📂Geometry

Cotangent Space and First-Order Differential Forms

Overview

We define the cotangent space and the differential 1-form. If differential manifolds are challenging, one can think of it as $M = \mathbb{R}^{n}$.

We use Einstein notation.

Cotangent Space¹

Let’s consider a $M$ as a $n$-dimensional differential manifold. Then, the tangent space $T_{p}M$ at point $p \in M$ becomes a $n$-dimensional vector space (function space), with the basis being $\left\{ \mathbf{e}_{i} = \left. \frac{\partial }{\partial x_{i}}\right|_{p} \right\}_{i}$.

At this time, the dual space $T_{p}^{\ast}M$ of the tangent space $T_{p} M$ is called the cotangent space.

$$T_{p}^{\ast}M := \left\{ \psi : T_{p}M \to \mathbb{R}\ |\ \psi \text{ is continuous and linear} \right\}$$

Description

Due to the properties of the dual space, $\dim T_{p}M = n = \dim T_{p}^{\ast}M$, and the dual basis $\left\{ (dx_{j})_{p} \right\}$ is defined as the following function.

$$(dx_{j})_{p} : T_{p}M \to \mathbb{R}$$

$$(dx_{j})_{p} \left(\textstyle \left. \frac{\partial }{\partial x_{k}}\right|_{p} \right) = \delta_{jk} = \begin{cases} 1, & j=k \\ 0, & j\ne k \end{cases}$$

Any $\omega_{p} \in T_{p}^{\ast}M$ can be expressed as follows against the basis $\left\{ (dx_{j})_{p} \right\}$.

$$\begin{align*} \omega_{p} =&\ (a_{p}^{1}, a_{p}^{2}, a_{p}^{3}),\quad a_{p}^{i} \in \mathbb{R} \\[1em] =&\ a_{p}^{1}(dx_{1})_{p} + a_{p}^{2}(dx_{1})_{p} + a_{p}^{3}(dx_{3})_{p} \end{align*}$$

Then, let’s consider the function $\omega$ mapping each point $p \in M$ to $\omega_{p} \in T_{p}^{\ast}M$.

Differential 1-Form

A $\omega$ mapping each point $p\in M$ on the differential manifold $M$ to the element $\omega_{p} \in T_{p}^{\ast}M$ of the cotangent space is called a 1-form.

$$\begin{align*} \omega : M &\to T^{\ast}M \\ p &\mapsto \omega_{p} \end{align*}$$

In this case, $T^{\ast}M = \bigcup \limits_{p \in M} T_{p}^{\ast}M$ is called a cotangent bundle.

Description

A 1-form is also known as a first-order form, and it is referred to as the exterior form of degree 1, field of linear form, etc., in English.

If a function is $a_{i}$, where $a_{i} : M \to \mathbb{R}$ and $a_{i}(p) = a_{p}^{i}$, then $\omega_{p}$ can be expressed as follows.

$$\begin{align*} \omega_{p} = \omega (p) =&\ (a_{1}(p), a_{2}(p), a_{3}(p)) \\ =&\ a_{1}(p)(dx_{1})_{p} + a_{2}(p)(dx_{1})_{p} + a_{3}(p)(dx_{3})_{p} \end{align*}$$

Then, $\omega$ is as follows. When using Einstein’s notation,

$$\omega = a_{1}dx_{1} + a_{2}dx_{2} + a_{3}dx_{3} = a_{i}dx_{i}$$

If each $a_{i}$ is a differentiable function, then $\omega$ is called a differential form of degree 1.

$\mathbb{R}^{n}$’s $1$-Form

This abstract talk might make it hard to understand its meaning. Differential forms provide the theoretical backdrop for dealing with $dx$ and $dy$ freely in calculus. Let’s look at an example in Euclidean space. Consider a function $f : \mathbb{R}^{n} \to \mathbb{R}$. Then, the differentiation of $f$ is $df_{p} : T_{p}\mathbb{R}^{n} \to T_{f(p)}\mathbb{R}$. If $v \in T_{p}\mathbb{R}^{n}$,

$$v = \sum\limits_{i} v_{i}\dfrac{\partial }{\partial x_{i}} = v_{i}\dfrac{\partial }{\partial x_{i}} = (v_{1}, \dots, v_{n})$$

If the coordinates of $\mathbb{R}$ are $y$, then the basis of $T_{f(p)}\mathbb{R}$ is $\left\{ \dfrac{\partial }{\partial y} \right\}$, and substituting $v$ into the differential yields,

$$\begin{align*} df_{p} (v) &= \begin{bmatrix}\dfrac{\partial f}{\partial x_{1}} & \cdots & \dfrac{\partial f}{\partial x_{n}}\end{bmatrix} \begin{bmatrix}v_{1} \\ \vdots \\ v_{n} \end{bmatrix} \\ &= \begin{bmatrix} v_{i} \dfrac{\partial f}{\partial x_{i}}\end{bmatrix} \\ &= \left( v_{i} \dfrac{\partial f}{\partial x_{i}} \right) \dfrac{\partial }{\partial y} \end{align*}$$

Now, let’s look at the $\mathbb{R}^{n}$’s $1$-form $\omega_{p} = \dfrac{\partial f}{\partial x_{i}}dx_{i}$. By substituting $v$,

$$\begin{align*} w_{p} (v) &= \dfrac{\partial f}{\partial x_{i}}dx_{i} (v) \\ &= \dfrac{\partial f}{\partial x_{i}}dx_{i} \left( v_{j}\dfrac{\partial }{\partial x_{j}} \right) \\ &= \dfrac{\partial f}{\partial x_{i}} v_{j}\delta_{ij} \\ &= v_{i}\dfrac{\partial f}{\partial x_{i}} \end{align*}$$

Then, in this case, since $\mathbb{R}$’s dimension is $1$ anyway, the following holds true.

$$df_{p}(v) = \begin{bmatrix} v_{i} \dfrac{\partial f}{\partial x_{i}}\end{bmatrix} = v_{i}\dfrac{\partial f}{\partial x_{i}} = \omega_{p}(v)$$

Therefore, it can be understood that the $1$-form $\omega_{p}$ on $\mathbb{R}^{n}$ and the differential of the function defined on $\mathbb{R}^{n}$ $df_{p}$ are the same. This is the essence of representing the total differential $df$ of scalar functions in calculus.

$$df = \dfrac{\partial f}{\partial x}dx + \dfrac{\partial f}{\partial y}dy + \dfrac{\partial f}{\partial z}dz$$

Examples

If $f(x,y) = x^{2} + y^{2}$, then,

$$df = \dfrac{\partial f}{\partial x}dx + \dfrac{\partial f}{\partial y}dy = 2xdx + 2ydy$$

If $f(x,y) = e^{xy} + 3x$, then,

$$df = (ye^{xy} + 3)dx + xe^{xy}dy$$

See Also

• 2nd order differential form
• k-th order differential form