# The automorphism group of a symplectic symmetric space

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

$$\\$$

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 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.