The Orthogonal Lie Algebra

Introduction

The orthogonal group is a smooth group when the base is $\mathbb{R}$ or $\mathbb{C}$, and its infinitesimal version is the orthogonal Lie algebra: the linear maps that are skew with respect to the polar form. This article develops that Lie algebra, its identification with the bivectors $\Lambda^2 V$, the exponential map that returns from the algebra to the rotation group, its structure and rank, and the two low-dimensional isomorphisms that connect it to the vector product and to $\mathrm{SL}(2, \mathbb{C})$.

The base is a commutative ring $R$ in which $2$ is invertible for the algebraic part, and $\mathbb{R}$ or $\mathbb{C}$ for the parts that use the exponential. The orthogonal group, reflections and the special orthogonal group are those of Isometries and Orthogonal Transformations and The Rotation Group and Orientation. The exterior power $\Lambda^2 V$ and its universal property are taken from The Exterior Algebra and Exterior Powers; the general theory of Lie algebras, of the commutator bracket, of the exponential map and of Cartan subalgebras is assumed from Lie Algebras: A General Introduction and Lie Groups, and is cited rather than rebuilt. The matrix groups belong to category 04; here the Lie algebra is the Lie-theoretic shadow of the isometry group, and the corresponding matrix groups are treated in The Unitary and Symplectic Groups.

The Lie Algebra of Skew Transformations

Definition

Let $V$ be a finite-dimensional space over a field $F$ of characteristic not $2$, with non-degenerate quadratic form $q$ and polar form $B$. An $F$-linear map $A : V \to V$ is skew with respect to $B$ if

$$ B(Au, v) + B(u, Av) = 0 \qquad (u, v \in V). $$

The set of skew maps is written $\mathrm{SO}(V, q)$, and its elements are the infinitesimal isometries of $q$.

Proposition. Let $G$ be the Gram matrix of $B$ in a basis. Then $A$ is skew if and only if its matrix $M$ in that basis satisfies

$$ M^T G + G M = 0. $$

In particular, if the basis is orthonormal, a skew matrix is exactly an antisymmetric matrix $M^T = -M$.

Proof. In coordinates, $B(u, v) = u^T G v$; the identity $B(Au, v) + B(u, Av) = 0$ becomes $u^T M^T G v + u^T G M v = 0$ for all $u, v$, that is $M^T G + GM = 0$. If $G = I$ this is $M^T = -M$.

Example. For $V = F^2$ with the standard form, the skew matrices are $M = \begin{pmatrix} 0 & a \\ -a & 0 \end{pmatrix}$, a one-dimensional space spanned by the rotation generator $J = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$.

Example. For $V = \mathbb{R}^3$ with the standard form, the skew matrices are the antisymmetric $3 \times 3$ matrices, a three-dimensional space, with basis

$$ L_1 = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & 1 & 0 \end{pmatrix}, \quad L_2 = \begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ -1 & 0 & 0 \end{pmatrix}, \quad L_3 = \begin{pmatrix} 0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}. $$

These are the infinitesimal rotations about the three coordinate axes.

The Bracket

Proposition. The space $\mathrm{SO}(V, q)$ is closed under the commutator $[A, C] = AC - CA$, and with this bracket it is a Lie algebra over $F$, a Lie subalgebra of $\mathrm{GL}(V)$.

Proof. If $A$ and $C$ are skew, then for all $u, v$,

$$ B([A, C]u, v) = B(ACu, v) - B(CAu, v) = -B(Cu, Av) + B(Au, Cv) = B(u, CAv) - B(u, ACv) = -B(u, [A, C]v), $$

so $[A, C]$ is skew. The commutator is bilinear and antisymmetric, and the Jacobi identity holds in any associative algebra, as in Lie Algebras: A General Introduction.

Dimension and Rank

Theorem. Let $V$ be a free module of rank $n$ over $R$ with a non-degenerate symmetric form $B$. Then $\mathrm{SO}(V, B)$ is a free $R$-module of rank

$$ \operatorname{rank}_R \mathrm{SO}(V, B) = \binom{n}{2} = \frac{n(n-1)}{2}. $$

Proof. Choose a basis and let $G$ be the Gram matrix, which is invertible because $B$ is non-degenerate. The condition $M^T G + GM = 0$ is equivalent to the antisymmetry of $N = GM$, since $N^T = M^T G^T = M^T G = -GM = -N$; the assignment $M \mapsto N$ is an isomorphism of $R$-modules, with inverse $N \mapsto G^{-1}N$. The antisymmetric matrices are freely generated by the $\binom{n}{2}$ entries above the diagonal, and $G$ is invertible, so the rank is $\binom{n}{2}$.

Remark. Non-degeneracy is needed. For $q = 0$ the condition becomes vacuous and $\mathrm{SO}(V, 0) = \mathrm{GL}(V)$, of rank $n^2$; there the "skew" condition imposes nothing because $B = 0$. Over a general ring the argument requires $G$ to be invertible, which is exactly the non-degeneracy hypothesis of Bilinear Forms.

The Identification with Bivectors

The Exterior Square

Let $\Lambda^2 V$ be the second exterior power of $V$, the quotient of $V \otimes V$ by the submodule generated by the elements $v \otimes v$, as in Exterior Powers. It is generated by the wedges $u \wedge v = -v \wedge u$, and if $V$ is free of rank $n$ then $\Lambda^2 V$ is free of rank $\binom{n}{2}$, with basis $e_i \wedge e_j$ for $i < j$. Its universal property is that alternating bilinear maps $V \times V \to W$ factor uniquely through it.

The Isomorphism with $\mathrm{SO}$

Theorem. Let $B$ be a symmetric bilinear form on $V$. There is a well-defined $F$-linear map

$$ \varphi : \Lambda^2 V \longrightarrow \operatorname{End}(V), \qquad \varphi(u \wedge v)(x) = B(v, x)\,u - B(u, x)\,v, $$

and its image lies in $\mathrm{SO}(V, B)$. When $B$ is non-degenerate, $\varphi$ is an isomorphism of vector spaces onto $\mathrm{SO}(V, B)$.

Proof. The right-hand side is alternating in $u, v$, because swapping $u$ and $v$ negates it; by the universal property of $\Lambda^2 V$ it therefore descends to a well-defined linear map, which we still call $\varphi$. For skewness, compute

$$ B\bigl(\varphi(u \wedge v)x, y\bigr) = B(v, x)B(u, y) - B(u, x)B(v, y), $$

$$ B\bigl(x, \varphi(u \wedge v)y\bigr) = B(v, y)B(u, x) - B(u, y)B(v, x), $$

and the two expressions sum to $0$. So $\varphi$ maps into $\mathrm{SO}(V, B)$. If $B$ is non-degenerate, the identification $V \to V^*$, $u \mapsto B(u, -)$, gives $\Lambda^2 V \cong V \otimes V^* \cong \operatorname{End}(V)$ after antisymmetrisation, and this is precisely $\varphi$; both sides have dimension $\binom{n}{2}$ over a field, so $\varphi$ is an isomorphism.

The isomorphism depends on $B$: a different non-degenerate form gives a different identification of $\Lambda^2 V$ with $\mathrm{SO}(V, B)$. It is the algebraic statement that bivectors are infinitesimal rotations.

The Bracket as a Commutator of Bivectors

Transporting the commutator along $\varphi$ makes $\Lambda^2 V$ a Lie algebra.

Definition. The bivector bracket on $\Lambda^2 V$ is

$$ [X, Y] = \varphi^{-1}\bigl([\varphi X, \varphi Y]\bigr), $$

where the bracket on the right is the commutator of endomorphisms. With this bracket and the identification $\varphi$, the space of bivectors $\Lambda^2 V$ is a Lie algebra isomorphic to $\mathrm{SO}(V, B)$.

Example (the plane). For $V = F^2$ with the standard form, $\Lambda^2 V$ is one-dimensional, spanned by $e_1 \wedge e_2$, and $\varphi(e_1 \wedge e_2) = -J$; the bracket is zero because $\mathrm{SO}(2)$ is abelian.

Example (space). For $V = F^3$ with the standard form, the bivectors $e_2 \wedge e_3$, $e_3 \wedge e_1$, $e_1 \wedge e_2$ are a basis, and $\varphi$ sends them to $-L_1, -L_2, -L_3$; the bracket of the generators $L_1, L_2, L_3$ is

$$ [L_1, L_2] = L_3, \qquad [L_2, L_3] = L_1, \qquad [L_3, L_1] = L_2, $$

which is the multiplication table of the vector product. This is the precise sense in which the commutator of bivectors in three dimensions is the vector product.

The Exponential Map

From the Algebra to the Group

Over $F = \mathbb{R}$ or $\mathbb{C}$ the exponential series $\exp(A) = \sum_{k \geq 0} A^k/k!$ converges for every endomorphism, and it carries skew maps to rotations.

Theorem. Let $q$ be a non-degenerate quadratic form on a real or complex space $V$, and let $A \in \mathrm{SO}(V, q)$. Then $\exp(A) \in \operatorname{SO}(V, q)$.

Proof. Consider $M(t) = \exp(tA)^T G \exp(tA)$, where $G$ is the Gram matrix. Differentiating gives

$$ M'(t) = \exp(tA)^T \bigl(A^T G + G A\bigr)\exp(tA) = 0, $$

since $A^T G + GA = 0$; hence $M(t) = M(0) = G$ and $\exp(tA) \in \operatorname{O}(V, q)$ for all $t$. Also $\operatorname{tr} A = 0$: from $A^T G = -GA$, transposing and using symmetry of $G$ gives $G A = -A^T G$, so $A = -G^{-1} A^T G$ and

$$ \operatorname{tr} A = -\operatorname{tr}\bigl(G^{-1}A^T G\bigr) = -\operatorname{tr}\bigl(A^T G G^{-1}\bigr) = -\operatorname{tr} A, $$

whence $\operatorname{tr}A = 0$ because $2 \neq 0$. Therefore $\det\exp(A) = e^{\operatorname{tr}A} = 1$ and $\exp(A) \in \operatorname{SO}(V, q)$.

Theorem (standard). If $q$ is positive definite on the real space $V$, then $\exp : \mathrm{SO}(V, q) \to \operatorname{SO}(V, q)$ is surjective onto the connected component. More precisely, every rotation of a Euclidean space is the exponential of a skew map, and the exponential is a local diffeomorphism near $0$.

The proof uses the spectral theorem for the positive definite form: a rotation is diagonalised over $\mathbb{C}$ into planar rotations, and each planar rotation block is the exponential of its skew generator. The statement is standard, and we cite it rather than reproduce the spectral argument.

The Cayley Transform

A rational alternative to the exponential avoids infinite series. If $A$ is skew and $I - A$ is invertible, the Cayley transform

$$ C(A) = (I + A)(I - A)^{-1} $$

is an orthogonal transformation. With $G$ the Gram matrix, skewness of $A$ reads $A^T G = -G A$, so that $(I + A)^T G = G(I - A)$ and $G(I + A) = (I - A)^T G$; hence

$$ C(A)^T G\,C(A) = \bigl((I - A)^{-1}\bigr)^T (I + A)^T G\,(I + A)(I - A)^{-1} = \bigl((I - A)^{-1}\bigr)^T G\,(I - A)(I + A)(I - A)^{-1} = \bigl((I - A)^{-1}\bigr)^T (I - A)^T G = G, $$

using that $I + A$ and $(I - A)^{-1}$ commute because both are polynomials in $A$. The Cayley transform is a birational map from the skew maps to the rotations that do not have $1$ as an eigenvalue; it is the algebraic analogue of the stereographic projection of the group onto its Lie algebra.

Structure and Rank

A Cartan Subalgebra

Assume $F = \mathbb{R}$ and $q$ definite, so that $\mathrm{SO}(V, q)$ is the compact orthogonal Lie algebra. Choose an orthonormal basis and pair the coordinates $(e_1, e_2), (e_3, e_4), \ldots$; let $\mathrm{H}$ be the space spanned by the block generators

$$ H_k = \varphi(e_{2k-1} \wedge e_{2k}), $$

each of which rotates the corresponding coordinate plane and annihilates the rest. Then $\mathrm{H}$ is a Cartan subalgebra: it is abelian, and it is maximal among the abelian subalgebras consisting of semisimple elements. Its dimension is

$$ \operatorname{rank}\mathrm{SO}(V, q) = \left\lfloor \frac{n}{2} \right\rfloor, $$

which is also the dimension of a maximal torus of $\operatorname{SO}(V, q)$. Concretely, a maximal torus consists of the block-diagonal matrices with $\lfloor n/2 \rfloor$ rotation blocks, so $\operatorname{SO}(2)$ is its own maximal torus, $\operatorname{SO}(3)$ has a circle of rotations about a fixed axis, and $\operatorname{SO}(4)$ has a two-dimensional torus, the product of two circles of independent planar rotations.

The Split and Compact Forms

Over $\mathbb{R}$ the isomorphism class of $\mathrm{SO}(V, q)$ depends on the signature, not only on the dimension, because the Gram matrix can be brought to the form $\operatorname{diag}(1^p, (-1)^r)$ by Sylvester's law. Writing

$$ \mathrm{SO}(p, r) = \mathrm{SO}(V, q), \qquad q \cong p\langle 1\rangle \perp r\langle -1\rangle, $$

the complexifications of all the $\mathrm{SO}(p, r)$ with $p + r = n$ are isomorphic to $\mathrm{SO}(n, \mathbb{C})$; over $\mathbb{R}$ they are the different real forms, of which the compact form $\mathrm{SO}(n) = \mathrm{SO}(n, 0)$ and the split form $\mathrm{SO}(m, m)$ or $\mathrm{SO}(m+1, m)$ are the extremes. The common complex dimension is $\binom{n}{2}$ and the common rank is $\lfloor n/2\rfloor$.

The Killing Form

Invariant Bilinear Forms

Definition. The Killing form of a Lie algebra $\mathrm{G}$ over a field is the symmetric bilinear form

$$ \kappa(X, Y) = \operatorname{tr}(\operatorname{ad}_X \circ \operatorname{ad}_Y), \qquad \operatorname{ad}_X(Z) = [X, Z], $$

and the trace form of a matrix Lie algebra is $\beta(X, Y) = \operatorname{tr}(XY)$.

Proposition. $\kappa$ and $\beta$ are symmetric and invariant:

$$ \kappa([X, Y], Z) = \kappa(X, [Y, Z]), \qquad \beta([X, Y], Z) = \beta(X, [Y, Z]). $$

Proof. Symmetry of $\kappa$ is the symmetry of the trace, and invariance follows from the identity $\operatorname{ad}_{[X,Y]} = [\operatorname{ad}_X, \operatorname{ad}_Y]$ together with the cyclicity of the trace:

$$ \kappa([X, Y], Z) = \operatorname{tr}(\operatorname{ad}_X\operatorname{ad}_Y\operatorname{ad}_Z - \operatorname{ad}_Y\operatorname{ad}_X\operatorname{ad}_Z) = \operatorname{tr}(\operatorname{ad}_X\operatorname{ad}_Y\operatorname{ad}_Z - \operatorname{ad}_X\operatorname{ad}_Z\operatorname{ad}_Y) = \kappa(X, [Y, Z]). $$

For the trace form, $\beta([X,Y],Z) + \beta(Y,[X,Z]) = \operatorname{tr}(XYZ - YXZ + YXZ - YZX) = \operatorname{tr}(XYZ - YZX) = 0$ by cyclicity, which is the stated invariance.

The two invariance properties are the hypothesis of the uniqueness statement used below: on a simple Lie algebra the invariant symmetric bilinear forms form a vector space of dimension one.

The Killing Form of $\mathrm{SO}(n)$

Theorem. Let $V = F^n$ carry the standard form, let $n \geq 3$, and let $\kappa$ be the Killing form of $\mathrm{SO}(n)$. Then

$$ \kappa(X, Y) = (n - 2)\operatorname{tr}(XY) \qquad (X, Y \in \mathrm{SO}(n)). $$

Proof. Both $\kappa$ and the trace form $\beta$ are symmetric bilinear and invariant under the adjoint action by the proposition above, so on a simple Lie algebra they are proportional. The algebra $\mathrm{SO}(n)$ is simple for $n = 3$ and for $n \geq 5$, so in those dimensions $\kappa = c\,\beta$. Let $L$ be the generator of the rotation in the $(1,2)$-plane, the matrix with $L_{12} = -1$, $L_{21} = 1$ and all other entries zero. Then $\beta(L, L) = \operatorname{tr}(L^2) = -2$, while the bracket relations of the basis $E_{ij} = e_ie_j^{T} - e_je_i^{T}$ give $\kappa(L, L) = -2(n - 2)$, so $c = n - 2$. For $n = 4$ a direct evaluation in the six-dimensional basis $E_{ij}$ gives the same identity.

Example. For $n = 3$ the isomorphism $\mathrm{SO}(3) \cong \mathbb{R}^3$ of the next section carries the Killing form to $\kappa(X, Y) = \operatorname{tr}([X]_{\times}[Y]_{\times}) = -2\,X \cdot Y$, twice the negative of the Euclidean product; the isomorphism is therefore not an isometry, and the negative sign is the reflection of the compactness of $\mathrm{SO}(3)$. For $n = 4$ the formula reads $\kappa = 2\beta$, the factor reflecting the splitting $\mathrm{SO}(4) \cong \mathrm{SU}(2) \oplus \mathrm{SU}(2)$, on which the invariant forms are the sum of the invariant forms of the two summands.

Corollary. For $n \geq 3$ the Killing form of $\mathrm{SO}(n)$ is negative definite, since for skew $X \neq 0$

$$ \operatorname{tr}(X^2) = -\operatorname{tr}(X^{T}X) = -\sum_{i,j} X_{ij}^2 < 0 . $$

Hence $\mathrm{SO}(n)$ is a compact semisimple Lie algebra, and by Cartan's criterion it is semisimple; it is simple for $n = 3$ and $n \geq 5$, while $\mathrm{SO}(4)$ decomposes as $\mathrm{SU}(2) \oplus \mathrm{SU}(2)$. For the indefinite forms the Killing form of $\mathrm{SO}(p, r)$ remains non-degenerate but is indefinite, and the signature distinguishes the real forms.

Low-Dimensional Isomorphisms

The Plane

For $n = 2$ the algebra $\mathrm{SO}(2)$ is one-dimensional and abelian, with basis $J$ satisfying $[J, J] = 0$; it is the Lie algebra of the circle group $\operatorname{SO}(2)$, and $\exp(\theta J) = R(\theta)$ recovers the plane rotations of The Rotation Group and Orientation.

The Space $\mathbb{R}^3$ and the Vector Product

Theorem. For the standard positive definite form on $\mathbb{R}^3$ there is an isomorphism of Lie algebras

$$ \mathrm{SO}(3) \cong \mathbb{R}^3, $$

where the bracket on the right is the vector product.

Proof. The map sends $L_1 \mapsto e_1$, $L_2 \mapsto e_2$, $L_3 \mapsto e_3$ in the basis of the previous example. The bracket relations $[L_1, L_2] = L_3$ and its cyclic permutations are exactly the relations $e_1 \times e_2 = e_3$ and its cyclic permutations, so the map is an isomorphism of Lie algebras.

Corollary. Under the isomorphism a skew map $A$ acts on $\mathbb{R}^3$ by $Ax = \omega \times x$, so a one-parameter group of rotations is generated by an element $\omega$ of $\mathbb{R}^3$, and $\omega$ is the axis of the rotation $\exp(tA)$. The double cover of $\operatorname{SO}(3)$ by the unit quaternions, of which this isomorphism is the infinitesimal shadow, belongs to the Clifford layer of the category.

The Signature $(3, 1)$ and $\mathrm{SL}(2, \mathbb{C})$

Theorem. For the form $q = \langle 1, 1, 1, -1\rangle$ on $\mathbb{R}^4$ there is an isomorphism of real Lie algebras

$$ \mathrm{SO}(3, 1) \cong \mathrm{SL}(2, \mathbb{C}), $$

where $\mathrm{SL}(2, \mathbb{C})$ is regarded as a real Lie algebra of dimension $6$. The same algebra is written $\mathrm{SO}(1, 3)$ in the custom that counts the negative signs first, since $\mathrm{SO}(p, r) \cong \mathrm{SO}(r, p)$ for all $p, r$.

Proof. Both sides have dimension $6$. In the basis of rotations and hyperbolic rotations, $\mathrm{SO}(3, 1)$ has generators $J_1, J_2, J_3$ for the compact part and $K_1, K_2, K_3$ for the complementary part, with brackets

$$ [J_i, J_j] = \sum_k \epsilon_{ijk} J_k, \qquad [J_i, K_j] = \sum_k \epsilon_{ijk} K_k, \qquad [K_i, K_j] = -\sum_k \epsilon_{ijk} J_k, $$

where $\epsilon_{ijk}$ is the Levi-Civita symbol. Setting $A_i = \tfrac{1}{2}(J_i + iK_i)$ and $B_i = \tfrac{1}{2}(J_i - iK_i)$ turns these into

$$ [A_i, A_j] = \sum_k \epsilon_{ijk} A_k, \qquad [B_i, B_j] = \sum_k \epsilon_{ijk} B_k, \qquad [A_i, B_j] = 0, $$

so the complexification is $\mathrm{SL}(2, \mathbb{C}) \oplus \mathrm{SL}(2, \mathbb{C})$ as a complex Lie algebra, each summand spanned by one of the triples. The real form $\mathrm{SO}(3, 1)$ is the fixed-point set of the conjugation of this complex algebra sending $(X, Y)$ to $(\bar Y, \bar X)$, that is the diagonal $\{(X, \bar X)\}$; the map $(X, \bar X) \mapsto X$ is an isomorphism of real Lie algebras onto $\mathrm{SL}(2, \mathbb{C})$. Hence $\mathrm{SO}(3, 1) \cong \mathrm{SL}(2, \mathbb{C})$ over $\mathbb{R}$. The same computation, with the negative signs counted first, is the usual derivation of the isomorphism $\mathrm{SO}(1, 3) \cong \mathrm{SL}(2, \mathbb{C})$; the two statements are the same one, since the two signatures are isometric by negation of the coordinates in which they differ.

Remark (the bracket relations). The first relation is the bracket of the compact rotations. For the last, using $K_i = iB_i - iA_i$ and $J_i = A_i + B_i$ gives $[K_i, K_j] = -[A_i,A_j] - [B_i,B_j] + [A_i,B_j] + [B_i,A_j] = -\sum\epsilon_{ijk}A_k - \sum\epsilon_{ijk}B_k = -\sum\epsilon_{ijk}J_k$, which is the stated sign; the middle relation likewise gives $\sum\epsilon_{ijk}K_k$.

Remark. The case $n = 4$ is the smallest in which the orthogonal Lie algebra has more than one isomorphism type over $\mathbb{R}$: $\mathrm{SO}(4) \cong \mathrm{SU}(2) \oplus \mathrm{SU}(2)$ is the compact form, while $\mathrm{SO}(3, 1)$ is a non-compact form with the same complexification. This is the Lie-theoretic form of the double-cover phenomena that the Clifford layer of the category describes.

Summary

The orthogonal Lie algebra $\mathrm{SO}(V, q)$ of a non-degenerate quadratic space is the space of skew maps $A$ with $B(Au, v) + B(u, Av) = 0$, equivalently $M^T G + GM = 0$ in a basis, and it is a Lie subalgebra of $\mathrm{GL}(V)$ under the commutator. It is free of rank $\binom{n}{2}$ when $V$ is free of rank $n$ and $B$ is non-degenerate; for $q = 0$ it degenerates to $\mathrm{GL}(V)$, so non-degeneracy is essential.

For any symmetric $B$ the formula $\varphi(u \wedge v)(x) = B(v, x)u - B(u, x)v$ gives a linear map $\Lambda^2 V \to \mathrm{SO}(V, B)$ which is an isomorphism when $B$ is non-degenerate. Transporting the commutator makes the bivectors $\Lambda^2 V$ a Lie algebra isomorphic to $\mathrm{SO}(V, B)$; the bracket of bivectors is their commutator as endomorphisms, and in dimension three it is the vector product.

Over $\mathbb{R}$ or $\mathbb{C}$ the exponential $\exp(A) = \sum_k A^k/k!$ sends $\mathrm{SO}(V, q)$ into $\operatorname{SO}(V, q)$; for a positive definite real form it is surjective onto the connected component. The Cayley transform $C(A) = (I + A)(I - A)^{-1}$ is a rational replacement. The algebra has dimension $\binom{n}{2}$ and rank $\lfloor n/2\rfloor$, a Cartan subalgebra being spanned by the block generators of the coordinate planes; the split and compact forms $\mathrm{SO}(p, r)$ have a common complexification $\mathrm{SO}(n, \mathbb{C})$ and differ as real forms. In low dimensions $\mathrm{SO}(2)$ is one-dimensional abelian, $\mathrm{SO}(3) \cong \mathbb{R}^3$ with the vector product as bracket, and $\mathrm{SO}(3, 1) \cong \mathrm{SO}(1, 3) \cong \mathrm{SL}(2, \mathbb{C})$ as real Lie algebras, the two signatures being the same up to the order of the signs.

The Killing form $\kappa(X, Y) = \operatorname{tr}(\operatorname{ad}_X\operatorname{ad}_Y)$ and the trace form $\beta(X, Y) = \operatorname{tr}(XY)$ are symmetric and invariant under the adjoint action. For the standard form on $F^n$ with $n \geq 3$ the two are proportional,

$$ \kappa(X, Y) = (n - 2)\operatorname{tr}(XY), $$

and $\kappa$ is negative definite because $\operatorname{tr}(X^2) = -\sum_{ij}X_{ij}^2$ for skew $X$; hence $\mathrm{SO}(n)$ is compact and semisimple, and simple for $n = 3$ and $n \geq 5$. The Killing form of an indefinite $\mathrm{SO}(p, r)$ is non-degenerate but indefinite.

Summary of Notation

Symbol	Meaning
$R$	Commutative ring with $2$ invertible
$F$	Field, characteristic not $2$ unless stated
$V$	Finite-dimensional free module or space
$q$, $B$	Quadratic form and its polar form, $q(v) = B(v, v)$
$G$	Gram matrix of $B$
$A$, $C$, $M$	Endomorphisms and their matrices
$[A, C] = AC - CA$	Commutator bracket
$\mathrm{GL}(V)$	Lie algebra of all endomorphisms
$\mathrm{SO}(V, q)$	Orthogonal Lie algebra of skew maps
$\mathrm{SO}(n)$	Compact orthogonal Lie algebra, signature $(n, 0)$
$\mathrm{SO}(p, r)$	Orthogonal Lie algebra of signature $(p, r)$, with $\mathrm{SO}(p, r) \cong \mathrm{SO}(r, p)$
$\mathrm{SL}(2, \mathbb{C})$	Lie algebra of traceless $2 \times 2$ complex matrices
$\mathrm{SU}(2)$	Lie algebra of traceless anti-Hermitian $2 \times 2$ matrices
$\Lambda^2 V$	Second exterior power, the bivectors
$\varphi$	Isomorphism $\Lambda^2 V \to \mathrm{SO}(V, B)$
$J$, $L_i$, $H_k$	Generators of skew maps
$E_{ij} = e_ie_j^{T} - e_je_i^{T}$	Standard basis of $\mathrm{SO}(n)$
$\kappa(X, Y)$	Killing form $\operatorname{tr}(\operatorname{ad}_X\operatorname{ad}_Y)$
$\beta(X, Y)$	Trace form $\operatorname{tr}(XY)$
$\operatorname{ad}_X(Y) = [X, Y]$	Adjoint map
$\operatorname{tr}$	Trace of an endomorphism or matrix
$\exp(A)$	Exponential of an endomorphism
$C(A)$	Cayley transform $(I+A)(I-A)^{-1}$
$\mathrm{H}$	Cartan subalgebra
$\operatorname{rank}\mathrm{SO}(V, q)$	$\lfloor n/2\rfloor$, dimension of a maximal torus
$\epsilon_{ijk}$	Levi-Civita symbol
$\times$	Vector product on $\mathbb{R}^3$
$\mathbb{R}, \mathbb{C}$	Real and complex numbers