Is the exponential map for $\text{Sp}(2n,{\mathbb R})$ surjective?

Question

For $\mathfrak{g} := {\mathfrak s}{\mathfrak p}(2n,\mathbb{R})$ and $G = \text{Sp}(2n,{\mathbb R})$, is the exponential map

\begin{equation} \text{exp} : \mathfrak{g} \to G \end{equation} surjective?

If not then is it possible to access all of the symplectic group $\text{Sp}(2n,\mathbb{R})$ via a product, i.e. $e^Xe^Y$, such that $X,Y \in {\mathfrak s}{\mathfrak p}(2n,\mathbb{R})$? If so what is the minimum number of products needed?

Answer 1

In his comment, user148177 already explained that the exponential function of $\text{Sp}_{2n}({\mathbb R})$ is not surjective.

Two factors, however, suffice:

Theorem (Polar decomposition) Any $g\in\text{Sp}_{2n}({\mathbb R})$ can uniquely be written as a product $g = h\cdot\text{exp}(X)$ with $h\in SO_{2n}({\mathbb R})\cap\text{Sp}_{2n}({\mathbb R})$ and $X\in\text{Sym}_{2n}({\mathbb R})\cap{{\mathfrak s}{\mathfrak p}}_{2n}({\mathbb R})$.

Reference: Hilgert-Neeb: The structure and geometry of Lie groups, Proposition 4.3.3.

As a compact Lie group $SO_{2n}({\mathbb R})\cap\text{Sp}_{2n}({\mathbb R})$ has surjective exponential, so $\text{Sp}_{2n}({\mathbb R})=\text{exp}({\mathfrak s}{\mathfrak p}_{2n}({\mathbb R}))^2$.

Answer 2

@Matta: The answers to both of your questions can alternatively be found in Dragt's book "Lie Methods for Nonlinear Dynamics with Applications to Accelerator Physics" https://www.physics.umd.edu/dsat/dsatliemethods.html

In summary:

1. No, a counterexample is a matrix whose spectrum has values in the branch cut $] -\infty, 0] \subset \mathbb{C}$:

$$ \begin{pmatrix} -1 & -1 \\ 0 & -1 \end{pmatrix} $$

2. Yes, a detailed proof can be found in the reference above in the section "Exponential Representation of Symplectic Matrices".