# When does $S \cap GL_4(\mathbb R)$ have precisely two connected components where $S \subset M_4(\mathbb R)$ is a linear subspace?

This is related to the question How many connected components for the intersection $S \cap GL_n(\mathbb R)$ where $S \subset M_n(\mathbb R)$ is a linear subspace? I asked. There is a nice example in the answer to show the intersection does not need to have two connected components as $GL_n(\mathbb R)$. The comment below by Travis is also very helpful.

Let $S \subset M_n(\mathbb R^n)$ be a linear subspace. What property should $S$ process such that $S \cap GL_n(\mathbb R)$ has two connected components?

I realized the question I asked before (crossed out above) might be too general to answer. Since it has not been answered, let me ask the specific question I am considering.

The question is now cross-posted here at MO.

Let $A \in M_4(\mathbb R)$ and $A = (e_2, x, e_4, y)$ where $e_2, e_4$ are standard basis in $\mathbb R^4$ and $x,y$ are undetermined variables. Let $\phi, \psi: M_4(\mathbb R) \to \mathbb R^4$ be linear maps defined by \begin{align*} &\phi: B \mapsto (AB-BA) e_1, \\ &\psi: B \mapsto (AB-BA)e_3. \end{align*} The subspace $S$ I am interested in is the intersection of kernels of the two linear maps, i.e., $S :=\text{ker}(\phi) \cap \text{ker}{\psi}$. In other words, the elements in $S \cap GL_4(\mathbb R)$ would preserve the structure of first and third columns of $A$ by conjugation, i.e., $(B^{-1}AB) e_1 = e_2, (B^{-1}AB)e_3 = e_4$ for $B \in S \cap GL_4(\mathbb R)$. I would like to determine:

1. whether there exists $A$ with eigenvalues all lying on the left open half plane of $\mathbb C$, i.e., with negative real parts ( we can freely choose $x, y$) such that $S \cap GL_4(\mathbb R)$ has precisely two connected components or precisely one component.
2. If there exists $A$, such that $\{V^{-1} A V: V \in S \cap GL_4(\mathbb R)\}$ is connected.

Edit 1: If $S \cap GL_4(\mathbb R)$ has precisely two connected components, I guess they should be $S \cap GL_4(\mathbb R)_+$ and $S \cap GL_4(\mathbb R)_-$. So if $V \in S$ and $\det(V) > 0$, $V$ should be path-connected with $I$. It is not hard to check the condition implies $V = (v_1, Av_1, v_3, Av_3)$. Since $e_2 = Ae_1, e_4 = Ae_3$, the question is can we continuously change $v_1, v_3$ to $e_1, e_3$ such that $(v_1, Av_1, v_3, Av_3)$ stay linearly independent during the process.

Edit 2: If the intersection only have one component, then the $2^{\text{nd}}$ question is immediate. Or if $A$ has two components but with a real eigenvalue, then $2$ should hold too. However, it is possible $2$ can be solve directly which I could not see.

Edit 3: I crossed out the restrictions I put on $A$ although I feel this should not matter too much. The second question is newly added which is actually my end question. Before I had a feeling there should be some "special" $A$ such that the intersection would only give $1$ or $2$ components. As mentioned above, it's highly possible we can directly attack the second question.

• To be clear, do you mean precisely two components? – Travis Jul 22 '18 at 0:52
• I didn't consider it would have one component. Actually I want to see when it has precisely $1$ component or precisely $2$ components. – user1101010 Jul 22 '18 at 0:59
• For $n$ odd, it's not possible to have just one component: If $A$ is in $S \cap \operatorname{GL}_n(\Bbb R)$, then so is $-A$, but $\det (-A) = - \det A$, so $A, -A$ lie in different components of $GL_n(\Bbb R)$. For $n$ even, it is possible: Consider $S := \left\{\pmatrix{a&-b\\b&a} : a, b, \in \Bbb R\right\}$. Then $\det \pmatrix{a&-b\\b&a} = a^2 + b^2$, so $S \cap \operatorname{GL}_n(\Bbb R) = S - \{0\}$, which is connected. – Travis Jul 22 '18 at 1:08