Understanding the set $M_1 \cap M_2$

Question

Let $ M_1= \left\lbrace \left( \begin{matrix} a & b & 0 \\ c & d & 0 \\ 0 & 0 & f \end{matrix} \right), a,d,f \in i\mathbb{R}, b = \overline{c} \right\rbrace $ be a subspace of skew-hermitian matrices.

Let $ M_2 = \lbrace g \lambda g ^{-1}, g \in SU(3)\rbrace $ such that $\lambda = \operatorname{diag}(i\lambda_1,i\lambda_2, i\lambda_3), \lambda_i \in \mathbb{R}.$

My first question is how can we describe the intersection of $M_1$ and $M_2$, in other words what are the matrices $g\in SU(3)$ s.t $g \lambda g ^{-1} \in M_2 $.

My second question is what are the connected components of $M_1 \cap M_2 $ ?

Answer 1

I am assuming $b = -\overline{c}$, so that the matrices in $M_1$ are indeed skew-Hermitian.

The matrices in $M_1$ are precisely the skew-Hermitian matrices with eigenvector $(0, 0, 1)$. Clearly, $(0, 0, 1)$ is an eigenvector of every matrix in $M_1$ (with eigenvalue $f$, as per the form in $M_1$'s definition). Conversely, skew-Hermitian matrices have orthogonal eigenspaces, and so if $\operatorname{span}(0, 0, 1)$ is invariant, then so is $\operatorname{span}((1, 0, 0), (0, 1, 0)) = \operatorname{span}(0, 0, 1)^\perp$. This gives us the required block-diagonal form.

So, we are looking for matrices $g\lambda g^{-1}$ in $M_2$ where one of the columns of $g$ is $(0, 0, 1)^\top$. We can permute the columns of $g$ provided we correspondingly permute the diagonal entries of $\lambda$ and we maintain an even permutation (to keep $\det g = 1$), hence we can assume without loss of generality that the third column of $g$ is $(0, 0, 1)^\top$.

The columns of $g$ are orthonormal, hence the other two columns will be orthogonal to the third column, giving us the familiar form $$g = \begin{pmatrix} p & q & 0 \\ r & s & 0 \\ 0 & 0 & 1 \end{pmatrix},$$ where $\begin{pmatrix} p & q \\ r & s\end{pmatrix} \in SU(2)$ (note that the determinant must be $1$, by simple calculation, so that $\det g = 1$). This seems like a fairly complete characterisation of the matrices $g$.

As for the connected components, the set is connected. The map \begin{align*} \Bbb{R}^3 \times SU(2) &\to M_1 \cap M_2 \\ (\lambda, h) &\mapsto \begin{pmatrix} h & \begin{matrix} 0 \\ 0 \end{matrix} \\ \begin{matrix} 0 & 0 \end{matrix} & 1 \end{pmatrix} \begin{pmatrix} i\lambda_1 & 0 & 0 \\ 0 & i\lambda_2 & 0 \\ 0 & 0 & i\lambda_3\end{pmatrix}\begin{pmatrix} h & \begin{matrix} 0 \\ 0 \end{matrix} \\ \begin{matrix} 0 & 0 \end{matrix} & 1 \end{pmatrix}^{-1} \end{align*} is continuous and surjective, making $M_1 \cap M_2$ connected.