Complex structures on the set A² = -I

Idea. Describe $\mathcal{J}$ by one equation in three variables and solve for $b$.

Step 1: $A\in\mathcal{J}$ has trace $0$ and determinant $1$. If $A^2=-I$, the eigenvalues satisfy $\lambda^2=-1$, so they are $\pm i$ (complex, not real, so the two eigenvalues of the real $2\times2$ matrix are $i$ and $-i$ as it has no real eigenvalue). Hence $\operatorname{tr}A=i+(-i)=0$ and $\det A=i\cdot(-i)=1$. (Alternatively: by Cayley-Hamilton $A^2-(\operatorname{tr}A)A+(\det A)I=0$, so $(\operatorname{tr}A)A=(\det A+1)I$; if $\operatorname{tr}A\ne0$, $A$ would be a scalar matrix with $A^2=-I$, impossible over $\mathbb{R}$.)

Step 2: equation. Hence $A=\begin{pmatrix}a&b\\c&-a\end{pmatrix}$ and $A^2=(a^2+bc)I$. So
$$\mathcal{J}=\{(a,b,c)\in\mathbb{R}^3:a^2+bc=-1\},$$
(a two-sheeted hyperboloid $bc=-1-a^2$ in the coordinates $(a,b,c)$).

Step 3: $c\ne0$. From $bc=-(1+a^2)<0$ we get $c\neq0$ and $b=-(1+a^2)/c$. So $\mathcal{J}=\mathcal{J}_+\sqcup\mathcal{J}_-$ with $\mathcal{J}_\pm=\mathcal{J}\cap\{\pm c>0\}$ (open in $\mathcal{J}$, disjoint, non-empty: they contain $\pm J$ with $J=\begin{pmatrix}0&-1\\1&0\end{pmatrix}$).

Step 4: each part is a plane. The map
$$\mathcal{J}_+\to\mathbb{R}\times(0,\infty),\quad A\mapsto(a,c)$$
is continuous, with continuous inverse $(a,c)\mapsto\begin{pmatrix}a&-(1+a^2)/c\\c&-a\end{pmatrix}$. So $\mathcal{J}_+\cong\mathbb{R}\times(0,\infty)\cong\mathbb{R}^2$, which is connected. Similarly $\mathcal{J}_-\cong\mathbb{R}\times(-\infty,0)\cong\mathbb{R}^2$.

Conclusion. $\mathcal{J}$ has exactly two connected components, each homeomorphic to $\mathbb{R}^2$.

Remark. The sign of $c$ is the *orientation* of the complex structure defined by $A$ (a map $A$ with $A^2=-I$ makes $\mathbb{R}^2$ a complex line with $i$ acting as $A$; the structure is compatible with the standard orientation iff $c>0$). This set is the "two-sheeted hyperboloid" $\simeq GL_2(\mathbb{R})/GL_1(\mathbb{C})$.