I'm looking at the definition of the Conley-Zehnder index, where it is important to look at the group $$\text{Sp}(2n)^* := \{ A \in \text{Sp}(2n) | \det (A-\text{Id}) \neq 0 \}$$and its complement $$\Sigma := \{ A \in \text{Sp}(2n) | \det (A - \text{Id}) = 0\}.$$I began to wonder whether $\Sigma$ is a hypersurface of $\text{Sp}(2n)$, so I considered the map $f: \text{Sp}(2n) \to \mathbb{R}$, $f(A) := \det ( A - \text{id})$, and I computed, using the Jacobi formula, that at a matrix $A \in \Sigma$, the differential $df_A : T_A \text{Sp}(2n) = A \cdot \mathfrak{sp}(2n) \to \mathbb{R}$ is $$df_A (A \cdot H) = \frac{d}{dt}|_{t = 0} f(A + tAH) = \frac{d}{dt}|_{t = 0} \det (A - \text{Id} + tAH ) = \text{tr} ( \text{adj}(A - \text{Id}) \cdot AH ),$$ for $H \in \mathfrak{sp}(2n).$ If $\text{rank}(A - \text{Id}) < 2n-1$, then $\text{adj}(A - \text{Id}) = 0$, so the differential is zero.
Therefore, it seemed logical to me that if $\text{rank}(A - \text{Id}) = 2n-1,$ then $df_A \neq 0$ and $A$ is a regular point of $\Sigma.$ I think Arnold says something along those lines in his paper "Characteristic Classes Entering in Quantization Conditions" for the Lagrangian Maslov Index (at the beginning of Section 2.1 - The Singular Cycle). However, I wasn't able to prove this because I know nothing about $\text{adj}(A - \text{Id})$ except that it's nonzero.
Is it true that if $\text{rank}(A - \text{Id}) = 2n-1,$ then $A$ is a regular point of $\Sigma$?
For what it's worth: I was able to boil the question down to pure linear algebra, though: if $K$ is a nonzero $2n \times 2n$ matrix, can we choose a matrix $$H = \begin{pmatrix} M & S \\ S' & -M^T \end{pmatrix},$$where $S, S'$ are symmetric, such that $\text{tr}(KH) \neq 0$? Unfortunately, I don't think that the question stated as is has an affirmative answer, because of the restraints on $H$ and the fact that $K$ can be pretty much anything.
