University Mathematics — Year 3 · Bachelor Year 3

20Submanifolds 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, and constrained optimization by Lagrange multipliers — 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’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 $V \ni a$ , $W \ni f(a)$ 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 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. 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) 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), with $f \colon V \to W$ bijective.

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) 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 of $y \mapsto D\varphi(y) = Df(\varphi(y))^{-1}$ : composition of continuous maps (inversion is continuous, Proposition 8.4): $\varphi \in \mathcal C^1$ ; 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 $A \ni a$ , $B \ni b$ 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 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 $a \in M$ ):

(Straightening) There is a $\mathcal C^k$ diffeomorphism $\Phi$ from an open $\Omega \ni a$ onto an open $\Omega' \subseteq \R^n$ with

$\Phi(M\cap\Omega) = \Omega' \cap \bigl(\R^d\times\{0\}\bigr).$
(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 $M\cap\Omega = F^{-1}(0)$ .
(Graph) Up to permuting coordinates, $M$ is locally the graph of a $\mathcal C^k$ map $\psi \colon A \subseteq \R^d \to \R^{n-d}$ .
(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 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) 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 onto the graph (inverse: the projection, continuous). (4) $\Rightarrow$ (1): let $\varphi(u_0) = a$ ; 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); its inverse $\Phi$ straightens: near $a$ , points of $M$ are exactly the $\varphi(u) = \Theta(u, 0)$ — for this, the homeomorphism 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 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). Matrix groups: $SL_n$ and $O_n$ are submanifolds of $M_n(\R)$ (Exercises 20.5 and 20.6) — 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:

if $M = F^{-1}(0)$ locally with $F$ a submersion: $T_aM = \ker DF(a)$ ;
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(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) 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). Now $\ker DF(a)^\perp = \operatorname{im}DF(a)^{\mathsf T}$ (the rank/orthogonality identity of Year 2, or Exercise 13.8 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: 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.

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: $d = n - q$ 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.

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. (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 of $\bigl(C\setminus\{0\}\bigr)\cap B(0,\varepsilon)$ , and compare with the number a straightening (Theorem 20.3(1)) would force for a plane minus a point.

Solution

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 (upper and lower punctured nappes, each path-connected: connect through circles and rays). If $C$ were a $2$ -submanifold at $0$ , a straightening would give a homeomorphism from $C\cap\Omega$ onto an open piece of a plane sending $0$ to a point $p$ ; small punctured plane-neighborhoods of $p$ have one component, and homeomorphisms preserve the number of components of punctured neighborhoods: contradiction.

Exercise 20.2 ★

Compute the tangent spaces: (a) $T_aS^{n-1}$ for any $a$ (answer: $a^\perp$ ); (b) the tangent plane to the torus of Example 20.4 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(2).

Solution

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(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 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 of $(1, 0)$ (the two square-root branches). (c) Same discussion for the polar coordinates map $(r, \theta) \mapsto (r\cos\theta, r\sin\theta)$ .

Solution

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

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 is a line, Proposition 20.6): $\mathcal C$ 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 $\mathcal C^\infty$ 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

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: closed ( $F$ continuous) 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).

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

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 and 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): a submanifold of full dimension $n^2$ (straightening: the identity chart).

Exercise 20.7 ★★

By Lagrange multipliers: (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

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: extrema exist).

(b) On the 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 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 critical point is the global maximum — uniform ignorance maximizes entropy.

(c) Minimize $(x-1)^2 + y^2$ on the 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 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 in finite dimension directly from this construction.

Solution

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 $S^{n-1}$ at some $v_1$ ; Lagrange (Theorem 20.7, 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: 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 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

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 on the 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(3)) explicitly at $(1, 0)$ , and explain in one sentence why some coordinate permutation is always sufficient but no single one always works.

Solution

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(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: 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 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

Solution of Exercise 20.11.

(a) The defining map $F(M) = M^{\mathsf T}M - I$ is continuous: $O_n = F^{-1}(0)$ is closed; columns of an orthogonal matrix are unit vectors, so the Euclidean (Frobenius) norm is exactly $\sqrt n$ : bounded. 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 by explicit curves — the Lie-group shortcut that Problem 20.1 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; 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

Solution of Exercise 20.12.

Existence: intersect $M$ with a large closed ball around $p$ to get a nonempty compact; the continuous 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 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’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 $S^3$ .

Verify that $\mathbb H$ is an associative $\R$ -algebra with center $\R$ , 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 $\det$ ).
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 $3$ -submanifold of $\R^4$ (Example 20.4).

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 $\rho_q(v) = q\,v\,\bar q$ .

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 $S^3 \to O(3)$ .
Compute the kernel: $\rho_q = \mathrm{id}$ iff $q$ commutes with $\mathrm i, \mathrm j, \mathrm k$ iff $q \in \R\cap S^3 = \{\pm1\}$ .
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 $uw + wu = -2\langle u, w\rangle$ ).
Conclude: $\operatorname{im}\rho \subseteq SO(3)$ (each $\rho_q$ is a rotation with axis and angle as computed — determinant $+1$ by continuity of $q \mapsto \det\rho_q$ on the 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.

Show that $SO(3)$ is a compact $3$ -dimensional submanifold of $M_3(\R)$ with $T_ISO(3) =$ antisymmetric matrices (Exercise 20.5; the determinant condition selects a union of components).
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 $SO(3)$ .
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 tell the same story at half and full speed respectively.

Part IV — The double cover, felt.

Show that the path $q(t) = \cos\frac t2 + \sin\frac t2\,\mathrm k$ , $t \in \intcc0{2\pi}$ , 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 undoable while a $4\pi$ rotation is (the belt trick), because $SO(3)$ ’s loops are detected in its double cover $S^3$ .
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 $a^2 + b^2 + c^2 + d^2 = 1$ .

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

$R_q = \begin{pmatrix} a^2 + b^2 - c^2 - d^2 & 2(bc - ad) & 2(bd + ac)\\ 2(bc + ad) & a^2 - b^2 + c^2 - d^2 & 2(cd - ab)\\ 2(bd - ac) & 2(cd + ab) & a^2 - b^2 - c^2 + d^2 \end{pmatrix}.$
(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 $R + I = 2\,uu^{\mathsf T}$ .
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, exponentials.

Show that the matrix representation 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).
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 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 $\theta \in \intcc0\pi$ .
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.
For pure quaternions $v, w$ prove the product rule $vw = -\langle v, w\rangle + v\wedge w$ , hence the commutator 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 to the matrix commutator.

Part VII — Global structure.

(No continuous section) Suppose $s \colon SO(3) \to S^3$ is continuous 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 with values in $\{\pm1\}$ , and derive a contradiction: there is no continuous global choice of a unit quaternion representing each rotation.
(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.
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 of $SO(3)$ is trivial.
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?
Conclude the topological summary: $SO(3)$ is compact and path-connected (give two proofs: continuous image of $S^3$ under $\rho$ ; image of $\exp$ ), while $O(3)$ has exactly two connected components, each homeomorphic to $SO(3)$ .
(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 $\frac1{\sqrt3}(1, -1, 1)$ .
(The Cayley transform) 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.)

Solution

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: 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 $3$ -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): $SO(3) \cong S^3/\{\pm1\}$ .

7. $O_3$ is a compact $3$ -dimensional submanifold (Exercise 20.5); $\det$ is continuous 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, hence itself a compact $3$ -submanifold, with the same tangent space 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 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 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, 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 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) 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 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 to the matrix commutator.

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 $t \mapsto s(R(t))$ and $t \mapsto q(t)^{-1} = \bar{q(t)}$ , $\varepsilon$ is continuous on the 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 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 and boundary never collide (trace $> -1$ versus $= -1$ ). So $E$ induces a continuous bijection from the ball-with-antipodal-gluing — compact — onto $SO(3)$ : a homeomorphism, 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: 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 on the 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 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: $SO(3) = \rho(S^3)$ is the continuous image of the path-connected sphere; alternatively, for $R = \eu^{A}$ with $A$ antisymmetric (question 8), $t \mapsto \eu^{tA}$ is a path in $SO(3)$ from $I$ to $R$ (orthogonal since $(\eu^{tA})^{\mathsf T} = \eu^{-tA}$ , determinant $1$ by continuity from $t = 0$ ). For $O(3)$ : $\det$ is continuous 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: 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 compact group by the simply connected $S^3$ — 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.

20.1 The inverse function theorem

20.2 Submanifolds: four definitions

20.3 Tangent spaces

20.4 Exercises

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

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