The cotangent space records the linear measurements that can be made at one point of a smooth space. The [tangent space](/page/Tangent%20Space) contains possible first-order velocities through the point; the cotangent space contains the linear functionals on those velocities. This dual viewpoint is the natural home of differentials of functions. If a smooth real-valued function is observed near a point, its first-order change is not a tangent vector. It is a linear rule that eats a tangent vector and returns a number. That rule is the differential of the function at the point. Cotangent spaces therefore connect local coordinates, gradients, differential forms, and symplectic geometry, and they provide the linear algebra behind one-forms and the cotangent bundle. The point of the definition is to separate two roles that are often blurred in Euclidean space. A velocity is a direction of motion. A covector is a device for measuring a direction of motion. An [inner product](/page/Inner%20Product) can identify these two objects, but a [smooth manifold](/page/Smooth%20Manifold) does not carry an inner product by default. The cotangent space is defined without choosing a metric, and that metric-free definition is why cotangent spaces appear throughout differential geometry. ## Definition ### Linear Measurements at a Point The cotangent space is the answer to a measurement problem. Once a point $p$ of a smooth manifold $M$ is fixed, the tangent space $T_pM$ describes possible first-order motions through $p$; what we also need is the [vector space](/page/Vector%20Space) of all linear tests that can be applied to those motions. A smooth manifold has no preferred inner product and no preferred way to identify directions with measurements, so the space of measurements has to be named on its own. Ordinary linear duality gives exactly the metric-free object we need. [definition: Cotangent Space] Let $M$ be a smooth manifold and let $p \in M$. The cotangent space of $M$ at $p$ is the dual vector space \begin{align*} T_p^*M = \operatorname{Hom}_{\mathbb{R}}(T_pM,\mathbb{R}). \end{align*} An element of $T_p^*M$ is called a covector at $p$. [/definition] The definition says that a covector is a linear functional on tangent vectors. If $\alpha \in T_p^*M$ and $v \in T_pM$, the value $\alpha(v)$ is a real number. The notation $T_p^*M$ is read as "T star p M" or "the dual of the tangent space at $p$." The notation $T_p^*M$ is only meaningful once the tangent space $T_pM$ has been fixed as a genuine vector space rather than as a picture of arrows in a chart. The next definition supplies that model. For cotangent spaces, the most useful version of $T_pM$ is the derivation model: it treats a tangent direction as an operator that differentiates germs of smooth functions at $p$. This is the model that makes a covector genuinely act on first-order directions, and it will later make the formula $df_p(v)=v(f)$ a structural identity rather than a mnemonic. [definition: Tangent Space at a Point] Let $M$ be a smooth manifold and let $p \in M$. Let $C_p^\infty(M)$ be the real algebra of germs at $p$ of smooth real-valued functions defined near $p$. The tangent space at $p$, denoted $T_pM$, is the real vector space of all maps $D:C_p^\infty(M)\to\mathbb{R}$ that are $\mathbb{R}$-linear and satisfy the Leibniz rule \begin{align*} D(fg)=f(p)D(g)+g(p)D(f) \end{align*} for all $f,g\in C_p^\infty(M)$. [/definition] The tangent-space definition explains what covectors measure, while the cotangent-space definition explains what all possible measurements are. The next missing ingredient is the act of measurement itself. Since a covector is a function whose inputs are tangent vectors, the most basic operation should simply evaluate that function; this operation must be stated explicitly because it is the coordinate-free source of the formulas that later look like dot products. [definition: Natural Pairing] Let $M$ be a smooth manifold and let $p\in M$. The natural pairing between $T_p^*M$ and $T_pM$ is the map \begin{align*} \operatorname{pair}_p:T_p^*M\times T_pM\to\mathbb{R} \end{align*} defined by \begin{align*} \operatorname{pair}_p(\alpha,v)=\alpha(v). \end{align*} [/definition] This pairing is canonical because it is ordinary function evaluation. It does not require a coordinate chart, a basis, or a Riemannian metric. ### Coordinate Covectors in Action The abstract definition becomes concrete once a coordinate basis is chosen. The next example shows how the dual coordinate covectors act as component-measuring devices rather than as new tangent directions. [example: A Covector on a Plane] Let $M$ be a two-dimensional smooth manifold, let $p\in M$, and choose local coordinates $(x,y)$ near $p$. Write a tangent vector at $p$ in the coordinate basis as \begin{align*} v=a\,\frac{\partial}{\partial x}\bigg|_p+b\,\frac{\partial}{\partial y}\bigg|_p. \end{align*} The coordinate covectors $dx|_p$ and $dy|_p$ form the [dual basis](/theorems/414) to $\partial/\partial x|_p$ and $\partial/\partial y|_p$, so their defining values are \begin{align*} dx|_p\left(\frac{\partial}{\partial x}\bigg|_p\right)=1,\quad dx|_p\left(\frac{\partial}{\partial y}\bigg|_p\right)=0,\quad dy|_p\left(\frac{\partial}{\partial x}\bigg|_p\right)=0,\quad dy|_p\left(\frac{\partial}{\partial y}\bigg|_p\right)=1. \end{align*} Because $dx|_p:T_pM\to\mathbb{R}$ is linear, evaluating it on $v$ gives \begin{align*} dx|_p(v)=dx|_p\left(a\,\frac{\partial}{\partial x}\bigg|_p+b\,\frac{\partial}{\partial y}\bigg|_p\right). \end{align*} Linearity in the input separates the two coordinate components: \begin{align*} dx|_p(v)=a\,dx|_p\left(\frac{\partial}{\partial x}\bigg|_p\right)+b\,dx|_p\left(\frac{\partial}{\partial y}\bigg|_p\right). \end{align*} Substituting the dual-basis values gives \begin{align*} dx|_p(v)=a\cdot 1+b\cdot 0. \end{align*} Since $a\cdot 1=a$ and $b\cdot 0=0$, this becomes \begin{align*} dx|_p(v)=a. \end{align*} The same calculation for $dy|_p$ starts from linearity: \begin{align*} dy|_p(v)=dy|_p\left(a\,\frac{\partial}{\partial x}\bigg|_p+b\,\frac{\partial}{\partial y}\bigg|_p\right). \end{align*} Separating the two terms gives \begin{align*} dy|_p(v)=a\,dy|_p\left(\frac{\partial}{\partial x}\bigg|_p\right)+b\,dy|_p\left(\frac{\partial}{\partial y}\bigg|_p\right). \end{align*} Using the dual-basis values gives \begin{align*} dy|_p(v)=a\cdot 0+b\cdot 1. \end{align*} Since $a\cdot 0=0$ and $b\cdot 1=b$, this becomes \begin{align*} dy|_p(v)=b. \end{align*} Now let $\alpha=3dx|_p-2dy|_p$. Since scalar multiples and sums of covectors evaluate pointwise, \begin{align*} \alpha(v)=(3dx|_p-2dy|_p)(v)=3\,dx|_p(v)-2\,dy|_p(v). \end{align*} Substituting $dx|_p(v)=a$ and $dy|_p(v)=b$ gives \begin{align*} \alpha(v)=3a-2b. \end{align*} Thus $dx|_p$ measures the $x$-component of the tangent vector, $dy|_p$ measures the $y$-component, and $3dx|_p-2dy|_p$ measures the signed linear combination $3a-2b$. [/example] To make these measurements independent of coordinates, the construction must name the whole vector space of linear measurements at once. This is why the cotangent space is defined as the whole [dual space](/page/Dual%20Space), not as a particular row-vector representation in a chosen chart. When $M$ has dimension $n$, the vector space $T_pM$ has dimension $n$. Its dual $T_p^*M$ also has dimension $n$. The equality of dimensions does not mean that tangent vectors and covectors are canonically the same. A choice of basis identifies a vector space with $\mathbb{R}^n$, and then the dual with row vectors. A different choice changes the coordinates of vectors and covectors in opposite ways. This opposite transformation law is the source of the name "co-vector." [example: Differential as a Linear Measurement] For $f(x,y)=x^2+y$ on $\mathbb{R}^2$, we compute the covector $df_{(1,3)}$ in the coordinate covector basis $dx|_{(1,3)},dy|_{(1,3)}$. Differentiating $x^2+y$ with respect to $x$, treating $y$ as constant, gives \begin{align*} \frac{\partial f}{\partial x}(x,y)=2x. \end{align*} Differentiating $x^2+y$ with respect to $y$, treating $x$ as constant, gives \begin{align*} \frac{\partial f}{\partial y}(x,y)=1. \end{align*} At $(1,3)$ these become \begin{align*} \frac{\partial f}{\partial x}(1,3)=2\cdot 1=2. \end{align*} and \begin{align*} \frac{\partial f}{\partial y}(1,3)=1. \end{align*} Using the coordinate formula $df_p=\frac{\partial f}{\partial x}(p)dx|_p+\frac{\partial f}{\partial y}(p)dy|_p$, we get \begin{align*} df_{(1,3)}=2dx|_{(1,3)}+1dy|_{(1,3)}. \end{align*} Since $1dy|_{(1,3)}=dy|_{(1,3)}$, this is \begin{align*} df_{(1,3)}=2dx|_{(1,3)}+dy|_{(1,3)}. \end{align*} Now let \begin{align*} v=a\,\frac{\partial}{\partial x}\bigg|_{(1,3)}+b\,\frac{\partial}{\partial y}\bigg|_{(1,3)}. \end{align*} The coordinate covectors are dual to the coordinate tangent vectors, so \begin{align*} dx|_{(1,3)}\left(\frac{\partial}{\partial x}\bigg|_{(1,3)}\right)=1 \end{align*} and \begin{align*} dx|_{(1,3)}\left(\frac{\partial}{\partial y}\bigg|_{(1,3)}\right)=0. \end{align*} By linearity of $dx|_{(1,3)}$, \begin{align*} dx|_{(1,3)}(v)=a\,dx|_{(1,3)}\left(\frac{\partial}{\partial x}\bigg|_{(1,3)}\right)+b\,dx|_{(1,3)}\left(\frac{\partial}{\partial y}\bigg|_{(1,3)}\right). \end{align*} Substituting the dual-basis values gives \begin{align*} dx|_{(1,3)}(v)=a\cdot 1+b\cdot 0. \end{align*} Since $a\cdot 1=a$ and $b\cdot 0=0$, \begin{align*} dx|_{(1,3)}(v)=a. \end{align*} Similarly, \begin{align*} dy|_{(1,3)}\left(\frac{\partial}{\partial x}\bigg|_{(1,3)}\right)=0 \end{align*} and \begin{align*} dy|_{(1,3)}\left(\frac{\partial}{\partial y}\bigg|_{(1,3)}\right)=1. \end{align*} By linearity of $dy|_{(1,3)}$, \begin{align*} dy|_{(1,3)}(v)=a\,dy|_{(1,3)}\left(\frac{\partial}{\partial x}\bigg|_{(1,3)}\right)+b\,dy|_{(1,3)}\left(\frac{\partial}{\partial y}\bigg|_{(1,3)}\right). \end{align*} Substituting the dual-basis values gives \begin{align*} dy|_{(1,3)}(v)=a\cdot 0+b\cdot 1. \end{align*} Since $a\cdot 0=0$ and $b\cdot 1=b$, \begin{align*} dy|_{(1,3)}(v)=b. \end{align*} Now evaluate the differential on $v$: \begin{align*} df_{(1,3)}(v)=(2dx|_{(1,3)}+dy|_{(1,3)})(v). \end{align*} Because sums and scalar multiples of covectors evaluate pointwise, \begin{align*} df_{(1,3)}(v)=2dx|_{(1,3)}(v)+dy|_{(1,3)}(v). \end{align*} Substituting $dx|_{(1,3)}(v)=a$ and $dy|_{(1,3)}(v)=b$ gives \begin{align*} df_{(1,3)}(v)=2a+b. \end{align*} Thus the covector $df_{(1,3)}$ measures a tangent vector with coordinate components $(a,b)$ by returning $2a+b$, the first-order rate of change of $f$ in that direction. [/example] ## Coordinate Description ### Dual Coordinate Bases Choose a coordinate chart $(U,\varphi)$ around $p$ with coordinates $(x_1,\ldots,x_n)$. The associated coordinate vector fields give a basis \begin{align*} \frac{\partial}{\partial x_1}\bigg|_p,\ldots,\frac{\partial}{\partial x_n}\bigg|_p \end{align*} for $T_pM$. The dual basis of $T_p^*M$ is denoted \begin{align*} dx_1|_p,\ldots,dx_n|_p. \end{align*} These covectors are characterized by their values on the coordinate basis vectors. For $1 \leq i,j \leq n$, they satisfy \begin{align*} dx_i|_p\left(\frac{\partial}{\partial x_j}\bigg|_p\right)=\delta_{ij}. \end{align*} Here $\delta_{ij}$ is the Kronecker delta. It equals $1$ when $i=j$ and $0$ when $i\neq j$. Every tangent vector has coordinate form \begin{align*} v=\sum_{j=1}^n v_j\frac{\partial}{\partial x_j}\bigg|_p. \end{align*} Every covector has coordinate form \begin{align*} \alpha=\sum_{i=1}^n \alpha_i dx_i|_p. \end{align*} The pairing is then \begin{align*} \alpha(v)=\sum_{i=1}^n \alpha_i v_i. \end{align*} The coordinate lists for $v$ and $\alpha$ are not two copies of the same data. They obey different coordinate transformation laws, which is why covectors should not be treated as tangent vectors with a different font. ### Change of Coordinates To make the transformation law precise, let $(U,\varphi)$ and $(V,\psi)$ be coordinate charts around $p$, with coordinates $(x_1,\ldots,x_n)$ and $(y_1,\ldots,y_n)$. Write a tangent vector in the two coordinate bases as \begin{align*} v=\sum_{j=1}^n v_{x,j}\frac{\partial}{\partial x_j}\bigg|_p=\sum_{i=1}^n v_{y,i}\frac{\partial}{\partial y_i}\bigg|_p. \end{align*} Write a covector in the two dual coordinate bases as \begin{align*} \alpha=\sum_{j=1}^n \alpha_{x,j} dx_j|_p=\sum_{i=1}^n \alpha_{y,i} dy_i|_p. \end{align*} Let $J$ be the [Jacobian matrix](/page/Jacobian%20Matrix) of the change from $x$-coordinates to $y$-coordinates, with entries \begin{align*} J_{ij}=\frac{\partial y_i}{\partial x_j}(p). \end{align*} Then \begin{align*} v_y=Jv_x,\qquad \alpha_y=(J^{-1})^\top\alpha_x. \end{align*} This inverse-transpose law is exactly what is needed for the scalar $\alpha(v)$ to remain independent of coordinates. [example: Cotangent Space of Euclidean Space] For $M=\mathbb{R}^n$ and $p\in\mathbb{R}^n$, use the standard coordinate tangent basis \begin{align*} \frac{\partial}{\partial x_1}\bigg|_p,\ldots,\frac{\partial}{\partial x_n}\bigg|_p \end{align*} for $T_p\mathbb{R}^n$. A tangent vector written in this basis has the form \begin{align*} v=\sum_{j=1}^n v_j\frac{\partial}{\partial x_j}\bigg|_p. \end{align*} Under the standard coordinate identification of $T_p\mathbb{R}^n$ with $\mathbb{R}^n$, this vector has coordinate column $(v_1,\ldots,v_n)$. The cotangent space is the dual vector space \begin{align*} T_p^*\mathbb{R}^n=\operatorname{Hom}_{\mathbb{R}}(T_p\mathbb{R}^n,\mathbb{R}). \end{align*} The standard coordinate covectors $dx_1|_p,\ldots,dx_n|_p$ are dual to the standard coordinate tangent vectors, so for $1\leq i,j\leq n$, \begin{align*} dx_i|_p\left(\frac{\partial}{\partial x_j}\bigg|_p\right)=\delta_{ij}. \end{align*} Here $\delta_{ij}=1$ when $i=j$, and $\delta_{ij}=0$ when $i\neq j$. Let \begin{align*} \alpha=\sum_{i=1}^n \alpha_i dx_i|_p. \end{align*} Evaluating $\alpha$ on $v$ gives \begin{align*} \alpha(v)=\left(\sum_{i=1}^n \alpha_i dx_i|_p\right)\left(\sum_{j=1}^n v_j\frac{\partial}{\partial x_j}\bigg|_p\right). \end{align*} Because sums and scalar multiples of covectors evaluate pointwise, this equals \begin{align*} \alpha(v)=\sum_{i=1}^n \alpha_i\, dx_i|_p\left(\sum_{j=1}^n v_j\frac{\partial}{\partial x_j}\bigg|_p\right). \end{align*} For each fixed $i$, the covector $dx_i|_p:T_p\mathbb{R}^n\to\mathbb{R}$ is linear, so \begin{align*} dx_i|_p\left(\sum_{j=1}^n v_j\frac{\partial}{\partial x_j}\bigg|_p\right)=\sum_{j=1}^n v_j\,dx_i|_p\left(\frac{\partial}{\partial x_j}\bigg|_p\right). \end{align*} Substituting this into the previous expression gives \begin{align*} \alpha(v)=\sum_{i=1}^n \alpha_i\sum_{j=1}^n v_j\,dx_i|_p\left(\frac{\partial}{\partial x_j}\bigg|_p\right). \end{align*} Using the dual-basis relation $dx_i|_p(\partial/\partial x_j|_p)=\delta_{ij}$, we get \begin{align*} \alpha(v)=\sum_{i=1}^n \alpha_i\sum_{j=1}^n v_j\delta_{ij}. \end{align*} For a fixed $i$, the terms in the inner sum are $0$ when $j\neq i$ and $v_i$ when $j=i$, so \begin{align*} \sum_{j=1}^n v_j\delta_{ij}=v_i. \end{align*} Therefore \begin{align*} \alpha(v)=\sum_{i=1}^n \alpha_i v_i. \end{align*} Writing out the finite sum gives \begin{align*} \alpha(v)=\alpha_1v_1+\cdots+\alpha_nv_n. \end{align*} Thus in standard coordinates a covector on $\mathbb{R}^n$ is represented by the coefficient list $(\alpha_1,\ldots,\alpha_n)$, while its coordinate-free meaning is the linear measurement $v\mapsto \sum_{i=1}^n \alpha_i v_i$. [/example] The Euclidean example can make covectors look like row vectors. That picture is useful in computations, but the definition is coordinate-free. The covector is the [linear map](/page/Linear%20Map) itself, not the row of numbers used to describe it in a basis. ## Differentials of Functions The most common source of covectors is the differential of a function. Let $f:M\to\mathbb{R}$ be smooth. At a point $p$, the differential $df_p$ measures the first-order change of $f$ in tangent directions. If $v\in T_pM$, then $df_p(v)$ is the [directional derivative](/page/Directional%20Derivative) of $f$ along $v$. In the derivation model of tangent vectors, this is simply \begin{align*} df_p(v)=v(f). \end{align*} Thus $df_p$ is a linear functional on $T_pM$. It is therefore an element of $T_p^*M$. [definition: Differential at a Point] Let $f:M\to\mathbb{R}$ be a smooth function and $p\in M$. The differential of $f$ at $p$ is the covector $df_p:T_pM\to\mathbb{R}$ defined by \begin{align*} df_p(v)=v(f). \end{align*} [/definition] In local coordinates $(x_1,\ldots,x_n)$, the differential is given by \begin{align*} df_p=\sum_{i=1}^n \frac{\partial f}{\partial x_i}(p) dx_i|_p. \end{align*} This formula is the coordinate expression of the familiar first-order approximation. For a tangent vector $v=\sum_{i=1}^n v_i\partial/\partial x_i|_p$, the pairing gives \begin{align*} df_p(v)=\sum_{i=1}^n \frac{\partial f}{\partial x_i}(p)v_i. \end{align*} On a Euclidean space with its standard inner product, this number can also be written as a dot product with the gradient. The gradient vector depends on the chosen inner product. The differential covector does not. This distinction matters on a general manifold. Without a Riemannian metric, the differential $df_p$ exists, but there is no canonical gradient vector. [example: Zero Differential] Suppose $f$ is locally constant near $p$, so the germ of $f$ at $p$ is the same as the germ of a constant function $c$. Let $v\in T_pM$. First, every tangent derivation kills constants. Since $1\cdot 1=1$, the Leibniz rule gives \begin{align*} v(1)=v(1\cdot 1). \end{align*} Applying the Leibniz rule to $1\cdot 1$ gives \begin{align*} v(1\cdot 1)=1(p)v(1)+1(p)v(1). \end{align*} Because $1(p)=1$, this becomes \begin{align*} v(1\cdot 1)=v(1)+v(1)=2v(1). \end{align*} Therefore \begin{align*} v(1)=2v(1). \end{align*} Subtracting $v(1)$ from both sides gives \begin{align*} 0=v(1). \end{align*} Since the constant function $c$ is $c\cdot 1$, linearity of $v$ gives \begin{align*} v(c)=v(c\cdot 1)=c\,v(1). \end{align*} Using $v(1)=0$, we get \begin{align*} v(c)=c\cdot 0=0. \end{align*} By the definition of the differential, \begin{align*} df_p(v)=v(f). \end{align*} The germ of $f$ at $p$ equals the germ of the constant function $c$, so $v(f)=v(c)$. Hence \begin{align*} df_p(v)=v(c)=0. \end{align*} Since this holds for every $v\in T_pM$, the covector $df_p:T_pM\to\mathbb{R}$ is the zero covector. The converse fails at a single point. On $M=\mathbb{R}$, let $f(x)=x^2$ and take $p=0$. A tangent vector at $0$ has the form \begin{align*} v=a\,\frac{\partial}{\partial x}\bigg|_0. \end{align*} The ordinary derivative of $f(x)=x^2$ is \begin{align*} f'(x)=2x. \end{align*} Evaluating at $0$ gives \begin{align*} f'(0)=2\cdot 0=0. \end{align*} Using the coordinate formula for the differential on $\mathbb{R}$, \begin{align*} df_0(v)=f'(0)a. \end{align*} Substituting $f'(0)=0$ gives \begin{align*} df_0(v)=0\cdot a=0. \end{align*} Thus $df_0=0$. However, $f$ is not locally constant near $0$: for every neighborhood of $0$, choose a nonzero $t$ in that neighborhood; then \begin{align*} f(0)=0^2=0. \end{align*} while \begin{align*} f(t)=t^2\neq 0. \end{align*} So a zero differential at one point records only vanishing first-order change at that point, not local constancy. [/example] ## Coordinate-Free Character The construction \begin{align*} T_p^*M=\operatorname{Hom}_{\mathbb{R}}(T_pM,\mathbb{R}) \end{align*} is functorial under linear maps. If $F:M\to N$ is smooth and $p\in M$, its derivative is a linear map \begin{align*} dF_p:T_pM\to T_{F(p)}N. \end{align*} Linear duality reverses the direction. Each covector $\beta\in T_{F(p)}^*N$ can be pulled back to a covector on $M$ at $p$. The pullback covector is defined by composition with $dF_p$. [definition: Pullback of a Covector] Let $F:M\to N$ be smooth and let $p\in M$. The pullback of covectors by $F$ at $p$ is the linear map \begin{align*} F_p^*:T_{F(p)}^*N\to T_p^*M \end{align*} defined by \begin{align*} F_p^*(\beta)=\beta\circ dF_p. \end{align*} For $\beta\in T_{F(p)}^*N$, the pullback covector $F_p^*\beta\in T_p^*M$ is characterized by \begin{align*} (F_p^*\beta)(v)=\beta(dF_p(v)) \end{align*} for every $v\in T_pM$. [/definition] This formula is the pointwise version of pullback for differential forms. It also explains why covectors transform contravariantly. Maps push tangent vectors forward, while they pull covectors back. The direction reversal is not a convention. It is forced by composition of linear maps. The following identity records how this reversal interacts with differentiating real-valued functions. It is the covector-level expression of the ordinary chain rule, and it is also the prototype for pullback identities for one-forms and higher differential forms. [quotetheorem:10162] The point of this identity is that pullback is compatible with taking differentials, so there is no ambiguity between first pulling a function back to $M$ and then differentiating it, or first differentiating on $N$ and then pulling the resulting covector back. This compatibility is what makes pullback a usable operation rather than just a pointwise linear-algebra construction. For example, coordinate functions on $N$ pull back to component functions of $F$, and their differentials become the basic first-order measurements of $F$ on $M$. Later, the same naturality principle extends from exact one-forms of the form $dg$ to arbitrary one-forms and then to higher differential forms. ## Relation with One-Forms A cotangent space is attached to one point. A differential one-form assigns a covector to each point in a smooth way. The collection of all cotangent spaces forms the cotangent bundle. [definition: Cotangent Bundle] Let $M$ be a smooth manifold. The cotangent bundle of $M$ is the disjoint union \begin{align*} T^*M=\bigsqcup_{p\in M}T_p^*M \end{align*} together with its natural [smooth vector bundle](/page/Smooth%20Vector%20Bundle) structure. [/definition] A smooth section of $T^*M$ is a differential one-form. For example, the differential $df$ of a smooth function is the one-form whose value at $p$ is $df_p$. In coordinates, a one-form has local expression \begin{align*} \omega=\sum_{i=1}^n \omega_i dx_i. \end{align*} Here each $\omega_i$ is a smooth function on the coordinate neighborhood. At a point $p$, the value of the one-form is the covector \begin{align*} \omega_p=\sum_{i=1}^n \omega_i(p) dx_i|_p. \end{align*} The distinction between a covector and a one-form is the distinction between pointwise linear algebra and a field of such linear data. The same distinction appears between a tangent vector at a point and a vector field on a region. ## Geometry of the Cotangent Bundle The cotangent bundle has more structure than the tangent bundle in several settings. It carries a canonical one-form, often called the tautological one-form. From it one obtains a canonical symplectic form. This is why cotangent bundles are the standard phase spaces in Hamiltonian mechanics. We write $\Omega^1(X)$ for the vector space of smooth differential one-forms on a smooth manifold $X$, and $\Omega^2(X)$ for the vector space of smooth differential two-forms on $X$. The canonical form is not added by a choice of coordinates; it is obtained from the projection $\pi:T^*M\to M$. At a point $\alpha\in T_p^*M$, a tangent vector to $T^*M$ projects to a tangent vector in $T_pM$, and the covector $\alpha$ can evaluate that projected vector. [definition: Canonical One-Form] Let $\pi:T^*M\to M$ be the cotangent bundle projection. The canonical one-form on $T^*M$ is the one-form $\theta\in\Omega^1(T^*M)$ whose value at $\alpha\in T_p^*M$ is the linear map \begin{align*} \theta_\alpha:T_\alpha(T^*M)\to\mathbb{R} \end{align*} defined by \begin{align*} \theta_\alpha(W)=\alpha(d\pi_\alpha(W)). \end{align*} [/definition] This definition explains why the form is called tautological: at the point $\alpha$, the form uses $\alpha$ itself to measure the base component of a tangent vector to the cotangent bundle. In local coordinates $(x_1,\ldots,x_n)$ on $M$, a point of $T^*M$ has coordinates $(x_1,\ldots,x_n,\xi_1,\ldots,\xi_n)$. The $x_i$ coordinates locate the base point. The $\xi_i$ coordinates describe the covector at that base point. The canonical one-form has local expression \begin{align*} \theta=\sum_{i=1}^n \xi_i dx_i. \end{align*} The local formula shows how the pointwise definition becomes the familiar phase-space expression. The next object isolates the part of this geometry that survives every coordinate change and every choice of local trivialisation: a two-form whose values pair horizontal and vertical first-order changes in the cotangent bundle. Taking the [exterior derivative](/theorems/1525) of the tautological form is exactly what turns the evaluation one-form into this invariant two-direction measurement. The sign convention $\omega_{\mathrm{can}}=-d\theta$ is the one used here because it gives the standard coordinate expression $\sum_i dx_i\wedge d\xi_i$. [definition: Canonical Symplectic Form] Let $\theta\in\Omega^1(T^*M)$ be the canonical one-form, and let $d:\Omega^1(T^*M)\to\Omega^2(T^*M)$ be the exterior derivative. The canonical symplectic form on $T^*M$ is the two-form \begin{align*} \omega_{\mathrm{can}}=-d\theta. \end{align*} [/definition] This two-form is called symplectic because it is closed and nondegenerate: closedness follows from $d\omega_{\mathrm{can}}=-d^2\theta=0$, and in local coordinates $(x_1,\ldots,x_n,\xi_1,\ldots,\xi_n)$ it has the form \begin{align*} \omega_{\mathrm{can}}=\sum_{i=1}^n dx_i\wedge d\xi_i, \end{align*} which pairs the base and fiber coordinate directions nondegenerately. Some authors use $d\theta$ instead of $-d\theta$; the sign convention changes formulas but not the fact that the form is built canonically from the cotangent bundle. This construction uses the dual nature of cotangent fibers. No Riemannian metric is required. ## Beyond and Connected Topics The cotangent space is the pointwise linear algebra behind differential forms, as developed in [Cambridge III Differential Geometry](/page/Cambridge%20III%20Differential%20Geometry). Exterior powers of $T_p^*M$ produce alternating covariant tensors, and smooth sections of those exterior powers give differential forms of higher degree. This page has kept the focus on degree $1$ because every higher-degree construction still starts from the same question: how do linear measurements of tangent directions combine? Covariant tensors allow several tangent inputs, while alternating forms impose the sign changes that make integration on manifolds possible. The cotangent bundle $T^*M$ is central in symplectic geometry because it has a canonical symplectic structure, and it also appears in complex-geometric settings such as [Several Complex Variables IV: Complex Geometry and Curvature](/page/Several%20Complex%20Variables%20IV%3A%20Complex%20Geometry%20and%20Curvature). In Riemannian geometry, a metric identifies $T_pM$ with $T_p^*M$, turning covectors into vectors and producing gradients from differentials. In analysis on manifolds, covectors appear as principal symbols of differential operators and as the natural variables in microlocal geometry. The metric dependence is not cosmetic: changing the metric changes which vector represents the same covector. The following example is a small warning against treating the Euclidean dot product as part of the definition. [example: Metric-Dependent Identification] Let $V=\mathbb{R}^2$, and let $\alpha\in V^*$ be the covector defined by \begin{align*} \alpha(v)=v_1 \end{align*} for every $v=(v_1,v_2)\in V$. First use the standard inner product \begin{align*} (u,v)=u_1v_1+u_2v_2. \end{align*} Take $e_1=(1,0)$ and let $v=(v_1,v_2)$ be arbitrary. Substituting $u=e_1$ into the formula for the standard inner product gives \begin{align*} (e_1,v)=1\cdot v_1+0\cdot v_2. \end{align*} Since $1\cdot v_1=v_1$ and $0\cdot v_2=0$, this is \begin{align*} (e_1,v)=v_1+0. \end{align*} Since $v_1+0=v_1$, we get \begin{align*} (e_1,v)=v_1. \end{align*} By the definition of $\alpha$, the same vector $v$ satisfies \begin{align*} \alpha(v)=v_1. \end{align*} Therefore \begin{align*} (e_1,v)=\alpha(v). \end{align*} Because $v\in V$ was arbitrary, $e_1$ represents the covector $\alpha$ under the standard inner product. Now change the inner product to \begin{align*} g(u,v)=2u_1v_1+u_2v_2. \end{align*} Let $w=(1/2,0)$ and again let $v=(v_1,v_2)$ be arbitrary. Substituting $u=w$ into the formula for $g$ gives \begin{align*} g(w,v)=2\cdot \frac{1}{2}\cdot v_1+0\cdot v_2. \end{align*} Since $2\cdot \frac{1}{2}=1$, this becomes \begin{align*} g(w,v)=1\cdot v_1+0\cdot v_2. \end{align*} Since $1\cdot v_1=v_1$ and $0\cdot v_2=0$, we have \begin{align*} g(w,v)=v_1+0. \end{align*} Since $v_1+0=v_1$, this gives \begin{align*} g(w,v)=v_1. \end{align*} By the definition of $\alpha$, \begin{align*} \alpha(v)=v_1. \end{align*} Therefore \begin{align*} g(w,v)=\alpha(v). \end{align*} Because $v\in V$ was arbitrary, the same covector $\alpha$ is represented by $w=(1/2,0)$ with respect to the inner product $g$. The covector is the fixed linear functional $v\mapsto v_1$; what changes is the vector used to represent that functional after an inner product has been chosen. [/example] ## Common Confusions A covector is not a tangent vector with a different notation. It is a linear functional on tangent vectors. An inner product can convert a tangent vector into a covector by sending $u$ to the functional $v\mapsto \langle u,v\rangle$. That conversion depends on the inner product. Changing the metric changes the covector associated to the same vector. Another common confusion concerns the symbol $dx_i$. It is not an infinitesimal number. At a point $p$, $dx_i|_p$ is the differential of the coordinate function $x_i$. It is the covector that extracts the $i$-th coordinate component of a tangent vector. The notation is compact because the same symbol is used for the coordinate one-form on a neighborhood. A final confusion concerns dimensions. The cotangent space and tangent space have the same dimension. They are dual spaces, not identical spaces. The natural pairing goes from one covector and one vector to a scalar. It does not turn two vectors into a scalar unless an additional [bilinear form](/page/Bilinear%20Form) has been chosen. ## Summary The cotangent space $T_p^*M$ is the dual of the tangent space $T_pM$. Its elements are covectors, meaning linear functionals on tangent vectors. Differentials $df_p$ of smooth real-valued functions are the basic examples. Coordinate covectors $dx_i|_p$ form the dual basis to coordinate tangent vectors. Covectors pull back along smooth maps because linear duality reverses arrows. One-forms are smooth choices of covectors across a manifold. The cotangent bundle collects all cotangent spaces and carries canonical geometry important in analysis and mechanics. ## References Androma, [Cambridge III Differential Geometry](/page/Cambridge%20III%20Differential%20Geometry). Androma, [Cambridge II Algebraic Geometry](/page/Cambridge%20II%20Algebraic%20Geometry). Androma, [Several Complex Variables IV: Complex Geometry and Curvature](/page/Several%20Complex%20Variables%20IV%3A%20Complex%20Geometry%20and%20Curvature). Androma, [Several Complex Variables V: CR Geometry and Boundary Behavior](/page/Several%20Complex%20Variables%20V%3A%20CR%20Geometry%20and%20Boundary%20Behavior). Lee, *Introduction to Smooth Manifolds* (2013). Lee, *Introduction to Riemannian Manifolds* (2018). Abraham and Marsden, *Foundations of Mechanics* (1978).