Lemma 6

Let $q$ be a non-degenerate bilinear form on a finite-dimensional complex vector space E of dimension $\geqslant 3$.

a) If $q$ is symmetric, then the adjoint representation of the orthogonal Lie algebra $\mathfrak{s}\mathfrak{o}(q)$ is isomorphic to the exterior square $\wedge^2E$ of the natural representation $\mathfrak{s}\mathfrak{o}(q)\rightarrow \mathfrak{g}\mathfrak{l}(E)$;

b) If $q$ is alternating, then the adjoint representation of the symplectic Lie algebra $\mathfrak{s}\mathfrak{p}(q)$ is isomorphic to the symmetric square $\mathsf{S}^2(E)$ of the natural representation $\mathfrak{s}\mathfrak{p}(q)\rightarrow \mathfrak{g}\mathfrak{l}(E)$.

Let $\mathbf{C}$ be provided with the identical involutive automorphism. The bilinear form $q$ is then $\varepsilon$-Hermitian (A, IX, § 3, n$^o1$, Def. 1) with $\varepsilon = 1$ in case a) and $\varepsilon =-1$ in case b). For every $u\in$ End(E), let $^tu$ denote the adjoint of $u$ with respect to $q$. Thus one has $q(x, u(y)) =q(^tu(x), y)$ for every $(x, y)\in E\times E$. Let $v$ denote the unique automorphism of $E\otimes E$ such that $v(x\otimes y) =y\otimes x$ for every $(x, y)\in E^2$.

There exists a unique linear mapping $w$ of $E\otimes E$ into End(E) such that $w(a\otimes b)$ is the linear mapping defined by $x\mapsto q(a, x)b$ for every $(a, b, x)\in E^3$. The mapping $w$ is an isomorphism. For every $s\in E\otimes E$, one has

In fact, for all $(x, y)$ and $(a, b)$ in $E\times E$, there follows

Finally, denote by $x\mapsto x^*$ the isomorphism of E onto $E'$ deduced from $q$, that is to say such that $\langle x^*, y\rangle =q(x, y)$ for every $(x, y)\in E^2$.

With these notations established, let H be the orthogonal group (resp. symplectic) of $q$. It is a Lie subgroup of $\mathbf{G}\mathbf{L}(E)$ whose Lie algebra $\mathfrak{h}$ is the subspace of End(E) consisting of the $u\in$ End(E) such that $^tu=-u$ (LIE, III, p. 146, cor. 1). Since the form $q$ is H-invariant, we have

for every $g\in H$ and every $(a, b)\in E\times E$.

Equip End(E) with the adjoint representation Ad of H. The linear mapping $w$ is a morphism of representations of H: in fact, for every $g\in H$ and every $(a, b, x)\in E^3$, we have

$=g(\langle a, g^{-1}x\rangle b) =$ Ad($g$)$(w(a\otimes b))x$,

whence the conclusion by linearity.

Since $v$ is likewise a morphism of representations of H, the same is true of the linear mapping $\theta =w-\varepsilon w\circ v$; the latter is therefore a morphism of $\mathfrak{h}$-modules.

Let F $\subset E\otimes E$ be the subspace of elements $s\in E\otimes E$ such that $v(s) =-\varepsilon s$. If $\varepsilon =-1$, the restriction to F of the canonical mapping of $E\otimes E$ into the symmetric square of E is an isomorphism of $\mathbf{C}$-vector spaces (A, III, p. 72); if $\varepsilon = 1$, the restriction to F of the canonical mapping of $E\otimes E$ into the exterior square of E is an isomorphism of $\mathbf{C}$-vector spaces (loc. cit., p. 82).

The image of F under $\theta$ is contained in $\mathfrak{h}:$ in fact, for every $s\in F$, formula (3) implies

By definition of F, the restriction of the mapping $\theta$ to F coincides with that of $2w$ and the linear mapping $\theta$ is therefore injective on F. Since dim(F) = dim($\mathfrak{h}$) (A, III, p. 75, th. 1 and p. 87, cor. 1 and LIE, VIII, p. 192 and p. 201), one concludes that $\theta$ induces by passing to the subspaces an isomorphism of $\mathfrak{h}$-modules of F onto $\mathfrak{h}$, whence the lemma.