Let $G = \mathrm{SP}(2n,\mathbb{R})$, $\mathfrak{g} = \mathfrak{sp}(2n,\mathbb{R})$, $\mathfrak{h} = \mathfrak{sp}(2m,\mathbb R) \leq \mathfrak{g}$ and $\mathfrak q= \mathfrak h^{\perp}$. We know that $\mathfrak h$ is of the form
\begin{align*}
\begin{bmatrix}
\ast & 0 & \ast & 0 \\
0 & 0 & 0 & 0 \\
\ast & 0 & \ast & 0 \\
0 & 0 & 0 & 0
\end{bmatrix}
\end{align*}
and so $\mathfrak q$ is of the form
\begin{align*}
\begin{bmatrix}
0 & \ast & 0 & \ast \\
\ast & \ast & \ast & \ast \\
0 & \ast & 0 & \ast \\
\ast & \ast & \ast & \ast
\end{bmatrix},
\end{align*}
with $0$ is a $m \times m$ matrix. Now i am interested in a condition on $m$ such that there is an elliptic and regular element in $\eta$ in $q$.
(An element is called regular if it is conjugatet to a compact subalgebra of $\mathfrak g$. An element ist called regular if the centraliser in G hase dimension rank of $\mathfrak g$ (in that case equal n).)
Here we will have a the CSA $\mathfrak b$ given by
\begin{align*}
\begin{bmatrix}
0 & D \\
-D & 0
\end{bmatrix}
\end{align*}
with a diagonal matrix $D$. (You can obtain this by notice that the maximal torus in $G$ is $\mathrm U(n)$)
There are the following known facts
if $D'$ is regular then the diagonal elements $d_i$ in the Diagonal Block $D$ of $D'$ are never zero and pairwise different.
the centraliser of a regular $D'$ in $\mathfrak b$ is given by \begin{align*} C_{D'}(G) = \left\{ \begin{bmatrix} 0&D \\ -D&0 \end{bmatrix}, \begin{bmatrix} D&0 \\ 0&D \end{bmatrix} : D \text{is diagonal} \right\} \end{align*}
the maximal possible rank of an element in $\mathfrak q$ is equal to $2n$ if $2m \leq n$.
For $X$ in $\mathfrak{b}$ I want to find an element $g$ in $G$, such that: $g X g^{-1} \in \mathfrak{g}$.
I have tried to compute such an element $g$, but I failed. Then I read something about Hamiltonian methods, but I could not find a helpful statement. And of cause I had a look at adjoint-orbits, but I was not able to find anything.
So now i hope you can help me finding such an element.