Connectedness of conjugacy orbits

1. A centraliser element of negative determinant. Let $Z\in C(A)$ with $\det Z<0$ and $P\in GL_n(\mathbb{R})$ with $\det P<0$. Then $\det(PZ)>0$ and
$$(PZ)A(PZ)^{-1}=PZAZ^{-1}P^{-1}=PAP^{-1},$$
because $ZAZ^{-1}=A$. So every element of $\mathcal{O}(A)$ can be written $QAQ^{-1}$ with $\det Q>0$, i.e.
$$\mathcal{O}(A)=\{QAQ^{-1}:Q\in GL_n^+(\mathbb{R})\}.$$
The map $Q\mapsto QAQ^{-1}$ is continuous on $GL_n^+$ (the inverse is a rational function of the entries with denominator $\det Q$), and $GL_n^+$ is path connected (Problem 8). Hence $\mathcal{O}(A)$ is path connected.

2. The orbit of $J$.

*Centraliser.* A matrix $Z=\begin{pmatrix}p&q\\r&s\end{pmatrix}$ commutes with $J$ iff $ZJ=JZ$, which gives $s=p$ and $r=-q$: $Z=pI+(-q)J$... precisely $Z=\begin{pmatrix}p&q\\-q&p\end{pmatrix}$, with $\det Z=p^2+q^2>0$. So *every* element of $C(J)$ has positive determinant, and part 1 does not apply.

*Lower-left entry.* For $P=\begin{pmatrix}p&q\\r&s\end{pmatrix}\in GL_2$, a direct computation gives
$$PJP^{-1}=\frac1{\det P}\begin{pmatrix}pr+qs&-(p^2+q^2)\\ r^2+s^2&-(pr+qs)\end{pmatrix}.$$
So the lower-left entry of $PJP^{-1}$ is $\dfrac{r^2+s^2}{\det P}$, which has the sign of $\det P$ (as $r^2+s^2>0$ for an invertible matrix).

*Conclusion.* The continuous function "lower-left entry" takes positive values on $\{PJP^{-1}:\det P>0\}$ and negative values on $\{PJP^{-1}:\det P<0\}$, both non-empty. So $\mathcal{O}(J)$ is the disjoint union of two non-empty open (relatively) sets $\{c>0\}$, $\{c<0\}$. Each is connected (image of $GL_2^+$, resp. $GL_2^-=D\cdot GL_2^+$). Hence $\mathcal{O}(J)$ has exactly two components. (This matches Problem 28, since $\mathcal{O}(J)=\{A:A^2=-I\}$: any real $2\times2$ matrix with $A^2=-I$ is conjugate to $J$.)

Remark. The criterion in part 1 is sufficient but not necessary for connectedness of an orbit. It captures the intuition that an orbit of the *connected-component group* $GL_n^+$ is connected, and the orbit of $GL_n^-$ is its mirror image; the orbit is connected exactly when these two images overlap.

Note. The source solution calls the orbit "$\{A:A^2=-I\}$ with two components (previous problem)"; the explicit formula for $PJP^{-1}$ above proves both the identification of the components with the orientation $\operatorname{sign}\det P$ and that conjugating by an orientation-reversing $P$ really changes the component.