Set of positive definite Hermitian matrices as quotient

Question

Let $$\mathcal{H}_n^{++} := \{ A \in \mathrm{Mat}_n(\mathbb{C}) \vert A = A^{\dagger} \text{ and } A > 0 \}$$ denote the space of positive definite Hermitian matrices. I have read in some references, that this space can be realized as the quotient $$ \mathcal{H}_n^{++} \cong \mathrm{GL}_n(\mathbb{C}) / \mathrm{U}(n) ,$$ where the action of $\mathrm{U}(n)$ is by conjugation. But I was not able to find a proof. Can anybody provide one?

Answer 1

In general, when a group acts on a space $\Omega$, for any $x\in\Omega$, the fibers of the map $G\to\mathrm{Orb}(x)$ given by $g\mapsto gx$ are the cosets of $\mathrm{Stab}(x)$. Thus, $g\mathrm{Stab}(x)\leftrightarrow gx$ is an isomorphism $G/\mathrm{Stab}(x)\cong\mathrm{Orb}(x)$ of $G$-sets (i.e. an equivariant/intertwining bijection). When $G$ is a nice topological group and the action is nice, this gives us a fiber bundle $\mathrm{Stab}(x)\to G\to\mathrm{Orb}(x)$. This is the Orbit-Stabilizer Theorem.

The group $\mathrm{GL}(n,\mathbb{C})$ acts on $\mathcal{H}_n^{++}$ on the right via $X\mapsto A^\dagger XA$. The stabilizer of $I_n$ is $\mathrm{U}(n)$. The orbit is all of $\mathcal{H}_n^{++}$; look up Cholesky decomposition. Indeed, the fibration $\mathrm{GL}(n,\mathbb{C})\to\mathcal{H}_n^{++}$ given by $A\mapsto A^\dagger A$ extracts $P^2$ from the polar decomposition $A=UP$.