Matrix Groups and Classical Groups

Introduction

A transformation group is determined by the structure it preserves, and the levels run from the bijections of a set to the automorphisms of a linear structure and then to the automorphisms of a linear structure together with a form. The automorphisms of a linear structure are the general linear groups of The General Linear Group, and the groups at the third level — orthogonal, unitary and symplectic — are the classical groups. This article develops them in their matrix form, which is the form in which they are computed with, and establishes the isomorphisms among the small ones that make the beginning of the theory recognisable.

The general theory of the groups $GL(V)$ and $SL(V)$, of transvections and elementary matrices, and of the way row and column operations generate them belongs to the linear companion articles; a bilinear form on a free module $V$ over a commutative ring $R$ is a map $\beta:V\times V\to R$ linear in each variable, and its isometry group is

$$ O(V,\beta) = \{\,g \in GL(V) : \beta(gx,gy) = \beta(x,y) \ \text{for all } x,y \in V\,\}, $$

the Hermitian case reading the same way with the form sesquilinear in the second variable and the alternating case with $\beta$ alternating; the symmetric, Hermitian and alternating forms give the three classical families. That is all that is taken from outside this article. The general construction of $O(V, g)$, $U(V, h)$ and $Sp(V, \omega)$ from a form, including the four families $A$, $B$, $C$, $D$ and the quaternionic unitary group, is not covered here. What is assumed here is the definition of a form and of its isometry group, and what is supplied here is: the explicit matrix conditions, the determinant constraints, the compactness and the centres, the realisation of $SU(2)$ as the unit quaternions, and the double covers $SU(2) \to SO(3)$ and $SU(2) \times SU(2) \to SO(4)$.

Throughout, $R$ is a commutative ring with identity $1 \neq 0$ and $K$ a field; $\mathbb{K}$ is $\mathbb{R}$ or $\mathbb{C}$ when only those are meant. The quaternion algebra is $\mathbb{H}$ with basis $e_0 = 1, e_1, e_2, e_3$, $e_k^2 = -e_0$, the norm is $N(q) = q q^{\natural}$ and $\mathrm{Sc}$, $\mathrm{Vect}$ are the scalar and vector parts. No physics is invoked: the rotation groups appear here as groups of linear transformations preserving a form, never as symmetry groups of a physical system.

Matrices, Determinants and the Two Linear Groups

Definition. For a commutative ring $R$ the general linear group $GL_n(R)$ is the group of invertible $n \times n$ matrices over $R$, that is, the group of units of the ring $M_n(R)$. Its special linear group is $SL_n(R) = \{A \in GL_n(R) : \det A = 1\}$.

Over a field the determinant classifies invertibility: $GL_n(K) = \{A \in M_n(K) : \det A \neq 0\}$, and $SL_n(K)$ is the kernel of the surjective homomorphism

$$ \det : GL_n(K) \longrightarrow K^\times. $$

Over a general ring the determinant still exists and is multiplicative, but $\det A$ need not be a unit, so $GL_n(R)$ is the set of matrices with $\det A \in R^\times$; the group is not determined by the determinant alone.

Remark (the determinant rules used below). The determinant is multiplicative, $\det(AB) = \det A \det B$, invariant under transposition, $\det(A^T) = \det A$, and equivariant with respect to conjugation of the entries, $\det(\overline A) = \overline{\det A}$. These are the standard properties of the alternating multilinear function of the columns, and they are what converts a form-preserving condition into a condition on the determinant.

Remark. For $n \geq 2$ and $K$ a field, $SL_n(K)$ is generated by transvections and is the commutator subgroup of $GL_n(K)$ away from the exceptional pair $(n, K) = (2, \mathbb{F}_2)$, so that $GL_n(K)^{ab} \cong K^\times$ via the determinant. This is the content of the companion article Transvections and Elementary Transformations, written in parallel.

The group $GL(V)$ of The General Linear Group acts on the set of subspaces of each dimension. The classical groups below are the subgroups of $GL(V)$ cut out by one further condition, the preservation of a form, and the whole of their structure is a consequence of that condition.

The Orthogonal Group

The Defining Conditions

Let $g$ be the standard symmetric bilinear form on $K^n$,

$$ g(x, y) = \sum_{i=1}^{n} x_i y_i . $$

Definition. The orthogonal group of $g$ is

$$ O(n, K) = \{A \in GL_n(K) : g(Ax, Ay) = g(x, y) \text{ for all } x, y \in K^n\}, $$

and $O(n) = O(n, \mathbb{R})$.

Proposition (matrix form). $O(n, K) = \{A \in GL_n(K) : A^T A = I\}$, and equivalently the columns of $A$, and also the rows, form an orthonormal basis of $K^n$ for $g$.

Proof. Computing in the standard basis, $g(Ae_i, Ae_j)$ is the $(i, j)$ entry of $A^T A$. The form is non-degenerate, so a map preserving it is determined by the Gram matrix of the images of a basis; hence $g(Ax, Ay) = g(x, y)$ for all $x, y$ if and only if $A^T A = I$. The entries of $A^T A$ are exactly the pairwise inner products of the columns, so $A^T A = I$ says the columns are orthonormal, and orthonormal columns form a basis because they are linearly independent over a field. The statement for rows is the same applied to $A^T$, which is orthogonal when $A$ is.

Proposition (compactness). $O(n)$ is a compact subgroup of $GL_n(\mathbb{R})$, and it is the isometry group of the unit sphere $S^{n-1} \subseteq \mathbb{R}^n$.

Proof. The condition $A^T A = I$ says that the columns are unit vectors with pairwise zero inner product, so the $n^2$ entries of $A$ satisfy $\sum_{i} A_{ij}^2 = 1$ for each $j$; hence $O(n)$ is bounded in $M_n(\mathbb{R}) \cong \mathbb{R}^{n^2}$. It is closed because the entries of $A^T A$ are continuous functions of the entries of $A$ and $O(n)$ is their preimage at the identity matrix. A closed bounded subset of $\mathbb{R}^{n^2}$ is compact. For the second statement, orthogonal maps preserve the norm and hence the sphere; conversely an isometry $f$ of the sphere extends to $\mathbb{R}^n$ by $x \mapsto |x|\, f(x/|x|)$ for $x \neq 0$ and $0 \mapsto 0$, which preserves norms and hence, by polarisation, the inner product, so the extension is a linear isometry of $\mathbb{R}^n$.

The Determinant and the Special Orthogonal Group

Proposition. If $A \in O(n, K)$ then $\det A = \pm 1$.

Proof. $1 = \det I = \det(A^T A) = (\det A)^2$ in $K$, and the solutions of $x^2 = 1$ in a field are $\pm 1$.

Definition. The special orthogonal group is $SO(n, K) = O(n, K) \cap SL_n(K) = \{A \in O(n, K) : \det A = 1\}$.

Theorem. $SO(n, K)$ is a normal subgroup of $O(n, K)$ of index $2$ when $-1 \neq 1$ in $K$, being the kernel of $\det : O(n, K) \to \{\pm 1\}$, and this homomorphism is surjective because $\operatorname{diag}(-1, 1, \ldots, 1)$ is orthogonal of determinant $-1$. Over $\mathbb{R}$, $SO(n)$ is connected and $O(n)$ has exactly two connected components; $SO(n)$ is the identity component.

Proof. The kernel of a homomorphism is normal and the surjectivity is exhibited. For connectedness, every element of $SO(n)$ is a product of rotations of coordinate planes, and each rotation of a plane can be joined to the identity by a path inside that plane; this is the standard generation of $SO(n)$ by plane rotations, and it exhibits a path. A closed subgroup of index $2$ is both open and closed, so the components of $O(n)$ are $SO(n)$ and the other coset.

For $n = 2$ the group $SO(2)$ consists of the matrices $\begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix}$ and is isomorphic to the circle group $S^1 = U(1)$; for $n = 1$, $O(1) = \{\pm 1\}$ and $SO(1) = \{1\}$. Every element of $O(n)$ is a product of at most $n$ reflections of hyperplanes, a theorem of Cartan and Dieudonné, proved in the cited literature.

The Unitary Group

Hermitian Forms and the Defining Conditions

Let $\mathbb{K} = \mathbb{R}$ or $\mathbb{C}$ and let $h$ be the standard Hermitian form on $\mathbb{K}^n$,

$$ h(x, y) = \sum_{i=1}^{n} \overline{x_i} y_i , $$

which is $\mathbb{C}$-linear in the second argument, conjugate-linear in the first, and satisfies $h(y, x) = (h(x,y))^{\natural}$. Over $\mathbb{R}$ it is the standard symmetric form again; the new case is $\mathbb{K} = \mathbb{C}$.

Definition. The unitary group is $U(n) = \{A \in GL_n(\mathbb{C}) : h(Ax, Ay) = h(x, y) \text{ for all } x, y\}$, and the special unitary group is $SU(n) = U(n) \cap SL_n(\mathbb{C})$.

Proposition (matrix form). With $A^* = \overline{A}^{\,T}$ the conjugate transpose, $U(n) = \{A \in GL_n(\mathbb{C}) : A^* A = I\}$; equivalently the columns of $A$ are orthonormal for $h$. Moreover $\det A \in S^1$ for $A \in U(n)$, the determinant $\det : U(n) \to S^1$ is surjective, and $U(n)/SU(n) \cong S^1$.

Proof. The $(i,j)$ entry of $A^* A$ is $h(Ae_i, Ae_j)$, so the argument of the orthogonal case applies with conjugation inserted: the form is non-degenerate and the Gram matrix determines the map. From $A^* A = I$ we get $1 = \det(A^*A) = \overline{\det A} \det A = |\det A|^2$, so $\det A \in S^1$. The map $\operatorname{diag}(\lambda, 1, \ldots, 1)$ is unitary of determinant $\lambda$ for every $\lambda \in S^1$, so the determinant is onto $S^1$, and $SU(n)$ is its kernel.

For $n = 1$, $U(1) = S^1$ and $SU(1) = \{1\}$.

Determinant, Centre and Decomposition

Theorem. $U(n)$ is a compact connected subgroup of $GL_n(\mathbb{C})$, and $U(n) \cong (SU(n) \times U(1))/\mu_n$, where $\mu_n \subseteq U(1)$ is the group of $n$-th roots of unity embedded diagonally as scalar matrices.

Proof. The entries satisfy the closed bounded conditions of the orthogonal case with conjugation and absolute values, so $U(n)$ is compact; connectedness follows because every unitary matrix is diagonalised by a unitary change of basis and the eigenvalues can be joined to $1$ along the circle. For the decomposition, let $A \in U(n)$ and $\lambda = \det A \in S^1$; choose $\mu \in S^1$ with $\mu^n = \lambda$ and write $A = \mu \cdot (\mu^{-1} A)$; the second factor is unitary of determinant $\mu^{-n}\lambda = 1$, hence lies in $SU(n)$. The pair $(\mu^{-1} A, \mu)$ is determined by $A$ up to multiplication by $(\omega^{-1}, \omega)$ with $\omega \in \mu_n$, which gives the asserted quotient.

Theorem (centres). $Z(U(n)) = \{\lambda I : \lambda \in S^1\} \cong S^1$ and $Z(SU(n)) = \{\lambda I : \lambda^n = 1\} \cong \mathbb{Z}/n\mathbb{Z}$. In particular $Z(SU(2)) = \{\pm I\}$.

Proof. If $A$ commutes with every unitary matrix it commutes with every diagonal unitary matrix, so $A$ is diagonal: comparing the $(i,j)$ entries of $AD = DA$ for a diagonal $D$ with distinct eigenvalues gives $A_{ij}(D_{jj} - D_{ii}) = 0$, hence $A_{ij} = 0$ for $i \neq j$, and diagonal unitaries with distinct eigenvalues exist in abundance. The same argument applies inside $SU(n)$, using diagonal unitaries of determinant $1$ with distinct entries, whose existence requires only that $n \geq 2$. A diagonal $A$ that commutes with a permutation matrix is unchanged by the corresponding permutation of its entries; for $n \geq 3$ a $3$-cycle of determinant $1$ lies in $SU(n)$ and moves any triple, so all diagonal entries of $A$ are equal, while for $n = 2$ the matrix $\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \in SU(2)$ forces $\operatorname{diag}(\lambda_1, \lambda_2)$ to have $\lambda_1 = \lambda_2$ by direct comparison. Hence $A = \lambda I$ in both groups, the unitarity gives $|\lambda| = 1$, and the additional condition $\det A = 1$ gives $\lambda^n = 1$. Conversely these matrices are central, and the maps $\lambda \mapsto \lambda I$ identify the centres with $S^1$ and with the group of $n$-th roots of unity.

The quotient $PSU(n) = SU(n)/Z(SU(n))$ is the projective special unitary group, and in the low-dimensional cases below it is a rotation group.

The Quaternionic Unitary Group

Quaternionic Hermitian Forms

Let $\mathbb{H}$ be the quaternion algebra, with conjugation $q^{\natural}$, norm $N(q) = q q^{\natural} = |q|^2$, and real part $\mathrm{Sc}(q) = \tfrac{1}{2}(q + q^{\natural})$. On $\mathbb{H}^n$ define

$$ h(x, y) = \sum_{i=1}^{n} \overline{x_i} y_i \in \mathbb{H}, $$

which is $\mathbb{H}$-linear in the second argument and conjugate-linear in the first in the sense $h(x \lambda, y) = \bar\lambda h(x,y)$. It takes values in $\mathbb{H}$, which is not commutative, so it is not a Hermitian form over a field; the property that survives is $h(y,x) = (h(x,y))^{\natural}$ together with the reality of $h(x,x) = \sum_i N(x_i) \in \mathbb{R}_{\geq 0}$.

Definition. The compact symplectic group or quaternionic unitary group is

$$ Sp(n) = \{A \in GL_n(\mathbb{H}) : A^* A = I\}, \qquad A^* = \overline{A}^{\,T}, $$

the group of $\mathbb{H}$-linear maps of $\mathbb{H}^n$ preserving $h$.

Proposition. $Sp(n)$ is a compact subgroup of $GL_n(\mathbb{H}) \cong GL_{4n}(\mathbb{R})$, the columns of $A \in Sp(n)$ are orthonormal for $h$, and $Sp(1)$ is the group of unit quaternions, hence is homeomorphic to the sphere $S^3$.

Proof. The condition $A^*A = I$ says the columns are orthonormal for $h$; it is a closed condition in the $4n^2$ real coordinates and the diagonal entries $h(Ae_j, Ae_j) = 1$ bound the entries by $1$ in norm, so $Sp(n)$ is closed and bounded. For $n = 1$ the condition is $N(q) = 1$, and the unit sphere of $\mathbb{H} \cong \mathbb{R}^4$ is $S^3$.

Sp(n) Inside SU(2n)

Since $\mathbb{H}$ is a two-dimensional complex vector space, with $i \in \mathbb{C}$ acting as left multiplication by $e_1$, the identification $\mathbb{H}^n \cong \mathbb{C}^{2n}$ carries $\mathbb{H}$-linear maps to $\mathbb{C}$-linear maps. The quaternionic form $h$ has real part a Hermitian form on $\mathbb{C}^{2n}$, so there is an injective homomorphism $Sp(n) \to U(2n)$.

Theorem. Under this identification $Sp(n) \subseteq SU(2n)$; that is, the complex determinant of the image of every element of $Sp(n)$ is $1$.

Proof. The complex determinant is a continuous homomorphism $\det : Sp(n) \to S^1$, since the image lies in $U(2n)$. Its derivative at the identity is the trace of the complex $2n \times 2n$ matrix, which for the image of a quaternionic matrix $X$ is $2\,\mathrm{Sc}(\operatorname{tr}_{\mathbb{H}} X)$: the diagonal $2 \times 2$ block of $X_{ii}$ contributes the trace $2\,\mathrm{Sc}(X_{ii})$. The Lie algebra of $Sp(n)$ consists of the quaternionic matrices with $X^* = -X$, and for such a matrix $\operatorname{tr}_{\mathbb{H}} X$ is a pure quaternion, because $(\operatorname{tr} X)^* = \operatorname{tr}(X^*) = -\operatorname{tr} X$. Hence the trace vanishes on the Lie algebra and the determinant has derivative zero; since $Sp(n)$ is connected and is generated by exponentials of its Lie algebra, $\det \equiv 1$.

So the compact symplectic group is a subgroup of the special unitary group: $Sp(1) = SU(2)$, and $Sp(n) \subseteq SU(2n)$ for every $n$.

The Two Uses of the Symbol Sp

The symbol $Sp$ carries two standard meanings that must be distinguished. In this article $Sp(n)$ is the compact quaternionic unitary group of dimension $n(2n+1)$, a compact group acting on $\mathbb{H}^n \cong \mathbb{C}^{2n}$. In, $Sp(2n, K)$ denotes the isometry group of a non-degenerate alternating form on $K^{2n}$, an algebraic group whose $\mathbb{C}$-points are a complex Lie group of the same dimension $n(2n+1)$ but which is not compact. The two are related by complexification: the complexification of the compact group $Sp(n)$ is $Sp(2n, \mathbb{C})$. What differs is therefore the indexing: $Sp(n)$ carries the rank $n$ and acts on a quaternionic space of $n$ coordinates, while $Sp(2n, \mathbb{C})$ acts on a complex space of $2n$ coordinates.

SU(2) and the Unit Quaternions

The Representation of the Quaternion Algebra

Write a quaternion as $q = z + w j$ with $z = a + b i$ and $w = c + d i$, where $i = e_1$, $j = e_2$ and $i j = e_3$; this is the splitting $\mathbb{H} = \mathbb{C} \oplus \mathbb{C} j$ with $j^2 = -1$ and $j z = \bar z j$.

Theorem. The map

$$ \rho : \mathbb{H} \longrightarrow M_2(\mathbb{C}), \qquad \rho(z + wj) = \begin{pmatrix} z & w \\ -\bar w & \bar z \end{pmatrix}, $$

is an injective homomorphism of $\mathbb{R}$-algebras, and

$$ \det \rho(q) = z \bar z + w \bar w = N(q) . $$

Proof. Multiplicativity is the identity $(z + wj)(z' + w'j) = (zz' - w\bar w') + (zw' + w\bar z')j$, obtained from $jz' = \bar z' j$ and $j^2 = -1$, which is exactly the matrix product $\rho(z + wj)\rho(z' + w'j) = \rho((z + wj)(z' + w'j))$; the entries of the product matrix are $zz' - w\bar w'$, $zw' + w\bar z'$, $-\overline{zw' + w\bar z'}$ and $\overline{zz' - w\bar w'}$. So $\rho$ is a homomorphism, and it is injective because $q \neq 0$ has an inverse, namely $q^{\natural} / N(q)$. The determinant is $z\bar z - w(-\bar w) = |z|^2 + |w|^2 = N(q)$.

Example. With the Pauli matrices

$$ \sigma_x = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \quad \sigma_y = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}, \quad \sigma_z = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}, $$

the images of the quaternion units are $\rho(e_1) = i\sigma_z$, $\rho(e_2) = i\sigma_y$ and $\rho(e_3) = i\sigma_x$. These satisfy $\rho(e_k)^2 = -I$ and $\rho(e_1)\rho(e_2) = \rho(e_3)$, so they realise the quaternion relations, and the general element of $\mathbb{H}$ is represented as $aI + i(b \sigma_z + c \sigma_y + d \sigma_x)$.

The Group of Unit Quaternions

Theorem. The restriction of $\rho$ to the unit sphere $\mathbb{H}^1 = \{q \in \mathbb{H} : N(q) = 1\}$ is an isomorphism of topological groups

$$ \mathbb{H}^1 \longrightarrow SU(2), \qquad \mathbb{H}^1 = S^3 . $$

Proof. For $q \in \mathbb{H}^1$ the matrix $\rho(q)$ is unitary, since $\rho(q)^* \rho(q) = \rho(q^{\natural})\rho(q) = \rho(q^{\natural} q) = \rho(1) = I$, and it has determinant $N(q) = 1$, so the image lies in $SU(2)$. Conversely, if $A = \begin{pmatrix} z & w \\ -\bar w & \bar z \end{pmatrix}$ is unitary of determinant $1$ then $|z|^2 + |w|^2 = 1$, so $A = \rho(q)$ with $q = z + wj$ of norm $1$. The map is a homeomorphism because it is a continuous bijection between compact Hausdorff spaces.

Corollary. $SU(2) \cong Sp(1) \cong \mathbb{H}^1 \cong S^3$; consequently $SU(2)$ is connected and simply connected, and its centre is $\{\pm I\}$, corresponding to the quaternions $\pm 1$.

Thus the three families $A$, $B$, $C$ meet at the bottom: the compact symplectic group of rank one, the special unitary group in two variables and the sphere of unit quaternions are the same group.

The Rotation Groups and the Double Covers

Conjugation by a Unit Quaternion

Identify $\mathbb{R}^3$ with the space of pure quaternions $\mathrm{Vect}(\mathbb{H}) = \{x \in \mathbb{H} : \mathrm{Sc}(x) = 0\}$.

Theorem. For $u \in \mathbb{H}^1$ the map $\varphi_u(x) = u x u^{\natural}$ restricts to a rotation of the pure quaternions, $\varphi_u \in SO(3)$, and

$$ \varphi : \mathbb{H}^1 \longrightarrow SO(3), \qquad u \longmapsto \varphi_u, $$

is a continuous surjective homomorphism with kernel $\{\pm 1\}$. Hence $SO(3) \cong \mathbb{H}^1/\{\pm 1\} \cong SU(2)/\{\pm I\} = PSU(2)$.

Proof. If $x$ is pure and $u$ is a unit then $u x u^{\natural}$ is pure, because $\overline{u x u^{\natural}} = u x^{\natural} u^{\natural} = -u x u^{\natural}$, conjugation reversing the scalar sign; the norms agree, $N(ux u^{\natural}) = N(u)N(x)N(u) = N(x)$, so $\varphi_u$ preserves the standard form of $\mathbb{R}^3$; and $\varphi$ is a homomorphism because the map is the conjugation action of the group of units on the algebra, which is multiplicative in $u$. The image lies in $SO(3)$ rather than merely in $O(3)$: the map $u \mapsto \varphi_u$ is continuous and $\mathbb{H}^1 = S^3$ is connected, so the image lies in the identity component. If $u$ lies in the kernel then $ux = xu$ for all pure $x$, in particular for $x = e_1, e_2, e_3$, so $u$ commutes with $e_1, e_2, e_3$ and hence lies in the centre of $\mathbb{H}$, which is $\mathbb{R}$; with $N(u) = 1$ this gives $u = \pm 1$. Thus the kernel has order two and the image is a compact connected subgroup of dimension $3$; as $SO(3)$ is connected of dimension $3$, the image is open and closed and therefore all of $SO(3)$.

Corollary. The homomorphism $\varphi$ is a two-to-one covering, the two preimages of a rotation being antipodal unit quaternions $u$ and $-u$; consequently $SO(3)$ is homeomorphic to the quotient of $S^3$ by the antipodal map, that is, to the real projective space $\mathbb{RP}^3$, and $\pi_1(SO(3)) \cong \mathbb{Z}/2\mathbb{Z}$.

The Rotation Angle

Proposition (Rodrigues form). If $u = \cos(\theta/2) + \sin(\theta/2) n$ with $n$ a pure unit quaternion, then $\varphi_u$ is the rotation about the axis $n$ through the angle $\theta$.

Proof. Conjugation by $u$ fixes $u$ itself and hence fixes $n = u - \mathrm{Sc}(u)$ up to normalisation, so the axis is $n$. For the angle, the trace of the matrix of $\varphi_u$ on the pure part is computed from the multiplication table: writing $a = \mathrm{Sc}(u)$ and $v = \mathrm{Vect}(u)$ with $N(v) = 1 - a^2$, the diagonal entries of the matrix are $a^2 + v_1^2 - v_2^2 - v_3^2$, $a^2 - v_1^2 + v_2^2 - v_3^2$ and $a^2 - v_1^2 - v_2^2 + v_3^2$, whose sum is $3a^2 - N(v) = 4a^2 - 1$. For a rotation through $\theta$ the trace is $1 + 2\cos \theta$, so $4a^2 - 1 = 1 + 2\cos \theta$, that is, $2a^2 - 1 = \cos \theta$, which is $\cos \theta = 2\cos^2(\theta/2) - 1$ with $a = \cos(\theta/2)$.

The proposition explains the two-to-one cover: the quaternions $u$ and $-u$ give the same rotation, and a rotation through $2\pi$ about any axis corresponds to $\theta = 2\pi$, hence to $u = \cos \pi + \sin \pi \, n = -1$, which is not the identity of $\mathbb{H}^1$; the path $t \mapsto \cos(\pi t) + \sin(\pi t) n$ in $S^3$ joins $1$ to $-1$ and closes only after two full turns.

Example. Take $u = \cos(\pi/4) + \sin(\pi/4)e_3 = (1 + e_3)/\sqrt 2$, so the expected rotation is through $\pi/2$ about the $e_3$-axis. Using $e_3 e_1 = e_2$, $e_1 e_3 = -e_2$, $e_2 e_3 = e_1$ and $e_3 e_2 = -e_1$, one computes $e_3 e_1 e_3 = e_2 e_3 = e_1$, hence

$$ u \, e_1 \, u^{\natural} = \tfrac{1}{2}(1 + e_3) e_1 (1 - e_3) = \tfrac{1}{2}(e_1 + e_2)(1 - e_3) = \tfrac{1}{2}(e_1 + e_2 + e_2 - e_1) = e_2 , $$

and similarly $u e_2 u^{\natural} = -e_1$ and $u e_3 u^{\natural} = e_3$. So $\varphi_u$ is the rotation of the $(e_1, e_2)$-plane through $\pi/2$, as predicted. Squaring, $u^2 = \tfrac{1}{2}(1 + e_3)^2 = e_3$, and indeed $\varphi_{e_3}$ is the rotation through $\pi$ about $e_3$, with trace $4 \cdot 0^2 - 1 = -1 = 1 + 2\cos \pi$. The example also shows that the quaternion product computes the composite of rotations: $\varphi_{u v} = \varphi_u \circ \varphi_v$ because $\varphi$ is a homomorphism.

SO(4) from Two Copies of SU(2)

Now let $\mathbb{H} \cong \mathbb{R}^4$ and consider the two-sided action.

Theorem. For $u, v \in \mathbb{H}^1$ the map $\psi_{(u,v)}(q) = u q v^{\natural}$ is an orthogonal transformation of $\mathbb{R}^4$ of determinant $1$, and

$$ \psi : \mathbb{H}^1 \times \mathbb{H}^1 \longrightarrow SO(4), \qquad (u,v) \longmapsto \psi_{(u,v)}, $$

is a surjective homomorphism with kernel $\{\pm(1,1)\}$ of order $2$. Hence

$$ SO(4) \cong (\mathbb{H}^1 \times \mathbb{H}^1)/\{\pm(1,1)\} \cong (SU(2) \times SU(2))/\{\pm(I,I)\}. $$

Proof. The map $\psi_{(u,v)}$ preserves the norm because $N(uqv^{\natural}) = N(u)N(q)N(v) = N(q)$, so it is orthogonal; it is the composite of left multiplication by $u$ and right multiplication by $ v^{\natural}$, and each of these has real determinant $1$: the maps $u \mapsto L_u$ and $v \mapsto R_v$ are continuous from the connected group $\mathbb{H}^1 = S^3$ into the orthogonal group $O(4)$ of $\mathbb{H} \cong \mathbb{R}^4$, so their images lie in the identity component $SO(4)$. The map is a homomorphism because each factor acts multiplicatively. Its kernel consists of the pairs with $uq = qv$ for all $q$; taking $q = 1$ gives $u = v$, and then $uq = qu$ for all $q$, so $u$ is central and, being a unit of norm $1$, is $\pm 1$. Thus the kernel has order $2$ and the image is a connected compact subgroup of dimension $6$, equal to $\dim SO(4)$; a connected subgroup of a connected group of the same dimension is open and closed, so the image is all of $SO(4)$.

Corollary. $SO(4)$ is not simple: the image of each factor is a normal subgroup isomorphic to $S^3/\{\pm 1\} = SU(2)/\{\pm I\} = SO(3)$, the two images intersect in $\{\pm I\}$ and together generate $SO(4)$, and the quotient by either of them is the other factor, $S^3 = SU(2)$; the quotient by the centre $\{\pm I\}$ is $SO(3) \times SO(3)$.

Example (isoclinic rotations). If $v = 1$ the map is left multiplication $q \mapsto uq$, and if $u = 1$ it is right multiplication; these two families of rotations commute with each other and each is a three-dimensional sphere inside $SO(4)$.

The Exceptional Isomorphisms

The double covers above are the first of a family of low-dimensional coincidences of the classical groups.

Isomorphism	Dimensions	Reason
$SO(2) \cong U(1) = S^1$	$1$	the rotation matrices are $e^{i\theta}$
$SU(2) \cong Sp(1) \cong \mathbb{H}^1 \cong S^3$	$3$	the representation $\rho$
$SO(3) \cong SU(2)/\{\pm I\} = PSU(2)$	$3$	conjugation by unit quaternions
$SO(4) \cong (SU(2) \times SU(2))/\{\pm(I,I)\}$	$6$	the two-sided action
$SO(5) \cong Sp(2)/\{\pm I\}$	$10$	the action of $Sp(2)$ on the trace-zero quaternionic Hermitian $2 \times 2$ matrices
$SO(6) \cong SU(4)/\{\pm I\}$	$15$	the action of $SU(4)$ on $\Lambda^2 \mathbb{C}^4 \cong \mathbb{C}^6$

The last two rows are the low-dimensional coincidences $B_2 \cong C_2$ and $A_3 \cong D_3$ of root systems, and the coincidence $SU(2) \cong Sp(1)$ of the rows above, together with the quotient relation $SO(3) \cong SU(2)/\{\pm I\}$ between them, reflects $A_1 \cong B_1 \cong C_1$ at the level of Lie algebras; all of them are cited here as standard, since their proof belongs with the classification of the simple Lie algebras. The dimension counts are $\dim SU(2) = 3$, $\dim SO(3) = 3$, $\dim SO(4) = 6$, $\dim Sp(2) = 10 = \dim SO(5)$, $\dim SU(4) = 15 = \dim SO(6)$.

The Classical Groups in One Table

The definitions are collected here, with the invariants used throughout. In the table $n \geq 3$ unless a smaller case is named explicitly; the Lie algebras, that is, the tangent spaces at the identity, are recorded in the last column, and their dimensions give the dimensions of the groups.

Group	Definition	Dimension	Tangent space at $I$
$O(n)$	$A^T A = I$, real	$n(n-1)/2$	skew-symmetric matrices
$SO(n)$	$A^T A = I$, $\det A = 1$	$n(n-1)/2$	skew-symmetric matrices
$U(n)$	$A^* A = I$, complex	$n^2$	skew-Hermitian matrices
$SU(n)$	$A^* A = I$, $\det A = 1$	$n^2 - 1$	skew-Hermitian, trace zero
$Sp(n)$	$A^* A = I$, quaternionic	$n(2n+1)$	quaternionic skew-Hermitian

Group	Compact	Connected	Centre	Fundamental group
$O(n)$	yes	no, two components	$\{\pm I\}$	$\mathbb{Z}/2\mathbb{Z}$ for $n \geq 3$
$SO(n)$	yes	yes	$\{\pm I\}$ for even $n \geq 4$, trivial for odd $n \geq 3$	$\mathbb{Z}/2\mathbb{Z}$ for $n \geq 3$
$U(n)$	yes	yes	$U(1)$	$\mathbb{Z}$
$SU(n)$	yes	yes	$\mathbb{Z}/n\mathbb{Z}$	trivial
$Sp(n)$	yes	yes	$\{\pm I\} \cong \mathbb{Z}/2\mathbb{Z}$	trivial

Remark. The determinant constraints read: $\det A = \pm 1$ for $A \in O(n)$; $|\det A| = 1$ for $A \in U(n)$, with $\det$ surjecting onto the circle; and $\det A = 1$ for $A \in Sp(n)$ in the sense of the complex determinant of its image in $SU(2n)$, the quaternionic determinant itself requiring the Dieudonné construction. Indeed the Dieudonné determinant of a quaternionic matrix takes values in the abelianization $\mathbb{H}^\times/[\mathbb{H}^\times, \mathbb{H}^\times] \cong \mathbb{R}_{>0}$, since the polar decomposition $\mathbb{H}^\times \cong \mathbb{R}_{>0} \times S^3$ together with the perfectness of $S^3 = SU(2)$ identifies the commutator subgroup with the unit sphere; on $Sp(n)$ its absolute value is $1$, because an isometry of $\mathbb{H}^n$ induces an isometry of the top exterior power.

Remark. All five families are closed subgroups of a general linear group over $\mathbb{R}$, $\mathbb{C}$ or $\mathbb{H}$, hence topological groups, and the five compact ones among them ($O$, $SO$, $U$, $SU$, $Sp$) are the basic examples of compact Lie groups. Their topological properties used here — compactness, connectedness, the covering maps and the fundamental groups — are stated as standard; the topological frame is from Topological Groups, written in parallel.

Summary

The classical groups are the transformation groups of a vector space equipped with a form, and their matrix forms are read off from the preservation of the form. The orthogonal group $O(n)$ is $\{A : A^TA = I\}$, compact, with $\det A = \pm 1$, and its kernel-of-determinant $SO(n)$ is the connected component of the identity, isomorphic to $S^1$ when $n = 2$. The unitary group $U(n) = \{A : A^*A = I\}$ is compact and connected with determinant of modulus one, and $U(n) \cong (SU(n) \times U(1))/\mu_n$, while the centres are the scalar matrices, so $Z(SU(n)) \cong \mathbb{Z}/n\mathbb{Z}$. The compact symplectic group $Sp(n)$ is the quaternionic unitary group, of dimension $n(2n+1)$, contained in $SU(2n)$; the two uses of the symbol $Sp$ in the literature are distinguished by rank versus the dimension of the space acted on.

The representations of the quaternion algebra make the lowest cases explicit. The map $\rho(z + wj) = \begin{pmatrix} z & w \\ -\bar w & \bar z \end{pmatrix}$ is an algebra isomorphism of $\mathbb{H}$ with a subalgebra of $M_2(\mathbb{C})$ and has determinant $N(q)$, so the unit quaternions are exactly $SU(2)$, a group isomorphic to $S^3$ and to $Sp(1)$. Conjugation by a unit quaternion is a rotation of the pure quaternions with trace $4a^2 - 1$, hence a rotation through the angle determined by $a = \cos(\theta/2)$ and an axis given by the vector part: the assignment $u \mapsto (x \mapsto ux u^{\natural})$ is a surjective homomorphism $\mathbb{H}^1 \to SO(3)$ with kernel $\{\pm 1\}$, exhibiting $SO(3) \cong \mathbb{RP}^3 = SU(2)/\{\pm I\}$ and the two-to-one cover. The two-sided action $(u,v) \mapsto (q \mapsto uqv^{\natural})$ gives the surjection $\mathbb{H}^1 \times \mathbb{H}^1 \to SO(4)$ with kernel $\{\pm(1,1)\}$, so $SO(4) \cong (SU(2) \times SU(2))/\{\pm(I,I)\}$ and $SO(4)$ is not simple. The remaining low-dimensional coincidences, $SO(5) \cong Sp(2)/\{\pm I\}$ and $SO(6) \cong SU(4)/\{\pm I\}$, follow the same pattern at the level of root systems and are cited as standard.

Summary of Notation

Symbol	Meaning
$M_n(R)$, $GL_n(R)$, $SL_n(R)$	Matrices, invertible matrices, determinant-one matrices over $R$
$\det$, $A^T$, $\overline A$, $A^* = \overline A^{\,T}$	Determinant, transpose, entrywise conjugate, conjugate transpose
$g(x, y) = \sum_i x_i y_i$	Standard symmetric bilinear form on $K^n$
$O(n, K)$, $O(n)$	Orthogonal group of $g$; $O(n) = O(n, \mathbb{R})$
$SO(n, K)$	Special orthogonal group, kernel of the determinant on $O(n,K)$
$h(x, y) = \sum_i \overline{x_i} y_i$	Standard Hermitian form
$U(n)$, $SU(n)$	Unitary and special unitary groups
$PSU(n)$	$SU(n)/Z(SU(n))$
$Sp(n)$	Compact symplectic (quaternionic unitary) group, $A^*A = I$ over $\mathbb{H}$
$Sp(2n, K)$	Symplectic group of an alternating form, as in the companion article
$\mathbb{H}$, $N(q) = q q^{\natural}$, $\mathrm{Sc}$, $\mathrm{Vect}$	Quaternions, norm, scalar and vector parts
$\mathbb{H}^1 = S^3$	Unit quaternions; also written $Sp(1)$
$\rho(z + wj) = \begin{pmatrix} z & w \\ -\bar w & \bar z \end{pmatrix}$	Representation of $\mathbb{H}$ in $M_2(\mathbb{C})$, determinant $N(q)$
$\sigma_x, \sigma_y, \sigma_z$	Pauli matrices; $\rho(e_1) = i\sigma_z$, $\rho(e_2) = i\sigma_y$, $\rho(e_3) = i\sigma_x$
$\varphi_u(x) = u x u^{\natural}$	Rotation by a unit quaternion; kernel $\{\pm 1\}$ of $\mathbb{H}^1 \to SO(3)$
$\psi_{(u,v)}(q) = u q v^{\natural}$	The map $\mathbb{H}^1 \times \mathbb{H}^1 \to SO(4)$, kernel $\{\pm(1,1)\}$
$S^{n-1}$, $\mathbb{RP}^3$	Unit sphere; $SO(3) \cong \mathbb{RP}^3$
$Z(G)$	Centre of a group
$\mathbb{Z}/n\mathbb{Z}$, $\mu_n$	Cyclic group of order $n$; $n$-th roots of unity
Dieudonné determinant	Determinant of a matrix over a division ring, valued in $\mathbb{H}^\times/[\mathbb{H}^\times, \mathbb{H}^\times]$