Let $\text{Sp}(N,\mathbb{R})$ denote the Lie group of $2N\times 2N$ symplectic matrices. Fix a particular $A\in \text{Sp}(N,\mathbb{R})$ and consider the following Lie subgroup:
$$\text{Sp}_A(N,\mathbb{R}) = \{M\in\text{Sp}(N,\mathbb{R}) : [A,M]=0\} = \ker \text{Ad}_A\cap \text{Sp}(N,\mathbb{R})$$ with $[X,Y]=XY-YX$ the usual commutator and $\text{Ad}_A:\text{GL}(2N,\mathbb{R})\to \text{GL}(2N,\mathbb{R})$, $X\mapsto [A,X]$.
Now let $\Pi_A:\text{Sp}_A(N,\mathbb{R})\to\text{Sp}_A(N,\mathbb{R})$ denote the standard representation of $\text{Sp}_A(N,\mathbb{R})$ on $\mathbb{R}^\text{2N}$.
Question: Is $\Pi_A$ completely reducible? That is, can $\Pi_A$ be decomposed into a finite sum of irreducible (finite-dimensional) representations. Does this depend on $A$?
If $\text{Sp}_A(N,\mathbb{R})$ were compact, the answer is of course yes. I am wondering if the non-compactness gets in the way.
I believe the answer is yes, but I want to be sure I am not missing an obvious issue (and moreover, I am interested if this follows from a more general result in representation theory). My reasoning is as follows. It is known (https://arxiv.org/abs/1307.2403) that $\mathbb{R}^{2N}$ may be decomposed into a direct sum of $A$-invariant symplectic subspaces. Following the recipe in that paper, I have been able to show that these invariant subspaces are also invariant subspaces of $\text{Sp}_A(N,\mathbb{R})$ (hence $\Pi_A$ is reducible). Thus, it appears that the restrictions of $\text{Sp}_A(N,\mathbb{R})$ to these subspaces would constitute irreducible summands of $\Pi_A$. Unfortunately, I'm not sure how to show that these summands are irreducible.
I'm semi-new to the technicalities of the representation theory of Lie groups and Lie algebras (only having experience through work in physics so far), so apologies if this is trivial!