Example 8

Spaces of hermitian forms.

In this example, K always denotes one of the fields $\mathbf{R}, \mathbf{C}, \mathbf{H}$. We write $\delta = \dim_{\mathbf{R}} K$ (thus $\delta = 1, 2$ or 4). A hermitian form $\Phi$ on the right vector space $K^n$ may be written

with $h_{ij} = \overline{h_{ji}}$ for all $i$ and $j$. We denote by $\mathfrak{H}$ the vector space over $\mathbf{R}$ formed by the hermitian matrices of $M_n(K)$. The mapping $(h_{ij}) \mapsto \Phi$ is an isomorphism of $\mathfrak{H}$ onto the vector space of hermitian forms on $K^n$, by means of which we shall identify these two spaces. Let $\mathfrak{H}_+^* \subset \mathfrak{H}$ be the set of nondegenerate positive hermitian forms on $K^n$. The set $\mathfrak{H}_+^*$ is convex in $\mathfrak{H}$; for, if $\Phi_1, \Phi_2$ are in $\mathfrak{H}_+^*$ and if $\lambda, \mu$ are two numbers $> 0$ such that $\lambda + \mu = 1$, it is clear that $\lambda \Phi_1 + \mu \Phi_2$ is a positive hermitian form; on the other hand, if $(\lambda \Phi_1 + \mu \Phi_2)(x, x) = 0$, then $\Phi_1(x, x) = \Phi_2(x, x) = 0$, therefore $x = 0$, so that $\lambda \Phi_1 + \mu \Phi_2$ is nondegenerate. Let us now show that $\mathfrak{H}_+^*$ is an open subset of $\mathfrak{H}$. Let S be the set of $x = (x_1, \ldots, x_n) \in K^n$ such that $x_1 \overline{x}_1 + \cdots + x_n \overline{x}_n = 1$; this is a compact subset of $K^n$; if $\Phi \in \mathfrak{H}_+^*$, the function $x \mapsto \Phi(x, x)$ is continuous and $> 0$ on S, hence its infimum is $> 0$; if $\Phi' \in \mathfrak{H}$ is sufficiently near $\Phi$, it follows that $\Phi'(x, x) > 0$ for all $x \in S$, so that $\Phi'$ is positive and nondegenerate.

The general linear group $\mathbf{GL}(n, K)$ operates continuously on the right in $\mathfrak{H}$ by $(X, \Phi) \mapsto \Phi \circ X$, that is, by $(X, H) \mapsto t \overline{X} \cdot H \cdot X$, where $H$ denotes the hermitian matrix corresponding to $\Phi$. It is clear that $\mathfrak{H}_+^*$ is stable under $\mathbf{GL}(n, K)$. More precisely, by Alg., Ch. IX, §6, No. 1, Cor. 1 of Th. 1,$^1$ $\mathfrak{H}_+^*$ is the orbit under $\mathbf{GL}(n, K)$ of the form $\sum_{i=1}^n \overline{x_i} y_i$ corresponding to the identity matrix $I_n$. The stabilizer of this form is $\mathbf{U}(n, K)$. By Lemma 2 of App. I, $\mathfrak{H}_+^*$ may be identified, as a topological homogeneous space, with $\mathbf{GL}(n, K)/\mathbf{U}(n, K)$.

For every $X \in \mathbf{GL}(n, K)$, let $\widetilde{X}$ be the automorphism $H \mapsto t \overline{X} \cdot H \cdot X$ of the real vector space $\mathfrak{H}$. If $\mu$ denotes the Haar measure of the additive group $\mathfrak{H}$, one has $\widetilde{X}^{-1}(\mu) = |\det \widetilde{X}| \cdot \mu$ (§ 1, No. 10, Cor. 1 of Prop. 15). Let us show that

where N denotes the norm in $M_n(K)$ regarded as an $\mathbf{R}\text{-algebra}$, and where $\lambda = 1 - \frac{\delta - 2}{\delta n}$. It suffices to verify (15) for $X$ running over a system of

(1) Cf. TVS, V, §2, No. 4, Cor. 1 of Th. 2.

generators of $\mathbf{GL}(n, K)$, hence (A, II, §10, No. 13, Cor. 2 of Prop. 14) for $X$ of the following types:

a) $X$ is the matrix of a mapping of the form

where $\sigma \in \mathfrak{S}_n$. In this case, a power of $X$ is equal to 1, therefore $|\det \widetilde{X}| = |N(X)| = 1$.

b) $X$ is the matrix of a mapping of the form

Then, if $(h_{ij}) \in \mathfrak{H}$, one has $\widetilde{X}((h_{ij})) = (h'_{ij})$ with $h'_{11} = \overline{a} h_{11} a = |a|^2 h_{11}$, $h'_{1i} = \overline{a} h_{1i}$ for $i > 1$, $h'_{ij} = h_{ij}$ for $i > 1, j > 1$; therefore On the other hand, if $Y = (y_{ij}) \in M_n(K)$, then $XY = (y'_{ij})$ with $y'_{1j} = ay_{1j}$ and $y'_{ij} = y_{ij}$ for $i > 1$; therefore $|N(X)| = |a|^{\delta n}$. The formula (15) is again verified.

c) $X$ is the matrix of a mapping of the form

Then $\widetilde{X}((h_{ij})) = (h'_{ij})$ with $h'_{11} = h_{11}$, $h'_{12} = h_{12} + h_{11} b$, $h'_{1i} = h_{1i}$ for $i > 2$, $h'_{22} = h_{22} + \overline{b} h_{12} + \overline{h}_{12} b + \overline{b} h_{11} b$, $h'_{2i} = h_{2i} + \overline{b} h_{1i}$ for $i > 2$, $h'_{ij} = h_{ij}$ for $i > 2, j > 2$. Taking into account Lemma 6, one sees that $|\det \widetilde{X}| = 1$. One verifies similarly that $|N(X)| = 1$, which completes the proof of formula (15).

This established, the measure $|N(H)|^{-\lambda/2} d\mu(H)$ on $\mathfrak{H}$ is invariant under $\mathbf{GL}(n, K)$, since

If $H \in \mathfrak{H}_+^*$, then $H = \widetilde{X}(I_n) = {}^t \overline{X} X$ for some $X \in \mathbf{GL}(n, K)$, therefore $N(H) = \overline{N(X)} N(X) > 0$. Consequently, on $\mathfrak{H}_+^*$, the unique (up to a constant factor) measure invariant under $\mathbf{GL}(n, K)$ (cf. §2, No. 6, Th. 3) is the measure

In particular,