Matrix groups and continuous symmetry

Lens

MotivationH

Symmetries that come in families

Put a book flat on a table. Turn it a quarter turn about a vertical axis, then a quarter turn about the axis pointing towards you. Start again and do the same two turns in the opposite order. The book ends up in two different positions. Rotations of the plane can be done in any order — they are just angles, and angles add — but rotations of space cannot.

Rotations are not isolated symmetries like the six symmetries of an equilateral triangle. They come in continuous families: you can turn a little, or a little more, and the rotations near the identity look like a small piece of three-dimensional space. A drone's flight controller, written perhaps by a robotics team in Bengaluru or Lagos, must store and combine such rotations hundreds of times a second without drifting, and almost all of them use quaternions, four numbers per rotation, rather than $3 \times 3$ matrices or three angles. A physicist adds a stranger fact: rotating an electron's state through $360°$ multiplies it by $-1$ , and only a rotation through $720°$ brings it back.

These are all statements about matrix groups: groups of invertible matrices that are at the same time geometric spaces, so that we can speak of paths, of tangent vectors and of size. This lesson studies the most important ones and the bridge between their algebra and their geometry, the matrix exponential.

IntuitionH

Groups that are also spaces

The rotations of the plane are the matrices $R(\theta) = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}$ . As $\theta$ varies they trace out a circle inside the four-dimensional space of $2 \times 2$ matrices, and $R(\theta)R(\varphi) = R(\theta + \varphi)$ : the group law is addition of angles. The velocity of the curve $\theta \mapsto R(\theta)$ at $\theta = 0$ is the matrix $J = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$ — an "infinitesimal rotation" — and the whole circle can be recovered from it as $R(\theta) = \exp(\theta J)$ , where $\exp$ is the familiar power series with matrices in it.

The same idea works for every matrix group $G$ . Curves in $G$ through the identity have velocity vectors, and these form a vector space, the Lie algebra $\mathfrak g$ — the linearisation of $G$ . The exponential map goes back from $\mathfrak g$ to $G$ . Linear algebra in $\mathfrak g$ then tells us about the group: the dimension of $\mathfrak g$ is the number of parameters of $G$ , and the failure of $G$ to be commutative is measured by the bracket $[X, Y] = XY - YX$ in $\mathfrak g$ .

The rotations of space form a three-dimensional group $SO(3)$ . It is covered twice by the group of unit quaternions, which is the three-dimensional sphere $S^3$ . Shrinking a path of rotations from $720°$ down to nothing is possible; from $360°$ it is not. That topological fact is the electron's sign.

DefinitionsH

Definitions

Definition— matrix group

$GL_n(\R)$ and $GL_n(\C)$ are the groups of invertible $n \times n$ real or complex matrices, viewed as open subsets of $\R^{n^2}$ and $\C^{n^2}$ . A matrix group (or linear Lie group) is a subgroup of $GL_n$ that is a closed subset of it. Topological words — continuous, compact, connected, path — refer to this topology.

Definition— the classical groups

$SL_n(k) = \{ A : \det A = 1 \}$ , the special linear group ( $k = \R$ or $\C$ ).
$O(n) = \{ A \in GL_n(\R) : A^{\mathsf T}A = I \}$ , the orthogonal group, and $SO(n) = \{ A \in O(n) : \det A = 1 \}$ , the rotation group.
$U(n) = \{ A \in GL_n(\C) : A^*A = I \}$ with $A^* = \overline A^{\mathsf T}$ , the unitary group, and $SU(n) = \{ A \in U(n) : \det A = 1 \}$ .
The Euclidean group $E(n)$ of isometries $x \mapsto Ax + b$ of $\R^n$ ( $A \in O(n)$ , $b \in \R^n$ ), realised as the matrices $\begin{pmatrix} A & b \\ 0 & 1 \end{pmatrix}$ acting on vectors $\begin{pmatrix} x \\ 1 \end{pmatrix}$ .

Definition— quaternions

The quaternions $\mathbb H$ are the real vector space with basis $1, i, j, k$ and the associative multiplication determined by Hamilton's rules $i^2 = j^2 = k^2 = ijk = -1$ (so $ij = k = -ji$ , $jk = i = -kj$ , $ki = j = -ik$ ). For $q = a + bi + cj + dk$ , the conjugate is $\bar q = a - bi - cj - dk$ and $|q|^2 = q\bar q = a^2 + b^2 + c^2 + d^2$ . A quaternion is pure if $a = 0$ ; we identify the pure quaternions with $\R^3$ .

Definition— matrix exponential and one-parameter subgroup

For a square matrix $X$ , $\exp X = \sum_{k \ge 0} X^k / k! = I + X + X^2/2 + \dots$ . A one-parameter subgroup of $G$ is a continuous homomorphism $\varphi : (\R, +) \to G$ , that is, $\varphi(s + t) = \varphi(s)\varphi(t)$ .

Definition— Lie algebra

The Lie algebra of a matrix group $G$ is $\mathfrak g = \{ X : \exp(tX) \in G \text{ for all } t \in \R \}$ , equipped with the bracket $[X, Y] = XY - YX$ . We write $\mathfrak{so}(n)$ , $\mathfrak{sl}_n$ , $\mathfrak u(n)$ , $\mathfrak{su}(n)$ for the Lie algebras of the classical groups.

Worked examplesH

Examples and non-examples

Example— The circle group, three times

$SO(2) = \{ R(\theta) \}$ is a circle. So is $U(1) = \{ z \in \C : |z| = 1 \}$ , and the map $e^{i\theta} \mapsto R(\theta)$ is an isomorphism of groups that is also a homeomorphism: multiplying by $e^{i\theta}$ is rotating by $\theta$ . Both are commutative. $O(2)$ is $SO(2)$ together with the reflections $\begin{pmatrix} \cos\theta & \sin\theta \\ \sin\theta & -\cos\theta \end{pmatrix}$ (determinant $-1$ ): two disjoint circles.

Example— Exponentials you can compute by hand

$\exp(\theta J) = R(\theta)$ : since $J^2 = -I$ , the even terms give $\cos\theta \, I$ and the odd terms $\sin\theta \, J$ .
$\exp\begin{pmatrix} 0 & t \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} 1 & t \\ 0 & 1 \end{pmatrix}$ , a shear: the series stops because the matrix squares to $0$ .
$\exp\begin{pmatrix} 0 & t \\ t & 0 \end{pmatrix} = \begin{pmatrix} \cosh t & \sinh t \\ \sinh t & \cosh t \end{pmatrix}$ , a "hyperbolic rotation" preserving $x^2 - y^2$ — a Lorentz boost in one space dimension.
$\exp(\operatorname{diag}(a, b)) = \operatorname{diag}(e^a, e^b)$ , and $\exp(PXP^{-1}) = P\exp(X)P^{-1}$ , so diagonalisable matrices are exponentiated by exponentiating eigenvalues.

Example— A non-compact connected group, and an element that is not an exponential

$SL_2(\R)$ contains $\operatorname{diag}(t, 1/t)$ for every $t > 0$ , so it is unbounded and not compact. It is connected: every $A \in SL_2(\R)$ factors (Gram–Schmidt on the columns) as $A = QR$ with $Q \in SO(2)$ and $R = \begin{pmatrix} a & b \\ 0 & 1/a \end{pmatrix}$ , $a > 0$ , and both factors can be slid continuously to $I$ . Yet $\exp : \mathfrak{sl}_2(\R) \to SL_2(\R)$ is not surjective: $A = \begin{pmatrix} -1 & 1 \\ 0 & -1 \end{pmatrix}$ is not $\exp X$ for any real $X$ . Indeed, $\det A = 1$ forces $\operatorname{tr} X = 0$ (Theorem 7(e) below), so $X$ has eigenvalues $\pm\mu$ with $\mu^2 = -\det X$ . If $\mu$ is real and non-zero, $\exp X$ has positive eigenvalues $e^{\pm\mu}$ ; if $\mu = 0$ , $X$ is nilpotent and $\exp X = I + X$ has eigenvalue $1$ ; if $\mu = i\nu$ is imaginary, $X$ is diagonalisable over $\C$ , so $\exp X$ is too — but $A$ is a Jordan block, not diagonalisable. No case gives $A$ .

Example— Groups that are not matrix groups in the sense above

The subgroup $\{ \exp(tX) : t \in \R \}$ of the torus $U(1) \times U(1)$ , with $X = \operatorname{diag}(i, i\sqrt2)$ , is a line winding densely around the torus: it is a subgroup but not closed, so we do not call it a matrix group. (Its closure is the whole torus.) Closedness is what makes the Lie algebra well behaved; von Neumann proved in 1929 that closed subgroups of $GL_n$ are automatically smooth manifolds.

ProofH

Orthogonal groups: compactness, components and Euler's theorem

Theorem— Proposition 1 (what orthogonal means)

For a real $n \times n$ matrix $A$ , the following are equivalent: (a) $A^{\mathsf T}A = I$ ; (b) $(Av) \cdot (Aw) = v \cdot w$ for all $v, w$ ; (c) the columns of $A$ are orthonormal. Every $A \in O(n)$ has $\det A = \pm 1$ , and $O(n)$ is a group.

Proof

(a) ⇒ (b): $(Av) \cdot (Aw) = v^{\mathsf T}A^{\mathsf T}Aw = v \cdot w$ . (b) ⇒ (c): the columns are $Ae_i$ , and $Ae_i \cdot Ae_j = e_i \cdot e_j$ . (c) ⇒ (a): the $(i, j)$ entry of $A^{\mathsf T}A$ is (column $i$ ) $\cdot$ (column $j$ ). Taking determinants in (a), $(\det A)^2 = \det A^{\mathsf T} \det A = 1$ . If $A, B \in O(n)$ then $(AB)^{\mathsf T}AB = B^{\mathsf T}A^{\mathsf T}AB = I$ , and $A^{-1} = A^{\mathsf T}$ is orthogonal.

∎

Theorem— Proposition 2 (compactness and components)

$O(n)$ and $SO(n)$ are compact. $O(n)$ is not connected: it is the disjoint union of the closed sets $SO(n)$ and $\{ \det = -1 \}$ . $U(n)$ and $SU(n)$ are compact; $GL_n$ and $SL_n$ ( $n \geq 2$ ) are not.

Proof

$O(n)$ is the preimage of $I$ under the continuous map $A \mapsto A^{\mathsf T}A$ , hence closed, and its entries satisfy $|a_{ij}| \le 1$ because the columns are unit vectors, so it is bounded; by Heine–Borel it is compact. $SO(n)$ is a closed subset of it. The determinant is continuous on $O(n)$ with values in $\{1, -1\}$ , so the two level sets are disjoint, closed and non-empty: $O(n)$ is disconnected, and no path of orthogonal matrices joins a reflection to $I$ . The same argument with $A^*A$ shows $U(n)$ and $SU(n)$ are compact. Finally $\operatorname{diag}(t, 1/t, 1, \dots, 1) \in SL_n(\R)$ is unbounded.

∎

Theorem— Proposition 3 (connectedness)

$SO(n)$ , $U(n)$ and $SU(n)$ are path-connected.

Proof

$U(n)$ . A unitary matrix is normal, so by the complex spectral theorem (see advanced linear algebra) $A = P \operatorname{diag}(e^{i\theta_1}, \dots, e^{i\theta_n}) P^*$ with $P$ unitary. The path $t \mapsto P \operatorname{diag}(e^{it\theta_1}, \dots, e^{it\theta_n})P^*$ , $t \in [0, 1]$ , runs inside $U(n)$ from $I$ to $A$ . If $\det A = 1$ we may choose the $\theta_j$ with $\sum \theta_j = 0$ , and then the path stays in $SU(n)$ .

$SO(n)$ . An orthogonal $A$ is normal with eigenvalues on the unit circle, in conjugate pairs $e^{\pm i\theta}$ . If $v = x + iy$ is an eigenvector for $e^{i\theta}$ with $\theta \notin \{0, \pi\}$ , then $\operatorname{span}(x, y)$ is an $A$ -invariant plane on which $A$ acts as a rotation by $\theta$ ; the real form of the spectral theorem gives an orthonormal basis (columns of $P \in O(n)$ ) in which $A$ is block diagonal with blocks $R(\theta_j)$ , $1$ and $-1$ . Since $\det A = 1$ the $-1$ s come in pairs, and a pair $\operatorname{diag}(-1, -1) = R(\pi)$ is a rotation block too. Shrinking every angle $\theta_j$ to $0$ gives a path in $SO(n)$ from $A$ to $I$ .

∎

So orientation is topological: the two components of $O(n)$ (and of $GL_n(\R)$ , by Gram–Schmidt) are told apart by the sign of the determinant. Here is the summary, including results stated without proof.

group	dimension	compact	connected
$GL_n(\R)$	$n^2$	no	no (two components)
$SL_n(\R)$ , $n \ge 2$	$n^2 - 1$	no	yes
$O(n)$	$n(n-1)/2$	yes	no (two components)
$SO(n)$	$n(n-1)/2$	yes	yes
$U(n)$	$n^2$	yes	yes
$SU(n)$	$n^2 - 1$	yes	yes
$E(n)$	$n(n+1)/2$	no	no (two components)

Theorem— Theorem 4 (Euler's rotation theorem)

Every $A \in SO(3)$ is a rotation about some axis: there is a unit vector $u$ with $Au = u$ , and $A$ rotates the plane $u^\perp$ by an angle $\theta$ with $1 + 2\cos\theta = \operatorname{tr} A$ .

Proof

Since $AA^{\mathsf T} = I$ and $\det A = 1$ ,

\det(A - I) = \det\big(A(I - A^{\mathsf T})\big) = \det(I - A^{\mathsf T}) = \det\big((I - A)^{\mathsf T}\big) = \det(I - A) = (-1)^3\det(A - I).

So $\det(A - I) = 0$ and there is a unit vector $u$ with $Au = u$ . As $A$ preserves dot products, it maps $u^\perp$ to itself; its restriction is an orthogonal map of a plane with determinant $1$ (the determinant of $A$ in a basis $u, v, w$ is $1 \cdot \det(A|_{u^\perp})$ ), hence a rotation $R(\theta)$ . In that basis $A = \operatorname{diag}(1, R(\theta))$ and $\operatorname{tr} A = 1 + 2\cos\theta$ .

∎

Notice where the dimension entered: $(-1)^3 = -1$ . In $SO(2)$ the rotation $R(\pi/2)$ fixes no non-zero vector, and in $SO(4)$ the matrix $\operatorname{diag}(R(\theta), R(\varphi))$ fixes none either. A surprising corollary: the composite of two rotations of space about different axes is again a rotation about a single axis. And the order matters: with $R_x$ and $R_z$ the quarter turns about the $x$ - and $z$ -axes, $R_xR_z \ne R_zR_x$ , so $SO(3)$ is not commutative, while $SO(2)$ is.

ProofH

Quaternions, SU(2) and the double cover

Theorem— Proposition 5 (unit quaternions)

$|pq| = |p||q|$ for all quaternions, so the unit quaternions form a group, which as a space is the sphere $S^3 \subset \R^4$ . The map

a + bi + cj + dk \;\longmapsto\; \begin{pmatrix} a + bi & -c - di \\ c - di & a - bi \end{pmatrix}

is an injective ring homomorphism $\mathbb H \to M_2(\C)$ with $\det = |q|^2$ , and it identifies the unit quaternions with $SU(2)$ .

Proof

A direct computation shows $\overline{pq} = \bar q\,\bar p$ , so $|pq|^2 = pq\,\bar q\,\bar p = p|q|^2\bar p = |p|^2|q|^2$ (real numbers commute with everything). So products and inverses ( $q^{-1} = \bar q$ ) of unit quaternions are unit quaternions. The images of $i, j, k$ are $\begin{pmatrix} i & 0 \\ 0 & -i \end{pmatrix}$ , $\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$ and $\begin{pmatrix} 0 & -i \\ -i & 0 \end{pmatrix}$ ; one checks that they satisfy Hamilton's rules, so the map is a ring homomorphism, and it is injective since the four matrices are independent. Writing $\alpha = a + bi$ and $\beta = c - di$ , the image is $\begin{pmatrix} \alpha & -\bar\beta \\ \beta & \bar\alpha \end{pmatrix}$ , with determinant $|\alpha|^2 + |\beta|^2 = |q|^2$ ; such a matrix with $|\alpha|^2 + |\beta|^2 = 1$ is unitary (its columns are orthonormal). Conversely, if $A = \begin{pmatrix} \alpha & \gamma \\ \beta & \delta \end{pmatrix} \in SU(2)$ then $A^{-1} = A^*$ reads $\begin{pmatrix} \delta & -\gamma \\ -\beta & \alpha \end{pmatrix} = \begin{pmatrix} \bar\alpha & \bar\beta \\ \bar\gamma & \bar\delta \end{pmatrix}$ , so $\delta = \bar\alpha$ and $\gamma = -\bar\beta$ .

∎

Theorem— Theorem 6 (the double cover SU(2) → SO(3))

For a unit quaternion $q$ define $\rho(q) : \R^3 \to \R^3$ by $\rho(q)v = qv\bar q$ , on pure quaternions $v$ . Then $\rho$ is a surjective homomorphism from the unit quaternions ( $\cong SU(2)$ ) onto $SO(3)$ , with kernel $\{\pm 1\}$ . Explicitly, if $u$ is a unit pure quaternion, then $\rho\big(\cos\tfrac\theta2 + \sin\tfrac\theta2\,u\big)$ is the rotation by $\theta$ about the axis $u$ .

Proof

$\rho(q) \in O(3)$ . The real part of $qv\bar q$ is $|q|^2$ times the real part of $v$ (a polynomial identity), so pure quaternions go to pure quaternions; $\rho(q)$ is linear, and $|qv\bar q| = |q||v||\bar q| = |v|$ , so it preserves lengths, hence (by polarisation) dot products.

Homomorphism. $\rho(pq)v = pqv\bar q\bar p = \rho(p)\rho(q)v$ , using $\overline{pq} = \bar q \bar p$ .

The axis and angle. For pure quaternions $u, w$ one has $uw = -u \cdot w + u \times w$ . Let $u$ be a unit pure quaternion, $c = \cos\frac\theta2$ , $s = \sin\frac\theta2$ , and $q = c + su$ . Then $q$ commutes with $u$ , so $\rho(q)u = u$ . If $w \perp u$ is a unit pure quaternion, put $w' = u \times w = uw$ , so that $wu = -w'$ and $w'u = w$ . Then

qw\bar q = (c + su)\,w\,(c - su) = c^2 w + 2cs\,w' - s^2 w = \cos\theta\, w + \sin\theta\, (u \times w),

which is the rotation by $\theta$ about $u$ . In particular $\det \rho(q) = 1$ , and since every unit quaternion has the form $c + su$ , $\rho$ lands in $SO(3)$ ; by Euler's theorem every element of $SO(3)$ is such a rotation, so $\rho$ is onto.

Kernel. $\rho(q) = I$ means $qv = vq$ for all pure $v$ ; commuting with $i$ , $j$ and $k$ forces $q$ to be real, so $q = \pm 1$ .

∎

So $SO(3) \cong S^3/\{\pm 1\}$ : a rotation is a pair of antipodal unit quaternions. As $\theta$ runs from $0$ to $2\pi$ , the rotations $\rho(q_\theta)$ make a closed loop in $SO(3)$ , but the quaternions $q_\theta = \cos\frac\theta2 + \sin\frac\theta2 u$ run from $1$ to $-1$ : half a great circle of $S^3$ . Only at $\theta = 4\pi$ does $q_\theta$ return to $1$ . This is the $360°$ versus $720°$ of the motivation, and it shows that $SO(3)$ has a loop that cannot be shrunk while its double has none: $\pi_1(SO(3)) \cong \Z/2$ (see the fundamental group).

ProofH

The exponential map and one-parameter subgroups

Fix any submultiplicative matrix norm ( $\|XY\| \le \|X\|\|Y\|$ ), for instance the operator norm. Then $\|X^k/k!\| \le \|X\|^k/k!$ , so the series for $\exp X$ converges absolutely for every $X$ .

Theorem— Theorem 7 (properties of exp)

For square matrices $X$ , $Y$ and invertible $P$ :

(a) if $XY = YX$ then $\exp(X + Y) = \exp X \exp Y$ ; (b) $\exp X$ is invertible with inverse $\exp(-X)$ ; (c) $\exp(PXP^{-1}) = P\exp(X)P^{-1}$ and $(\exp X)^{\mathsf T} = \exp(X^{\mathsf T})$ ; (d) $t \mapsto \exp(tX)$ is a one-parameter subgroup with $\frac{d}{dt}\exp(tX) = X\exp(tX)$ , so its tangent vector at $t = 0$ is $X$ ; (e) $\det(\exp X) = e^{\operatorname{tr} X}$ .

Proof

(a) For commuting $X, Y$ the binomial theorem holds, and multiplying absolutely convergent series term by term,

\exp X \exp Y = \sum_{m} \sum_{j + l = m} \frac{X^j Y^l}{j!\,l!} = \sum_m \frac{(X + Y)^m}{m!} = \exp(X + Y).

(b) $X$ and $-X$ commute, so $\exp X \exp(-X) = \exp 0 = I$ . (c) $(PXP^{-1})^k = PX^kP^{-1}$ and $(X^{\mathsf T})^k = (X^k)^{\mathsf T}$ , term by term. (d) $sX$ and $tX$ commute, so (a) gives the homomorphism property; differentiating the power series term by term (allowed for power series) gives $\sum_{k \ge 1} t^{k-1}X^k/(k-1)! = X\exp(tX)$ . (e) Over $\C$ write $X = PTP^{-1}$ with $T$ upper triangular (Schur, or Jordan form). Powers of $T$ are upper triangular with diagonal entries $t_{ii}^k$ , so $\exp T$ is upper triangular with diagonal $e^{t_{ii}}$ , and $\det \exp X = \det \exp T = \prod e^{t_{ii}} = e^{\operatorname{tr} T} = e^{\operatorname{tr} X}$ .

∎

Theorem— Theorem 8 (every one-parameter subgroup is exponential)

If $\varphi : \R \to GL_n$ is a continuous homomorphism, then there is a unique matrix $X$ with $\varphi(t) = \exp(tX)$ for all $t$ .

Proof

First we show $\varphi$ is differentiable. Since $\varphi(0) = I$ and $\varphi$ is continuous, $B = \frac1\varepsilon\int_0^\varepsilon \varphi(s)\,ds$ is close to $I$ , hence invertible, for small $\varepsilon > 0$ . Now $\varphi(t)\,\varepsilon B = \int_0^\varepsilon \varphi(t + s)\,ds = \int_t^{t + \varepsilon}\varphi(s)\,ds$ , which is differentiable in $t$ by the fundamental theorem of calculus; multiplying by $(\varepsilon B)^{-1}$ shows $\varphi$ is differentiable. Differentiating $\varphi(t + s) = \varphi(t)\varphi(s)$ in $s$ at $s = 0$ gives $\varphi'(t) = \varphi(t)X$ with $X = \varphi'(0)$ . Then $\psi(t) = \varphi(t)\exp(-tX)$ has $\psi' = \varphi X\exp(-tX) - \varphi X \exp(-tX) = 0$ (as $X$ commutes with $\exp(-tX)$ ), so $\psi \equiv \psi(0) = I$ . Uniqueness: $X = \varphi'(0)$ .

∎

The flow of x′ = Jx is exp(tJ): rotations

Arrows show where the system moves next. Click to start a trajectory, or choose an example.

Examples

x′ = −y, y′ = x

darker, longer arrows = faster forward in time backward in timeequilibrium (hollow = stable or centre)

Click the plane to start a trajectory, or type a start point below.

Start pointx₀y₀

Trajectories: 3. They are numerical approximations (fourth-order Runge–Kutta), not exact formulas. The latest trajectory closes up: it is a periodic orbit.

Linear systemA = [0−110]

T = 0, D = 1, T² − 4D = −4; λ = 0 ± 1i

Equilibrium at (0, 0): centre: closed orbits around it

Trace T and determinant D: below the T-axis, saddles; above the dashed parabola D = T²/4, spirals; between them, nodes; on the upper D-axis, centres. Left of the D-axis is stable, right is unstable.

ProofH

Lie algebras: the tangent space at the identity

Theorem— Theorem 9 (classical Lie algebras)

(a) $\mathfrak{so}(n) = \mathfrak o(n) = \{ X : X^{\mathsf T} = -X \}$ , of dimension $n(n-1)/2$ . (b) $\mathfrak{sl}_n = \{ X : \operatorname{tr} X = 0 \}$ . (c) $\mathfrak u(n) = \{ X : X^* = -X \}$ and $\mathfrak{su}(n) = \{ X \in \mathfrak u(n) : \operatorname{tr} X = 0 \}$ . (d) In each case $\mathfrak g$ is exactly the set of velocity vectors $\gamma'(0)$ of differentiable curves $\gamma$ in $G$ with $\gamma(0) = I$ .

Proof

(a) If $X^{\mathsf T} = -X$ then $(\exp tX)^{\mathsf T}\exp tX = \exp(-tX)\exp(tX) = I$ by Theorem 7(b, c), and $\det \exp(tX) = e^{t \operatorname{tr} X} = 1$ since a skew matrix has zero diagonal. Conversely, let $\gamma$ be a differentiable curve in $O(n)$ with $\gamma(0) = I$ and $\gamma'(0) = X$ — for example $\gamma(t) = \exp(tX)$ when $X \in \mathfrak o(n)$ , by Theorem 7(d). Differentiating $\gamma(t)^{\mathsf T}\gamma(t) = I$ at $t = 0$ by the product rule gives $X^{\mathsf T} + X = 0$ . A skew matrix is determined by its $n(n-1)/2$ entries above the diagonal. (b) By Theorem 7(e), $\det\exp(tX) = 1$ for all $t$ if and only if $t\operatorname{tr} X \in 2\pi i\Z$ for all $t$ , i.e. $\operatorname{tr} X = 0$ . (c) is (a) with $X^*$ in place of $X^{\mathsf T}$ , combined with (b). (d) is the curve argument above for $O(n)$ and $U(n)$ ; for $SL_n$ , differentiate $\det\gamma(t) = 1$ using $\frac{d}{dt}\det\gamma(t)\big|_{t=0} = \operatorname{tr}\gamma'(0)$ when $\gamma(0) = I$ .

∎

Two facts complete the picture; we state them with references. For any matrix group, $\mathfrak g$ is a real vector space closed under the bracket. Closure under addition follows from the Lie product formula $\exp(X + Y) = \lim_{m \to \infty}\big(\exp\frac Xm \exp\frac Ym\big)^m$ ; closure under the bracket follows by differentiating the curve $t \mapsto \exp(tX)\,Y\exp(-tX)$ in $\mathfrak g$ at $t = 0$ , which gives $XY - YX$ (Hall, Lie Groups, Lie Algebras, and Representations, chapters 2–3). And the bracket measures non-commutativity to second order: $\exp(tX)\exp(tY) = \exp\big(t(X + Y) + \tfrac{t^2}{2}[X, Y] + O(t^3)\big)$ , the start of the Baker–Campbell–Hausdorff formula, which the first experiment checks numerically.

Example— Infinitesimal rotations and the cross product

A skew $3 \times 3$ matrix is $X_v = \begin{pmatrix} 0 & -v_3 & v_2 \\ v_3 & 0 & -v_1 \\ -v_2 & v_1 & 0 \end{pmatrix}$ for a unique $v \in \R^3$ , and $X_v w = v \times w$ . One checks $[X_u, X_v] = X_{u \times v}$ : the Lie algebra $\mathfrak{so}(3)$ is $\R^3$ with the cross product. $\exp(\theta X_u)$ , for a unit vector $u$ , is the rotation by $\theta$ about $u$ (Rodrigues' formula $\exp(\theta X_u) = I + \sin\theta\,X_u + (1 - \cos\theta)X_u^2$ , from $X_u^3 = -X_u$ ), so by Euler's theorem $\exp : \mathfrak{so}(3) \to SO(3)$ is surjective. A spinning top's angular velocity is a vector in $\mathfrak{so}(3)$ . Similarly $\mathfrak{su}(2)$ has basis $\tfrac{i}{2}\sigma_1, \tfrac i2\sigma_2, \tfrac i2\sigma_3$ (Pauli matrices) with the same brackets up to sign, and the derivative of $\rho$ is an isomorphism $\mathfrak{su}(2) \cong \mathfrak{so}(3)$ — the double cover is a local isomorphism.

Formal mathematicsF

Matrix groups in Lean and Mathlib

Mathlib defines Matrix.orthogonalGroup n R, Matrix.specialOrthogonalGroup, Matrix.unitaryGroup and Matrix.specialUnitaryGroup as submonoids of matrices carrying a group structure, the exponential NormedSpace.exp, and the quaternions ℍ[ℝ]. Proposition 1, the determinant, and the commutativity of $SO(2)$ :

orthogonal_preserves_dotLean 4Proved here from basic factsFormally verified

/-- **Orthogonal matrices preserve the dot product**, hence lengths and angles: if `AᵀA = I`
then `(A v) · (A w) = v · (Aᵀ A w) = v · w`. -/
theorem orthogonal_preserves_dot {n : Type*} [Fintype n] [DecidableEq n] {A : Matrix n n ℝ}
    (hA : Aᵀ * A = 1) (v w : n → ℝ) : (A *ᵥ v) ⬝ᵥ (A *ᵥ w) = v ⬝ᵥ w := by
  rw [dotProduct_mulVec, ← mulVec_transpose, mulVec_mulVec, hA, one_mulVec]