**Why is the automorphism group of a sympelctic symmetric space a Lie group?**


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A symplectic symmetric space is a triple $(M, \omega, s)$, where $(M, \omega)$ is a symplectic manifold and $ s \; \colon M \times M \to M $, $(x, y) \mapsto s\_x(y)$, is such that $s\_x$ is an involutive symplectic diffeomorphism with an isolated fixed point at $x$ and $s\_xs\_ys\_x = s\_{s\_x(y)} \; \forall \; x, y \in M\; $ $\big($this can be read as $s\_x s = ss\_x \; \forall x \in M$$\big)$.


An interesting fact about a symplectic symmetric space is the one that once $x$ is an isolated fixed point of $s\_x$ and $s\_x^2 = Id$, $ds\_{x (x)} = -Id\_{T\_xM}$. Furthermore, a symplectic symmetric space admits a unique affine connection $\nabla$ on it such that $s\_x$ is an affinity for every $x \in M$ and such that $\nabla \omega = 0$. This connection has no torsion and is thus a symplectic connection.


The automorphism group $Aut(M, \omega, s)$ of $(M, \omega, s)$ is the group of symplectic automorphisms $\varphi$ of $(M, \omega)$ that satisfy $\varphi \circ s\_x = s\_{\varphi(x)} \circ \varphi \; \forall \; x \in M$ $\big($ or, in other words, $\varphi \circ s = s\circ \varphi$ $\big)$. It's simple to check that this group is the intersection of the symplectic automorphism group of $(M, \omega)$ and the affine group of $(M, \nabla)$.


My question then is: why is $Aut(M, \omega, s)$ a Lie group?



The automorphism group of a symplectic symmetric space
The affine group of $(M,\nabla)$ is a Lie group $G$ by Kobayashi's theorem that shows that the automorphism group of any affine connection is a Lie group (see Kobayashi and Nomizu's *Foundations of Differential Geometry*). The dimension of $G$ is at most $n+n^2$ (where $n=\dim M$). 


The subgroup $H$ of $G$ consisting of those elements of $G$ that preserve the symplectic structure on $M$ is clearly a closed subgroup, and since every closed subgroup of a Lie group is a Lie group (see, for example, Helgason's proof in his *Differential Geometry, Lie Groups, and Symmetric Spaces*), it follows that $H$ is a Lie group as well.


