---
title: "Differential Forms and Stokes’ Theorem"
book: "University Mathematics — Year 3"
subject: math
language: en
chapter: 21
exercises: 12
source: https://one-course.com/books/math/5/en/chapter/21-differential-forms-and-stokes-theorem
---

# Chapter 21 — Differential Forms and Stokes’ Theorem

One theorem of analysis has, over two centuries, absorbed all the others of its kind: the fundamental theorem of calculus, Green–Riemann (proved in the Year 2 volume), the divergence theorem of Gauss, the curl theorem of Kelvin–Stokes — each says that the integral of some derivative over a region equals the integral of the original object over the [boundary](#def-b3-forms-boundary). The language of *[differential forms](#def-b3-forms-diffform)* makes them one statement, $\int_M\dd\omega = \int_{\partial M}\omega$, and makes that statement provable in one stroke. This chapter builds the language honestly — alternating multilinear algebra, the [exterior derivative](#def-b3-forms-d), [pullbacks](#def-b3-forms-pullback), [orientation](#def-b3-forms-orientation), integration on the submanifolds of [Chapter 20](https://one-course.com/books/math/5/en/chapter/20-submanifolds-of-rn#ch-b3-submanifolds) — proves Stokes’ theorem, and cashes the first cheques: the classical integral theorems, the [winding number](#def-b3-forms-winding) that secretly ran [Chapter 17](https://one-course.com/books/math/5/en/chapter/17-laurent-series-and-the-residue-theorem#ch-b3-residues), and, in the weekend problem, Brouwer’s fixed-point theorem. Throughout, *smooth* means $\mathcal
C^\infty$; every map and form is smooth unless stated otherwise. This costs no generality worth having at this level and frees the hands.

## 21.1 Alternating multilinear algebra

**Definition 21.1.**

Let $E$ be a real vector space of dimension $n$. A *$k$-linear alternating form* on $E$ is a map $\alpha\colon E^k \to \R$, linear in each variable, with $\alpha(v_1, \dots, v_k) = 0$ whenever two arguments are equal. Their space is written $\Lambda^k E^*$; by convention $\Lambda^0E^* = \R$. Alternation forces antisymmetry: exchanging two arguments changes the sign (expand $\alpha(\dots, v + w, \dots, v + w, \dots) = 0$), and more generally $\alpha(v_{\sigma(1)}, \dots, v_{\sigma(k)}) =
\varepsilon(\sigma)\,\alpha(v_1, \dots, v_k)$ for every permutation $\sigma$.

**Example 21.2.**

On $E = \R^n$: $\Lambda^1E^* = E^*$ is the [dual space](https://one-course.com/books/math/5/en/chapter/8-banach-spaces-and-the-fundamental-theorems#def-b3-banach-operator); the determinant in the canonical basis is an $n$-linear [alternating form](#def-b3-forms-alternating), and [Proposition 21.4](#prop-b3-forms-basis) will show it spans $\Lambda^nE^*$ — the deep reason the determinant is unique up to scale. For $k > n$, $\Lambda^kE^* = \{0\}$: $k$ vectors are dependent, and expanding one along the others kills $\alpha$ by alternation.

**Definition 21.3.**

For $\ell_1, \dots, \ell_k \in E^*$, their *exterior product* is the $k$-linear [alternating form](#def-b3-forms-alternating)

$$
(\ell_1 \wedge \dots \wedge \ell_k)(v_1, \dots, v_k)
= \det\bigl(\ell_i(v_j)\bigr)_{1 \leq i, j \leq k} .
$$

Alternation and multilinearity are those of the determinant in its columns.

**Proposition 21.4 (Basis of Λk\Lambda^kΛk).**

Let $(e_1, \dots, e_n)$ be a basis of $E$ with [dual](https://one-course.com/books/math/5/en/chapter/8-banach-spaces-and-the-fundamental-theorems#def-b3-banach-operator) basis $(e_1^*, \dots, e_n^*)$. The forms

$$
e_I^* = e_{i_1}^* \wedge \dots \wedge e_{i_k}^*,
\qquad I = \{i_1 < \dots < i_k\} \subseteq \{1, \dots, n\},
$$

form a basis of $\Lambda^kE^*$; hence $\dim\Lambda^kE^* =
\binom nk$. Explicitly, $\alpha =
\sum_{\abs I = k}\alpha(e_{i_1}, \dots, e_{i_k})\,e_I^*$.

**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 $c_J = 0$. ∎

**Definition 21.5.**

The [exterior product](#def-b3-forms-wedge) extends to a bilinear map $\Lambda^kE^* \times \Lambda^\ell E^* \to
\Lambda^{k+\ell}E^*$, determined by bilinearity and $(e_I^*) \wedge (e_J^*) = e_I^* \wedge e_J^*$ (concatenate and reorder; the product is $0$ if $I \cap J \neq
\varnothing$). It is associative, and *graded-anticommutative*:

$$
\beta \wedge \alpha = (-1)^{k\ell}\,\alpha \wedge \beta
\qquad (\alpha \in \Lambda^k, \ \beta \in \Lambda^\ell).
$$

**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](#def-b3-forms-wedge) (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. ∎

**Proposition 21.6 (Pullback, linear case).**

A linear map $u\colon E \to F$ induces, for each $k$, the linear map $u^*\colon \Lambda^kF^* \to \Lambda^kE^*$, $(u^*\alpha)(v_1, \dots, v_k) = \alpha(u(v_1), \dots,
u(v_k))$. It satisfies $u^*(\alpha \wedge \beta) = u^*\alpha
\wedge u^*\beta$ and $(u \circ w)^* = w^* \circ u^*$. Moreover:

1. If $\dim E = n$ and $u\colon E \to E$ , then on the line $\Lambda^nE^*$ : $u^*\alpha = (\det u)\,\alpha$ .
2. If $\operatorname{rk}u < k$ , then $u^* = 0$ on $\Lambda^kF^*$ .

**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](#prop-b3-forms-basis)) 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](#def-b3-forms-alternating) vanishes on a dependent family (expand the dependent vector along the others). ∎

## 21.2 Differential forms and the exterior derivative

**Definition 21.7.**

Let $U \subseteq \R^n$ be open. A *differential $k$-form* on $U$ is a smooth map $\omega\colon U \to \Lambda^k(\R^n)^*$; in the basis of [Proposition 21.4](#prop-b3-forms-basis) (with $\dd x_i$ written for $e_i^*$),

$$
\omega = \sum_{\abs I = k} a_I\,\dd x_I,
\qquad
\dd x_I = \dd x_{i_1} \wedge \dots \wedge \dd x_{i_k},
$$

with smooth coefficients $a_I \in \mathcal C^\infty(U)$. Their space is $\Omega^k(U)$; $\Omega^0(U) = \mathcal
C^\infty(U)$. A $0$-form is a function; a $1$-form is a field of linear forms (e.g. the differential $\dd f$ of a function); an $n$-form is $a\,\dd x_1\wedge\dots\wedge\dd
x_n$, the natural integrand of [Chapter 11](https://one-course.com/books/math/5/en/chapter/11-product-measures-fubini-change-of-variables#ch-b3-product).

**Definition 21.8 (Exterior derivative).**

The *exterior derivative* is the linear map $\dd\colon \Omega^k(U) \to \Omega^{k+1}(U)$ defined by

$$
\dd\Bigl(\sum_I a_I\,\dd x_I\Bigr)
= \sum_I \dd a_I \wedge \dd x_I
= \sum_I\sum_{j=1}^n \frac{\partial a_I}{\partial
x_j}\,\dd x_j \wedge \dd x_I .
$$

On $0$-forms it is the usual differential.

**Theorem 21.9.**

(a) $\dd(\omega \wedge \eta) = \dd\omega \wedge \eta +
(-1)^k\,\omega \wedge \dd\eta$ for $\omega \in \Omega^k$ (the *graded Leibniz rule*). (b) $\dd \circ \dd = 0$.

**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

$$
\dd(\omega\wedge\eta) = (b\,\dd a + a\,\dd b) \wedge \dd x_I
\wedge \dd x_J
= (\dd a \wedge \dd x_I) \wedge (b\,\dd x_J) +
a\,\dd b \wedge \dd x_I \wedge \dd x_J,
$$

and moving the $1$-form $\dd b$ past the $k$ factors of $\dd
x_I$ costs $(-1)^k$ ([Definition 21.5](#def-b3-forms-wedgegeneral)): 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). ∎

**Definition 21.10 (Pullback).**

Let $\varphi\colon U \to V$ be smooth ($U \subseteq \R^m$, $V \subseteq \R^n$ open). The *pullback* $\varphi^*\colon \Omega^k(V) \to \Omega^k(U)$ is defined pointwise by the linear pullback along the differential: $(\varphi^*\omega)_x = (D\varphi(x))^*\,\omega_{\varphi(x)}$. Concretely, $\varphi^*$ substitutes: $\varphi^*f = f \circ
\varphi$ on functions, $\varphi^*(\dd y_i) =
\dd\varphi_i = \sum_j\frac{\partial\varphi_i}{\partial
x_j}\dd x_j$, and $\varphi^*(a\,\dd y_{i_1}\wedge\dots\wedge
\dd y_{i_k}) =
(a\circ\varphi)\,\dd\varphi_{i_1}\wedge\dots\wedge
\dd\varphi_{i_k}$.

**Theorem 21.11.**

(a) $\varphi^*(\omega \wedge \eta) = \varphi^*\omega \wedge
\varphi^*\eta$ and $(\psi \circ \varphi)^* = \varphi^* \circ
\psi^*$. (b) $\varphi^*(\dd\omega) = \dd(\varphi^*\omega)$: the [exterior derivative](#def-b3-forms-d) commutes with every smooth substitution — the identity that makes it *the* derivative of the theory. (c) If $\varphi\colon U \to V$ is smooth between opens of $\R^n$ and $\omega = a\,\dd y_1 \wedge \dots \wedge \dd
y_n$, then $\varphi^*\omega = (a \circ
\varphi)\,\det\bigl(D\varphi\bigr)\,\dd x_1 \wedge \dots
\wedge \dd x_n$.

**Proof.** (a) Pointwise statements about linear [pullbacks](#def-b3-forms-pullback) ([Proposition 21.6](#prop-b3-forms-linearpullback)), 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](#thm-b3-forms-dsquare)(a)) and $\dd\dd\varphi_{i_r} =
0$:

$$
\dd(\varphi^*\omega) = \dd(a\circ\varphi) \wedge
\dd\varphi_{i_1}\wedge\dots\wedge\dd\varphi_{i_k}
= \varphi^*(\dd a) \wedge \varphi^*(\dd y_I)
= \varphi^*(\dd a \wedge \dd y_I) =
\varphi^*(\dd\omega).
$$

(c) Pointwise this is exactly $u^*\alpha = (\det u)\alpha$ on top-degree forms ([Proposition 21.6](#prop-b3-forms-linearpullback)(1) with $u = D\varphi(x)$). ∎

**Example 21.12 (Polar coordinates).**

For $\varphi(r, \theta) = (r\cos\theta, r\sin\theta)$: $\varphi^*\dd x = \cos\theta\,\dd r -
r\sin\theta\,\dd\theta$, $\varphi^*\dd y = \sin\theta\,\dd r
+ r\cos\theta\,\dd\theta$, so

$$
\varphi^*(\dd x \wedge \dd y) =
(\cos\theta\,\dd r - r\sin\theta\,\dd\theta) \wedge
(\sin\theta\,\dd r + r\cos\theta\,\dd\theta)
= r\,\dd r \wedge \dd\theta,
$$

the Jacobian of [Example 11.12](https://one-course.com/books/math/5/en/chapter/11-product-measures-fubini-change-of-variables#ex-b3-product-polar) appearing by pure algebra — no [measure](https://one-course.com/books/math/5/en/chapter/9-measure-theory#def-b3-measure-measure) theory. [Exercise 21.9](#exo-b3-forms-9) turns this remark into a statement: for oriented integrals, the change-of-variables formula *is* the [pullback](#def-b3-forms-pullback) formula.

## 21.3 Closed and exact forms; the Poincaré lemma

**Definition 21.13.**

$\omega \in \Omega^k(U)$ is *closed* if $\dd\omega = 0$, *exact* if $\omega = \dd\eta$ for some $\eta \in \Omega^{k-1}(U)$ (a *primitive* of $\omega$). Exact $\Rightarrow$ closed by $\dd^2 = 0$; the converse is a question about the *shape* of $U$.

**Example 21.14 (The angular form).**

On $U = \R^2 \setminus \{0\}$,

$$
\omega_\theta = \frac{x\,\dd y - y\,\dd x}{x^2 + y^2}
$$

is [closed](#def-b3-forms-closedexact) (direct computation: [Exercise 21.4](#exo-b3-forms-4)) but not [exact](#def-b3-forms-closedexact): its integral along the unit circle is $2\pi \neq 0$, while integrals of [exact forms](#def-b3-forms-closedexact) along [closed](#def-b3-forms-closedexact) curves vanish ([Proposition 21.28](#prop-b3-forms-exactloop)). Locally, $\omega_\theta = \dd\theta$ for any smooth determination $\theta$ of the polar angle — whence the name and the obstruction: no such determination exists on all of $U$. This single form runs the [winding number](#def-b3-forms-winding) ([Section 21.6](#sec-21-6)) and, through it, the [residue](https://one-course.com/books/math/5/en/chapter/17-laurent-series-and-the-residue-theorem#def-b3-residues-singularities) theorem of [Chapter 17](https://one-course.com/books/math/5/en/chapter/17-laurent-series-and-the-residue-theorem#ch-b3-residues).

**Theorem 21.15 (Poincaré lemma).**

Let $U \subseteq \R^n$ be open and *star-shaped* with respect to $0$. Every [closed](#def-b3-forms-closedexact) $k$-form on $U$ ($k \geq 1$) is [exact](#def-b3-forms-closedexact).

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

$$
\dd(h\omega) + h(\dd\omega) = \omega
\qquad (k \geq 1);\tag{21.1}
$$

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

$$
h\omega = \sum_{\abs I = k}\ \sum_{r=1}^{k}(-1)^{r-1}
\Bigl(\int_0^1 t^{k-1}a_I(tx)\,\dd t\Bigr)\,
x_{i_r}\,\dd x_{i_1}\wedge\dots\wedge
\widehat{\dd x_{i_r}}\wedge\dots\wedge\dd x_{i_k}
$$

(the hat deletes a factor; the integrals are smooth in $x$ by differentiation under the integral, [Theorem 10.15](https://one-course.com/books/math/5/en/chapter/10-the-lebesgue-integral#thm-b3-lebesgue-paramdiff), all derivatives being dominated on [compacts](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact)). Checking [(21.1)](#eq-b3-forms-homotopy) 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,

$$
\dd(h\omega) = k\Bigl(\int_0^1t^{k-1}a(tx)\dd t\Bigr)\dd x_I
+ \sum_{r=1}^k(-1)^{r-1}\sum_{j=1}^n
\Bigl(\int_0^1t^{k}\,\partial_ja(tx)\,\dd t\Bigr)
x_{i_r}\,\dd x_j\wedge\dd x_{I\setminus i_r} :
$$

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:

$$
h(\dd\omega) = \sum_{j=1}^n\Bigl(\int_0^1t^{k}
\partial_ja(tx)\dd t\Bigr)x_j\,\dd x_I
- \sum_{j=1}^n\sum_{r=1}^k(-1)^{r-1}
\Bigl(\int_0^1t^{k}\partial_ja(tx)\dd t\Bigr)
x_{i_r}\,\dd x_j\wedge\dd x_{I\setminus i_r} .
$$

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

$$
\Bigl(\int_0^1\bigl(k\,t^{k-1}a(tx) +
t^k{\textstyle\sum_j}x_j\,\partial_ja(tx)\bigr)\dd t\Bigr)
\dd x_I
= \Bigl(\int_0^1\frac{\dd}{\dd t}\bigl(t^ka(tx)\bigr)\dd
t\Bigr)\dd x_I
= a(x)\,\dd x_I = \omega,
$$

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. ∎

**Remark 21.16.**

For $k = 1$ and $\omega = \sum_j a_j\dd x_j$, the primitive is $f(x) = \int_0^1\sum_ja_j(tx)\,x_j\,\dd t$ — the line integral of $\omega$ along the segment $[0, x]$: the theorem is the several-variables “a field with symmetric Jacobian is a gradient” of Year 2, now in every degree. The angular form ([Example 21.14](#ex-b3-forms-angular)) shows the hypothesis on $U$ is not decorative: $\R^2\setminus\{0\}$ is not star-shaped, and there closedness does not imply exactness. What survives on a general [open set](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-topology) is measured by the *de Rham cohomology* $H^k(U) =
\ker\dd/\operatorname{im}\dd$ — see [Exercise 21.12](#exo-b3-forms-12) for the first nontrivial computation.

## 21.4 Orientation and integration on submanifolds

Integrating a $k$-form requires $k$-dimensional oriented territory. Recall from [Chapter 20](https://one-course.com/books/math/5/en/chapter/20-submanifolds-of-rn#ch-b3-submanifolds) ([Theorem 20.3](https://one-course.com/books/math/5/en/chapter/20-submanifolds-of-rn#thm-b3-submanifolds-characterizations)) that a $k$-submanifold $M \subseteq \R^n$ is locally the image of a regular parametrization $\gamma\colon V \to M \cap W$ ($V
\subseteq \R^k$ open, $\gamma$ a [homeomorphism](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) onto its image with injective differential).

**Definition 21.17.**

An *orientation* of $M$ is a choice, for each $p \in M$, of one of the two orientation classes of bases of the [tangent space](https://one-course.com/books/math/5/en/chapter/20-submanifolds-of-rn#def-b3-submanifolds-tangent) $T_pM$, which is *locally coherent*: around each point there is a parametrization $\gamma$ whose coordinate frame $(\partial_1\gamma, \dots,
\partial_k\gamma)$ is positively oriented at every point of its domain. Such parametrizations are called *direct*. $M$ is *orientable* if an orientation exists; the Möbius band shows this can fail. All submanifolds in this chapter are oriented.

**Definition 21.18 (Integral of a form).**

Let $M$ be an oriented $k$-submanifold and $\omega$ a $k$-form defined on a [neighborhood](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-topology) of $M$, with $\operatorname{supp}\omega \cap M$ [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact). (a) If $\operatorname{supp}\omega \cap M \subseteq
\gamma(V)$ for a single direct parametrization, set

$$
\int_M\omega = \int_V\gamma^*\omega
$$

— the right side being the Lebesgue integral over $V$ ([Chapter 11](https://one-course.com/books/math/5/en/chapter/11-product-measures-fubini-change-of-variables#ch-b3-product)) of the coefficient $g$ of $\gamma^*\omega = g\,\dd u_1\wedge\dots\wedge\dd u_k$, which is [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) with [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) support. (b) In general, choose finitely many direct parametrizations $\gamma_i(V_i)$ covering the [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) $\operatorname{supp}\omega \cap M$ and a subordinate [partition of unity](#lem-b3-forms-partition) $(\chi_i)$ ([Lemma 21.20](#lem-b3-forms-partition)), and set $\int_M\omega =
\sum_i\int_M\chi_i\,\omega$, each term computed by (a). For a curve ($k = 1$) parametrized by $\gamma\colon\intcc ab\to\R^n$ we write $\int_\gamma\omega = \int_a^b\gamma^*\omega$, no injectivity required.

**Lemma 21.19 (Consistency).**

Definition (a) does not depend on the direct parametrization, and definition (b) depends neither on the cover nor on the [partition of unity](#lem-b3-forms-partition). Moreover, if $\Phi$ is a diffeomorphism of [neighborhoods](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-topology) of two oriented submanifolds with $\Phi(M) = M'$, carrying a direct frame of $M$ to a direct frame of $M'$ at some point of each component of $M$, then $\int_{M'}\omega =
\int_M\Phi^*\omega$.

**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](https://one-course.com/books/math/5/en/chapter/20-submanifolds-of-rn#thm-b3-submanifolds-characterizations), via the local graph description), and $\gamma = \delta\circ\tau$ there, so $\gamma^*\omega = \tau^*(\delta^*\omega)$ ([Theorem 21.11](#thm-b3-forms-pullback)(a)). Write $\delta^*\omega =
g\,\dd u_1\wedge\dots\wedge\dd u_k$; then ([Theorem 21.11](#thm-b3-forms-pullback)(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](https://one-course.com/books/math/5/en/chapter/11-product-measures-fubini-change-of-variables#thm-b3-product-changeofvar)) gives $\int(g\circ\tau)\abs{\det D\tau} = \int g$: the two integrals agree. *Here is the whole reason [orientation](#def-b3-forms-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](#def-b3-forms-orientation) comparison is locally constant, and fixed at one point per component), and $(\Phi\circ\gamma)^*\omega =
\gamma^*(\Phi^*\omega)$. ∎

**Lemma 21.20 (Partitions of unity, compact case).**

Let $K \subseteq \R^n$ be [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) and $W_1, \dots, W_m$ [open sets](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-topology) covering $K$. There exist $\chi_1, \dots, \chi_m \in
\mathcal C^\infty(\R^n)$ with $0 \leq \chi_i \leq 1$, $\operatorname{supp}\chi_i \subseteq W_i$ [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact), and $\sum\chi_i = 1$ on a [neighborhood](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-topology) of $K$.

**Proof.** Each $x \in K$ lies in some $W_{i(x)}$ with a [closed](#def-b3-forms-closedexact) 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](https://one-course.com/books/math/5/en/chapter/12-the-lp-spaces#thm-b3-lp-regularization)). 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](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-topology) $\Omega_0 = \{\sum_j\Theta_j > \tfrac12\} \supseteq K$, 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

$$
\chi_i = \frac{\Theta_i}{\rho + \sum_j\Theta_j} .
$$

The denominator is everywhere $> 0$ and equals $\sum_j\Theta_j$ on $\{\sum_j\Theta_j \geq 1\}$, an open [neighborhood](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-topology) 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. ∎

## 21.5 Stokes’ theorem

**Definition 21.21.**

A *$k$-submanifold with boundary* $M \subseteq \R^n$ is a set covered by regular parametrizations of two kinds: *[interior](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-interior) charts* $\gamma\colon V \to M \cap W$ with $V \subseteq \R^k$ open, and *boundary charts* $\gamma\colon V \cap H^k \to M
\cap W$, where $H^k = \{u \in \R^k : u_k \geq 0\}$ and $\gamma$ extends smoothly and regularly to the open $V$. The *boundary* $\partial M$ is the set of points reached at $u_k = 0$; it is a $(k-1)$-submanifold without boundary, parametrized by the maps $u' \mapsto \gamma(u',
0)$. An [orientation](#def-b3-forms-orientation) of $M$ induces one on $\partial M$ by the *outward-normal-first* rule: at $p \in \partial M$, a basis $(w_1, \dots, w_{k-1})$ of $T_p\partial M$ is positive iff $(\nu, w_1, \dots, w_{k-1})$ is a positive basis of $T_pM$, where $\nu \in T_pM \setminus T_p\partial M$ points out of $M$ (in a boundary chart: $\nu =
-\partial_k\gamma$, up to adding tangential components — the [orientation](#def-b3-forms-orientation) class does not see them).

![The outward-normal-first rule: at each boundary point, put the outward vector first; the bases that complete it to a positive frame of M orient M. For a plane domain with the standard orientation this is the counterclockwise rule of Green–Riemann.](https://one-course.com/images/onecourse/chapters/math-5/b3-forms/fig-f4f14ea3a97a.svg)

*The outward-normal-first rule: at each [boundary](#def-b3-forms-boundary) point, put the outward vector $\nu$ first; the bases that [complete](https://one-course.com/books/math/5/en/chapter/7-complete-spaces-baire-ascoli-stoneweierstrass#def-b3-complete-complete) it to a positive frame of $M$ orient $\partial M$. For a plane domain with the standard [orientation](#def-b3-forms-orientation) this is the counterclockwise rule of Green–Riemann.*

**Lemma 21.22 (Stokes on the half-space).**

Let $\eta$ be a smooth $(k-1)$-form on $\R^k$ with [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) support. Then

$$
\int_{H^k}\dd\eta = \int_{\partial H^k}\eta,
$$

where $H^k$ carries the standard [orientation](#def-b3-forms-orientation) of $\R^k$ and $\partial H^k = \{u_k = 0\} \cong \R^{k-1}$ the induced one, which is $(-1)^k$ times the standard [orientation](#def-b3-forms-orientation) of $\R^{k-1}$.

**Proof.** First the [orientation](#def-b3-forms-orientation) bookkeeping: the outward normal along $\partial H^k$ is $-e_k$, and

$$
\det(-e_k, e_1, \dots, e_{k-1}) =
-\det(e_k, e_1, \dots, e_{k-1}) = -(-1)^{k-1}\det(e_1,
\dots, e_k) = (-1)^k :
$$

the frame $(e_1, \dots, e_{k-1})$ of $\partial H^k$ is positive for the induced [orientation](#def-b3-forms-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](https://one-course.com/books/math/5/en/chapter/11-product-measures-fubini-change-of-variables#thm-b3-product-fubini)) 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](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-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 $\intco0\infty$:

$$
\begin{align*}
\int_{H^k}\dd\eta &= (-1)^{k-1}\int_{\R^{k-1}}
\Bigl(\int_0^\infty\partial_kf\,\dd u_k\Bigr)\dd u' \\
&= (-1)^{k-1}\int_{\R^{k-1}}\bigl(0 - f(u', 0)\bigr)\dd u'
= (-1)^k\int_{\R^{k-1}}f(u',0)\,\dd u' .
\end{align*}
$$

On the [boundary](#def-b3-forms-boundary) side, $\eta$ restricts to $f(u',
0)\,\dd u_1\wedge\dots\wedge\dd u_{k-1}$, and the induced [orientation](#def-b3-forms-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. ∎

**Theorem 21.23 (Stokes).**

Let $M \subseteq \R^n$ be a [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) oriented $k$-submanifold with [boundary](#def-b3-forms-boundary), $\partial M$ carrying the induced [orientation](#def-b3-forms-orientation), and let $\omega$ be a smooth $(k-1)$-form on a [neighborhood](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-topology) of $M$. Then

$$
\int_M\dd\omega = \int_{\partial M}\omega .
$$

In particular, if $\partial M = \varnothing$: $\int_M\dd\omega = 0$.

**Proof.** Cover the [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) $M$ by finitely many images of direct charts ([interior](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-interior) or [boundary](#def-b3-forms-boundary)), and take a [partition of unity](#lem-b3-forms-partition) $(\chi_i)$ subordinate to the corresponding [open sets](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-topology) $W_i$ of $\R^n$ ([Lemma 21.20](#lem-b3-forms-partition)), with $\sum\chi_i
= 1$ on a [neighborhood](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-topology) of $M$. On that [neighborhood](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-topology) $\dd(\sum\chi_i) = 0$, so

$$
\int_M\dd\omega = \sum_i\int_M\dd(\chi_i\omega),
\qquad
\int_{\partial M}\omega =
\sum_i\int_{\partial M}\chi_i\omega :
$$

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

*[Interior](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-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](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) support in $V$) and apply [Lemma 21.22](#lem-b3-forms-halfspace) 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](#def-b3-forms-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](#thm-b3-forms-pullback)(b)), so by [Definition 21.18](#def-b3-forms-integral):

$$
\int_M\dd\omega = \int_{H^k}\dd\eta
\overset{\text{\text{Lemma 21.22}}}{=}
\int_{\partial H^k}\eta .
$$

It remains to identify the right side with $\int_{\partial
M}\omega$. The [boundary](#def-b3-forms-boundary) is parametrized by $\beta(u') =
\gamma(u', 0)$, and $\beta^*\omega$ is the restriction of $\eta$ to $\{u_k = 0\}$ ([pullback](#def-b3-forms-pullback) under the inclusion $u'
\mapsto (u', 0)$ composed with $\gamma$). The [orientation](#def-b3-forms-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](#def-b3-forms-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](#def-b3-forms-orientation) to the standard $\R^{k-1}$ ([Lemma 21.22](#lem-b3-forms-halfspace)). The two sign conventions cancel: $\int_{\partial H^k}\eta = \int_{\partial M}\omega$. ∎

**Example 21.24 (The classical theorems).**

Let $D \subseteq \R^2$ be a [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) domain with smooth [boundary](#def-b3-forms-boundary), standardly oriented. For $\omega = P\,\dd x +
Q\,\dd y$: $\dd\omega = \bigl(\partial_xQ -
\partial_yP\bigr)\dd x\wedge\dd y$, and Stokes reads

$$
\int_D\Bigl(\frac{\partial Q}{\partial x} -
\frac{\partial P}{\partial y}\Bigr)\dd x\,\dd y
= \oint_{\partial D}P\,\dd x + Q\,\dd y :
$$

*Green–Riemann*, proved for elementary domains in the Year 2 volume and now in natural generality. In $\R^3$, Stokes applied to the flux $2$-form of a vector field on a [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) domain gives the *divergence theorem* $\int_\Omega\operatorname{div}F =
\int_{\partial\Omega}\langle F, \nu\rangle\,\dd S$, and applied to a $1$-form on a surface-with-boundary, the classical *Kelvin–Stokes* curl theorem; [Exercise 21.7](#exo-b3-forms-7) spells out both dictionaries.

## 21.6 The winding number

**Definition 21.25.**

Let $\gamma\colon\intcc01\to\R^2\setminus\{a\}$ be a smooth [closed](#def-b3-forms-closedexact) curve. Its *winding number* around $a$ is

$$
\operatorname{Ind}_\gamma(a) =
\frac1{2\pi}\int_\gamma\omega_\theta^a,
\qquad
\omega_\theta^a = \frac{(x - a_1)\,\dd y - (y - a_2)\,\dd
x}{(x - a_1)^2 + (y - a_2)^2} .
$$

**Proposition 21.26.**

$\operatorname{Ind}_\gamma(a) \in \Z$; as a function of $a$ it is constant on each [connected component](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-components) of $\R^2\setminus\gamma(\intcc01)$ and zero on the unbounded component. For $\gamma(t) = a + r(\cos2\pi t, \sin2\pi t)$: $\operatorname{Ind}_\gamma(a) = 1$.

**Proof.** Take $a = 0$ and write $\gamma = (x, y)$, $\rho =
\norm\gamma > 0$, $\vartheta(t) =
\int_0^t\gamma^*\omega_\theta$, so that $\vartheta' =
\frac{xy' - yx'}{x^2 + y^2}$. In complex notation let $u(t)
= \gamma(t)\,\rho(t)^{-1}\eu^{-\iu\vartheta(t)}$; then $\abs u = 1$ and a direct computation gives

$$
\frac{u'}{u} = \frac{\gamma'}{\gamma} - \frac{\rho'}{\rho}
- \iu\vartheta'
= \frac{\bar\gamma\gamma'}{\abs\gamma^2} -
\frac{\rho'}{\rho} - \iu\,\frac{xy' -
yx'}{\rho^2}
= \frac{(xx' + yy')}{\rho^2} - \frac{\rho'}{\rho} = 0,
$$

since $\bar\gamma\gamma' = (xx' + yy') + \iu(xy' - yx')$ and $\rho\rho' = xx' + yy'$. So $u$ is constant: $\gamma(t) =
\rho(t)\,c\,\eu^{\iu\vartheta(t)}$ with $\abs c = 1$, and $\gamma(1) = \gamma(0)$ with $\rho(1) = \rho(0)$ forces $\eu^{\iu\vartheta(1)} = \eu^{\iu\vartheta(0)} = 1$: $\vartheta(1) \in 2\pi\Z$, i.e. $\operatorname{Ind}_\gamma(0) \in \Z$. As a function of $a$ on the open complement of the [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) curve, the defining integral is [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) ([Theorem 10.14](https://one-course.com/books/math/5/en/chapter/10-the-lebesgue-integral#thm-b3-lebesgue-paramcont), domination on a [neighborhood](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-topology) of each $a$); a [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) integer-valued function is locally constant, hence constant on components. For $\norm a$ large the integrand is $O(1/\norm a)$ uniformly in $t$, so the index tends to $0$ and vanishes on the unbounded component. For the circle: $\gamma^*\omega_\theta^a = 2\pi\,\dd t$ directly. ∎

**Remark 21.27.**

Under $\C \cong \R^2$, $\frac{\dd z}{z} = \frac{x\,\dd x +
y\,\dd y}{x^2 + y^2} + \iu\,\omega_\theta$, so $\operatorname{Ind}_\gamma(a) =
\frac1{2\iu\pi}\oint_\gamma\frac{\dd z}{z - a}$: this is the index of [Chapter 17](https://one-course.com/books/math/5/en/chapter/17-laurent-series-and-the-residue-theorem#ch-b3-residues), and [Proposition 21.26](#prop-b3-forms-windinginteger) re-proves its integrality and local constancy by real-variable means — the [topological](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-topology) half of the [residue](https://one-course.com/books/math/5/en/chapter/17-laurent-series-and-the-residue-theorem#def-b3-residues-singularities) theorem, now standing on Stokes.

**Proposition 21.28.**

If $\omega = \dd f$ is [exact](#def-b3-forms-closedexact) on the open $U$ and $\gamma\colon\intcc01\to U$ is a [closed](#def-b3-forms-closedexact) curve, then $\int_\gamma\omega = 0$.

**Proof.** $\int_\gamma\dd f = \int_0^1(f\circ\gamma)'(t)\,\dd t =
f(\gamma(1)) - f(\gamma(0)) = 0$ — the chain rule identifies $\gamma^*(\dd f)$ with $(f\circ\gamma)'\,\dd t$. ∎

**Method 21.29.**

Computing with forms: (1) mechanize — wedges reorder with signs, $\dd$ differentiates coefficients into fresh $\dd
x_j$’s, [pullbacks](#def-b3-forms-pullback) substitute; trust the algebra, it encodes every Jacobian. (2) To integrate a form over a submanifold: parametrize directly, pull back, integrate the coefficient; [orientation](#def-b3-forms-orientation) is the only trap — check one frame. (3) To prove an integral identity, look for a Stokes shape: is the integrand [exact](#def-b3-forms-closedexact)? is the domain a [boundary](#def-b3-forms-boundary)? (4) To compare integrals over two “parallel” submanifolds, apply Stokes to the region between them (the deformation argument, [Exercise 21.10](#exo-b3-forms-10)). (5) A nonzero integral of a [closed form](#def-b3-forms-closedexact) certifies a [topological](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-topology) obstruction — no primitive, no retraction, no zero-free extension: this is how the weekend problem kills retractions of the ball.

## 21.7 Exercises

**Exercise 21.1 ★.**

On $\R^3$, let $\omega = x\,\dd y - z\,\dd x$ and $\eta =
\dd x + y\,\dd z$. Compute $\omega\wedge\eta$, $\dd\omega$, $\dd\eta$, and $\dd(\omega\wedge\eta)$, and verify the graded Leibniz rule on this example.

**Solution of Exercise 21.1.**

Expanding and killing repeated factors:

$$
\omega\wedge\eta = (x\,\dd y - z\,\dd x)\wedge(\dd x +
y\,\dd z)
= -x\,\dd x\wedge\dd y + xy\,\dd y\wedge\dd z +
yz\,\dd z\wedge\dd x
$$

(using $\dd y\wedge\dd x = -\dd x\wedge\dd y$ and $-zy\,\dd
x\wedge\dd z = yz\,\dd z\wedge\dd x$). Next $\dd\omega =
\dd x\wedge\dd y - \dd z\wedge\dd x = \dd x\wedge\dd y +
\dd x\wedge\dd z$ and $\dd\eta = \dd y\wedge\dd z$. Finally

$$
\dd(\omega\wedge\eta) = y\,\dd x\wedge\dd y\wedge\dd z +
z\,\dd y\wedge\dd z\wedge\dd x = (y + z)\,\dd x\wedge\dd
y\wedge\dd z
$$

(the first term of $\omega\wedge\eta$ contributes $\dd(-x)
\wedge\dd x\wedge\dd y = 0$; cyclic permutations of three factors are even). Leibniz check: $\dd\omega\wedge\eta =
(\dd x\wedge\dd y + \dd x\wedge\dd z)\wedge(\dd x + y\,\dd
z) = y\,\dd x\wedge\dd y\wedge\dd z$, and $(-1)^1\omega\wedge\dd\eta = -(x\,\dd y - z\,\dd
x)\wedge\dd y\wedge\dd z = z\,\dd x\wedge\dd y\wedge\dd z$; the sum matches.

**Exercise 21.2 ★.**

Identify, on $\R^3$, the three incarnations of $\dd$: for $f \in \Omega^0$, $\dd f \leftrightarrow \nabla f$; for the work form $\omega_F = F_1\dd x + F_2\dd y + F_3\dd z$, $\dd\omega_F \leftrightarrow \operatorname{curl}F$; for the flux form $\sigma_F = F_1\,\dd y\wedge\dd z + F_2\,\dd
z\wedge\dd x + F_3\,\dd x\wedge\dd y$, $\dd\sigma_F
\leftrightarrow \operatorname{div}F$. Deduce from $\dd^2 =
0$ the identities $\operatorname{curl}\nabla f = 0$ and $\operatorname{div}\operatorname{curl}F = 0$.

**Solution of Exercise 21.2.**

$\dd f = \sum_i\partial_if\,\dd x_i$ has the coefficients of $\nabla f$. For the work form,

$$
\dd\omega_F = (\partial_yF_3 - \partial_zF_2)\,\dd y\wedge\dd
z + (\partial_zF_1 - \partial_xF_3)\,\dd z\wedge\dd x +
(\partial_xF_2 - \partial_yF_1)\,\dd x\wedge\dd y =
\sigma_{\operatorname{curl}F},
$$

the flux form of the curl (collect the six terms of $\sum_i\dd F_i\wedge\dd x_i$). For the flux form, $\dd\sigma_F = (\partial_xF_1 + \partial_yF_2 +
\partial_zF_3)\,\dd x\wedge\dd y\wedge\dd z$: the divergence. Then $\dd^2f = 0$ reads $\sigma_{\operatorname{curl}\nabla f} = 0$, i.e. $\operatorname{curl}\nabla f = 0$, and $\dd^2\omega_F = 0$ reads $(\operatorname{div}\operatorname{curl}F)\,\dd
x\wedge\dd y\wedge\dd z = 0$: the two vector identities are one identity, $\dd^2 = 0$, read in two degrees.

**Exercise 21.3 ★★.**

Decide whether each $1$-form is [closed](#def-b3-forms-closedexact), [exact](#def-b3-forms-closedexact) on its domain, and compute a primitive when one exists: (a) $(2xy + z^2)\,\dd x + x^2\,\dd y + 2xz\,\dd z$ on $\R^3$; (b) $\dfrac{x\,\dd x + y\,\dd y}{x^2 + y^2}$ on $\R^2\setminus\{0\}$; (c) $\dfrac{-y\,\dd x + x\,\dd y}{x^2 + y^2}$ on the half-plane $\{x > 0\}$.

**Solution of Exercise 21.3.**

(a) Closedness is the symmetry of the cross-partials: $\partial_y(2xy + z^2) = 2x = \partial_x(x^2)$, $\partial_z(2xy + z^2) = 2z = \partial_x(2xz)$, $\partial_z(x^2) = 0 = \partial_y(2xz)$. The domain $\R^3$ is star-shaped: [exact](#def-b3-forms-closedexact) ([Theorem 21.15](#thm-b3-forms-poincare)), with primitive $f = x^2y + xz^2$ (check $\dd f$). (b) $\dfrac{x\,\dd x + y\,\dd y}{x^2 + y^2} =
\tfrac12\,\dd\log(x^2 + y^2)$: [exact](#def-b3-forms-closedexact) on all of $\R^2\setminus\{0\}$ (hence [closed](#def-b3-forms-closedexact)) — the radial cousin of the angular form is harmless. (c) On $\{x > 0\}$ the form is $\omega_\theta$, [closed](#def-b3-forms-closedexact) ([Exercise 21.4](#exo-b3-forms-4)); the half-plane is convex, so it is [exact](#def-b3-forms-closedexact) there, and indeed $f = \arctan(y/x)$ satisfies $\dd f
= \frac{-y\,\dd x + x\,\dd y}{x^2 + y^2}$. [Exact](#def-b3-forms-closedexact) on the half-plane, [non-exact](#def-b3-forms-closedexact) on the punctured plane: the obstruction lives in the hole, not in the formula.

**Exercise 21.4 ★★.**

(The angular form) Verify that $\omega_\theta$ ([Example 21.14](#ex-b3-forms-angular)) is [closed](#def-b3-forms-closedexact); compute $\int_\gamma\omega_\theta$ for $\gamma$ the circle of radius $r$ around $0$; conclude that $\omega_\theta$ is not [exact](#def-b3-forms-closedexact) on $\R^2\setminus\{0\}$, and that $\R^2\setminus\{0\}$ is star-shaped with respect to none of its points (two routes: via [Theorem 21.15](#thm-b3-forms-poincare), and directly from the geometry).

**Solution of Exercise 21.4.**

Closedness: with $\rho^2 = x^2 + y^2$,

$$
\partial_x\Bigl(\frac{x}{\rho^2}\Bigr) = \frac{\rho^2 -
2x^2}{\rho^4} = \frac{y^2 - x^2}{\rho^4} =
\partial_y\Bigl(\frac{-y}{\rho^2}\Bigr),
$$

so $\dd\omega_\theta = \bigl(\partial_x(x/\rho^2) -
\partial_y(-y/\rho^2)\bigr)\,\dd x\wedge\dd y = 0$. On $\gamma(t) = (r\cos2\pi t, r\sin2\pi t)$:

$$
\gamma^*\omega_\theta = \frac{2\pi r^2(\cos^22\pi t +
\sin^22\pi t)}{r^2}\,\dd t = 2\pi\,\dd t,
\qquad\text{so}\qquad
\int_\gamma\omega_\theta = 2\pi .
$$

If $\omega_\theta$ were [exact](#def-b3-forms-closedexact) this integral would vanish ([Proposition 21.28](#prop-b3-forms-exactloop)): it is not [exact](#def-b3-forms-closedexact). Were $\R^2\setminus\{0\}$ star-shaped with respect to some $p$, the Poincaré lemma (translated to $p$) would make every [closed form](#def-b3-forms-closedexact) [exact](#def-b3-forms-closedexact) — contradiction. Directly: for any $p
\neq 0$, the segment from $p$ to the point $-p$ of the domain passes through $0$: star-shapedness fails at every point.

**Exercise 21.5 ★★.**

(Area form of a hypersurface) Let $M \subseteq \R^n$ be a [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) oriented hypersurface whose [orientation](#def-b3-forms-orientation) is given by a unit normal field $\nu$ ($(w_1, \dots, w_{n-1})$ positive iff $(\nu, w_1, \dots, w_{n-1})$ positive in $\R^n$). Show that the $(n-1)$-form $\sigma_\nu =
\sum_i(-1)^{i-1}\nu_i\,\dd x_1\wedge\dots\wedge\widehat{\dd
x_i}\wedge\dots\wedge\dd x_n$ restricts on $M$ to the *area form*: for a direct parametrization $\gamma$, $\gamma^*\sigma_\nu = \sqrt{\det G}\,\dd u_1\wedge\dots
\wedge\dd u_{n-1}$, where $G = ({}^t D\gamma)(D\gamma)$ is the Gram matrix. *(Note that $(\gamma^*\sigma_\nu)(e_1, \dots, e_{n-1}) = \det(\nu,
\partial_1\gamma, \dots, \partial_{n-1}\gamma)$ and square this determinant.)* Compute $\int_{S^2}\sigma_\nu = 4\pi$.

**Solution of Exercise 21.5.**

For vectors $v_1, \dots, v_{n-1}$, expanding the determinant along its first column gives

$$
\det(\nu, v_1, \dots, v_{n-1}) =
\sum_{i=1}^n(-1)^{i-1}\nu_i\,
\det\bigl(\text{rows} \neq i \text{ of } (v_1, \dots,
v_{n-1})\bigr)
= \sigma_\nu(v_1, \dots, v_{n-1}),
$$

since $(\dd x_1\wedge\dots\wedge\widehat{\dd
x_i}\wedge\dots\wedge\dd x_n)(v_1, \dots, v_{n-1})$ is exactly the $i$-th deleted minor ([Definition 21.3](#def-b3-forms-wedge)). Applying this to $v_j =
\partial_j\gamma$: $(\gamma^*\sigma_\nu)(e_1, \dots,
e_{n-1}) = \det A$ with $A = (\nu\ \partial_1\gamma\ \cdots\
\partial_{n-1}\gamma)$. Now ${}^t\!AA$ is block diagonal: $\langle\nu, \nu\rangle = 1$ and $\langle\nu,
\partial_j\gamma\rangle = 0$ ($\nu$ normal, the $\partial_j\gamma$ tangent), so $(\det A)^2 =
\det({}^t\!AA) = \det G$; and $\det A > 0$ for a direct parametrization (that is what the $\nu$-orientation means): $\gamma^*\sigma_\nu = \sqrt{\det G}\,\dd
u_1\wedge\dots\wedge\dd u_{n-1}$, the Gram area element. For $S^2$, $\nu(x) = x$ and the spherical parametrization $\gamma(\theta, \varphi) = (\sin\varphi\cos\theta,
\sin\varphi\sin\theta, \cos\varphi)$ on $(0,2\pi)\times(0,\pi)$ (direct; it misses one meridian, a set that carries no area): $\det G = \sin^2\varphi$, so $\int_{S^2}\sigma_\nu =
\int_0^{2\pi}\!\!\int_0^\pi\sin\varphi\,\dd\varphi\,
\dd\theta = 4\pi$.

**Exercise 21.6 ★★.**

Compute $\int_{S^2}\omega$ for $\omega = x\,\dd y\wedge\dd z
+ y\,\dd z\wedge\dd x + z\,\dd x\wedge\dd y$: (a) directly in spherical coordinates; (b) via Stokes on the unit ball. Deduce $\operatorname{vol}(B^3) = \frac{4\pi}3$ from the area of $S^2$, and generalize: $n\operatorname{vol}(B^n) = \operatorname{area}(S^{n-1})$, consistent with [Theorem 11.13](https://one-course.com/books/math/5/en/chapter/11-product-measures-fubini-change-of-variables#thm-b3-product-ballvolume).

**Solution of Exercise 21.6.**

The given $\omega$ is $\sigma_\nu$ for $\nu(x) = x$ on $S^2$, so (a) is the computation just done: $\int_{S^2}\omega = 4\pi$. (b) $\dd\omega = 3\,\dd
x\wedge\dd y\wedge\dd z$, and Stokes on the unit ball gives $\int_{S^2}\omega = 3\operatorname{vol}(B^3)$: hence $\operatorname{vol}(B^3) = \frac{4\pi}3$. In general, the form $\sigma = \sum_i(-1)^{i-1}x_i\,\dd
x_1\wedge\dots\wedge\widehat{\dd x_i}\wedge\dots\wedge\dd
x_n$ restricts on $S^{n-1}$ to the area form ($\nu = x$ in [Exercise 21.5](#exo-b3-forms-5)), $\dd\sigma = n\,\dd
x_1\wedge\dots\wedge\dd x_n$, and Stokes yields $\operatorname{area}(S^{n-1}) = n\operatorname{vol}(B^n)$ — consistent with the Gamma-function formulas of [Theorem 11.13](https://one-course.com/books/math/5/en/chapter/11-product-measures-fubini-change-of-variables#thm-b3-product-ballvolume).

**Exercise 21.7 ★★.**

(The dictionaries) Derive carefully from [Theorem 21.23](#thm-b3-forms-stokes): (a) the divergence theorem in $\R^3$ (combine [Exercise 21.2](#exo-b3-forms-2) and [Exercise 21.5](#exo-b3-forms-5)); (b) the Kelvin–Stokes theorem $\int_S\langle\operatorname{curl}F, \nu\rangle\,\dd S =
\oint_{\partial S}\langle F, \tau\rangle\,\dd\ell$ for a [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) oriented surface with [boundary](#def-b3-forms-boundary) in $\R^3$. Check the [orientation](#def-b3-forms-orientation) conventions agree on the upper half-sphere bounded by the equator.

**Solution of Exercise 21.7.**

(a) Stokes applied to the flux form $\sigma_F$ on the [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) domain $\Omega$: $\int_\Omega\operatorname{div}F\,\dd x =
\int_{\partial\Omega}\sigma_F$ ([Exercise 21.2](#exo-b3-forms-2) for the [interior](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-interior) side). Identify the [boundary](#def-b3-forms-boundary) integrand: for tangent $v_1, v_2$ at a point of $\partial\Omega$, the first-column expansion of [Exercise 21.5](#exo-b3-forms-5) gives $\sigma_F(v_1, v_2) = \det(F, v_1, v_2)$; writing $F =
\langle F, \nu\rangle\nu + T$ with $T$ tangent, the column $T$ is a combination of $v_1, v_2$’s plane, so $\det(T,
v_1, v_2) = 0$ and $\sigma_F\vert_{\partial\Omega} = \langle F,
\nu\rangle\,\sigma_\nu = \langle F, \nu\rangle\,\dd S$: the divergence theorem, with $\nu$ the outward normal (outward-normal-first is exactly the induced [orientation](#def-b3-forms-orientation)). (b) Stokes applied to $\omega_F$ on the surface-with-boundary $S$: $\dd\omega_F =
\sigma_{\operatorname{curl}F}$ restricts to $\langle\operatorname{curl}F, \nu\rangle\,\dd S$ by the same identification, while on the [boundary](#def-b3-forms-boundary) curve $\gamma^*\omega_F = \langle F\circ\gamma,
\gamma'\rangle\,\dd t$, i.e. $\oint\langle F,
\tau\rangle\,\dd\ell$. Upper half-sphere with outward (radial) $\nu$: at the equator point $p = (1,0,0)$ the outward-within-the-surface vector is $-e_3$; completing it to positive frames shows the equator is traversed counterclockwise seen from above ($+e_3$): the right-hand rule, same convention on both sides of the identity.

**Exercise 21.8 ★★.**

(Green’s identities) For $u, v$ smooth on a [neighborhood](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-topology) of a [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) domain $\Omega \subseteq \R^n$ with smooth [boundary](#def-b3-forms-boundary), prove

$$
\int_\Omega\bigl(u\,\Delta v + \langle\nabla u, \nabla
v\rangle\bigr) = \int_{\partial\Omega}u\,\partial_\nu
v\,\dd S,
\qquad
\int_\Omega\bigl(u\,\Delta v - v\,\Delta u\bigr) =
\int_{\partial\Omega}\bigl(u\,\partial_\nu v -
v\,\partial_\nu u\bigr)\dd S .
$$

Deduce: a [harmonic function](https://one-course.com/books/math/5/en/chapter/18-conformal-maps-and-the-riemann-mapping-theorem#prop-b3-conformal-harmonicholo) on $\Omega$ vanishing on $\partial\Omega$ vanishes identically, and two [harmonic functions](https://one-course.com/books/math/5/en/chapter/18-conformal-maps-and-the-riemann-mapping-theorem#prop-b3-conformal-harmonicholo) with the same [boundary](#def-b3-forms-boundary) values coincide — uniqueness in the Dirichlet problem of [Chapter 18](https://one-course.com/books/math/5/en/chapter/18-conformal-maps-and-the-riemann-mapping-theorem#ch-b3-conformal).

**Solution of Exercise 21.8.**

Apply the divergence theorem ([Exercise 21.7](#exo-b3-forms-7)(a), whose proof is dimension-free) to $F = u\nabla v$: $\operatorname{div}(u\nabla v) = u\,\Delta v + \langle\nabla
u, \nabla v\rangle$ and $\langle F, \nu\rangle =
u\,\partial_\nu v$: the first identity. Swapping $u, v$ and subtracting cancels the symmetric term: the second. If $\Delta u = 0$ on $\Omega$ and $u = 0$ on $\partial\Omega$: the first identity with $v = u$ gives $\int_\Omega\norm{\nabla u}^2 = 0$, so $\nabla u \equiv 0$ and $u$ is constant on each component; every component’s [closure](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-interior) meets $\partial\Omega$ (boundedness), where $u = 0$: $u \equiv 0$. Two [harmonic functions](https://one-course.com/books/math/5/en/chapter/18-conformal-maps-and-the-riemann-mapping-theorem#prop-b3-conformal-harmonicholo) with equal [boundary](#def-b3-forms-boundary) values differ by such a $u$: they coincide — uniqueness for the Dirichlet problem, complementing the existence theory on the disc of [Chapter 18](https://one-course.com/books/math/5/en/chapter/18-conformal-maps-and-the-riemann-mapping-theorem#ch-b3-conformal).

**Exercise 21.9 ★★.**

(Change of variables, oriented form) Let $\varphi\colon U
\to V$ be a diffeomorphism of opens of $\R^n$ with $\det
D\varphi > 0$, and $f$ [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) with [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) support in $V$. Show that the [pullback](#def-b3-forms-pullback) identity $\int_U\varphi^*(f\,\dd x_1\wedge\dots\wedge\dd x_n) =
\int_Vf\,\dd x_1\wedge\dots\wedge\dd x_n$ is equivalent to the change-of-variables theorem ([Theorem 11.11](https://one-course.com/books/math/5/en/chapter/11-product-measures-fubini-change-of-variables#thm-b3-product-changeofvar)) for such $\varphi$, and explain exactly where the absolute value on the Jacobian went.

**Solution of Exercise 21.9.**

By [Theorem 21.11](#thm-b3-forms-pullback)(c), $\varphi^*(f\,\dd y_1\wedge\dots\wedge\dd y_n) =
(f\circ\varphi)\,\det(D\varphi)\,\dd
x_1\wedge\dots\wedge\dd x_n$, so the [pullback](#def-b3-forms-pullback) identity reads

$$
\int_U(f\circ\varphi)\,\det(D\varphi)\,\dd x = \int_Vf\,\dd
y .
$$

Since $\det D\varphi > 0$ everywhere, $\det D\varphi =
\abs{\det D\varphi}$, and this is verbatim the change-of-variables formula ([Theorem 11.11](https://one-course.com/books/math/5/en/chapter/11-product-measures-fubini-change-of-variables#thm-b3-product-changeofvar)) for [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) compactly supported integrands: each statement is the other. The absolute value went into the *hypothesis*: [orientation](#def-b3-forms-orientation). For orientation-reversing $\varphi$ the form identity acquires a global minus sign (forms feel [orientation](#def-b3-forms-orientation)), while the [measure](https://one-course.com/books/math/5/en/chapter/9-measure-theory#def-b3-measure-measure) formula keeps $\abs{\det}$ ([measures](https://one-course.com/books/math/5/en/chapter/9-measure-theory#def-b3-measure-measure) do not): two bookkeepings of one Jacobian.

**Exercise 21.10 ★★★.**

(Deformation) Let $\omega$ be a *[closed](#def-b3-forms-closedexact)* $2$-form on $\R^3\setminus\{0\}$ and $S_r$ the sphere of radius $r$ centered at $0$. Show that $\int_{S_r}\omega$ does not depend on $r > 0$ *(apply Stokes to the shell between two radii; mind the two [boundary](#def-b3-forms-boundary) [orientations](#def-b3-forms-orientation))*. Apply to the *solid-angle form*

$$
\omega = \frac{x\,\dd y\wedge\dd z + y\,\dd z\wedge\dd x +
z\,\dd x\wedge\dd y}{(x^2 + y^2 + z^2)^{3/2}} :
$$

check it is [closed](#def-b3-forms-closedexact), compute $\int_{S_r}\omega = 4\pi$, and conclude it is [closed](#def-b3-forms-closedexact) but not [exact](#def-b3-forms-closedexact) on $\R^3\setminus\{0\}$ — the two-dimensional sibling of $\omega_\theta$, and the geometric content of Gauss’s law in electrostatics.

**Solution of Exercise 21.10.**

The shell $A = \{r_1 \leq \norm x \leq r_2\}$ is a [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) $3$-submanifold with [boundary](#def-b3-forms-boundary) $S_{r_2}\cup S_{r_1}$; the induced [orientations](#def-b3-forms-orientation) are the usual sphere [orientation](#def-b3-forms-orientation) on $S_{r_2}$ (outward from $A$ = away from $0$) and the *opposite* on $S_{r_1}$ (outward from $A$ = toward $0$). Stokes with $\dd\omega = 0$:

$$
0 = \int_A\dd\omega = \int_{S_{r_2}}\omega -
\int_{S_{r_1}}\omega .
$$

Solid-angle form: with $\rho = \norm x$ and $\sigma =
x\,\dd y\wedge\dd z + y\,\dd z\wedge\dd x + z\,\dd
x\wedge\dd y$, $\omega = \rho^{-3}\sigma$ and

$$
\dd\omega = \sum_i\partial_i\bigl(x_i\rho^{-3}\bigr)\,
\dd x\wedge\dd y\wedge\dd z
= \bigl(3\rho^{-3} - 3\rho^{-5}\cdot\rho^2\bigr)\dd
x\wedge\dd y\wedge\dd z = 0 .
$$

On $S_r$, for tangent $v_1, v_2$: $\omega(v_1, v_2) =
r^{-3}\det(x, v_1, v_2) = r^{-3}\,r\det(\nu, v_1, v_2) =
r^{-2}\,\dd S(v_1, v_2)$, so $\int_{S_r}\omega =
r^{-2}\cdot4\pi r^2 = 4\pi$: constant in $r$, as deformation predicts, and nonzero — so $\omega$ is [closed](#def-b3-forms-closedexact) but not [exact](#def-b3-forms-closedexact) on $\R^3\setminus\{0\}$ (an [exact form](#def-b3-forms-closedexact) integrates to $0$ over the boundaryless $S_r$ by Stokes). This is Gauss’s law: the flux of the field of a unit charge through any enclosing sphere is $4\pi$, whatever the radius.

**Exercise 21.11 ★★.**

Let $\gamma$ be a smooth [closed](#def-b3-forms-closedexact) curve in $\R^2\setminus\{0\}$ with $n = \operatorname{Ind}_\gamma(0)$. Show $\int_\gamma\omega = n\int_{S^1}\omega$ for *every* [closed](#def-b3-forms-closedexact) $1$-form $\omega$ on $\R^2\setminus\{0\}$ *(write $\omega = c\,\omega_\theta
+ \dd f$ by [Exercise 21.12](#exo-b3-forms-12))*. Interpretation: on the punctured plane, the [winding number](#def-b3-forms-winding) is the only obstruction to the vanishing of periods.

**Solution of Exercise 21.11.**

Write $\omega = c\,\omega_\theta + \dd f$ ([Exercise 21.12](#exo-b3-forms-12)) with $c =
\frac1{2\pi}\int_{S^1}\omega$. Then

$$
\int_\gamma\omega = c\int_\gamma\omega_\theta +
\int_\gamma\dd f
= c\cdot2\pi\operatorname{Ind}_\gamma(0) + 0
= n\int_{S^1}\omega,
$$

by [Proposition 21.28](#prop-b3-forms-exactloop) and the definition of the index. The single integer $n$ controls every period on the punctured plane: [closed](#def-b3-forms-closedexact) $1$-forms cannot distinguish two loops with the same [winding number](#def-b3-forms-winding).

**Exercise 21.12 ★★★.**

(First de Rham computation) Show that every [closed](#def-b3-forms-closedexact) $1$-form $\omega$ on $U = \R^2\setminus\{0\}$ is uniquely

$$
\omega = c\,\omega_\theta + \dd f,
\qquad c = \frac1{2\pi}\int_{S^1}\omega,\quad f \in
\mathcal C^\infty(U) :
$$

define $f(p)$ by integrating $\omega - c\,\omega_\theta$ along a [path](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-pathconnected) from $(1,0)$ to $p$ (radial piece then circular arc), show the result is independent of the choices precisely because the $S^1$-period vanishes, and check $\dd
f = \omega - c\,\omega_\theta$. Conclude: $H^1(\R^2\setminus\{0\}) \cong \R$, generated by the angular form.

**Solution of Exercise 21.12.**

Let $\alpha = \omega - c\,\omega_\theta$: [closed](#def-b3-forms-closedexact), and $\int_{S^1}\alpha = 0$ by the choice of $c$ ($\int_{S^1}\omega_\theta = 2\pi$). Pull back by the polar map $\Phi(\rho, \theta) = (\rho\cos\theta, \rho\sin\theta)$, a surjective local diffeomorphism $(0, \infty)\times\R \to
U$: $\Phi^*\alpha$ is [closed](#def-b3-forms-closedexact) ([Theorem 21.11](#thm-b3-forms-pullback)(b)) on the *convex* open $(0,\infty)\times\R$, hence [exact](#def-b3-forms-closedexact) ([Theorem 21.15](#thm-b3-forms-poincare)): $\Phi^*\alpha = \dd g$. For fixed $\rho$: $g(\rho, \theta + 2\pi) - g(\rho, \theta)
= \int_\theta^{\theta + 2\pi}\partial_\theta g\,\dd s$ is the integral of $\alpha$ around the circle of radius $\rho$, which equals $\int_{S^1}\alpha = 0$ (the annulus between the two circles is a [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) surface with [boundary](#def-b3-forms-boundary); Stokes as in [Exercise 21.10](#exo-b3-forms-10), one dimension down). So $g$ is $2\pi$-periodic in $\theta$ and descends to a well-defined function $f$ on $U$ with $f\circ\Phi = g$; $f$ is smooth ($\Phi$ is a local diffeomorphism) and $\Phi^*(\dd f) = \dd g = \Phi^*\alpha$ forces $\dd f =
\alpha$. Hence $\omega = c\,\omega_\theta + \dd f$. Uniqueness: integrating over $S^1$ fixes $c$, since [exact forms](#def-b3-forms-closedexact) have zero period; and $f$ is unique up to an additive constant. The map $[\omega] \mapsto
\frac1{2\pi}\int_{S^1}\omega$ is therefore a linear isomorphism $H^1(\R^2\setminus\{0\}) \to \R$, and the class of $\omega_\theta$ generates: the hole is exactly one-dimensional, cohomologically speaking.

## 21.8 Problem: Brouwer’s fixed-point theorem

**Problem 21.1.**

Weekend problem — no retraction, no escape

Brouwer’s theorem states that *every [continuous map](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) of the [closed](#def-b3-forms-closedexact) unit ball $\bar B = \bar B^n \subseteq \R^n$ into itself has a fixed point* — one of mathematics’ great theorems, with consequences from game theory (Nash equilibria) to matrix analysis. The differential-form proof is the cleanest known: Stokes shows the sphere is not a retract of the ball, and everything follows. Throughout, $S = S^{n-1} = \partial\bar B$, $n \geq 2$, and

$$
\sigma = \sum_{i=1}^n(-1)^{i-1}x_i\,
\dd x_1\wedge\dots\wedge\widehat{\dd x_i}\wedge\dots\wedge
\dd x_n
$$

(the hat deletes the factor).

**Part I — The measuring instrument.**

1. Compute $\dd\sigma$ , and deduce from Stokes ( [Theorem 21.23](#thm-b3-forms-stokes) ) that $\int_S\sigma =  n\operatorname{vol}(\bar B) > 0$ , where $S$ carries the [boundary](#def-b3-forms-boundary) [orientation](#def-b3-forms-orientation) of the ball.
2. For $n = 2$ and $n = 3$ , identify the restriction of $\sigma$ to $S$ with the arclength and area forms ( [Exercise 21.5](#exo-b3-forms-5) with $\nu(x) = x$ ) and recompute $\int_S\sigma$ directly.
3. Let $W \subseteq \R^N$ be open and $\varphi\colon W  \to \R^n$ smooth with $\norm{\varphi(x)} = 1$ for all $x \in W$ . Show that $\varphi^*(\dd\sigma) = 0$ . *(Differentiate $\norm\varphi^2 = 1$: the image of $D\varphi(x)$ lies in the hyperplane $\varphi(x)^\perp$, of dimension $n - 1$; then apply [Proposition 21.6](#prop-b3-forms-linearpullback)(2).)*
4. Where is the flaw in the following “proof” that $\int_S\sigma = 0$ : “ $S$ is [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) without [boundary](#def-b3-forms-boundary) , and $\sigma$ restricted to $S$ is a top form on it, hence [closed](#def-b3-forms-closedexact) , hence $\int_S\sigma =  \int_S\dd(\text{something}) = 0$ by Stokes”? (Pinpoint the wrong word.)
5. Explain in one paragraph the strategy of Part II: what will be integrated, over what, and where the contradiction will come from.

**Part II — No smooth retraction.** Suppose, for contradiction, that $r$ is a *smooth retraction* of the ball onto its sphere: $r$ is smooth on a [neighborhood](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-topology) of $\bar B$, $r(\bar B) \subseteq S$, and $r(x) = x$ for all $x \in S$.

6. Justify $\int_Sr^*\sigma = \int_S\sigma$ *(on $S$, $r$ restricts to the identity: if $\gamma$ is a direct parametrization of a piece of $S$, then $r\circ\gamma = \gamma$)* .
7. Using Stokes on $\bar B$ , show $\int_Sr^*\sigma =  \int_{\bar B}\dd(r^*\sigma)$ .
8. Show $\dd(r^*\sigma) = r^*(\dd\sigma) = 0$ (question 3 applied to $\varphi = r$ ), and conclude: *there is no smooth retraction $\bar B \to S$.*
9. Settle the excluded case $n = 1$ by hand: show directly that no [continuous map](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) $\intcc{-1}1 \to  \{-1, 1\}$ fixes both endpoints, and name the theorem you used.

**Part III — Smooth Brouwer.** Let $g$ be smooth on a [neighborhood](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-topology) of $\bar B$ with $g(\bar B)
\subseteq \bar B$ and *no* fixed point in $\bar B$.

10. Show $\delta = \min_{x\in\bar B}\norm{g(x) - x} >  0$ .
11. For $x \in \bar B$ let $u(x) = \frac{x -  g(x)}{\norm{x - g(x)}}$ and let $r(x) = x +  t(x)\,u(x)$ be the intersection of the ray $\{x + tu(x) : t \geq 0\}$ with $S$. Solve the quadratic $\norm{x + tu}^2 = 1$ and obtain $$t(x) = -\langle x, u(x)\rangle +  \sqrt{1 - \norm x^2 + \langle x, u(x)\rangle^2}  \;\geq\; 0 .$$
12. Show that the radicand is *strictly* positive on $\bar B$ : if $1 - \norm x^2 + \langle x,  u(x)\rangle^2 = 0$ then $\norm x = 1$ and $\langle  x, u(x)\rangle = 0$ , i.e. $\langle x, x -  g(x)\rangle = 0$ , i.e. $\langle x, g(x)\rangle =  1$ ; by Cauchy–Schwarz with $\norm x = 1$ , $\norm{g(x)} \leq 1$ , this forces $g(x) = x$ — excluded. Deduce that $r$ is smooth on a [neighborhood](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-topology) of $\bar B$ .
13. Show $r(\bar B) \subseteq S$ and $r(x) = x$ for $x  \in S$ *(for $\norm x = 1$, check $t(x) = 0$ using $\langle x, u(x)\rangle \geq 0$, which itself follows from $\langle x, x - g(x)\rangle = 1 -  \langle x, g(x)\rangle \geq 0$)* . Conclude with Part II: *every smooth self-map of $\bar B$ has a fixed point.*

**Part IV — [Continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) Brouwer.** Let $f\colon\bar B\to\bar B$ be [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) without fixed point.

14. Show $\varepsilon = \min_{\bar B}\norm{f - x} > 0$ , and produce a *polynomial* map $p\colon\R^n\to  \R^n$ with $\sup_{\bar B}\norm{p - f} <  \varepsilon/2$ (Stone–Weierstrass, [Theorem 7.15](https://one-course.com/books/math/5/en/chapter/7-complete-spaces-baire-ascoli-stoneweierstrass#thm-b3-complete-stoneweierstrass) , coordinate by coordinate — justify the passage from scalar to vector approximation).
15. The map $p$ may leave the ball; set $g = \frac{p}{1  + \varepsilon/2}$ . Show $g(\bar B) \subseteq \bar B$ and $\sup_{\bar B}\norm{g - f} < \varepsilon$ .
16. Derive a contradiction with Part III and conclude: *every [continuous map](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) $\bar B^n \to \bar B^n$ has a fixed point.*
17. Show by example that the theorem fails on: the open ball; the sphere $S$ ; a [closed](#def-b3-forms-closedexact) annulus. Which property of $\bar B$ does each counterexample lose?

**Part V — Dividends.**

18. (Perron–Frobenius, existence) Let $A$ be an $n\times n$ matrix with all entries $> 0$ , and $\Delta = \{x \in \R^n : x_i \geq 0,\ \sum x_i =  1\}$ . Show the map $x \mapsto Ax/\norm{Ax}_1$ is well defined and [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) on $\Delta$ , that $\Delta$ is [homeomorphic](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) to a [closed](#def-b3-forms-closedexact) ball of $\R^{n-1}$ (radial [homeomorphism](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) from a convex [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) with nonempty [interior](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-interior) in its affine span), and conclude that $A$ has an eigenvector with strictly positive entries and eigenvalue $> 0$ .
19. Deduce that every stochastic matrix with positive entries (columns summing to $1$ ) has a stationary probability vector $\pi = A\pi$ — the PageRank-type vector. (Uniqueness holds too but needs other tools.)
20. (Hairy ball, setup) Let $v$ be smooth on a [neighborhood](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-topology) of $S = S^{n-1}$ with $\langle v(x),  x\rangle = 0$ and $\norm{v(x)} = 1$ for $x \in S$ (a *unit tangent field* ). For $t \in \R$ set $F_t(x) = x + t\,v(x)$ . Show $\norm{F_t(x)} = \sqrt{1 + t^2}$ on $S$ : $F_t$ maps $S$ into the sphere $\sqrt{1+t^2}\,S$ .
21. Show that $P(t) = \int_SF_t^*\sigma$ is a *polynomial* in $t$ *(each coefficient function of $F_t^*\sigma$ in a chart is polynomial in $t$, with coefficients smooth in the chart variable; integration is linear)* .
22. Show that for $\abs t$ small, $F_t$ is a diffeomorphism from $S$ onto $\sqrt{1+t^2}\,S$: injectivity for $t\operatorname{Lip}(v) < 1$; local diffeomorphism by the inverse function theorem ([Theorem 20.1](https://one-course.com/books/math/5/en/chapter/20-submanifolds-of-rn#thm-b3-submanifolds-ift) in charts); image open and [closed](#def-b3-forms-closedexact) in the [connected](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-connected) target sphere. Deduce, using [Lemma 21.19](#lem-b3-forms-consistency) and the scaling $\sigma_{\lambda x} =  \lambda^{n}\,\sigma_x$ under $x \mapsto \lambda x$ (check it), that $$P(t) = \pm(1 + t^2)^{n/2}\int_S\sigma,  \qquad\text{with the sign } + \text{ for small } t$$ ([orientation](#def-b3-forms-orientation) preserved by [continuity](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) from $t = 0$).
23. Conclude (Milnor): if $n$ is *odd* , $(1 +  t^2)^{n/2}$ is not a polynomial in $t$ , yet it agrees with the polynomial $P(t)/\int_S\sigma$ near $0$ — contradiction. Hence *the even-dimensional spheres $S^{n-1}$ ($n$ odd) carry no unit tangent field* , and, by normalizing and smoothing (convolve componentwise and project — justify both steps), no [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) nowhere-vanishing tangent field at all: every wind on Earth leaves a calm point.
24. (Odd spheres comb freely) Exhibit on $S^{2m-1}  \subseteq \R^{2m} \cong \C^m$ an explicit smooth unit tangent field: $v(x) = \iu x$ in complex notation, i.e. $$v(x_1, y_1, \dots, x_m, y_m) =  (-y_1, x_1, \dots, -y_m, x_m) .$$ Verify tangency and unit length, and conclude that the parity dichotomy of question 23 is sharp: a sphere is combable exactly when its dimension is odd. Where does the polynomial argument of question 22 break for even $n$?
25. (Zeros from [boundary](#def-b3-forms-boundary) behavior) Let $f \colon \bar  B^n \to \R^n$ be [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) with $\langle f(x),  x\rangle \geq 0$ for every $x \in S^{n-1}$ . Show that $f$ vanishes somewhere in $\bar B^n$ . *(If not, $g(x) = -f(x)/\norm{f(x)}$ maps $\bar B$ [continuously](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) into $S \subseteq \bar B$; apply Brouwer to $g$ and contradict the [boundary](#def-b3-forms-boundary) hypothesis.)* Deduce the *surjectivity criterion* : a [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) $F\colon\R^n\to\R^n$ with $\frac{\langle F(x), x\rangle}{\norm x} \to  +\infty$ as $\norm x \to \infty$ is surjective — the finite-dimensional ancestor of the coercivity arguments of nonlinear analysis.

**Solution of Problem 21.1.**

**1.** $\dd\sigma = \sum_i(-1)^{i-1}\,\dd x_i\wedge\dd
x_1\wedge\dots\wedge\widehat{\dd x_i}\wedge\dots\wedge\dd
x_n$; carrying $\dd x_i$ across the $i-1$ preceding factors costs $(-1)^{i-1}$, which cancels the prefactor: each of the $n$ terms equals $\dd x_1\wedge\dots\wedge\dd x_n$, so $\dd\sigma = n\,\dd x_1\wedge\dots\wedge\dd x_n$. Stokes on the ball: $\int_S\sigma = \int_{\bar B}\dd\sigma =
n\operatorname{vol}(\bar B) > 0$.

**2.** $n = 2$: $\sigma = x\,\dd y - y\,\dd x$; on $\gamma(t) = (\cos t, \sin t)$, $\gamma^*\sigma = (\cos^2t +
\sin^2t)\,\dd t = \dd t$, the arclength form: $\int_S\sigma = 2\pi = 2\operatorname{vol}(\bar B^2)$. $n = 3$: $\sigma\vert_S$ is the area form ([Exercise 21.5](#exo-b3-forms-5) with $\nu(x) = x$): $\int_S\sigma = 4\pi = 3\cdot\tfrac{4\pi}3$. Both match question 1.

**3.** Differentiating $\norm\varphi^2 = 1$: $2\langle
D\varphi(x)h, \varphi(x)\rangle = 0$ for every $h$, so $\operatorname{im}D\varphi(x) \subseteq \varphi(x)^\perp$, a hyperplane: $\operatorname{rk}D\varphi(x) \leq n - 1$. Since $\dd\sigma$ is an $n$-form (question 1), pointwise $\varphi^*(\dd\sigma)_x =
(D\varphi(x))^*(\dd\sigma)_{\varphi(x)} = 0$ by [Proposition 21.6](#prop-b3-forms-linearpullback)(2): a map into the sphere has no room to pull back a volume.

**4.** The flaw is the second “hence”: on the $(n-1)$-dimensional $S$, every $(n-1)$-form is trivially [closed](#def-b3-forms-closedexact) (there are no nonzero $n$-forms on an $(n-1)$-manifold), but *[closed](#def-b3-forms-closedexact) does not mean [exact](#def-b3-forms-closedexact)*, and $\int_S\dd\eta = 0$ requires an actual primitive $\eta$ defined on $S$. The restriction of $\sigma$ is precisely not [exact](#def-b3-forms-closedexact) — its integral is $n\operatorname{vol}(\bar B) \neq
0$ — and this non-exactness powers the entire problem.

**5.** We shall integrate $r^*\sigma$ over the sphere and count two ways. Because $r$ fixes $S$ pointwise, the integral equals $\int_S\sigma = n\operatorname{vol}(\bar B)
\neq 0$. Because $r$ is defined on the ball, Stokes converts the same integral into $\int_{\bar B}\dd(r^*\sigma) =
\int_{\bar B}r^*(\dd\sigma)$; and because $r$ takes values in the sphere, question 3 makes that integrand vanish. One number, two values: the retraction cannot exist.

**6.** Both integrals are computed through direct parametrizations $\gamma$ of pieces of $S$ ([Definition 21.18](#def-b3-forms-integral)); since $r\circ\gamma =
\gamma$ (the parametrization lands in $S$, where $r$ is the identity), $\gamma^*(r^*\sigma) = (r\circ\gamma)^*\sigma =
\gamma^*\sigma$ ([Theorem 21.11](#thm-b3-forms-pullback)(a)): the local integrands coincide, and any [partition of unity](#lem-b3-forms-partition) gives $\int_Sr^*\sigma = \int_S\sigma$.

**7.** $r^*\sigma$ is a smooth $(n-1)$-form on a [neighborhood](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-topology) of the [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) oriented $\bar B$, whose [boundary](#def-b3-forms-boundary) with the induced [orientation](#def-b3-forms-orientation) is $S$: Stokes ([Theorem 21.23](#thm-b3-forms-stokes)) gives exactly $\int_Sr^*\sigma = \int_{\bar B}\dd(r^*\sigma)$.

**8.** $\dd(r^*\sigma) = r^*(\dd\sigma)$ ([Theorem 21.11](#thm-b3-forms-pullback)(b)), which vanishes by question 3 applied to $\varphi = r$. Chaining questions 6–8:

$$
0 < n\operatorname{vol}(\bar B) = \int_S\sigma =
\int_Sr^*\sigma = \int_{\bar B}\dd(r^*\sigma) = 0 :
$$

absurd. *There is no smooth retraction of $\bar B^n$ onto $S^{n-1}$* ($n \geq 2$).

**9.** A [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) $r\colon\intcc{-1}1\to\{-1,1\}$ with $r(\pm1) = \pm1$ would map a [connected](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-connected) set onto the disconnected $\{-1, 1\}$, impossible: [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) images of [connected](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-connected) sets are [connected](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-connected) — equivalently, the intermediate value theorem would force $r$ to take the value $0$. The same statement in every dimension is exactly Part II; [connectedness](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-connected) is the $1$-dimensional shadow of the cohomological obstruction $\int_S\sigma \neq 0$.

**10.** $x \mapsto \norm{g(x) - x}$ is [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) and everywhere $> 0$ on the [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) $\bar B$: its minimum $\delta$ is attained, hence $> 0$.

**11.** $\norm{x + tu}^2 = 1$ reads $t^2 + 2t\langle x,
u\rangle + \norm x^2 - 1 = 0$, whose roots are

$$
t_\pm = -\langle x, u\rangle \pm \sqrt{\langle x,
u\rangle^2 + 1 - \norm x^2} .
$$

Their product is $\norm x^2 - 1 \leq 0$: the roots straddle $0$ (or one vanishes), so the ray parameter — the nonnegative root — is $t(x) = t_+$.

**12.** If the radicand vanished at $x \in \bar B$: both its terms being nonnegative, $\norm x = 1$ and $\langle x, u(x)\rangle = 0$, i.e. $\langle x, x -
g(x)\rangle = 0$, i.e. $\langle x, g(x)\rangle = 1$. By Cauchy–Schwarz, $1 = \langle x, g(x)\rangle \leq
\norm x\,\norm{g(x)} \leq 1$: equality throughout, which forces $g(x)$ collinear with $x$, of norm $1$, positively: $g(x) = x$ — excluded. So the radicand is [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) and $> 0$ on $\bar B$, hence bounded below by some $c > 0$ there and on a [neighborhood](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-topology) (uniform [continuity](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity)). On that [neighborhood](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-topology), $u$ is smooth ($\norm{x - g(x)} \geq
\delta/2$ shrinking if needed), the radicand stays $\geq
c/2$, and the square root is smooth on $\intoo0\infty$: $r$ is smooth near $\bar B$.

**13.** $\norm{r(x)} = 1$ by construction of $t(x)$: $r(\bar B) \subseteq S$. For $x \in S$: $\langle x,
u(x)\rangle = \frac{1 - \langle x, g(x)\rangle}{\norm{x -
g(x)}} \geq 0$ (Cauchy–Schwarz once more), and $\norm x =
1$ reduces the radicand to $\langle x, u\rangle^2$, whose square root is $\langle x, u\rangle$ itself (it is $\geq
0$): $t(x) = 0$ and $r(x) = x$. So $r$ is a smooth retraction of the ball onto the sphere — contradicting Part II. *Every smooth self-map of $\bar B$ has a fixed point.*

**14.** $\varepsilon > 0$ exactly as in question 10. The polynomials form a subalgebra of $\mathcal C(\bar B,
\R)$ containing the constants and separating points ($x
\mapsto x_i$ do), so Stone–Weierstrass ([Theorem 7.15](https://one-course.com/books/math/5/en/chapter/7-complete-spaces-baire-ascoli-stoneweierstrass#thm-b3-complete-stoneweierstrass)) approximates each coordinate: pick polynomials $p_i$ with $\sup_{\bar
B}\abs{p_i - f_i} < \frac{\varepsilon}{2\sqrt n}$; the vector map $p = (p_1, \dots, p_n)$ then satisfies $\sup_{\bar B}\norm{p - f} \leq
\bigl(\sum_i\sup_{\bar B}\abs{p_i - f_i}^2\bigr)^{1/2} <
\varepsilon/2$.

**15.** On $\bar B$: $\norm p \leq \norm f +
\frac\varepsilon2 \leq 1 + \frac\varepsilon2$, so $\norm{g} = \frac{\norm p}{1 + \varepsilon/2} \leq 1$: $g(\bar B) \subseteq \bar B$. Moreover $\norm{g - p} =
\frac{\varepsilon/2}{1 + \varepsilon/2}\norm p \leq
\frac\varepsilon2$, hence $\norm{g - f} \leq \norm{g - p} +
\norm{p - f} < \varepsilon$ on $\bar B$.

**16.** $g$ is polynomial, hence smooth, and maps $\bar B$ into itself: Part III provides $x_0 = g(x_0)$. Then $\norm{f(x_0) - x_0} = \norm{f(x_0) - g(x_0)} <
\varepsilon = \min_{\bar B}\norm{f - \operatorname{id}}$: contradiction. *Every [continuous map](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) $\bar B^n \to
\bar B^n$ has a fixed point.*

**17.** *Open ball:* $f(x) = \frac{x + e_1}2$ maps $B$ into $B$ ($\norm{f(x)} < 1$ strictly) and its only fixed point $e_1$ lies on the sphere: compactness lost. *Sphere:* the antipodal map $x \mapsto -x$ is fixed-point free; $S$ is [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) but has the “wrong” [topology](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-topology) — it is exactly the non-retract of Part II. *Annulus:* a rotation by any angle $\not\equiv 0$ fixes nothing; the hole shelters the rotation — convexity (more precisely, the ball-like [topology](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-topology)) lost. Brouwer’s theorem is really about *[compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) convex* sets, as question 18 exploits.

**18.** For $x \in \Delta$: some $x_j > 0$, so $(Ax)_i
\geq A_{ij}x_j > 0$ for every $i$; hence $\norm{Ax}_1 > 0$ and $T(x) = Ax/\norm{Ax}_1$ is well defined, [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity), and lands in $\Delta$ (positive entries summing to $1$). $\Delta$ is convex, [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact), with nonempty [interior](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-interior) in the affine hyperplane $\{\sum x_i = 1\} \cong \R^{n-1}$; the radial map from its barycenter — each ray from the barycenter meets $\partial\Delta$ in exactly one point, by convexity and compactness, and the corresponding gauge function is [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) — is a [homeomorphism](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) $\Delta \to
\bar B^{n-1}$. Transporting Brouwer through it: $T$ has a fixed point $x^*$, i.e. $Ax^* = \lambda x^*$ with $\lambda
= \norm{Ax^*}_1 > 0$; and $x^* = Ax^*/\lambda$ has strictly positive entries by the opening computation. *A positive matrix has a positive eigenvector.*

**19.** Question 18 gives $A\pi = \lambda\pi$, $\pi \in
\Delta$, $\pi > 0$. Sum the coordinates: $\sum_i(A\pi)_i =
\sum_j\pi_j\sum_iA_{ij} = \sum_j\pi_j = 1$ (columns sum to $1$), while $\sum_i\lambda\pi_i = \lambda$. So $\lambda = 1$ and $A\pi = \pi$: a stationary probability vector — the equilibrium of the Markov chain, PageRank’s mathematical core.

**20.** On $S$: $\norm{F_t(x)}^2 = \norm x^2 +
2t\langle x, v(x)\rangle + t^2\norm{v(x)}^2 = 1 + 0 + t^2$ by tangency and $\norm v = 1$: $F_t(S) \subseteq
\sqrt{1+t^2}\,S$.

**21.** Fix a finite atlas of direct parametrizations $\gamma$ and a [partition of unity](#lem-b3-forms-partition), independent of $t$. In a chart, $F_t\circ\gamma = \gamma + t(v\circ\gamma)$, so each coefficient of $(F_t\circ\gamma)^*\sigma$ is a sum of products of one factor $(x_i + tv_i)\circ\gamma$ (affine in $t$) and an $(n-1)\times(n-1)$ determinant with entries affine in $t$: a polynomial in $t$ of degree $\leq n$ with coefficients smooth in the chart variable. Multiplying by the $t$-independent partition functions and integrating termwise: $P(t) = \sum_{k=0}^n c_kt^k$, a polynomial.

**22.** *Injectivity:* $v$ is Lipschitz on $S$ (smooth on a [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact)), say with constant $L$; for $\abs t <
1/L$, $\norm{F_t(x) - F_t(y)} \geq (1 - \abs
tL)\norm{x - y} > 0$. *Diffeomorphism:* $G_t =
F_t/\sqrt{1 + t^2}$ maps $S$ to $S$; in charts its Jacobians converge uniformly to those of $G_0 = \operatorname{id}$ as $t \to 0$, so for small $t$ they are invertible and $G_t$ is a local diffeomorphism ([Theorem 20.1](https://one-course.com/books/math/5/en/chapter/20-submanifolds-of-rn#thm-b3-submanifolds-ift) in charts), injective, with image open (local diffeo) and [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) in the *[connected](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-connected)* $S^{n-1}$ ($n \geq 2$): image $= S$, so $G_t$ is a diffeomorphism of $S$. *Scaling:* under $s_\lambda(x) = \lambda x$, each coefficient $x_i$ gains $\lambda$ and each of the $n - 1$ differentials gains $\lambda$: $s_\lambda^*\sigma =
\lambda^n\sigma$. Since $F_t = s_{\sqrt{1+t^2}}\circ G_t$:

$$
P(t) = \int_SG_t^*\bigl(s_{\sqrt{1+t^2}}^{\,*}\sigma\bigr)
= (1 + t^2)^{n/2}\int_SG_t^*\sigma
= (1 + t^2)^{n/2}\int_S\sigma
\quad\text{for small } t,
$$

the last equality by [Lemma 21.19](#lem-b3-forms-consistency) ($G_t$ a diffeomorphism of $S$, orientation-preserving for small $t$: its chart Jacobian determinants vary [continuously](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity), never vanish, and are positive at $t = 0$).

**23.** If $n$ is odd and a smooth unit tangent field exists, questions 21–22 make the polynomial $P(t)/\int_S\sigma$ (legitimate: $\int_S\sigma \neq 0$ by question 1) agree with $(1 + t^2)^{n/2}$ near $0$; two smooth functions agreeing near $0$ with one a polynomial force $(1 + t^2)^{n/2}$ to be that polynomial on all of $\R$. But if $Q(t)^2 = (1 + t^2)^n$ with $Q \in \R[t]$, unique factorization in $\R[t]$ ([Chapter 2](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#ch-b3-rings)) gives the [irreducible](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#def-b3-rings-divisibility) $t^2 + 1$ an even multiplicity in $Q^2$ and the odd multiplicity $n$ in $(1 + t^2)^n$: impossible. So no smooth unit tangent field exists on $S^{n-1}$ for $n$ odd — the even-dimensional spheres. Finally, a merely [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) nowhere-vanishing tangent field $w$ would produce one: extend it to a [neighborhood](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-topology) by $\tilde w(x) = w(x/\norm x)$, mollify componentwise ([Theorem 12.9](https://one-course.com/books/math/5/en/chapter/12-the-lp-spaces#thm-b3-lp-regularization)) to a smooth $v_0$ with $\sup_S\norm{v_0 - \tilde w} <
\frac12\min_S\norm w$, project tangentially $v_1(x) = v_0(x)
- \langle v_0(x), x\rangle x$ — on $S$ this changes $v_0$ by at most its normal component, itself at most $\norm{v_0 -
\tilde w}$ since $\tilde w$ is tangent, so $\norm{v_1 - \tilde
w} \leq 2\norm{v_0 - \tilde w} < \min\norm w$ and $v_1$ never vanishes on $S$ — and normalize: $v =
v_1/\norm{v_1}$ is a smooth unit tangent field. Hence on every even-dimensional sphere, *every [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) tangent field has a zero*: every wind on Earth leaves a calm point.

**24.** $\langle v(x), x\rangle = \sum_j(-y_jx_j +
x_jy_j) = 0$: tangent; $\norm{v(x)} = \norm x = 1$: unit. Smoothness is clear (linear map). So every odd-dimensional sphere carries a smooth unit tangent field — multiplication by $\iu$ along the complex lines — and question 23’s obstruction is exactly the parity of the dimension. In the polynomial argument, for even $n$ the function $(1 +
t^2)^{n/2}$ *is* a polynomial, and no contradiction arises: the proof does not merely fail to apply, its conclusion is genuinely false, as $v$ witnesses.

**25.** Suppose $f$ never vanishes on $\bar B$. Then $g(x) = -f(x)/\norm{f(x)}$ is [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) $\bar B \to S
\subseteq \bar B$, and Brouwer (question 16) provides $x^* =
g(x^*)$. Since $g$ takes values in $S$, $x^* \in S$, and

$$
\langle f(x^*), x^*\rangle = \bigl\langle f(x^*),\
-\tfrac{f(x^*)}{\norm{f(x^*)}}\bigr\rangle
= -\norm{f(x^*)} < 0,
$$

contradicting the [boundary](#def-b3-forms-boundary) hypothesis. So $f$ has a zero. Surjectivity: given $y \in \R^n$, apply the above to $f(x) =
F(x) - y$ on a ball $\bar B(0, R)$ with $R$ so large that $\langle F(x), x\rangle \geq \norm y\,\norm x$ on the sphere of radius $R$ (coercivity); then $\langle f(x), x\rangle =
\langle F(x), x\rangle - \langle y, x\rangle \geq 0$ there (Cauchy–Schwarz), and the rescaled statement gives a zero of $f$: $F(x) = y$. Every coercive [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) field is onto — the degree-free shadow of the variational existence theorems, delivered by pure [topology](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-topology).