21Differential Forms and Stokes’ Theorem

Proof. Generating. Let $\alpha \in \Lambda^kE^*$ and $\beta = \sum_I\alpha(e_I)\,e_I^*$, where $\alpha(e_I)$ abbreviates $\alpha(e_{i_1}, \dots, e_{i_k})$. Both sides are $k$-linear and alternating, so they agree as soon as they agree on all $k$-tuples $(e_{j_1}, \dots, e_{j_k})$ with $j_1 < \dots < j_k$ (multilinearity reduces to tuples of basis vectors, alternation to strictly increasing ones). And $e_I^*(e_{j_1}, \dots, e_{j_k}) = \det(e_{i_r}^*(e_{j_s})) = \delta_{IJ}$: for $I = J$ the matrix is the identity; for $I \neq J$ some row is zero. So $\beta(e_J) = \alpha(e_J)$ for all $J$: $\beta = \alpha$. Freeness. If $\sum_I c_Ie_I^* = 0$, evaluating on $(e_{j_1}, \dots, e_{j_k})$ gives $c_J = 0$. ∎

Proof. Both properties are checked on basis elements and extended by bilinearity. Associativity: both parenthesisings of $e_I^* \wedge e_J^* \wedge e_K^*$ equal the wedge of the concatenated family of $1$-forms, by the determinant formula of Definition 21.3 (Laplace expansion by blocks). The sign rule: moving each of the $\ell$ factors of $\beta$ past the $k$ factors of $\alpha$ costs a sign per adjacent transposition (a swap of two rows of the determinant), hence $(-1)^{k\ell}$ in total. ∎

Proof. The functorial identities are immediate from the definitions (for the product rule, check on wedges of $1$-forms with the determinant formula — $\det(\ell_i(uv_j)) = \det((u^*\ell_i)(v_j))$ — then extend bilinearly). (1) $u^*$ maps the one-dimensional $\Lambda^nE^*$ (Proposition 21.4) to itself, so $u^*\alpha = c\,\alpha$ with $c$ independent of $\alpha \neq 0$; testing on $\alpha = e_1^* \wedge \dots \wedge e_n^*$ and $(v_j) = (e_j)$ gives $c = \det(e_i^*(ue_j)) = \det u$. (2) For $v_1, \dots, v_k \in E$, the vectors $u(v_1), \dots, u(v_k)$ lie in the image of $u$, of dimension $< k$: they are linearly dependent, and an alternating form vanishes on a dependent family (expand the dependent vector along the others). ∎

Proof. (a) By bilinearity it suffices to treat $\omega = a\,\dd x_I$, $\eta = b\,\dd x_J$. Then $\omega \wedge \eta = ab\,\dd x_I \wedge \dd x_J$ and

and moving the $1$-form $\dd b$ past the $k$ factors of $\dd x_I$ costs $(-1)^k$ (Definition 21.5): the second term is $(-1)^k\,\omega \wedge \dd\eta$. (b) For a function: $\dd(\dd a) = \sum_{i,j} \frac{\partial^2a}{\partial x_j\partial x_i}\,\dd x_j \wedge \dd x_i$. The coefficient is symmetric in $(i,j)$ by Schwarz’s theorem on mixed partials (proved in the Year 2 volume; $a$ is $\mathcal C^\infty$), while $\dd x_j \wedge \dd x_i$ is antisymmetric: pairing the terms $(i,j)$ and $(j,i)$, everything cancels. For a general $\omega = \sum a_I\dd x_I$: $\dd\dd\omega = \sum\dd(\dd a_I \wedge \dd x_I) = \sum(\dd\dd a_I\wedge\dd x_I - \dd a_I\wedge\dd(\dd x_I))$ by (a), and both terms vanish ($\dd(\dd x_I) = 0$ since the coefficient is constant). ∎

Proof. (a) Pointwise statements about linear pullbacks (Proposition 21.6), plus the chain rule $D(\psi\circ\varphi)(x) = D\psi(\varphi(x))\,D\varphi(x)$. (b) For a $0$-form $f$: $\varphi^*(\dd f) = \dd f \circ D\varphi = \dd(f \circ \varphi)$ is the chain rule. For $\omega = a\,\dd y_I$: using (a), $\varphi^*\omega = (a\circ\varphi)\,\dd\varphi_{i_1}\wedge\dots\wedge \dd\varphi_{i_k}$, so by the Leibniz rule (Theorem 21.9(a)) and $\dd\dd\varphi_{i_r} = 0$:

(c) Pointwise this is exactly $u^*\alpha = (\det u)\alpha$ on top-degree forms (Proposition 21.6(1) with $u = D\varphi(x)$). ∎

Proof. We build a linear homotopy operator $h\colon \Omega^k(U) \to \Omega^{k-1}(U)$ with

if $\dd\omega = 0$, then $\omega = \dd(h\omega)$ and we are done. For $\omega = \sum_I a_I\,\dd x_I$ set

(the hat deletes a factor; the integrals are smooth in $x$ by differentiation under the integral, Theorem 10.15, all derivatives being dominated on compacts). Checking (21.1) is a computation done once in a lifetime, so we do it in full. Fix $I$ and take $\omega = a\,\dd x_I$ (linearity). First,

the first group collects the terms in which $\dd$ hits the factor $x_{i_r}$ — the wedge $\dd x_{i_r}\wedge\dd x_{I\setminus i_r}$ reassembles $\dd x_I$ with a sign $(-1)^{r-1}$ that cancels the prefactor, and the $k$ values of $r$ give the factor $k$ — while the second group collects the terms where $\dd$ hits the integral (chain rule brings out $t\,\partial_ja(tx)$). Next, $\dd\omega = \sum_j\partial_ja\,\dd x_j\wedge\dd x_I$, and applying the definition of $h$ in degree $k+1$, the index $j$ occupying the first slot:

The double sums cancel in $\dd(h\omega) + h(\dd\omega)$, which therefore equals

by the fundamental theorem of calculus. Star-shapedness entered where it had to: $tx \in U$ for $t \in \intcc01$, so that $a(tx)$ makes sense. ∎

Proof. (a) Let $\gamma\colon V \to M$, $\delta\colon V' \to M$ be direct parametrizations whose images contain $\operatorname{supp}\omega\cap M$. The transition $\tau = \delta^{-1}\circ\gamma$ is a diffeomorphism between the relevant open subsets of $V, V'$ (smoothness of transitions: Theorem 20.3, via the local graph description), and $\gamma = \delta\circ\tau$ there, so $\gamma^*\omega = \tau^*(\delta^*\omega)$ (Theorem 21.11(a)). Write $\delta^*\omega = g\,\dd u_1\wedge\dots\wedge\dd u_k$; then (Theorem 21.11(c)) $\tau^*(\delta^*\omega) = (g\circ\tau)\,\det(D\tau)\,\dd u_1\wedge\dots\wedge\dd u_k$. Both frames being direct, $D\tau$ maps a positive basis to a positive basis: $\det D\tau > 0$, so $\det D\tau = \abs{\det D\tau}$ and the change-of-variables theorem (Theorem 11.11) gives $\int(g\circ\tau)\abs{\det D\tau} = \int g$: the two integrals agree. Here is the whole reason orientation exists: without the sign control, the Jacobian and its absolute value differ and the integral is ill-defined. (b) If $(\chi_i)$ and $(\tilde\chi_j)$ are two admissible partitions (covers included), then by (a) and finite additivity, $\sum_i\int\chi_i\omega = \sum_{i,j}\int\chi_i\tilde\chi_j\omega = \sum_j\int\tilde\chi_j\omega$, each double term computable in either chart. The last statement: if $\gamma$ runs over direct parametrizations of $M$, then $\Phi\circ\gamma$ runs over direct parametrizations of $M'$ (the orientation comparison is locally constant, and fixed at one point per component), and $(\Phi\circ\gamma)^*\omega = \gamma^*(\Phi^*\omega)$. ∎

Proof. Each $x \in K$ lies in some $W_{i(x)}$ with a closed ball $\bar B(x, 2r_x) \subseteq W_{i(x)}$; compactness extracts $x_1, \dots, x_N$ with the balls $B(x_s, r_{x_s})$ covering $K$. For each $s$ take a bump $\theta_s \in \mathcal C^\infty$, $0 \leq \theta_s \leq 1$, $\theta_s = 1$ on $\bar B(x_s, r_{x_s})$, $\operatorname{supp}\theta_s \subseteq B(x_s, 2r_{x_s})$ (mollify the indicator of the ball of radius $\frac32r_{x_s}$, Theorem 12.9). Assign each $s$ to one index $i(s)$ with $B(x_s, 2r_{x_s}) \subseteq W_{i(s)}$ and set $\Theta_i = \sum_{i(s) = i}\theta_s$. On the open set $\Omega_0 = \{\sum_j\Theta_j > \tfrac12\} \supseteq K$, the functions $\Theta_i/\sum_j\Theta_j$ do the job but are defined only there; to globalize, let $\rho \in \mathcal C^\infty(\R^n)$ satisfy $\rho = 0$ where $\sum_j\Theta_j \geq 1$ and $\rho > 0$ where $\sum_j\Theta_j \leq \tfrac12$ (mollify a suitable cut-off of $1 - \sum_j\Theta_j$), and set

The denominator is everywhere $> 0$ and equals $\sum_j\Theta_j$ on $\{\sum_j\Theta_j \geq 1\}$, an open neighborhood of $K$ (each point of $K$ lies in some ball $B(x_s, r_{x_s})$ where $\theta_s = 1$); there $\sum_i\chi_i = 1$. Supports and bounds are clear. ∎

Proof. First the orientation bookkeeping: the outward normal along $\partial H^k$ is $-e_k$, and

the frame $(e_1, \dots, e_{k-1})$ of $\partial H^k$ is positive for the induced orientation exactly when $k$ is even, whence the stated comparison. By linearity take $\eta = f\,\dd u_1\wedge\dots\wedge\widehat{\dd u_i}\wedge\dots\wedge\dd u_k$, $f \in \mathcal C^\infty_c(\R^k)$; then $\dd\eta = (-1)^{i-1}\,\partial_if\,\dd u_1\wedge\dots\wedge\dd u_k$ (moving $\dd u_i$ into its slot costs $i - 1$ swaps). Two cases, both by Tonelli–Fubini (Theorem 11.6) and the one-variable fundamental theorem of calculus.

Case $i < k$. Integrating first in $u_i$ over $\R$: $\int_\R\partial_if\,\dd u_i = 0$ (compact support), so $\int_{H^k}\dd\eta = 0$. And the restriction of $\eta$ to $\{u_k = 0\}$ contains the factor $\dd u_k$, which restricts to $0$ ($u_k$ is constant there): $\int_{\partial H^k}\eta = 0$ too.

Case $i = k$. Integrating first in $u_k$ over $\intco0\infty$:

On the boundary side, $\eta$ restricts to $f(u', 0)\,\dd u_1\wedge\dots\wedge\dd u_{k-1}$, and the induced orientation being $(-1)^k$ times the standard one, $\int_{\partial H^k}\eta = (-1)^k\int_{\R^{k-1}}f(u',0)\,\dd u'$. The two sides agree. ∎

Proof. Cover the compact $M$ by finitely many images of direct charts (interior or boundary), and take a partition of unity $(\chi_i)$ subordinate to the corresponding open sets $W_i$ of $\R^n$ (Lemma 21.20), with $\sum\chi_i = 1$ on a neighborhood of $M$. On that neighborhood $\dd(\sum\chi_i) = 0$, so

both sides are additive, and it suffices to prove the theorem for a form supported in a single chart image.

Interior chart. If $\operatorname{supp}\omega \cap M \subseteq \gamma(V)$ with $V$ open in $\R^k$: extend $\eta = \gamma^*\omega$ by zero to $\R^k$ (smooth, compact support in $V$) and apply Lemma 21.22 with the support away from $\partial H^k$ (translate $V$ into the open upper half-space — or simply repeat the case $i < k$ computation over all of $\R^k$): $\int_M\dd\omega = \int_{\R^k}\dd\eta = 0$, and $\int_{\partial M}\omega = 0$ since $\omega$ vanishes near $\partial M$.

Boundary chart. If $\operatorname{supp}\omega\cap M \subseteq \gamma(V \cap H^k)$: with $\eta = \gamma^*\omega$ extended by zero, $\gamma^*(\dd\omega) = \dd\eta$ (Theorem 21.11(b)), so by Definition 21.18:

It remains to identify the right side with $\int_{\partial M}\omega$. The boundary is parametrized by $\beta(u') = \gamma(u', 0)$, and $\beta^*\omega$ is the restriction of $\eta$ to $\{u_k = 0\}$ (pullback under the inclusion $u' \mapsto (u', 0)$ composed with $\gamma$). The orientation comparison is the same $(-1)^k$ on both sides: the frame $(\partial_1\beta, \dots, \partial_{k-1}\beta)$ sits in the induced orientation of $\partial M$ with the sign $\det(-e_k, e_1, \dots, e_{k-1}) = (-1)^k$ computed in the chart (the outward vector pulls back to $-e_k$), which is exactly the sign relating $\partial H^k$’s induced orientation to the standard $\R^{k-1}$ (Lemma 21.22). The two sign conventions cancel: $\int_{\partial H^k}\eta = \int_{\partial M}\omega$. ∎