---
title: "Submanifolds of ℝn"
book: "University Mathematics — Year 3"
subject: math
language: en
chapter: 20
exercises: 12
source: https://one-course.com/books/math/5/en/chapter/20-submanifolds-of-rn
---

# Chapter 20 — Submanifolds of ℝn

Spheres, tori, rotation groups: the natural habitats of geometry and mechanics are not vector spaces but *curved sets that look flat up close*. This chapter gives that phrase a precise meaning — submanifolds of $\R^n$ — and the calculus to work on them. The foundation is the inverse function theorem, proved here by the Banach fixed point; everything else is change of coordinates: the four equivalent descriptions of a submanifold (local straightening, level sets, graphs, parametrizations), [tangent spaces](#def-b3-submanifolds-tangent), and constrained optimization by [Lagrange multipliers](#thm-b3-submanifolds-lagrange) — which, as a parting demonstration, re-proves the spectral theorem for symmetric matrices in three lines of geometry. The weekend problem builds the rotation group $SO(3)$ and its quaternionic double cover: algebra ([Chapter 1](https://one-course.com/books/math/5/en/chapter/1-group-theory#ch-b3-groups)’s $Q_8$ grown up) meeting geometry.

## 20.1 The inverse function theorem

**Theorem 20.1 (Inverse function theorem).**

Let $U \subseteq \R^n$ be open, $f \colon U \to \R^n$ of class $\mathcal C^1$, and $a \in U$ with $Df(a)$ invertible. Then there are [open sets](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-topology) $V \ni a$, $W \ni f(a)$ such that $f \colon V \to W$ is a bijection with $\mathcal C^1$ inverse, and

$$
D(f^{-1})(y) = \bigl(Df(f^{-1}(y))\bigr)^{-1}
\qquad (y \in W).
$$

If $f$ is $\mathcal C^k$, so is $f^{-1}$.

**Proof.** Normalize: replacing $f$ by $x \mapsto Df(a)^{-1}\bigl(f(a +
x) - f(a)\bigr)$, we may assume $a = 0$, $f(0) = 0$, $Df(0) =
I$ (the general statement follows by composing with the affine bijections). Write $f(x) = x + g(x)$: $Dg(0) = 0$, and by [continuity](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) of $Dg$ choose $r > 0$ with $\vertiii{Dg(x)} \leq \frac12$ on $\bar B(0, r)$; the mean value inequality gives $\norm{g(x) - g(x')} \leq \frac12
\norm{x - x'}$ there.

*Bijectivity onto a [neighborhood](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-topology).* For $y \in B(0,
\frac r2)$, solving $f(x) = y$ means finding a fixed point of $\Phi_y(x) = y - g(x)$; $\Phi_y$ maps $\bar B(0, r)$ into itself ($\norm{\Phi_y(x)} \leq \norm y + \frac12\norm x \leq
r$) and is $\frac12$-Lipschitz: Banach ([Theorem 7.4](https://one-course.com/books/math/5/en/chapter/7-complete-spaces-baire-ascoli-stoneweierstrass#thm-b3-complete-banach)) gives a unique solution $x = \varphi(y) \in \bar B(0,r)$. Moreover $f$ is injective on $\bar B(0,r)$:

$$
\norm{f(x) - f(x')} \geq \norm{x - x'} - \norm{g(x) - g(x')}
\geq \tfrac12\norm{x - x'} .
\tag{$*$}
$$

Set $W = B(0, \frac r2)$ and $V = f^{-1}(W)\cap B(0, r)$: open ([continuity](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity)), with $f \colon V \to W$ bijective.

*[Continuity](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) and differentiability of the inverse.* $(*)$ says $\varphi = f^{-1}$ is $2$-Lipschitz. Fix $y_0 = f(x_0)
\in W$; invertibility of $A = Df(x_0)$ (its distance to $I$ is $\leq \frac12$: Neumann, [Proposition 8.4](https://one-course.com/books/math/5/en/chapter/8-banach-spaces-and-the-fundamental-theorems#prop-b3-banach-neumann)) and differentiability of $f$ give, for $y = f(x)$ near $y_0$:

$$
\varphi(y) - \varphi(y_0) - A^{-1}(y - y_0)
= -A^{-1}\bigl(f(x) - f(x_0) - A(x - x_0)\bigr)
= -A^{-1}\,o(\norm{x - x_0}) = o(\norm{y - y_0}),
$$

using $(*)$ to convert $\norm{x - x_0} \leq 2\norm{y - y_0}$: $\varphi$ is differentiable at $y_0$ with the inverse differential. [Continuity](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) of $y \mapsto D\varphi(y) =
Df(\varphi(y))^{-1}$: composition of [continuous maps](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) (inversion is [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity), [Proposition 8.4](https://one-course.com/books/math/5/en/chapter/8-banach-spaces-and-the-fundamental-theorems#prop-b3-banach-neumann)): $\varphi \in \mathcal C^1$; bootstrapping the same formula gives $\mathcal C^k$. ∎

**Theorem 20.2 (Implicit function theorem).**

Let $F \colon U \subseteq \R^p\times\R^q \to \R^q$ be $\mathcal C^1$ near $(a, b)$, $F(a,b) = 0$, and suppose the partial differential $D_yF(a,b) \in \mathcal L(\R^q)$ is invertible. Then there are [neighborhoods](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-topology) $A \ni a$, $B \ni b$ and a $\mathcal C^1$ map $\psi \colon A \to B$ with

$$
\bigl\{(x, y)\in A\times B : F(x,y) = 0\bigr\}
= \{(x, \psi(x)) : x \in A\},
$$

and $D\psi(x) = -D_yF(x, \psi(x))^{-1}\,D_xF(x,
\psi(x))$.

**Proof.** Apply [Theorem 20.1](#thm-b3-submanifolds-ift) to $\Theta(x, y) = (x,
F(x,y))$: its differential at $(a,b)$, block-triangular with invertible diagonal blocks $I$ and $D_yF$, is invertible. The local inverse has the form $\Theta^{-1}(x, z) = (x, h(x,
z))$; set $\psi(x) = h(x, 0)$: then $F(x, y) = 0$ iff $\Theta(x,y) = (x, 0)$ iff $y = \psi(x)$, locally. The formula: differentiate $F(x, \psi(x)) = 0$ by the chain rule. ∎

## 20.2 Submanifolds: four definitions

**Theorem 20.3 (Equivalent characterizations).**

Let $M \subseteq \R^n$, $d \in \{0, \dots, n\}$, and $k \geq
1$. The following are equivalent, for each point $a \in M$ (and $M$ is a *$d$-dimensional submanifold of class $\mathcal C^k$* if they hold at every $a
\in M$):

1. (*Straightening*) There is a $\mathcal C^k$ diffeomorphism $\Phi$ from an open $\Omega \ni a$ onto an open $\Omega' \subseteq \R^n$ with $$\Phi(M\cap\Omega) = \Omega' \cap  \bigl(\R^d\times\{0\}\bigr).$$
2. ( *Level set* ) There is a $\mathcal C^k$ *submersion* $F \colon \Omega \to \R^{n-d}$ (i.e. $DF(x)$ surjective) on an open $\Omega \ni a$ with $M\cap\Omega = F^{-1}(0)$ .
3. ( *Graph* ) Up to permuting coordinates, $M$ is locally the graph of a $\mathcal C^k$ map $\psi  \colon A \subseteq \R^d \to \R^{n-d}$ .
4. ( *Parametrization* ) There is a $\mathcal C^k$ *immersion* $\varphi \colon A \subseteq \R^d \to  \R^n$ ( $D\varphi(u)$ injective) with $A$ open, $\varphi$ a [homeomorphism](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) from $A$ onto $M \cap  \Omega$ for some open $\Omega \ni a$ .

**Proof.** (1)$\Rightarrow$(2): $F = (\Phi_{d+1}, \dots, \Phi_n)$ (last coordinates of $\Phi$): a submersion ($D\Phi$ invertible). (2)$\Rightarrow$(3): $DF(a)$ surjective: some $q \times q$ minor of the Jacobian is invertible ($q = n - d$); after permuting coordinates, $D_yF(a)$ is invertible, and the implicit function theorem ([Theorem 20.2](#thm-b3-submanifolds-implicit)) expresses $M$ locally as a graph $y = \psi(x)$. (3)$\Rightarrow$(4): $\varphi(x) = (x, \psi(x))$: an immersion (differential $\bigl(\begin{smallmatrix}I\\
D\psi\end{smallmatrix}\bigr)$ injective), a [homeomorphism](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) onto the graph (inverse: the projection, [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity)). (4)$\Rightarrow$(1): let $\varphi(u_0) = a$; [complete](https://one-course.com/books/math/5/en/chapter/7-complete-spaces-baire-ascoli-stoneweierstrass#def-b3-complete-complete) $\operatorname{im}D\varphi(u_0)$ by a supplement $E$ ($\dim E = n - d$) and define $\Theta(u, v) = \varphi(u) + v$ on $A\times E$: $D\Theta(u_0, 0)$ is bijective (image contains $\operatorname{im}D\varphi(u_0)$ and $E$), so $\Theta$ is a local diffeomorphism ([Theorem 20.1](#thm-b3-submanifolds-ift)); its inverse $\Phi$ straightens: near $a$, points of $M$ are exactly the $\varphi(u) = \Theta(u, 0)$ — for this, the [homeomorphism](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) hypothesis in (4) guarantees that $M\cap\Omega$, for small $\Omega$, contains no other sheets ($\varphi(u') + v = m \in
M$ close to $a$ with small $v \neq 0$ must be excluded: $m = \varphi(u'')$ for some $u''$ near $u_0$ by the [homeomorphism](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) property, and local injectivity of $\Theta$ forces $v = 0$). Then $\Phi(M\cap\Omega) = (A\times\{0\})
\cap\Phi(\Omega)$ up to shrinking. ∎

**Example 20.4.**

The sphere $S^{n-1} = \{\norm x^2 = 1\}$: level set of the submersion $F(x) = \norm x_2^2 - 1$ on $\R^n\setminus\{0\}$ ($DF(x) = 2x^{\mathsf T} \ne 0$): a $\mathcal C^\infty$ submanifold of dimension $n - 1$. The torus in $\R^3$: level set of $\bigl(\sqrt{x^2 + y^2} -
R\bigr)^2 + z^2 - r^2$ ($0 < r < R$). The cone $\{x^2 + y^2 =
z^2\}$ is *not* a submanifold at $0$ ([Exercise 20.1](#exo-b3-submanifolds-1)). Matrix groups: $SL_n$ and $O_n$ are submanifolds of $M_n(\R)$ (Exercises [20.5](#exo-b3-submanifolds-5) and [20.6](#exo-b3-submanifolds-6)) — the starting point of Lie theory.

## 20.3 Tangent spaces

**Definition 20.5.**

Let $M$ be a $d$-submanifold and $a \in M$. The *tangent space* $T_aM$ is the set of velocity vectors $\gamma'(0)$ of $\mathcal C^1$ curves $\gamma \colon
\intoo{-\varepsilon}\varepsilon \to M$ with $\gamma(0) = a$.

**Proposition 20.6.**

$T_aM$ is a $d$-dimensional vector subspace of $\R^n$, and:

1. if $M = F^{-1}(0)$ locally with $F$ a submersion: $T_aM = \ker DF(a)$ ;
2. if $M$ is parametrized by the immersion $\varphi$ ( $\varphi(u_0) = a$ ): $T_aM =  \operatorname{im}D\varphi(u_0)$ .

**Proof.** Curves in $M$ satisfy $F(\gamma(t)) = 0$; the chain rule at $0$ gives $DF(a)\gamma'(0) = 0$: $T_aM \subseteq \ker DF(a)$. Conversely, straightening ([Theorem 20.3](#thm-b3-submanifolds-characterizations)(1)) transports lines of $\R^d\times\{0\}$ to curves of $M$: every vector of a $d$-dimensional subspace is realized; comparing dimensions ($\dim\ker DF(a) = n - (n - d) = d$) forces equality in (1), and the same transport argument gives (2) ($D\varphi(u_0)$ applied to straight lines in $A$; dimensions again). ∎

**Theorem 20.7 (Lagrange multipliers).**

Let $M = F^{-1}(0)$ with $F = (F_1, \dots, F_q) \colon \Omega
\to \R^q$ a $\mathcal C^1$ submersion, and let $f \colon
\Omega \to \R$ be $\mathcal C^1$. If the restriction $f\restriction_M$ has a local extremum at $a \in M$, then there are unique reals $\lambda_1, \dots, \lambda_q$ (*[Lagrange multipliers](#thm-b3-submanifolds-lagrange)*) with

$$
\nabla f(a) = \lambda_1\nabla F_1(a) + \dots +
\lambda_q\nabla F_q(a) .
$$

**Proof.** For every curve $\gamma$ in $M$ through $a$: $t \mapsto
f(\gamma(t))$ has a local extremum at $0$, so $0 =
\frac{\dd}{\dd t}f(\gamma(t))\big|_0 = \langle\nabla f(a),
\gamma'(0)\rangle$: $\nabla f(a) \perp T_aM = \ker DF(a)$ ([Proposition 20.6](#prop-b3-submanifolds-tangent)). Now $\ker DF(a)^\perp
= \operatorname{im}DF(a)^{\mathsf T}$ (the rank/orthogonality identity of Year 2, or [Exercise 13.8](https://one-course.com/books/math/5/en/chapter/13-hilbert-spaces#exo-b3-hilbert-8) in finite dimension: $(\ker T)^\perp = \operatorname{im}T^*$), which is spanned by the gradients $\nabla F_i(a)$ — independent, as $DF(a)$ is surjective: the multipliers exist and are unique. ∎

**Example 20.8 (The spectral theorem, geometrically).**

Let $A$ be a real symmetric $n\times n$ matrix and maximize $f(x) = \langle Ax, x\rangle$ on the sphere $S^{n-1}$ ([compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact): the maximum is attained, at some $v_1$). Lagrange with $F(x) = \norm x^2 - 1$: $\nabla f = 2Ax$ and $\nabla F =
2x$ give $Av_1 = \lambda_1v_1$ — an eigenvector, with $\lambda_1 = \max_{S^{n-1}}\langle Ax, x\rangle$. Restrict $A$ to $v_1^\perp$ (invariant: $\langle Av, v_1\rangle =
\langle v, Av_1\rangle = \lambda_1\langle v, v_1\rangle = 0$) and iterate: an orthonormal basis of eigenvectors. The spectral theorem of Year 2, re-proved by pure optimization — and the infinite-dimensional shadow of the same argument proved [Lemma 15.6](https://one-course.com/books/math/5/en/chapter/15-compact-operators-and-the-spectral-theorem#lem-b3-spectral-existence).

**Method 20.9.**

To prove a set is a submanifold: exhibit it locally as $F^{-1}(0)$ with $DF$ surjective *on the set* (the most common route), or as a graph. To compute its dimension and [tangent space](#def-b3-submanifolds-tangent): $d = n - q$ and $T_a = \ker DF(a)$. To optimize on it: Lagrange — always check compactness (or coercivity) first, so an extremum exists to which the theorem can apply, and remember that the multiplier equation is only *necessary*: collect all critical points, then compare values. For matrix groups, differentiate curves at the identity to identify [tangent spaces](#def-b3-submanifolds-tangent).

## 20.4 Exercises

**Exercise 20.1 ★.**

(a) Verify that the following are $\mathcal C^\infty$ submanifolds and give their dimensions: $S^{n-1}$; the hyperboloid $\{x^2 + y^2 - z^2 = 1\}$; the torus of [Example 20.4](#ex-b3-submanifolds-examples). (b) Show that the cone $C = \{x^2 + y^2 = z^2\} \subseteq
\R^3$ is not a $2$-submanifold at $0$: determine the number of [connected components](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-components) of $\bigl(C\setminus\{0\}\bigr)\cap B(0,\varepsilon)$, and compare with the number a straightening ([Theorem 20.3](#thm-b3-submanifolds-characterizations)(1)) would force for a plane minus a point.

**Solution of Exercise 20.1.**

(a) Each is $F^{-1}(0)$ for a submersion: $\norm x^2 - 1$ on $\R^n\setminus\{0\}$ (gradient $2x \neq 0$): dimension $n-1$; $x^2 + y^2 - z^2 - 1$ (gradient $(2x, 2y, -2z) \neq 0$ on the hyperboloid, where $x^2 + y^2 = 1 + z^2 > 0$): dimension $2$; the torus function $G = (\rho - R)^2 + z^2 - r^2$, $\rho =
\sqrt{x^2+y^2}$, is $\mathcal C^\infty$ near the torus ($\rho
\geq R - r > 0$ there) with $\nabla G \neq 0$ (its $z$-component is $2z$, and where $z = 0$ the radial component is $2(\rho - R)\ne0$ since $\abs{\rho - R} = r$): dimension $2$.

(b) For small $\varepsilon$, $(C\setminus\{0\})\cap B(0,\varepsilon)$ has exactly $2$ [connected components](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-components) (upper and lower punctured nappes, each [path-connected](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-pathconnected): connect through circles and rays). If $C$ were a $2$-submanifold at $0$, a straightening would give a [homeomorphism](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) from $C\cap\Omega$ onto an open piece of a plane sending $0$ to a point $p$; small punctured [plane-neighborhoods](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-topology) of $p$ have *one* component, and [homeomorphisms](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) preserve the number of components of punctured [neighborhoods](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-topology): contradiction.

**Exercise 20.2 ★.**

Compute the [tangent spaces](#def-b3-submanifolds-tangent): (a) $T_aS^{n-1}$ for any $a$ (answer: $a^\perp$); (b) the tangent plane to the torus of [Example 20.4](#ex-b3-submanifolds-examples) at an arbitrary point of the outer equator $\{z = 0,\ x^2 + y^2 = (R + r)^2\}$; (c) the tangent line to the helix $\varphi(t) = (\cos t,
\sin t, t)$ at $\varphi(t_0)$, checking [Proposition 20.6](#prop-b3-submanifolds-tangent)(2).

**Solution of Exercise 20.2.**

(a) $T_aS^{n-1} = \ker\bigl(2a^{\mathsf T}\bigr) = a^\perp$. (b) At $p = ((R+r)\cos\theta, (R+r)\sin\theta, 0)$: $\nabla G
= (2r\cos\theta, 2r\sin\theta, 0)$, so the tangent plane is $\operatorname{Vect}\bigl((-\sin\theta, \cos\theta, 0),\ (0,
0, 1)\bigr)$: the vertical plane tangent to the outer equator. (c) The helix is an embedded curve with $\varphi'(t_0) =
(-\sin t_0, \cos t_0, 1) \neq 0$: the tangent line at $\varphi(t_0)$ is $\varphi(t_0) + \R\,\varphi'(t_0)$, as [Proposition 20.6](#prop-b3-submanifolds-tangent)(2) prescribes.

**Exercise 20.3 ★★.**

Let $f(x, y) = (x^2 - y^2,\ 2xy)$ (i.e. $z \mapsto z^2$). (a) At which points does [Theorem 20.1](#thm-b3-submanifolds-ift) apply? (b) Show that $f$ is locally but not globally invertible on $\R^2\setminus\{0\}$, and exhibit explicitly the two local inverses defined on a [neighborhood](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-topology) of $(1, 0)$ (the two square-root branches). (c) Same discussion for the [polar coordinates](https://one-course.com/books/math/5/en/chapter/11-product-measures-fubini-change-of-variables#ex-b3-product-polar) map $(r,
\theta) \mapsto (r\cos\theta, r\sin\theta)$.

**Solution of Exercise 20.3.**

(a) $Df(x,y) = \bigl(\begin{smallmatrix}2x & -2y\\ 2y &
2x\end{smallmatrix}\bigr)$, $\det = 4(x^2 + y^2)$: the theorem applies at every point except the origin. (b) $f(-z) = f(z)$: never injective on a set symmetric about $0$; on $\R^2\setminus\{0\}$ it is a local diffeomorphism everywhere yet $2$-to-$1$ globally. Near $(1, 0) = f(\pm(1,
0))$, the two inverses are the two square-root branches: in complex notation $w \mapsto \pm\sqrt w$ (principal branch), i.e.

$$
(u, v) \longmapsto \pm\Bigl(\sqrt{\tfrac{u +
\sqrt{u^2+v^2}}{2}},\ \frac{v}{2\sqrt{(u +
\sqrt{u^2+v^2})/2}}\Bigr).
$$

(c) Jacobian $r > 0$: local diffeomorphism on $\intoo0\infty\times\R$, but $\theta \mapsto \theta + 2\pi$ gives the same point: locally invertible (angle determined up to $2\pi$ on a half-plane), never globally.

**Exercise 20.4 ★★.**

(Folium) Let $F(x,y) = x^3 + y^3 - 3xy$ and $\mathcal C =
F^{-1}(0)$. (a) Show that near every point of $\mathcal C$ other than the origin, $\mathcal C$ is a $1$-submanifold, locally a graph in $x$ or in $y$ (which, where?). (b) Compute the tangent line at $(\frac32, \frac32)$. (c) What happens at the origin? (Two branches cross: exhibit two $\mathcal C^1$ curves in $\mathcal C$ through $0$ with independent velocities, and conclude that no straightening exists.)

**Solution of Exercise 20.4.**

(a) $\nabla F = 3(x^2 - y,\ y^2 - x)$ vanishes iff $y = x^2$ and $x = y^2$, i.e. $x^4 = x$: at $(0,0)$ and $(1,1)$; only $(0,0)$ lies on $\mathcal C$ ($F(1,1) = -1$). So on $\mathcal
C\setminus\{0\}$, $F$ is a submersion: a $1$-submanifold, locally a graph $y = \psi(x)$ where $F_y = 3(y^2 - x) \neq 0$ and $x = \chi(y)$ where $F_x = 3(x^2 - y) \neq 0$ (at least one holds off the origin).

(b) At $(\frac32, \frac32)$: $\nabla F = 3(\frac94 - \frac32)
(1, 1) = \frac94(1,1)$: tangent line $x + y = 3$.

(c) The rational parametrization $x = \frac{3t}{1 + t^3}$, $y = \frac{3t^2}{1+t^3}$ passes through $0$ at $t = 0$ with velocity $(3, 0)$; exchanging $x \leftrightarrow y$ (the curve is symmetric, or reparametrize by $1/t$) gives a second $\mathcal C^1$ curve through $0$ with velocity $(0, 3)$. Two independent tangent directions are impossible for a $1$-submanifold (its [tangent space](#def-b3-submanifolds-tangent) is a line, [Proposition 20.6](#prop-b3-submanifolds-tangent)): $\mathcal C$ is not a submanifold at the origin — a transverse self-crossing.

**Exercise 20.5 ★★.**

Let $F(M) = M^{\mathsf T}M$ from $M_n(\R)$ to the space $S_n$ of symmetric matrices. (a) Show $DF(M)(H) = M^{\mathsf T}H + H^{\mathsf T}M$ and that $DF(M)$ is surjective onto $S_n$ at every $M \in O_n$ *(given $S \in S_n$, try $H = \frac12MS$)*. (b) Conclude that $O_n = F^{-1}(I)$ is a [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) $\mathcal C^\infty$ submanifold of dimension $\frac{n(n-1)}2$, with $T_IO_n = \{H : H^{\mathsf T} = -H\}$, the antisymmetric matrices. (c) Show that $\eu^{tH} \in O_n$ for every antisymmetric $H$: the tangent directions integrate to curves in the group.

**Solution of Exercise 20.5.**

(a) $F(M + H) = M^{\mathsf T}M + M^{\mathsf T}H + H^{\mathsf
T}M + H^{\mathsf T}H$: $DF(M)(H) = M^{\mathsf T}H +
H^{\mathsf T}M$. For $M \in O_n$ and $S$ symmetric, $H =
\frac12MS$ gives $DF(M)(H) = \frac12(S + S^{\mathsf T}) = S$: surjective onto $S_n$.

(b) $O_n = F^{-1}(I)$ with $F$ a submersion (onto $S_n$, of dimension $\frac{n(n+1)}2$) at each of its points: a submanifold of dimension $n^2 - \frac{n(n+1)}2 =
\frac{n(n-1)}2$. [Compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact): closed ($F$ [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity)) and bounded (columns are unit vectors). Tangent at $I$: $\ker DF(I) =
\{H : H + H^{\mathsf T} = 0\}$.

(c) $\bigl(\eu^{tH}\bigr)^{\mathsf T}\eu^{tH} =
\eu^{tH^{\mathsf T}}\eu^{tH} = \eu^{-tH}\eu^{tH} = I$ (transpose passes through the series; the exponentials of commuting matrices multiply, [Theorem 19.8](https://one-course.com/books/math/5/en/chapter/19-ordinary-differential-equations#thm-b3-ode-matrixexp)).

**Exercise 20.6 ★★.**

(a) Show that $\det \colon M_n(\R) \to \R$ has differential $D\det(M)(H) = \operatorname{tr}\bigl(\operatorname{com}(M)
^{\mathsf T}H\bigr)$, nonzero at every $M \in SL_n$. (b) Conclude that $SL_n(\R)$ is a submanifold of dimension $n^2 - 1$ with $T_ISL_n = \{H : \operatorname{tr}H = 0\}$. (c) Is $GL_n(\R)$ a submanifold? Of what dimension?

**Solution of Exercise 20.6.**

(a) $\det(M + H) = \det M\,\det(I + M^{-1}H) = \det M\bigl(1
+ \operatorname{tr}(M^{-1}H) + O(\norm H^2)\bigr)$ for invertible $M$ (expansion of $\det$ near $I$: the linear term of $\prod(1 + \lambda_i)$); with $\det M\cdot M^{-1} =
\operatorname{com}(M)^{\mathsf T}$: $D\det(M)(H) = \operatorname{tr}\bigl(\operatorname{com}
(M)^{\mathsf T}H\bigr)$, and the formula extends to all $M$ by [density](https://one-course.com/books/math/5/en/chapter/10-the-lebesgue-integral#ex-b3-lebesgue-gamma) and [continuity](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity). On $SL_n$, $\det M = 1$: $D\det(M) \ne 0$ (its value on $H = M$ is $\operatorname{tr}(I)\cdot\dots = n\det M = n$).

(b) $SL_n = \det^{-1}(1)$ with $\det$ a submersion there (values in $\R$): dimension $n^2 - 1$; $T_ISL_n = \ker
D\det(I) = \{H : \operatorname{tr}H = 0\}$.

(c) $GL_n$ is an *open* subset of $M_n(\R)$ ([Exercise 6.8](https://one-course.com/books/math/5/en/chapter/6-general-topology#exo-b3-topology-8)): a submanifold of full dimension $n^2$ (straightening: the identity chart).

**Exercise 20.7 ★★.**

By [Lagrange multipliers](#thm-b3-submanifolds-lagrange): (a) find the extrema of $f(x,y) = xy$ on the circle $x^2 +
y^2 = 1$; (b) show that among all probability vectors $(p_1, \dots,
p_n)$ (positive, summing to $1$), the entropy $-\sum
p_i\ln p_i$ is maximized exactly at the uniform distribution; (c) find the point of the ellipse $\{x^2/4 + y^2 = 1\}$ closest to $(1, 0)$, and check the multiplier equation geometrically (normal alignment).

**Solution of Exercise 20.7.**

(a) $(y, x) = \lambda(2x, 2y)$ and $x^2 + y^2 = 1$: $y =
2\lambda x$, $x = 2\lambda y$ give $x^2 = y^2 = \frac12$. Values of $xy$: $\pm\frac12$: maximum $\frac12$ at $\pm\frac1{\sqrt2}(1,1)$, minimum $-\frac12$ at $\pm\frac1{\sqrt2}(1,-1)$ (the constraint set is [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact): extrema exist).

(b) On the [interior](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-interior) of the simplex ($p_i > 0$), Lagrange for $H(p) = -\sum p_i\ln p_i$ with constraint $\sum p_i = 1$: $-\ln p_i - 1 = \lambda$ for all $i$: all $p_i$ equal, $p_i =
\frac1n$, with $H = \ln n$. The maximum over the [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) simplex is attained; if it were attained on the boundary (some $p_i = 0$), the distribution lives on $\leq n - 1$ points and by induction $H \leq \ln(n-1) < \ln n$: the [interior](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-interior) critical point is the global maximum — uniform ignorance maximizes entropy.

(c) Minimize $(x-1)^2 + y^2$ on the [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) ellipse: $(2(x{-}1), 2y) = \lambda(\frac x2, 2y)$. If $y \neq 0$: $\lambda = 1$, then $2(x - 1) = \frac x2$ gives $x =
\frac43$, $y^2 = 1 - \frac49 = \frac59$: distance$^2$ $=
\frac19 + \frac59 = \frac23$. If $y = 0$: $x = \pm2$, distances $1$ and $3$. Closest points: $\bigl(\frac43,
\pm\frac{\sqrt5}3\bigr)$, at distance $\sqrt{2/3} < 1$. The multiplier equation says the segment from $(1,0)$ to the closest point is parallel to $\nabla$(ellipse): it meets the ellipse orthogonally, as geometry demands.

**Exercise 20.8 ★★★.**

Write out [Example 20.8](#ex-b3-submanifolds-rayleigh) in full: prove by induction that a real symmetric matrix admits an orthonormal basis of eigenvectors, with $\lambda_1 \geq \dots \geq \lambda_n$ the successive constrained maxima of the Rayleigh quotient. Then deduce the Courant–Fischer formulas of [Exercise 15.8](https://one-course.com/books/math/5/en/chapter/15-compact-operators-and-the-spectral-theorem#exo-b3-spectral-8) in finite dimension directly from this construction.

**Solution of Exercise 20.8.**

Induction on $n$; $n = 1$ trivial. The Rayleigh function $f(x) = \langle Ax, x\rangle$ attains its maximum $\lambda_1$ on the [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) $S^{n-1}$ at some $v_1$; Lagrange ([Theorem 20.7](#thm-b3-submanifolds-lagrange), sphere as level set) gives $2Av_1 = 2\lambda v_1$, and $\lambda = \langle Av_1,
v_1\rangle = \lambda_1$. The hyperplane $v_1^\perp$ is $A$-invariant (symmetry: $\langle Av, v_1\rangle = \langle
v, Av_1\rangle = 0$); the restriction is symmetric, and induction yields an orthonormal eigenbasis $v_2, \dots, v_n$ of $v_1^\perp$ with eigenvalues $\lambda_2 \geq \dots \geq
\lambda_n$, each the maximum of $f$ on the sphere of the remaining orthocomplement. Courant–Fischer follows exactly as in [Exercise 15.8](https://one-course.com/books/math/5/en/chapter/15-compact-operators-and-the-spectral-theorem#exo-b3-spectral-8): expand $x = \sum c_iv_i$; on a $k$-dimensional test space intersect with $\operatorname{Vect}(v_k, \dots, v_n)$ (dimension count in $\R^n$) to get $\min \leq \lambda_k$, and $\operatorname{Vect}(v_1, \dots, v_k)$ achieves $\min =
\lambda_k$.

**Exercise 20.9 ★★★.**

(Hadamard’s inequality) For $M \in GL_n(\R)$ with columns $c_1, \dots, c_n$:

$$
\abs{\det M} \;\leq\; \prod_{i=1}^n\norm{c_i}_2 ,
$$

with equality iff the columns are orthogonal. *(Reduce to columns of norm $1$ by scaling; maximize $\det$ on the [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) product of spheres $(S^{n-1})^n$; at a maximizer, Lagrange in each column separately gives $\nabla_{c_i}\det =
\lambda_ic_i$, and $\nabla_{c_i}\det$ is the $i$-th column of $\operatorname{com}(M)$: deduce $M^{\mathsf T}M$ diagonal, hence $= I$, hence $\det = \pm1$.)* Geometric reading: the volume of a parallelepiped is at most the product of its edge lengths.

**Solution of Exercise 20.9.**

Scaling each column to unit norm divides $\abs{\det}$ by $\prod\norm{c_i}$: it suffices to prove $\abs{\det M} \leq 1$ when all columns are unit, with equality iff $M \in O_n$. The function $\det$ is [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) on the [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) $(S^{n-1})^n$: it attains a maximum $m \geq \det I = 1 > 0$ at some $M$. Fixing all columns but the $i$-th, $\det$ is linear in $c_i$ with gradient the $i$-th column of $\operatorname{com}(M)$; Lagrange on the $i$-th sphere: $\operatorname{com}(M)_{\cdot i} = \lambda_i c_i$. The identity $M^{\mathsf T}\operatorname{com}(M) = \det(M)\,I$ reads $\langle c_j, \operatorname{com}(M)_{\cdot i}\rangle =
\det(M)\,\delta_{ij}$, i.e. $\lambda_i\langle c_j,
c_i\rangle = \det(M)\delta_{ij}$; taking $j = i$: $\lambda_i
= \det M = m \neq 0$, and then $j \neq i$ gives $\langle c_i,
c_j\rangle = 0$: the columns are orthonormal, $M \in O_n$, $m = \abs{\det M} = 1$. Hence $\abs{\det} \leq
\prod\norm{c_i}$ always, with equality exactly for orthogonal columns (rescale back): a parallelepiped’s volume is largest, for given edge lengths, when the edges are perpendicular.

**Exercise 20.10 ★★.**

Near which of its points is the circle $S^1$ a graph $y =
\psi(x)$? A graph $x = \chi(y)$? Verify the graph characterization ([Theorem 20.3](#thm-b3-submanifolds-characterizations)(3)) explicitly at $(1, 0)$, and explain in one sentence why *some* coordinate permutation is always sufficient but no single one always works.

**Solution of Exercise 20.10.**

$y = \pm\sqrt{1 - x^2}$ works near every point with $y \neq
0$; $x = \pm\sqrt{1 - y^2}$ near every point with $x \neq 0$; at $(1, 0)$: the graph $x = \sqrt{1 - y^2}$ over $y \in
\intoo{-1}1$, which is [Theorem 20.3](#thm-b3-submanifolds-characterizations)(3) with the coordinates swapped. Some permutation always works because the tangent line, being one-dimensional, cannot be simultaneously vertical and horizontal — but it can be either, so no fixed choice of “dependent” coordinate serves at every point.

**Exercise 20.11 ★★.**

(The orthogonal group as a submanifold, quantitatively) (a) Show that $O_n = \{M : M^{\mathsf T}M = I\}$ is [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact): bounded (each column is a unit vector, so $\norm M \leq
\sqrt n$ for the Euclidean matrix norm) and closed. (b) Show that its [tangent space](#def-b3-submanifolds-tangent) at $I$ is the space of antisymmetric matrices, of dimension $\frac{n(n-1)}2$, and at a general $A \in O_n$: $T_AO_n = \{AK : K^{\mathsf T} =
-K\}$. (c) Deduce that the map $t \mapsto A\exp(tK)$ is, for each antisymmetric $K$, a curve in $O_n$ through $A$ with velocity $AK$ *(verify $\exp(tK) \in O_n$ using $\exp(X)^{\mathsf T} = \exp(X^{\mathsf T})$ and $\exp(-X)\exp(X) = I$)*: every tangent vector is realized by an explicit curve, with no implicit function theorem needed.

**Solution of Exercise 20.11.**

(a) The defining map $F(M) = M^{\mathsf T}M - I$ is [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity): $O_n = F^{-1}(0)$ is closed; columns of an orthogonal matrix are unit vectors, so the Euclidean ([Frobenius](https://one-course.com/books/math/5/en/chapter/4-field-extensions-and-galois-theory#thm-b3-galois-finitefields)) norm is exactly $\sqrt n$: bounded. [Compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) by Heine–Borel in $M_n(\R) \cong \R^{n^2}$.

(b) $O_n$ is the level set $F = 0$ studied in the chapter: $DF(A)H = A^{\mathsf T}H + H^{\mathsf T}A$, surjective onto symmetric matrices at each $A \in O_n$ (given symmetric $S$, take $H = \frac12AS$), so $O_n$ is a submanifold of dimension $n^2 - \frac{n(n+1)}2 = \frac{n(n-1)}2$ with

$$
T_AO_n = \ker DF(A) = \{H : A^{\mathsf T}H \text{
antisymmetric}\} = \{AK : K^{\mathsf T} = -K\} ;
$$

at $A = I$ these are the antisymmetric matrices.

(c) $\exp(tK)^{\mathsf T}\exp(tK) = \exp(tK^{\mathsf T})
\exp(tK) = \exp(-tK)\exp(tK) = I$ (the two matrices $\pm tK$ commute, so the product of exponentials is the exponential of the sum): $\exp(tK) \in O_n$, and $\gamma(t) =
A\exp(tK)$ is a curve in $O_n$ with $\gamma(0) = A$, $\gamma'(0) = AK$. As $K$ runs over antisymmetric matrices, $AK$ sweeps $T_AO_n$: the exponential realizes the whole [tangent space](#def-b3-submanifolds-tangent) by explicit curves — the Lie-group shortcut that [Problem 20.1](#pb-b3-submanifolds-1) exploits for $SO(3)$.

**Exercise 20.12 ★★.**

(Critical points of the distance) Let $M \subseteq \R^n$ be a submanifold and $p \notin M$. Show that if $x_0 \in M$ minimizes the distance to $p$ (such a point exists when $M$ is closed and nonempty — why?), then

$$
p - x_0 \;\perp\; T_{x_0}M
$$

*(differentiate $t \mapsto \norm{\gamma(t) - p}^2$ along curves in $M$)*. Deduce: the closest point on a sphere lies on the ray through the [center](https://one-course.com/books/math/5/en/chapter/1-group-theory#ex-b3-groups-actions); and use the condition to compute the distance from $p = (2, 0)$ to the parabola $y = x^2$ (reduce to a cubic and solve it numerically to three digits).

**Solution of Exercise 20.12.**

Existence: intersect $M$ with a large closed ball around $p$ to get a nonempty [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact); the [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) distance attains its minimum there, and points outside the ball are farther. First-order condition: for a curve $\gamma$ in $M$ with $\gamma(0) = x_0$, the function $h(t) = \norm{\gamma(t)
- p}^2$ is differentiable with a minimum at $0$:

$$
0 = h'(0) = 2\,\langle\gamma'(0),\ x_0 - p\rangle,
$$

and $\gamma'(0)$ sweeps $T_{x_0}M$: $p - x_0 \perp
T_{x_0}M$. Sphere $S(c, r)$: the [tangent space](#def-b3-submanifolds-tangent) at $x_0$ is $(x_0 - c)^\perp$, so $p - x_0 \parallel x_0 - c$: $x_0$ lies on the line through $c$ and $p$, at distance $r$ from $c$ — the ray point, as geometry insists. Parabola: at $x_0 = (x, x^2)$ the tangent is spanned by $(1, 2x)$; orthogonality to $p - x_0 = (2 - x, -x^2)$ reads

$$
(2 - x) - 2x^3 = 0,
\qquad\text{i.e.}\qquad 2x^3 + x - 2 = 0,
$$

with unique real root ($x \mapsto 2x^3 + x$ is strictly increasing) $x \approx 0.835$; then $x_0 \approx (0.835,
0.698)$ and $d(p, M) = \sqrt{(2 - 0.835)^2 + 0.698^2}
\approx 1.358$.

## 20.5 Problem: $SO(3)$ and the quaternions

**Problem 20.1.**

Weekend problem — rotations, the group $S^3$, and the double cover

The quaternions $\mathbb H = \{t + x\mathrm i + y\mathrm j +
z\mathrm k\}$ — the algebra whose unit group contains [Problem 1.1](https://one-course.com/books/math/5/en/chapter/1-group-theory#pb-b3-groups-1)’s $Q_8$ — parametrize three-dimensional rotations twice over: the map “conjugate by a unit quaternion” is a surjective morphism $S^3 \to
SO(3)$ with kernel $\{\pm1\}$. We build everything. Recall/define: multiplication is $\R$-bilinear with $\mathrm i^2 = \mathrm j^2 = \mathrm k^2 =
\mathrm{ijk} = -1$; the conjugate of $q = t + x\mathrm i +
y\mathrm j + z\mathrm k$ is $\bar q = t - x\mathrm i -
y\mathrm j - z\mathrm k$; $N(q) = q\bar q = t^2 + x^2 + y^2
+ z^2$.

**Part I — The algebra $\mathbb H$ and the group $S^3$.**

1. Verify that $\mathbb H$ is an associative $\R$ -algebra with [center](https://one-course.com/books/math/5/en/chapter/1-group-theory#ex-b3-groups-actions) $\R$ , that $\overline{pq} =  \bar q\,\bar p$ , and that $N(pq) = N(p)N(q)$ *(one clean route: represent $q$ as the $2\times2$ complex matrix $\bigl(\begin{smallmatrix}\alpha & \beta\\  -\bar\beta & \bar\alpha\end{smallmatrix}\bigr)$, $q =  \alpha + \beta\mathrm j$, and use $\det$)* .
2. Deduce that every $q \neq 0$ is invertible ( $q^{-1} =  \bar q/N(q)$ ): $\mathbb H$ is a (noncommutative) field, and $S^3 = \{N(q) = 1\}$ is a group — and a [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) $3$ -submanifold of $\R^4$ ( [Example 20.4](#ex-b3-submanifolds-examples) ).

**Part II — The rotation morphism.** Identify $\R^3$ with the *pure quaternions* $P = \{x\mathrm i +
y\mathrm j + z\mathrm k\}$, and for $q \in S^3$ define $\rho_q(v) = q\,v\,\bar q$.

3. Show that $\rho_q$ maps $P$ to $P$ *(pure quaternions are those with $\bar v = -v$)* , is $\R$ -linear, preserves the norm, and that $\rho  \colon q \mapsto \rho_q$ is a group morphism $S^3  \to O(3)$ .
4. Compute the kernel: $\rho_q = \mathrm{id}$ iff $q$ commutes with $\mathrm i, \mathrm j, \mathrm k$ iff $q \in \R\cap S^3 = \{\pm1\}$ .
5. Write $q = \cos\frac\theta2 + \sin\frac\theta2\,u$ with $u \in P$ , $N(u) = 1$ (why is this always possible for $q \in S^3$ ?). Show that $\rho_q$ fixes $u$ and, on the plane $u^\perp\cap P$ , acts as the rotation of angle $\theta$ *(compute $\rho_q(w)$ for $w \perp u$ using $uw = -wu$ for orthogonal pure units — prove this identity from the multiplication table, or from $uw + wu =  -2\langle u, w\rangle$)* .
6. Conclude: $\operatorname{im}\rho \subseteq SO(3)$ (each $\rho_q$ is a rotation with axis and angle as computed — determinant $+1$ by [continuity](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) of $q  \mapsto \det\rho_q$ on the [connected](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-connected) $S^3$, or directly), and $\rho$ is *onto* $SO(3)$: every rotation of $\R^3$ has an axis (prove: a real $3\times3$ orthogonal matrix with $\det = 1$ has eigenvalue $1$ — consider the characteristic polynomial) and is therefore some $\rho_q$. Summary: $$SO(3) \;\cong\; S^3/\{\pm 1\} .$$

**Part III — $SO(3)$ as a submanifold; Rodrigues.**

7. Show that $SO(3)$ is a [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) $3$ -dimensional submanifold of $M_3(\R)$ with $T_ISO(3) =$ antisymmetric matrices ( [Exercise 20.5](#exo-b3-submanifolds-5) ; the determinant condition selects a union of components).
8. For the antisymmetric matrix $A_u$ associated with $u \in \R^3$ ($A_uv = u\wedge v$, the cross product), prove *Rodrigues’ formula*: $$\eu^{\theta A_u} = I + \sin\theta\,A_u + (1 -  \cos\theta)\,A_u^2 \qquad (\norm u = 1)$$ *(from $A_u^3 = -A_u$: split the exponential series along $A_u$’s powers)*, and identify it as the rotation of axis $u$ and angle $\theta$. Deduce that $\exp$ maps the antisymmetric matrices *onto* $SO(3)$.
9. Relate the two parametrizations: show that $t  \mapsto \rho_{q(t)}$ with $q(t) = \cos\frac t2 +  \sin\frac t2\,u$ is a one-parameter group of rotations whose derivative at $t = 0$ is $A_u$ — the quaternionic and [matrix exponentials](https://one-course.com/books/math/5/en/chapter/19-ordinary-differential-equations#thm-b3-ode-matrixexp) tell the same story at half and full speed respectively.

**Part IV — The double cover, felt.**

10. Show that the [path](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-pathconnected) $q(t) = \cos\frac t2 + \sin\frac  t2\,\mathrm k$ , $t \in \intcc0{2\pi}$ , is a loop in $SO(3)$ (its image $\rho_{q(t)}$ returns to the identity) whose quaternionic lift is *not* a loop: $q(2\pi) = -q(0)$ . Continuing to $t = 4\pi$ closes the lift. Explain in a short paragraph what this says: a $2\pi$ rotation is not [continuously](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) undoable while a $4\pi$ rotation is (the belt trick), because $SO(3)$ ’s loops are detected in its double cover $S^3$ .
11. Deduce also the practical dividend: composition of rotations = multiplication of quaternions ( $4$ multiplications’ worth of data instead of $9$ , no drift from orthogonality) — verify on the composition of two quarter-turns about $\mathrm i$ and $\mathrm j$ : compute the axis and angle of the product.

**Part V — The explicit matrix: Euler–Rodrigues.** Write $q = a + b\mathrm i + c\mathrm j +
d\mathrm k \in S^3$, so that $a^2 + b^2 + c^2 + d^2 = 1$.

12. Compute $\rho_q(\mathrm i)$ in full from the multiplication table; then obtain $\rho_q(\mathrm  j)$ and $\rho_q(\mathrm k)$ by the cyclic substitution $\mathrm i \to \mathrm j \to \mathrm k  \to \mathrm i$, $(b, c, d) \to (c, d, b)$ *(justify it: cycling $\mathrm i, \mathrm j,  \mathrm k$ extends to an automorphism of $\mathbb  H$, because the defining relations are cyclically symmetric)*. Conclude that the matrix of $\rho_q$ in the basis $(\mathrm i, \mathrm j, \mathrm k)$ is the *Euler–Rodrigues matrix* $$R_q = \begin{pmatrix}  a^2 + b^2 - c^2 - d^2 & 2(bc - ad) & 2(bd + ac)\\  2(bc + ad) & a^2 - b^2 + c^2 - d^2 & 2(cd - ab)\\  2(bd - ac) & 2(cd + ab) & a^2 - b^2 - c^2 + d^2  \end{pmatrix}.$$
13. (Reading a rotation backwards) Show that $$\operatorname{tr}R_q = 4a^2 - 1 = 1 + 2\cos\theta,  \qquad  \tfrac12\bigl(R_q - R_q^{\mathsf T}\bigr) =  \sin\theta\,A_u,$$ in the notation of questions 5 and 8. Deduce an algorithm recovering $\pm q$ from a rotation matrix $R$: the angle from the trace; the axis from the antisymmetric part when $0 < \theta < \pi$; and, when $\theta = \pi$, prove and use the identity $R  + I = 2\,uu^{\mathsf T}$.
14. Evaluate $R_q$ for question 11’s product $q =  \frac12(1 + \mathrm i + \mathrm j + \mathrm k)$ : a permutation matrix appears. Identify the rotation and reconcile with the axis and angle found in question 11.

**Part VI — Inside $S^3$: $SU(2)$, [conjugacy classes](https://one-course.com/books/math/5/en/chapter/1-group-theory#ex-b3-groups-actions), exponentials.**

15. Show that the matrix [representation](https://one-course.com/books/math/5/en/chapter/5-representations-of-finite-groups#def-b3-representations-rep) of question 1 (call it $\Phi$) restricts to a group isomorphism from $S^3$ onto the *special unitary group* $$SU(2) = \bigl\{U \in M_2(\C) : U^*U = I,\ \det U =  1\bigr\}$$ *(for surjectivity, write out the equations $U^{-1} = U^*$ and $\det U = 1$ for a general $2\times2$ complex matrix)*.
16. Show that the real part is a conjugation invariant on $S^3$ — $\operatorname{Re}(pq\bar p) =  \operatorname{Re}q$ for all $p \in S^3$ — and, conversely, that two unit quaternions with the same real part are conjugate in $S^3$ *(reduce to moving one unit pure axis onto another, which Part II provides)* . Describe the [conjugacy classes](https://one-course.com/books/math/5/en/chapter/1-group-theory#ex-b3-groups-actions) of $S^3$ geometrically; translate into $SU(2)$ (level sets of the trace); and project by $\rho$ : two rotations are conjugate in $SO(3)$ if and only if they have the same angle $\theta \in \intcc0\pi$ .
17. Define $\exp$ on $\mathbb H$ by the exponential series; check absolute convergence, using $\abs{pq} = \abs p\,\abs q$ for $\abs q =  \sqrt{N(q)}$. Show, for a unit pure $u$ and $\theta  \in \R$, $$\exp(\theta u) = \cos\theta + \sin\theta\,u ,$$ deduce that $\exp$ maps the hyperplane $P$ onto $S^3$, and check that $\rho_{\exp(su)} =  \eu^{2sA_u}$: the half-angle phenomenon of question 9 again.
18. For pure quaternions $v, w$ prove the product rule $vw = -\langle v, w\rangle + v\wedge w$ , hence the [commutator](https://one-course.com/books/math/5/en/chapter/1-group-theory#def-b3-groups-derived) identity $vw - wv = 2\,v\wedge w$ ; prove also $[A_v, A_w] = A_{v\wedge w}$ for the matrices of question 8. Conclude that the derivative of $\rho$ at $1$ along the curves $t \mapsto \exp(tv)$ is the linear isomorphism $v \mapsto 2A_v$ from $P$ onto the antisymmetric matrices, and that it transports the quaternion [commutator](https://one-course.com/books/math/5/en/chapter/1-group-theory#def-b3-groups-derived) to the matrix [commutator](https://one-course.com/books/math/5/en/chapter/1-group-theory#def-b3-groups-derived) .

**Part VII — Global structure.**

19. (No [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) section) Suppose $s \colon SO(3) \to  S^3$ is [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) with $\rho \circ s =  \operatorname{id}$ . For the loop $R(t) =  \rho_{q(t)}$ of question 10, set $\varepsilon(t) =  s(R(t))\,q(t)^{-1}$ for $t \in \intcc0{2\pi}$ . Show that $\varepsilon$ is [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) with values in $\{\pm1\}$ , and derive a contradiction: there is no [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) global choice of a unit quaternion representing each rotation.
20. (The ball model) Let $\bar B \subseteq \R^3$ be the closed ball of radius $\pi$ and $E(v) = \eu^{A_v}$ , with $E(0) = I$ . Show that $E$ maps $\bar B$ onto $SO(3)$ , is injective on the open ball, and on the boundary sphere identifies exactly antipodes: $E(\pi u) = E(-\pi u) = 2uu^{\mathsf T} - I$ , with no other coincidences. Thus $SO(3)$ is the ball with antipodal boundary points glued — the projective space $\mathbb{RP}^3$ — and a diameter becomes question 10’s non-contractible loop.
21. Show that $\rho_p\rho_q\rho_p^{-1} = \rho_{pq\bar  p}$ ; that the involutions of $SO(3)$ (the $R \neq  I$ with $R^2 = I$ ) are exactly the half-turns $\rho_w$ with $w$ a unit pure quaternion; and that the [center](https://one-course.com/books/math/5/en/chapter/1-group-theory#ex-b3-groups-actions) of $SO(3)$ is trivial.
22. Show that every rotation is a product of two half-turns: for $q = \cos\frac\theta2 +  \sin\frac\theta2\,u$ , choose a unit pure $w \perp  u$ , check that $w' = qw$ is again a unit pure quaternion, and verify $\rho_q =  \rho_{w'}\rho_w$ . Where do the two axes lie, and what angle do they make?
23. Conclude the [topological](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-topology) summary: $SO(3)$ is [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) and [path-connected](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-pathconnected) (give two proofs: [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) image of $S^3$ under $\rho$ ; image of $\exp$ ), while $O(3)$ has exactly two [connected components](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-components) , each [homeomorphic](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) to $SO(3)$ .
24. (A composition, three ways) Let $R_1$ be the rotation by $\frac\pi2$ about the $z$ -axis and $R_2$ the rotation by $\frac\pi2$ about the $x$ -axis. Compute the axis and angle of $R_2R_1$ : (i) by multiplying the two $3\times3$ matrices and using trace/antisymmetric part (Part V); (ii) by multiplying the corresponding unit quaternions $q_2q_1$ . Check the two answers agree: angle $\frac{2\pi}3$ , axis $\frac1{\sqrt3}(1, -1, 1)$ .
25. (The [Cayley transform](https://one-course.com/books/math/5/en/chapter/18-conformal-maps-and-the-riemann-mapping-theorem#ex-b3-conformal-mobius)) For $K$ antisymmetric, show that $I + K$ is invertible and $$C(K) = (I - K)(I + K)^{-1} \in SO(n),$$ with $-1$ never an eigenvalue of $C(K)$; show that $K \mapsto C(K)$ is a bijection from antisymmetric matrices onto $\{R \in SO(n) : -1 \notin  \operatorname{Sp}R\}$, with inverse $R \mapsto (I -  R)(I + R)^{-1}$. (A rational chart of $SO(n)$, companion to the transcendental $\exp$ of [Exercise 20.11](#exo-b3-submanifolds-11).)

**Solution of Problem 20.1.**

**1.** Map $q = t + x\mathrm i + y\mathrm j + z\mathrm k
\mapsto \bigl(\begin{smallmatrix}\alpha & \beta\\ -\bar\beta
& \bar\alpha\end{smallmatrix}\bigr)$ with $\alpha = t + \iu
x$, $\beta = y + \iu z$: one checks that $1, \mathrm i,
\mathrm j, \mathrm k$ go to $I$, $\bigl(\begin{smallmatrix}
\iu & 0\\ 0 & -\iu\end{smallmatrix}\bigr)$, $\bigl(\begin{smallmatrix}0 & 1\\ -1 &
0\end{smallmatrix}\bigr)$, $\bigl(\begin{smallmatrix}0 &
\iu\\ \iu & 0\end{smallmatrix}\bigr)$, whose products reproduce the quaternion table: the map is an injective algebra morphism, so $\mathbb H$ inherits associativity; $N(q) = \abs\alpha^2 + \abs\beta^2 = \det$ is multiplicative, and conjugation corresponds to the adjugate-transpose, giving $\overline{pq} = \bar q\bar p$. [Center](https://one-course.com/books/math/5/en/chapter/1-group-theory#ex-b3-groups-actions): commuting with $\mathrm i$ forces $y = z = 0$, with $\mathrm j$ forces $x = 0$: $\R$.

**2.** $q\bar q = N(q)$: for $q \neq 0$, $q^{-1} = \bar
q/N(q)$: a division algebra. On $S^3$: $N(pq) = 1$ and $N(q^{-1}) = 1$: a group; and $S^3 \subseteq \R^4$ is the unit sphere: a [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) $3$-submanifold.

**3.** $v$ is pure iff $\bar v = -v$; then $\overline{qv\bar q} = q\bar v\bar q = -qv\bar q$: $\rho_q$ preserves $P$. Linearity is clear; $N(qv\bar q) =
N(q)N(v)N(q) = N(v)$: an isometry of $(P, N) \cong (\R^3,
\norm\cdot^2)$: $\rho_q \in O(3)$. And $\rho_{pq}(v) =
pqv\overline{pq} = p(qv\bar q)\bar p =
\rho_p(\rho_q(v))$: a morphism.

**4.** $\rho_q = \mathrm{id}$ iff $qv = vq$ for all pure $v$, iff $q$ commutes with $\mathrm i, \mathrm j, \mathrm k$, iff $q$ is central (question 1): $q \in \R\cap S^3 =
\{\pm1\}$.

**5.** Write $q = t + p$ ($t \in \R$, $p$ pure): $1 =
N(q) = t^2 + N(p)$, so $t = \cos\frac\theta2$ and $p =
\sin\frac\theta2\,u$ with $N(u) = 1$ for some $\theta$ (if $p = 0$, $q = \pm1$ acts trivially). Since $u^2 = -N(u)
= -1$, $q$ and $u$ commute, and $\rho_q(u) = qu\bar q =
uq\bar q = u$: the axis. For pure units $w \perp u$: the product rule $vw = -\langle v, w\rangle + v\wedge w$ (expand in coordinates from the table) gives $uw = u\wedge w
= -wu$. Then

$$
\rho_q(w) = \bigl(\cos\tfrac\theta2 +
\sin\tfrac\theta2u\bigr)\,w\,\bigl(\cos\tfrac\theta2 -
\sin\tfrac\theta2u\bigr)
= \cos\theta\,w + \sin\theta\,(u\wedge w),
$$

using $uwu = -u^2w = w$ and the double-angle formulas: the rotation of angle $\theta$ in the oriented plane $(w,
u\wedge w)$.

**6.** Each $\rho_q$ is a rotation about $u$ by $\theta$: in the orthonormal basis $(u, w, u\wedge w)$ its matrix has determinant $+1$: $\operatorname{im}\rho \subseteq
SO(3)$. Surjectivity: a matrix $R \in SO(3)$ has $1$ as an eigenvalue, since

$$
\det(R - I) = \det R\,\det(I - R^{\mathsf T}) = \det(I - R)
= (-1)^3\det(R - I),
$$

so $\det(R - I) = 0$. Take a unit eigenvector $u$; $R$ preserves $u^\perp$ and restricts there to a rotation of some angle $\theta$ (planar orthogonal, determinant $1$): $R = \rho_q$ for $q = \cos\frac\theta2 +
\sin\frac\theta2\,u$. With question 4 and the first isomorphism theorem ([Theorem 1.3](https://one-course.com/books/math/5/en/chapter/1-group-theory#thm-b3-groups-firstiso)): $SO(3)
\cong S^3/\{\pm1\}$.

**7.** $O_3$ is a [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) $3$-dimensional submanifold ([Exercise 20.5](#exo-b3-submanifolds-5)); $\det$ is [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) on it with values in $\{\pm1\}$, so $SO(3) = O_3\cap\{\det = 1\}$ is open and closed in $O_3$: a union of [connected components](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-components), hence itself a [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) $3$-submanifold, with the same [tangent space](#def-b3-submanifolds-tangent) at $I$: the antisymmetric matrices.

**8.** $A_u^2v = u\wedge(u\wedge v) = \langle u,
v\rangle u - v$ (unit $u$), so $A_u^3v = u\wedge(\langle
u,v\rangle u - v) = -u\wedge v$: $A_u^3 = -A_u$. Splitting the exponential series by [residues](https://one-course.com/books/math/5/en/chapter/17-laurent-series-and-the-residue-theorem#def-b3-residues-singularities) of powers mod the relation $A^3 = -A$:

$$
\eu^{\theta A_u} = I + \Bigl(\theta - \frac{\theta^3}{3!} +
\cdots\Bigr)A_u + \Bigl(\frac{\theta^2}{2!} -
\frac{\theta^4}{4!} + \cdots\Bigr)A_u^2
= I + \sin\theta\,A_u + (1 - \cos\theta)\,A_u^2 .
$$

On $u$: $A_uu = 0$: fixed. On $w \perp u$: $\eu^{\theta
A_u}w = w + \sin\theta\,u\wedge w + (1 - \cos\theta)(-w) =
\cos\theta\,w + \sin\theta\,u\wedge w$: the rotation of axis $u$, angle $\theta$ — Rodrigues. Every rotation has this form (question 6): $\exp$ is onto $SO(3)$ from the antisymmetric matrices.

**9.** With $q(t) = \cos\frac t2 + \sin\frac t2\,u$: question 5 shows $\rho_{q(t)}$ is the rotation of axis $u$ and angle $t$, i.e. $\rho_{q(t)} = \eu^{tA_u}$, whose derivative at $t = 0$ is $A_u$. The quaternion runs at half the angle — the analytic trace of the double cover.

**10.** $\rho_{q(t)}$ is the rotation about $\mathrm k$ by angle $t$: at $t = 2\pi$ it returns to the identity — a loop in $SO(3)$. Its lift satisfies $q(2\pi) = \cos\pi =
-1 = -q(0)$: the lifted [path](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-pathconnected) is *not* closed; only at $t = 4\pi$ does $q$ return to $1$. Interpretation: the loop of full rotations is not contractible in $SO(3)$ — its lift ends at the other sheet of the cover — while the double loop is; a body attached to its surroundings by straps (the belt trick) returns to an untwisted state after $4\pi$ but not after $2\pi$. Rotation groups remember the parity of full turns; $S^3$, being [simply connected](https://one-course.com/books/math/5/en/chapter/18-conformal-maps-and-the-riemann-mapping-theorem#def-b3-conformal-simplyconnected), is where that memory lives.

**11.** Quarter turns: $q_{\mathrm i} =
\cos\frac\pi4 + \sin\frac\pi4\,\mathrm i$, $q_{\mathrm j} =
\cos\frac\pi4 + \sin\frac\pi4\,\mathrm j$. Product (applying the $\mathrm j$-turn first):

$$
q_{\mathrm i}q_{\mathrm j} = \tfrac12(1 + \mathrm i)(1 +
\mathrm j) = \tfrac12\bigl(1 + \mathrm i + \mathrm j +
\mathrm k\bigr),
$$

of norm $1$, with $\cos\frac\theta2 = \frac12$: $\theta =
\frac{2\pi}3$, and axis $u = \frac{\mathrm i + \mathrm j +
\mathrm k}{\sqrt3}$ (the pure part normalized). Two successive quarter-turns about orthogonal axes make a $120^\circ$ rotation about the cube’s main diagonal — four real multiplications’ worth of bookkeeping, orthogonality preserved exactly: why flight software and graphics engines compose rotations through quaternions.

**12.** From the table, $\mathrm{ji} = -\mathrm k$ and $\mathrm{ki} = \mathrm j$, so

$$
q\,\mathrm i = a\mathrm i - b + c(\mathrm{ji}) +
d(\mathrm{ki}) = -b + a\mathrm i + d\mathrm j - c\mathrm k .
$$

Multiplying by $\bar q = a - b\mathrm i - c\mathrm j -
d\mathrm k$ with the scalar–vector rule $(t_1 + p_1)(t_2 +
p_2) = t_1t_2 - \langle p_1, p_2\rangle + t_1p_2 + t_2p_1 +
p_1\wedge p_2$, where $p_1 = (a, d, -c)$ and $p_2 = (-b, -c,
-d)$: the scalar part is $-ab - (-ab) = 0$ (pure, as it must be), and the vector part is

$$
b(b, c, d) + a(a, d, -c) + (-c^2 - d^2,\ bc + ad,\ bd - ac)
= \bigl(a^2 + b^2 - c^2 - d^2,\ 2(bc + ad),\ 2(bd -
ac)\bigr):
$$

the first column of $R_q$. The cyclic map $\sigma(\mathrm i)
= \mathrm j$, $\sigma(\mathrm j) = \mathrm k$, $\sigma(\mathrm k) = \mathrm i$ preserves the relations $\mathrm i^2 = \mathrm j^2 = \mathrm k^2 = \mathrm{ijk} =
-1$ (the word $\mathrm{ijk}$ is cyclically invariant up to the relation $\mathrm{ijk} = \mathrm{jki}$, which holds in any ring: conjugating $\mathrm{ijk} = -1$ by the invertible $\mathrm i$), so $\sigma$ extends to an $\R$-algebra automorphism, and $\sigma(\rho_q(v)) =
\rho_{\sigma(q)}(\sigma(v))$. Unwinding, the image of $\mathrm j$ is the first-column formula after the substitution $(b, c, d) \to (c, d, b)$ with the basis relabeled $\mathrm i \to \mathrm j \to \mathrm k \to \mathrm
i$, which is exactly the second column displayed; one more turn gives the third.

**13.** Summing the diagonal, $\operatorname{tr}R_q =
3a^2 - (b^2 + c^2 + d^2) = 4a^2 - 1$ (unit norm), and with $a = \cos\frac\theta2$: $4\cos^2\frac\theta2 - 1 = 1 +
2\cos\theta$. Antisymmetric part: the three independent entries of $R_q - R_q^{\mathsf T}$ are $4ab, 4ac, 4ad$ (in positions $(3,2), (1,3), (2,1)$), so $\frac12(R_q -
R_q^{\mathsf T}) = A_m$ with $m = 2a\,(b, c, d) =
2\cos\frac\theta2\sin\frac\theta2\,u = \sin\theta\,u$. Algorithm: $\theta = \arccos\frac{\operatorname{tr}R -
1}{2} \in \intcc0\pi$; if $0 < \theta < \pi$, read $u$ off $\frac{R - R^{\mathsf T}}{2\sin\theta}$ and set $q =
\pm(\cos\frac\theta2 + \sin\frac\theta2 u)$; if $\theta =
0$, $q = \pm1$. For $\theta = \pi$: $a = 0$, and Rodrigues (question 8) gives $R = I + 2A_u^2 = I + 2(uu^{\mathsf T} -
I) = 2uu^{\mathsf T} - I$, i.e. $R + I = 2uu^{\mathsf T}$; any nonzero column of $R + I$, normalized, is $\pm u$, and $q = \pm u$.

**14.** With $a = b = c = d = \frac12$: all diagonal entries vanish, $2(bc - ad) = 0$, $2(bd + ac) = 1$, $2(bc +
ad) = 1$, $2(cd - ab) = 0$, $2(bd - ac) = 0$, $2(cd + ab) =
1$:

$$
R_q = \begin{pmatrix} 0 & 0 & 1\\ 1 & 0 & 0\\ 0 & 1 & 0
\end{pmatrix},
$$

the cyclic permutation $e_1 \to e_2 \to e_3 \to e_1$. Its trace is $0 = 1 + 2\cos\theta$, so $\theta = \frac{2\pi}3$, and it fixes $(1,1,1)$: the rotation by $120^\circ$ about the main diagonal — precisely question 11’s answer, now visible as the matrix that cycles the coordinate axes.

**15.** On the basis one checks $\Phi(\bar q) =
\Phi(q)^*$ (the matrix of $\bar q$ has $\alpha' =
\bar\alpha$, $\beta' = -\beta$, which is the conjugate transpose of $\bigl(\begin{smallmatrix}\alpha & \beta\\
-\bar\beta & \bar\alpha\end{smallmatrix}\bigr)$). Hence $\Phi(q)^*\Phi(q) = \Phi(\bar qq) = N(q)I$ and $\det\Phi(q)
= \abs\alpha^2 + \abs\beta^2 = N(q)$: for $q \in S^3$, $\Phi(q) \in SU(2)$, and $\Phi$ is an injective morphism (question 1). Surjectivity: let $U =
\bigl(\begin{smallmatrix}\alpha & \beta\\ \gamma &
\delta\end{smallmatrix}\bigr)$ with $\det U = 1$; then $U^{-1} = \bigl(\begin{smallmatrix}\delta & -\beta\\
-\gamma & \alpha\end{smallmatrix}\bigr)$, and $U^{-1} = U^*
= \bigl(\begin{smallmatrix}\bar\alpha & \bar\gamma\\
\bar\beta & \bar\delta\end{smallmatrix}\bigr)$ forces $\delta = \bar\alpha$, $\gamma = -\bar\beta$, and then $1 =
\det U = \abs\alpha^2 + \abs\beta^2$: $U = \Phi(q)$ for the unit quaternion $q$ with coordinates $\alpha = a + \iu b$, $\beta = c + \iu d$. So $S^3 \cong SU(2)$.

**16.** Real scalars are central and $N(p) = 1$ gives $\overline{pq\bar p} = p\bar q\bar p$, so $pq\bar p +
\overline{pq\bar p} = p(q + \bar q)\bar p = q + \bar q$: the real part is invariant. Conversely let $\operatorname{Re}q = \operatorname{Re}q' = a$; then the pure parts have the same norm $\sqrt{1 - a^2} = s$. If $s =
0$, $q = q' = \pm1$. If $s > 0$, write $q = a + su$, $q' =
a + su'$ with $u, u'$ unit pure; question 6 provides a rotation carrying $u'$ to $u$, i.e. $p \in S^3$ with $\rho_p(u') = u$, and then $pq'\bar p = a + s\rho_p(u') =
q$. The classes of $S^3$ are therefore $\{1\}$, $\{-1\}$, and for each $a \in \intoo{-1}1$ the $2$-sphere $\{a + su :
u \text{ unit pure}\}$ of radius $s$. Under $\Phi$, $\operatorname{tr}\Phi(q) = \alpha + \bar\alpha =
2\operatorname{Re}q$: the classes of $SU(2)$ are the level sets of the trace. Projecting: if $q' = pq\bar p$ then $\rho_{q'} = \rho_p\rho_q\rho_p^{-1}$; conversely $\rho_{q'} = \rho_p\rho_q\rho_p^{-1} = \rho_{pq\bar p}$ forces $q' = \pm pq\bar p$ (kernel), so $\operatorname{Re}q' = \pm\operatorname{Re}q$, i.e. $\cos\frac{\theta'}2 = \abs{\operatorname{Re}q'} =
\abs{\operatorname{Re}q} = \cos\frac\theta2$ for the angles in $\intcc0\pi$: conjugate rotations have equal angles. Conversely, equal angles allow representatives with the same nonnegative real part, conjugate by the above: in $SO(3)$, the [conjugacy class](https://one-course.com/books/math/5/en/chapter/1-group-theory#ex-b3-groups-actions) of a rotation is exactly its angle.

**17.** $N$ is multiplicative, so $\abs\cdot$ is a multiplicative norm on $\mathbb H \cong \R^4$ and $\abs{q^k} = \abs q^k$: the series $\sum q^k/k!$ converges absolutely in the finite-dimensional (hence [complete](https://one-course.com/books/math/5/en/chapter/7-complete-spaces-baire-ascoli-stoneweierstrass#def-b3-complete-complete)) space, dominated by $\sum\abs q^k/k! = \eu^{\abs q}$. For a unit pure $u$: $u^2 = -1$, so $(\theta u)^{2m} =
(-1)^m\theta^{2m}$ and $(\theta u)^{2m+1} =
(-1)^m\theta^{2m+1}u$; splitting the series,

$$
\exp(\theta u) = \sum_m\frac{(-1)^m\theta^{2m}}{(2m)!} +
u\sum_m\frac{(-1)^m\theta^{2m+1}}{(2m+1)!} = \cos\theta +
\sin\theta\,u .
$$

Any $q \in S^3$ is $\cos\alpha + \sin\alpha\,u$ with $\alpha \in \intcc0\pi$ (question 5): $q = \exp(\alpha u)$, so $\exp(P) = S^3$. Finally $\exp(su) = \cos s + \sin s\,u
= q(2s)$ in question 9’s notation, and $\rho_{q(t)} =
\eu^{tA_u}$ there: $\rho_{\exp(su)} = \eu^{2sA_u}$.

**18.** Expanding $vw$ coordinatewise with the table: the products $\mathrm i\cdot\mathrm i = -1$, … give the scalar $-(v_1w_1 + v_2w_2 + v_3w_3)$, and the mixed products ($\mathrm{ij} = \mathrm k$, $\mathrm{ji} =
-\mathrm k$, …) give the vector $(v_2w_3 - v_3w_2,\
v_3w_1 - v_1w_3,\ v_1w_2 - v_2w_1)$: $vw = -\langle v,
w\rangle + v\wedge w$. Subtracting the reversed product: $vw - wv = 2\,v\wedge w$ (the scalar parts cancel, the cross products add). For the matrices, with $a\wedge(b\wedge
c) = b\langle a, c\rangle - c\langle a, b\rangle$:

$$
[A_v, A_w]x = v\wedge(w\wedge x) - w\wedge(v\wedge x)
= w\langle v, x\rangle - v\langle w, x\rangle
= (v\wedge w)\wedge x = A_{v\wedge w}x .
$$

Derivative: $\overline{\exp(tv)} = \exp(-tv)$ (conjugation is [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) and negates pure quaternions), so

$$
\frac{\dd}{\dd t}\Bigr|_{t=0}\exp(tv)\,x\,\exp(-tv) = vx -
xv = 2\,v\wedge x = 2A_vx :
$$

the differential is $v \mapsto 2A_v$, a linear bijection from $P$ onto the antisymmetric matrices, and $[2A_v, 2A_w]
= 4A_{v\wedge w} = 2A_{2v\wedge w} = 2A_{[v,w]}$ shows it carries the quaternion [commutator](https://one-course.com/books/math/5/en/chapter/1-group-theory#def-b3-groups-derived) to the matrix [commutator](https://one-course.com/books/math/5/en/chapter/1-group-theory#def-b3-groups-derived).

**19.** Applying $\rho$: $\rho(\varepsilon(t)) =
\rho(s(R(t)))\,\rho(q(t))^{-1} = R(t)R(t)^{-1} =
\operatorname{id}$, so $\varepsilon(t) \in \ker\rho =
\{\pm1\}$ (question 4). As a product of the [continuous maps](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) $t \mapsto s(R(t))$ and $t \mapsto q(t)^{-1} =
\bar{q(t)}$, $\varepsilon$ is [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) on the [connected](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-connected) interval $\intcc0{2\pi}$ with values in the discrete pair $\{\pm1\}$: it is constant, say $\varepsilon(t) \equiv
\varepsilon$. But $R(0) = R(2\pi) = I$, so $s(R(0)) =
s(R(2\pi))$, while $s(R(0)) = \varepsilon\,q(0) =
\varepsilon$ and $s(R(2\pi)) = \varepsilon\,q(2\pi) =
-\varepsilon$: contradiction. No [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) section exists: the sign ambiguity $\pm q$ is global, not a defect of a particular formula.

**20.** *Onto:* every $R \in SO(3)$ is $\eu^{\theta A_u}$ for some unit $u$ and $\theta \in
\intcc0{2\pi}$ (questions 6 and 8); if $\theta > \pi$, Rodrigues gives $\eu^{\theta A_u} = \eu^{(2\pi -
\theta)A_{-u}}$ (both equal $I + \sin\theta A_u + (1 -
\cos\theta)A_u^2$, and $A_{-u} = -A_u$ with $\sin(2\pi -
\theta) = -\sin\theta$, $\cos(2\pi - \theta) =
\cos\theta$), so $R = E(v)$ with $\norm v \leq \pi$. *Injective inside:* if $E(v) = E(v') \neq I$ with $\norm v, \norm{v'} < \pi$, question 13 recovers the same angle $\theta = \norm v = \norm{v'} \in \intoo0\pi$ from the trace and, since $\sin\theta \neq 0$, the same axis from the antisymmetric part: $v = v'$; and $E(v) = I$ forces $\theta \in \{0\}$ on the open ball. *Boundary:* $E(\pi u) = I + 2A_u^2 = 2uu^{\mathsf T} - I$ depends on $u$ only through $uu^{\mathsf T}$, whence $E(\pi u) =
E(-\pi u)$; conversely $2uu^{\mathsf T} - I =
2u'u'^{\mathsf T} - I$ applied to $u$ gives $u = \langle
u', u\rangle u'$, so $u' = \pm u$. [Interior](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-interior) and boundary never collide (trace $> -1$ versus $= -1$). So $E$ induces a [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) bijection from the ball-with-antipodal-gluing — [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) — onto $SO(3)$: a [homeomorphism](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity), and $SO(3)
\cong \mathbb{RP}^3$. A diameter from $\pi u$ to $-\pi u$ has glued endpoints: it is a loop in $SO(3)$, and its $E$-description matches question 10’s family of rotations about $u$ sweeping a full turn.

**21.** $\rho$ is a morphism and $\rho_p^{-1} =
\rho_{p^{-1}} = \rho_{\bar p}$, so $\rho_p\rho_q\rho_p^{-1}
= \rho_{pq\bar p}$; by question 16, conjugating a rotation preserves its angle and rotates its axis by $\rho_p$. Involutions: $\rho_q^2 = \rho_{q^2} = \operatorname{id}$ iff $q^2 = \pm1$. If $q^2 = 1$ then $(q - 1)(q + 1) = q^2 -
1 = 0$ (central scalars, so this factorization is valid) and $q = \pm1$ in the division ring $\mathbb H$, giving $\rho_q
= I$, excluded; $q^2 = -1$ with $q = a +
su$ gives $a^2 - s^2 + 2as\,u = -1$, so $a = 0$, $s = 1$: $q$ is a unit pure $w$, and $\rho_w$ is the half-turn about $w$ (angle $\pi$, question 5). [Center](https://one-course.com/books/math/5/en/chapter/1-group-theory#ex-b3-groups-actions): if $\rho_q$ commutes with every $\rho_p$, then $\rho_{pq\bar p} = \rho_q$, so $pq\bar p = \varepsilon(p)\,q$ with $\varepsilon(p) \in
\{\pm1\}$; $p \mapsto \varepsilon(p) = (pq\bar p)q^{-1}$ is [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) on the [connected](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-connected) $S^3$ and equals $1$ at $p =
1$, hence $\equiv 1$: $q$ commutes with all of $S^3$, hence with all of $\mathbb H$ (rescale), so $q \in \R \cap S^3 =
\{\pm1\}$ (question 1) and $\rho_q = I$: the [center](https://one-course.com/books/math/5/en/chapter/1-group-theory#ex-b3-groups-actions) is trivial.

**22.** Since $u \perp w$ are unit pures, $uw =
u\wedge w$ is pure (question 18), so

$$
w' = qw = \cos\tfrac\theta2\,w + \sin\tfrac\theta2\,u\wedge
w
$$

is pure, of norm $\abs q\abs w = 1$. Then $w'w = qw\cdot w
= qw^2 = -q$, and

$$
\rho_{w'}\rho_w = \rho_{w'w} = \rho_{-q} = \rho_q .
$$

Both axes $w$ and $w' = \cos\frac\theta2 w +
\sin\frac\theta2(u\wedge w)$ lie in the plane $u^\perp$ orthogonal to the rotation axis, and $\langle w', w\rangle
= \cos\frac\theta2$: they make the half-angle $\frac\theta2$. This is the classical generation: two half-turns about axes meeting at angle $\frac\theta2$ compose to the rotation of angle $\theta$ about their common perpendicular.

**23.** Compactness is question 7. [Path-connectedness](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-connected): $SO(3) = \rho(S^3)$ is the [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) image of the [path-connected](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-pathconnected) sphere; alternatively, for $R = \eu^{A}$ with $A$ antisymmetric (question 8), $t \mapsto \eu^{tA}$ is a [path](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-pathconnected) in $SO(3)$ from $I$ to $R$ (orthogonal since $(\eu^{tA})^{\mathsf T} = \eu^{-tA}$, determinant $1$ by [continuity](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) from $t = 0$). For $O(3)$: $\det$ is [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) onto $\{\pm1\}$, so $O(3)$ is disconnected, $O(3) = SO(3)
\sqcup D\,SO(3)$ for any fixed $D$ with $\det D = -1$ (e.g. $D = -I$), and left multiplication by $D$ is a [homeomorphism](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity): exactly two components, each a copy of $SO(3)$. The two-to-one $\rho$, section-free by question 19, is thus an honest double cover of a [connected](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-connected) [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) group by the [simply connected](https://one-course.com/books/math/5/en/chapter/18-conformal-maps-and-the-riemann-mapping-theorem#def-b3-conformal-simplyconnected) $S^3$ — the geometry behind the belt trick.

**24.** (i) Matrices:

$$
R_1 = \begin{pmatrix} 0 & -1 & 0\\ 1 & 0 & 0\\ 0 & 0 & 1
\end{pmatrix},
\quad
R_2 = \begin{pmatrix} 1 & 0 & 0\\ 0 & 0 & -1\\ 0 & 1 & 0
\end{pmatrix},
\quad
R_2R_1 = \begin{pmatrix} 0 & -1 & 0\\ 0 & 0 & -1\\ 1 & 0 &
0\end{pmatrix}.
$$

Trace $0 = 1 + 2\cos\theta$ gives $\cos\theta = -\frac12$: $\theta = \frac{2\pi}3$. Antisymmetric part $\frac{R -
R^{\mathsf T}}2$ has entries encoding $\sin\theta\,(v_3,
-v_2, v_1)$-wise the axis: here $\frac{R - R^{\mathsf T}}2 =
\frac12\bigl(\begin{smallmatrix}0 & -1 & -1\\ 1 & 0 & -1\\ 1
& 1 & 0\end{smallmatrix}\bigr)$, which reads (Part V’s dictionary $A_v$) $v\sin\theta = \frac12(1, -1, 1)$; with $\sin\frac{2\pi}3 = \frac{\sqrt3}2$: $v = \frac1{\sqrt3}(1,
-1, 1)$. (ii) Quaternions: $q_1 = \cos\frac\pi4 + \sin\frac\pi4\,k =
\frac{\sqrt2}2(1 + k)$, $q_2 = \frac{\sqrt2}2(1 + i)$, and

$$
q_2q_1 = \tfrac12(1 + i)(1 + k) = \tfrac12(1 + k + i + ik)
= \tfrac12\bigl(1 + i - j + k\bigr)
$$

($ik = -j$). So $\cos\frac\theta2 = \frac12$: $\theta =
\frac{2\pi}3$, and the vector part $\frac12(i - j + k)$ has direction $\frac1{\sqrt3}(1, -1, 1)$ — the same answer, with the quaternion route requiring one line of multiplication instead of a matrix product: the practical reason flight software composes attitudes in $S^3$.

**25.** $I + K$ invertible: $(I + K)v = 0$ gives $0 =
\langle v, v\rangle + \langle v, Kv\rangle = \norm v^2$ (antisymmetry kills the second term): $v = 0$. Orthogonality of $C = C(K)$: using $(I \pm K)^{\mathsf T} = I \mp K$ and the fact that all four matrices $I \pm K$, $(I \pm K)^{-1}$ commute (polynomial expressions in $K$, plus limits):

$$
C^{\mathsf T}C = (I + K)^{-\mathsf T}(I - K)^{\mathsf T}
(I - K)(I + K)^{-1}
= (I - K)^{-1}(I + K)(I - K)(I + K)^{-1} = I .
$$

Determinant: $\det(I - K) = \det\bigl((I - K)^{\mathsf
T}\bigr) = \det(I + K)$, so $\det C = 1$: $C \in SO(n)$. No eigenvalue $-1$: $Cv = -v$ means $(I - K)w = -(I + K)w$ for $w = (I + K)^{-1}v$, i.e. $2w = 0$: $v = 0$. Inversion: from $C(I + K) = I - K$, solve $K(I + C) = I -
C$; since $-1 \notin \operatorname{Sp}C$, $I + C$ is invertible and $K = (I - C)(I + C)^{-1}$, which is antisymmetric whenever $C$ is orthogonal without eigenvalue $-1$ (transpose the expression and use $C^{\mathsf T} =
C^{-1}$: $K^{\mathsf T} = (I - C^{-1})(I + C^{-1})^{-1} =
(C - I)(C + I)^{-1} = -K$). The two maps are mutually inverse by construction: a global rational parametrization of the dense open piece of $SO(n)$ avoiding eigenvalue $-1$ — no series, no trigonometry, and in dimension $3$ it is the half-angle substitution $K = \tan\frac\theta2\,A_v$ in disguise.