus de Rham cohomology records the global obstruction left over after local exactness has done all it can. ## The Topological Meaning of de Rham Cohomology Why should a quotient of differential forms know anything about holes, connected components, or cycles? The answer is that closed forms can be integrated over cycles, exact forms integrate to zero over boundaries, and Stokes theorem makes this pairing depend only on cohomology classes. The final part of the course develops this idea until it reaches de Rham theorem. [definition: Singular Chain] A singular $k$-simplex in a topological space $X$ is a continuous map $\sigma:\Delta^k\to X$. A singular $k$-chain is a finite formal integer linear combination of singular $k$-simplices. [/definition] Singular cohomology is built from these combinatorial pieces, while de Rham cohomology is built from smooth forms. De Rham theorem says that, for smooth manifolds, these two constructions agree after passing to real coefficients. The point of introducing singular chains here is to name the topological object that forms can test by integration. The theorem below is needed to turn that pairing into a precise comparison between analytic cohomology classes and topological cohomology classes. [quotetheorem:3596] [citeproof:3596] This theorem is the conceptual endpoint of the course. It identifies an analytic object, built from smooth differential forms and derivatives, with a topological object, built from continuous simplices and coboundaries. The smooth-manifold hypothesis matters because integration of forms over chains and the local Poincare lemma both use smooth local models. The real coefficients also matter: de Rham cohomology does not detect torsion phenomena that can appear in integral singular cohomology. Finally, the theorem gives an isomorphism of cohomology theories, but it does not by itself compute the groups; computations still require tools such as Mayer-Vietoris, homotopy invariance, and explicit generators. ## Prerequisites and Working Conventions What should the reader already know, and what conventions will the course use? The notes assume fluency with multivariable calculus, linear algebra, differential forms on open subsets of $\mathbb R^n$, the wedge product, pullback, and the classical Stokes theorem. They also use basic point-set topology and elementary group theory, especially for examples involving quotient spaces and matrix groups. [remark: Notation for Manifolds] Smooth manifolds are denoted by $M,N$, coordinate charts by $(U,\varphi)$, tangent spaces by $T_pM$, cotangent spaces by $T_p^*M$, and differential forms by $\Omega^k(M)$. For a smooth map $F:M\to N$, the differential at $p$ is written $dF_p:T_pM\to T_{F(p)}N$. [/remark] These symbols are coordinate-free, but computations often pass through coordinates. The next convention separates coordinate calculations from geometric structure that is actually present on the manifold. [remark: Inner Products and Coordinates] A smooth manifold does not come with an inner product on tangent spaces unless a Riemannian metric has been specified. Coordinate calculations may use Euclidean notation inside a chart, but any statement about the manifold must be invariant under transition maps. [/remark] Orientation will later determine signs in integration. It is helpful to fix the sign convention before boundary terms appear in Stokes theorem. [remark: Orientation Convention] For an oriented manifold with boundary, the boundary orientation is the one used in Stokes theorem \begin{align*} \int_M d\omega = \int_{\partial M}\omega. \end{align*} Local sign conventions will be fixed when manifolds with boundary are introduced. [/remark] ## Roadmap of the Course How do the lectures fit together? The first part builds smooth manifolds, smooth maps, tangent spaces, cotangent spaces, and vector bundles. The second part constructs differential forms on manifolds, develops pullback and exterior derivative, introduces orientation, and proves Stokes theorem. The third part studies closed and exact forms through de Rham cohomology. Computations begin with contractible spaces, spheres, tori, and projective spaces, then move to Mayer-Vietoris sequences and homotopy invariance. The course culminates by comparing de Rham cohomology with singular cohomology, completing the passage from local calculus to global topology. The first chapter begins with the spaces on which differential forms live. We need a notion of space that is locally modeled on Euclidean space but not confined to it. # 1. Smooth Manifolds This opening chapter answers the question: what kind of space can support differential calculus without being a subset of Euclidean space? Differential Forms I worked on open subsets of $\mathbb R^n$, where coordinates are already present. A manifold is a space which is locally Euclidean, but whose global shape may require many coordinate systems patched together. The aim is to define the spaces, maps, and subspaces on which differential forms will later live. ## Local Euclidean Spaces What topological hypotheses are needed before local coordinates behave like the coordinate patches used in analysis? Being locally modelled on $\mathbb R^n$ is not enough by itself: the space must also have enough separation and countability for limits, partitions, and local constructions to work well. [definition: Topological Manifold] An $n$-dimensional topological manifold is a topological space $M$ such that: 1. $M$ is Hausdorff. 2. $M$ is second-countable. 3. For every $p \in M$, there exist an open set $U \subset M$ with $p \in U$ and a homeomorphism $\varphi: U \to V$, where $V \subset \mathbb R^n$ is open. [/definition] A pair $(U,\varphi)$ as in the definition is a coordinate chart. The open set $U$ is the coordinate neighbourhood, and the map $\varphi$ assigns coordinates $(x_1,\dots,x_n)$ to points of $U$. [definition: Dimension Of A Topological Manifold] The dimension of a topological manifold $M$ is the integer $n$ such that every point of $M$ has a neighbourhood homeomorphic to an open subset of $\mathbb R^n$. [/definition] The invariance of domain theorem ensures that this integer is well-defined: no nonempty open subset of $\mathbb R^m$ is homeomorphic to a nonempty open subset of $\mathbb R^n$ when $m \ne n$. [example: Coordinate Chart On Euclidean Space] Let $M=\mathbb R^n$ with its standard topology. We show that $M$ is an $n$-dimensional topological manifold by verifying the three conditions in the definition. First, $\mathbb R^n$ is Hausdorff: if $p,q\in\mathbb R^n$ and $p\ne q$, then $r=\frac{1}{3}|p-q|>0$, and the open balls $B(p,r)$ and $B(q,r)$ are disjoint because any $x$ lying in both would satisfy \begin{align*} |p-q| \le |p-x|+|x-q| < r+r=\frac{2}{3}|p-q|, \end{align*} which is impossible. Second, $\mathbb R^n$ is second-countable: the balls $B(a,r)$ with $a\in\mathbb Q^n$ and $r\in\mathbb Q_{>0}$ form a countable basis for the Euclidean topology, since every Euclidean open ball contains a smaller ball with rational center and rational radius around each of its points. For the local Euclidean condition, use the single chart \begin{align*} (\mathbb R^n,\operatorname{id}_{\mathbb R^n}). \end{align*} The set $\mathbb R^n$ is open in itself, and the identity map satisfies \begin{align*} \operatorname{id}_{\mathbb R^n}(x)=x,\qquad \operatorname{id}_{\mathbb R^n}^{-1}(x)=x. \end{align*} Both maps are continuous, so $\operatorname{id}_{\mathbb R^n}$ is a homeomorphism from $\mathbb R^n$ onto the open subset $\mathbb R^n\subset\mathbb R^n$. Thus every point has a coordinate neighbourhood modelled on $\mathbb R^n$, so $\mathbb R^n$ is an $n$-dimensional topological manifold. [/example] The Euclidean example checks the definition in the simplest case. The following remark records why the extra topological hypotheses in the definition are not optional decoration. [remark: Why Hausdorff And Second-Countable] The Hausdorff condition lets limits be unique, which is needed for geometry based on local data. Second-countability prevents pathological spaces with too many disconnected local pieces and gives a manageable countable basis for later analytic constructions. [/remark] ## Smooth Atlases How do we decide whether a function on a locally Euclidean space is differentiable when different coordinate charts may describe the same point? The answer is that changes of coordinates must be smooth maps between open subsets of Euclidean space. [definition: Smoothly Compatible Charts] Let $(U,\varphi)$ and $(V,\psi)$ be coordinate charts on a topological manifold $M$. They are smoothly compatible if either $U \cap V=\varnothing$, or the transition map \begin{align*} \psi \circ \varphi^{-1}: \varphi(U \cap V) \to \psi(U \cap V) \end{align*} is a diffeomorphism between open subsets of $\mathbb R^n$. [/definition] The transition map is the coordinate change from the $\varphi$-coordinates to the $\psi$-coordinates. Smooth compatibility means that calculus performed in one chart agrees with calculus performed in another. A manifold needs more than isolated compatible pairs: it needs enough charts to cover every point while keeping all overlaps compatible. This global collection is the data that lets us do calculus everywhere on the space. [definition: Smooth Atlas] A smooth atlas on an $n$-dimensional topological manifold $M$ is a collection $\mathcal A=\{(U_i,\varphi_i)\}_{i \in I}$ of coordinate charts such that: 1. $M=\bigcup_{i \in I}U_i$. 2. Every pair of charts in $\mathcal A$ is smoothly compatible. [/definition] A smooth atlas specifies which coordinate systems are allowed. Enlarging an atlas by adding all charts compatible with it produces the maximal smooth atlas determined by the original one. To make smoothness an intrinsic structure rather than a choice of a particular list of charts, we package the underlying topological manifold together with the full compatible coordinate system it determines. That package is the object on which later tangent spaces, forms, and maps will be defined. [definition: Smooth Manifold] A smooth $n$-manifold is a pair $(M,\mathcal A)$ where $M$ is an $n$-dimensional topological manifold and $\mathcal A$ is a maximal smooth atlas on $M$. [/definition] In practice the maximal atlas is rarely written out. It is enough to give any smooth atlas, since it determines the same maximal atlas by compatibility. This practical shortcut still needs justification. If two proposed completions added different compatible charts, then the phrase "the smooth structure determined by an atlas" would be ambiguous, and later notions such as smooth maps and tangent vectors would depend on a hidden choice. The obstruction to rule out is exactly this possible non-uniqueness of the maximal compatible enlargement. [quotetheorem:3901] [citeproof:3901] This theorem is why finite or countable practical atlases can define a full smooth structure. The hypotheses are needed because smoothness is only meaningful after transition maps have been controlled: if a proposed chart has a transition map whose inverse is not smooth, then it changes the differentiable structure rather than merely enlarging the old one. For example, on $\mathbb R$ the coordinate $x \mapsto x^3$ is a homeomorphism, but its inverse fails to be smooth at $0$, so it is not smoothly compatible with the standard chart. The maximal atlas records exactly the coordinate systems that preserve the same smooth notion of differentiability; it does not say that a topological manifold has only one possible smooth structure. ## Standard Examples Of Smooth Manifolds Which spaces should count as manifolds in geometry? The definition is designed to include familiar curved spaces, quotient spaces, and matrix groups, while still allowing calculus to be checked in coordinates. [example: The Sphere With Stereographic Charts] Write points of $\mathbb R^{n+1}$ as $(x,t)$ with $x\in\mathbb R^n$ and $t\in\mathbb R$. Let \begin{align*} U_N=S^n\setminus\{N\},\qquad U_S=S^n\setminus\{S\}. \end{align*} These two open sets cover $S^n$. Since $S^n$ has the subspace topology from $\mathbb R^{n+1}$, it is Hausdorff and second-countable. We define stereographic coordinate maps \begin{align*} \pi_N(x,t)=\frac{x}{1-t}\quad\text{on }U_N,\qquad \pi_S(x,t)=\frac{x}{1+t}\quad\text{on }U_S. \end{align*} The denominators are nonzero on their domains: $t=1$ on $S^n$ implies $(x,t)=N$, and $t=-1$ on $S^n$ implies $(x,t)=S$. For $u\in\mathbb R^n$, put $r^2=|u|^2$. The inverse of $\pi_N$ is \begin{align*} \pi_N^{-1}(u)=\left(\frac{2u}{r^2+1},\frac{r^2-1}{r^2+1}\right). \end{align*} Indeed, \begin{align*} \left|\frac{2u}{r^2+1}\right|^2+\left(\frac{r^2-1}{r^2+1}\right)^2 &=\frac{4r^2}{(r^2+1)^2}+\frac{(r^2-1)^2}{(r^2+1)^2}\\ &=\frac{4r^2+r^4-2r^2+1}{(r^2+1)^2}\\ &=\frac{r^4+2r^2+1}{(r^2+1)^2}\\ &=\frac{(r^2+1)^2}{(r^2+1)^2}\\ &=1, \end{align*} so the formula lands in $S^n$. Its last coordinate is never $1$, because \begin{align*} \frac{r^2-1}{r^2+1}=1 \quad\Longrightarrow\quad r^2-1=r^2+1, \end{align*} which would imply $-1=1$. Also, \begin{align*} \pi_N\left(\frac{2u}{r^2+1},\frac{r^2-1}{r^2+1}\right) &=\frac{\frac{2u}{r^2+1}}{1-\frac{r^2-1}{r^2+1}}\\ &=\frac{\frac{2u}{r^2+1}}{\frac{r^2+1-r^2+1}{r^2+1}}\\ &=\frac{\frac{2u}{r^2+1}}{\frac{2}{r^2+1}}\\ &=u. \end{align*} Conversely, if $(x,t)\in U_N$, then $|x|^2+t^2=1$, so $|x|^2=1-t^2=(1-t)(1+t)$. Therefore \begin{align*} \pi_N^{-1}(\pi_N(x,t)) &=\left(\frac{2\frac{x}{1-t}}{\frac{|x|^2}{(1-t)^2}+1}, \frac{\frac{|x|^2}{(1-t)^2}-1}{\frac{|x|^2}{(1-t)^2}+1}\right)\\ &=\left(\frac{2x(1-t)}{|x|^2+(1-t)^2}, \frac{|x|^2-(1-t)^2}{|x|^2+(1-t)^2}\right). \end{align*} Using $|x|^2=(1-t)(1+t)$, \begin{align*} |x|^2+(1-t)^2 &=(1-t)(1+t)+(1-t)^2\\ &=(1-t)((1+t)+(1-t))\\ &=2(1-t), \end{align*} and \begin{align*} |x|^2-(1-t)^2 &=(1-t)(1+t)-(1-t)^2\\ &=(1-t)((1+t)-(1-t))\\ &=2t(1-t). \end{align*} Thus $\pi_N^{-1}(\pi_N(x,t))=(x,t)$. The same calculation with $t$ replaced by $-t$ gives \begin{align*} \pi_S^{-1}(u)=\left(\frac{2u}{|u|^2+1},\frac{1-|u|^2}{|u|^2+1}\right), \end{align*} so both $\pi_N$ and $\pi_S$ are homeomorphisms onto $\mathbb R^n$. It remains to check smooth compatibility. On $U_N\cap U_S=S^n\setminus\{N,S\}$, the corresponding points in $\mathbb R^n$ exclude $0$: in the formula for $\pi_N^{-1}(u)$, $u=0$ gives $(0,\dots,0,-1)=S$. For $u\in\mathbb R^n\setminus\{0\}$ and $r^2=|u|^2$, we have \begin{align*} (\pi_S\circ \pi_N^{-1})(u) &=\pi_S\left(\frac{2u}{r^2+1},\frac{r^2-1}{r^2+1}\right)\\ &=\frac{\frac{2u}{r^2+1}}{1+\frac{r^2-1}{r^2+1}}\\ &=\frac{\frac{2u}{r^2+1}}{\frac{r^2+1+r^2-1}{r^2+1}}\\ &=\frac{\frac{2u}{r^2+1}}{\frac{2r^2}{r^2+1}}\\ &=\frac{u}{|u|^2}. \end{align*} Each coordinate of $u\mapsto u/|u|^2$ is $u_i/(u_1^2+\cdots+u_n^2)$, a rational function with nonzero denominator on $\mathbb R^n\setminus\{0\}$, hence smooth there. Applying the same formula twice gives \begin{align*} \frac{\frac{u}{|u|^2}}{\left|\frac{u}{|u|^2}\right|^2} &=\frac{\frac{u}{|u|^2}}{\frac{|u|^2}{|u|^4}}\\ &=\frac{\frac{u}{|u|^2}}{\frac{1}{|u|^2}}\\ &=u, \end{align*} so the transition map is its own inverse. Therefore the two stereographic charts form a smooth atlas on $S^n$, and $S^n$ is a smooth $n$-manifold. [/example] [illustration:dfii-stereographic-projection] [example: The Torus As A Quotient] [claim]The quotient $T^n=\mathbb R^n/\mathbb Z^n$ is a smooth $n$-manifold.[/claim] [proof]Let $q:\mathbb R^n\to T^n$ be the quotient map, so $q(x)=q(y)$ exactly when $y-x\in\mathbb Z^n$. Fix $x\in\mathbb R^n$ and choose $0<\varepsilon<\frac12$. If $u,v\in B(x,\varepsilon)$ and $q(u)=q(v)$, then $v-u=m$ for some $m\in\mathbb Z^n$. If $m\ne 0$, then at least one coordinate of $m$ is a nonzero integer, so $|m|\ge 1$. But \begin{align*} |m| &=|v-u|\\ &\le |v-x|+|x-u|\\ &<\varepsilon+\varepsilon\\ &<1, \end{align*} a contradiction. Hence $m=0$, so $u=v$. Therefore $q$ is injective on $B(x,\varepsilon)$. The set $q(B(x,\varepsilon))$ is open in the quotient topology because \begin{align*} q^{-1}(q(B(x,\varepsilon))) =\bigcup_{m\in\mathbb Z^n}(B(x,\varepsilon)+m), \end{align*} which is open in $\mathbb R^n$. Define \begin{align*} \varphi_x:q(B(x,\varepsilon))\to B(0,\varepsilon), \qquad \varphi_x(q(u))=u-x. \end{align*} This is well-defined because $q$ is injective on $B(x,\varepsilon)$. It is bijective: for $w\in B(0,\varepsilon)$, the point $u=x+w$ lies in $B(x,\varepsilon)$ and \begin{align*} \varphi_x(q(x+w))=(x+w)-x=w. \end{align*} The inverse is $w\mapsto q(x+w)$, which is continuous since it is the restriction of the continuous map $q$ after the translation $w\mapsto x+w$. Also, if $W\subset B(x,\varepsilon)$ is open, then \begin{align*} q^{-1}(q(W))=\bigcup_{m\in\mathbb Z^n}(W+m), \end{align*} so $q(W)$ is open in $T^n$; hence $\varphi_x$ is continuous as well. Thus $\varphi_x$ is a coordinate chart. These charts cover $T^n$ because every point of $T^n$ has the form $q(x)$. Now take two charts $\varphi_x$ and $\varphi_y$. On their overlap, suppose \begin{align*} \varphi_x(q(a))=a-x=u. \end{align*} Then $a=x+u$. If the same quotient point is also represented in the $y$-chart, there is a unique $m\in\mathbb Z^n$ such that \begin{align*} x+u=y+v+m, \end{align*} where $v=\varphi_y(q(a))$. Solving for $v$ gives \begin{align*} v=u+x-y-m. \end{align*} Therefore each connected piece of the transition map has the form \begin{align*} u\longmapsto u+x-y-m, \end{align*} a restriction of a translation of $\mathbb R^n$. Translations are smooth and have smooth inverses, so all transition maps are diffeomorphisms between open subsets of $\mathbb R^n$. Hence the quotient charts define a smooth atlas on $T^n$.[/proof] The torus is therefore locally indistinguishable from $\mathbb R^n$, while globally points differing by integer translations have been identified. [/example] The torus is a quotient of Euclidean space by translations. Projective space is a different quotient, where lines through the origin replace translation orbits and the charts come from choosing a nonzero homogeneous coordinate. [example: Real Projective Space] [claim]Real projective space $\mathbb RP^n$ is a smooth $n$-manifold.[/claim] [proof]Real projective space is the set of equivalence classes \begin{align*} [x_0:\cdots:x_n]=\{\lambda(x_0,\dots,x_n):\lambda\in\mathbb R\setminus\{0\}\} \end{align*} of nonzero vectors in $\mathbb R^{n+1}$. For each $i\in\{0,\dots,n\}$, let \begin{align*} U_i=\{[x_0:\cdots:x_n]\in\mathbb RP^n:x_i\ne 0\}. \end{align*} These sets cover $\mathbb RP^n$, because every nonzero vector has at least one nonzero coordinate. Define \begin{align*} \varphi_i:U_i\to\mathbb R^n,\qquad \varphi_i([x_0:\cdots:x_n]) = \left(\frac{x_0}{x_i},\dots,\widehat{\frac{x_i}{x_i}},\dots,\frac{x_n}{x_i}\right), \end{align*} where the hat means that the $i$th entry is omitted. This is well-defined: if $(y_0,\dots,y_n)=\lambda(x_0,\dots,x_n)$ with $\lambda\ne 0$, then \begin{align*} \frac{y_j}{y_i} = \frac{\lambda x_j}{\lambda x_i} = \frac{x_j}{x_i} \end{align*} for every $j\ne i$. The inverse map inserts a $1$ in the $i$th position. Explicitly, for $u=(u_j)_{j\ne i}\in\mathbb R^n$, set \begin{align*} \psi_i(u)=[y_0:\cdots:y_n], \qquad y_i=1,\quad y_j=u_j\ \text{for }j\ne i. \end{align*} Then \begin{align*} (\varphi_i\circ\psi_i)(u) &=\left(\frac{y_0}{y_i},\dots,\widehat{\frac{y_i}{y_i}},\dots,\frac{y_n}{y_i}\right)\\ &=(y_0,\dots,\widehat{y_i},\dots,y_n)\\ &=(u_j)_{j\ne i}\\ &=u. \end{align*} Conversely, if $[x]\in U_i$, then \begin{align*} \psi_i(\varphi_i([x])) &= \left[\frac{x_0}{x_i}:\cdots:\widehat{\frac{x_i}{x_i}}:\cdots:\frac{x_n}{x_i}\ \text{with }1\text{ inserted in position }i\right]\\ &= \left[\frac{x_0}{x_i}:\cdots:\frac{x_i}{x_i}:\cdots:\frac{x_n}{x_i}\right]\\ &= [x_0:\cdots:x_n], \end{align*} because multiplying the displayed representative by $x_i\ne 0$ gives $(x_0,\dots,x_n)$. Thus each $\varphi_i$ is a bijection from $U_i$ to $\mathbb R^n$, and the quotient topology makes these maps homeomorphisms onto their images. It remains to compute the coordinate changes. Fix $i\ne j$. In the $i$th chart, write a point as \begin{align*} [y_0:\cdots:y_n], \qquad y_i=1,\quad y_k=u_k\ \text{for }k\ne i. \end{align*} This point lies in $U_j$ exactly when $y_j=u_j\ne 0$. Applying $\varphi_j$ gives the coordinates \begin{align*} (\varphi_j\circ\varphi_i^{-1})(u) = \left(\frac{y_0}{y_j},\dots,\widehat{\frac{y_j}{y_j}},\dots,\frac{y_n}{y_j}\right). \end{align*} Since $y_j=u_j$, the entries are \begin{align*} \frac{y_i}{y_j}&=\frac{1}{u_j},\\ \frac{y_k}{y_j}&=\frac{u_k}{u_j}\qquad(k\ne i,j). \end{align*} Hence every coordinate of the transition map is either $1/u_j$ or $u_k/u_j$, defined on the open set where $u_j\ne 0$. These are rational functions with nonzero denominator on their domain, so the transition maps are smooth. The same formula with $i$ and $j$ exchanged gives the inverse transition map, so the projective coordinate charts form a smooth atlas on $\mathbb RP^n$.[/proof] The projective charts show that a line can be described locally by normalizing one nonzero coordinate to $1$; the only coordinate changes needed are ratios of these normalized coordinates. [/example] For $n=2$, $\mathbb RP^2$ can also be viewed as $S^2$ with antipodal points identified. This description is geometrically useful, but the projective coordinate charts are the most direct way to verify the smooth structure. [example: Matrix Lie Groups] [claim]$GL(n,\mathbb R)$ is a smooth manifold of dimension $n^2$, and $SO(n)$ is a smooth manifold of dimension $n(n-1)/2$.[/claim] [proof]Identify $M_n(\mathbb R)$ with $\mathbb R^{n^2}$ by listing the matrix entries. The determinant is a polynomial in those entries, hence continuous, and \begin{align*} GL(n,\mathbb R) &=\{A\in M_n(\mathbb R):\det A\ne 0\}\\ &=\det^{-1}(\mathbb R\setminus\{0\}). \end{align*} Since $\mathbb R\setminus\{0\}$ is open in $\mathbb R$, the set $GL(n,\mathbb R)$ is open in $\mathbb R^{n^2}$. Therefore the identity chart on this open subset gives $GL(n,\mathbb R)$ the structure of a smooth manifold of dimension $n^2$. Now let \begin{align*} F:GL(n,\mathbb R)\to \operatorname{Sym}_n(\mathbb R), \qquad F(A)=A^\top A. \end{align*} The codomain is the vector space of symmetric matrices, with dimension \begin{align*} \dim \operatorname{Sym}_n(\mathbb R) &=n+\frac{n(n-1)}{2}\\ &=\frac{2n+n^2-n}{2}\\ &=\frac{n(n+1)}{2}, \end{align*} because a symmetric matrix has $n$ freely chosen diagonal entries and one freely chosen entry for each unordered pair $\{i,j\}$ with $i<j$. The derivative of $F$ at $A$ is computed from the first-order term in $F(A+tH)$. For $H\in M_n(\mathbb R)$, \begin{align*} F(A+tH) &=(A+tH)^\top(A+tH)\\ &=(A^\top+tH^\top)(A+tH)\\ &=A^\top A+tH^\top A+tA^\top H+t^2H^\top H\\ &=F(A)+t(H^\top A+A^\top H)+t^2H^\top H. \end{align*} Hence \begin{align*} dF_A(H)=H^\top A+A^\top H. \end{align*} At a point $A\in O(n)=F^{-1}(\{I\})$, we have $A^\top A=I$. To prove that $dF_A$ is surjective, let $S\in\operatorname{Sym}_n(\mathbb R)$ and choose \begin{align*} H=\frac12 AS. \end{align*} Then \begin{align*} dF_A(H) &=H^\top A+A^\top H\\ &=\left(\frac12 AS\right)^\top A+A^\top\left(\frac12 AS\right)\\ &=\frac12 S^\top A^\top A+\frac12 A^\top A S\\ &=\frac12 S^\top+\frac12 S\\ &=\frac12 S+\frac12 S\\ &=S, \end{align*} using $S^\top=S$ and $A^\top A=I$. Thus $I$ is a regular value of $F$. By *Regular Value Theorem*, $O(n)=F^{-1}(\{I\})$ is a smooth submanifold of $GL(n,\mathbb R)$ of dimension \begin{align*} \dim O(n) &=\dim GL(n,\mathbb R)-\dim \operatorname{Sym}_n(\mathbb R)\\ &=n^2-\frac{n(n+1)}{2}\\ &=\frac{2n^2-n^2-n}{2}\\ &=\frac{n(n-1)}{2}. \end{align*} Finally, if $A\in O(n)$, then \begin{align*} 1 &=\det I\\ &=\det(A^\top A)\\ &=\det(A^\top)\det(A)\\ &=\det(A)\det(A)\\ &=(\det A)^2. \end{align*} Therefore $\det A=\pm 1$ on $O(n)$. The determinant map restricted to $O(n)$ is continuous, and \begin{align*} SO(n) &=\{A\in O(n):\det A=1\}\\ &=(\det|_{O(n)})^{-1}\left(\left(\frac12,\frac32\right)\right), \end{align*} so $SO(n)$ is open in $O(n)$. An open subset of a smooth manifold is a smooth manifold of the same dimension, hence $SO(n)$ is a smooth manifold of dimension $n(n-1)/2$.[/proof] Thus $GL(n,\mathbb R)$ is a manifold because invertibility is an open condition, while $SO(n)$ is a manifold because the orthogonality equations impose exactly the independent symmetric-matrix constraints. [/example] These examples show two recurring constructions. Open subsets inherit smooth structures directly, while level sets and quotients require theorems guaranteeing that the local Euclidean property survives. ## Smooth Maps And Diffeomorphisms Once manifolds have smooth structures, what should it mean for a map between them to be smooth? Since manifolds are not usually vector spaces, smoothness must be tested after choosing charts in the source and target. [definition: Smooth Map] Let $M$ and $N$ be smooth manifolds, and let $F:M\to N$ be a map. The map $F$ is smooth if for every $p \in M$, every chart $(U,\varphi)$ around $p$, and every chart $(V,\psi)$ around $F(p)$ with $F(U)\subset V$ after possibly shrinking $U$, the coordinate representation \begin{align*} \psi \circ F \circ \varphi^{-1}: \varphi(U) \to \psi(V) \end{align*} is smooth as a map between open subsets of Euclidean spaces. [/definition] Smooth compatibility of charts makes this definition independent of the chosen coordinates. Replacing either chart composes the coordinate representation with smooth transition maps. Once smooth maps are intrinsic, we need a notion of sameness that preserves all smooth structure, not just the underlying topology. The correct equivalences are the smooth maps that can be reversed smoothly. [definition: Diffeomorphism] A diffeomorphism between smooth manifolds $M$ and $N$ is a bijective smooth map $F:M\to N$ whose inverse $F^{-1}:N\to M$ is smooth. [/definition] Diffeomorphic manifolds are the same object for smooth geometry. They may be presented by different equations, quotients, or atlases, but their calculus is equivalent. [example: A Smooth Map Between A Sphere And Projective Space] Let $q:S^n\to \mathbb RP^n$ send $x=(x_0,\dots,x_n)$ to the line $[x_0:\cdots:x_n]$. We show that $q$ is smooth and that its fibres are exactly antipodal pairs. First, $q$ identifies antipodal points because \begin{align*} q(-x) &=[-x_0:\cdots:-x_n]\\ &=[x_0:\cdots:x_n], \end{align*} since multiplying a homogeneous representative by the nonzero scalar $-1$ does not change its projective class. Conversely, if $q(x)=q(y)$, then $[x]=[y]$, so there is some $\lambda\in\mathbb R\setminus\{0\}$ with \begin{align*} y=\lambda x. \end{align*} Since $x,y\in S^n$, we have $|x|^2=|y|^2=1$, and therefore \begin{align*} 1 &=|y|^2\\ &=|\lambda x|^2\\ &=\lambda^2|x|^2\\ &=\lambda^2. \end{align*} Thus $\lambda=\pm 1$, so $y=x$ or $y=-x$. To check smoothness, work near a point $x\in S^n$. Choose an index $i$ with $x_i\ne 0$, and restrict to the open subset \begin{align*} W_i=\{(z_0,\dots,z_n)\in S^n:z_i\ne 0\}. \end{align*} The image of $W_i$ lies in the projective chart \begin{align*} U_i=\{[z_0:\cdots:z_n]\in\mathbb RP^n:z_i\ne 0\}. \end{align*} In this chart, \begin{align*} \varphi_i([z_0:\cdots:z_n]) = \left(\frac{z_0}{z_i},\dots,\widehat{\frac{z_i}{z_i}},\dots,\frac{z_n}{z_i}\right). \end{align*} Hence the coordinate representation of $q$ is \begin{align*} (\varphi_i\circ q)(z_0,\dots,z_n) &=\varphi_i([z_0:\cdots:z_n])\\ &= \left(\frac{z_0}{z_i},\dots,\widehat{\frac{z_i}{z_i}},\dots,\frac{z_n}{z_i}\right). \end{align*} Each component is the restriction to $W_i$ of the Euclidean function \begin{align*} (z_0,\dots,z_n)\longmapsto \frac{z_j}{z_i}\qquad (j\ne i), \end{align*} defined on the open set $\{z_i\ne 0\}\subset\mathbb R^{n+1}$. This function is smooth there because it is a quotient of coordinate functions with nonzero denominator. Therefore the coordinate representation of $q$ is smooth near every point of $S^n$, so $q:S^n\to\mathbb RP^n$ is smooth. Thus the projection from the sphere to projective space is a smooth map whose only identifications are $x\sim -x$. [/example] The definition of a smooth map should not depend on the particular charts used to test it. Once chart changes are smooth, the coordinate formula for a composite map is obtained by composing the coordinate formulas for the two maps, so smoothness ought to be stable under composition. This stability is what allows smooth manifolds and smooth maps to form a usable category rather than just a collection of chart-dependent objects. The sphere-to-projective-space map illustrates one smooth map, but the theory needs a general rule saying that smooth maps can be combined without leaving the category. The result below records that coordinate-level chain rule as an intrinsic statement about manifolds. [quotetheorem:3902] [citeproof:3902] This categorical viewpoint will matter throughout the course: tangent bundles, cotangent bundles, differential forms, and cohomology will all be functorial constructions attached to smooth maps. The smooth-compatibility hypothesis is what makes the statement coordinate-independent. Without it, a map could look smooth in one chart and fail to be smooth after changing coordinates, so composition would not be an intrinsic operation on the underlying spaces. A useful warning is that the identity map between the same topological manifold equipped with two different smooth atlases need not be a diffeomorphism. ## Submanifolds And Regular Values How do curved spaces arise inside a known manifold? The most useful answer is that they appear as images of well-behaved parametrisations or as level sets of maps with full-rank derivative. [definition: Immersion] Let $F:M\to N$ be a smooth map between smooth manifolds. The map $F$ is an immersion if for every $p \in M$, the derivative \begin{align*} dF_p:T_pM\to T_{F(p)}N \end{align*} is injective. [/definition] Immersions capture the first-order part of being a submanifold: infinitesimally, no tangent direction collapses. First-order injectivity alone does not control the global image. A parametrisation can be locally nondegenerate while still identifying distant points or giving the image a topology different from the one on the source. To model an actual parametrised submanifold, we also require the map to place the source homeomorphically onto its image. [definition: Embedding] Let $F:M\to N$ be a smooth map between smooth manifolds. The map $F$ is an embedding if $F$ is an immersion and $F:M\to F(M)$ is a homeomorphism onto its image, where $F(M)$ has the subspace topology from $N$. [/definition] An immersion has the right infinitesimal dimension but may fail to give a well-behaved subset. An embedding is the version of a parametrised submanifold whose topology and smooth structure agree with the image. Submanifolds also arise as solution sets of equations rather than as parametrised images. For a level set $F^{-1}(\{y\})$ to be smooth, the defining equations must not become redundant or singular along the solution set. The local condition that detects this is a surjectivity requirement on the derivative at every point lying over the chosen value. This condition needs its own name because it is the hypothesis that turns equations into geometry. Instead of describing the whole level set directly, we single out values whose fibres have independent first-order constraints at every solution. [definition: Regular Value] Let $F:M\to N$ be a smooth map and let $y \in N$. The point $y$ is a regular value of $F$ if for every $p \in F^{-1}(\{y\})$, the derivative \begin{align*} dF_p:T_pM\to T_yN \end{align*} is surjective. [/definition] The regular value condition says that the equations defining $F^{-1}(\{y\})$ have independent first-order constraints at every solution. The geometric question is whether this infinitesimal independence is enough to make the solution set into a genuine manifold, with dimension reduced by the number of independent equations. The theorem below is the bridge from a rank condition on the derivative to actual local coordinates on the level set. [quotetheorem:3903] [citeproof:3903] This theorem is the main practical tool for recognizing submanifolds without writing down charts by hand. It turns a rank computation into a smooth structure, but the regularity hypothesis is essential. For example, the level set of $F:\mathbb R^2\to\mathbb R$ given by $F(x,y)=xy$ at $0$ is the union of the two coordinate axes, which crosses itself at the origin and is not a $1$-dimensional submanifold there. The theorem also does not classify arbitrary subsets defined by equations; it applies only at levels where the derivative has full rank at every solution. [example: The Two-Sphere As A Regular Level Set] Let $f:\mathbb R^3\to\mathbb R$ be given by \begin{align*} f(x,y,z)=x^2+y^2+z^2. \end{align*} Then \begin{align*} f^{-1}(\{1\}) &=\{(x,y,z)\in\mathbb R^3:f(x,y,z)=1\}\\ &=\{(x,y,z)\in\mathbb R^3:x^2+y^2+z^2=1\}\\ &=S^2. \end{align*} We show that $1$ is a regular value of $f$. For $(x,y,z)\in\mathbb R^3$ and $(h_1,h_2,h_3)\in\mathbb R^3$, expand $f((x,y,z)+t(h_1,h_2,h_3))$: \begin{align*} f(x+th_1,y+th_2,z+th_3) &=(x+th_1)^2+(y+th_2)^2+(z+th_3)^2\\ &=x^2+2txh_1+t^2h_1^2+y^2+2tyh_2+t^2h_2^2+z^2+2tzh_3+t^2h_3^2\\ &=f(x,y,z)+t(2xh_1+2yh_2+2zh_3)+t^2(h_1^2+h_2^2+h_3^2). \end{align*} Hence the derivative is the linear map \begin{align*} df_{(x,y,z)}(h_1,h_2,h_3)=2xh_1+2yh_2+2zh_3. \end{align*} Now let $(x,y,z)\in f^{-1}(\{1\})=S^2$. Then \begin{align*} x^2+y^2+z^2=1, \end{align*} so at least one of $x,y,z$ is nonzero. To prove that $df_{(x,y,z)}:\mathbb R^3\to\mathbb R$ is surjective, let $a\in\mathbb R$ be arbitrary and choose \begin{align*} (h_1,h_2,h_3)=\frac{a}{2(x^2+y^2+z^2)}(x,y,z). \end{align*} Since $x^2+y^2+z^2=1$, this vector is well-defined, and \begin{align*} df_{(x,y,z)}(h_1,h_2,h_3) &=2x\left(\frac{ax}{2(x^2+y^2+z^2)}\right) +2y\left(\frac{ay}{2(x^2+y^2+z^2)}\right) +2z\left(\frac{az}{2(x^2+y^2+z^2)}\right)\\ &=\frac{a(x^2+y^2+z^2)}{x^2+y^2+z^2}\\ &=a. \end{align*} Thus $df_{(x,y,z)}$ is surjective at every point of $f^{-1}(\{1\})$, so $1$ is a regular value of $f$. By *Regular Value Theorem*, $S^2=f^{-1}(\{1\})$ is a smooth submanifold of $\mathbb R^3$ of dimension \begin{align*} \dim \mathbb R^3-\dim \mathbb R &=3-1\\ &=2. \end{align*} Thus the usual sphere is a smooth surface because it is cut out by one equation whose derivative is nonzero along the level set. [/example] The sphere example uses one scalar equation. Orthogonal matrices give a matrix-valued version of the same regular-level-set idea, where the independent equations are the symmetric entries of $A^\top A=I$. [example: Orthogonal Matrices As A Regular Level Set] [claim]For $F:GL(n,\mathbb R)\to \operatorname{Sym}_n(\mathbb R)$ defined by $F(A)=A^\top A$, the level set $O(n)=F^{-1}(\{I\})$ is a smooth submanifold of $GL(n,\mathbb R)$ of dimension $n(n-1)/2$.[/claim] [proof]First, $F(A)$ is symmetric because \begin{align*} (A^\top A)^\top &=A^\top (A^\top)^\top\\ &=A^\top A. \end{align*} The entries of $A^\top A$ are polynomial functions of the entries of $A$, since \begin{align*} (A^\top A)_{ij} &=\sum_{k=1}^n (A^\top)_{ik}A_{kj}\\ &=\sum_{k=1}^n A_{ki}A_{kj}. \end{align*} Thus $F$ is smooth. For $H\in M_n(\mathbb R)$, expand $F(A+tH)$: \begin{align*} F(A+tH) &=(A+tH)^\top(A+tH)\\ &=(A^\top+tH^\top)(A+tH)\\ &=A^\top A+tA^\top H+tH^\top A+t^2H^\top H\\ &=F(A)+t(H^\top A+A^\top H)+t^2H^\top H. \end{align*} Therefore \begin{align*} dF_A(H)=H^\top A+A^\top H. \end{align*} Now let $A\in F^{-1}(\{I\})=O(n)$, so \begin{align*} A^\top A=I. \end{align*} To prove that $dF_A:M_n(\mathbb R)\to \operatorname{Sym}_n(\mathbb R)$ is surjective, let $S\in \operatorname{Sym}_n(\mathbb R)$ be arbitrary and choose \begin{align*} H=\frac12 AS. \end{align*} Since $S^\top=S$, we get \begin{align*} dF_A(H) &=H^\top A+A^\top H\\ &=\left(\frac12 AS\right)^\top A+A^\top\left(\frac12 AS\right)\\ &=\frac12 S^\top A^\top A+\frac12 A^\top A S\\ &=\frac12 S^\top I+\frac12 I S\\ &=\frac12 S+\frac12 S\\ &=S. \end{align*} Hence $dF_A$ is surjective at every $A\in O(n)$, so $I$ is a regular value of $F$. By *Regular Value Theorem*, $O(n)=F^{-1}(\{I\})$ is a smooth submanifold of $GL(n,\mathbb R)$ of dimension \begin{align*} \dim GL(n,\mathbb R)-\dim \operatorname{Sym}_n(\mathbb R). \end{align*} Since $GL(n,\mathbb R)$ is an open subset of $M_n(\mathbb R)\cong \mathbb R^{n^2}$, \begin{align*} \dim GL(n,\mathbb R)=n^2. \end{align*} A symmetric matrix has $n$ free diagonal entries and one free off-diagonal entry for each pair $1\le i<j\le n$, so \begin{align*} \dim \operatorname{Sym}_n(\mathbb R) &=n+\frac{n(n-1)}{2}\\ &=\frac{2n+n^2-n}{2}\\ &=\frac{n(n+1)}{2}. \end{align*} Therefore \begin{align*} \dim O(n) &=n^2-\frac{n(n+1)}{2}\\ &=\frac{2n^2-n^2-n}{2}\\ &=\frac{n(n-1)}{2}. \end{align*} [/proof] The orthogonal group is smooth because the equations $A^\top A=I$ impose exactly the independent symmetric-matrix constraints near every orthogonal matrix. [/example] These submanifold examples complete the local model-building part of the chapter. The next remark marks the shift from spaces themselves to the linear algebra attached to each point. [remark: From Manifolds To Differential Forms] This chapter supplies the spaces and maps. The next step is to attach linear algebra to each point by defining tangent and cotangent spaces, then to assemble alternating covectors into differential forms on a manifold. [/remark] A smooth manifold is the setting in which calculus can be done intrinsically, without relying on a particular embedding in ℝ^n. Once charts and smooth transition maps are in place, the next task is to extract the linear data at each point that makes derivatives meaningful. # 2. Tangent and Cotangent Bundles This chapter turns the local coordinate picture of a smooth manifold into the intrinsic linear algebra that supports differential forms. We use the preceding chapter's definitions of smooth manifolds, charts, smooth maps, and transition maps, together with the ordinary chain rule from multivariable calculus. Each point $p \in M$ receives a vector space $T_pM$ measuring first-order motion through $p$, and these vector spaces assemble into the tangent bundle $TM$. The cotangent spaces $T_p^*M$ then provide the fibres in which differential $1$-forms live, so this chapter supplies the geometric stage for pullbacks, exterior derivatives, flows, and phase spaces developed later. ## Tangent Vectors from Curves How should we describe a velocity vector at a point of a manifold when the manifold may not sit inside any ambient Euclidean space? Writing a vector as an arrow in $\mathbb R^n$ fails because different charts give different coordinate arrows, and there is no preferred ambient vector space for an abstract manifold. The geometric repair is to regard a tangent vector as the first-order behaviour of a smooth curve passing through the point, then quotient out the chart-dependent parametrisations that have the same first-order effect. [definition: Curve Germ Tangency] Let $M$ be a smooth $n$-manifold and let $p \in M$. Two smooth curves $\gamma_1,\gamma_2:(-\varepsilon,\varepsilon)\to M$ with $\gamma_1(0)=\gamma_2(0)=p$ are tangent at $p$ if for one, hence every, chart $(U,\varphi)$ about $p$, \begin{align*} \frac{d}{dt}\Big|_{t=0}(\varphi\circ\gamma_1)(t)=\frac{d}{dt}\Big|_{t=0}(\varphi\circ\gamma_2)(t) \end{align*} in $\mathbb R^n$. [/definition] The phrase "for one, hence every" is justified by the smooth change-of-coordinates map. If $\psi\circ\varphi^{-1}$ is the transition map, the ordinary chain rule sends equal coordinate velocities in the $\varphi$-chart to equal coordinate velocities in the $\psi$-chart. Having identified when two curves represent the same first-order motion, we still need an actual object that stores all possible first-order directions at the point. The construction should remember velocity data but forget the particular curve used to produce it, since many curves have the same first-order behaviour. Taking equivalence classes under tangency gives precisely that object. [definition: Tangent Space by Curves] Let $M$ be a smooth manifold and let $p\in M$. The tangent space $T_pM$ is the set of equivalence classes $[\gamma]$ of smooth curves $\gamma:(-\varepsilon,\varepsilon)\to M$ with $\gamma(0)=p$, under the relation of tangency at $p$. [/definition] A chart turns this set into a vector space: add and scale curve classes by adding and scaling their coordinate velocity vectors in $\mathbb R^n$, then return to $M$ using the inverse coordinate map near $\varphi(p)$. The compatibility of these operations under transition maps means the resulting vector space does not depend on the chosen chart. This construction uses coordinates to define algebraic operations, so it must be checked that the answer is intrinsic. The theorem below is the point where the curve definition becomes a genuine coordinate-independent vector space of dimension equal to the manifold dimension. [quotetheorem:3904] [citeproof:3904] The smoothness of the transition maps is essential here: if charts were merely homeomorphic, a coordinate velocity need not transform linearly or even exist after changing charts. The theorem does not identify $T_pM$ with a preferred copy of $\mathbb R^n$; it says that every chart gives a temporary coordinate description, and transition derivatives translate between them. This is the first place where first-order calculus, rather than topology alone, enters the structure of a smooth manifold. [example: Nonlinear Reparametrisation Does Not Change the Tangent Vector] Take $M=\mathbb R$ with its standard smooth structure and $p=0$. Let $\gamma(t)=t$ and $\eta(t)=t+t^2$. Both curves pass through $0$ because \begin{align*} \gamma(0)&=0, & \eta(0)&=0+0^2=0. \end{align*} Using the identity chart $\operatorname{id}_{\mathbb R}:\mathbb R\to\mathbb R$, their coordinate representatives are the same functions: \begin{align*} (\operatorname{id}_{\mathbb R}\circ\gamma)(t)&=t, & (\operatorname{id}_{\mathbb R}\circ\eta)(t)&=t+t^2. \end{align*} Their coordinate velocities at $0$ are \begin{align*} \frac{d}{dt}\Big|_{t=0}(\operatorname{id}_{\mathbb R}\circ\gamma)(t) &=\frac{d}{dt}\Big|_{t=0}t =1,\\ \frac{d}{dt}\Big|_{t=0}(\operatorname{id}_{\mathbb R}\circ\eta)(t) &=\frac{d}{dt}\Big|_{t=0}(t+t^2) =1+2\cdot 0 =1. \end{align*} Thus $\gamma$ and $\eta$ are tangent at $0$, so they determine the same tangent vector $[\gamma]=[\eta]\in T_0\mathbb R$. The curves are not equal near $0$, since $\eta(t)-\gamma(t)=t^2$ is nonzero for every $t\ne 0$, but their first-order motion at $0$ is identical. [/example] ## Tangent Vectors as Derivations Curves describe motion, but differential forms interact with functions. If we kept only curve classes, every later construction would have to choose curves in order to evaluate a form or compare maps. The algebraic description avoids this dependence: a tangent vector is exactly an operator that differentiates smooth functions at a point and satisfies the product rule. [definition: Derivation at a Point] Let $M$ be a smooth manifold and let $p\in M$. A derivation at $p$ is a linear map \begin{align*} D:C^\infty(M)\to\mathbb R \end{align*} satisfying the Leibniz rule \begin{align*} D(fg)=f(p)D(g)+g(p)D(f) \end{align*} for all $f,g\in C^\infty(M)$. [/definition] The Leibniz rule forces a derivation to depend only on first-order information near $p$. Constants are killed by any derivation, and if a function vanishes to second order at $p$, every derivation sends it to zero. We now need to compare this algebraic description with the earlier geometric one by curves. If both give the same tangent space, then tangent vectors can be used interchangeably as velocities of curves or as first-order differential operators on smooth functions. [quotetheorem:3905] [citeproof:3905] The smoothness hypothesis on functions is necessary: for continuous functions there is no coordinate Taylor expansion, and the product rule alone does not recover a finite-dimensional tangent space. The theorem does not say that tangent vectors differentiate functions away from $p$; they are first-order operators depending only on germs at $p$. This reformulation is what makes the differential of a smooth map expressible by precomposition with functions. [example: Coordinate Coefficients Are Measured by Test Functions] Let $(U,\varphi)$ be a chart with coordinate functions $(x_1,\dots,x_n)$, and write $a_i=x_i(p)$ and $y_i=x_i-a_i$. Suppose $v\in T_pM$ satisfies $v(x_j)=0$ for every $j$. Since derivations kill constants, this also gives \begin{align*} v(y_j)=v(x_j-a_j)=v(x_j)-v(a_j)=0. \end{align*} Indeed, if $D$ is any derivation, then \begin{align*} D(1)=D(1\cdot 1)=1\cdot D(1)+1\cdot D(1)=2D(1), \end{align*} so $D(1)=0$, and therefore $D(c)=cD(1)=0$ for every constant $c$. We show that $v(f)=0$ for every smooth function $f$, which proves $v=0$ as a derivation. Work in coordinates near $p$. Put $F=f\circ\varphi^{-1}$ and $a=\varphi(p)$. After shrinking the coordinate neighbourhood, assume the segment $a+tz$ stays in the coordinate image for $0\leq t\leq 1$. For $z=(z_1,\dots,z_n)$ near $0$, the one-variable fundamental theorem of calculus applied to $t\mapsto F(a+tz)$ gives \begin{align*} F(a+z)-F(a) &=\int_0^1 \frac{d}{dt}F(a+tz)\,dt\\ &=\int_0^1 \sum_{i=1}^n z_i\,\frac{\partial F}{\partial x_i}(a+tz)\,dt\\ &=\sum_{i=1}^n z_i\frac{\partial F}{\partial x_i}(a) +\sum_{i=1}^n z_i\int_0^1\left(\frac{\partial F}{\partial x_i}(a+tz)-\frac{\partial F}{\partial x_i}(a)\right)dt. \end{align*} For each $i$, applying the same theorem to $s\mapsto \frac{\partial F}{\partial x_i}(a+sz)$ gives \begin{align*} \frac{\partial F}{\partial x_i}(a+tz)-\frac{\partial F}{\partial x_i}(a) &=\int_0^t \frac{d}{ds}\frac{\partial F}{\partial x_i}(a+sz)\,ds\\ &=\int_0^t \sum_{j=1}^n z_j\,\frac{\partial^2 F}{\partial x_j\partial x_i}(a+sz)\,ds. \end{align*} Substituting this into the previous identity yields \begin{align*} F(a+z) =F(a)+\sum_{i=1}^n \frac{\partial F}{\partial x_i}(a)z_i +\sum_{i,j=1}^n z_i z_j A_{ij}(z), \end{align*} where \begin{align*} A_{ij}(z)=\int_0^1\int_0^t \frac{\partial^2 F}{\partial x_j\partial x_i}(a+sz)\,ds\,dt \end{align*} is smooth near $0$. Translating back to $M$, near $p$ we have \begin{align*} f =f(p)+\sum_{i=1}^n \frac{\partial F}{\partial x_i}(a)y_i +\sum_{i,j=1}^n y_i y_j B_{ij}, \end{align*} with $B_{ij}=A_{ij}\circ(\varphi-a)$ smooth near $p$. Applying $v$ gives \begin{align*} v(f) &=v(f(p)) +\sum_{i=1}^n \frac{\partial F}{\partial x_i}(a)v(y_i) +\sum_{i,j=1}^n v(y_i y_j B_{ij})\\ &=0+\sum_{i=1}^n \frac{\partial F}{\partial x_i}(a)\cdot 0 +\sum_{i,j=1}^n \left(y_i(p)v(y_jB_{ij})+(y_jB_{ij})(p)v(y_i)\right)\\ &=0+0+\sum_{i,j=1}^n \left(0\cdot v(y_jB_{ij})+0\cdot 0\right)\\ &=0. \end{align*} Thus every smooth test function has derivative $0$ in the direction $v$, so the coordinate functions $x_1,\dots,x_n$ detect all first-order components of a tangent vector. [/example] ## Differentials of Smooth Maps If a smooth map sends points of $M$ to points of $N$, what is its first-order effect on tangent vectors? Applying the map only to coordinate representatives is not intrinsic, because the representative vector changes when the chart changes. The differential solves this by pushing forward curves, or equivalently by letting tangent derivations act on functions pulled back from the target. [definition: Differential of a Smooth Map] Let $F:M\to N$ be a smooth map and let $p\in M$. The differential of $F$ at $p$ is the map \begin{align*} dF_p:T_pM\to T_{F(p)}N \end{align*} defined on curve classes by \begin{align*} dF_p([\gamma])=[F\circ\gamma]. \end{align*} [/definition] Using derivations, the same map is characterised by its action on functions: for $v\in T_pM$ and $h\in C^\infty(N)$, \begin{align*} (dF_pv)(h)=v(h\circ F). \end{align*} This formula is often the most efficient way to compute with $dF_p$ without introducing coordinates. The definition by curve classes is only useful if it really produces a linear map between tangent spaces, independent of representatives and compatible with the vector-space operations already placed on tangent spaces. The needed check is that smooth maps preserve first-order equivalence of curves and send sums and scalar multiples of velocities to the corresponding sums and scalar multiples after applying the map. Before using $dF_p$ as a genuine derivative between manifolds, we must know that this construction is well defined and linear. This is the first structural test for tangent maps: it guarantees that the notation $dF_p$ denotes an honest linear transformation, so later constructions can use it without returning to representatives of curves each time. The next formal step is therefore not another definition, but a validation theorem. It answers the practical question of whether the curve-class formula really gives a linear map $T_pM\to T_{F(p)}N$ for every smooth map $F$. [quotetheorem:3906] [citeproof:3906] Smoothness of $F$ is necessary: a merely continuous map can send smooth curves to curves with no well-defined first derivative at the image point. The theorem does not say that $dF_p$ determines $F$ globally; maps with the same first-order behaviour at $p$ may differ away from $p$ or to higher order at $p$. Its role is local and first-order, and this is exactly the level at which pullbacks of covectors and forms will operate. There is also a compatibility question for composing smooth maps. If tangent maps are to behave like the first-order part of ordinary functions, then differentiating $G\circ F$ should give the same result as first differentiating $F$ and then differentiating $G$ at the image point. This functorial behaviour is what makes tangent spaces usable across maps rather than only at isolated manifolds. [quotetheorem:3907] [citeproof:3907] The matching of base points is essential: $dF_p$ lands in $T_{F(p)}N$, so there is no single fixed target tangent space unless the image point is fixed. The theorem does not make tangent spaces independent of points; it says that smooth maps carry first-order data at $p$ functorially to first-order data at $F(p)$. More precisely, this is the pointwise form of the tangent functor, which later becomes the bundle map $dF:TM\to TN$ lying over $F$. [example: Pushforward of a Rotation Vector Field] Let \begin{align*} R_\theta(x,y)=(x\cos\theta-y\sin\theta,\;x\sin\theta+y\cos\theta). \end{align*} Write \begin{align*} (x',y')=R_\theta(x,y), \end{align*} so \begin{align*} x'&=x\cos\theta-y\sin\theta, & y'&=x\sin\theta+y\cos\theta. \end{align*} The coordinate vector of $X_{(x,y)}$ in the basis $(\partial_x,\partial_y)$ is $(-y,x)$. Since $R_\theta$ is linear, its Jacobian matrix is constant: \begin{align*} J R_\theta = \begin{pmatrix} \cos\theta & -\sin\theta\\ \sin\theta & \cos\theta \end{pmatrix}. \end{align*} Therefore \begin{align*} d(R_\theta)_{(x,y)}X_{(x,y)} &= \begin{pmatrix} \cos\theta & -\sin\theta\\ \sin\theta & \cos\theta \end{pmatrix} \begin{pmatrix} -y\\ x \end{pmatrix}\\ &= \begin{pmatrix} -y\cos\theta-x\sin\theta\\ -y\sin\theta+x\cos\theta \end{pmatrix}\\ &= \begin{pmatrix} -(x\sin\theta+y\cos\theta)\\ x\cos\theta-y\sin\theta \end{pmatrix}\\ &= \begin{pmatrix} -y'\\ x' \end{pmatrix}. \end{align*} Thus, as a tangent vector at $R_\theta(x,y)=(x',y')$, \begin{align*} d(R_\theta)_{(x,y)}X_{(x,y)} &=-y'\partial_x|_{(x',y')}+x'\partial_y|_{(x',y')}\\ &=X_{(x',y')}\\ &=X_{R_\theta(x,y)}. \end{align*} So the rotation vector field is carried to itself by every rotation: rotating the point and then taking the infinitesimal rotation vector gives the same result as first taking the infinitesimal rotation vector and then pushing it forward. [/example] ## The Tangent Bundle and Vector Fields So far $T_pM$ has been attached to a single point $p$. Treating the family $\{T_pM\}_{p\in M}$ as unrelated vector spaces would make it impossible to define a smooth vector field, a flow, or a smoothly varying differential equation on $M$. The tangent bundle assembles all tangent spaces into one manifold lying over $M$, with coordinates that track both the base point and the vector in the fibre. [illustration:dfii-tangent-bundle-projection] [definition: Tangent Bundle] Let $M$ be a smooth manifold. The tangent bundle of $M$ is the disjoint union \begin{align*} TM=\bigsqcup_{p\in M}T_pM. \end{align*} The projection map $\pi:TM\to M$ sends $v\in T_pM$ to $p$. [/definition] A chart $(U,\varphi)$ on $M$ with coordinates $(x_1,\dots,x_n)$ gives a bundle chart \begin{align*} \widetilde{\varphi}:\pi^{-1}(U)&\to \varphi(U)\times\mathbb R^n, & v&\mapsto (\varphi(p),v_1,\dots,v_n), \end{align*} where $v=\sum_i v_i\partial_{x_i}|_p$. The base coordinates record the point, and the fibre coordinates record the tangent vector at that point. These local descriptions must agree smoothly on overlaps before $TM$ can itself be treated as a manifold. The issue is that changing charts changes both the base point coordinates and the vector coordinates, with the latter transformed by a Jacobian depending on the base point. The next structural result verifies that these overlap maps are smooth and therefore assemble the disjoint union into a smooth bundle. [quotetheorem:3908] [citeproof:3908] The smooth dependence of the Jacobian on the base point is necessary: if transition functions were only once differentiable with discontinuous derivative, the displayed transition maps for $TM$ need not be smooth. The theorem does not make $TM$ a product $M\times\mathbb R^n$ in general; it gives local product charts whose fibre coordinates may twist from chart to chart. This local product structure is the setting in which vector fields become smooth sections and later generate flows. [definition: Vector Field] Let $M$ be a smooth manifold. A vector field on $M$ is a section of the tangent bundle, that is, a map $X:M\to TM$ such that $\pi\circ X=\operatorname{id}_M$. [/definition] A vector field is smooth when it is smooth as a map of manifolds. In local coordinates it has the form \begin{align*} X=\sum_{i=1}^n X_i\partial_{x_i}, \end{align*} where each coefficient function $X_i:U\to\mathbb R$ is smooth. [example: Tangent Space to the Sphere] [claim]For $S^n=\{x\in\mathbb R^{n+1}:|x|=1\}$ and $p\in S^n$, the tangent space is \begin{align*} T_pS^n=\{v\in\mathbb R^{n+1}:v\cdot p=0\}. \end{align*}[/claim] [proof]Let $\gamma:(-\varepsilon,\varepsilon)\to S^n$ be a smooth curve with $\gamma(0)=p$. Since $\gamma(t)\in S^n$ for every $t$, we have \begin{align*} \gamma(t)\cdot\gamma(t)=|\gamma(t)|^2=1. \end{align*} Differentiating both sides gives \begin{align*} \frac{d}{dt}\bigl(\gamma(t)\cdot\gamma(t)\bigr) &=\frac{d}{dt}(1)\\ \gamma'(t)\cdot\gamma(t)+\gamma(t)\cdot\gamma'(t) &=0\\ 2\gamma(t)\cdot\gamma'(t)&=0. \end{align*} At $t=0$, this becomes \begin{align*} 2p\cdot\gamma'(0)=0, \end{align*} so \begin{align*} p\cdot\gamma'(0)=0. \end{align*} Thus every tangent vector to $S^n$ at $p$, viewed as an ambient velocity in $\mathbb R^{n+1}$, lies in $\{v:v\cdot p=0\}$. Conversely, let $v\in\mathbb R^{n+1}$ satisfy $v\cdot p=0$. Define \begin{align*} \gamma(t)=\frac{p+tv}{|p+tv|} \end{align*} for $t$ near $0$. The denominator is nonzero near $0$ because \begin{align*} |p+tv|^2 &=(p+tv)\cdot(p+tv)\\ &=p\cdot p+2t\,p\cdot v+t^2v\cdot v\\ &=1+0+t^2|v|^2\\ &=1+t^2|v|^2. \end{align*} Also, \begin{align*} |\gamma(t)|^2 &=\frac{|p+tv|^2}{|p+tv|^2}\\ &=1, \end{align*} so $\gamma(t)\in S^n$, and \begin{align*} \gamma(0)=\frac{p}{|p|}=p. \end{align*} Using $|p+tv|=(1+t^2|v|^2)^{1/2}$, we write \begin{align*} \gamma(t)=(p+tv)(1+t^2|v|^2)^{-1/2}. \end{align*} Differentiating at $0$ gives \begin{align*} \gamma'(0) &=v(1+0^2|v|^2)^{-1/2} +(p+0v)\left(-\frac12\right)(1+0^2|v|^2)^{-3/2}(2\cdot 0\cdot |v|^2)\\ &=v\cdot 1+p\cdot 0\\ &=v. \end{align*} Therefore every vector orthogonal to $p$ occurs as the velocity of a curve in $S^n$ through $p$.[/proof] The tangent space to the sphere at $p$ is exactly the hyperplane through the origin perpendicular to the radius vector $p$. [/example] ## Cotangent Spaces and the Tautological Form Differential forms are built from covectors rather than vectors. Vectors act on functions, but a $1$-form must eat a vector and return a scalar in a way that varies smoothly with the base point; this cannot be expressed using tangent vectors alone. Once $T_pM$ is available, the cotangent space is its dual, and the cotangent bundle assembles these dual spaces over all points of $M$. [illustration:dfii-cotangent-fibre] [definition: Cotangent Space] Let $M$ be a smooth manifold and let $p\in M$. The cotangent space at $p$ is the dual vector space \begin{align*} T_p^*M=(T_pM)^*. \end{align*} Its elements are called covectors at $p$. [/definition] If $(x_1,\dots,x_n)$ are local coordinates, the coordinate covectors $dx_i|_p$ are defined by \begin{align*} dx_i|_p\left(\partial_{x_j}|_p\right)=\delta_{ij}. \end{align*} Thus $(dx_1|_p,\dots,dx_n|_p)$ is the basis dual to $(\partial_{x_1}|_p,\dots,\partial_{x_n}|_p)$. To compare covectors at nearby points, and eventually to speak about smooth one-forms, the pointwise dual spaces must be assembled over the whole manifold. This is not just a set-theoretic union: the base point has to vary together with the covector coordinates, and chart changes will use inverse-transpose Jacobian data rather than the tangent-bundle rule. The cotangent bundle is the global object that records this moving family of dual spaces. [definition: Cotangent Bundle] Let $M$ be a smooth manifold. The cotangent bundle is the disjoint union \begin{align*} T^*M=\bigsqcup_{p\in M}T_p^*M, \end{align*} with projection $\pi:T^*M\to M$ sending $\alpha\in T_p^*M$ to $p$. [/definition] The same transition-map construction used for $TM$ makes $T^*M$ into a smooth $2n$-manifold. In coordinates a covector has the form $\alpha=\sum_i \xi_i dx_i|_p$, so a cotangent bundle chart records $(x_1,\dots,x_n,\xi_1,\dots,\xi_n)$. The next object extracts geometry directly from the projection $\pi:T^*M\to M$. A point of $T^*M$ is itself a covector on the base, so any tangent vector to $T^*M$ can be projected to $M$ and then evaluated by that covector. Packaging this evaluation at every point gives the canonical one-form that later produces the standard symplectic form on a cotangent bundle. [definition: Tautological One-Form] Let $M$ be a smooth manifold and let $\pi:T^*M\to M$ be the cotangent bundle projection. The tautological one-form is the smooth section \begin{align*} \lambda:T^*M\to T^*(T^*M) \end{align*} of the cotangent bundle of $T^*M$ such that, for each $\alpha\in T_p^*M$, the covector \begin{align*} \lambda_\alpha:T_\alpha(T^*M)\to\mathbb R \end{align*} is defined by \begin{align*} \lambda_\alpha(V)=\alpha(d\pi_\alpha(V)) \end{align*} for every $V\in T_\alpha(T^*M)$. [/definition] The tautological form evaluates a tangent vector to the cotangent bundle by first projecting it down to a tangent vector on $M$, then applying the covector that labels the point of $T^*M$. In canonical coordinates $(x_1,\dots,x_n,\xi_1,\dots,\xi_n)$ on $T^*M$, it is written \begin{align*} \lambda=\sum_{i=1}^n \xi_i dx_i. \end{align*} This formula is the starting point for the canonical symplectic form in later geometry courses. Its definition also explains why cotangent bundles are natural phase spaces: a point of $T^*M$ already contains both a base position and a covector that can pair with base velocities. [illustration:dfii-tautological-one-form] [example: Vertical Vectors Are Invisible to the Tautological Form] For $M=\mathbb R^n$, identify \begin{align*} T^*\mathbb R^n=\mathbb R^n\times(\mathbb R^n)^* \end{align*} and write a point of the cotangent bundle as $(x,\xi)$, where $x\in\mathbb R^n$ and $\xi\in(\mathbb R^n)^*$. The cotangent bundle projection is \begin{align*} \pi:\mathbb R^n\times(\mathbb R^n)^*&\to\mathbb R^n, & \pi(x,\xi)&=x. \end{align*} A tangent vector at $(x,\xi)$ is represented in these product coordinates by a pair \begin{align*} (v,\eta)\in \mathbb R^n\times(\mathbb R^n)^*, \end{align*} where $v$ is the base component and $\eta$ is the fibre component. To compute $d\pi_{(x,\xi)}(v,\eta)$, use the curve \begin{align*} \Gamma(t)=(x+tv,\xi+t\eta). \end{align*} Then \begin{align*} \Gamma(0)&=(x,\xi),\\ \pi(\Gamma(t))&=\pi(x+tv,\xi+t\eta)\\ &=x+tv, \end{align*} so \begin{align*} d\pi_{(x,\xi)}(v,\eta) &=\frac{d}{dt}\Big|_{t=0}\pi(\Gamma(t))\\ &=\frac{d}{dt}\Big|_{t=0}(x+tv)\\ &=v. \end{align*} By the definition of the tautological one-form, \begin{align*} \lambda_{(x,\xi)}(v,\eta) &=\xi\bigl(d\pi_{(x,\xi)}(v,\eta)\bigr)\\ &=\xi(v). \end{align*} Therefore a vertical tangent vector satisfies \begin{align*} \lambda_{(x,\xi)}(0,\eta) &=\xi\bigl(d\pi_{(x,\xi)}(0,\eta)\bigr)\\ &=\xi(0)\\ &=0, \end{align*} because $\xi$ is linear, while a horizontal tangent vector satisfies \begin{align*} \lambda_{(x,\xi)}(v,0) &=\xi\bigl(d\pi_{(x,\xi)}(v,0)\bigr)\\ &=\xi(v). \end{align*} Thus $\lambda$ ignores motion purely in the covector fibre and records exactly the base velocity paired with the covector labelling the point of $T^*\mathbb R^n$. [/example] ## Local Frames, Coframes, and Change of Coordinates Calculations with tangent and cotangent bundles require local bases that vary smoothly. A basis chosen separately in each fibre gives no control over smoothness, so its coefficients cannot be differentiated or transformed reliably. Local frames and coframes provide smoothly varying bases, and the important point is that tangent vectors and covectors transform in dual ways under a change of coordinates. [definition: Local Frame and Coframe] Let $M$ be a smooth $n$-manifold and let $U\subset M$ be open. A local frame on $U$ is an ordered $n$-tuple $(E_1,\dots,E_n)$ of smooth maps \begin{align*} E_i:U\to TU \end{align*} such that $\pi\circ E_i=\operatorname{id}_U$ for each $i$ and $(E_1(p),\dots,E_n(p))$ is a basis of $T_pM$ for every $p\in U$. A local coframe on $U$ is an ordered $n$-tuple $(\varepsilon^1,\dots,\varepsilon^n)$ of smooth maps \begin{align*} \varepsilon^i:U\to T^*U \end{align*} such that $\pi\circ \varepsilon^i=\operatorname{id}_U$ for each $i$ and $(\varepsilon^1(p),\dots,\varepsilon^n(p))$ is a basis of $T_p^*M$ for every $p\in U$. [/definition] Every coordinate chart supplies a local frame $(\partial_{x_1},\dots,\partial_{x_n})$ and its dual coframe $(dx_1,\dots,dx_n)$. General frames are useful because many geometric constructions are not tied to a coordinate system. When two coordinate charts overlap, their associated frames and coframes give two bases for the same tangent and cotangent spaces. Calculations with vector fields and one-forms are intrinsic only if we know exactly how the component functions change from one basis to the other. The transformation rule below records the Jacobian relation for tangent vectors and the dual inverse relation for covectors. [quotetheorem:3909] [citeproof:3909] The invertibility of the coordinate change is necessary: without a genuine chart transition, there is no inverse Jacobian relating the two bases. The theorem does not say that vector and covector components transform in the same way; it says precisely that they transform by dual inverse rules. This dual transformation law is the reason covectors, and hence differential forms, have coordinate expressions that are independent of the chart once their coefficients transform correctly. The next chapter uses these local coframes to define $k$-forms as alternating covariant tensors on each tangent space. [example: Covector Transformation in Polar Coordinates] On a local polar coordinate branch in $\mathbb R^2\setminus\{0\}$, we have $r>0$ and \begin{align*} x&=r\cos\theta, & y&=r\sin\theta. \end{align*} Since \begin{align*} r=(x^2+y^2)^{1/2}, \end{align*} the partial derivatives of $r$ in Cartesian coordinates are \begin{align*} \frac{\partial r}{\partial x} &=\frac12(x^2+y^2)^{-1/2}\cdot 2x\\ &=\frac{x}{(x^2+y^2)^{1/2}}\\ &=\frac{x}{r}\\ &=\cos\theta, \\ \frac{\partial r}{\partial y} &=\frac12(x^2+y^2)^{-1/2}\cdot 2y\\ &=\frac{y}{(x^2+y^2)^{1/2}}\\ &=\frac{y}{r}\\ &=\sin\theta. \end{align*} Therefore the coframe transformation law gives \begin{align*} dr &=\frac{\partial r}{\partial x}dx+\frac{\partial r}{\partial y}dy\\ &=\cos\theta\,dx+\sin\theta\,dy. \end{align*} On a coordinate branch where $\theta=\arctan(y/x)$, the partial derivatives of $\theta$ are \begin{align*} \frac{\partial \theta}{\partial x} &=\frac{1}{1+(y/x)^2}\cdot\frac{\partial}{\partial x}\left(\frac{y}{x}\right)\\ &=\frac{1}{1+y^2/x^2}\left(-\frac{y}{x^2}\right)\\ &=\frac{x^2}{x^2+y^2}\left(-\frac{y}{x^2}\right)\\ &=-\frac{y}{x^2+y^2}\\ &=-\frac{r\sin\theta}{r^2}\\ &=-\frac{\sin\theta}{r}, \\ \frac{\partial \theta}{\partial y} &=\frac{1}{1+(y/x)^2}\cdot\frac{\partial}{\partial y}\left(\frac{y}{x}\right)\\ &=\frac{1}{1+y^2/x^2}\cdot\frac1x\\ &=\frac{x^2}{x^2+y^2}\cdot\frac1x\\ &=\frac{x}{x^2+y^2}\\ &=\frac{r\cos\theta}{r^2}\\ &=\frac{\cos\theta}{r}. \end{align*} Hence \begin{align*} d\theta &=\frac{\partial \theta}{\partial x}dx+\frac{\partial \theta}{\partial y}dy\\ &=-\frac{\sin\theta}{r}dx+\frac{\cos\theta}{r}dy. \end{align*} Thus the polar coframe is obtained from the Cartesian coframe by the dual Jacobian transformation, with $dr$ measuring radial change and $d\theta$ measuring angular change scaled by $1/r$. [/example] With tangent and cotangent spaces defined pointwise, we can assemble the local linear algebra into the language of differential forms on a manifold. That is the natural next step, because forms are the objects on which exterior differentiation, pullback, and integration will act. # 3. Differential Forms on Manifolds This chapter turns the tangent and cotangent constructions of the preceding chapter into a calculus on manifolds. On Euclidean space, a differential form is written using the coordinate symbols $dx_i$ and the exterior derivative is computed by differentiating coefficients. The point of the manifold theory is to explain why these coordinate formulae define intrinsic objects, how they transform under smooth maps, and why the identities $d^2=0$ and $\varphi^*d=d\varphi^*$ survive on every smooth manifold. ## Alternating Covectors and Differential Forms What should replace the constant covectors $dx_i$ from $\mathbb R^n$ when the underlying space has no preferred global coordinates? At each point $p\in M$, the cotangent space $T_p^*M$ supplies the linear measurements of tangent vectors at $p$. A $k$-form assigns, smoothly in $p$, an alternating $k$-linear function on $T_pM$. [definition: Alternating Covector] Let $V$ be a finite-dimensional real vector space. An alternating $k$-covector on $V$ is a multilinear map \begin{align*} \alpha: V^k \to \mathbb R \end{align*} such that \begin{align*} \alpha(v_{\sigma(1)},\dots,v_{\sigma(k)})=\operatorname{sgn}(\sigma)\alpha(v_1,\dots,v_k) \end{align*} for every permutation $\sigma\in S_k$ and all $v_1,\dots,v_k\in V$. [/definition] The vector space of alternating $k$-covectors on $V$ is denoted $\Lambda^k V^*$. For $k=0$ this is interpreted as $\mathbb R$, and for $k>\dim V$ it is the zero vector space. On a manifold, the relevant vector space for alternating covectors changes from point to point because the tangent space changes from point to point. To make a geometric object vary smoothly, these pointwise spaces must be bundled over the manifold in the same way that tangent and cotangent spaces were bundled. This produces the natural home for degree-$k$ covariant data. [definition: Exterior Power of the Cotangent Bundle] Let $M$ be a smooth manifold. The bundle of alternating $k$-covectors on $M$ is \begin{align*} \Lambda^k(T^*M)=\bigsqcup_{p\in M}\Lambda^k(T_p^*M), \end{align*} with projection sending an element of $\Lambda^k(T_p^*M)$ to $p$. [/definition] This bundle is a smooth vector bundle of rank $\binom{n}{k}$ when $M$ has dimension $n$. The geometric object we want is not a single alternating covector at one point, but a smoothly varying choice of such a covector at every point. Smoothness is what lets the coefficients in local coordinates be differentiated, wedged, pulled back, and integrated. A differential form is therefore defined as a smooth section of this exterior-power bundle. [definition: Differential Form] Let $M$ be a smooth manifold. A differential $k$-form on $M$ is a smooth section \begin{align*} \omega: M\to \Lambda^k(T^*M). \end{align*} The space of differential $k$-forms on $M$ is denoted $\Omega^k(M)$. [/definition] A $0$-form is a smooth function $f\in C^\infty(M)$. A $1$-form assigns to each point $p$ a covector on $T_pM$, while an $n$-form on an $n$-manifold assigns an oriented volume density with signs. [example: Basic Forms on Euclidean Space] On an open set $U\subset \mathbb R^n$ with coordinates $(x_1,\dots,x_n)$, each tangent space $T_xU$ has coordinate basis $\partial/\partial x_1|_x,\dots,\partial/\partial x_n|_x$, and the covectors $(dx_1)_x,\dots,(dx_n)_x$ are the dual basis, so \begin{align*} (dx_i)_x\left(\frac{\partial}{\partial x_j}\Big|_x\right)=\delta_{ij}. \end{align*} For a strictly increasing multi-index $I=(i_1<\cdots<i_k)$, write \begin{align*} (dx_I)_x=(dx_{i_1})_x\wedge\cdots\wedge(dx_{i_k})_x. \end{align*} The alternating $k$-covectors $(dx_I)_x$ form a basis of $\Lambda^k(T_x^*U)$, so for every $\omega\in\Omega^k(U)$ and every $x\in U$ there are unique real numbers $f_I(x)$ such that \begin{align*} \omega_x=\sum_{1\le i_1<\cdots<i_k\le n} f_{i_1\cdots i_k}(x)\,(dx_{i_1})_x\wedge\cdots\wedge(dx_{i_k})_x. \end{align*} Equivalently, as a form on $U$, \begin{align*} \omega=\sum_{1\le i_1<\cdots<i_k\le n} f_{i_1\cdots i_k}\,dx_{i_1}\wedge\cdots\wedge dx_{i_k}. \end{align*} The coefficients are determined by evaluating $\omega_x$ on coordinate tangent vectors. If $I=(i_1<\cdots<i_k)$, then \begin{align*} \omega_x\left(\frac{\partial}{\partial x_{i_1}}\Big|_x,\dots,\frac{\partial}{\partial x_{i_k}}\Big|_x\right) &=\sum_J f_J(x)\,(dx_J)_x\left(\frac{\partial}{\partial x_{i_1}}\Big|_x,\dots,\frac{\partial}{\partial x_{i_k}}\Big|_x\right)\\ &=f_I(x), \end{align*} because $(dx_J)_x$ gives $1$ on the ordered coordinate vectors indexed by $J$ and gives $0$ on this ordered list when $J\ne I$. Since $\omega$ is a smooth section and the coordinate vector fields are smooth, each function $f_I$ is smooth. Thus coordinate $k$-forms on Euclidean open sets are exactly smooth linear combinations of the wedge products $dx_{i_1}\wedge\cdots\wedge dx_{i_k}$. [/example] ## Coordinate Expressions and Change of Chart How can a form be written down in a coordinate chart without losing its intrinsic meaning? Coordinates identify a small piece of $M$ with an open subset of $\mathbb R^n$, but different charts give different coordinate covectors. The definition of a form is arranged so that the coefficient functions and the covectors transform together. Let $(U,x)$ be a coordinate chart on an $n$-manifold $M$, with coordinate functions $x_i:U\to\mathbb R$. The differentials $dx_1,\dots,dx_n$ are local smooth $1$-forms on $U$. [quotetheorem:3910] [citeproof:3910] The theorem is the local bridge from the manifold definition to the computational rules from exterior calculus on $\mathbb R^n$. The hypotheses are local: without a chart there is no distinguished list $dx_1,\dots,dx_n$, and without smoothness of $\omega$ the pointwise coefficients need not be smooth functions. The theorem does not say that the coefficient functions are globally defined, since a form on a circle or sphere may require several coordinate descriptions. Its role is to make every later construction reducible to a coordinate calculation followed by a check that the result transforms correctly. A coordinate expression is useful only if it can be translated across overlaps of charts. On an overlap, the same intrinsic one-form can be written using the $x$-coframe or the $y$-coframe, and the obstruction to consistency is whether these coframes transform by a smooth Jacobian rule. The overlap formula below gives the precise relation needed to compare local expressions. [quotetheorem:3911] [citeproof:3911] The smoothness of the transition map is essential: the coefficients $\partial y_a/\partial x_i$ are the entries of its Jacobian matrix, so a merely continuous change of coordinates would not produce a smooth transformation law for covectors. The formula does not say that $dy_a=dx_a$ on an overlap; it says that the two coframes are related by the derivative of the coordinate change. This is the first instance of the general principle that forms are invariant objects whose coordinate representatives vary by Jacobian data. For higher-degree forms, the change-of-chart formula is obtained by wedging these expressions and collecting terms. The determinant minors appearing in this expansion are the alternating analogue of the Jacobian matrix. [example: Transformation of a Two-Form] Let $M$ be a surface, and work on an overlap where both coordinate systems $(x_1,x_2)$ and $(y_1,y_2)$ are defined. We compute $dy_1\wedge dy_2$ in the $x$-coordinates. The coordinate transformation rule for $1$-forms gives \begin{align*} dy_1&=\frac{\partial y_1}{\partial x_1}dx_1+\frac{\partial y_1}{\partial x_2}dx_2,\\ dy_2&=\frac{\partial y_2}{\partial x_1}dx_1+\frac{\partial y_2}{\partial x_2}dx_2. \end{align*} Using bilinearity of the wedge product, \begin{align*} dy_1\wedge dy_2 &=\left(\frac{\partial y_1}{\partial x_1}dx_1+\frac{\partial y_1}{\partial x_2}dx_2\right) \wedge \left(\frac{\partial y_2}{\partial x_1}dx_1+\frac{\partial y_2}{\partial x_2}dx_2\right)\\ &=\frac{\partial y_1}{\partial x_1}\frac{\partial y_2}{\partial x_1}\,dx_1\wedge dx_1 +\frac{\partial y_1}{\partial x_1}\frac{\partial y_2}{\partial x_2}\,dx_1\wedge dx_2\\ &\quad+\frac{\partial y_1}{\partial x_2}\frac{\partial y_2}{\partial x_1}\,dx_2\wedge dx_1 +\frac{\partial y_1}{\partial x_2}\frac{\partial y_2}{\partial x_2}\,dx_2\wedge dx_2. \end{align*} Since the wedge product is alternating, $dx_1\wedge dx_1=0$, $dx_2\wedge dx_2=0$, and $dx_2\wedge dx_1=-dx_1\wedge dx_2$. Hence \begin{align*} dy_1\wedge dy_2 &=\frac{\partial y_1}{\partial x_1}\frac{\partial y_2}{\partial x_2}\,dx_1\wedge dx_2 -\frac{\partial y_1}{\partial x_2}\frac{\partial y_2}{\partial x_1}\,dx_1\wedge dx_2\\ &=\left( \frac{\partial y_1}{\partial x_1}\frac{\partial y_2}{\partial x_2} -\frac{\partial y_1}{\partial x_2}\frac{\partial y_2}{\partial x_1} \right)dx_1\wedge dx_2\\ &=\det\left(\frac{\partial(y_1,y_2)}{\partial(x_1,x_2)}\right)dx_1\wedge dx_2. \end{align*} Thus the coefficient multiplying the local area form $dx_1\wedge dx_2$ is exactly the Jacobian determinant of the coordinate change. [/example] ## The Exterior Derivative The next problem is to differentiate a form in a way that does not require choosing coordinates once and for all. In Euclidean space, the exterior derivative differentiates coefficient functions and appends the corresponding coordinate covector. On a manifold, this formula can be used in charts, but it must agree on overlaps. [definition: Exterior Derivative in Coordinates] Let $(U,x)$ be a coordinate chart on a smooth $n$-manifold $M$. In this chart, the exterior derivative is the operator \begin{align*} d:\Omega^k(U)\to\Omega^{k+1}(U) \end{align*} defined as follows. If \begin{align*} \omega=\sum_I f_I\,dx_{i_1}\wedge\cdots\wedge dx_{i_k} \end{align*} is a $k$-form on $U$, then \begin{align*} d\omega=\sum_I\sum_{j=1}^n \frac{\partial f_I}{\partial x_j}\,dx_j\wedge dx_{i_1}\wedge\cdots\wedge dx_{i_k}. \end{align*} [/definition] A naive derivative of each component of a form would depend on the chosen coordinates: after changing chart, derivatives of the transition functions also appear. The alternating placement of $dx_j$ is exactly what makes those extra terms cancel. This formula is familiar from $\mathbb R^n$, but the definition has only local content until we prove that changing coordinates gives the same $(k+1)$-form. The next obstruction is therefore independence of charts. To use exterior differentiation on a manifold rather than inside one coordinate patch, the coordinate formula must glue to a single global form on overlaps. [quotetheorem:3912] [citeproof:3912] The chart hypothesis is not an extra structure on the final operator; it is only a device for computing the local formula. Smooth coordinate changes are needed because the proof differentiates transition functions, and the statement would fail in a nonsmooth atlas. The theorem does not identify $d$ with an ordinary directional derivative or a connection, since it differentiates forms without choosing vector fields, metrics, or Christoffel symbols. Its purpose is to make exterior differentiation an intrinsic operation, which is what allows the identity $d^2=0$ to be stated globally. Once exterior differentiation is intrinsic, the next structural question is what happens when it is applied twice. Alternation suggests that second derivative terms should cancel in pairs, and the resulting identity is the algebraic constraint that makes closed and exact forms meaningful. [quotetheorem:3913] [citeproof:3913] Smoothness is needed here because the proof uses equality of mixed partial derivatives in coordinate charts. The theorem does not say that every closed form is exact; that converse is false on manifolds such as the circle, where the angle form gives a closed form with nonzero integral around the loop. The identity $d^2=0$ is the analytic origin of de Rham cohomology. It says that every exact form is closed, creating the quotient spaces studied later in the course. [example: Gradient Curl Divergence Pattern] On $\mathbb R^3$ with coordinates $(x,y,z)$, identify a vector field $X=(P,Q,R)$ with the $1$-form \begin{align*} \alpha=P\,dx+Q\,dy+R\,dz, \end{align*} and identify a $2$-form \begin{align*} A\,dy\wedge dz+B\,dz\wedge dx+C\,dx\wedge dy \end{align*} with the vector field $(A,B,C)$. For a smooth function $f$, the exterior derivative is \begin{align*} df=f_x\,dx+f_y\,dy+f_z\,dz. \end{align*} Applying $d$ again gives \begin{align*} d(df) &=d(f_x)\wedge dx+d(f_y)\wedge dy+d(f_z)\wedge dz\\ &=(f_{xx}\,dx+f_{xy}\,dy+f_{xz}\,dz)\wedge dx\\ &\quad +(f_{yx}\,dx+f_{yy}\,dy+f_{yz}\,dz)\wedge dy\\ &\quad +(f_{zx}\,dx+f_{zy}\,dy+f_{zz}\,dz)\wedge dz\\ &=f_{xy}\,dy\wedge dx+f_{xz}\,dz\wedge dx +f_{yx}\,dx\wedge dy+f_{yz}\,dz\wedge dy\\ &\quad +f_{zx}\,dx\wedge dz+f_{zy}\,dy\wedge dz\\ &=(f_{yx}-f_{xy})\,dx\wedge dy+(f_{xz}-f_{zx})\,dz\wedge dx+(f_{zy}-f_{yz})\,dy\wedge dz\\ &=0, \end{align*} because mixed partial derivatives agree and $dy\wedge dx=-dx\wedge dy$, $dx\wedge dz=-dz\wedge dx$, and $dz\wedge dy=-dy\wedge dz$. Under the identification of $2$-forms with vector fields, the coefficients of $d(df)$ are exactly \begin{align*} \operatorname{curl}(\nabla f) =(f_{zy}-f_{yz},\,f_{xz}-f_{zx},\,f_{yx}-f_{xy}), \end{align*} so $d^2f=0$ is the identity $\operatorname{curl}(\nabla f)=0$. Now let $X=(P,Q,R)$ and let $\alpha=P\,dx+Q\,dy+R\,dz$. Then \begin{align*} d\alpha &=dP\wedge dx+dQ\wedge dy+dR\wedge dz\\ &=(P_x\,dx+P_y\,dy+P_z\,dz)\wedge dx\\ &\quad +(Q_x\,dx+Q_y\,dy+Q_z\,dz)\wedge dy\\ &\quad +(R_x\,dx+R_y\,dy+R_z\,dz)\wedge dz\\ &=(Q_x-P_y)\,dx\wedge dy+(P_z-R_x)\,dz\wedge dx+(R_y-Q_z)\,dy\wedge dz. \end{align*} Thus $d\alpha$ corresponds to \begin{align*} \operatorname{curl}X=(R_y-Q_z,\,P_z-R_x,\,Q_x-P_y). \end{align*} Applying $d$ to this $2$-form gives \begin{align*} d(d\alpha) &=d(R_y-Q_z)\wedge dy\wedge dz +d(P_z-R_x)\wedge dz\wedge dx\\ &\quad +d(Q_x-P_y)\wedge dx\wedge dy\\ &=(R_{yx}-Q_{zx})\,dx\wedge dy\wedge dz\\ &\quad +(P_{zy}-R_{xy})\,dy\wedge dz\wedge dx\\ &\quad +(Q_{xz}-P_{yz})\,dz\wedge dx\wedge dy\\ &=\bigl(R_{xy}-Q_{xz}+P_{yz}-R_{xy}+Q_{xz}-P_{yz}\bigr)\,dx\wedge dy\wedge dz\\ &=0, \end{align*} where $dy\wedge dz\wedge dx=dx\wedge dy\wedge dz$ and $dz\wedge dx\wedge dy=dx\wedge dy\wedge dz$. The scalar coefficient is \begin{align*} \frac{\partial}{\partial x}(R_y-Q_z) +\frac{\partial}{\partial y}(P_z-R_x) +\frac{\partial}{\partial z}(Q_x-P_y) =\operatorname{div}(\operatorname{curl}X), \end{align*} so $d^2\alpha=0$ is the identity $\operatorname{div}(\operatorname{curl}X)=0$. [/example] ## Pullback of Differential Forms How should forms move along a smooth map $\varphi:M\to N$? Tangent vectors push forward from $M$ to $N$, so covectors and alternating covectors pull back from $N$ to $M$. This direction reversal is essential: functions, covectors, and differential forms all pull back by precomposition with the appropriate derivative. [definition: Pullback of a Differential Form] Let $\varphi:M\to N$ be a smooth map, and let $\omega\in\Omega^k(N)$. The pullback $\varphi^*\omega\in\Omega^k(M)$ is defined by \begin{align*} (\varphi^*\omega)_p(v_1,\dots,v_k)=\omega_{\varphi(p)}(d\varphi_p(v_1),\dots,d\varphi_p(v_k)) \end{align*} for $p\in M$ and $v_1,\dots,v_k\in T_pM$. [/definition] For $0$-forms this reduces to $\varphi^*f=f\circ\varphi$. For $1$-forms it says that a covector on $N$ is evaluated after sending tangent vectors forward by $d\varphi_p$. The pointwise definition now has to be checked against the algebra of forms. Later calculations build complicated forms from sums, scalar multiples, and wedge products, so pullback is useful only if it preserves those operations with the expected degree and signs. The following structural result is what lets pullback be used as an algebra homomorphism on differential forms, not merely as a pointwise formula. [quotetheorem:3914] [citeproof:3914] The smoothness of $\varphi$ is essential because the definition uses $d\varphi_p$ at every point. The theorem does not say that pullback is invertible: if $\varphi$ collapses a manifold to a point, all positive-degree forms pull back to zero. This compatibility makes coordinate computations efficient and prepares the naturality identity for the exterior derivative. If $\omega=\sum_I f_I dy_{i_1}\wedge\cdots\wedge dy_{i_k}$ in coordinates on $N$, then \begin{align*} \varphi^*\omega=\sum_I (f_I\circ\varphi)\,d(y_{i_1}\circ\varphi)\wedge\cdots\wedge d(y_{i_k}\circ\varphi). \end{align*} [quotetheorem:3574] [citeproof:3574] Naturality requires smoothness of $\varphi$ and smoothness of the form because both sides differentiate pulled-back coefficient functions. It does not say that exterior derivative commutes with pushforward; forms are contravariant, and pushforward of forms is not available for an arbitrary smooth map. Naturality is the main functorial property of differential forms. It is the mechanism behind change of variables, invariance of closed forms under pullback, and the induced maps on de Rham cohomology. [example: Pullback of the Standard Area Form on the Sphere] Let \begin{align*} s=1+u^2+v^2. \end{align*} Then 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*} We compute the pullback by substituting these functions into \begin{align*} \omega_{S^2}=X\,dY\wedge dZ+Y\,dZ\wedge dX+Z\,dX\wedge dY. \end{align*} Since \begin{align*} ds=2u\,du+2v\,dv, \end{align*} the differentials of the component functions are \begin{align*} dX &=d(2u\,s^{-1})\\ &=2s^{-1}\,du-2u\,s^{-2}\,ds\\ &=\frac{2}{s}\,du-\frac{2u}{s^2}(2u\,du+2v\,dv)\\ &=\frac{2s-4u^2}{s^2}\,du-\frac{4uv}{s^2}\,dv\\ &=\frac{2(1-u^2+v^2)}{s^2}\,du-\frac{4uv}{s^2}\,dv, \end{align*} \begin{align*} dY &=d(2v\,s^{-1})\\ &=2s^{-1}\,dv-2v\,s^{-2}\,ds\\ &=\frac{2}{s}\,dv-\frac{2v}{s^2}(2u\,du+2v\,dv)\\ &=-\frac{4uv}{s^2}\,du+\frac{2s-4v^2}{s^2}\,dv\\ &=-\frac{4uv}{s^2}\,du+\frac{2(1+u^2-v^2)}{s^2}\,dv, \end{align*} and \begin{align*} dZ &=d\left((u^2+v^2-1)s^{-1}\right)\\ &=\frac{2u\,du+2v\,dv}{s}-\frac{u^2+v^2-1}{s^2}(2u\,du+2v\,dv)\\ &=\frac{s-(u^2+v^2-1)}{s^2}(2u\,du+2v\,dv)\\ &=\frac{2}{s^2}(2u\,du+2v\,dv)\\ &=\frac{4u}{s^2}\,du+\frac{4v}{s^2}\,dv. \end{align*} For two $1$-forms $a\,du+b\,dv$ and $c\,du+d\,dv$, bilinearity and alternation give \begin{align*} (a\,du+b\,dv)\wedge(c\,du+d\,dv) &=ac\,du\wedge du+ad\,du\wedge dv+bc\,dv\wedge du+bd\,dv\wedge dv\\ &=(ad-bc)\,du\wedge dv. \end{align*} Therefore \begin{align*} dY\wedge dZ &=\left[ \left(-\frac{4uv}{s^2}\right)\left(\frac{4v}{s^2}\right) -\left(\frac{2(1+u^2-v^2)}{s^2}\right)\left(\frac{4u}{s^2}\right) \right]du\wedge dv\\ &=\frac{-16uv^2-8u(1+u^2-v^2)}{s^4}\,du\wedge dv\\ &=\frac{-8u-8u^3-8uv^2}{s^4}\,du\wedge dv\\ &=-\frac{8u}{s^3}\,du\wedge dv, \end{align*} \begin{align*} dZ\wedge dX &=\left[ \left(\frac{4u}{s^2}\right)\left(-\frac{4uv}{s^2}\right) -\left(\frac{4v}{s^2}\right)\left(\frac{2(1-u^2+v^2)}{s^2}\right) \right]du\wedge dv\\ &=\frac{-16u^2v-8v(1-u^2+v^2)}{s^4}\,du\wedge dv\\ &=\frac{-8v-8u^2v-8v^3}{s^4}\,du\wedge dv\\ &=-\frac{8v}{s^3}\,du\wedge dv, \end{align*} and \begin{align*} dX\wedge dY &=\left[ \left(\frac{2(1-u^2+v^2)}{s^2}\right)\left(\frac{2(1+u^2-v^2)}{s^2}\right) -\left(-\frac{4uv}{s^2}\right)\left(-\frac{4uv}{s^2}\right) \right]du\wedge dv\\ &=\frac{4(1-u^2+v^2)(1+u^2-v^2)-16u^2v^2}{s^4}\,du\wedge dv\\ &=\frac{4\left(1-(u^2-v^2)^2\right)-16u^2v^2}{s^4}\,du\wedge dv\\ &=\frac{4-4u^4+8u^2v^2-4v^4-16u^2v^2}{s^4}\,du\wedge dv\\ &=\frac{4-4u^4-8u^2v^2-4v^4}{s^4}\,du\wedge dv\\ &=\frac{4\left(1-(u^2+v^2)^2\right)}{s^4}\,du\wedge dv. \end{align*} Substituting into the three terms of $\sigma^*\omega_{S^2}$ gives \begin{align*} \sigma^*\omega_{S^2} &=X\,dY\wedge dZ+Y\,dZ\wedge dX+Z\,dX\wedge dY\\ &=\frac{2u}{s}\left(-\frac{8u}{s^3}\right)du\wedge dv +\frac{2v}{s}\left(-\frac{8v}{s^3}\right)du\wedge dv\\ &\quad+\frac{u^2+v^2-1}{s}\cdot \frac{4\left(1-(u^2+v^2)^2\right)}{s^4}\,du\wedge dv\\ &=\frac{-16u^2-16v^2}{s^4}\,du\wedge dv +\frac{4(u^2+v^2-1)\left(1-(u^2+v^2)^2\right)}{s^5}\,du\wedge dv. \end{align*} Writing $r=u^2+v^2$, this coefficient is \begin{align*} \frac{-16r}{s^4}+\frac{4(r-1)(1-r^2)}{s^5} &=\frac{-16r(1+r)+4(r-1)(1-r)(1+r)}{(1+r)^5}\\ &=\frac{-16r(1+r)-4(r-1)^2(1+r)}{(1+r)^5}\\ &=\frac{-4(1+r)\left(4r+(r-1)^2\right)}{(1+r)^5}\\ &=\frac{-4(1+r)(r^2+2r+1)}{(1+r)^5}\\ &=-\frac{4}{(1+r)^2}. \end{align*} Since $r=u^2+v^2$, we obtain \begin{align*} \sigma^*\omega_{S^2} =-\frac{4}{(1+u^2+v^2)^2}\,du\wedge dv. \end{align*} The negative sign records that this stereographic parametrisation orients the plane oppositely to the outward radial area form on the sphere; at $(u,v)=(0,0)$, the image point is $(0,0,-1)$ while the parametrised tangent orientation points in the positive $Z$-direction. [/example] ## Interior Product and Cartan Formula Differentiation of forms has another companion operation: insertion of a vector field into the first slot. This operation lowers degree by one and records how a form reacts along a chosen direction. Together with $d$, it describes the infinitesimal change of forms under the flow of a vector field. [definition: Interior Product] Let $M$ be a smooth manifold and let $V\in\mathfrak X(M)$ be a smooth vector field. The interior product with $V$ is the map \begin{align*} \iota_V:\Omega^k(M)\to\Omega^{k-1}(M) \end{align*} defined for $k\ge 1$ by \begin{align*} (\iota_V\omega)_p(v_1,\dots,v_{k-1})=\omega_p(V_p,v_1,\dots,v_{k-1}). \end{align*} For $k=0$, set $\iota_V f=0$. [/definition] The interior product is algebraic rather than differential: it uses the value of $V$ and $\omega$ at the same point and takes no derivatives of coefficients. Alternation is essential here; inserting $V$ into a non-alternating tensor would not interact with products through the same signed rule. The operation lowers degree, so it must be paired with a sign convention whenever it passes across a $k$-form. The operation must next be compatible with the wedge product, since most forms are built from simpler ones by wedging. Inserting one vector into an alternating product forces that vector to pass across the first factor, and the resulting sign determines whether $\iota_V$ behaves like a graded derivation. This compatibility is the algebraic rule needed before combining insertion with $d$ in Cartan's formula. [quotetheorem:3915] [citeproof:3915] This identity should be read as the graded Leibniz rule for inserting a vector field into a wedge product. Smoothness of $V$ is still part of the setting: without it, $\iota_V\omega$ need not be a smooth form even when $\omega$ is smooth. The operation itself is pointwise and algebraic, while derivatives of $V$ enter only when forms are compared along the flow of $V$, which leads to the Lie derivative. A vector field should also differentiate forms in the direction of its flow, but forms at different points cannot be subtracted directly. The flow solves this comparison problem by moving the form and then pulling it back to the original point, where an ordinary derivative in the parameter $t$ can be taken. This gives a coordinate-free way to measure infinitesimal change of a differential form along $V$, and it is the operator that Cartan's formula will relate to $d$ and $\iota_V$. [definition: Lie Derivative of a Differential Form] Let $V\in\mathfrak X(M)$ have local flow $\Phi_t^V$. For each $k\ge 0$, the Lie derivative along $V$ is the operator \begin{align*} \mathcal L_V:\Omega^k(M)\to\Omega^k(M) \end{align*} defined by \begin{align*} \mathcal L_V\omega=\frac{d}{dt}\Big|_{t=0}(\Phi_t^V)^*\omega \end{align*} for $\omega\in\Omega^k(M)$. [/definition] This definition says that $\mathcal L_V\omega$ measures the infinitesimal change of $\omega$ as points are transported along the vector field $V$ and covectors are pulled back to the starting point. The use of pullback is forced by variance: the flow sends tangent vectors forward, so forms must be compared by pulling them back to the original point. The definition does not require a globally defined flow for all time; the derivative at $t=0$ only needs the local flow. [quotetheorem:1535] [citeproof:1535] The flow hypothesis is needed because $\mathcal L_V$ is defined by infinitesimal transport; for a nonsmooth vector field there may be no smooth local flow and no smooth Lie derivative on forms. The formula does not say that $d$ and $\iota_V$ commute separately, since their graded commutator is exactly the Lie derivative. Cartan formula is a compact way to remember the relation between topology, calculus, and dynamics. In later chapters it becomes a basic tool for showing that flows generated by vector fields act as the identity on de Rham cohomology in many situations. [example: Liouville One-Form on the Cotangent Bundle of Euclidean Space] On $T^*\mathbb R^n$ use coordinates $(q_1,\dots,q_n,p_1,\dots,p_n)$, where the $q_i$ are base coordinates and the $p_i$ are fibre coordinates, and define \begin{align*} \lambda=\sum_{i=1}^n p_i\,dq_i. \end{align*} We compute $d\lambda$ from the coordinate formula for the exterior derivative. For each $i$, the $1$-form $p_i\,dq_i$ has coefficient function $p_i$ multiplying the coordinate $1$-form $dq_i$, so \begin{align*} d(p_i\,dq_i) &=dp_i\wedge dq_i+p_i\,d(dq_i). \end{align*} Since $q_i$ is a smooth coordinate function, $dq_i=dq_i$ is its exterior derivative, and nilpotence of the exterior derivative gives \begin{align*} d(dq_i)=d^2q_i=0. \end{align*} Therefore \begin{align*} d(p_i\,dq_i) &=dp_i\wedge dq_i+p_i\cdot 0\\ &=dp_i\wedge dq_i. \end{align*} Using linearity of $d$, \begin{align*} d\lambda &=d\left(\sum_{i=1}^n p_i\,dq_i\right)\\ &=\sum_{i=1}^n d(p_i\,dq_i)\\ &=\sum_{i=1}^n dp_i\wedge dq_i. \end{align*} Thus the Liouville form differentiates to $\sum_i dp_i\wedge dq_i$. With the common convention $\omega=-d\lambda$, the associated canonical symplectic form is \begin{align*} \omega &=-\sum_{i=1}^n dp_i\wedge dq_i\\ &=\sum_{i=1}^n dq_i\wedge dp_i, \end{align*} because $dp_i\wedge dq_i=-dq_i\wedge dp_i$ for each $i$. [/example] The chapter has built the operational core of differential forms on manifolds: local coordinate expansions, the coordinate-independent exterior derivative, pullback, contraction, and Lie differentiation. The identities $d^2=0$ and $\varphi^*d=d\varphi^*$ are the algebraic backbone for integration on manifolds and for the cohomology theory that follows. The tangent and cotangent bundles provide the stage on which differential forms become a calculus rather than just pointwise linear algebra. After that, the theory shifts from describing infinitesimal data to organizing it into operations that can be differentiated, pulled back, and compared across charts. # 4. Orientations and Integration on Manifolds This chapter builds the integration theory needed after the course has introduced tangent spaces, differential forms, pullbacks, and coordinate changes. In Euclidean space an $n$-form can be integrated because coordinates carry a preferred sign convention, but on a manifold those signs must agree from chart to chart. The main goals are to define orientations intrinsically, to construct enough partitions of unity to reduce global integrals to local coordinate integrals, and to prove that the resulting integral of a compactly supported top-degree form is independent of all auxiliary choices. The chapter ends with manifolds with boundary, where the induced orientation is the sign convention needed for Stokes theorem in the next chapter. ## Orientations of Vector Spaces and Manifolds The first question is what extra choice is needed before a top-degree form can be interpreted as a signed density. At a point of an $n$-manifold, changing coordinates by a map with negative Jacobian determinant reverses the sign of an $n$-form, so integration requires a coherent choice of which bases count as positive. [definition: Orientation of a Vector Space] Let $V$ be a real vector space of dimension $n$. An orientation of $V$ is an equivalence class of ordered bases, where two ordered bases $(v_1,\dots,v_n)$ and $(w_1,\dots,w_n)$ are equivalent if the change-of-basis matrix from the $v_i$ to the $w_i$ has positive determinant. [/definition] There are exactly two orientations on $V$ when $n \ge 1$. Choosing one of them lets us call ordered bases positive or negative. For $V=\mathbb R^n$, the standard orientation is the one represented by $(e_1,\dots,e_n)$. [example: Orientation of the Plane] Take $V=\mathbb R^2$ with the standard ordered basis $(e_1,e_2)$. For the ordered basis $(e_2,e_1)$, the columns of the change-of-basis matrix relative to $(e_1,e_2)$ are the coordinates of $e_2$ and $e_1$: \begin{align*} e_2&=0e_1+1e_2,\\ e_1&=1e_1+0e_2, \end{align*} so \begin{align*} P=\begin{pmatrix}0&1\\ 1&0\end{pmatrix}, \qquad \det P=(0)(0)-(1)(1)=-1. \end{align*} Since the determinant is negative, $(e_2,e_1)$ lies in the orientation class opposite to the standard one. For the ordered basis $(e_1,e_1+e_2)$, the coordinate identities are \begin{align*} e_1&=1e_1+0e_2,\\ e_1+e_2&=1e_1+1e_2, \end{align*} hence the change-of-basis matrix is \begin{align*} Q=\begin{pmatrix}1&1\\ 0&1\end{pmatrix}, \qquad \det Q=(1)(1)-(1)(0)=1. \end{align*} Thus $(e_1,e_1+e_2)$ represents the standard orientation. This is the linear model for oriented coordinate changes: the sign of the determinant records whether the chosen orientation is preserved or reversed. [/example] A manifold orientation asks for this choice in every tangent space, with smooth variation. The smoothness condition is most conveniently expressed by a nowhere-vanishing top-degree form. [definition: Orientation Form] Let $M$ be a smooth $n$-manifold. An orientation form on $M$ is a 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] An orientation form declares an ordered basis $(v_1,\dots,v_n)$ of $T_pM$ to be positive when $\omega_p(v_1,\dots,v_n)>0$. Multiplying $\omega$ by a positive smooth function does not change the resulting orientation. For integration, the data we need is the consistent sign choice itself, not a particular top-degree form used to present it. This leads to two separate notions: whether such a smooth choice exists at all, and whether one of the two possible choices has been fixed. The terminology below records that distinction. [definition: Oriented Manifold] A smooth $n$-manifold $M$ is orientable if it admits an orientation form. An oriented manifold is a pair consisting of a smooth manifold $M$ and a choice of orientation. [/definition] Choosing an orientation separately in each tangent space is not enough: those choices could jump discontinuously from point to point and would not interact correctly with smooth coordinate changes. A nowhere-vanishing top-degree form supplies exactly the missing smooth compatibility. The point that remains is equivalence: does every smooth orientation arise from such a form, and does every such form determine a smooth orientation? Establishing this lets us pass freely between geometric sign choices and analytic top-degree forms. [quotetheorem:3916] [citeproof:3916] This theorem converts a geometric sign choice into an analytic object. The nowhere-vanishing hypothesis is essential, because a top-degree form that vanishes at a point cannot distinguish positive from negative bases there. The result does not say that every top-degree form is an orientation form; multiplying an orientation form by a smooth function that changes sign or vanishes destroys the orientation data. It also foreshadows the role of partitions of unity: they are the device that turns locally defined forms into global ones while preserving positivity in oriented charts. ## Oriented Atlases and Non-Orientability The next problem is how to recognise orientability from charts alone. Since the Jacobian determinant measures the sign change between coordinate volume forms, the right condition is that all transition maps preserve orientation. [definition: Oriented Atlas] Let $M$ be a smooth $n$-manifold. An atlas $\{(U_i,\varphi_i)\}_{i\in I}$ is oriented if for every pair $i,j$ and every point of $\varphi_i(U_i\cap U_j)$, 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. [/definition] Local coordinate charts always carry a sign convention inherited from $\mathbb R^n$, but incompatible overlaps can reverse that convention after travelling around the manifold. The obstruction is not visible inside one chart; it appears when transition maps around overlaps force a positive basis to become negative. To recognise orientability from an atlas, we need a criterion saying exactly when the chartwise signs can be made compatible everywhere. [quotetheorem:3577] [citeproof:3577/differential-forms-ii-manifolds-and-cohomology] The positivity condition on every overlap cannot be weakened to nonzero determinant, since negative determinants exactly encode orientation reversal. The theorem also does not make orientability a property of a single chart; the obstruction appears only when charts are compared around loops. In practice this characterisation is often the easiest way to prove non-orientability, while orientation forms are often the easiest way to compute integrals once an orientation is known. [example: The Sphere Is Orientable] View $S^2\subset\mathbb R^3$ as the unit sphere with outward normal $N(p)=p$. Let $\Omega=dx\wedge dy\wedge dz$ be the standard volume form on $\mathbb R^3$, and define a $2$-form on $S^2$ by \begin{align*} \omega_p(v,w)=\Omega_p(p,v,w) \end{align*} for $v,w\in T_pS^2$. Since $p$ is orthogonal to $T_pS^2$ and $|p|=1$, if $(v,w)$ is a basis of $T_pS^2$, then $(p,v,w)$ is a basis of $\mathbb R^3$, so $\omega_p(v,w)\ne 0$. Thus $\omega$ is nowhere vanishing and declares $(v,w)$ positive exactly when $(p,v,w)$ is positively oriented in $\mathbb R^3$. In spherical coordinates \begin{align*} p(\theta,\phi)=(\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta), \qquad 0<\theta<\pi,\quad 0<\phi<2\pi, \end{align*} the coordinate tangent vectors are \begin{align*} p_\theta&=(\cos\theta\cos\phi,\cos\theta\sin\phi,-\sin\theta),\\ p_\phi&=(-\sin\theta\sin\phi,\sin\theta\cos\phi,0). \end{align*} Therefore \begin{align*} \omega(p_\theta,p_\phi) &=\det \begin{pmatrix} \sin\theta\cos\phi & \cos\theta\cos\phi & -\sin\theta\sin\phi\\ \sin\theta\sin\phi & \cos\theta\sin\phi & \sin\theta\cos\phi\\ \cos\theta & -\sin\theta & 0 \end{pmatrix}\\ &=(\sin\theta\cos\phi)\bigl((\cos\theta\sin\phi)(0)-(\sin\theta\cos\phi)(-\sin\theta)\bigr)\\ &\quad-(\cos\theta\cos\phi)\bigl((\sin\theta\sin\phi)(0)-(\sin\theta\cos\phi)(\cos\theta)\bigr)\\ &\quad+(-\sin\theta\sin\phi)\bigl((\sin\theta\sin\phi)(-\sin\theta)-(\cos\theta\sin\phi)(\cos\theta)\bigr)\\ &=\sin^3\theta\cos^2\phi+\sin\theta\cos^2\theta\cos^2\phi +\sin\theta\sin^2\phi(\sin^2\theta+\cos^2\theta)\\ &=\sin\theta\cos^2\phi(\sin^2\theta+\cos^2\theta)+\sin\theta\sin^2\phi\\ &=\sin\theta(\cos^2\phi+\sin^2\phi)\\ &=\sin\theta. \end{align*} Since $\sin\theta>0$ for $0<\theta<\pi$, the chart orientation determined by $(\partial_\theta,\partial_\phi)$ agrees with the outward orientation, and the orientation form is \begin{align*} \omega=\sin\theta\,d\theta\wedge d\phi \end{align*} away from the poles. This identifies the intrinsic orientation with the usual spherical-coordinate area orientation. [/example] Orientability can fail when a loop in the manifold reverses local orientation. The model example is the Mobius band, where moving once around the central circle flips the transverse direction. [illustration:dfii-mobius-orientation-obstruction] [example: Mobius Band as a Non-Orientable Manifold] Construct the Mobius band as \begin{align*} M=([0,2\pi]\times(-1,1))/\sim, \qquad (0,t)\sim(2\pi,-t). \end{align*} Along the central circle $\gamma(\theta)=[(\theta,0)]$, use the local ordered frame \begin{align*} E_\theta=dq_{(\theta,0)}(\partial_\theta), \qquad T_\theta=dq_{(\theta,0)}(\partial_t), \end{align*} where $q:[0,2\pi]\times(-1,1)\to M$ is the quotient map. The gluing map across the identified ends is \begin{align*} G(0,t)=(2\pi,-t), \end{align*} so its differential sends the coordinate directions to \begin{align*} dG(\partial_\theta)&=\partial_\theta,\\ dG(\partial_t)&=-\partial_t. \end{align*} Therefore, after one circuit around $\gamma$, the frame $(E_0,T_0)$ returns as $(E_0,-T_0)$. Relative to the original ordered basis $(E_0,T_0)$, the returned ordered basis has change-of-basis matrix \begin{align*} P= \begin{pmatrix} 1&0\\ 0&-1 \end{pmatrix}, \qquad \det P=(1)(-1)-(0)(0)=-1. \end{align*} Thus one trip around the central circle reverses orientation: the angular vector returns unchanged, while the transverse vector returns with the opposite sign. A globally consistent choice of positive ordered bases would have to assign the same orientation to the starting frame and to the frame obtained by transporting it around this closed loop, but the determinant computation shows that these two frames lie in opposite orientation classes. Hence the Mobius band admits no oriented atlas and no orientation form. [/example] The Mobius band example isolates the global obstruction. The next remark states the general lesson: local orientation choices always exist, and the difficulty is making them agree around loops. [remark: Orientation Is Global] Every smooth manifold is locally orientable because each coordinate chart inherits an orientation from $\mathbb R^n$. The issue is whether these local choices can be made compatible on all overlaps. Non-orientability is therefore not a local singularity; it is a global twisting phenomenon. [/remark] ## Partitions of Unity Integration will be defined by reducing an $n$-form to coordinate patches and summing the resulting Euclidean integrals. The difficulty is that a form may meet many charts, and the answer must not depend on arbitrary choices. Partitions of unity solve this by decomposing a global object into controlled local pieces. [definition: Locally Finite Cover] Let $X$ be a topological space. A cover $\{U_i\}_{i\in I}$ of $X$ is locally finite if every point $x\in X$ has a neighbourhood meeting only finitely many of the sets $U_i$. [/definition] Local finiteness is the finiteness condition that lets infinitely many local functions behave like a finite sum near each point. To use a chosen open cover in integration, we need more than local finiteness of the sets: we need smooth weights that are supported inside those sets and add to one everywhere. Such weights let a global form be decomposed into chart-sized pieces without changing its total contribution. This is the role of a partition of unity subordinate to the cover. [definition: Subordinate Partition of Unity] Let $M$ be a smooth manifold and let $\{U_i\}_{i\in I}$ be an open cover of $M$. A smooth partition of unity subordinate to this cover is a family of functions $\{\rho_i\}_{i\in I}$ with $\rho_i\in C^\infty(M;\mathbb R)$ such that: 1. $0\le \rho_i\le 1$ for all $i$; 2. $\operatorname{supp}\rho_i\subset U_i$ for all $i$; 3. the family $\{\operatorname{supp}\rho_i\}_{i\in I}$ is locally finite; 4. $\sum_{i\in I}\rho_i(p)=1$ for every $p\in M$. [/definition] The local finiteness condition is part of the definition because it makes the sum smooth: near any point only finitely many terms appear. Without it, the pointwise sum could fail to have good differentiability properties. Under the course convention, smooth manifolds are second-countable and Hausdorff, hence paracompact. That topological input is what makes the following existence theorem valid for arbitrary open covers. [quotetheorem:3917] [citeproof:3917] The paracompactness hypothesis is the hidden topological engine of the theorem; without it, arbitrary open covers need not admit locally finite refinements. The smoothness of the bump functions uses the coordinate structure, so this is stronger than a purely topological partition of unity. For integration, the key consequence is that compactly supported forms can be decomposed into finitely many coordinate-supported pieces even when the manifold itself is non-compact. [example: Bump Functions on a Coordinate Ball] Let $\beta:\mathbb R\to\mathbb R$ be \begin{align*} \beta(s)= \begin{cases} e^{-1/s},& s>0,\\ 0,& s\le 0. \end{cases} \end{align*} For $s>0$, each derivative has the form \begin{align*} \beta^{(k)}(s)=P_k(1/s)e^{-1/s} \end{align*} for some polynomial $P_k$, by induction from the product and chain rules. Since $u^m e^{-u}\to 0$ as $u\to\infty$ for every $m$, substituting $u=1/s$ gives $\beta^{(k)}(s)\to 0$ as $s\downarrow 0$. Thus all one-sided derivatives match the zero derivatives on $s\le 0$, so $\beta$ is smooth. Define \begin{align*} \rho(x)=\beta(4-|x|^2),\qquad x\in\mathbb R^n. \end{align*} If $|x|<2$, then $4-|x|^2>0$, so \begin{align*} \rho(x)=e^{-1/(4-|x|^2)}>0. \end{align*} If $|x|\ge 2$, then $4-|x|^2\le 0$, so $\rho(x)=0$. In particular, $\rho$ is positive on $B(0,1)$ and vanishes outside $B(0,2)$. Now suppose $(U,\varphi)$ is a coordinate chart and $B(0,2)\subset \varphi(U)$. Define a function $\widetilde\rho$ on $M$ by \begin{align*} \widetilde\rho(p)= \begin{cases} \rho(\varphi(p)),& p\in U,\\ 0,& p\notin U. \end{cases} \end{align*} On $U$ this is smooth because it is the composition of the smooth functions $\rho$ and $\varphi$. On $M\setminus \varphi^{-1}(\overline{B(0,2)})$ it is identically zero, so it is smooth there as well. Its support is contained in $\varphi^{-1}(\overline{B(0,2)})\subset U$, and it is positive on $\varphi^{-1}(B(0,1))$. These are the local bump functions that are rescaled and summed in the partition-of-unity construction. [/example] This local bump is the ingredient used to build global partitions of unity. Compact support matters because it keeps sums finite on compact regions even when the cover itself is infinite. [remark: Compact Supports Simplify Sums] If $K\subset M$ is compact and $\{\rho_i\}$ is locally finite, then only finitely many supports $\operatorname{supp}\rho_i$ meet $K$. Thus compactly supported forms are split into finitely many nonzero pieces by a subordinate partition of unity. [/remark] ## Integration of Top-Degree Forms The central question of this chapter is how to integrate an $n$-form on an oriented $n$-manifold. The construction must agree with ordinary integration in a single oriented chart and must be independent of the particular atlas and partition of unity used. Start with the local case. If $(U,\varphi)$ is an oriented coordinate chart on an oriented $n$-manifold and $\alpha\in\Omega_c^n(U)$, write \begin{align*} \alpha = f\,dx_1\wedge\cdots\wedge dx_n \end{align*} in the coordinates $x_i$ induced by $\varphi$. Here $\mathcal L^n$ denotes Lebesgue measure on $\mathbb R^n$. Define \begin{align*} \int_U \alpha := \int_{\varphi(U)} (f\circ\varphi^{-1})(u)\,d\mathcal L^n(u). \end{align*} The support condition ensures that this is an ordinary compactly supported integral on an open subset of $\mathbb R^n$. [quotetheorem:3918] [citeproof:3918] The orientation hypothesis is essential: if the coordinate change has negative determinant, the same top-degree form acquires the opposite sign in the new chart. Compact support is also part of the local definition used here, because it avoids questions about improper integrals on non-compact coordinate domains. This result is the local consistency statement that makes the global partition-of-unity definition meaningful. Now let $\omega\in\Omega_c^n(M)$. Choose an oriented atlas $\{(U_i,\varphi_i)\}$ and a partition of unity $\{\rho_i\}$ subordinate to the cover. The integral is defined by \begin{align*} \int_M \omega := \sum_i \int_{U_i} \rho_i\omega. \end{align*} Only finitely many terms are nonzero because $\omega$ has compact support and the partition is locally finite. [quotetheorem:1529] [citeproof:1529] This theorem is the point where all earlier hypotheses come together: orientation fixes signs, partitions of unity localise the form, and compact support keeps the sum finite. It does not define integrals of arbitrary non-compactly supported forms; those require additional convergence or density assumptions. The construction is nevertheless intrinsic, so later results such as Stokes theorem can be stated without reference to a chosen atlas. [example: Integrating a Two-Form over the Sphere] Let $S^2$ have the outward orientation. On the coordinate patch away from one meridian and the poles, use spherical coordinates \begin{align*} p(\theta,\phi) &=(\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta), \qquad 0<\theta<\pi,\quad 0<\phi<2\pi. \end{align*} From the earlier computation of the outward orientation form on $S^2$, \begin{align*} \operatorname{vol}_{S^2} =\sin\theta\,d\theta\wedge d\phi \end{align*} on this oriented chart. Therefore the coordinate definition of integration gives \begin{align*} \int_{S^2}\operatorname{vol}_{S^2} &=\int_0^{2\pi}\int_0^\pi \sin\theta\,d\theta\,d\phi\\ &=\int_0^{2\pi}\bigl[-\cos\theta\bigr]_{\theta=0}^{\theta=\pi}\,d\phi\\ &=\int_0^{2\pi}\bigl(-\cos\pi-(-\cos 0)\bigr)\,d\phi\\ &=\int_0^{2\pi}\bigl(-(-1)+1\bigr)\,d\phi\\ &=\int_0^{2\pi}2\,d\phi\\ &=\bigl[2\phi\bigr]_{\phi=0}^{\phi=2\pi}\\ &=4\pi. \end{align*} The omitted poles and meridian form a union of coordinate curves and points, so they contribute zero to the two-dimensional coordinate integral. The partition-of-unity definition of integration then gives the same intrinsic value on the whole oriented sphere. [/example] The sphere calculation uses a compact surface without boundary. The annulus gives a complementary test case where polar coordinates meet boundary components, preparing the sign conventions used later. [example: Integrating over an Annulus] Let $A=\{(x,y)\in\mathbb R^2:1<x^2+y^2<4\}$ with the standard orientation, and let $\omega=f(x,y)\,dx\wedge dy$ be compactly supported in $A$. On the polar coordinate domain $(1,2)\times(0,2\pi)$, set \begin{align*} \Phi(r,\theta)=(r\cos\theta,r\sin\theta). \end{align*} Then \begin{align*} \Phi^*x&=r\cos\theta, & \Phi^*y&=r\sin\theta, \end{align*} so \begin{align*} d(\Phi^*x)&=\cos\theta\,dr-r\sin\theta\,d\theta,\\ d(\Phi^*y)&=\sin\theta\,dr+r\cos\theta\,d\theta. \end{align*} Therefore \begin{align*} \Phi^*(dx\wedge dy) &=d(\Phi^*x)\wedge d(\Phi^*y)\\ &=(\cos\theta\,dr-r\sin\theta\,d\theta) \wedge(\sin\theta\,dr+r\cos\theta\,d\theta)\\ &=\cos\theta\sin\theta\,dr\wedge dr +r\cos^2\theta\,dr\wedge d\theta -r\sin^2\theta\,d\theta\wedge dr -r^2\sin\theta\cos\theta\,d\theta\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*} Here $dr\wedge dr=d\theta\wedge d\theta=0$ and $d\theta\wedge dr=-dr\wedge d\theta$ were used. Thus \begin{align*} \Phi^*\omega &=\Phi^*\bigl(f(x,y)\,dx\wedge dy\bigr)\\ &=f(r\cos\theta,r\sin\theta)\,\Phi^*(dx\wedge dy)\\ &=f(r\cos\theta,r\sin\theta)\,r\,dr\wedge d\theta. \end{align*} Since $r>0$ on $(1,2)$, the polar chart preserves the standard orientation. Hence the coordinate definition of integration gives \begin{align*} \int_A\omega =\int_0^{2\pi}\int_1^2 f(r\cos\theta,r\sin\theta)\,r\,dr\,d\theta. \end{align*} The factor $r$ is exactly the determinant factor in the coordinate expression of the pulled-back form. [/example] ## Manifolds with Boundary and Induced Orientation Stokes theorem needs an orientation not only on $M$ but also on its boundary. The question is which sign convention on $\partial M$ makes the boundary integral match the interior exterior derivative. [definition: Manifold with Boundary] A smooth $n$-manifold with boundary is a second-countable Hausdorff space $M$ equipped with charts to open subsets of the closed half-space \begin{align*} \mathbb H^n=\{x\in\mathbb R^n:x_n\ge 0\} \end{align*} with smooth transition maps in the sense of extensions to open subsets of $\mathbb R^n$. [/definition] The half-space model separates interior points from points lying on the hyperplane face. Stokes theorem will integrate over exactly that face, so we need an intrinsic definition of which points of $M$ count as boundary points, independent of the particular boundary chart used to see them. [definition: Boundary of a Manifold with Boundary] Let $M$ be a smooth $n$-manifold with boundary. The boundary $\partial M$ is the set of points that are mapped by boundary charts to points of $\{x_n=0\}\subset\mathbb H^n$. [/definition] The boundary is itself a smooth $(n-1)$-manifold. Its smooth charts are obtained by restricting boundary charts to the hyperplane $x_n=0$. Once $\partial M$ is identified as a manifold, its orientation must still be fixed by a convention. The sign cannot be chosen independently if Stokes theorem is to generalize the endpoint signs in one-variable calculus. The outward-normal-first rule is the convention that transfers the orientation of $M$ to its boundary with the correct sign. [definition: Outward Normal First Boundary Orientation] Let $M$ be an oriented smooth $n$-manifold with boundary. The induced orientation on $\partial M$ is defined as follows: an ordered basis $(v_1,\dots,v_{n-1})$ of $T_p\partial M$ is positive if $(\nu,v_1,\dots,v_{n-1})$ 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] [illustration:dfii-boundary-orientation] The choice of outward transverse vector does not affect the induced orientation, because any two outward transverse vectors differ by a positive normal component plus a tangent vector. This is the outward-normal-first convention. [example: Boundary Orientation of an Interval] Give $[a,b]$ the standard orientation determined by $dx$, so a nonzero tangent vector $v=c\,\partial_x$ is positive exactly when \begin{align*} dx(v)=dx(c\,\partial_x)=c\,dx(\partial_x)=c>0. \end{align*} At the left endpoint $a$, an outward-pointing transverse vector is $\nu_a=-\partial_x$. Since \begin{align*} dx(\nu_a)=dx(-\partial_x)=-dx(\partial_x)=-1<0, \end{align*} the one-vector basis $(\nu_a)$ is negatively oriented in $T_a[a,b]$. The boundary is $0$-dimensional, so its ordered basis is the empty basis; by the outward-normal-first convention, this empty basis is positive at $a$ only if $(\nu_a)$ is positive in $T_a[a,b]$. It is not, so the point $a$ receives sign $-1$. At the right endpoint $b$, an outward-pointing transverse vector is $\nu_b=\partial_x$. Here \begin{align*} dx(\nu_b)=dx(\partial_x)=1>0, \end{align*} so $(\nu_b)$ is positively oriented in $T_b[a,b]$. Hence the empty basis of $T_b\partial[a,b]$ is positive, and the point $b$ receives sign $+1$. Therefore the induced oriented boundary is \begin{align*} \partial[a,b]=\{b\}-\{a\}. \end{align*} [/example] Intervals show the boundary-orientation convention in dimension one. The annulus shows the same outward-normal rule when the boundary has two components with opposite induced directions. [example: Boundary Orientation of the Annulus] Let \begin{align*} A=\{(x,y)\in\mathbb R^2:1\le x^2+y^2\le 4\} \end{align*} with the standard orientation $dx\wedge dy$. On the polar coordinate chart \begin{align*} \Phi(r,\theta)=(r\cos\theta,r\sin\theta), \end{align*} we have \begin{align*} d(\Phi^*x)&=\cos\theta\,dr-r\sin\theta\,d\theta,\\ d(\Phi^*y)&=\sin\theta\,dr+r\cos\theta\,d\theta, \end{align*} and therefore \begin{align*} \Phi^*(dx\wedge dy) &=d(\Phi^*x)\wedge d(\Phi^*y)\\ &=(\cos\theta\,dr-r\sin\theta\,d\theta) \wedge(\sin\theta\,dr+r\cos\theta\,d\theta)\\ &=\cos\theta\sin\theta\,dr\wedge dr +r\cos^2\theta\,dr\wedge d\theta -r\sin^2\theta\,d\theta\wedge dr -r^2\sin\theta\cos\theta\,d\theta\wedge d\theta\\ &=r\cos^2\theta\,dr\wedge d\theta +r\sin^2\theta\,dr\wedge d\theta\\ &=r\,dr\wedge d\theta. \end{align*} Thus the ordered basis $(\partial_r,\partial_\theta)$ is positive for $r>0$, because \begin{align*} (r\,dr\wedge d\theta)(\partial_r,\partial_\theta) &=r\bigl(dr(\partial_r)d\theta(\partial_\theta) -dr(\partial_\theta)d\theta(\partial_r)\bigr)\\ &=r(1\cdot 1-0\cdot 0)\\ &=r>0. \end{align*} On the outer boundary $r=2$, the outward normal is $\nu=\partial_r$. Hence the boundary tangent $\partial_\theta$ is positive, since \begin{align*} (dx\wedge dy)(\partial_r,\partial_\theta)=r>0. \end{align*} The parametrization $\theta\mapsto(2\cos\theta,2\sin\theta)$ moves counterclockwise as $\theta$ increases, so the outer boundary inherits the counterclockwise orientation. On the inner boundary $r=1$, the outward normal for the annulus points toward decreasing $r$, so $\nu=-\partial_r$. The tangent $\partial_\theta$ is negative there, because \begin{align*} (dx\wedge dy)(-\partial_r,\partial_\theta) &=-(dx\wedge dy)(\partial_r,\partial_\theta)\\ &=-r<0. \end{align*} Instead $-\partial_\theta$ is positive, since \begin{align*} (dx\wedge dy)(-\partial_r,-\partial_\theta) &=(dx\wedge dy)(\partial_r,\partial_\theta)\\ &=r>0. \end{align*} Thus increasing $\theta$ gives the wrong boundary direction on the inner circle, and the induced orientation is clockwise. The two boundary components have opposite orientations because their outward normals point in opposite radial directions. [/example] The annulus computation explains the sign convention geometrically. The following remark connects that convention to the global Stokes formula it is designed to make true. [remark: Preparation for Stokes] The sign convention for $\partial M$ is chosen so that the general Stokes formula will read \begin{align*} \int_M d\eta=\int_{\partial M}\eta \end{align*} for compactly supported $(n-1)$-forms $\eta$ on an oriented manifold with boundary. The interval computation is the one-dimensional model for the sign. [/remark] Once forms can be integrated locally and related by coordinate changes, the remaining issue is how to make those integrals independent of the chart choices. Orientations solve that sign problem, and they prepare the ground for the global Stokes theorem that turns local differential data into boundary information. # 5. The General Stokes' Theorem The preceding chapters built the language needed to integrate differential forms on oriented manifolds. This chapter proves the theorem that makes that language powerful: the exterior derivative measures the boundary contribution of a form. The general Stokes theorem is the common source of Green theorem, the divergence theorem, and the classical Stokes theorem from vector calculus, and it also explains why closed and exact forms record topological information. ## The Boundary Term in Integration What should replace the endpoint term $F(b)-F(a)$ from the one-variable fundamental theorem of calculus when the domain is an oriented manifold? The answer is that the derivative of an $(n-1)$-form integrates over an $n$-manifold by becoming the original form integrated over the oriented boundary. The main point is not the formula alone, but the compatibility of three choices: the orientation on $M$, the induced orientation on $\partial M$, and the sign convention in $d$. [definition: Manifold With Boundary] An $n$-dimensional smooth manifold with boundary is a Hausdorff, second-countable topological space $M$ equipped with a smooth atlas whose charts take values in the closed half-space \begin{align*} \mathbb H^n = \{x=(x_1,\dots,x_n)\in \mathbb R^n : x_n\ge 0\}, \end{align*} with smooth transition maps in the manifold-with-boundary sense. The boundary $\partial M$ is the set of points whose local chart image lies in $\{x_n=0\}$. [/definition] The boundary $\partial M$ is itself a smooth $(n-1)$-manifold. To state Stokes theorem with a definite sign, the boundary orientation has to be determined from the orientation of $M$. The convention is chosen so that the outward transverse direction is placed first, leaving the remaining ordered basis to orient the boundary. This gives the sign rule used in the boundary integral. [definition: Boundary Orientation] Let $M$ be an oriented smooth $n$-manifold with boundary. The induced 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 $\nu$ is an outward-pointing tangent vector transverse to $\partial M$ at $p$. [/definition] For the model half-space $\mathbb H^n$ with coordinates $(x_1,\dots,x_n)$ and standard orientation $dx_1\wedge\cdots\wedge dx_n$, the outward normal along $\{x_n=0\}$ points in the $-\partial_{x_n}$ direction. The induced boundary orientation is therefore represented by $(-1)^n dx_1\wedge\cdots\wedge dx_{n-1}$ on the boundary chart: indeed $(-\partial_{x_n},v_1,\dots,v_{n-1})$ must be positive in $T_p\mathbb H^n$. With the boundary sign fixed, the central analytic question can now be stated globally. Exterior differentiation should turn an $(n-1)$-form on $M$ into an $n$-form whose integral depends only on the boundary values of the original form. The following theorem asserts that the local normal-variable calculation survives passage through charts and partitions of unity. [quotetheorem:3555] [citeproof:3555] This is the global form of the fundamental theorem of calculus. Compact support is the condition that makes the integral finite and prevents unwanted terms from appearing at infinity when $M$ is non-compact. The orientation of $M$ fixes the sign of the left-hand side, while the outward-normal-first convention fixes the sign of the right-hand side; changing either convention changes the formula by a sign. Smoothness is used so that exterior derivatives, restrictions to the boundary, and partitions of unity behave well in charts. The theorem does not say that every closed form is exact, and it does not remove boundary terms; rather, it identifies exactly which boundary term a derivative produces. The rest of the chapter uses this identity first as a local analytic calculation, then as the source of the classical integral theorems and the period invariants of closed forms. Before moving to higher-dimensional examples, it is worth checking that the sign convention reproduces the familiar endpoint formula rather than its negative. [example: Interval Fundamental Theorem] Let $M=[a,b]$ carry the standard orientation, so the positive coordinate form is $dx$, and let $\omega=f$ be a smooth $0$-form. Since the exterior derivative of a function is \begin{align*} df=f'(x)\,dx, \end{align*} integration over the oriented interval gives \begin{align*} \int_{[a,b]} df &=\int_a^b f'(x)\,dx \\ &=f(b)-f(a), \end{align*} by the one-variable fundamental theorem of calculus. It remains to check that the boundary integral has the same sign. At the left endpoint $a$, the outward-pointing tangent vector is $-\partial_x$, which is negative relative to the orientation of $[a,b]$; hence $a$ receives sign $-1$. At the right endpoint $b$, the outward-pointing tangent vector is $\partial_x$, which is positive; hence $b$ receives sign $+1$. Therefore integration of the $0$-form $f$ over the oriented boundary is \begin{align*} \int_{\partial [a,b]} f &=(-1)f(a)+(+1)f(b) \\ &=f(b)-f(a). \end{align*} Thus *General Stokes Theorem* on the oriented $1$-manifold $[a,b]$ is exactly the usual endpoint formula. [/example] ## Reduction to the Half-Space Why does a theorem about arbitrary manifolds reduce to a calculation in $\mathbb H^n$? Integration of forms is defined by charts, and partitions of unity let us localise compactly supported forms without changing the global sum. The geometric content of the proof is that orientation and boundary orientation are local notions compatible with changes of coordinates. [explanation: Local Form of the Proof] Suppose $\omega$ is supported in a single boundary chart $(U,\varphi)$, with $\varphi(U)\subset \mathbb H^n$. Pulling back by $\varphi^{-1}$ turns the desired identity into an identity for compactly supported forms on the half-space. Since exterior derivative commutes with pullback, the analytic part of the argument is the Euclidean calculation. A general compactly supported form is written as $\omega=\sum_i \rho_i\omega$, where $(\rho_i)$ is a partition of unity on a neighbourhood of $\operatorname{supp}\omega$. Each summand has support in a chart, so the local identity applies to $\rho_i\omega$. The equality for $\omega$ follows after summing, because only finitely many summands meet the compact support. [/explanation] The local model is therefore the half-space calculation, with the boundary hyperplane carrying the induced orientation described above. At this stage the global geometry has been stripped away, leaving one concrete analytic question: what does the exterior derivative contribute when an $n$-form is integrated over $\mathbb H^n$? The answer must produce exactly the integral over the boundary hyperplane, with the sign determined by the boundary orientation convention. [quotetheorem:3919] [citeproof:3919] This local theorem is the only analytic calculation in the proof. Compact support is essential here because the integration-by-parts argument in tangential directions has no end contribution only when the coefficient vanishes at infinity; without it, an extra boundary-at-infinity term may remain. The orientation on $\partial\mathbb H^n$ is also essential: using the standard orientation of $\mathbb R^{n-1}$ instead of the induced one would change the formula by the factor $(-1)^n$. By itself, the half-space theorem is not yet the global theorem, since globalisation still requires coordinate invariance of form integration and a partition of unity to assemble the local identities. This proof also shows where the theorem can fail if its hypotheses are weakened: non-compact support may leave an end contribution, and an inconsistent boundary orientation reverses the sign of the boundary integral. [remark: Compact Support and Non-Compact Manifolds] On a compact manifold, every smooth form is compactly supported. On a non-compact manifold, compact support is part of the theorem hypotheses unless decay or integrability conditions control behaviour at infinity. [/remark] ## Classical Integral Theorems as Special Cases How do the vector-calculus theorems fit into the same formula? In low dimensions, forms encode scalar functions, vector fields, fluxes, and circulations through the Euclidean volume form and the Hodge star. Stokes theorem says that all of these identities are expressions of the same boundary principle. [example: Green Theorem in the Plane] Let $D\subset \mathbb R^2$ be a compact oriented region with smooth boundary, oriented by the area form $dx\wedge dy$, and let \begin{align*} \omega = P\,dx + Q\,dy \end{align*} for $P,Q\in C^\infty(D)$. We compute the exterior derivative term by term. Since \begin{align*} dP &= \frac{\partial P}{\partial x}\,dx+\frac{\partial P}{\partial y}\,dy,\\ dQ &= \frac{\partial Q}{\partial x}\,dx+\frac{\partial Q}{\partial y}\,dy, \end{align*} we have \begin{align*} d(P\,dx) &= 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*} because $dx\wedge dx=0$ and $dy\wedge dx=-dx\wedge dy$. Similarly, \begin{align*} d(Q\,dy) &= 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*} Adding the two parts gives \begin{align*} d\omega &= d(P\,dx)+d(Q\,dy) \\ &= \left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dx\wedge dy. \end{align*} Applying *General Stokes Theorem* to the $1$-form $\omega$ gives \begin{align*} \int_D d\omega &= \int_{\partial D}\omega, \end{align*} so, since integration against $dx\wedge dy$ is ordinary planar area integration, \begin{align*} \int_D \left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)d\mathcal L^2 &= \int_{\partial D} P\,dx+Q\,dy. \end{align*} Thus Green theorem is exactly Stokes theorem for the $1$-form $P\,dx+Q\,dy$, with $\partial D$ carrying the induced positive boundary orientation. [/example] The next example uses a $2$-form rather than a $1$-form, so the boundary integral is a flux integral instead of a circulation integral. [example: Divergence Theorem in Three Dimensions] Let $M\subset \mathbb R^3$ be a compact oriented solid region with smooth boundary, oriented by $dx\wedge dy\wedge dz$, and let the boundary carry the induced outward-normal-first orientation. For a smooth vector field $F=(F_1,F_2,F_3)$, define \begin{align*} \omega = F_1\,dy\wedge dz - F_2\,dx\wedge dz + F_3\,dx\wedge dy. \end{align*} We compute $d\omega$ term by term. Since \begin{align*} dF_i=\frac{\partial F_i}{\partial x}\,dx+\frac{\partial F_i}{\partial y}\,dy+\frac{\partial F_i}{\partial z}\,dz, \end{align*} we have \begin{align*} d(F_1\,dy\wedge dz) &=dF_1\wedge dy\wedge dz\\ &=\left(\frac{\partial F_1}{\partial x}\,dx+\frac{\partial F_1}{\partial y}\,dy+\frac{\partial F_1}{\partial z}\,dz\right)\wedge dy\wedge dz\\ &=\frac{\partial F_1}{\partial x}\,dx\wedge dy\wedge dz +\frac{\partial F_1}{\partial y}\,dy\wedge dy\wedge dz +\frac{\partial F_1}{\partial z}\,dz\wedge dy\wedge dz\\ &=\frac{\partial F_1}{\partial x}\,dx\wedge dy\wedge dz. \end{align*} Similarly, \begin{align*} d(-F_2\,dx\wedge dz) &=-dF_2\wedge dx\wedge dz\\ &=-\left(\frac{\partial F_2}{\partial x}\,dx+\frac{\partial F_2}{\partial y}\,dy+\frac{\partial F_2}{\partial z}\,dz\right)\wedge dx\wedge dz\\ &=-\frac{\partial F_2}{\partial y}\,dy\wedge dx\wedge dz\\ &=\frac{\partial F_2}{\partial y}\,dx\wedge dy\wedge dz, \end{align*} because $dy\wedge dx=-dx\wedge dy$, and \begin{align*} d(F_3\,dx\wedge dy) &=dF_3\wedge dx\wedge dy\\ &=\left(\frac{\partial F_3}{\partial x}\,dx+\frac{\partial F_3}{\partial y}\,dy+\frac{\partial F_3}{\partial z}\,dz\right)\wedge dx\wedge dy\\ &=\frac{\partial F_3}{\partial z}\,dz\wedge dx\wedge dy\\ &=\frac{\partial F_3}{\partial z}\,dx\wedge dy\wedge dz. \end{align*} Adding the three identities gives \begin{align*} d\omega &=\left(\frac{\partial F_1}{\partial x}+\frac{\partial F_2}{\partial y}+\frac{\partial F_3}{\partial z}\right)dx\wedge dy\wedge dz\\ &=(\operatorname{div}F)\,dx\wedge dy\wedge dz. \end{align*} Now let $\nu=(\nu_1,\nu_2,\nu_3)$ be the outward unit normal on $\partial M$, and let $(u,v)$ be a positive orthonormal basis for $T_p\partial M$. By the boundary orientation convention, $(\nu,u,v)$ is positive in $\mathbb R^3$, so $u\times v=\nu$. Evaluating $\omega$ on $(u,v)$ gives \begin{align*} \omega(u,v) &=F_1\bigl(dy\wedge dz\bigr)(u,v)-F_2\bigl(dx\wedge dz\bigr)(u,v)+F_3\bigl(dx\wedge dy\bigr)(u,v)\\ &=F_1(u_2v_3-u_3v_2)-F_2(u_1v_3-u_3v_1)+F_3(u_1v_2-u_2v_1)\\ &=F_1(u\times v)_1+F_2(u\times v)_2+F_3(u\times v)_3\\ &=F_1\nu_1+F_2\nu_2+F_3\nu_3\\ &=F\cdot \nu. \end{align*} Thus the restriction of $\omega$ to the oriented boundary is the flux density $(F\cdot\nu)\,dS$. Applying *General Stokes Theorem* gives \begin{align*} \int_M \operatorname{div}F\,dV &=\int_M d\omega\\ &=\int_{\partial M}\omega\\ &=\int_{\partial M} F\cdot \nu\,dS. \end{align*} So the divergence theorem is Stokes theorem applied to the $2$-form that encodes the flux of $F$. [/example] For surfaces in $\mathbb R^3$, the same general theorem appears in its most familiar curl form. [example: Classical Stokes Theorem for a Surface] Let $S\subset \mathbb R^3$ be an oriented compact smooth surface with boundary, and let $F=(F_1,F_2,F_3)$ be a smooth vector field. Put \begin{align*} \omega = F_1\,dx+F_2\,dy+F_3\,dz. \end{align*} We compute $d\omega$ on $S$. Since \begin{align*} dF_i=\frac{\partial F_i}{\partial x}\,dx+\frac{\partial F_i}{\partial y}\,dy+\frac{\partial F_i}{\partial z}\,dz, \end{align*} we get \begin{align*} d(F_1\,dx) &=dF_1\wedge dx\\ &=\left(\frac{\partial F_1}{\partial x}\,dx+\frac{\partial F_1}{\partial y}\,dy+\frac{\partial F_1}{\partial z}\,dz\right)\wedge dx\\ &=-\frac{\partial F_1}{\partial y}\,dx\wedge dy-\frac{\partial F_1}{\partial z}\,dx\wedge dz, \end{align*} because $dx\wedge dx=0$, $dy\wedge dx=-dx\wedge dy$, and $dz\wedge dx=-dx\wedge dz$. Similarly, \begin{align*} d(F_2\,dy) &=dF_2\wedge dy\\ &=\frac{\partial F_2}{\partial x}\,dx\wedge dy-\frac{\partial F_2}{\partial z}\,dy\wedge dz, \end{align*} and \begin{align*} d(F_3\,dz) &=dF_3\wedge dz\\ &=\frac{\partial F_3}{\partial x}\,dx\wedge dz+\frac{\partial F_3}{\partial y}\,dy\wedge dz. \end{align*} Adding the three parts gives \begin{align*} d\omega &=\left(\frac{\partial F_2}{\partial x}-\frac{\partial F_1}{\partial y}\right)dx\wedge dy +\left(\frac{\partial F_3}{\partial x}-\frac{\partial F_1}{\partial z}\right)dx\wedge dz +\left(\frac{\partial F_3}{\partial y}-\frac{\partial F_2}{\partial z}\right)dy\wedge dz. \end{align*} If \begin{align*} \operatorname{curl}F =\left(\frac{\partial F_3}{\partial y}-\frac{\partial F_2}{\partial z}, \frac{\partial F_1}{\partial z}-\frac{\partial F_3}{\partial x}, \frac{\partial F_2}{\partial x}-\frac{\partial F_1}{\partial y}\right), \end{align*} then this is exactly the flux $2$-form of $\operatorname{curl}F$: \begin{align*} d\omega &=(\operatorname{curl}F)_1\,dy\wedge dz -(\operatorname{curl}F)_2\,dx\wedge dz +(\operatorname{curl}F)_3\,dx\wedge dy. \end{align*} Thus, if $\nu$ is the oriented unit normal on $S$, the restriction of $d\omega$ to $S$ is $(\operatorname{curl}F\cdot \nu)\,dS$ by the same flux-form calculation used for the divergence theorem. On the boundary curve $\partial S$, let $T$ be the positive unit tangent determined by the induced boundary orientation. Evaluating $\omega$ on $T=(T_1,T_2,T_3)$ gives \begin{align*} \omega(T) &=F_1\,dx(T)+F_2\,dy(T)+F_3\,dz(T)\\ &=F_1T_1+F_2T_2+F_3T_3\\ &=F\cdot T. \end{align*} Applying *General Stokes Theorem* to the restricted $1$-form $\omega|_S$ gives \begin{align*} \int_{\partial S} F\cdot T\,ds &=\int_{\partial S}\omega\\ &=\int_S d\omega\\ &=\int_S \operatorname{curl}F\cdot \nu\,dS. \end{align*} So the classical Stokes theorem is the general Stokes theorem applied to the circulation $1$-form associated to $F$. [/example] These examples are not separate phenomena. The difference between circulation and flux is encoded by the degree of the form and by the identification of vector fields with forms using the Euclidean metric and orientation. ## Closed Manifolds and Vanishing Boundary Integrals What changes when the manifold has no boundary? Stokes theorem then says that the integral of an exact top-degree form over the whole manifold is zero. This is the first signal that exactness is a global restriction, not just a local differential equation. [definition: Closed Manifold] A closed manifold is a compact smooth manifold without boundary. [/definition] For a closed oriented manifold, Stokes theorem has no boundary term to measure. This makes exact top-degree forms globally constrained: although they may be nonzero pointwise, their total signed integral should vanish. The following result records this boundary-free consequence in the form most often used to detect non-exact forms. [quotetheorem:3920] [citeproof:3920] The result is often the fastest way to recognize that a top-degree form on a closed manifold is not exact: if its integral is non-zero, no global primitive can exist. The geometric point is that a closed manifold has no boundary contribution available to cancel the total integral. Compactness and absence of boundary are both doing work. If $M$ has boundary, the integral of $d\omega$ is controlled by the boundary value of $\omega$ rather than being forced to vanish. If $M$ is non-compact, compact support or a substitute decay condition is needed to avoid an end contribution. In de Rham language, the theorem says that integration over a closed oriented manifold descends from top-degree forms to top-degree cohomology classes, because exact representatives integrate to zero. [example: Area Form on the Sphere] Let $S^2\subset\mathbb R^3$ be the unit sphere with the outward orientation, and let $\sigma$ be its Riemannian area form. On the usual spherical parametrization \begin{align*} \Phi(\varphi,\theta) =(\sin\varphi\cos\theta,\sin\varphi\sin\theta,\cos\varphi), \qquad 0\le \varphi\le \pi,\quad 0\le \theta\le 2\pi, \end{align*} we have \begin{align*} \frac{\partial\Phi}{\partial\varphi} &=(\cos\varphi\cos\theta,\cos\varphi\sin\theta,-\sin\varphi),\\ \frac{\partial\Phi}{\partial\theta} &=(-\sin\varphi\sin\theta,\sin\varphi\cos\theta,0), \end{align*} and hence \begin{align*} \frac{\partial\Phi}{\partial\varphi}\times \frac{\partial\Phi}{\partial\theta} &=(\sin^2\varphi\cos\theta,\sin^2\varphi\sin\theta,\sin\varphi\cos\varphi)\\ &=\sin\varphi\,(\sin\varphi\cos\theta,\sin\varphi\sin\theta,\cos\varphi)\\ &=\sin\varphi\,\Phi(\varphi,\theta). \end{align*} For $0<\varphi<\pi$, this cross product points in the outward normal direction and has length $\sin\varphi$, so \begin{align*} \Phi^*\sigma=\sin\varphi\,d\varphi\wedge d\theta. \end{align*} Therefore \begin{align*} \int_{S^2}\sigma &=\int_0^{2\pi}\int_0^\pi \sin\varphi\,d\varphi\,d\theta\\ &=\int_0^{2\pi}\left[-\cos\varphi\right]_{\varphi=0}^{\varphi=\pi}\,d\theta\\ &=\int_0^{2\pi}\bigl(-\cos\pi+\cos 0\bigr)\,d\theta\\ &=\int_0^{2\pi}2\,d\theta\\ &=4\pi. \end{align*} If $\sigma$ were exact, say $\sigma=d\eta$ for some $\eta\in\Omega^1(S^2)$, then *Exact Top-Degree Forms Integrate to Zero* would give \begin{align*} 4\pi &=\int_{S^2}\sigma\\ &=\int_{S^2}d\eta\\ &=0, \end{align*} which is impossible. Thus the area form $\sigma$ is not exact. [/example] The same vanishing principle is useful whenever a geometric integral is unchanged by adding exact error terms. [example: Gauss Bonnet Integrand as a Two-Form] Let $\Sigma$ be a closed oriented Riemannian surface with Gaussian curvature $K$ and area form $dA$. Since $K$ is a smooth function and $dA$ is a smooth $2$-form, the product $K\,dA$ is a smooth $2$-form on $\Sigma$. The Gauss-Bonnet theorem states that \begin{align*} \int_\Sigma K\,dA = 2\pi\chi(\Sigma). \end{align*} Now let $\eta\in\Omega^1(\Sigma)$, so $d\eta$ is an exact $2$-form. If we replace the curvature form by $K\,dA+d\eta$, then linearity of integration gives \begin{align*} \int_\Sigma (K\,dA+d\eta) &=\int_\Sigma K\,dA+\int_\Sigma d\eta \\ &=2\pi\chi(\Sigma)+\int_\Sigma d\eta. \end{align*} Because $\Sigma$ is closed, Stokes theorem gives \begin{align*} \int_\Sigma d\eta &=\int_{\partial\Sigma}\eta \\ &=\int_{\varnothing}\eta \\ &=0. \end{align*} Therefore \begin{align*} \int_\Sigma (K\,dA+d\eta) &=2\pi\chi(\Sigma)+0 \\ &=2\pi\chi(\Sigma). \end{align*} Thus adding an exact $2$-form changes the integrand but not its integral over a closed surface. [/example] ## Periods of Closed Forms How can an integral detect a cohomology class rather than a particular representative? If $\alpha$ is closed and $c$ is a cycle, the integral of $\alpha$ over $c$ depends only on the de Rham cohomology class of $\alpha$ and the homology class of $c$. Stokes theorem is the mechanism behind both invariances. [definition: Period of a Closed Form] Let $M$ be a smooth manifold, let $\alpha\in\Omega^k(M)$ be closed, and let $c$ be a smooth oriented $k$-cycle in $M$. The period of $\alpha$ over $c$ is the integral \begin{align*} \int_c \alpha. \end{align*} [/definition] Periods turn differential forms into numbers attached to cycles, so Stokes theorem immediately predicts what happens to exact forms. If a form is a derivative $d\beta$, then integrating it over a cycle should depend only on the boundary of that cycle. Since cycles have zero boundary, the period has no place to receive a contribution from the primitive. This is the first precise link between exactness and vanishing numerical invariants. [quotetheorem:3921] [citeproof:3921] This is the cohomological version of the vanishing boundary integral. Exact forms are invisible to integration over cycles. The cycle hypothesis is essential. If $c$ is a chain with non-zero boundary, Stokes theorem gives $\int_c d\beta=\int_{\partial c}\beta$, which need not vanish. Thus exact forms vanish on cycles not because their primitives are zero, but because the boundary of the integration domain is zero. This is precisely what allows periods to depend on homology classes rather than on chosen cycle representatives. [quotetheorem:3922] [citeproof:3922] The theorem turns exactness into a test against every cycle. It is not enough to test one convenient loop or surface unless that cycle is known to generate the relevant homology. The result is therefore a global exactness criterion: periods detect the obstruction that remains after all local primitive-finding has been attempted. [example: Winding Number as a Period on the Circle] Write the angular form on $\mathbb C\setminus\{0\}\cong \mathbb R^2\setminus\{0\}$ as \begin{align*} \alpha &=\frac{-y\,dx+x\,dy}{x^2+y^2}. \end{align*} Its restriction to $S^1$ is denoted $d\theta$. For the positively oriented parametrization \begin{align*} c(t)=(\cos t,\sin t),\qquad 0\le t\le 2\pi, \end{align*} we have \begin{align*} c^*dx&=-\sin t\,dt,\\ c^*dy&=\cos t\,dt,\\ c^*(x^2+y^2)&=\cos^2t+\sin^2t=1. \end{align*} Therefore \begin{align*} c^*\alpha &=\frac{-(\sin t)(-\sin t\,dt)+(\cos t)(\cos t\,dt)}{1}\\ &=(\sin^2t+\cos^2t)\,dt\\ &=dt. \end{align*} Thus \begin{align*} \int_{S^1}d\theta &=\int_0^{2\pi}c^*\alpha\\ &=\int_0^{2\pi}dt\\ &=2\pi. \end{align*} The form $d\theta$ is closed on $S^1$ because every $2$-form on a $1$-manifold is zero. It is not exact: if $d\theta=df$ for some smooth single-valued function $f:S^1\to\mathbb R$, then the integral of the exact form $df$ over the closed cycle $S^1$ would be $0$, contradicting the computed value $2\pi$. For a smooth loop $\gamma:S^1\to\mathbb C\setminus\{0\}$, written in a parameter $t\in[0,2\pi]$ as $\gamma(t)=(x(t),y(t))$, the pullback is \begin{align*} \gamma^*\alpha &=\frac{-y(t)\,x'(t)+x(t)\,y'(t)}{x(t)^2+y(t)^2}\,dt. \end{align*} The winding number is therefore \begin{align*} \operatorname{wind}(\gamma,0) &=\frac{1}{2\pi}\int_{S^1}\gamma^*d\theta\\ &=\frac{1}{2\pi}\int_0^{2\pi} \frac{x(t)y'(t)-y(t)x'(t)}{x(t)^2+y(t)^2}\,dt. \end{align*} So the winding number is a period of the closed angular form, normalized by the basic period $\int_{S^1}d\theta=2\pi$. [/example] This example turns non-exactness into a measurable period. The next remark makes the general point: closedness is local, while exactness is obstructed by global cycles. [remark: Why Closed Forms Are Not Always Exact] The equation $d\alpha=0$ is local, while the equation $\alpha=d\beta$ asks for a global primitive. Stokes theorem supplies tests for failure of global exactness: a closed form with a non-zero period over some cycle cannot be exact. De Rham cohomology packages exactly this obstruction. [/remark] Stokes' theorem is the bridge from local calculus to global topology. It shows that the exterior derivative detects boundary behavior, and that observation leads directly to the cohomological viewpoint where closed forms are studied modulo exact ones. # 6. de Rham Cohomology This chapter turns the calculus of differential forms into a topological invariant. The exterior derivative gives a cochain complex, and the failure of a closed form to be exact measures global information about the manifold. The guiding question is: which features of a manifold can be detected by solving the equation $\omega=d\eta$ for closed forms $\omega$? The previous chapters built forms, pullbacks, orientations, and Stokes' theorem. The prerequisites for this chapter are the exterior derivative, the identity $d^2=0$, pullback of forms, integration of top forms, and Stokes' theorem on oriented manifolds. We now use those tools to define de Rham cohomology, compute its first basic examples, and prove that it depends only on the smooth homotopy type of the manifold. ## Closed Forms, Exact Forms, and the de Rham Complex When a differential form has zero exterior derivative, it satisfies an infinitesimal compatibility condition. The central problem is whether this local condition comes from a global potential: given $\omega \in \Omega^k(M)$ with $d\omega=0$, does there exist $\eta \in \Omega^{k-1}(M)$ such that $\omega=d\eta$? [definition: Closed Differential Form] Let $M$ be a smooth manifold and let $k \ge 0$. A form $\omega \in \Omega^k(M)$ is closed if \begin{align*} d\omega=0. \end{align*} The vector space of closed $k$-forms is \begin{align*} Z^k(M):=\ker\left(d:\Omega^k(M)\to\Omega^{k+1}(M)\right). \end{align*} [/definition] Closedness is a differential equation. In degree $0$, it says that a smooth function has zero differential; in degree $1$, it generalises the condition that a vector field has vanishing curl in Euclidean vector calculus. Closed forms become especially meaningful when compared with forms that are visibly derivatives of lower-degree forms. Those derivative forms have no local obstruction to being closed because applying $d$ twice gives zero. The definition below names this more restrictive class, which will become the denominator in de Rham cohomology. [definition: Exact Differential Form] Let $M$ be a smooth manifold. For $k\ge 1$, a form $\omega \in \Omega^k(M)$ is exact if there exists $\eta \in \Omega^{k-1}(M)$ such that \begin{align*} \omega=d\eta. \end{align*} The vector space of exact $k$-forms is \begin{align*} B^k(M):=\operatorname{im}\left(d:\Omega^{k-1}(M)\to\Omega^k(M)\right). \end{align*} In degree $0$, set $B^0(M):=0$. [/definition] The identity $d^2=0$ implies every exact form is closed, so $B^k(M)\subset Z^k(M)$. De Rham cohomology records the quotient: closed forms are the cycles, exact forms are the boundaries. [definition: de Rham Cohomology Group] Let $M$ be a smooth manifold. The $k$-th de Rham cohomology group of $M$ is the real vector space \begin{align*} H^k_{\mathrm{dR}}(M):=Z^k(M)/B^k(M). \end{align*} If $\omega\in Z^k(M)$, its cohomology class is denoted by $[\omega]$. [/definition] The quotient means that two closed forms represent the same cohomology class precisely when they differ by an exact form. Thus $H^k_{\mathrm{dR}}(M)$ vanishes exactly when every closed $k$-form has a global primitive. [example: Exact and Closed One-Forms on the Line] On $M=\mathbb R$, let $\omega=f(x)\,dx$ with $f\in C^\infty(\mathbb R)$. Since there are no nonzero $2$-forms on a $1$-dimensional manifold, $d\omega=0$, so every smooth $1$-form on $\mathbb R$ is closed. We show that it is also exact by constructing a primitive. Define \begin{align*} F(x):=\int_0^x f(t)\,dt. \end{align*} By the fundamental theorem of calculus, $F$ is smooth and \begin{align*} F'(x)=f(x). \end{align*} The exterior derivative of the $0$-form $F$ is \begin{align*} dF &=\frac{dF}{dx}\,dx\\ &=F'(x)\,dx\\ &=f(x)\,dx\\ &=\omega. \end{align*} Thus $\omega=dF$, so $\omega\in B^1(\mathbb R)$. Since every smooth $1$-form on $\mathbb R$ lies in $B^1(\mathbb R)$, we have \begin{align*} Z^1(\mathbb R)=B^1(\mathbb R), \qquad H^1_{\mathrm{dR}}(\mathbb R)=Z^1(\mathbb R)/B^1(\mathbb R)=0. \end{align*} The primitive is obtained by integrating from the fixed basepoint $0$, and on the line there is no loop that can create a path-dependence obstruction. [/example] This computation is more than a sample primitive: it identifies the topological feature whose absence makes exactness automatic. On a line, every point can be reached from the basepoint by a canonical interval, so integrating a closed $1$-form has no path-dependence obstruction. The circle example below shows precisely what breaks when a noncontractible loop is present. [remark: Cohomology as an Obstruction] A nonzero class $[\omega]\in H^k_{\mathrm{dR}}(M)$ is not merely a form. It is an obstruction to solving $d\eta=\omega$ globally. Local primitives may exist in charts, while global primitives fail to glue because the topology of $M$ prevents a consistent choice. [/remark] ## Low-Degree de Rham Cohomology The first computations ask what the quotient definition says in degrees where the forms have familiar meanings. Degree $0$ concerns functions, while top degree on a closed oriented manifold is controlled by integration. [quotetheorem:3587] [citeproof:3587] This theorem explains why $H^0_{\mathrm{dR}}$ is the cohomological shadow of connectedness. The hypothesis on connected components is not cosmetic: without separating components, a locally constant function can choose a different value on each piece. The result also has a limitation: degree $0$ sees only connectedness and cannot detect loops, holes, or orientation. Those features first appear in positive degrees, where exactness becomes a global question rather than just local constancy. [example: Two Disjoint Circles] Let $M=S^1\sqcup S^1$, and write its two connected components as $S^1_1$ and $S^1_2$. A $0$-form on $M$ is a smooth function $f:M\to\mathbb R$, equivalently a pair of smooth functions \begin{align*} f_1:=f|_{S^1_1},\qquad f_2:=f|_{S^1_2}. \end{align*} The form $f$ is closed exactly when $df=0$, which means \begin{align*} df_1=0,\qquad df_2=0. \end{align*} If $\gamma:[a,b]\to S^1_j$ is a smooth path in one component, then \begin{align*} \frac{d}{dt}(f_j\circ\gamma)(t) &=df_j|_{\gamma(t)}(\gamma'(t))\\ &=0, \end{align*} so $f_j\circ\gamma$ is constant on $[a,b]$. Since each circle is path-connected, $f_j$ is constant on $S^1_j$. Thus every closed $0$-form has the form \begin{align*} f|_{S^1_1}=a,\qquad f|_{S^1_2}=b \end{align*} for a unique ordered pair $(a,b)\in\mathbb R^2$. Define \begin{align*} \Phi:Z^0(M)\to\mathbb R^2,\qquad \Phi(f)=(a,b), \end{align*} where $a$ and $b$ are the two componentwise constants. This map is linear, and its inverse sends $(a,b)$ to the smooth function that is constantly $a$ on $S^1_1$ and constantly $b$ on $S^1_2$. Hence \begin{align*} Z^0(M)\cong\mathbb R^2. \end{align*} By definition $B^0(M)=0$, so \begin{align*} H^0_{\mathrm{dR}}(M) &=Z^0(M)/B^0(M)\\ &=Z^0(M)/0\\ &\cong Z^0(M)\\ &\cong \mathbb R^2. \end{align*} The two coordinates record the independently chosen constant values on the two connected components. [/example] Top-degree cohomology is tied to integration because an $n$-form on an oriented $n$-manifold can be integrated, while exact top-degree forms have integral zero on closed manifolds by Stokes' theorem. This gives a well-defined linear map from top-degree de Rham cohomology to $\mathbb R$, but it is not yet clear whether this map loses information. On a connected closed oriented manifold, the central question is whether total integral is the only invariant of a top-degree cohomology class. [quotetheorem:3923] [citeproof:3923] The theorem says that for closed connected oriented manifolds, the top-degree class remembers total signed integral against a chosen orientation, not the particular density used to compute it. Connectedness matters because otherwise each connected component has its own independent top-degree integral. The closedness assumption is also essential: on noncompact manifolds or manifolds with boundary, exact top-degree forms need not be invisible to integration in the same way without extra support or boundary conditions. This result is the first appearance of the integration pairing between cohomology classes and fundamental cycles. [remark: Orientation and Compactness] The statement uses both orientation and compactness. Orientation is needed to integrate global top forms with signs. Compactness without boundary, expressed here by the word closed, is what makes Stokes' theorem give zero for integrals of exact top forms. [/remark] ## Euclidean Space and the Poincare Lemma The next question is whether de Rham cohomology detects ordinary Euclidean space as topologically featureless. Since $\mathbb R^n$ is contractible, every closed form of positive degree should have a primitive. [quotetheorem:3586] [citeproof:3586] The theorem isolates the local nature of exactness. Straight-line contractibility removes the topological obstruction that appears on spaces such as $S^1$, where closed forms can have nonzero periods around a loop. The conclusion does not choose a canonical primitive; it only says that at least one primitive exists in positive degree. [example: A Top-Degree Primitive on the Plane] On $\mathbb R^2$, consider the $2$-form \begin{align*} \omega=e^{x+y}\,dx\wedge dy. \end{align*} It is closed because $d\omega$ is a $3$-form on the $2$-manifold $\mathbb R^2$, and all $3$-forms on $\mathbb R^2$ vanish. We show that $\omega$ is exact by exhibiting the primitive \begin{align*} \eta=(e^{x+y}-e^y)\,dy. \end{align*} Let $g(x,y)=e^{x+y}-e^y$. Then $\eta=g\,dy$, so \begin{align*} d\eta &=d(g\,dy)\\ &=dg\wedge dy+g\,d(dy)\\ &=dg\wedge dy\\ &=\left(\frac{\partial g}{\partial x}\,dx+\frac{\partial g}{\partial y}\,dy\right)\wedge dy\\ &=\frac{\partial g}{\partial x}\,dx\wedge dy+\frac{\partial g}{\partial y}\,dy\wedge dy\\ &=\frac{\partial g}{\partial x}\,dx\wedge dy. \end{align*} Here $dy\wedge dy=0$ by alternatingness of the wedge product. Since \begin{align*} \frac{\partial g}{\partial x} &=\frac{\partial}{\partial x}(e^{x+y}-e^y)\\ &=e^{x+y}-0\\ &=e^{x+y}, \end{align*} we obtain \begin{align*} d\eta=e^{x+y}\,dx\wedge dy=\omega. \end{align*} Thus $\omega$ is exact even though it is a top-degree form; the primitive is a $1$-form whose exterior derivative recovers the prescribed area density. [/example] ## The Circle and a Closed Non-Exact Form The first non-Euclidean computation should show what a genuine cohomology class looks like. The circle has a loop that cannot be contracted to a point, and de Rham cohomology detects this through a closed $1$-form whose integral around the loop is nonzero. [definition: Angular One-Form on the Circle] Let $S^1\subset\mathbb R^2$ be the unit circle. The angular $1$-form $d\theta\in\Omega^1(S^1)$ is the restriction to $S^1$ of the form \begin{align*} \alpha=-y\,dx+x\,dy \end{align*} on $\mathbb R^2$. [/definition] The notation $d\theta$ is traditional: there is no globally defined smooth angle function $\theta:S^1\to\mathbb R$ whose differential is this form. That failure is exactly what the cohomology class measures. To prove that this form represents a nonzero cohomology class, closedness alone is not enough; we must rule out the existence of a global primitive. The circle offers a numerical test for this: integrate a closed $1$-form once around the loop. A nonzero period cannot occur for an exact form, so the period detects the obstruction. [quotetheorem:3924] [citeproof:3924] The theorem identifies the first genuine obstruction: a closed $1$-form is exact exactly when its period around the circle vanishes. The one-dimensional hypothesis explains the quick closedness argument, while the nonzero integral supplies the global obstruction. This computation also previews the later comparison with singular cohomology: integration over loops pairs de Rham classes with homology classes. [illustration:dfii-universal-cover-circle] [example: The Period Detects Non-Exactness] Let $\omega=d\theta$, where $d\theta$ is the restriction to $S^1$ of $\alpha=-y\,dx+x\,dy$. Since $S^1$ is one-dimensional, every $2$-form on $S^1$ is zero, so \begin{align*} d\omega\in\Omega^2(S^1)=0. \end{align*} Thus $\omega$ is closed. To show that $\omega$ is not exact, use the parametrisation \begin{align*} \gamma:[0,2\pi]\to S^1,\qquad \gamma(t)=(\cos t,\sin t). \end{align*} Along $\gamma$ we have \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^*\omega &=\gamma^*(-y\,dx+x\,dy)\\ &=-(\sin t)(-\sin t\,dt)+(\cos t)(\cos t\,dt)\\ &=(\sin^2 t+\cos^2 t)\,dt\\ &=dt. \end{align*} Hence \begin{align*} \int_{S^1}\omega &=\int_0^{2\pi}\gamma^*\omega\\ &=\int_0^{2\pi}dt\\ &=2\pi. \end{align*} If $\omega$ were exact, then $\omega=df$ for some smooth function $f:S^1\to\mathbb R$. Pulling back along $\gamma$ would give \begin{align*} \gamma^*\omega &=\gamma^*(df)\\ &=d(f\circ\gamma)\\ &=\frac{d}{dt}(f\circ\gamma)(t)\,dt. \end{align*} Thus \begin{align*} \int_{S^1}\omega &=\int_0^{2\pi}\frac{d}{dt}(f\circ\gamma)(t)\,dt\\ &=(f\circ\gamma)(2\pi)-(f\circ\gamma)(0)\\ &=f(1,0)-f(1,0)\\ &=0, \end{align*} contradicting $\int_{S^1}\omega=2\pi$. Therefore $d\theta$ is closed but not exact, and its nonzero period around the circle is the obstruction to a global angular potential. [/example] ## Pullback and Functoriality A cohomology theory should compare different manifolds. The natural operation on differential forms is pullback, so the question is whether a smooth map $\varphi:M\to N$ sends closed forms to closed forms and exact forms to exact forms in a way that descends to cohomology. [quotetheorem:3925] [citeproof:3925] This construction reverses arrows: a map $M\to N$ gives a cohomology map from $N$ back to $M$. The commuting identity $d\varphi^*=\varphi^*d$ is the essential hypothesis behind the construction; without it, pullback would not respect closed and exact forms. The theorem does not assert that $\varphi^*$ is injective or surjective, since a map may collapse topological information or miss part of the target. Functoriality is the next step because it checks that these induced maps behave coherently under composition. [quotetheorem:3926] [citeproof:3926] A direct consequence is diffeomorphism invariance. Functoriality is stronger than a bookkeeping rule: it is what lets inverse maps produce inverse maps on cohomology. The statement is still purely smooth at this stage, so it does not yet identify manifolds that are merely deformable into one another. Homotopy invariance will weaken the required relationship from diffeomorphism to smooth homotopy equivalence. [quotetheorem:3927] [citeproof:3927] The diffeomorphism hypothesis is much stronger than needed for an isomorphism on cohomology, but it gives the cleanest first invariance result. It also preserves dimension and smooth structure, whereas de Rham cohomology alone cannot recover either of those data. Even among diffeomorphisms, the induced map can record orientation behaviour, as the next example shows on the circle. [example: Reflection of the Circle] Let $r:S^1\to S^1$ be reflection across the $x$-axis, so $r(x,y)=(x,-y)$. We compute the induced map on the generator $[d\theta]\in H^1_{\mathrm{dR}}(S^1)$. The angular form $d\theta$ is the restriction to $S^1$ of \begin{align*} \alpha=-y\,dx+x\,dy. \end{align*} Since \begin{align*} r^*x&=x,& r^*y&=-y,\\ r^*dx&=d(r^*x)=dx,& r^*dy&=d(r^*y)=d(-y)=-dy, \end{align*} we get \begin{align*} r^*\alpha &=r^*(-y\,dx+x\,dy)\\ &=-(r^*y)\,r^*dx+(r^*x)\,r^*dy\\ &=-(-y)\,dx+x(-dy)\\ &=y\,dx-x\,dy\\ &=-(-y\,dx+x\,dy)\\ &=-\alpha. \end{align*} Restricting to $S^1$ gives \begin{align*} r^*(d\theta)=-d\theta. \end{align*} Therefore on cohomology classes, \begin{align*} r^*[d\theta] &=[r^*(d\theta)]\\ &=[-d\theta]\\ &=-[d\theta]. \end{align*} Since $[d\theta]$ spans $H^1_{\mathrm{dR}}(S^1)$, the induced map $r^*:H^1_{\mathrm{dR}}(S^1)\to H^1_{\mathrm{dR}}(S^1)$ is multiplication by $-1$. Reflection is a diffeomorphism, but its action on first cohomology records the reversal of orientation around the circle. [/example] ## Homotopy Invariance Diffeomorphism invariance is still a smooth classification statement. The deeper question is whether de Rham cohomology survives continuous deformation: if two maps can be joined by a smooth homotopy, do they induce the same map on cohomology? [definition: Smooth Homotopy] Let $M$ and $N$ be smooth manifolds. Two smooth maps $f_0,f_1:M\to N$ are smoothly homotopic if there exists a smooth map \begin{align*} F:[0,1]\times M\to N \end{align*} such that \begin{align*} F(0,x)=f_0(x),\qquad F(1,x)=f_1(x) \end{align*} for all $x\in M$. [/definition] The key question is whether de Rham cohomology sees the exact position of a smooth map, or only its deformation class. A homotopy gives a cylinder $[0,1]\times M$ connecting the two endpoint maps, and the theorem below identifies the change in pullback along that cylinder as an exact error on closed forms. [quotetheorem:3928] [citeproof:3928] The result concerns maps on cohomology, not equality of the pullback forms themselves: the endpoint pullbacks may differ as differential forms, but they define the same cohomology class on closed representatives. This is the bridge from differential geometry to topology, since cohomology ignores deformation through smooth homotopies. Its limitation is equally important: homotopy equivalent manifolds can have very different geometry, dimension, or smooth embeddings, so de Rham cohomology is not a complete invariant of manifolds. The practical value is computational: replace a complicated manifold by a homotopy equivalent model whose cohomology is easier to compute. [illustration:dfii-homotopy-cylinder] [example: The Punctured Plane Deformation Retracts to the Circle] [claim]For $M=\mathbb R^2\setminus\{0\}$, one has \begin{align*} H^0_{\mathrm{dR}}(M)\cong\mathbb R,\qquad H^1_{\mathrm{dR}}(M)\cong\mathbb R,\qquad H^k_{\mathrm{dR}}(M)=0\quad\text{for }k\ge2. \end{align*} [/claim] [proof]Let $i:S^1\hookrightarrow M$ be the inclusion and define \begin{align*} r:M\to S^1,\qquad r(x)=\frac{x}{|x|}. \end{align*} The map $r$ is smooth on $M$ because $|x|=\sqrt{x_1^2+x_2^2}$ is positive there. If $p\in S^1$, then $|p|=1$, so \begin{align*} (r\circ i)(p) &=r(p)\\ &=\frac{p}{|p|}\\ &=p. \end{align*} Hence $r\circ i=\operatorname{id}_{S^1}$. It remains to compare $i\circ r$ with $\operatorname{id}_M$. Define \begin{align*} H:[0,1]\times M\to M,\qquad H(t,x)=\left((1-t)+\frac{t}{|x|}\right)x. \end{align*} For $x\in M$, the scalar \begin{align*} (1-t)+\frac{t}{|x|} \end{align*} is positive for every $t\in[0,1]$, so $H(t,x)\ne0$ and $H$ really takes values in $M$. The map is smooth because it is built from smooth functions on $M$. At the endpoints, \begin{align*} H(0,x) &=\left(1+\frac{0}{|x|}\right)x\\ &=x \end{align*} and \begin{align*} H(1,x) &=\left(0+\frac{1}{|x|}\right)x\\ &=\frac{x}{|x|}\\ &=(i\circ r)(x). \end{align*} Thus $i\circ r$ is smoothly homotopic to $\operatorname{id}_M$. By *de Rham Cohomology is a Homotopy Invariant*, the maps $i$ and $r$ induce inverse isomorphisms on de Rham cohomology, so \begin{align*} H^k_{\mathrm{dR}}(M)\cong H^k_{\mathrm{dR}}(S^1) \end{align*} for every $k\ge0$. By *de Rham Cohomology of the Circle*, \begin{align*} H^k_{\mathrm{dR}}(S^1)\cong \begin{cases} \mathbb R, & k=0,\\ \mathbb R, & k=1,\\ 0, & k\ge2. \end{cases} \end{align*} Substituting these values gives \begin{align*} H^0_{\mathrm{dR}}(M)&\cong\mathbb R,\\ H^1_{\mathrm{dR}}(M)&\cong\mathbb R,\\ H^k_{\mathrm{dR}}(M)&=0\quad\text{for }k\ge2. \end{align*} [/proof] The missing point leaves one essential loop around the origin, and the first cohomology group records exactly the same period obstruction as the circle. [/example] The punctured-plane computation is the model use case for homotopy invariance: remove geometric excess while preserving the obstruction measured by periods. Instead of constructing all closed forms on $\mathbb R^2\setminus\{0\}$ directly, we compute on a deformation retract. This is the same philosophy that later connects de Rham classes to singular cohomology classes and integration over cycles. [remark: What Cohomology Can and Cannot See] de Rham cohomology is insensitive to smooth deformations, so it cannot distinguish homotopy equivalent manifolds. It does detect features that survive homotopy, such as connected components and the essential loop in $S^1$. Later chapters compare these analytic invariants with singular cohomology, where the same information is built from chains and simplices rather than differential forms. [/remark] De Rham cohomology records the global obstructions that remain after all local primitives have been accounted for. To understand those obstructions efficiently, we first need the local converse to exactness, namely that on contractible regions closed forms are already exact. # 7. The Poincare Lemma The chapter begins with the local exactness problem for differential forms on Euclidean domains. It then turns the geometric contraction of a star-shaped set into an explicit homotopy operator, before comparing this local result with global obstructions on the circle and punctured three-space. ## Closed Forms and the Local Exactness Problem Suppose $\beta \in \Omega^k(U)$ satisfies $d\beta = 0$. The equation $\beta = d\alpha$ asks for a potential $(k-1)$-form $\alpha$, generalising the question of whether a curl-free vector field has a scalar potential or a divergence-free vector field has a vector potential. The Poincare lemma says that, on a star-shaped open set, no further analytic obstruction appears. [definition: Star-Shaped Open Set] Let $U \subset \mathbb R^n$ be open. The set $U$ is star-shaped with centre $a \in U$ if for every $t \in [0,1]$ the map \begin{align*} H_t:U&\to U, & H_t(x)&=a+t(x-a) \end{align*} is well-defined. [/definition] The map $H_t$ contracts $U$ linearly onto the point $a$. This is stronger than mere contractibility because it gives a particular contraction through straight line segments, and that extra structure is what produces the explicit primitive. The local exactness problem has two separate conditions that should not be conflated. The equation $d\omega=0$ is a compatibility condition forced by $d^2=0$, while the equation $\omega=d\eta$ asks for an actual potential. The obstruction is that compatibility need not produce a potential globally, so we name the two conditions before stating when the obstruction disappears. [definition: Closed and Exact Differential Form] Let $U \subset \mathbb R^n$ be open and let $k \ge 0$. A form $\omega \in \Omega^k(U)$ is closed if $d\omega = 0$. For $k \ge 1$, it is exact if there exists $\eta \in \Omega^{k-1}(U)$ such that $\omega = d\eta$. [/definition] For $k=0$, closed means locally constant, while exactness is not part of the same pattern because there are no $(-1)$-forms. The Poincare lemma is therefore stated for positive degrees. With this terminology fixed, the central local question becomes precise: on a domain that can be collapsed by straight lines, does every closed positive-degree form actually have a potential? The Poincare lemma gives the Euclidean model for that answer. [quotetheorem:3615] [citeproof:3615] The theorem is local in nature. It explains why sufficiently small Euclidean coordinate balls have no positive-degree de Rham cohomology, but it does not remove global obstructions produced by holes. The star-shaped hypothesis is stronger than contractibility in a way that matters here: every point must be visible from a common centre by a straight segment lying inside the domain. For example, $\mathbb R^2_0$ is not star-shaped, and the angular form introduced below is closed but not globally exact there. ## The Homotopy Operator To use the Poincare lemma concretely, it is helpful to name the standard operator associated with the radial contraction of a star-shaped domain. This operator will also make clear why the hypothesis is tied to straight line segments inside the domain. [definition: Radial Homotopy Operator] Let $U \subset \mathbb R^n$ be star-shaped with centre $a$, and define $H_t(x)=a+t(x-a)$. For $k \ge 1$ and $\omega \in \Omega^k(U)$, the radial homotopy operator $K: \Omega^k(U) \to \Omega^{k-1}(U)$ is \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*} [/definition] The factor $t^{k-1}$ comes from the $k-1$ tangent vectors $v_1,\dots,v_{k-1}$ being pushed forward by $dH_t$, while the remaining slot is occupied by the velocity of the homotopy. This formula is the coordinate-level version of the chain homotopy induced by the contraction. [quotetheorem:3929] [citeproof:3929] When $\omega$ is closed, the term $Kd\omega$ drops out and $K\omega$ is a primitive. When $\omega$ is not closed, the same identity says exactly how far $K\omega$ is from being a primitive. The identity depends on star-shapedness because the radial path $t\mapsto a+t(x-a)$ must remain inside $U$ for every $x\in U$. It also depends on the chosen centre $a$: changing $a$ changes $K\omega$, although the resulting primitives differ by a closed form of degree $k-1$ when $\omega$ is closed. Finally, this is a Euclidean radial construction, so on a general manifold it must be replaced by a homotopy operator coming from an actual smooth contraction rather than by straight-line integration. [example: Primitive of a Constant Two-Form] On $U=\mathbb R^3$, take $\omega=dx_1\wedge dx_2$ and choose the star-centre $a=0$. Since $d(dx_i)=0$ for coordinate $1$-forms, \begin{align*} d\omega &=d(dx_1\wedge dx_2)\\ &=d(dx_1)\wedge dx_2-dx_1\wedge d(dx_2)\\ &=0\wedge dx_2-dx_1\wedge 0\\ &=0, \end{align*} so $\omega$ is closed. For $k=2$, the radial homotopy operator gives, for $v=v_1\partial_{x_1}+v_2\partial_{x_2}+v_3\partial_{x_3}$, \begin{align*} (K\omega)_x(v) &=\int_0^1 t\,\omega_{tx}(x,v)\,dt\\ &=\int_0^1 t\,(dx_1\wedge dx_2)(x,v)\,dt\\ &=\int_0^1 t\,(dx_1(x)dx_2(v)-dx_2(x)dx_1(v))\,dt\\ &=\int_0^1 t\,(x_1v_2-x_2v_1)\,dt\\ &=\left[\frac{t^2}{2}\right]_{0}^{1}(x_1v_2-x_2v_1)\\ &=\frac{1}{2}(x_1v_2-x_2v_1). \end{align*} Thus \begin{align*} K\omega=\frac{1}{2}(x_1\,dx_2-x_2\,dx_1). \end{align*} Set $\eta=\frac{1}{2}(x_1\,dx_2-x_2\,dx_1)$. Then \begin{align*} d\eta &=\frac{1}{2}\left(d(x_1\,dx_2)-d(x_2\,dx_1)\right)\\ &=\frac{1}{2}\left(dx_1\wedge dx_2+x_1\,d(dx_2)-dx_2\wedge dx_1-x_2\,d(dx_1)\right)\\ &=\frac{1}{2}\left(dx_1\wedge dx_2-dx_2\wedge dx_1\right)\\ &=\frac{1}{2}\left(dx_1\wedge dx_2+dx_1\wedge dx_2\right)\\ &=dx_1\wedge dx_2. \end{align*} So $\eta$ is a primitive of the constant two-form $\omega$, obtained by integrating the contraction of $\omega$ with the radial vector field. [/example] This computation is the differential-form version of building a vector potential from a radial gauge condition. Different choices of star-centre produce different primitives, differing by an exact lower-degree correction. ## Contractible Manifolds and de Rham Cohomology The local result suggests a global question: if a manifold can be contracted to a point, should all closed positive-degree forms be exact? For smooth manifolds the answer is yes when the contraction is smooth, and the statement is naturally expressed using de Rham cohomology. [definition: de Rham Cohomology Group] Let $M$ be a smooth manifold. For $k\ge 0$, the $k$-th de Rham cohomology group is \begin{align*} H^k_{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*} [/definition] The numerator records closed forms, and the denominator records the forms already known to be exact. Thus $H^k_{dR}(M)$ measures the failure of the converse to $d^2=0$ in degree $k$. To connect this quotient to geometry, we need a smooth version of the idea that the whole manifold can be shrunk to one point. The definition below records exactly the kind of contraction that can be differentiated and integrated against forms. [definition: Smoothly Contractible Manifold] A smooth manifold $M$ is smoothly contractible if there exist a point $p\in M$ and a smooth map $H:[0,1]\times M\to M$ such that $H(1,x)=x$ and $H(0,x)=p$ for every $x\in M$. [/definition] The smooth contraction supplies the same kind of homotopy operator as in Euclidean space. The straight-line formula is replaced by integration along the parameter of $H$. The point of introducing smooth contractibility is to force a cohomological consequence: closed forms should have no global obstruction to being exact in positive degree. The theorem makes that consequence precise in the language of de Rham groups. [quotetheorem:3586] [citeproof:3586] This theorem explains why de Rham cohomology cannot detect fine geometric shape such as curvature or length. It detects whether closed forms are prevented from having global primitives by the topology of the manifold. The hypothesis is a smooth contraction, not merely an informal continuous shrinking, because the proof differentiates the homotopy and integrates differential forms along its parameter. The conclusion is also one-directional: vanishing positive de Rham cohomology does not by itself say that the manifold is contractible. The punctured examples below show the opposite phenomenon from the theorem: when no global contraction exists, closed forms can acquire nonzero periods and hence fail to be exact. [remark: Contractibility and Cohomology] Contractibility implies $H^k_{dR}(M)=0$ for $k\ge 1$, but the reverse implication is not a definition of contractibility. There are spaces whose cohomology groups vanish in positive degree but which are not contractible. In this course, the slogan that contractible spaces have no positive-degree de Rham cohomology should be read as the direction proved by the homotopy operator, together with the intuition that de Rham cohomology records part of the obstruction to contraction. [/remark] ## The Circle as the First Obstruction Where does the Poincare lemma fail? It fails when the domain contains loops that cannot be filled in while staying inside the domain. The circle gives the basic model: there is a closed $1$-form measuring angular change, but no globally single-valued angle function whose differential it is. [definition: Angular One-Form on the Punctured Plane] On $\mathbb R^2_0=\mathbb R^2\setminus\{0\}$, define \begin{align*} d\theta = \frac{-x_2\,dx_1+x_1\,dx_2}{x_1^2+x_2^2}. \end{align*} [/definition] The notation $d\theta$ is traditional: locally it is the differential of an angle coordinate. Globally, however, $\theta$ cannot be chosen as a smooth real-valued function on all of $\mathbb R^2_0$ or on $S^1$. The theorem below measures exactly this obstruction. If $d\theta$ were globally exact on the circle, its integral around the closed loop would have to vanish by the fundamental theorem/Stokes principle for exact forms. The puncture should therefore show up as a nonzero period: one full turn accumulates angular change even though the path returns to its starting point. [quotetheorem:3616] [citeproof:3616] This is the first appearance of periods: integrals of closed forms over closed cycles. Exact forms have zero periods, so a nonzero period detects a nonzero de Rham cohomology class. The puncture is essential: on any small open arc of $S^1$, or on any simply connected coordinate patch in $\mathbb R^2_0$ avoiding a branch cut, the form is the differential of a genuine angle function. Thus the theorem does not contradict the Poincare lemma, because the obstruction appears only after trying to choose compatible local primitives around the whole loop. This motivates the cohomological viewpoint: periods are numerical tests for whether a closed form represents a nonzero global class. [illustration:dfii-angular-form-punctured-plane] [example: Winding Number from the Angular Form] [claim]For every smooth closed curve $\gamma:S^1\to\mathbb R^2_0$, the number \begin{align*} \frac{1}{2\pi}\oint_\gamma d\theta \end{align*} is the winding number of $\gamma$ about the origin.[/claim] [proof]Parametrize $S^1$ by $t\in[0,1]$ with endpoints identified, and write $\gamma(t)=(x_1(t),x_2(t))$. Since $\gamma([0,1])$ is compact and avoids the origin, we can choose a partition \begin{align*} 0=t_0<t_1<\cdots<t_N=1 \end{align*} such that on each interval $[t_{j-1},t_j]$ there are smooth functions $r_j(t)>0$ and $\varphi_j(t)$ with \begin{align*} x_1(t)&=r_j(t)\cos\varphi_j(t),& x_2(t)&=r_j(t)\sin\varphi_j(t). \end{align*} On such an interval, \begin{align*} dx_1 &=\left(r_j'\cos\varphi_j-r_j\sin\varphi_j\,\varphi_j'\right)dt,\\ dx_2 &=\left(r_j'\sin\varphi_j+r_j\cos\varphi_j\,\varphi_j'\right)dt. \end{align*} Therefore the pullback of the numerator of $d\theta$ is \begin{align*} -x_2\,dx_1+x_1\,dx_2 &=-r_j\sin\varphi_j\left(r_j'\cos\varphi_j-r_j\sin\varphi_j\,\varphi_j'\right)dt\\ &\quad+r_j\cos\varphi_j\left(r_j'\sin\varphi_j+r_j\cos\varphi_j\,\varphi_j'\right)dt\\ &=\left(-r_jr_j'\sin\varphi_j\cos\varphi_j+r_j^2\sin^2\varphi_j\,\varphi_j'\right)dt\\ &\quad+\left(r_jr_j'\cos\varphi_j\sin\varphi_j+r_j^2\cos^2\varphi_j\,\varphi_j'\right)dt\\ &=r_j^2(\sin^2\varphi_j+\cos^2\varphi_j)\varphi_j'\,dt\\ &=r_j^2\varphi_j'\,dt. \end{align*} The denominator is \begin{align*} x_1^2+x_2^2 &=r_j^2\cos^2\varphi_j+r_j^2\sin^2\varphi_j\\ &=r_j^2. \end{align*} Hence on $[t_{j-1},t_j]$, \begin{align*} \gamma^*(d\theta)=\varphi_j'(t)\,dt, \end{align*} and so \begin{align*} \int_{t_{j-1}}^{t_j}\gamma^*(d\theta) &=\int_{t_{j-1}}^{t_j}\varphi_j'(t)\,dt\\ &=\varphi_j(t_j)-\varphi_j(t_{j-1}). \end{align*} Adding over the partition gives \begin{align*} \oint_\gamma d\theta &=\sum_{j=1}^N\left(\varphi_j(t_j)-\varphi_j(t_{j-1})\right). \end{align*} At each interior point $t_j$, the two angles $\varphi_j(t_j)$ and $\varphi_{j+1}(t_j)$ describe the same nonzero vector in $\mathbb R^2$, so \begin{align*} \varphi_j(t_j)-\varphi_{j+1}(t_j)=2\pi m_j \end{align*} for some integer $m_j$. Since $\gamma(1)=\gamma(0)$, also \begin{align*} \varphi_N(1)-\varphi_1(0)=2\pi m_N \end{align*} for some integer $m_N$. Rearranging the sum by inserting these endpoint differences, \begin{align*} \oint_\gamma d\theta &=\varphi_N(1)-\varphi_1(0) +\sum_{j=1}^{N-1}\left(\varphi_j(t_j)-\varphi_{j+1}(t_j)\right)\\ &=2\pi m_N+\sum_{j=1}^{N-1}2\pi m_j\\ &=2\pi\sum_{j=1}^N m_j. \end{align*} Thus \begin{align*} \frac{1}{2\pi}\oint_\gamma d\theta =\sum_{j=1}^N m_j\in\mathbb Z. \end{align*} This integer is exactly the total change of any locally chosen angle along $\gamma$, divided by $2\pi$, which is the winding number of $\gamma$ about the origin.[/proof] The angular form therefore converts the geometric act of circling the missing origin into the period of a closed $1$-form. [/example] The phrase "linking number" in this low-dimensional example means that the loop detects the puncture by circling it. In higher dimensions, analogous closed forms measure how cycles wrap around deleted submanifolds. ## Punctured Space and the Sphere The example on the punctured plane raises a useful test question: when does removing a point create a cohomological obstruction? The answer depends on degree. In $\mathbb R^3_0$, the obstruction appears in degree $2$, and it is detected by integration over spheres surrounding the origin. [definition: Solid Angle Two-Form] On $\mathbb R^3_0$, define \begin{align*} \omega = \frac{1}{(x_1^2+x_2^2+x_3^2)^{3/2}} \left(x_1\,dx_2\wedge dx_3+x_2\,dx_3\wedge dx_1+x_3\,dx_1\wedge dx_2\right). \end{align*} [/definition] This form is the area form on the unit sphere pulled back by the radial projection $r:\mathbb R^3_0\to S^2$, up to orientation convention. It is closed on the punctured space, but its integral over any sphere enclosing the origin is nonzero. [quotetheorem:3930] [citeproof:3930] This theorem should be compared with the star-shaped Poincare lemma. The form has no singularity on its domain, and it is closed there; the obstruction is that $\mathbb R^3_0$ is not contractible and contains a sphere that cannot be filled inside the domain. Puncturing $\mathbb R^3$ matters because the missing point prevents the surrounding $S^2$ from being the boundary of a compact region contained in the domain. The theorem does not say that primitives fail locally: every point of $\mathbb R^3_0$ has a small star-shaped neighbourhood, so the Poincare lemma gives local primitives. What fails is global compatibility, and the sphere detects that failure by carrying a nonzero integral of the closed form. [illustration:dfii-solid-angle-form] [example: Primitive Locally but Not Globally on Punctured Three-Space] Let $\rho=(x_1^2+x_2^2+x_3^2)^{1/2}$, and on the usual spherical coordinates \begin{align*} x_1&=\rho\sin\varphi\cos\theta,& x_2&=\rho\sin\varphi\sin\theta,& x_3&=\rho\cos\varphi. \end{align*} On $U_+$, use \begin{align*} \eta_+=(1-\cos\varphi)\,d\theta, \end{align*} which is smooth there because \begin{align*} 1-\cos\varphi &=1-\frac{x_3}{\rho} =\frac{\rho-x_3}{\rho} =\frac{x_1^2+x_2^2}{\rho(\rho+x_3)},\\ d\theta &=\frac{-x_2\,dx_1+x_1\,dx_2}{x_1^2+x_2^2}, \end{align*} so \begin{align*} \eta_+ =\frac{-x_2\,dx_1+x_1\,dx_2}{\rho(\rho+x_3)}, \end{align*} whose denominator is nonzero exactly away from the nonpositive $x_3$-axis. In spherical coordinates the solid angle form restricts to the standard oriented area form on the unit sphere: \begin{align*} \omega=\sin\varphi\,d\varphi\wedge d\theta. \end{align*} On the northern chart, the usual potential can be written $\eta_+=(1-\cos\varphi)d\theta$. Hence \begin{align*} d\eta_+ &=d\bigl((1-\cos\varphi)d\theta\bigr)\\ &=d(1-\cos\varphi)\wedge d\theta+(1-\cos\varphi)d(d\theta)\\ &=\sin\varphi\,d\varphi\wedge d\theta+0\\ &=\omega. \end{align*} Similarly, on $U_-$ set \begin{align*} \eta_-=-(1+\cos\varphi)\,d\theta. \end{align*} Since \begin{align*} 1+\cos\varphi =\frac{\rho+x_3}{\rho} =\frac{x_1^2+x_2^2}{\rho(\rho-x_3)}, \end{align*} we have \begin{align*} \eta_-=\frac{x_2\,dx_1-x_1\,dx_2}{\rho(\rho-x_3)}, \end{align*} which is smooth away from the nonnegative $x_3$-axis. Its exterior derivative is \begin{align*} d\eta_- &=-d\bigl((1+\cos\varphi)d\theta\bigr)\\ &=-\left(-\sin\varphi\,d\varphi\wedge d\theta+(1+\cos\varphi)d(d\theta)\right)\\ &=\sin\varphi\,d\varphi\wedge d\theta\\ &=\omega. \end{align*} On the overlap $U_+\cap U_-$, \begin{align*} \eta_+-\eta_- &=(1-\cos\varphi)d\theta+(1+\cos\varphi)d\theta\\ &=2\,d\theta. \end{align*} For a circle of latitude $\gamma(t)=(\sin\varphi_0\cos t,\sin\varphi_0\sin t,\cos\varphi_0)$ with $0<\varphi_0<\pi$, \begin{align*} \int_\gamma(\eta_+-\eta_-) &=\int_0^{2\pi}2\,dt\\ &=4\pi. \end{align*} Thus the two local primitives differ by a closed $1$-form with nonzero period on the overlap, so they cannot be glued into one global primitive on $\mathbb R^3_0$. This matches the global obstruction: if $\omega=d\eta$ on all of $\mathbb R^3_0$, then integrating over the unit sphere would give \begin{align*} \int_{S^2}\omega=\int_{S^2}d\eta=0, \end{align*} but the solid angle form has integral $4\pi$ over $S^2$. [/example] This is the geometric content of the Poincare lemma in practice: local primitives exist on contractible patches, while global primitives require compatibility across overlaps. De Rham cohomology measures the obstruction to making those local choices agree everywhere. The Poincaré lemma explains why de Rham cohomology is trivial on sufficiently small or contractible sets, but nontrivial globally. The next step is to organize local triviality across overlaps, so that the failure of primitives to match can be measured systematically. # 8. Mayer-Vietoris Sequence The previous chapters built de Rham cohomology from differential forms and showed that it is invariant under smooth homotopy. The next problem is computational: a manifold is often too large to understand at once, but it may be covered by open pieces whose cohomology is already known. The Mayer-Vietoris sequence is the mechanism that turns local information on two overlapping open sets into global information on the union. ## From Restrictions to an Exact Sequence of Complexes Suppose a smooth manifold $M$ is written as the union of two open subsets $U$ and $V$. If a differential form is known on all of $M$, then it restricts to forms on $U$ and on $V$ which agree on the overlap $U \cap V$. The first question is whether this compatibility condition is the only obstruction to gluing forms. [definition: Restriction Maps for an Open Cover] Let $M$ be a smooth manifold and let $M = U \cup V$, where $U,V \subset M$ are open. For each $k \ge 0$, define \begin{align*} r &: \Omega^k(M) \to \Omega^k(U) \oplus \Omega^k(V), & r(\omega) &= (\omega|_U,\omega|_V),\\ s &: \Omega^k(U) \oplus \Omega^k(V) \to \Omega^k(U \cap V), & s(\alpha,\beta) &= \alpha|_{U \cap V}-\beta|_{U \cap V}. \end{align*} [/definition] The sign in $s$ records agreement rather than difference-free notation: $s(\alpha,\beta)=0$ precisely when the two local forms have the same restriction to the overlap. The fact that exterior differentiation commutes with restriction will make these maps compatible with the de Rham complexes. The remaining issue is whether this compatibility condition is complete: can every matching pair of local forms be glued, and is every global form exactly the same thing as such compatible local data? The exact sequence below packages that sheaf property in a form usable for cohomology. [quotetheorem:3931] [citeproof:3931] The openness hypothesis is doing real work. Differential forms are naturally defined on open submanifolds, and the partition of unity argument needs smooth cutoff functions whose supports stay away from the boundary where zero extension would otherwise create nonsmooth behaviour. The theorem does not say that arbitrary compatible data on closed pieces glues smoothly: for example, on $S^1$ two smooth functions on closed semicircles can agree at the endpoints while their derivatives fail to match there, so the glued function need not be smooth. What the theorem gives is an exact algebraic expression of the sheaf property for smooth forms, which is exactly the input needed before passing to cohomology. [example: Gluing Functions on an Open Cover] Let $M=U\cup V$ with $U,V$ open, and consider degree $0$, so $\Omega^0(W)=C^\infty(W)$. Suppose $f_U\in C^\infty(U)$ and $f_V\in C^\infty(V)$ satisfy \begin{align*} f_U|_{U\cap V}=f_V|_{U\cap V}. \end{align*} Define $f:M\to \mathbb R$ by \begin{align*} f(p)= \begin{cases} f_U(p), & p\in U,\\ f_V(p), & p\in V. \end{cases} \end{align*} This is well-defined: if $p\in U\cap V$, then the two possible values are equal because $f_U(p)=f_V(p)$ on the overlap. To check smoothness, let $p\in M$. Since $M=U\cup V$, either $p\in U$ or $p\in V$. If $p\in U$, then $U$ is an open neighbourhood of $p$ and \begin{align*} f|_U=f_U, \end{align*} so $f$ is smooth near $p$. If $p\in V$, the same argument gives \begin{align*} f|_V=f_V, \end{align*} so $f$ is smooth near $p$. Smoothness is local on open neighbourhoods, hence $f\in C^\infty(M)$. In the notation of the restriction maps, the compatibility condition is exactly \begin{align*} s(f_U,f_V) &=f_U|_{U\cap V}-f_V|_{U\cap V}\\ &=0. \end{align*} The glued function satisfies \begin{align*} r(f)=(f|_U,f|_V)=(f_U,f_V), \end{align*} so every element of $\ker s$ in degree $0$ lies in $\operatorname{im} r$. Thus middle exactness in degree $0$ is precisely the usual gluing property for smooth functions. In a coordinate chart, a differential $k$-form is a finite sum of smooth coefficient functions times basis forms $dx^{i_1}\wedge\cdots\wedge dx^{i_k}$, so the same gluing argument applies coefficient by coefficient. [/example] This short exact sequence is the algebraic input. Passing from complexes to cohomology changes short exactness into a long exact sequence, and the new map that appears is the connecting homomorphism. ## The Long Exact Sequence in de Rham Cohomology The next question is what the exact sequence of complexes says about closed forms modulo exact forms. Exactness at the cochain level does not give a short exact sequence on cohomology, because a local primitive may fail to glue. The failure is measured by a boundary map into the next cohomological degree. [quotetheorem:3589] [citeproof:3589] The open-cover hypotheses are again essential because the proof uses the short exact sequence of complexes, whose surjectivity came from partitions of unity on open sets. Without openness, there may be no smooth extension-by-zero construction, so a closed form on an overlap need not admit the required lift. For instance, decomposing a circle into two closed arcs is geometrically tempting, but smooth data on the closed overlap endpoints does not control all derivatives at those endpoints, so the cochain-level exact sequence fails. Thus the theorem is not a general gluing statement for arbitrary subsets; it is a theorem about open covers in the category where de Rham forms and partitions of unity behave correctly. [remark: Exactness as a Computation Tool] Exactness means that the image of each map is the kernel of the next. In computations, this converts geometric restriction questions into algebraic constraints on vector spaces, often reducing an unknown cohomology group of $M$ to known groups of $U$, $V$, and $U \cap V$. [/remark] For connected open pieces with contractible intersections, the sequence usually collapses to short fragments. These fragments are where most low-dimensional computations happen. ## The Connecting Homomorphism The formal long exact sequence is useful only after the connecting map is made concrete. Given a closed form on $U \cap V$, the question is how it produces a closed form of one higher degree on $M$. [definition: Connecting Homomorphism] Let $M = U \cup V$ be an open cover and let $[\gamma] \in H^k_{\mathrm{dR}}(U \cap V)$ be represented by a closed form $\gamma \in \Omega^k(U \cap V)$. Choose $(\alpha,\beta) \in \Omega^k(U) \oplus \Omega^k(V)$ such that \begin{align*} \beta|_{U \cap V}-\alpha|_{U \cap V}=\gamma. \end{align*} Then $d\alpha$ and $d\beta$ agree on $U \cap V$, so they glue to a form $\eta \in \Omega^{k+1}(M)$. The connecting homomorphism is \begin{align*} \delta[\gamma] = [\eta] \in H^{k+1}_{\mathrm{dR}}(M). \end{align*} [/definition] The definition is independent of the chosen lift, but in practice a partition of unity gives a standard lift. If $\rho_U+\rho_V=1$ is subordinate to $U,V$, then one may take $\alpha=\rho_V\gamma$ on $U \cap V$, extended by zero where appropriate, and $\beta=-\rho_U\gamma$ with the corresponding extension. [quotetheorem:3932] [citeproof:3932] This theorem says that the boundary map is canonical on cohomology, not that there is a canonical differential form representative of its value. Different partitions of unity or different lifts usually give different global closed forms, but their difference is exact. The construction also depends on the open cover: changing $U$ and $V$ changes the intermediate overlap group, even when the resulting cohomology group of $M$ is the same. In computations below, $\delta$ is useful because it converts a mismatch on $U\cap V$ into an actual global cohomology class. [example: Boundary Map in Degree Zero] Let $M=S^1$, and choose connected open arcs $U,V \subset S^1$ with $U\cup V=S^1$ and \begin{align*} U\cap V=A\sqcup B, \end{align*} where $A$ and $B$ are the two connected components of the overlap. Since a closed $0$-form is a locally constant function, a class in $H^0_{\mathrm{dR}}(U\cap V)$ is represented uniquely by its two constant values: \begin{align*} \gamma|_A=a,\qquad \gamma|_B=b. \end{align*} Thus we identify $H^0_{\mathrm{dR}}(U\cap V)$ with $\mathbb R^2$ by writing $\gamma=(a,b)$. Under this identification, the map \begin{align*} s^*:H^0_{\mathrm{dR}}(U)\oplus H^0_{\mathrm{dR}}(V)\to H^0_{\mathrm{dR}}(U\cap V) \end{align*} has a simple form. Because $U$ and $V$ are connected, a class in $H^0_{\mathrm{dR}}(U)\oplus H^0_{\mathrm{dR}}(V)$ is represented by two constants $(x,y)$. On both overlap components, the restriction from $U$ is $x$ and the restriction from $V$ is $y$, so \begin{align*} s^*(x,y) &=\bigl(x-y,\ x-y\bigr). \end{align*} Therefore \begin{align*} \operatorname{im}(s^*) &=\{(t,t):t\in \mathbb R\}. \end{align*} The diagonal classes vanish under the connecting map because exactness gives \begin{align*} \ker(\delta)=\operatorname{im}(s^*). \end{align*} For an arbitrary overlap class $(a,b)$, decompose it as \begin{align*} (a,b) &=(b,b)+(a-b,0)\\ &=b(1,1)+(a-b)(1,0). \end{align*} Since $(1,1)\in \operatorname{im}(s^*)=\ker(\delta)$ and $\delta$ is linear, \begin{align*} \delta(a,b) &=\delta\bigl(b(1,1)+(a-b)(1,0)\bigr)\\ &=b\,\delta(1,1)+(a-b)\,\delta(1,0)\\ &=(a-b)\,\delta(1,0). \end{align*} Thus the connecting homomorphism depends only on the jump $a-b$ between the two overlap components. Once a generator of $H^1_{\mathrm{dR}}(S^1)$ is chosen, $\delta(a,b)$ is a scalar multiple of that generator, with scalar proportional to $a-b$. [/example] This example captures the geometric meaning of $\delta$: it records the obstruction to choosing local primitives or local data in a way that matches globally. ## Computing the Cohomology of the Circle The first substantial computation is $S^1$. The question is whether the Mayer-Vietoris sequence detects the single loop even though each open arc is contractible. [example: de Rham Cohomology of the Circle] Let $S^1=U\cup V$, where $U$ and $V$ are connected open arcs slightly longer than semicircles. Write \begin{align*} U\cap V=A\sqcup B \end{align*} for the two connected components of the overlap. The sets $U$, $V$, $A$, and $B$ are contractible, so by homotopy invariance and the computation of de Rham cohomology of a point, \begin{align*} H^0_{\mathrm{dR}}(U) &\cong \mathbb R, & H^0_{\mathrm{dR}}(V) &\cong \mathbb R, & H^0_{\mathrm{dR}}(U\cap V) &\cong H^0_{\mathrm{dR}}(A)\oplus H^0_{\mathrm{dR}}(B)\cong \mathbb R^2,\\ H^k_{\mathrm{dR}}(U) &=0, & H^k_{\mathrm{dR}}(V)&=0, & H^k_{\mathrm{dR}}(U\cap V)&=0 \quad (k\ge 1). \end{align*} The degree-zero part of the Mayer-Vietoris sequence is therefore \begin{align*} 0 \longrightarrow H^0_{\mathrm{dR}}(S^1) \longrightarrow \mathbb R\oplus\mathbb R \xrightarrow{s^*} \mathbb R^2 \xrightarrow{\delta} H^1_{\mathrm{dR}}(S^1) \longrightarrow 0. \end{align*} Under the identifications above, an element of $H^0_{\mathrm{dR}}(U)\oplus H^0_{\mathrm{dR}}(V)$ is a pair of constants $(x,y)$. On both components $A$ and $B$, the restriction of the class from $U$ is $x$, and the restriction of the class from $V$ is $y$. Hence \begin{align*} s^*(x,y) &=\bigl(x|_A-y|_A,\ x|_B-y|_B\bigr)\\ &=(x-y,\ x-y). \end{align*} Thus \begin{align*} \operatorname{im}(s^*) &=\{(x-y,x-y):x,y\in \mathbb R\}\\ &=\{(t,t):t\in \mathbb R\}, \end{align*} the diagonal line in $\mathbb R^2$. Exactness gives \begin{align*} \ker(\delta)=\operatorname{im}(s^*)=\{(t,t):t\in \mathbb R\}. \end{align*} Since the next term after $H^1_{\mathrm{dR}}(S^1)$ is $H^1_{\mathrm{dR}}(U)\oplus H^1_{\mathrm{dR}}(V)=0$, exactness also gives \begin{align*} \operatorname{im}(\delta)=H^1_{\mathrm{dR}}(S^1). \end{align*} Therefore $\delta$ induces an isomorphism \begin{align*} \mathbb R^2/\operatorname{im}(s^*) \cong H^1_{\mathrm{dR}}(S^1). \end{align*} To compute the quotient, define $L:\mathbb R^2\to\mathbb R$ by \begin{align*} L(a,b)=a-b. \end{align*} Then \begin{align*} \ker L &=\{(a,b):a-b=0\}\\ &=\{(a,a):a\in\mathbb R\}\\ &=\operatorname{im}(s^*), \end{align*} and $L$ is surjective because $L(t,0)=t$. Hence \begin{align*} \mathbb R^2/\operatorname{im}(s^*)\cong \mathbb R, \end{align*} so \begin{align*} H^1_{\mathrm{dR}}(S^1)\cong \mathbb R. \end{align*} The same exact segment also identifies $H^0_{\mathrm{dR}}(S^1)$ with the kernel of $s^*$. Since \begin{align*} \ker(s^*) &=\{(x,y):s^*(x,y)=(0,0)\}\\ &=\{(x,y):(x-y,x-y)=(0,0)\}\\ &=\{(x,x):x\in\mathbb R\}\\ &\cong \mathbb R, \end{align*} we get $H^0_{\mathrm{dR}}(S^1)\cong \mathbb R$. Finally, since $S^1$ is one-dimensional, $\Omega^k(S^1)=0$ for every $k\ge 2$, and therefore $H^k_{\mathrm{dR}}(S^1)=0$ for every $k\ge 2$. The single nonzero positive-degree class is exactly the class detected by the difference between the two overlap components. [/example] The computation expresses the loop as a mismatch between two overlap components. No individual open arc contains the topology of the circle; the topology appears in how the arcs are glued. ## Spheres and Induction The same idea computes the de Rham cohomology of all spheres. The question is how to cover $S^n$ by simple pieces so that the overlap has the homotopy type of a smaller sphere. [quotetheorem:3590] [citeproof:3590] The condition $n\ge 1$ excludes the zero-sphere, whose two connected components would make the degree-zero statement different. For $n>1$, connectedness of the overlap is the reason no degree-one class appears; the only new class is pushed upward through the connecting homomorphism until it reaches degree $n$. Mayer-Vietoris computes the vector spaces here, but it does not by itself choose preferred generators or normalisations of top-degree forms. Those choices usually come from orientation and integration, which enter later. [example: The Case of the Two-Sphere] For $S^2$, take \begin{align*} U=S^2\setminus\{N\},\qquad V=S^2\setminus\{S\}. \end{align*} Stereographic projection identifies each of $U$ and $V$ with $\mathbb R^2$, so both are contractible. Their intersection is the twice-punctured sphere \begin{align*} U\cap V=S^2\setminus\{N,S\}, \end{align*} which is diffeomorphic to $S^1\times \mathbb R$ by recording the angular coordinate around the equator and the height coordinate. The deformation retraction \begin{align*} S^1\times \mathbb R &\longrightarrow S^1\times \{0\},& (\theta,t)&\longmapsto (\theta,0) \end{align*} shows that $U\cap V$ has the same de Rham cohomology as $S^1$. The relevant part of the Mayer-Vietoris sequence is \begin{align*} H^1_{\mathrm{dR}}(U)\oplus H^1_{\mathrm{dR}}(V) \longrightarrow H^1_{\mathrm{dR}}(U\cap V) \xrightarrow{\delta} H^2_{\mathrm{dR}}(S^2) \longrightarrow H^2_{\mathrm{dR}}(U)\oplus H^2_{\mathrm{dR}}(V). \end{align*} Since $U$ and $V$ are contractible, \begin{align*} H^1_{\mathrm{dR}}(U)&=0,& H^1_{\mathrm{dR}}(V)&=0,& H^2_{\mathrm{dR}}(U)&=0,& H^2_{\mathrm{dR}}(V)&=0. \end{align*} Substituting these groups gives the exact segment \begin{align*} 0 \longrightarrow H^1_{\mathrm{dR}}(U\cap V) \xrightarrow{\delta} H^2_{\mathrm{dR}}(S^2) \longrightarrow 0. \end{align*} Exactness at $H^1_{\mathrm{dR}}(U\cap V)$ gives \begin{align*} \ker(\delta)=\operatorname{im}(0)=0, \end{align*} so $\delta$ is injective. Exactness at $H^2_{\mathrm{dR}}(S^2)$ gives \begin{align*} \operatorname{im}(\delta)=\ker(0)=H^2_{\mathrm{dR}}(S^2), \end{align*} so $\delta$ is surjective. Hence \begin{align*} H^2_{\mathrm{dR}}(S^2) &\cong H^1_{\mathrm{dR}}(U\cap V)\\ &\cong H^1_{\mathrm{dR}}(S^1)\\ &\cong \mathbb R. \end{align*} Thus the top cohomology class of $S^2$ is produced by applying the connecting homomorphism to the angular class on the equatorial overlap; after choosing an orientation, it may be represented by a global area form. [/example] This induction is the prototype for using Mayer-Vietoris with a cover whose pieces are contractible and whose intersections are already understood. ## Good Covers and the General Computation Strategy For a general manifold, two open sets are rarely enough. The problem becomes how to organise many local calculations without losing control of the algebra. Mayer-Vietoris is used repeatedly, adding one open set at a time. [definition: Good Cover] A good cover of a smooth manifold $M$ is an open cover $\{U_i\}_{i \in I}$ such that every nonempty finite intersection \begin{align*} U_{i_0} \cap \cdots \cap U_{i_p} \end{align*} is contractible. [/definition] Good covers reduce local de Rham cohomology to constants in degree $0$. The global information is therefore carried by the pattern of intersections, which is the point where de Rham theory begins to meet combinatorial topology. [explanation: Induction on a Good Cover] Suppose $M$ has a finite good cover $U_1,\dots,U_m$. Set $M_j=U_1\cup\cdots\cup U_j$. If $H^*_{\mathrm{dR}}(M_{j-1})$ is known, then Mayer-Vietoris applied to \begin{align*} M_j=M_{j-1}\cup U_j \end{align*} relates $H^*_{\mathrm{dR}}(M_j)$ to $H^*_{\mathrm{dR}}(M_{j-1})$, $H^*_{\mathrm{dR}}(U_j)$, and $H^*_{\mathrm{dR}}(M_{j-1}\cap U_j)$. The set $U_j$ is contractible, and $M_{j-1}\cap U_j$ is a union of intersections from the good cover. Repeating this process turns the cohomology computation into a finite sequence of exact-sequence calculations. [/explanation] The calculation method is important because it explains why Mayer-Vietoris is useful in practice. The following remark distills that method into the local-to-global engine used throughout de Rham computations. [remark: Why Mayer-Vietoris Is the Main Engine] Homotopy invariance tells us the cohomology of contractible pieces, and the Poincare lemma supplies the local vanishing behind that fact. Mayer-Vietoris explains how the pieces assemble. In de Rham theory, many computations consist of choosing a cover, simplifying the pieces by homotopy, and reading the remaining global classes from exactness. [/remark] The conceptual message of this chapter is that cohomology is local-to-global data. Differential forms glue as a sheaf, but closed forms modulo exact forms remember the obstructions created by overlap geometry. The Mayer-Vietoris sequence packages those obstructions into an exact algebraic object that can be used directly in computations. Mayer-Vietoris turns local-to-global patching into an exact sequence, making it possible to compute de Rham cohomology from simpler pieces. With that machinery in hand, we can move from abstract structure to explicit calculations on the standard spaces that test the theory. # 9. Computations of de Rham Cohomology This chapter turns the formal machinery of de Rham cohomology into explicit calculations. The preceding chapters supplied the tools: homotopy invariance, the Mayer--Vietoris sequence, orientations, integration, and products of forms. We now apply them to the standard test spaces of topology: spheres, tori, compact surfaces, and real projective spaces. The guiding theme is that cohomology records global obstructions to writing closed forms as exact forms, and the computations below make those obstructions visible. ## Spheres and Mayer--Vietoris Induction How can the cohomology of a sphere be computed without writing down all closed forms on it? A naive attempt would try to classify every closed $k$-form on $S^n$ directly, but this quickly becomes unmanageable because the condition $d\alpha=0$ is a system of differential equations depending on coordinates and transition maps. The efficient method is to cut the sphere into two contractible open pieces whose intersection deformation retracts onto a smaller sphere. This converts the computation of $S^n$ into the computation of $S^{n-1}$ through the Mayer--Vietoris long exact sequence. [quotetheorem:3590] [citeproof:3590] The calculation says that a connected sphere has no intermediate de Rham cohomology. The hypotheses and endpoints matter: $S^0$ is not connected, so degree $0$ has two independent constant classes, while the induction step for $n\ge2$ relies on the intersection of the two punctured hemispheres retracting onto $S^{n-1}$. The theorem does not say that every closed form on a sphere is zero; it says that in intermediate degrees every closed form is exact, and in top degree closed forms are classified up to exact forms by their total integral. This is the model computation for later uses of Mayer--Vietoris, where a difficult global space is replaced by simpler pieces and their overlap. [example: The Circle] Let $S^1=\mathbb R/2\pi\mathbb Z$, and let $t$ denote the angular coordinate modulo $2\pi$. The form $dt$ is invariant under the translations $t\mapsto t+2\pi m$, so it descends from $\mathbb R$ to a global $1$-form on $S^1$, and it is closed because \begin{align*} d(dt)=0. \end{align*} We show that $[dt]$ is the nonzero degree-one generator. Suppose, for contradiction, that $dt=df$ for some smooth function $f:S^1\to\mathbb R$. Pulling back along the quotient parametrization $\gamma:[0,2\pi]\to S^1$, $\gamma(t)=[t]$, gives \begin{align*} \gamma^*(dt)&=dt,\\ \gamma^*(df)&=d(f\circ\gamma). \end{align*} Thus $dt=d(f\circ\gamma)$ on $[0,2\pi]$. Integrating both sides, \begin{align*} \int_0^{2\pi} dt &=\int_0^{2\pi} d(f\circ\gamma)\\ 2\pi &=(f\circ\gamma)(2\pi)-(f\circ\gamma)(0). \end{align*} Since $\gamma(0)=\gamma(2\pi)$ in $S^1$, the function $f\circ\gamma$ has the same endpoint value: \begin{align*} (f\circ\gamma)(2\pi)-(f\circ\gamma)(0)=0. \end{align*} This gives $2\pi=0$, a contradiction, so $dt$ is closed but not exact and $[dt]\ne 0$. By *De Rham Cohomology of Spheres*, $H^1_{\mathrm{dR}}(S^1)\cong\mathbb R$, so this nonzero class spans $H^1_{\mathrm{dR}}(S^1)$. [/example] For $S^n$ with $n\ge2$, a volume form represents the top class. The precise representative depends on a choice of Riemannian metric and orientation, but the cohomology class is determined up to a nonzero scalar by its total integral. [remark: Reduced Shape of the Answer] The sphere computation is often the first place where cohomology behaves like a dimension counter. There is one class for each connected component in degree $0$, one top-degree class for an orientable connected compact $n$-manifold, and no middle classes for $S^n$. [/remark] ## Tori and the Kunneth Formula What happens when a manifold is built as a product? A first guess might be that closed forms on $M\times N$ are just sums of pullbacks from the two factors, but this misses mixed forms such as $dt_1\wedge dt_2$ on $S^1\times S^1$. The torus $T^n=(S^1)^n$ has many closed forms coming from its circle factors, and the product structure explains all of them through both pullback and wedge product. The relevant mechanism is the Kunneth formula for de Rham cohomology. [quotetheorem:3933] [citeproof:3933] The theorem makes products computable: once bases for the factors are known, wedge products of their pullbacks give bases for the product. The coefficient field is important in this form of the statement: over $\mathbb R$, the algebraic Kunneth theorem has no torsion correction, while integral cohomology can contain additional Tor terms. The product hypothesis is also essential; a fibre bundle may be locally a product without having the cohomology of a product, as the Hopf fibration below will show. In this chapter the formula is used only for smooth manifolds whose de Rham groups are finite-dimensional, which covers the compact examples under discussion and avoids analytic issues from infinite-dimensional direct products. [quotetheorem:3934] [citeproof:3934] The theorem packages the torus as the cleanest product example. The degree-one classes come from the circle coordinates, and all higher classes are obtained by wedging distinct degree-one generators. This is a ring statement, not just a list of Betti numbers: the product remembers orientation signs and the top-degree generator. The normalization of generators matters in examples, because choosing $dt_j$ rather than a rescaled form fixes the numerical periods around the standard coordinate circles. [example: Betti Numbers of the Three-Torus] Let $T^3=S^1\times S^1\times S^1$ with angular coordinates $t_1,t_2,t_3$. By *Cohomology Ring of the Torus*, \begin{align*} H^*_{\mathrm{dR}}(T^3)\cong \Lambda^*(\mathbb R^3), \end{align*} with degree-one generators represented by $dt_1,dt_2,dt_3$. Thus the degree $k$ basis elements are indexed by $k$-element subsets of $\{1,2,3\}$. In degree $0$, the only $0$-fold wedge is the unit class $1$, so \begin{align*} \dim H^0_{\mathrm{dR}}(T^3)=1. \end{align*} In degree $1$, the $1$-element subsets are $\{1\},\{2\},\{3\}$, giving the basis \begin{align*} [dt_1],\quad [dt_2],\quad [dt_3], \end{align*} so \begin{align*} \dim H^1_{\mathrm{dR}}(T^3)=3. \end{align*} In degree $2$, the $2$-element subsets are $\{1,2\},\{1,3\},\{2,3\}$, giving the basis \begin{align*} [dt_1\wedge dt_2],\quad [dt_1\wedge dt_3],\quad [dt_2\wedge dt_3], \end{align*} so \begin{align*} \dim H^2_{\mathrm{dR}}(T^3)=3. \end{align*} In degree $3$, the only $3$-element subset is $\{1,2,3\}$, giving the basis \begin{align*} [dt_1\wedge dt_2\wedge dt_3], \end{align*} so \begin{align*} \dim H^3_{\mathrm{dR}}(T^3)=1. \end{align*} Therefore \begin{align*} (b_0,b_1,b_2,b_3)=(1,3,3,1). \end{align*} The three middle degree-two classes record the three independent coordinate two-tori inside $T^3$, while the single degree-three class is the top product class. [/example] The exterior algebra description also remembers the ring structure, not only the dimensions. For instance, on $T^2$ the product of the two degree-one generators is the top-degree orientation class. [remark: Forms Detecting Cycles] On $T^n$, the class $[dt_i]$ detects winding around the $i$th circle factor. Wedge products record simultaneous winding in several independent directions, which is why the indexing set for $H^k(T^n)$ consists of $k$-element subsets of $\{1,\dots,n\}$. [/remark] ## Compact Oriented Surfaces How does genus appear in de Rham cohomology? Counting closed $1$-forms directly is again the wrong starting point: a handle contributes global periods around loops, and these periods are not visible in any single coordinate chart. A compact connected oriented surface $\Sigma_g$ of genus $g$ has one $0$-dimensional class because it is connected, one $2$-dimensional class because it is compact and oriented, and $2g$ independent degree-one classes corresponding to the standard handle cycles. [quotetheorem:3935] [citeproof:3935] The middle cohomology records the handles. Each handle contributes two independent degree-one classes, corresponding to the two basic loops on that handle. The assumptions cannot simply be removed: if the surface is disconnected, $H^0$ has one copy of $\mathbb R$ for each connected component; if the surface is noncompact, top compact-orientation arguments no longer give $H^2\cong\mathbb R$; and if the surface is nonorientable, there is no global orientation class in top degree. For example, the punctured plane is a noncompact connected surface with vanishing top de Rham cohomology, and $\mathbb RP^2$ is compact but nonorientable with $H^2_{\mathrm{dR}}(\mathbb RP^2)=0$. The theorem therefore classifies only the compact connected oriented case, and it prepares the later interpretation of these dimensions through Poincare duality and singular cohomology. [definition: Euler Characteristic] For a smooth manifold $M$ with finite-dimensional de Rham cohomology groups and only finitely many nonzero cohomology groups, the Euler characteristic is \begin{align*} \chi(M) := \sum_{k\ge0} (-1)^k \dim H^k_{\mathrm{dR}}(M). \end{align*} [/definition] For a compact oriented surface, the cohomology table gives \begin{align*} \chi(\Sigma_g)=1-2g+1=2-2g. \end{align*} This agrees with the cell decomposition count $1-2g+1$ and shows how the cohomological definition recovers the usual topological invariant. [example: Low-Genus Surfaces] Consider the sphere $S^2=\Sigma_0$ and the torus $T^2=\Sigma_1$. By *De Rham Cohomology of a Compact Oriented Surface*, a compact connected oriented surface $\Sigma_g$ has \begin{align*} \dim H^0_{\mathrm{dR}}(\Sigma_g)&=1,\\ \dim H^1_{\mathrm{dR}}(\Sigma_g)&=2g,\\ \dim H^2_{\mathrm{dR}}(\Sigma_g)&=1. \end{align*} For the sphere, $g=0$, so \begin{align*} \dim H^0_{\mathrm{dR}}(S^2)&=1,\\ \dim H^1_{\mathrm{dR}}(S^2)&=2\cdot 0=0,\\ \dim H^2_{\mathrm{dR}}(S^2)&=1. \end{align*} Hence the Betti numbers of $S^2$ are \begin{align*} (b_0,b_1,b_2)=(1,0,1). \end{align*} For the torus, $g=1$, so \begin{align*} \dim H^0_{\mathrm{dR}}(T^2)&=1,\\ \dim H^1_{\mathrm{dR}}(T^2)&=2\cdot 1=2,\\ \dim H^2_{\mathrm{dR}}(T^2)&=1. \end{align*} Thus the Betti numbers of $T^2$ are \begin{align*} (b_0,b_1,b_2)=(1,2,1). \end{align*} The difference is concentrated in degree $1$: the sphere has no handle cycles, while the torus has two independent degree-one classes, represented by the angular forms from its two circle factors. [/example] The Euler characteristic is not merely a summary of the dimensions; in dimension two it distinguishes the genus of a compact connected oriented surface. Later, de Rham's theorem will identify this real cohomological invariant with the corresponding singular cohomology invariant. ## Real Projective Space What cohomology remains after identifying antipodal points on a sphere? A naive quotient argument might expect the cohomology of $\mathbb RP^n$ to look like the cohomology of $S^n$ with the antipodal points merely renamed, but forms must descend through the quotient, so they must be invariant under the deck transformation. Real projective space $\mathbb RP^n$ is the quotient of $S^n$ by the free action $x\mapsto -x$. Over real coefficients, most of its cohomology vanishes; the only possible top-degree class depends on orientability. [definition: Real Projective Space] For $n\ge0$, real projective space $\mathbb RP^n$ is the set of lines through the origin in $\mathbb R^{n+1}$, equipped with its standard smooth structure. [/definition] The quotient map $q:S^n\to\mathbb RP^n$ identifies antipodal points. It is a two-sheeted covering map, and the antipodal map controls which forms on $S^n$ descend to projective space. The computation now turns the quotient description into cohomology. Since projective space is covered by a sphere with a finite deck group, the theorem determines which real de Rham classes survive the antipodal identification. [quotetheorem:3936] [citeproof:3936] The top-degree statement is the cohomological form of the orientability criterion: $\mathbb RP^n$ is orientable exactly for odd $n$. The finite-cover hypothesis in the proof is doing real work: averaging over the two deck transformations divides by $2$, so the argument belongs naturally to real coefficients, or more generally to characteristic zero coefficients. The theorem also does not compute integral cohomology; torsion classes in intermediate degrees disappear after tensoring with $\mathbb R$, which is why the real answer is smaller than the full topological answer. This limitation points toward singular cohomology with other coefficient rings, while the orientability conclusion connects the top de Rham group with integration of top-degree forms. [example: Non-Orientability of the Projective Plane] The projective plane $\mathbb RP^2$ is the quotient of $S^2$ by the antipodal identification $x\sim -x$, so it is a compact connected smooth surface. We compute its top de Rham cohomology using *De Rham Cohomology of Real Projective Space*. Substituting $n=2$ into the theorem gives \begin{align*} H^2_{\mathrm{dR}}(\mathbb RP^2) &\cong \begin{cases} \mathbb R, & 2 \text{ odd},\\ 0, & 2 \text{ even}. \end{cases} \end{align*} Since $2$ is even, the second case applies, and therefore \begin{align*} H^2_{\mathrm{dR}}(\mathbb RP^2)=0. \end{align*} Now suppose $\mathbb RP^2$ were orientable. Because it is compact and connected, *De Rham Cohomology of a Compact Oriented Surface* would apply with $\mathbb RP^2=\Sigma_g$ for some genus $g$, and would give \begin{align*} H^2_{\mathrm{dR}}(\mathbb RP^2)\cong \mathbb R. \end{align*} This contradicts the computation \begin{align*} H^2_{\mathrm{dR}}(\mathbb RP^2)=0. \end{align*} Hence $\mathbb RP^2$ is not orientable. In de Rham terms, the obstruction is exactly the absence of a nonzero top-degree cohomology class represented by a global orientation form. [/example] This example also shows a limitation of de Rham theory. Because the coefficients are real vector spaces, torsion phenomena disappear from the calculation. [remark: Why Real Coefficients Miss Torsion] With integer coefficients, real projective spaces have additional torsion information in intermediate degrees. De Rham cohomology uses real vector spaces, so torsion is invisible. This is why the real-coefficient answer for $\mathbb RP^n$ is much smaller than the full integral cohomology calculation. [/remark] ## Compact Lie Groups and Fibre Bundles How far do these computations extend beyond spaces built from cells or products? Compact connected Lie groups and smooth fibre bundles give the next source of computable examples. The full theory requires spectral sequences or characteristic classes, but several structural features are already visible from differential forms. [definition: Smooth Fibre Bundle] A smooth fibre bundle with fibre $F$ is a smooth surjective map $\pi:E\to B$ such that every point $b\in B$ has an open neighbourhood $U\subset B$ and a diffeomorphism \begin{align*} \pi^{-1}(U) \cong U\times F \end{align*} commuting with projection to $U$. [/definition] The point of the definition is local product structure. Cohomology, however, is global, so twisting of the local products can change the answer. [example: The Hopf Fibration as a Warning] The Hopf fibration is a smooth fibre bundle \begin{align*} S^1 \longrightarrow S^3 \longrightarrow S^2. \end{align*} We show that its total space $S^3$ does not have the same de Rham cohomology as the product $S^1\times S^2$. By the *Kunneth Formula for De Rham Cohomology*, \begin{align*} H^1_{\mathrm{dR}}(S^1\times S^2) &\cong \left(H^1_{\mathrm{dR}}(S^1)\otimes H^0_{\mathrm{dR}}(S^2)\right) \oplus \left(H^0_{\mathrm{dR}}(S^1)\otimes H^1_{\mathrm{dR}}(S^2)\right). \end{align*} Using the sphere computation, \begin{align*} H^1_{\mathrm{dR}}(S^1)&\cong \mathbb R,& H^0_{\mathrm{dR}}(S^2)&\cong \mathbb R,& H^0_{\mathrm{dR}}(S^1)&\cong \mathbb R,& H^1_{\mathrm{dR}}(S^2)&=0. \end{align*} Therefore \begin{align*} H^1_{\mathrm{dR}}(S^1\times S^2) &\cong (\mathbb R\otimes \mathbb R)\oplus(\mathbb R\otimes 0)\\ &\cong \mathbb R\oplus 0\\ &\cong \mathbb R. \end{align*} Similarly, \begin{align*} H^2_{\mathrm{dR}}(S^1\times S^2) &\cong \left(H^2_{\mathrm{dR}}(S^1)\otimes H^0_{\mathrm{dR}}(S^2)\right) \oplus \left(H^1_{\mathrm{dR}}(S^1)\otimes H^1_{\mathrm{dR}}(S^2)\right) \oplus \left(H^0_{\mathrm{dR}}(S^1)\otimes H^2_{\mathrm{dR}}(S^2)\right)\\ &\cong (0\otimes \mathbb R)\oplus(\mathbb R\otimes 0)\oplus(\mathbb R\otimes \mathbb R)\\ &\cong 0\oplus 0\oplus \mathbb R\\ &\cong \mathbb R. \end{align*} For the total space of the Hopf fibration, the sphere computation gives \begin{align*} H^1_{\mathrm{dR}}(S^3)=0,\qquad H^2_{\mathrm{dR}}(S^3)=0, \end{align*} because both $1$ and $2$ are intermediate degrees between $0$ and $3$. Hence \begin{align*} H^1_{\mathrm{dR}}(S^3)&\not\cong H^1_{\mathrm{dR}}(S^1\times S^2),\\ H^2_{\mathrm{dR}}(S^3)&\not\cong H^2_{\mathrm{dR}}(S^1\times S^2). \end{align*} Thus the local product condition in the definition of a fibre bundle does not force the total space to have the cohomology of the global product $S^1\times S^2$. [/example] The computations so far suggest that cohomology can see global structure invisible from local product charts. Compact Lie groups provide a rich class where this question becomes systematic: multiplication imposes strong algebraic constraints, while examples such as tori and $SU(2)$ show that the resulting cohomology need not look like that of a product of circles in any naive way. The next quoted result previews the general shape of the answer. [quotetheorem:3937] [citeproof:3937] The abelian case is already contained in the torus computation, since every compact connected abelian Lie group is a torus. Nonabelian compact Lie groups require new tools that organise how cohomology behaves in fibrations, so the theorem is best read here as a structural landmark rather than a computational recipe. [example: The Group SU(2)] An element of $SU(2)$ is a complex matrix \begin{align*} A= \begin{pmatrix} a & b\\ c & d \end{pmatrix} \end{align*} satisfying $A^*A=I$ and $\det A=1$. The first column has length $1$, so \begin{align*} |a|^2+|c|^2=1. \end{align*} The second column must have length $1$ and be orthogonal to $(a,c)$, so it has the form \begin{align*} (b,d)=\lambda(-\overline c,\overline a) \end{align*} for some $\lambda\in\mathbb C$ with $|\lambda|=1$. Then \begin{align*} \det A &= a(\lambda\overline a)-(\lambda(-\overline c))c\\ &= \lambda |a|^2+\lambda |c|^2\\ &= \lambda(|a|^2+|c|^2)\\ &= \lambda. \end{align*} Since $\det A=1$, we get $\lambda=1$, and hence every element of $SU(2)$ has the form \begin{align*} A= \begin{pmatrix} a & -\overline c\\ c & \overline a \end{pmatrix} \quad\text{with}\quad |a|^2+|c|^2=1. \end{align*} Thus the map \begin{align*} SU(2)&\longrightarrow S^3\subset\mathbb C^2,\\ \begin{pmatrix} a & -\overline c\\ c & \overline a \end{pmatrix} &\longmapsto (a,c) \end{align*} is bijective, with inverse \begin{align*} (a,c)\longmapsto \begin{pmatrix} a & -\overline c\\ c & \overline a \end{pmatrix}. \end{align*} Both maps are smooth because their coordinate functions are built from the real and imaginary parts of $a$ and $c$ using only conjugation and projection coordinates. Therefore $SU(2)$ is diffeomorphic to $S^3$. By diffeomorphism invariance of de Rham cohomology and *De Rham Cohomology of Spheres*, \begin{align*} H^k_{\mathrm{dR}}(SU(2)) &\cong H^k_{\mathrm{dR}}(S^3)\\ &\cong \begin{cases} \mathbb R, & k=0,3,\\ 0, & 0<k<3. \end{cases} \end{align*} Since $SU(2)$ is $3$-dimensional, $H^k_{\mathrm{dR}}(SU(2))=0$ also for $k>3$. Hence $SU(2)$ has exactly one degree-zero class, one top-degree class, and no intermediate de Rham cohomology. [/example] These examples mark the transition from computation by cutting and products to computation by structure. Mayer--Vietoris, Kunneth, and orientability already determine many standard spaces; fibre bundles explain why further invariants are needed for more complicated manifolds. The computations show how de Rham cohomology reflects the topology of spheres, tori, projective spaces, and related examples. The final step is to identify this analytic invariant with the more classical singular cohomology built from simplices and chains. # 10. de Rham's Theorem The preceding chapters built differential forms, integration on manifolds, Stokes theorem, and de Rham cohomology. This chapter connects that analytic theory to singular cohomology with real coefficients. The main construction sends a closed form to the cochain obtained by integrating it over singular cycles, and the main theorem says that this construction loses no information. The point is not only computational: it says that a theory built from smooth calculus recovers a purely topological invariant. ## Singular Cohomology with Real Coefficients Which topological theory has the right shape to receive integrals of differential forms? Since forms integrate over parametrised simplices and their boundaries, we begin with singular chains. [definition: Singular Simplex] Let $M$ be a topological space. A singular $k$-simplex in $M$ is a continuous map $\sigma:\Delta^k\to M$, where \begin{align*} \Delta^k=\{(t_0,\dots,t_k)\in\mathbb R^{k+1}:t_i\ge0,\ \sum_{i=0}^k t_i=1\}. \end{align*} [/definition] A singular simplex need not be embedded. It is a flexible parametrised piece of the space, which is why singular homology works for arbitrary continuous maps. [definition: Singular Chain Complex] Let $C_k(M;\mathbb R)$ be the real vector space freely generated by singular $k$-simplices in $M$. The boundary operator is the linear map \begin{align*} \partial_k:C_k(M;\mathbb R)\to C_{k-1}(M;\mathbb R) \end{align*} defined on a singular simplex $\sigma:\Delta^k\to M$ by \begin{align*} \partial_k\sigma=\sum_{i=0}^k(-1)^i\sigma\circ\iota_i, \end{align*} where $\iota_i:\Delta^{k-1}\to\Delta^k$ is the inclusion of the face $t_i=0$, and extended linearly to all chains. [/definition] The alternating signs make every codimension-two face appear twice with opposite signs, so $\partial^2=0$. Chains record how simplices sit inside the space, but integration of forms will produce numbers from those chains. To compare forms with topology, we therefore pass to cochains: linear functionals on chains whose coboundary operation is dual to the boundary map. [definition: Singular Cohomology] Let $C^k(M;\mathbb R)=\operatorname{Hom}_{\mathbb R}(C_k(M;\mathbb R),\mathbb R)$. For $\varphi\in C^k(M;\mathbb R)$ define $\delta\varphi\in C^{k+1}(M;\mathbb R)$ by $(\delta\varphi)(c)=\varphi(\partial c)$. The $k$-th singular cohomology group is \begin{align*} H^k(M;\mathbb R)=\ker(\delta:C^k\to C^{k+1})/\operatorname{im}(\delta:C^{k-1}\to C^k). \end{align*} [/definition] Cocycles are cochains that vanish on boundaries, while coboundaries are cochains already explained by one lower degree. Since a cochain is a linear functional on chains, it is natural to ask how much of cohomology is simply the linear dual of homology. With real coefficients, the answer has the clean vector-space form needed here: a cohomology class can be evaluated on homology classes, and this pairing identifies real cohomology with the dual vector space of real homology in the finite-dimensional situations used in these notes. [remark: Real Coefficients Avoid Torsion Corrections] The statement above is the only part of the universal-coefficient philosophy used in this chapter. Over $\mathbb R$, the chain and cochain groups are vector spaces, so the period pairing between cohomology and homology behaves like ordinary linear duality. Over $\mathbb Z$, torsion in homology contributes extra information, so cohomology is not merely the dual of homology. De Rham theory works over real vector spaces, which is why periods of differential forms naturally land in this simpler real-coefficient setting. [/remark] The simplest singular cohomology computation is the one-point space. It fixes the convention for the coboundary and gives a baseline for later comparisons with de Rham cohomology. [example: Cohomology of a Point] Let $M=\{p\}$, and let $e_k:\Delta^k\to M$ denote the unique singular $k$-simplex, the constant map with value $p$. Since $C_k(M;\mathbb R)$ is freely generated by $e_k$, we have $C_k(M;\mathbb R)\cong\mathbb R e_k$ for every $k\ge0$. The boundary is \begin{align*} \partial e_k &=\sum_{i=0}^k(-1)^i e_{k-1}\\ &=\left(\sum_{i=0}^k(-1)^i\right)e_{k-1}. \end{align*} The alternating sum is $0$ when $k$ is odd and $1$ when $k$ is even, so \begin{align*} \partial e_k= \begin{cases} 0,& k\text{ odd},\\ e_{k-1},& k\text{ even}. \end{cases} \end{align*} Now identify $C^k(M;\mathbb R)=\operatorname{Hom}_{\mathbb R}(C_k(M;\mathbb R),\mathbb R)$ with $\mathbb R$ by sending a cochain $\varphi$ to $\varphi(e_k)$. Under this identification, \begin{align*} (\delta\varphi)(e_{k+1}) &=\varphi(\partial e_{k+1})\\ &=\left(\sum_{i=0}^{k+1}(-1)^i\right)\varphi(e_k). \end{align*} Thus $\delta:C^k\to C^{k+1}$ is the zero map when $k$ is even and the identity map when $k$ is odd: \begin{align*} 0\longrightarrow \mathbb R \xrightarrow{0}\mathbb R \xrightarrow{\operatorname{id}}\mathbb R \xrightarrow{0}\mathbb R \xrightarrow{\operatorname{id}}\mathbb R \xrightarrow{0}\cdots . \end{align*} Therefore \begin{align*} H^0(M;\mathbb R)&=\ker(\delta:C^0\to C^1)\cong\mathbb R,\\ H^{2m+1}(M;\mathbb R)&=\ker(\operatorname{id})/\operatorname{im}(0)=0,\\ H^{2m}(M;\mathbb R)&=\ker(0)/\operatorname{im}(\operatorname{id})=\mathbb R/\mathbb R=0 \end{align*} for every $m\ge0$ in the positive-degree cases. Hence the point has one degree-zero cohomology class, represented by the cochain taking $e_0$ to $1$, and no positive-degree real cohomology. [/example] ## The de Rham Homomorphism How does a differential form become a singular cohomology class? The answer is to evaluate the form on each smooth singular simplex by integration. [definition: Integration Cochain] Let $M$ be a smooth manifold, and let $C^k_{\mathrm{sm}}(M;\mathbb R)$ be the real vector space of linear maps from smooth singular $k$-chains in $M$ to $\mathbb R$. The integration construction is the map \begin{align*} I:\Omega^k(M)\to C^k_{\mathrm{sm}}(M;\mathbb R) \end{align*} defined by \begin{align*} I(\omega)(\sigma)=\int_{\Delta^k}\sigma^*\omega \end{align*} for every $\omega\in\Omega^k(M)$ and every smooth singular simplex $\sigma:\Delta^k\to M$, and extended linearly to smooth singular chains. [/definition] Smooth singular cohomology agrees with ordinary singular cohomology through the standard smoothing comparison for singular chains. This lets the course work with smooth simplices when integrating forms while still computing the usual singular cohomology groups. The key compatibility is Stokes theorem on the standard simplex. The possible obstruction is that integration might not respect the quotient relation defining cohomology: replacing a form by a cohomologous one should replace the resulting cochain by a cohomologous cochain. This requires exterior differentiation on forms to match the singular coboundary after integration over every simplex. [quotetheorem:3938] [citeproof:3938] The smoothness requirement is doing real work here: pullbacks of differential forms and Stokes theorem are available for smooth singular simplices, not for arbitrary continuous ones. The alternating signs in the singular boundary agree with the boundary orientation convention on the standard simplex, so the algebraic coboundary and exterior derivative line up exactly. This compatibility does not yet say that every singular cohomology class comes from a form, or that periods detect exactness; it only proves that integration descends to a cochain map. That cochain map is the bridge from de Rham cohomology to singular cohomology. We need a named comparison map between the two cohomology theories. Integration sends closed forms to cocycles and exact forms to coboundaries, so it descends from forms and cochains to the corresponding quotient groups. [definition: de Rham Homomorphism] Let $M$ be a smooth manifold. The de Rham homomorphism is \begin{align*} \mathcal I_k:H^k_{dR}(M)\to H^k(M;\mathbb R),\qquad [\omega]\mapsto[I(\omega)]. \end{align*} [/definition] If $d\omega=0$, then $I(\omega)$ is a cocycle. If $\omega=d\eta$, then $I(\omega)=\delta I(\eta)$, so the map is well-defined on cohomology. [example: Generator of the Circle] Let $\alpha=\frac{1}{2\pi}d\theta$ on $S^1$, and let $\gamma:[0,1]\to S^1$ be the positively oriented singular $1$-simplex \begin{align*} \gamma(t)=e^{2\pi i t}. \end{align*} The endpoint values agree, $\gamma(0)=1=\gamma(1)$, so \begin{align*} \partial\gamma=\gamma(1)-\gamma(0)=0, \end{align*} and $\gamma$ is a cycle representing the positive fundamental class of $S^1$. Along this parametrisation, the angular coordinate satisfies $\theta(\gamma(t))=2\pi t$ modulo $2\pi$, so on the interval $[0,1]$ its differential pulls back as \begin{align*} \gamma^*(d\theta) &=d(\theta\circ\gamma)\\ &=d(2\pi t)\\ &=2\pi\,dt. \end{align*} Therefore \begin{align*} \gamma^*\alpha &=\gamma^*\left(\frac{1}{2\pi}d\theta\right)\\ &=\frac{1}{2\pi}\gamma^*(d\theta)\\ &=\frac{1}{2\pi}(2\pi\,dt)\\ &=dt. \end{align*} By the definition of the integration cochain, \begin{align*} I(\alpha)(\gamma) &=\int_{[0,1]}\gamma^*\alpha\\ &=\int_0^1 dt\\ &=\left[t\right]_{0}^{1}\\ &=1. \end{align*} Thus the de Rham class $[\alpha]$ maps to a singular cohomology class whose evaluation on the positively oriented fundamental cycle is $1$, so it is the standard generator of $H^1(S^1;\mathbb R)$. [/example] ## The Isomorphism Theorem Could there be closed forms whose periods all vanish but which are not exact? Could there be singular cohomology classes not represented by forms? The theorem says that both possibilities are excluded. [quotetheorem:3596] [citeproof:3596] Conceptually, the result says that smooth differential data and real singular cohomology carry the same global information. The hypotheses still matter: the source theory uses smooth forms on a smooth manifold, while the target is singular cohomology with real coefficients. The isomorphism does not recover integral cohomology or torsion, since real vector spaces cannot see finite-order classes. Although standard constructions of the comparison map use auxiliary choices such as covers and partitions of unity, the resulting cohomology isomorphism is the canonical one induced by integration. ## Mayer-Vietoris and the Five Lemma Why does local agreement imply global agreement? The Mayer-Vietoris sequence records how cohomology classes on two open sets match on their overlap. If a manifold is covered by $U$ and $V$, then a class on the whole manifold restricts to compatible classes on the pieces, and the obstruction to gluing lives on $U\cap V$. This exact sequence is the mechanism that turns local computations into global ones while keeping track of those compatibility conditions. [quotetheorem:3589] [citeproof:3589] Openness is essential because partitions of unity are subordinate to open covers; for arbitrary closed covers, the restriction and gluing argument need not produce smooth global forms with the required support control. The sequence does not compute cohomology by itself; it relates unknown groups for $M$ to groups for $U$, $V$, and $U\cap V$ together with connecting maps. In the proof of de Rham theorem, this relation is used in parallel for de Rham and singular cohomology, and the integration map gives a morphism between the two long exact sequences. [quotetheorem:1938] [citeproof:1938] Exactness and commutativity are the essential hypotheses. Exactness is what lets information move from a kernel to an image, while commutativity ensures that moving horizontally and then vertically gives the same result as moving vertically and then horizontally. Knowing only that most vertical maps are isomorphisms would not control whether the middle map preserves the obstruction represented by the adjacent arrows. This lemma is therefore the algebraic engine that turns local de Rham isomorphisms into global ones through Mayer-Vietoris sequences. [example: Computing the Circle by Gluing] Cover $S^1$ by two connected open arcs $U,V$ such that $U\cap V=W_1\sqcup W_2$ is the disjoint union of two connected open arcs. Since each of $U$, $V$, $W_1$, and $W_2$ is diffeomorphic to an open interval, the *Poincare lemma* gives \begin{align*} H^0_{dR}(U)&\cong\mathbb R,& H^0_{dR}(V)&\cong\mathbb R,\\ H^0_{dR}(U\cap V)&\cong H^0_{dR}(W_1)\oplus H^0_{dR}(W_2)\cong\mathbb R^2, \end{align*} and \begin{align*} H^1_{dR}(U)=0,\qquad H^1_{dR}(V)=0. \end{align*} The relevant part of the Mayer-Vietoris sequence for de Rham cohomology is \begin{align*} H^0_{dR}(U)\oplus H^0_{dR}(V) \longrightarrow H^0_{dR}(U\cap V) \longrightarrow H^1_{dR}(S^1) \longrightarrow H^1_{dR}(U)\oplus H^1_{dR}(V). \end{align*} Under the identifications above, the first map sends a pair of constants $(a,b)\in\mathbb R\oplus\mathbb R$ to the difference of their restrictions on each component of $U\cap V$: \begin{align*} (a,b)\longmapsto (a-b,a-b)\in\mathbb R^2. \end{align*} Hence its image is \begin{align*} \operatorname{im}=\{(c,c):c\in\mathbb R\}\subset\mathbb R^2, \end{align*} the diagonal line. Since the next group is \begin{align*} H^1_{dR}(U)\oplus H^1_{dR}(V)=0\oplus0=0, \end{align*} exactness gives \begin{align*} H^1_{dR}(S^1) &\cong H^0_{dR}(U\cap V)/\operatorname{im}\bigl(H^0_{dR}(U)\oplus H^0_{dR}(V)\bigr)\\ &\cong \mathbb R^2/\{(c,c):c\in\mathbb R\}\\ &\cong \mathbb R, \end{align*} where the last isomorphism is induced by $(x,y)\mapsto x-y$. Thus the two overlap components differ by one independent constant, and that mismatch is exactly the one-dimensional class in $H^1_{dR}(S^1)$. [/example] ## Pairings, Winding Number, and Consequences What information does the isomorphism preserve? It preserves the pairing between cohomology and homology, so integrals of closed forms over cycles are topological invariants. [definition: de Rham Pairing] Let $M$ be a smooth manifold. The de Rham pairing is \begin{align*} H^k_{dR}(M)\times H_k(M;\mathbb R)\to\mathbb R,\qquad ([\omega],[c])\mapsto\int_c\omega. \end{align*} [/definition] Stokes theorem makes the pairing well-defined: exact forms integrate to zero over cycles, and closed forms integrate to zero over boundaries. Thus the value depends only on the de Rham cohomology class of the form and the homology class of the cycle. Under de Rham theorem, this pairing is the usual evaluation pairing between singular cohomology and singular homology with real coefficients. This is the precise sense in which periods of closed forms are topological invariants rather than artifacts of a chosen parametrisation. [example: Winding Number as a Pairing] Let $\widetilde\gamma:[0,1]\to\mathbb C\setminus\{0\}$ be a smooth parametrisation of the loop, with $\widetilde\gamma(0)=\widetilde\gamma(1)$. Since $\widetilde\gamma(t)\ne0$ for all $t$, choose smooth functions $r:[0,1]\to(0,\infty)$ and $\theta:[0,1]\to\mathbb R$ such that \begin{align*} \widetilde\gamma(t)=r(t)e^{i\theta(t)}. \end{align*} Equivalently, \begin{align*} x(t)=r(t)\cos\theta(t),\qquad y(t)=r(t)\sin\theta(t). \end{align*} We compute the pullback of \begin{align*} \omega=\frac{1}{2\pi}\frac{x\,dy-y\,dx}{x^2+y^2}. \end{align*} Along $\widetilde\gamma$, \begin{align*} dx &=d(r\cos\theta)\\ &=\cos\theta\,dr-r\sin\theta\,d\theta, \end{align*} and \begin{align*} dy &=d(r\sin\theta)\\ &=\sin\theta\,dr+r\cos\theta\,d\theta. \end{align*} Therefore \begin{align*} x\,dy-y\,dx &=(r\cos\theta)(\sin\theta\,dr+r\cos\theta\,d\theta) -(r\sin\theta)(\cos\theta\,dr-r\sin\theta\,d\theta)\\ &=r\cos\theta\sin\theta\,dr+r^2\cos^2\theta\,d\theta -r\sin\theta\cos\theta\,dr+r^2\sin^2\theta\,d\theta\\ &=r^2(\cos^2\theta+\sin^2\theta)\,d\theta\\ &=r^2\,d\theta. \end{align*} Also \begin{align*} x^2+y^2 &=r^2\cos^2\theta+r^2\sin^2\theta\\ &=r^2. \end{align*} Since $r(t)>0$, division by $r^2$ gives \begin{align*} \widetilde\gamma^*\omega &=\frac{1}{2\pi}\frac{r^2\,d\theta}{r^2}\\ &=\frac{1}{2\pi}d\theta. \end{align*} Hence \begin{align*} \int_\gamma\omega &=\int_{0}^{1}\widetilde\gamma^*\omega\\ &=\frac{1}{2\pi}\int_0^1 d\theta\\ &=\frac{1}{2\pi}\bigl(\theta(1)-\theta(0)\bigr). \end{align*} Because $\widetilde\gamma(1)=\widetilde\gamma(0)$, we have \begin{align*} r(1)e^{i\theta(1)}=r(0)e^{i\theta(0)}. \end{align*} Taking absolute values gives $r(1)=r(0)$, and then \begin{align*} e^{i(\theta(1)-\theta(0))}=1. \end{align*} Thus $\theta(1)-\theta(0)=2\pi N$ for a unique integer $N$, and \begin{align*} \int_\gamma\omega &=\frac{1}{2\pi}(2\pi N)\\ &=N. \end{align*} This integer $N$ is the winding number of $\gamma$ around $0$: the form $\omega$ records total change of argument divided by $2\pi$, and the resulting value depends only on the homology class of the loop in $\mathbb C\setminus\{0\}$. [/example] The de Rham theorem lets smooth forms detect real cohomology, and singular cohomology is invariant under homeomorphism. This raises an important consequence: although the definition of de Rham cohomology uses smooth forms and smooth maps, the resulting vector spaces should depend only on the underlying topological space. The theorem below states this topological invariance over real coefficients. [quotetheorem:3597] [citeproof:3597] This result is stronger than what is visible from the definition of differential forms alone, since a homeomorphism need not pull back smooth forms. The theorem gives an isomorphism of cohomology groups through singular cohomology, not a literal pullback map on forms for nonsmooth maps. It also remains a statement over real coefficients, so torsion phenomena in integral cohomology remain invisible to de Rham cohomology. [quotetheorem:3939] [citeproof:3939] Closedness is part of the statement because boundary changes the top-degree story: on a compact oriented manifold with nonempty boundary, ordinary top homology over $\mathbb R$ vanishes, while relative homology carries the fundamental class. Connectedness is also used to make the top group either one-dimensional or zero; for a disconnected closed orientable manifold, one copy of $\mathbb R$ appears for each connected component. The theorem is therefore best read as the closed connected case of the broader relationship between orientation, fundamental classes, and top-degree cohomology. [example: Volume Forms Detect the Top Class] Let $M$ be a connected closed oriented smooth $n$-manifold, and let $\Omega\in\Omega^n(M)$ satisfy $\int_M\Omega\ne0$. We show that $[\Omega]\ne0$ in $H^n_{dR}(M)$ and then that it spans the whole top-degree cohomology group. Suppose, for contradiction, that $[\Omega]=0$. By the definition of de Rham cohomology, this means that there is some $(n-1)$-form $\eta\in\Omega^{n-1}(M)$ such that \begin{align*} \Omega=d\eta. \end{align*} Integrating both sides over $M$ gives \begin{align*} \int_M\Omega &=\int_M d\eta\\ &=\int_{\partial M}\eta \end{align*} by *Stokes theorem*. Since $M$ is closed, it has empty boundary, so \begin{align*} \int_{\partial M}\eta &=\int_{\varnothing}\eta\\ &=0. \end{align*} Therefore \begin{align*} \int_M\Omega=0, \end{align*} contradicting the assumption $\int_M\Omega\ne0$. Hence $[\Omega]\ne0$. Because $M$ is connected, closed, and oriented, *Top Cohomology and Orientability* gives \begin{align*} H^n_{dR}(M)\cong\mathbb R. \end{align*} Thus $H^n_{dR}(M)$ is a one-dimensional real vector space. In a one-dimensional vector space, every nonzero vector spans: if $v\ne0$ and $w$ is any element, then after identifying the space with $\mathbb R$, we have $v=a$ and $w=b$ with $a\ne0$, so \begin{align*} w=b=\frac{b}{a}a=\frac{b}{a}v. \end{align*} Applying this to $v=[\Omega]$, the nonzero class $[\Omega]$ spans $H^n_{dR}(M)$. Thus a top-degree form with nonzero total integral detects the fundamental top cohomology class. [/example] De Rham theorem is the bridge from calculus to topology. It says that computations with forms, periods, Mayer-Vietoris sequences, and singular cycles are different languages for the same cohomological invariant. De Rham's theorem completes the passage from differential forms to topological invariants. With that bridge established, the remaining applications interpret cohomology classes in geometric terms and show how the theory organizes the structure of manifolds. # 11. Applications and Further Directions The final chapter gathers the structural consequences of de Rham cohomology developed earlier in the course. It assumes the basic language of differential forms, Stokes theorem, Mayer--Vietoris, and de Rham theorem, together with the interpretation of closed forms as cohomology classes. We now use those invariants to detect global geometry: duality between complementary degrees, numerical invariants such as the Euler characteristic, obstructions to geometric structures, and the analytic refinement supplied by Hodge theory. ## Poincare Duality The basic question is how cohomology in degree $k$ is related to cohomology in complementary degree $n-k$ on an oriented $n$-manifold. Integration of top-degree forms gives a natural way to pair these groups: wedge a closed $k$-form with a closed $(n-k)$-form and integrate the resulting $n$-form over the manifold. [definition: Poincare Pairing] Let $M$ be a closed oriented smooth $n$-manifold. The Poincare pairing is the bilinear map \begin{align*} H^k_{dR}(M) \times H^{n-k}_{dR}(M) &\longrightarrow \mathbb R,\\ ([\omega],[\eta]) &\longmapsto \int_M \omega \wedge \eta. \end{align*} [/definition] This is well-defined because changing either representative by an exact form changes the integrand by an exact $n$-form. Since $M$ has no boundary, Stokes theorem makes the integral of that exact form vanish. Well-definedness alone leaves a serious possibility: a nonzero cohomology class might pair to zero with every complementary class, making integration too weak to detect it. The duality question is whether closedness and orientation rule out this hidden radical in the pairing. [quotetheorem:3598] [citeproof:3598/differential-forms-ii-manifolds-and-cohomology] The theorem says that Betti numbers are symmetric: $\beta_k(M)=\beta_{n-k}(M)$ for closed oriented $n$-manifolds. The hypotheses are essential. If $M=[0,1]$, then $H^0_{dR}(M)\cong\mathbb R$ but $H^1_{dR}(M)=0$, so the closed-manifold pairing between degrees $0$ and $1$ cannot be non-degenerate. If $M$ is a closed non-orientable surface such as $\mathbb RP^2$, there is no real top-degree fundamental class giving integration against an orientation form, and the top de Rham cohomology vanishes. Non-degeneracy also does not give a canonical isomorphism $H^k_{dR}(M)\cong H^{n-k}_{dR}(M)$ by itself; it gives a canonical identification with the dual of the complementary group. This duality is the structural input behind the symmetry of Betti numbers, the middle-dimensional intersection pairing, and the compatibility with Hodge star later in the chapter. [example: Poincare Duality On A Surface] Let $\Sigma_g$ be a closed oriented surface of genus $g$. Its de Rham Betti numbers are \begin{align*} \beta_0(\Sigma_g) &= 1, & \beta_1(\Sigma_g) &= 2g, & \beta_2(\Sigma_g) &= 1. \end{align*} Choose an orientation form $\mu$ normalized by $\int_{\Sigma_g}\mu=1$. If $c\in H^0_{dR}(\Sigma_g)$ is represented by the constant function $c$ and $t[\mu]\in H^2_{dR}(\Sigma_g)$, then the degree $0$--degree $2$ Poincare pairing is \begin{align*} \langle c,t[\mu]\rangle &=\int_{\Sigma_g} c\,t\mu\\ &=ct\int_{\Sigma_g}\mu\\ &=ct. \end{align*} Thus the pairing between $H^0_{dR}(\Sigma_g)$ and $H^2_{dR}(\Sigma_g)$ is the ordinary product pairing on $\mathbb R\times\mathbb R$. For the middle degree, choose a standard symplectic basis \begin{align*} a_1,\ldots,a_g,b_1,\ldots,b_g \end{align*} of $H^1_{dR}(\Sigma_g)$ satisfying \begin{align*} \int_{\Sigma_g} a_i\wedge a_j&=0,\\ \int_{\Sigma_g} b_i\wedge b_j&=0,\\ \int_{\Sigma_g} a_i\wedge b_j&=\delta_{ij}. \end{align*} Since $a_i$ and $b_j$ are $1$-forms, graded commutativity gives \begin{align*} b_j\wedge a_i=-a_i\wedge b_j, \end{align*} so \begin{align*} \int_{\Sigma_g} b_j\wedge a_i &=-\int_{\Sigma_g} a_i\wedge b_j\\ &=-\delta_{ij}. \end{align*} Therefore, in the ordered basis $(a_1,\ldots,a_g,b_1,\ldots,b_g)$, the matrix of the middle pairing is \begin{align*} \begin{pmatrix} 0 & I_g \\ -I_g & 0 \end{pmatrix}. \end{align*} The example shows the Betti-number symmetry $\beta_0=\beta_2$ and also shows that middle-dimensional Poincare duality records the signed intersection structure, not just the dimensions of the cohomology groups. [/example] The surface case is the first place where duality contains more than dimension-counting information. It packages the intersection behaviour of cycles into a bilinear form on cohomology. [remark: Boundary Warning] Poincare duality in this form requires $M$ to be closed. For manifolds with boundary, the correct statement relates absolute and relative cohomology groups, with the boundary condition accounting for the Stokes term. [/remark] ## The Euler Characteristic Once cohomology has become computable, a natural next question is how much information survives after we compress all Betti numbers into a single integer. The Euler characteristic is the alternating sum of the dimensions of the de Rham cohomology groups, and it connects cohomology with vector fields and zeros. [definition: Euler Characteristic] Let $M$ be a smooth manifold whose de Rham cohomology groups are finite-dimensional. The Euler characteristic of $M$ is \begin{align*} \chi(M) := \sum_{k \ge 0} (-1)^k \beta_k(M), \qquad \beta_k(M):=\dim H^k_{dR}(M). \end{align*} [/definition] For a closed $n$-manifold the sum only runs from $0$ to $n$. By de Rham theorem, this definition agrees with the usual topological Euler characteristic computed from singular cohomology or from a finite CW-decomposition. The remaining issue is whether this alternating sum is merely a definition in de Rham language or whether it is forced by a broader homological principle. To use $\chi(M)$ as a genuine topological invariant, the alternating Betti-number formula must be stable under the standard comparison with singular cohomology. [quotetheorem:2266] [citeproof:2266/differential-forms-ii-manifolds-and-cohomology] The formula requires finite-dimensional cohomology groups; for arbitrary non-compact manifolds the alternating sum may be infinite or undefined. For closed manifolds this finiteness is guaranteed, and the formula becomes a topological invariant independent of the particular smooth structure used to compute de Rham cohomology. The Euler characteristic is much coarser than the full list of Betti numbers: for example, different spaces can have the same alternating sum while having different cohomology groups. Its usefulness comes from being computable in several independent ways, which is why it can connect cohomology to vector fields in Poincare--Hopf. [example: Euler Characteristic Of A Genus Surface] For the closed oriented genus $g$ surface $\Sigma_g$, the de Rham Betti numbers are \begin{align*} \beta_0(\Sigma_g)&=1,\\ \beta_1(\Sigma_g)&=2g,\\ \beta_2(\Sigma_g)&=1. \end{align*} Since $\Sigma_g$ is $2$-dimensional, its Euler characteristic is the alternating sum \begin{align*} \chi(\Sigma_g) &=\sum_{k=0}^{2}(-1)^k\beta_k(\Sigma_g)\\ &=(-1)^0\beta_0(\Sigma_g)+(-1)^1\beta_1(\Sigma_g)+(-1)^2\beta_2(\Sigma_g)\\ &=1\cdot 1+(-1)\cdot 2g+1\cdot 1\\ &=1-2g+1\\ &=2-2g. \end{align*} For $g=0$, this gives $\chi(S^2)=2-2\cdot 0=2$; for $g=1$, this gives $\chi(T^2)=2-2\cdot 1=0$. Replacing $g$ by $g+1$ changes the value from $2-2g$ to \begin{align*} 2-2(g+1)&=2-2g-2\\ &=(2-2g)-2, \end{align*} so each added handle decreases the Euler characteristic by $2$. [/example] The surface computation shows that $\chi$ changes when handles are added, but it does not yet explain why this number should constrain geometric objects on the manifold. A vector field with isolated zeros carries local index data, and the obstruction question is whether those local signed contributions can be adjusted independently or whether their total is forced by the topology. [quotetheorem:3940] [citeproof:3940] Compactness is used to make the global signed count finite and to prevent zeros from escaping to infinity; on a non-compact manifold such as $\mathbb R^2$, the constant vector field has no zeros while the ordinary compact-manifold conclusion is not the relevant statement. The isolated-zero hypothesis is also substantive, because a vector field may vanish along a curve or region, where local indices are not defined without perturbation or extra machinery. The theorem gives a necessary condition for a nowhere-zero vector field, not a complete classification of all vector fields. Its immediate use is that a nowhere-zero vector field has empty zero set, so the right-hand side must vanish. [example: The Hairy Ball Theorem] Take $M=S^2$. Its de Rham cohomology groups are \begin{align*} H^0_{dR}(S^2)&\cong \mathbb R,\\ H^1_{dR}(S^2)&=0,\\ H^2_{dR}(S^2)&\cong \mathbb R, \end{align*} so the Betti numbers are $\beta_0(S^2)=1$, $\beta_1(S^2)=0$, and $\beta_2(S^2)=1$. Therefore the Euler characteristic is \begin{align*} \chi(S^2) &=\sum_{k=0}^{2}(-1)^k\beta_k(S^2)\\ &=(-1)^0\beta_0(S^2)+(-1)^1\beta_1(S^2)+(-1)^2\beta_2(S^2)\\ &=1\cdot 1+(-1)\cdot 0+1\cdot 1\\ &=1+0+1\\ &=2. \end{align*} Now let $X$ be a smooth vector field on $S^2$ with isolated zeros. By the *Poincare--Hopf Theorem*, \begin{align*} \sum_{p\in Z(X)}\operatorname{ind}_p(X)=\chi(S^2)=2. \end{align*} If $X$ had no zeros, then $Z(X)=\varnothing$, and the left-hand side would be the empty sum: \begin{align*} \sum_{p\in \varnothing}\operatorname{ind}_p(X)=0. \end{align*} This would force $0=2$, which is impossible. Hence every smooth vector field on $S^2$ has at least one zero. In particular, $TS^2$ cannot be trivial. Indeed, if $TS^2$ were trivial as a rank-two bundle, there would be a global smooth frame $(e_1,e_2)$ for $TS^2$, and the first frame vector $e_1$ would be a smooth nowhere-zero vector field on $S^2$. The preceding argument rules this out, so the tangent bundle of the sphere is non-trivial. [/example] ## Cohomological Obstructions The general problem is to decide when a desired geometric object exists globally after it exists locally. Cohomology measures failures of local data to patch: exactness, orientability, and non-vanishing sections all become questions about whether certain cohomology classes vanish. [explanation: Closed Forms As Obstructions] A closed form represents no obstruction when it is exact, because it has a global primitive. If $\omega \in \Omega^k(M)$ is closed and $[\omega]\ne 0$ in $H^k_{dR}(M)$, then local primitives cannot be chosen so that they agree globally. For instance, the angular form on $S^1$ is locally $d\theta$ but represents the generator of $H^1_{dR}(S^1)$, so there is no globally defined real-valued angle function on the circle. The same pattern appears throughout the course. Mayer--Vietoris detects how local primitives disagree on overlaps; de Rham theorem identifies the obstruction with singular cohomology; integration over cycles gives concrete tests for non-vanishing. [/explanation] The following example turns this obstruction language into a concrete integration test. [example: Magnetic Monopole Form] On $\mathbb R^3\setminus\{0\}$, let $r=(x^2+y^2+z^2)^{1/2}$ and \begin{align*} \omega=\frac{x\,dy\wedge dz+y\,dz\wedge dx+z\,dx\wedge dy}{r^3}. \end{align*} We first verify that $\omega$ is closed. Since $d(dy)=d(dz)=d(dx)=0$, \begin{align*} d\omega &=d(xr^{-3})\wedge dy\wedge dz+d(yr^{-3})\wedge dz\wedge dx+d(zr^{-3})\wedge dx\wedge dy\\ &=\frac{\partial(xr^{-3})}{\partial x}\,dx\wedge dy\wedge dz +\frac{\partial(yr^{-3})}{\partial y}\,dy\wedge dz\wedge dx +\frac{\partial(zr^{-3})}{\partial z}\,dz\wedge dx\wedge dy\\ &=\left(\frac{\partial(xr^{-3})}{\partial x} +\frac{\partial(yr^{-3})}{\partial y} +\frac{\partial(zr^{-3})}{\partial z}\right)dx\wedge dy\wedge dz. \end{align*} Here \begin{align*} \frac{\partial(xr^{-3})}{\partial x} &=r^{-3}+x(-3r^{-4})\frac{\partial r}{\partial x}\\ &=r^{-3}-3x^2r^{-5}, \end{align*} and similarly \begin{align*} \frac{\partial(yr^{-3})}{\partial y}&=r^{-3}-3y^2r^{-5},\\ \frac{\partial(zr^{-3})}{\partial z}&=r^{-3}-3z^2r^{-5}. \end{align*} Therefore \begin{align*} d\omega &=\left(3r^{-3}-3(x^2+y^2+z^2)r^{-5}\right)dx\wedge dy\wedge dz\\ &=\left(3r^{-3}-3r^2r^{-5}\right)dx\wedge dy\wedge dz\\ &=0. \end{align*} On the unit sphere, use the spherical parametrization \begin{align*} x=\sin\theta\cos\phi,\qquad y=\sin\theta\sin\phi,\qquad z=\cos\theta, \end{align*} so $r=1$. Then \begin{align*} dx&=\cos\theta\cos\phi\,d\theta-\sin\theta\sin\phi\,d\phi,\\ dy&=\cos\theta\sin\phi\,d\theta+\sin\theta\cos\phi\,d\phi,\\ dz&=-\sin\theta\,d\theta. \end{align*} Hence \begin{align*} x\,dy\wedge dz &=\sin\theta\cos\phi\left(\sin^2\theta\cos\phi\,d\theta\wedge d\phi\right)\\ &=\sin^3\theta\cos^2\phi\,d\theta\wedge d\phi,\\ y\,dz\wedge dx &=\sin\theta\sin\phi\left(\sin^2\theta\sin\phi\,d\theta\wedge d\phi\right)\\ &=\sin^3\theta\sin^2\phi\,d\theta\wedge d\phi,\\ z\,dx\wedge dy &=\cos\theta\left(\sin\theta\cos\theta\,d\theta\wedge d\phi\right)\\ &=\sin\theta\cos^2\theta\,d\theta\wedge d\phi. \end{align*} Adding these three terms gives \begin{align*} \omega|_{S^2} &=\left(\sin^3\theta\cos^2\phi+\sin^3\theta\sin^2\phi+\sin\theta\cos^2\theta\right)d\theta\wedge d\phi\\ &=\sin\theta\left(\sin^2\theta(\cos^2\phi+\sin^2\phi)+\cos^2\theta\right)d\theta\wedge d\phi\\ &=\sin\theta\,d\theta\wedge d\phi, \end{align*} which is the usual oriented area form on the unit sphere. Therefore \begin{align*} \int_{S^2}\omega &=\int_0^{2\pi}\int_0^\pi \sin\theta\,d\theta\,d\phi\\ &=\int_0^{2\pi}\left[-\cos\theta\right]_{\theta=0}^{\theta=\pi}\,d\phi\\ &=\int_0^{2\pi}2\,d\phi\\ &=4\pi. \end{align*} If $\omega$ were exact on $\mathbb R^3\setminus\{0\}$, say $\omega=d\alpha$, then its restriction to $S^2$ would satisfy $\omega|_{S^2}=d(\alpha|_{S^2})$. Since $\partial S^2=\varnothing$, *Stokes theorem* would give \begin{align*} \int_{S^2}\omega=\int_{S^2}d(\alpha|_{S^2})=\int_{\partial S^2}\alpha|_{S^2}=0, \end{align*} contradicting $\int_{S^2}\omega=4\pi$. Thus $\omega$ is closed but not exact, and its non-zero de Rham class detects the missing origin. [/example] The missing-origin example detects a nonzero topological class by integrating over a sphere. The same idea becomes an orientability test when the top-degree cohomology of a closed manifold is interpreted through integration. [explanation: Orientability As A Cohomological Condition] For a connected smooth $n$-manifold, orientability can be phrased as the existence of a consistent choice of local generators for top-degree forms. On a closed oriented manifold, this produces a fundamental integration functional on $H^n_{dR}(M)$. If a compact connected $n$-manifold is orientable, then $H^n_{dR}(M)\cong \mathbb R$; if it is non-orientable, the top de Rham cohomology with real coefficients vanishes. This gives a cohomological test for orientability in the closed connected case. The test is not merely numerical: the non-zero top class is represented by a volume form compatible with the orientation. The word closed matters here; compact manifolds with boundary, such as $[0,1]$ or the closed disk, have vanishing top absolute de Rham cohomology even though they are orientable. [/explanation] Projective space is a standard family where the orientability test changes with dimension. [example: Projective Space And Orientation] View $\mathbb RP^n$ as the quotient of $S^n$ by the antipodal map \begin{align*} A:S^n&\longrightarrow S^n,\\ A(x)&=-x. \end{align*} The antipodal map is the restriction of the linear map $-\operatorname{id}_{\mathbb R^{n+1}}$ on $\mathbb R^{n+1}$. Its determinant is \begin{align*} \det(-I_{n+1}) &=(-1)^{n+1}\det(I_{n+1})\\ &=(-1)^{n+1}. \end{align*} Thus $A$ preserves the orientation of $S^n$ exactly when $(-1)^{n+1}=1$, which is exactly when $n+1$ is even, equivalently when $n$ is odd. Since $\mathbb RP^n=S^n/\{x\sim -x\}$, an orientation on $S^n$ descends to the quotient exactly when the deck transformation $A$ preserves orientation. Therefore $\mathbb RP^n$ is orientable exactly when $n$ is odd. For a closed connected $n$-manifold, the top de Rham cohomology is $\mathbb R$ in the orientable case and $0$ in the non-orientable case. Hence \begin{align*} H^n_{dR}(\mathbb RP^n) &\cong \mathbb R \qquad &&\text{if } n \text{ is odd},\\ H^n_{dR}(\mathbb RP^n) &=0 &&\text{if } n \text{ is even}. \end{align*} In particular, for $\mathbb RP^2$, the antipodal map on $S^2$ has orientation sign \begin{align*} (-1)^{2+1}=(-1)^3=-1, \end{align*} so it reverses orientation. Local coordinate volume forms can be chosen on small charts, but transporting such a choice around the identification $x\sim -x$ changes its sign, so the local choices cannot be patched into one global orientation form. [/example] ## Hodge Theory Preview The final question is whether each de Rham cohomology class has a preferred representative. Before choosing a metric there is no canonical gauge: if $\omega$ is closed, then every $\omega+d\alpha$ represents the same class, and this freedom can change the representative substantially. On a compact Riemannian manifold, analysis supplies a preferred representative by choosing the unique harmonic form in each class. This is the bridge from the topological theory of this course to elliptic partial differential equations and spectral geometry. [definition: Hodge Star] Let $(M,g)$ be an oriented Riemannian $n$-manifold. The Hodge star is the bundle map \begin{align*} *: \Lambda^kT^*M \longrightarrow \Lambda^{n-k}T^*M \end{align*} determined by \begin{align*} \alpha \wedge *\beta = (\alpha,\beta)_g\,dV_g \end{align*} for $k$-forms $\alpha,\beta$, where $(\cdot,\cdot)_g$ is the pointwise inner product induced by $g$ and $dV_g$ is the Riemannian volume form. [/definition] The Hodge star turns wedge products into inner products under integration. It lets us define the formal adjoint of $d$, and hence an elliptic operator on forms. Cohomology classes are represented by many closed forms, so analysis needs a way to choose a preferred representative from each class. The obstruction is gauge freedom: adding an exact form changes the representative without changing the class. The codifferential imposes the complementary co-closed condition that leads to harmonic representatives. [definition: Codifferential And Harmonic Form] Let $(M,g)$ be an oriented Riemannian $n$-manifold. The codifferential is the formal adjoint $d^*:\Omega^k(M)\to\Omega^{k-1}(M)$ of $d$ with respect to the $L^2$ inner product on forms. The Hodge Laplacian on $k$-forms is the operator \begin{align*} \Delta : \Omega^k(M) &\longrightarrow \Omega^k(M),\\ \Delta &:= dd^*+d^*d. \end{align*} A form $\omega\in\Omega^k(M)$ is harmonic if $\Delta \omega=0$. [/definition] On a compact manifold without boundary, harmonicity is equivalent to the pair of equations $d\omega=0$ and $d^*\omega=0$. Thus harmonic forms are closed forms satisfying an additional co-closed gauge condition. For this analytic gauge to be useful, arbitrary forms must split into pieces that separate exact variation, coexact variation, and the harmonic remainder. The problem is not just to define harmonic forms, but to know that they occur as the finite-dimensional residual part after the exact and coexact directions have been accounted for. [quotetheorem:3941] [citeproof:3941] Closedness is part of the clean statement: on manifolds with boundary, exact and coexact pieces interact with boundary conditions, and one must impose absolute or relative boundary conditions to recover a correct decomposition. Compactness is what makes the analytic inverse theory Fredholm; on non-compact manifolds, harmonic forms can behave badly at infinity and need not represent de Rham cohomology in this simple way. The decomposition does not say that every form is closed or cohomologically meaningful; rather, it separates an arbitrary form into exact, coexact, and harmonic components, and the harmonic component is the part that survives in cohomology. [quotetheorem:3942] [citeproof:3942] This theorem says that de Rham cohomology can be represented analytically: each class has a unique harmonic representative. The vector space $H^k_{dR}(M)$ is independent of the metric, while the distinguished representative depends on the metric. The closedness hypothesis cannot simply be dropped: on an interval, boundary conditions determine which harmonic forms are allowed to represent absolute or relative cohomology. Non-compact examples also show the limitation; on $\mathbb R$, the constant $1$-form $dx$ is harmonic but represents the zero de Rham class because $dx=d x$, so harmonicity alone is not the same as cohomological non-triviality without the compact closed setting. In applications, the workflow is to compute the cohomology class topologically, choose a metric when analytic control is useful, and then solve the harmonic representative problem inside that fixed class. [example: Harmonic Forms On The Circle] Let $S^1=\mathbb R/2\pi\mathbb Z$ with its standard metric $d\theta^2$ and orientation $d\theta$. In dimension $1$, the Hodge star is determined by \begin{align*} *1&=d\theta,\\ *d\theta&=1, \end{align*} because $1\wedge *1=d\theta$ and $d\theta\wedge *d\theta=d\theta$. We compute the harmonic $0$-forms and $1$-forms explicitly. For a smooth function $f(\theta)$, harmonicity on a closed Riemannian manifold is equivalent to $df=0$ and $d^*f=0$. The second condition is automatic because $d^*$ lowers degree, while \begin{align*} df=f'(\theta)\,d\theta. \end{align*} Thus $df=0$ exactly when \begin{align*} f'(\theta)=0 \end{align*} for every $\theta$, so $f$ is constant on the connected circle. Hence \begin{align*} \mathcal H^0(S^1)=\{c:c\in\mathbb R\}. \end{align*} Now let $\alpha=a(\theta)d\theta$ be a $1$-form. Since $S^1$ is $1$-dimensional, \begin{align*} d\alpha=d(a(\theta)d\theta)=a'(\theta)d\theta\wedge d\theta=0. \end{align*} For the co-closed condition, using $d^*=-*d*$ on $1$-forms in dimension $1$, \begin{align*} d^*\alpha &=-*d*(a(\theta)d\theta)\\ &=-*d(a(\theta))\\ &=-*(a'(\theta)d\theta)\\ &=-a'(\theta). \end{align*} Thus $d^*\alpha=0$ exactly when $a'(\theta)=0$, so $a$ is constant. Therefore \begin{align*} \mathcal H^1(S^1)=\{c\,d\theta:c\in\mathbb R\}. \end{align*} These harmonic forms represent the two de Rham cohomology groups: \begin{align*} H^0_{dR}(S^1)&\cong \mathcal H^0(S^1)\cong \mathbb R,\\ H^1_{dR}(S^1)&\cong \mathcal H^1(S^1)\cong \mathbb R. \end{align*} In this example, Hodge theory selects the constant representative in degree $0$ and the constant angular form representative in degree $1$, which is the precise sense in which it chooses the least oscillatory representative of each class. [/example] Together, the tools in this chapter give a reusable pattern: pair complementary classes to extract global information, compress Betti numbers when an Euler characteristic obstruction is enough, and use harmonic representatives when a metric turns the topological question into an elliptic equation. [remark: What Comes Next] Hodge theory opens several further directions. In geometry, the Hodge star and Laplacian lead to curvature identities and index theorems. In topology, harmonic representatives make Poincare duality compatible with an inner product. In mathematical physics, closed non-exact forms model conserved fluxes such as the monopole form above. [/remark]