Given $\mathbb{R}^{2N}$ equipped with a symplectic form $\Omega^{ab}$ and a compatible symmetric, positive definite, bilinear form $G^{ab}$, we can look at the symplectic group $\mathrm{Sp}(2N,\mathbb{R})$ together with its subgroup $\mathrm{U}(N)$ given by those elements $M\in\mathrm{Sp}(2N,\mathbb{R})$ that also preserve the metric $G^{ab}$. In short, we have $\mathrm{U}(N)=\mathrm{Sp}(2N,\mathbb{R})\cup\mathrm{O}(N)$.

I would like to get a better understanding of the quotient $\mathrm{Sp}(2N,\mathbb{R})/\mathrm{U}(N)$. I read that as a manifold it is isomorphic to the so called Siegel disk, but could not find a good introduction on the geometry.

In order to understand this a little bit better, I tried to study it from the perspective of the algebra. I defined a positive-definite metric on the Lie algebra given by $\langle A,B\rangle=\mathrm{tr}(AB^\intercal)$ which is not invariant, but useful. In particular, I can define the orthogonal complement $\perp\!\!\mathrm{u}(N)\subset\mathrm{sp}(2N,\mathbb{R})$ containing those elements that are orthogonal to the generators $\mathrm{u}(N)$ of $\mathrm{U}(N)$.

At this point, I'm trying to interpret $\mathrm{Sp}(2N,\mathbb{R})$ as fiber bundle over $\exp(\perp\!\!\mathrm{u}(N))$ with fibers isomorphic to $\mathrm{U}(N)$. This means I'm thinking of a general element $M\in \mathrm{Sp}(2N,\mathbb{R})$ as being expressed as product \begin{align} M=e^Au\quad\text{where}\quad A\in\perp\!\!\mathrm{u}(N)\quad\text{and}\quad u\in\mathrm{U}(N)\,. \end{align} I believe that this provides a suitable set of "generalized coordinates" in a neighborhood around the identity, but I don't have any intuition for global obstructions/etc.

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    $\begingroup$ This is all quite standard. Check Helgasson's book on symmetric spaces. $\endgroup$ – Moishe Kohan Nov 23 '17 at 4:40
  • $\begingroup$ Ok, great. This was helpful. $\endgroup$ – LFH Dec 13 '17 at 7:36

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