I am reading V. I. Arnol’d, A. B. Givental’ , Symplectic Geometry $\S$4.4 about the topology of the symplectic group Sp(2n). It is stated that
The manifold $Sp(2n, \mathbb{R})$ is diffeomorphic to the Cartesian product of the unitary group $U(n)$ with a vector space of dimension $n(n+ 1)$.
And then the authors explained that
The key to the proof is given by the polar decomposition: an invertible operator $A$ on a Euclidean space can be represented uniquely in the form of a product $S \cdot U$ of an invertible symmetric positive operator $S=\left(A A^*\right)^{1 / 2}$ and an orthogonal operator $U=S^{-1} A$. For symmetric operators $A$ acting on the underlying real space $\mathbb{R}^{2 n}$ of the Hermitian space $\mathbb{C}^n$, the operators $U$ turn out to be unitary, and the logarithms $\ln S$ of the operators $S$ fill out the $n(n+1)$ dimensional space of symmetric Hamiltonian operators.
I actually cannot quite understand what the authors are talking about here. It seems that they want to show that after the polar decomposition, a symplectic matrix $A$ is decomposed into a unitary matrix $U$ and a symmetric positive definite matrix $S$, but $U$ actually satisfies $$ U J U^{-1} = S^{-1}A J A^{\top} S^{-\top} = S^{-1} J S^{-\top} $$ which is not the condition of being a complex operator. So in what sense is $U$ "unitary"?
And at the same time, I don't know why "space of symmetric Hamiltonian operators" is of dimension $n(n+1)$. Does this actually mean "symmetric complex matrix" instead of Hermitian matrix?