Does anyone know any reference that treat $Sp(2n,\mathbb{R})$ detailedly? For $Sp(2n,\mathbb{R})$, I mean the subgroup of $GL(2n,\mathbb{R})$ that preserves the standard symplectic form $\sum_{i=0}^n dx_i \wedge dy_i$ on $\mathbb{R}^{2n}$. The Wikipedia or nLab pages state that $\pi_1 (Sp(2n,\mathbb{R}))\cong \mathbb{Z}$ without a proof. I guess this probably follows from some fibration but I cannot find reference for it. Also, does anyone know if there's a concrete geometric construction of the universal covering $\widetilde{Sp(2n,\mathbb{R})}$?
Thank you.
In Chapter 2 of McDuff and Salamon's Introduction to Symplectic Topology, a polar decomposition is proved for symplectic matrices (see also this MSE question.). Namely, any symplectic matrix $\Psi$ is expressible as a product of two symplectic matrices $S$ and $U$, $\Psi = S.U$, where $S$ is positive real symmetric and $U$ is orthogonal (hence unitary as it is also symplectic). Consequently, as topological spaces (and introducing some notations), $Sp(2n, \mathbb{R}) \cong S_+Sp(2n, \mathbb{R}) \times U(n, \mathbb{C})$.
As I explain below, it follows that $\pi_1(Sp(2n)) \cong \pi_1(S_+Sp(2n)) \times \pi_1(U(n)) \cong \{ e \} \times \mathbb{Z} \cong \mathbb{Z}$.
It also follows that the universal cover of $Sp(2n)$ is isomorphic to the product of the universal covers of $S_+Sp(2n)$ (which is itself) and of $U(n, \mathbb{C})$. That is, as topological spaces, $\widetilde{U(n)} \cong S_+Sp(2n) \times SU(n) \times \mathbb{R}$.
The group $S_+ Sp(2n, \mathbb{R})$ happens to be contractible. Indeed, given $S \in S_+Sp(2n)$, the spectral theorem for symmetric matrices assures that there exists a diagonal matrix $D$ and an orthonormal matrix $Q$ such that $S = QDQ^{-1}$. As $S$ is invertible (because symplectic), none of its eigenvalues (i.e. the diagonal entries in $D$) is $0$. As $S$ is positive, all its eigenvalues are (strictly) positive. Hence for any $r \in \mathbb{R}$, $D^r$ is well-defined by taking the $r$-th power of each of its entries, and we set $S^r = QD^rQ^{-1}$. As the map $(r, S) \mapsto S^r$ is continuous and as $S^0 = Id$, the group $S_+Sp(2n)$ is indeed contractible. In particular, it is simply-connected, hence equal to its universal cover.
The unitary group $U(n, \mathbb{C})$ is compact and rather well-understood. It is a classical result (outlined in the aforementioned book) that $\pi_1(U(n)) \cong \mathbb{Z}$. Namely, think of $U \in Sp(2n, \mathbb{R}) \cap O(2n, \mathbb{R})$ as a complex matrix $U \in U(n, \mathbb{C})$, the determinant function $det : U(n) \to U(1) \cong S^1 \subset \mathbb{C}^{\times}_*$ is surjective, with $SU(n) = det^{-1}(1)$. Hence, $U(n)$ is a $SU(n)$-bundle over $U(1)$, hence the exact sequence $1 \to SU(n) \to U(n) \to U(1) \to 1$. The associated long exact sequence of homotopy groups allows to prove that $\pi_1(U(n)) \cong \pi_1(SU(n)) \times \pi_1(U(1))$. It is a classical result (also presented in the book) that $\pi_1(SU(n)) \cong 1$ and that $\pi_1(U(1)) \cong \mathbb{Z}$.
In fact, $U(n)$ is (topologically, not as a group) the trivial $SU(n)$-bundle over $U(1)$. Indeed, given $U \in U(n)$, let $e^{i\theta} := det(U) \in U(1)$. The determinant map has a right inverse $s : U(1) \to U(n) : e^{i\theta} \mapsto diag(e^{i\theta}, 1, \dots, 1)$. Set $U_1 := (s \circ det)(U)$ and $U_2 := U_1^{-1}U \in SU(n)$. Then $U = U_1.U_2$, which gives a topological decomposition $U(n) \cong U(1) \times SU(n)$. Consequently, the universal cover of $U(n)$ is the product of that of $SU(n)$ (which is itself as it is simply-connected) with that of $U(1)$ (which is $\mathbb{R})$.