Differential forms, integration and the generalized Stokes theorem (part II)

Today let's see how to integrate differential forms. We begin with $\mathbb{R}^n$ and then move on to manifolds. The $n$-form $dx^1 \wedge \cdots \wedge dx^n$ is a basis for the space of linear $n$-forms on $\mathbb{R}^n$; therefore any differential $n$-form on $\mathbb{R}^n$ is given by $$\omega = f(x^1, \ldots, x^n) dx^1 \wedge \cdots \wedge dx^n$$ To recover the function $f$ from $\omega$ we can just evaluate it on the standard basis: $$f(p) = \omega_p(e_1, \ldots, e_n)$$ Therefore $n$-forms in $\mathbb{R}^n$ are in direct correspondence to scalar functions. Integrating $\omega$ is easy: we set $$\int_{\mathbb{R}^n} \omega = \int_{\mathbb{R}^n} f(x^1, \ldots, x^n) dx^1 \cdots dx^n$$ where the integral on the right hand side is the ordinary Riemann integral of the function $f$^[1]. The same definition holds for integrating $\omega$ over any domain $D \subseteq \mathbb{R}^n$ (some reasonably well-behaved subset, possibly with boundary).

Now let $M$ be an oriented $n$-dimensional manifold^[2]. For simplicity we'll assume given a (properly oriented) parametrization^[3] $\phi : D \to M$ of $M$. If $\omega$ is an $n$-form on $M$, we define $$\int_M \omega = \int_D \phi^* \omega$$ where $\phi^* \omega$ is the pullback form defined by, given tangent vectors $v_1, \ldots, v_n$ at the point $p \in D$^[4], $$(\phi^* \omega)_p(v_1, \ldots, v_n) = \omega_{\phi(p)}(D_p \phi \cdot v_1, \ldots, D_p \phi \cdot v_n)$$ This integral can be proven to be independent of the parametrization, provided it induces the given orientation on $M$.

Example: Consider a vector field $X$ on $\mathbb{R}^n$. We can re-interpret it as a differential $1$-form $X^\flat$ defined by $$X^\flat_p(v) = X(p) \cdot v$$ Exercise: Show that, if $X = (X^1(x^1, \ldots, x^n), \ldots, X^n(x^1, \ldots, x^n))$, then $$X^\flat = X^1 dx^1 + \cdots + X^n dx^n$$ Now suppose $\gamma : [0, 1] \to \mathbb{R}^n$ parametrizes a curve. We have $$\int_{\gamma} X^\flat = \int_{[0,1]} \gamma^* X^\flat$$ By definition, $$(\gamma^* X^\flat)_t(v) = X^\flat(\gamma'(t)v) = X(\gamma(t)) \cdot \gamma'(t)v$$ where $v$ is a tangent vector at $t \in [0, 1]$. Since this is just a number, $dt(v) = v$. The expression in coordinates is then $$\gamma^* X^\flat = X(\gamma(t)) \cdot \gamma'(t)\ dt$$ Therefore $$\int_\gamma X^\flat = \int_{0}^{1} X(\gamma(t)) \cdot \gamma'(t)\ dt$$ is the line integral of the vector field $X$ along the curve $\gamma$. Observe that, indeed, such line integrals get a flipped sign if you switch the direction of the parametrization of $\gamma$.

Example: Consider a vector field $X$ on $\mathbb{R}^3$. We can re-interpret it as a $2$-form $\omega$^[5] given by $$\omega_p(v, w) = \det(X(p), v, w)$$ Exercise: Show that, for $X = (X^1(x,y,z),X^2(x,y,z),X^3(x,y,z))$, $$\omega = X^1 dy \wedge dz - X^2 dx \wedge dz + X^3 dx \wedge dy$$ Suppose $\phi : D \to \mathbb{R}^3$, where $\phi = \phi(u,v)$ and $D \subseteq \mathbb{R}^2$, parametrizes a surface $\Sigma$ oriented by the normal $\hat{n}$, so that $$\frac{\partial \phi}{\partial u} \times \frac{\partial \phi}{\partial v} = \left| \frac{\partial \phi}{\partial u} \times \frac{\partial \phi}{\partial v} \right| \hat{n}$$ Taking $e_1 = (1, 0)$ and $e_2 = (0, 1)$ we find $$\begin{equation*} \begin{split} (\phi^* \omega)_p(e_1, e_2) & = \omega_{\phi(p)} \left( D_p \phi \cdot e_1, D_p \phi \cdot e_2 \right) \\ & = \omega_{\phi(p)} \left( \frac{\partial \phi}{\partial u}, \frac{\partial \phi}{\partial v} \right) \\ & = \det \left( X(\phi(p)), \frac{\partial \phi}{\partial u}, \frac{\partial \phi}{\partial v} \right) \\ & = X(\phi(p)) \cdot \left( \frac{\partial \phi}{\partial u} \times \frac{\partial \phi}{\partial v} \right) \\ & = X(\phi(p)) \cdot \hat{n} \left| \frac{\partial \phi}{\partial u} \times \frac{\partial \phi}{\partial v} \right| \end{split} \end{equation*}$$ Therefore $$\phi^* \omega = \left( X(\phi(p)) \cdot \hat{n} \right) \left| \frac{\partial \phi}{\partial u} \times \frac{\partial \phi}{\partial v} \right| du \wedge dv$$ and thus $$\int_\Sigma \omega = \int_D \left( X(\phi(p)) \cdot \hat{n} \right) \left| \frac{\partial \phi}{\partial u} \times \frac{\partial \phi}{\partial v} \right| du dv$$ is the flux of $X$ through $\Sigma$. Again, we would've gotten a flipped sign by choosing the opposite orientation $-\hat{n}$.

This previous example can be generalized to arbitrary dimensions: a vector field $X$ on $\mathbb{R}^n$ can be re-interpreted as the $(n-1)$-form $\omega$ given by $$\omega_p(v_1, \ldots, v_{n-1}) = \det(X(p), v_1, \ldots, v_{n-1})$$ In coordinates, $$\omega = X^1 dx^2 \wedge \cdots \wedge dx^n - X^2 dx^1 \wedge dx^3 \wedge \cdots \wedge dx^n + \cdots + (-1)^{n-1} X^n dx^1 \wedge dx^2 \wedge \cdots \wedge dx^{n-1}$$ Then the integral of $\omega$ over an oriented hypersurface $\Sigma \subset \mathbb{R}^n$ is equal to the flux of $X$ through $\Sigma$.

Next time we'll be ready to see how it all fits together into the generalized Stokes theorem.

Notes

[1] Of course this may or may not converge -- we typically integrate compactly supported forms only, i.e. those where $f$ is zero outside a compact set ↩

[2] If you don't know the precise definitions, don't worry -- it'll be clear in the examples. ↩

[3] More specifically, $\phi$ is a surjective map $D \to M$ that is injective and $C^1$ with invertible derivative within the interior of the domain $D \subseteq \mathbb{R}^n$, such that the boundary of $D$ maps to a set of $n$-dimensional volume equal to zero. (As an example, think of $D = [0, 2\pi] \times [0, \pi]$ and $\phi : D \to S^2$ the spherical coordinates parametrization). ↩

[4] Here $D_p \phi \cdot v_1$ is the directional derivative of $\phi$ at $p$ acting on the vector $v_1$ and so forth. ↩

[5] This is also known as $*X^b$, where $*$ is the Hodge star. ↩