Differential Forms

Differential forms result from the application of the tools of exterior algebra to the cotangent spaces of a smooth manifold.

This post introduces the basics of differential forms. Other aspects of differential forms will be discussed in different posts.

Introduction

For each point \(p \in M\) of a smooth manifold \(M\), the \(k\)-th exterior power of the cotangent space \(T^*_pM\) is denoted \(\Lambda^k(T^*_pM)\), which, according to the concrete definition of the \(k\)-th exterior power, is the space of all alternating, covariant \(k\)- tensors on \(T^*_pM\). These vector spaces can be collected into a smooth vector bundle \(\Lambda^k(T^*M)\) over \(M\):

\[\Lambda^k(T^*M) = \bigsqcup_{p \in M}\Lambda^k(T^*_pM).\]

A section of the vector bundle \(\Lambda^kT^*M\) is called a differential \(\mathbf{k}\)-form (or a \(\mathbf{k}\)-form). The integer \(k\) is called the degree of the differential form. The vector space of all smooth \(k\)-forms is denoted \(\Omega^k(M)\):

\[\Omega^k(M) = \Gamma(\Lambda^k(T^*M)).\]

The wedge product operation can be defined point-wise for differential forms. Given a \(k\)-form \(\omega\) and an \(l\)-form \(\eta\), the wedge product \(\omega \wedge \eta\) is the \((k+l)\)-form defined such that, at each point \(p \in M\),

\[(\omega \wedge \eta)_p = \omega_p \wedge \eta_p.\]

A \(0\)-form is a \(0\)-tensor field, i.e. a smooth function \(f : M \rightarrow \mathbb{R}\), and, by convention, the wedge product of a \(0\)-form with a \(k\)-form \(\omega\) is the \(k\)-form \(f \wedge \omega = f\omega\), i.e. \((f \wedge \omega)_p = f(p)\omega_p\).

Differential Forms in Coordinates

From the general theory of exterior algebra, we know how to express \(k\)-covectors in terms of a basis. Thus, given a smooth chart on the manifold \(M\), it follows that every \(k\)-form \(\omega\) can be written in local coordinates in terms of the respective coordinate frame and dual frame as

\[\omega = \omega_I dx^{I}\]

where \(I \in \mathcal{I}_k\), \(\mathcal{I}_k\) is the set of all increasing multi-indices of length \(k\), the summation convention is implied, and \(dx^{I}\) is an abbreviation defined such that

\[dx^{I} = dx^{i_1} \wedge \dots \wedge dx^{i_k},\]

and the coefficients \(\omega_I : M \rightarrow \mathbb{R}\) are component functions defined such that

\[\omega_I = \omega\left(\frac{\partial}{\partial x^{i_1}},\dots,\frac{\partial}{\partial x^{i_k}}\right),\]

that is,

\[\omega_I(p) = \omega_p\left(\frac{\partial}{\partial x^{i_1}}\bigg\rvert_p,\dots,\frac{\partial}{\partial x^{i_k}}\bigg\rvert_p\right)\]

for all \(p \in M\).

A Few Examples

Consider a few examples on Euclidean spaces \(\mathbb{R}^n\):

• \(1\)-form on \(\mathbb{R}^2\): \(\frac{1}{\sqrt{x^2 + y^2}}(-y ~dx + x ~dy)\); this form can be used to compute line integrals over circles (see the article on line and surface integrals for a discussion).
• \(2\)-form on \(\mathbb{R}^3\): \(\frac{x}{r} ~dy \wedge dz + \frac{y}{r} ~dz \wedge dx + \frac{z}{r} ~dx \wedge dy\); this form can be used to compute surface integrals over spheres (see the article on line and surface integrals for a discussion).

Pullbacks

The pullback operation on differential forms is one of the most important operations on differential forms. The pullback of a differential form is defined the same way as the pullback of any tensor field. Given a smooth map \(F : M \rightarrow N\) between smooth manifold \(M\) and \(N\) and a differential \(k\)-form on \(N\), the pullback \(F^*\omega\) is a differential form on \(M\) defined as follows for any point \(p\) and tangent vectors \((v_1,\dots,v_k)\) in \(T_pM\):

\[(F^*\omega)_p(v_1,\dots,v_k) = \omega_{F(p)}\left(dF_p(v_1),\dots,dF_p(v_k)\right).\]

Preservation under Pullback

Many operations on differential forms are preserved under pullback.

Wedge Product

The wedge product is preserved under pullback. Given smooth manifolds \(M\) and \(N\), a \(k\)-form \(\omega \in \Omega^k(N)\) and an \(l\)-form \(\eta \in \Omega^l(N)\) and a smooth map \(F : M \rightarrow N\), we compute, for any point \(p \in M\) and tangent vectors \(v_1,\dots,v_k,v_{k+1},\dots,v_{k+l} \in T_pM\), the following:

\begin{align}\left(F^*(\omega \wedge \eta)\right)_p(v_1,\dots,v_k,v_{k+1},\dots,v_{k+l}) &= (\omega \wedge \eta)_{F(p)}\left(dF_p(v_1),\dots,dF_p(v_k),dF_p(v_{k+1}),\dots,dF_p(v_{k+l})\right) \\&= (\omega_{F(p)} \wedge \eta_{F(p)})\left(dF_p(v_1),\dots,dF_p(v_k),dF_p(v_{k+1}),\dots,dF_p(v_{k+l})\right) \\&= \frac{(k+l)!}{k!l!}\mathrm{Alt}\left(\omega_{F(p)} \otimes \eta_{F(p)}\right)\left(dF_p(v_1),\dots,dF_p(v_k),dF_p(v_{k+1}),\dots,dF_p(v_{k+l})\right) \\&= \frac{1}{k!l!}\sum_{\sigma \in S_{k+l}}(\mathrm{sgn} ~\sigma)\omega_{F(p)}\left(dF_p(v_{\sigma(1)}),\dots,dF_p(v_{\sigma(k)})\right)\eta_{F(p)}\left(dF_p(v_{\sigma(k+1)}),\dots,dF_p(v_{\sigma(k+l)})\right) \\&= \frac{1}{k!l!}\sum_{\sigma \in S_{k+l}}(\mathrm{sgn} ~\sigma)(F^* \omega)_p\left(v_{\sigma(1)},\dots,v_{\sigma(k)}\right)(F^* \eta)_p\left(v_{\sigma(k+1)},\dots,v_{\sigma(k+l)}\right) \\&= \frac{1}{k!l!}\sum_{\sigma \in S_{k+l}}(\mathrm{sgn} ~\sigma)(F^* \omega)_p \otimes (F^* \eta)_p\left(v_{\sigma(1)},\dots,v_{\sigma(k)},v_{\sigma(k+1)},\dots,v_{\sigma(k+l)}\right) \\&= \frac{(k+l)!}{k!l!}\mathrm{Alt}\left((F^* \omega)_p \otimes (F^* \eta)_p\right)\left(v_1,\dots,v_k,v_{k+1},\dots,v_{k+l}\right) \\&= (F^* \omega)_p \wedge (F^* \eta)_p\left(v_1,\dots,v_k,v_{k+1},\dots,v_{k+l}\right) \\&= ((F^* \omega) \wedge (F^* \eta))_p\left(v_1,\dots,v_k,v_{k+1},\dots,v_{k+l}\right) .\end{align}

Thus, it follows that

\[F^*(\omega \wedge \eta) = (F^* \omega) \wedge (F^* \eta).\]

Function Multiplication

Next, we'll establish a few general facts about the behavior of operations under pullback which are true for pullbacks of any tensor fields, not just pullbacks of differential forms.

First, note that, for any smooth manifolds \(M\) and \(N\), function \(f : N \rightarrow \mathbb{R}\), smooth map \(F : M \rightarrow N\), and differential form \(\omega \in \Omega^k(N)\),

\begin{align}F^*(f\omega)_p(v_1,\dots,v_k) &= (f\omega)_{F(p)}\left(dF_p(v_1),\dots,dF_p(v_k)\right) \\&= f(F(p))\omega_{F(p)}\left(dF_p(v_1),\dots,dF_p(v_k)\right) \\&= (f \circ F)(p)\omega_{F(p)}\left(dF_p(v_1),\dots,dF_p(v_k)\right) \\&= (f \circ F)(p)(F^* \omega)_p(v_1,\dots,v_k).\end{align}

Thus, it follows that

\[F^*(f\omega) = (f \circ F)F^* \omega.\]

Addition

Likewise

\begin{align}F^*(\omega + \eta)_p(v_1,\dots,v_k) &= (\omega + \eta)_{F(p)}\left(dF_p(v_1),\dots,dF_p(v_k)\right) \\&= (\omega_{F(p)} + \eta_{F(p)})\left(dF_p(v_1),\dots,dF_p(v_k)\right) \\&= \omega_{F(p)}\left(dF_p(v_1),\dots,dF_p(v_k)\right) + \eta_{F(p)})\left(dF_p(v_1),\dots,dF_p(v_k)\right).\end{align}

Thus, it follows that

\[F^*(\omega + \eta) = F^* \omega + F^* \eta.\]

Scalar Multiplication

Similarly, for any \(a \in \mathbb{R}\),

\begin{align}F^*(a\omega)(v_1,\dots,v_k) &= (a\omega)_{F(p)}\left(dF_p(v_1),\dots,dF_p(v_k)\right) \\&= a \omega_{F(p)}\left(dF_p(v_1),\dots,dF_p(v_k)\right) \\&= a F^* \omega (v_1,\dots,v_k).\end{align}

Thus, it follows that

\[F^*(a\omega) = aF^*\omega.\]

The previous two facts establish that \(F^* : \Omega^k(N) \rightarrow \Omega^k(M)\) is linear over \(\mathbb{R}\).

Differential Forms in Coordinates under Pullback

Also note that

\begin{align}(F^*(df))_p(v) &= df_{F(p)}\left(dF_p(v)\right) \\&= d(f \circ F)_p(v).\end{align}

Thus, it follows that

\[F^*(df) = d(f \circ F).\]

Putting all these previous facts together, we can establish the behavior of the coordinate representations of differential forms under pullback. In any smooth chart with coordinate functions \((y^{i_1},\dots,y^{i_k})\), we compute

\begin{align}F^*\left(\sum_{I \in \mathcal{I}_k}\omega_I dy^{i_1} \wedge \dots \wedge dy^{i_k}\right) &= \sum_{I \in \mathcal{I}_k}F^*\left(\omega_I dy^{i_1} \wedge \dots \wedge dy^{i_k}\right) & \text{(Addition under Pullback)} \\&= \sum_{I \in \mathcal{I}_k}(\omega_I \circ F)F^*\left(dy^{i_1} \wedge \dots \wedge dy^{i_k}\right) & \text{(Function Multiplication under Pullback)} \\&= \sum_{I \in \mathcal{I}_k}(\omega_I \circ F)F^*\left(dy^{i_1}\right) \wedge \dots \wedge F^*\left(dy^{i_k}\right) & \text{(Wedge Product under Pullback)} \\&= \sum_{I \in \mathcal{I}_k}(\omega_I \circ F)d\left(y^{i_1} \circ F\right) \wedge \dots \wedge d\left(y^{i_k} \circ F\right) & \text{(Differentials under Pullback)}.\end{align}

This, and the other preceding facts, indicate that pullbacks are computed by substitution.

Example

Consider an example of the pullback of a differential form in coordinates. The circle of radius \(r \in \mathbb{R}\) can be defined as a one-dimensional implicit surface as follows:

\[S^1 = \{(x,y) \mid x^2 + y^2 = r\}.\]

The circle is a 1-dimensional manifold which is a sub-manifold of \(\mathbb{R}^2\). The circle can be parameterized using polar coordinates (with the value of \(r\) constant) by the map \(F : \mathbb{R} \rightarrow S^1\):

\[F(\theta) = (r \cos \theta, r \sin \theta).\]

Consider the following differential \(1\)-form on \(\mathbb{R}^2\):

\[\omega = \frac{-y~dx + x~dy}{\sqrt{x^2+y^2}}.\]

Since \(S^1\) is a sub-manifold of \(\mathbb{R}^2\), this differential form is also defined when restricted to \(S^1\).

We can then compute the pullback as follows:

\begin{align}F^*\omega &= F^*\left(\frac{-y~dx + x~dy}{\sqrt{x^2+y^2}}\right) \\&= \frac{1}{\sqrt{r^2 \cos^2 \theta + r^2 \sin^2 \theta}}F^*(-y~dx + x~dy) & \text{(Function Multiplication under Pullback)} \\&= \frac{1}{\sqrt{r^2 \cos^2 \theta + r^2 \sin^2 \theta}}(F^*(-y~dx) + F^*(x~dy)) & \text{(Addition under Pullback)} \\&= \frac{1}{\sqrt{r^2 \cos^2 \theta + r^2 \sin^2 \theta}}((-r~ \sin \theta)F^*(dx) + (r~ \cos \theta)F^*(dy)) & \text{(Function Multiplication under Pullback)} \\&= \frac{1}{\sqrt{r^2 \cos^2 \theta + r^2 \sin^2 \theta}}((-r~ \sin \theta)d(r~ \cos \theta) + (r~ \cos \theta)d(r~ \sin \theta)) & \text{(Differentials under Pullback)} \\&= \frac{1}{\sqrt{r^2 \cos^2 \theta + r^2 \sin^2 \theta}}((-r~ \sin \theta)(-r~ \sin \theta) + (r~ \cos \theta)(r~ \cos \theta)) \\&= r.\end{align}

In other words, pullbacks are computed by substitution. In this example, this amounts to substituting \(r \cos \theta\) for \(x\), \(r \sin \theta\) for \(y\) and then simplifying.

Example

For another example in polar coordinates, consider the following differential \(2\)-form on \(\mathbb{R}^2\):

\[\omega = dx \wedge dy.\]

Using the map

\[F(r,\theta) = (r \cos \theta, r \sin \theta),\]

we compute the pullback as follows:

\begin{align}F^*(\omega) &= F^*(dx \wedge dy) \\&= d(r \cos \theta) \wedge d(r \sin \theta) \\&= (\cos \theta ~dr - r \sin \theta ~d\theta) \wedge (\sin \theta ~dr + r \cos \theta ~d\theta) \\&= \cos \theta \sin \theta ~dr \wedge dr - r \sin^2 \theta ~d\theta \wedge dr + r \cos^2 \theta ~dr \wedge d\theta - r^2 \sin \theta \cos \theta ~d\theta \wedge d \theta \\&= r \sin^2 \theta ~dr \wedge d\theta + r \cos^2 \theta ~dr \wedge d\theta \\&= r(\sin^2 \theta + \cos^2 \theta)~dr \wedge d\theta\\&= r ~dr \wedge d\theta.\end{align}

Example

Now, consider an example in spherical coordinates.

The sphere with radius \(r\) in \(\mathbb{R}^3\) can be defined as an implicit surface:

\[S^2 = \{(x, y, z) \mid x^2 + y^2 + z^2 = r\}.\]

We use the parametric curve \(F : [0,\pi] \times [0,2\pi] \rightarrow S^2\) defined using spherical coordinates as follows:

\[F(\varphi, \theta) = (r\sin \varphi \cos \theta, r\sin \varphi \sin \theta, r\cos \varphi).\]

We define the following differential form:

\[\omega = \frac{x}{r}dy \wedge dz + \frac{y}{r} dz \wedge dx + \frac{z}{r} dx \wedge dy.\]

We first compute the following by substitution:

• \(F^*dx = d(r\sin \varphi \cos \theta) = r\cos \varphi \cos \theta ~d\varphi - r\sin \varphi \sin \theta ~d\theta,\)
• \(F^*dy = d(r\sin \varphi \sin \theta) = r\cos \varphi \sin \theta ~d\varphi + r\sin \varphi \cos \theta ~d\theta,\)
• \(F^*dz = d(r\cos \varphi) = -r\sin \varphi ~d\varphi.\)

Next, we compute each of the wedge products:

• \(F^*(dy \wedge dz) = F^*(dy) \wedge F^*(dz) = -r^2\sin^2 \varphi \cos \theta ~d\theta \wedge d\varphi,\)
• \(F^*(dz \wedge dx) = F^*(dz) \wedge F^*(dx) = r^2\sin^2 \varphi \sin \theta ~d\varphi \wedge d\theta,\)
• \(F^*(dx \wedge dy) = F^*(dx) \wedge F^*(dy) = r^2\cos \varphi \sin \varphi \cos^2 \theta ~d\varphi \wedge d\theta - r^2\sin \varphi \cos \varphi \sin^2 \theta ~d\theta \wedge d\varphi.\)

We then normalize the wedge products, using \(d\varphi \wedge d\theta\) consistently, and exploiting the fact that \(d\theta \wedge d\varphi = - d\varphi \wedge d\theta\):

• \(F^*(dy \wedge dz) = r^2\sin^2 \varphi \cos \theta ~d\varphi \wedge d\theta\)
• \(F^*(dz \wedge dx) = F^*(dz) \wedge F^*(dx) = r^2\sin^2 \varphi \sin \theta ~d\varphi \wedge d\theta,\)
• \(F^*(dx \wedge dy) = F^*(dx) \wedge F^*(dy) = r^2\cos \varphi \sin \varphi \cos^2 \theta ~d\varphi \wedge d\theta + r^2\sin \varphi \cos \varphi \sin^2 \theta ~d\varphi \wedge d\theta.\)

Next, we pullback the components of the differential form:

• \(F^*((x/r) dy \wedge dz) = r\sin^3 \varphi \cos^2 \theta ~d\varphi \wedge d\theta\)
• \(F^*((y/r) dz \wedge dx) = r\sin^3 \varphi \sin^2 \theta ~d\varphi \wedge d\theta,\)
• \(F^*((z/r) dx \wedge dy) = r\cos^2 \varphi \sin \varphi \cos^2 \theta ~d\varphi \wedge d\theta + r\sin \varphi \cos^2 \varphi \sin^2 \theta ~d\varphi \wedge d\theta.\)

Finally, we add and group using trigonometric identities to simplify:

\begin{align}F^*\omega &= (r\sin^3 \varphi \cos^2 \theta + r\sin^3 \varphi \sin^2 \theta + r\cos^2 \varphi \sin \varphi \cos^2 \theta + r\sin \varphi \cos^2 \varphi \sin^2 \theta)~d\varphi \wedge d\theta \\&= r\sin^3 \varphi (\cos^2 \theta + \sin^2 \theta) + r\sin \varphi \cos^2 \theta (\cos^2 \theta + \sin^2 \theta) \\&= r\sin^3 \varphi + r\sin \varphi \cos^2 \theta \\&= r\sin \varphi (\cos^2 \varphi + \sin^2 \varphi) \\&= r \sin \varphi.\end{align}

Example

Now consider the example of a generic paramterization, which highlights a particular pattern in pullbacks of differential forms. Suppose we have a parameterization of a \(2\)-dimensional surface embedded in \(\mathbb{R}^3\):

\[F(s,t) = (x(s,t), y(s,t), z(s,t)).\]

The formula for a covector field in coordinates indicates the following:

\[dx = \frac{\partial x}{\partial s}ds + \frac{\partial x}{\partial t}dt\]

\[dy = \frac{\partial y}{\partial s}ds + \frac{\partial y}{\partial t}dt\]

\[dz = \frac{\partial z}{\partial s}ds + \frac{\partial z}{\partial t}dt\]

Then, under pullback:

\begin{align}F^*(dx \wedge dy) &= \left(\frac{\partial x}{\partial s}ds + \frac{\partial x}{\partial t}dt\right) \wedge \left(\frac{\partial y}{\partial s}ds + \frac{\partial y}{\partial t}dt\right)\\&= \left(\frac{\partial x}{\partial s} \frac{\partial y}{\partial t} - \frac{\partial x}{\partial t} \frac{\partial y}{\partial s}\right)ds \wedge dt \\&= \frac{\partial(x,y)}{\partial(s,t)}ds \wedge dt\end{align}

\begin{align}F^*(dy \wedge dz) &= \left(\frac{\partial y}{\partial s}ds + \frac{\partial y}{\partial t}dt\right) \wedge \left(\frac{\partial z}{\partial s}ds + \frac{\partial z}{\partial t}dt\right)\\&= \left(\frac{\partial y}{\partial s} \frac{\partial z}{\partial t} - \frac{\partial y}{\partial t} \frac{\partial z}{\partial s}\right)ds \wedge dt \\&= \frac{\partial(y,z)}{\partial(s,t)}ds \wedge dt\end{align}

\begin{align}F^*(dz \wedge dx) &= \left(\frac{\partial z}{\partial s}ds + \frac{\partial z}{\partial t}dt\right) \wedge \left(\frac{\partial x}{\partial s}ds + \frac{\partial x}{\partial t}dt\right)\\&= \left(\frac{\partial z}{\partial s} \frac{\partial x}{\partial t} - \frac{\partial z}{\partial t} \frac{\partial x}{\partial s}\right)ds \wedge dt \\&= \frac{\partial(z,x)}{\partial(s,t)}ds \wedge dt\end{align}

Thus, the differential form

\[f_x ~ dy \wedge dz + f_y ~ dz \wedge dx + f_z ~ dx \wedge dy\]

transforms as

\begin{align}F^*(f_x ~ dy \wedge dz + f_y ~ dz \wedge dx + f_z ~ dx \wedge dy) &= F^*(f_x ~ dy \wedge dz) \\&+ F^*(f_y ~ dz \wedge dx) \\&+ F^*(f_z ~ dx \wedge dy) \\&= (f_x \circ F)(\frac{\partial(y,z)}{\partial(s,t)}ds \wedge dt) \\&+ (f_y \circ F)(\frac{\partial(z,x)}{\partial(s,t)}ds \wedge dt) \\&+ (f_z \circ F)(\frac{\partial(x,y)}{\partial(s,t)}ds \wedge dt).\end{align}

In every case, the pullbacks of the top-level forms had a Jacobian determinant as their coefficient. This is a general pattern of differential forms: the pullback of a top-level form has a coefficient of the determinant of the respective Jacobian matrix.

Next, let's prove this as a general fact.

Pullbacks of Top-Degree Forms

Consider any smooth map \(F : M \rightarrow N\) between smooth manifolds \(M\) and \(N\) of dimension \(n\) together with smooth charts \((U, (x^i))\) and \((V, (y^j))\) on \(M\) and \(N\) respectively. Then, for any continuous function \(u : V \rightarrow \mathbb{R}\), we compute (on \(U \cap F^{-1}(V)\)) the following:

\begin{align}F^*(u ~dy^1 \wedge \dots \wedge dy^n)\left(\frac{\partial}{\partial x^1},\dots,\frac{\partial}{\partial x^n}\right) &= (u \circ F)dF^1 \wedge \dots \wedge dF^n\left(\frac{\partial}{\partial x^1},\dots,\frac{\partial}{\partial x^n}\right) & \text{(Pullback in Coordinates)} \\&= (u \circ F)\mathrm{det}\left(dF^j\left(\frac{\partial}{\partial x^i}\right)\right) & \text{(Wedge Product of Covector Fields)} \\&= (u \circ F)\mathrm{det}\left(\frac{\partial F^j}{\partial x^i}\right) \\&= (u \circ F)\mathrm{det}DF \\&= (u \circ F)\mathrm{det}DF ~dx^1 \wedge \dots \wedge dx^n\left(\frac{\partial}{\partial x^1},\dots,\frac{\partial}{\partial x^n}\right) & \text{(Since \(dx^1 \wedge \dots \wedge dx^n\left(\frac{\partial}{\partial x^1},\dots,\frac{\partial}{\partial x^n}\right) = 1\))}.\end{align}

This computation used several facts about the wedge product detailed in the post on exterior algebra.

Since differential forms are completely determined by their action on the coordinate basis vector fields (in analogy to how tensors are completely determined by their actions on basis vectors), this suffices to demonstrate the following equality:

\[F^*(u ~dy^1 \wedge \dots \wedge dy^n) = (u \circ F)\mathrm{det}DF ~dx^1 \wedge \dots \wedge dx^n.\]

Thus, when pulling back a differential form expressed in local coordinates, it transforms by a factor of the determinant of the Jacobian matrix of the coordinate representation of the differential \(dF\) of the map \(F\).