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

The last ingredient we need for the generalized Stokes theorem is the exterior derivative of a differential form. Consider a $k$-form $\omega$ in $\mathbb{R}^n$. Its exterior derivative $d\omega$ will be a $(k+1)$-form, i.e. a function of a point $p$ and $k+1$ vectors $v_1, \ldots, v_{k+1}$, which we can think of as forming the parallelogram $$P_p(v_1, \ldots, v_{k+1}) = \{p + t_1 v_1 + \cdots + t_{k+1} v_{k+1}\ |\ 0 \leq t_i \leq 1, \ i = 1, \ldots, k+1\}$$ By this point it should not be particularly surprising to see the definition: $$d\omega_p(v_1, \ldots, v_{k+1}) = \lim_{h \to 0} \frac{1}{h^{k+1}} \int_{\partial P_p(hv_1, \ldots, hv_{k+1})} \omega$$ Observe that the boundary of the $(k+1)$-parallelogram $P_p(v_1, \ldots, v_{k+1})$ is a union of $k$-dimensional parallelograms, and so it makes sense to integrate the $k$-form $\omega$ over it. Furthermore note that, if $v_1, \ldots, v_{k+1}$ is an orthonormal set, then $$h^{k+1} = \text{vol}(P_p(hv_1, \ldots, hv_{k+1}))$$ Example: A function $f : \mathbb{R} \to \mathbb{R}$ is a $0$-form. Taking $e_1 = 1$, we have $P_p(he_1) = [p, p+h]$ and so $$df_p(e_1) = \lim_{h \to 0} \frac{1}{h} \int_{\partial [p, p+h]} f = f'(p)$$ Therefore $$df = f'(x) dx$$ Exercise: Show that, for a function $f : \mathbb{R}^n \to \mathbb{R}$, thought of as a $0$-form like in the previous example, one has $$df = \frac{\partial f}{\partial x^1} dx^1 + \cdots + \frac{\partial f}{\partial x^n} dx^n$$ Example: Consider a vector field $V$ on the plane $\mathbb{R}^2$, given by $V(x, y) = (V^1(x,y), V^2(x,y))$. Recall that it can be re-interpreted as the $1$-form $$V^\flat = V^1 dx + V^2 dy$$ The exterior derivative $dV^\flat$, being a $2$-form in $\mathbb{R}^2$, is a function times $dx \wedge dy$. To recover this function, we evaluate on the standard basis: $$dV^\flat_p(e_1, e_2) = \lim_{h \to 0} \frac{1}{h^2} \int_{\partial P_p(he_1, he_2)} V^\flat$$ Note that $P_p(he_1, he_2)$ is a shape with piecewise differentiable boundary that shrinks down to $p$ as $h \to 0$, and $h^2$ is its volume. Finally, the integral of the $1$-form $V^\flat$ is the line integral of the vector field $V$ along the boundary. We can conclude that $$dV^\flat_p(e_1, e_2) = (\text{curl}\ V)(p) = \frac{\partial V^2}{\partial x}(p) - \frac{\partial V^1}{\partial y}(p)$$ Thus $$dV^\flat = \left( \frac{\partial V^2}{\partial x} - \frac{\partial V^1}{\partial y} \right) dx \wedge dy$$ Example: Let's see how this previous example plays out in $\mathbb{R}^3$. Again, a vector field $V(x,y,z)$ given by $$V(x,y,z) = (V^1(x,y,z), V^2(x,y,z), V^3(x,y,z))$$ can be promoted to the $1$-form $$V^\flat = V^1 dx + V^2 dy + V^3 dz$$ The same reasoning as the previous example will show that $$dV^\flat_p(e_1, e_2) = (\text{curl}\ V)_z(p) = \frac{\partial V^2}{\partial x}(p) - \frac{\partial V^1}{\partial y}(p)$$ since $e_1, e_2$ span a counter-clockwise oriented parallelogram in the plane perpendicular to $\hat{z}$. Similarly, $$dV^\flat_p(e_2, e_3) = (\text{curl}\ V)_x(p) = \frac{\partial V^3}{\partial y}(p) - \frac{\partial V^2}{\partial z}(p)$$ and (note we choose $e_3, e_1$ and not $e_1, e_3$ to achieve the correct $\hat{y}$ normal) $$dV^\flat_p(e_3, e_1) = (\text{curl}\ V)_y(p) = \frac{\partial V^1}{\partial z}(p) - \frac{\partial V^3}{\partial x}(p)$$ We can thus conclude that $$dV^\flat = \left( \frac{\partial V^2}{\partial x} - \frac{\partial V^1}{\partial y} \right) dx \wedge dy - \left( \frac{\partial V^1}{\partial z} - \frac{\partial V^3}{\partial x} \right) dx \wedge dz + \left( \frac{\partial V^1}{\partial z} - \frac{\partial V^3}{\partial x} \right) dy \wedge dz$$ The minus sign is because we wrote $dx \wedge dz$ instead of $dz \wedge dx$.

Exercise: Guess, or work out, the formula for $dV^\flat$ in $n$ dimensions, where $V^\flat = V^1 dx^1 + \cdots + V^n dx^n$.

Example: Consider a vector field $V(x,y,z) = (V^1(x,y,z), V^2(x,y,z), V^3(x,y,z))$ in $\mathbb{R}^3$. We also re-interpreted it as the $2$-form $$\omega = V^1 dy \wedge dz - V^2 dx \wedge dz + V^3 dx \wedge dy$$ whose integral over a surface equals the flux of $V$ through it. We can figure out the $3$-form $dw$ by evaluating on the standard basis: $$d\omega_p(e_1, e_2, e_3) = \lim_{h \to 0} \frac{1}{h^3} \int_{\partial P_p(he_1, he_2, he_3)} \omega$$ As before, $P_p(he_1, he_2, he_3)$ shrinks down to $p$ as $h \to 0$, and $h^3$ is its volume. The integral of $\omega$ is the flux of $V$ through the boundary, and hence $$d\omega_p(e_1, e_2, e_3) = (\text{div}\ V)(p) = \frac{\partial V^1}{\partial x}(p) + \frac{\partial V^2}{\partial y}(p) + \frac{\partial V^3}{\partial z}(p)$$ Therefore $$d\omega = \left( \frac{\partial V^1}{\partial x} + \frac{\partial V^2}{\partial y} + \frac{\partial V^3}{\partial z} \right) dx \wedge dy \wedge dz$$ Exercise: Generalize the previous example to arbitrary dimensions.

The fundamental tool for computing exterior derivatives is the following fact:

Proposition: For a $C^2$ function $f : \mathbb{R}^n \to \mathbb{R}$, $$d(f dx^{i_1} \wedge \cdots \wedge dx^{i_k}) = df \wedge dx^{i_1} \wedge \cdots \wedge dx^{i_k}$$ Proof: This can be computed directly by using the Taylor approximation as we did previously. You can find the proof in appendix A22 of Hubbard & Hubbard. $\square$

This allows us to compute any exterior derivative symbolically. For instance, $$d(x^2 dy + yz dx) = 2x dx \wedge dy + (z dy + y dz) \wedge dx = (2x - z) dx \wedge dy - y dx \wedge dz$$ Exercise: Go back and compute the exterior derivatives in the examples symbolically as above.

We have defined the exterior derivative of a differential form on $\mathbb{R}^n$. For a differential form on a general manifold $M$, we can compute its exterior derivative at any point by using local coordinates and the technique as shown above. It's possible to show this doesn't depend on the choice of coordinates (see for instance Lee's Introduction to Smooth Manifolds), and so the exterior derivative of forms on manifolds is equally well-defined.

Without further ado, we can now admire the austere, glacial beauty of the generalized Stokes theorem:

Theorem (Stokes): Let $M$ be a $k$-dimensional oriented manifold (possibly with boundary), and $\omega$ a $(k-1)$-form (of class $C^2$) on $M$. Then $$\int_{M} d\omega = \int_{\partial M} \omega$$ Examples: By taking $\omega$ to be the $0$-form $f : \mathbb{R} \to \mathbb{R}$ and $M = [a, b]$, this is the fundamental theorem of calculus. If we take $\omega = V^1 dx + V^2 dy$ on the plane, we get Green's theorem, and similarly if $$\omega = V^1 dx + V^2 dy + V^3 dz$$ in $\mathbb{R}^3$ we get the classical Stokes theorem. Finally, by taking $$\omega = V^1 dy \wedge dz - V^2 dx \wedge dz + V^3 dx \wedge dy$$ we obtain Gauss's theorem.

The heuristic idea for Stokes's theorem is the same as it was each time: by subdividing $M$ into many small parallelograms, each integral of $d\omega$ is approximated as the integral of $\omega$ over the boundary, and such integrals cancel out except for sides belonging to $\partial M$. You can find a rigorous proof along these lines as appendix A23 in Hubbard & Hubbard. However, there is a much easier (albeit much less intuitive) proof using a trick with partitions of unity; see, for instance, Bott & Tu's Differential Forms in Algebraic Topology.

Exercise: Convince yourself that $d(d\omega) = 0$ by looking at the following picture:

This fact is the beginning of the theory of de Rham cohomology; but, for now, I've said enough!