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

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

$$\text{exp} : \mathfrak{g} \to G$$ 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?

• Exponential map is always surjective on a neighborhood of the identity, and any neighborhood of the identity generates the entire group (for a connected group), so the answer to the second question is at least yes. – user148177 Dec 4 '14 at 10:49
• @user148177: I think Uther wants to know if one can strengthen your argument to say not only that a possibly varying finite number of exponential factors will do for every individual $g\in\text{Sp}(2n,{\mathbb R})$, but whether there is a universal bound on the number of factors. – Hanno Dec 4 '14 at 10:53
• Yes, I understand my comment is only a partial answer. – user148177 Dec 4 '14 at 10:57
• The number of products necessary is what I'm interested in. The answer to the second question is of course, yes. Sorry for the poor wording. – Matta Dec 4 '14 at 10:59
• Also, the answer to the first question is no, since Sp(2, R) = SL(2, R). For the exponential map to be surjective implies existence of square roots, and for example diag(-4, -1/4) has no square root in R. In general you can take a matrix diag(1/a, 1/b, 1/c, ..., a, b, c, ...) where the a, b, c are distinct and negative. – user148177 Dec 4 '14 at 11:09

In his comment, user148177 already explained that the exponential function of $\text{Sp}_{2n}({\mathbb R})$ is not surjective.
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 $X\in\text{Sym}_{2n}({\mathbb R})\cap{{\mathfrak s}{\mathfrak p}}_{2n}({\mathbb R})$.
As a compact Lie group $SO_{2n}({\mathbb R})\cap\text{Sp}_{2n}({\mathbb R})$ has surjective exponential, so $\text{Sp}_{2n}({\mathbb R})=\text{exp}({\mathfrak s}{\mathfrak p}_{2n}({\mathbb R}))^2$.