Proposition 2.9.18D01DFDB6AE9A

Proof. Identify the space of complex $n\times n$ matrices with $\mathbb R^{2n^2}$ through the real and imaginary parts of the entries, and let $\gamma $ be the measure on it with density $\mathrm e^{-\pi \operatorname {tr}(X^*X)}$ with respect to Lebesgue measure. Since $\operatorname {tr}(X^*X)=\sum _{j,k}|X_{jk}|^2$, this density is the Gaussian $\mathrm e^{-\pi |x|^2}$ on $\mathbb R^{2n^2}$, so $\gamma $ is a probability measure by lemma 2.9.2 (with $\xi =0$ and $t=1$). The singular matrices form a $\gamma $-null set: they are the zero set of the real polynomial $|\det X|^2$, which is not identically zero since $\det 1=1$, and the zero set of such a polynomial is Lebesgue null.

To prove the last fact, let $P$ be a real polynomial in $m\geq 1$ variables that is not identically zero; we argue by induction on $m$. If $m=1$, the zero set is finite. If $m\geq 2$, write $P=\sum _ kc_ k(x')x_ m^ k$ as a polynomial in the last variable, with coefficients $c_ k$ that are polynomials in $x'=(x_1,\ldots ,x_{m-1})$. Some $c_ k$ is not identically zero, so by induction the common zero set $Z'$ of the coefficients is null in $\mathbb R^{m-1}$. For $x'\notin Z'$, the polynomial $P(x',\cdot )$ is not identically zero and has finitely many zeros. By Tonelli's theorem applied to the indicator function of the closed set $\{ P=0\} $, this set is null.

For invertible $X$ define

where $(X^*X)^{1/2}$ is the positive square root of the positive definite matrix $X^*X$. By the spectral theorem for Hermitian matrices, this square root exists and is unique. It depends continuously on the matrix. Indeed, let $A_ k\to A$ be positive definite matrices. The square roots $A_ k^{1/2}$ are bounded, since $\operatorname {tr}(A_ k^{1/2}A_ k^{1/2})=\operatorname {tr}A_ k$. Every convergent subsequence of them has a positive semidefinite limit $S$ with $S^2=A$; then $S$ is invertible, hence positive definite, and $S=A^{1/2}$ by uniqueness. A bounded sequence of matrices all of whose convergent subsequences have the same limit converges to it, so $A_ k^{1/2}\to A^{1/2}$. Since inversion is continuous on invertible matrices, $Q$ is continuous on the open set of invertible matrices, and $Q(X)^*Q(X)=(X^*X)^{-1/2}X^*X(X^*X)^{-1/2}=1$. Let $\eta $ be the image under $Q$ of the restriction of $\gamma $ to the invertible matrices; it is a Borel probability measure on $\mathrm U(n)$.

Fix $U,V\in \mathrm U(n)$. The map $X\mapsto UXV$ is real linear and preserves $\operatorname {tr}(X^*X)$, so it is orthogonal on $\mathbb R^{2n^2}$ and preserves $\gamma $. By uniqueness of positive square roots, $(V^*X^*XV)^{1/2}=V^*(X^*X)^{1/2}V$, and therefore $Q(UXV)=UQ(X)V$. Consequently the image of $\eta $ under $g\mapsto UgV$ is the image of $\gamma $ under $X\mapsto Q(UXV)$, which is $\eta $ because $X\mapsto UXV$ preserves $\gamma $. This proves the invariance of $\eta $ on both sides.

Now let $Y$ be as in the statement and fix $y_0\in Y$. By transitivity, $Y$ is the image of the compact space $\mathrm U(n)$ under the continuous map $g\mapsto gy_0$, so $Y$ is compact, and hence separable. Let $\sigma $ be the image of $\eta $ under $g\mapsto gy_0$. For $h\in \mathrm U(n)$, the image of $\sigma $ under $h$ is the image of $\eta $ under $g\mapsto hgy_0$, which is $\sigma $ by left invariance; thus $\sigma $ is invariant. Let $f$ be continuous on $Y$. Every $y\in Y$ has the form $hy_0$, by transitivity, and right invariance gives

Let $\tau $ be any invariant Borel probability measure on $Y$. We integrate the last identity against $\tau $ and apply Fubini's theorem to the bounded continuous function $(g,y)\mapsto f(gy)$ on $\mathrm U(n)\times Y$, which is product-measurable by lemma 2.2.8 because $\mathrm U(n)$ and $Y$ are separable metric spaces. We obtain

Thus $\sigma $ and $\tau $ agree on continuous functions. For a nonempty closed set $F\subset Y$, the continuous functions $\max \{ 0,1-j\operatorname {dist}(y,F)\} $ decrease to $\mathbf1_ F$, so dominated convergence gives $\sigma (F)=\tau (F)$. The closed sets are stable under finite intersections and generate the Borel $\sigma $-algebra, so lemma 2.2.6(2) gives $\sigma =\tau $. In particular $\sigma $ does not depend on $y_0$.

Finally, $\mathrm U(n)$ acts continuously on the unit sphere of $\mathbb C^ n$, and transitively: every unit vector is the first column of a unitary matrix, obtained by completing it to an orthonormal basis with the Gram–Schmidt process. For the Grassmannians, these properties are verified below, where they are introduced. $\square$