Proposition 9

Let V be a vector space of finite dimension n over K and β a non-degenerate symmetric (resp. alternating) bilinear form on V. Let g be the Lie algebra consisting of the x ∈ gl(V) such that β(xm, m') + β(m, xm') = 0 for all m, m' in V. Then g is reductive; g is even semi-simple except in the case where β is symmetric and n = 2.

For all u ∈ gl(V) let u* denote its adjoint relative to β; then Tr(u) = Tr(u*) by Proposition 7 of Algebra, Chapter IX, § 1, no. 8. The condition

for all m, m' in V means that $u + u^* = 0$. In particular, if $v \in \mathrm{gl}(V)$ then $(v - v^*)^* = v^* - v$ and hence $v - v^* \in g$. Then let u be an element of g orthogonal to g with respect to the bilinear form φ associated with the identity representation of g. For all $v \in \mathrm{gl}(V)$, $\mathrm{Tr}\,u(v - v^*) = 0$, hence

and hence $\mathrm{Tr}(uw) = 0$. It follows that $u = 0$, so that φ is non-degenerate. Hence g is reductive (Proposition 5). It remains to show that the centre of g is zero (except when β is symmetric and $n = 2$). By extending the base field, we can assume that K is algebraically closed.

(a) When β is symmetric, it can be identified with the bilinear form on $K^n$ with matrix $I_n$ with respect to the canonical basis (Algebra, Chapter IX, § 6, Corollary 1 to Theorem 1). Under these conditions g is identified with the Lie algebra of skew-symmetric matrices ($§ 3$, no. 4, Example 1). Let $U = (u_{ij}) \in g$; we use the fact that U commutes with the matrix $(v_{ij}) \in g$ all of whose elements are zero except $v_{i_0j_0}$ and $v_{j_0i_0}$ ($i_0 \neq j_0$) which are equal respectively to 1 and −1. We find that $u_{i_0j} = u_{j_0i} = u_{i_0} = u_{j_0} = 0$ for $i \neq i_0, j_0$ and $j \neq i_0, j_0$. If $n > 2$, there exist, for all distinct indices $i_0$ and $j$, distinct indices i and $j_0$ such that $i \neq i_0, j_0 \neq j, j_0 \neq i_0$; hence $u_{i_0j} = 0$. This proves that an element of the centre of g is zero.

(b) When β is alternating and $n = 2m$, β can be identified with the bilinear form on $K^{2m}$ with matrix $\begin{pmatrix} 0 & I_m \\ -I_m & 0 \end{pmatrix}$ with respect to the canonical basis (Algebra, Chapter IX, § 5, Corollary 1 to Theorem 1). Under these conditions g is identified with the Lie algebra of matrices of the form $U = \begin{pmatrix} A & B \\ C & D \end{pmatrix}$ where $D = -{}^tA$, B and C are symmetric ($A, B, C, D$ in $\mathbf{M}_m(K)$) ($§ 3$, no. 4, Example 1). We use first the fact that U commutes with the matrix $\begin{pmatrix} X & 0 \\ 0 & -{}^tX \end{pmatrix}$, where $X \in \mathbf{M}_m(K)$. Then $AX = XA, CX = -{}^tXC, XB = -B\cdot{}^tX$; as these equalities must hold for all X, it follows that A is a scalar matrix $\lambda I_m$. We now use the fact that U commutes with the matrix $\begin{pmatrix} 0 & Y \\ 0 & 0 \end{pmatrix}$, where

Y is a symmetric matrix of $\mathbf{M}_m(\mathbf{K})$. Then $\lambda Y = YC = CY = 0$. This proves first that $\lambda = 0$. Moreover, for all $X \in \mathbf{M}_m(\mathbf{K})$, $X + {}^tX$ is symmetric and hence $XC = -{}^tXC$. Using the equation $CX = -{}^tXC$ obtained above, we see that $C$ commutes with every element of $\mathbf{M}_m(\mathbf{K})$ and hence that $C$ is a scalar matrix, necessarily zero since $YC = 0$. It is similarly shown that $B = 0$.

For $\beta$ symmetric and $n = 2$, $g$ is of dimension 1 and hence commutative. For the other cases cf. Exercises 25 and 26.