Is the following subset of space of matrices connected, open?

Question

If $S=\left\{ A=\begin{bmatrix}A_1&0 \\ 0&A_2 \end{bmatrix} \in \mathbb{M}_4(\mathbb{C}): det A_1=detA_2 \right\}$. Then is this set open in $\mathbb{M}_{4}(\mathbb{C})$ with usual topology. If not then what are the interior points. I would like some hints towards this. Also if someone could suggest a book to study topology of space of complex matrices, it will be helpfull to me. Thank you.

I have an analytic map from open unit disc to the norm unit ball of space of matrices having image contained in $S$. That is why i am wondering about the interior of this set.

Answer 1

An idea: denote by $\;\mathcal B\;$ the set of all blocks-matrices in $\;M_4(\Bbb C)\;$, and define

$$f:\mathcal B\to\Bbb C\;,\;\;f\begin{pmatrix}A_1&0 \\ 0&A_2 \end{pmatrix}:=\det A_1-\det A_2$$

The above map is continuous (wrt the usual topologies in domain and codomain: the Euclidean one in $\;\Bbb C\;$ and the one in $\;\mathcal B\subset M_4(\Bbb C)\;$ inherited from the Euclidean one in $\;\Bbb C^4\;$), as it is apolynomial map in the entries of the matrix $\;A=\begin{pmatrix}A_1&0 \\ 0&A_2 \end{pmatrix}\;$.

Finally, just observe that $\;S=f^{-1}(\{0\})\;$ and thus $\;S\;$ is closed...

Answer 2

The usual topology on $M = \mathbb{M}_4(\mathbb{C})$ is the topology induced by any norm on $M$. We may take $\lVert (a_{ij}) \rVert = \max \lvert a_{ij} \rvert$. This shows that $S$ does not have any interior point because if $A \in S$, then $A_\epsilon = A + \epsilon R$, where $R = (r_{ij})$ is the matrix having $r_{14} =1$ and $r_{ij} =0$ else, satisfies $A_\epsilon \notin S$ and $\lVert A_\epsilon - A \rVert = \epsilon$.

The set $S$ is closed. Write $A=\begin{bmatrix}A_1&B_1 \\ B_2&A_2 \end{bmatrix}$ and define $$f : M \to \mathbb R, f(A) = \lVert B_1\rVert + \lVert B_2\rVert + \det A_1 - \det A_2 .$$ This is a continuous map and $S = f^{-1}(0)$.

The set $S$ is pathwise connected. In fact, it is star-shaped set with center $0$. If $A \in S$, then the line segment $u :[0,1] \to M$ connecting $0$ and $A$ (given by $u(t) = t A$) is contained in $S$.