Rank-one idempotents and involutions of M_2(R) are cylinders

Idea. In $M_2$ a matrix is "almost determined" by its trace and determinant. Fixing them cuts $\mathbb{R}^4$ down to a surface, which the coordinates $(z,w)$ describe explicitly.

Step 0: the dictionary. Writing $x=x_1+ix_2$, one checks that the real part of $zx+w\bar x$ is $(z_1+w_1)x_1+(w_2-z_2)x_2$ and the imaginary part is $(z_2+w_2)x_1+(z_1-w_1)x_2$, where $z=z_1+iz_2$, $w=w_1+iw_2$. With the given $z,w$ these coefficients are $a,b,c,d$. Moreover, directly from the formulas,
$$\operatorname{tr}A=a+d=2\operatorname{Re}z,\qquad |z|^2-|w|^2=\tfrac14\bigl[(a+d)^2+(c-b)^2-(a-d)^2-(b+c)^2\bigr]=ad-bc=\det A.$$
Conversely, $a=\operatorname{Re}z+\operatorname{Re}w$, $d=\operatorname{Re}z-\operatorname{Re}w$, $c=\operatorname{Im}z+\operatorname{Im}w$, $b=\operatorname{Im}w-\operatorname{Im}z$, so $(z,w)\mapsto A$ is the inverse linear map.

Step 1: $\mathcal I$ in terms of trace and determinant. We claim $A\in\mathcal I\iff\operatorname{tr}A=1$ and $\det A=0$.
- ($\Rightarrow$) $A$ has rank $1$, so $\det A=0$ and $A\ne0$. Cayley-Hamilton gives $A^2-(\operatorname{tr}A)A+(\det A)I=0$, i.e. $A^2=(\operatorname{tr}A)A$. Since $A^2=A$ and $A\ne0$ we get $\operatorname{tr}A=1$.
- ($\Leftarrow$) Cayley-Hamilton with $\operatorname{tr}A=1$, $\det A=0$ gives $A^2=A$. Also $A\neq0$ (its trace is $1$) and $\det A=0$, so the rank is exactly $1$.

Step 2: $\mathcal J$ in terms of trace and determinant. We claim $A\in\mathcal J\iff\operatorname{tr}A=0$ and $\det A=-1$.
- ($\Rightarrow$) The minimal polynomial of $A$ divides $t^2-1=(t-1)(t+1)$. It cannot be $t-1$ or $t+1$ because $A\ne\pm I$. So it equals $t^2-1$, which has degree $2$, hence it is the characteristic polynomial: $\operatorname{tr}A=0$, $\det A=-1$.
- ($\Leftarrow$) Cayley-Hamilton gives $A^2-0\cdot A+(-1)I=0$, i.e. $A^2=I$, and $A\ne\pm I$ since $\operatorname{tr}(\pm I)=\pm2\ne0$.

Step 3: coordinates on $\mathcal I$. By Steps 0 and 1, $A\in\mathcal I$ iff $\operatorname{Re}z=\frac12$ and $|z|^2=|w|^2$. Write $z=\frac12+iq$ with $q\in\mathbb{R}$. Then $|w|=|z|=\sqrt{\frac14+q^2}>0$, so $w\ne0$ and $u=w/|w|\in S^1$ is defined. Define
$$\Phi:\mathcal I\to\mathbb{R}\times S^1,\qquad A\mapsto\Bigl(\operatorname{Im}z,\ \frac w{|w|}\Bigr).$$
$\Phi$ is continuous (coordinates are linear in $A$ and $w\neq0$). Its inverse sends $(q,u)$ to the matrix with $z=\frac12+iq$, $w=\sqrt{\frac14+q^2}\,u$; this is continuous, and the matrix obtained satisfies $\operatorname{Re}z=\frac12$, $|z|=|w|$, hence lies in $\mathcal I$. So $\Phi$ is a homeomorphism: $\mathcal I\cong\mathbb{R}\times S^1$.

Step 4: coordinates on $\mathcal J$. Now $\operatorname{Re}z=0$ and $|z|^2-|w|^2=-1$. With $z=iq$ this reads $|w|^2=q^2+1$, so $|w|\ge1$ and again $w\ne0$. The same argument with $|w|=\sqrt{1+q^2}$ gives a homeomorphism $\mathcal J\to\mathbb{R}\times S^1$, $A\mapsto(\operatorname{Im}z,w/|w|)$.

Check. The matrix $\operatorname{diag}(1,0)$ has $a=1$, $b=c=d=0$, hence $z=w=\frac12$: it corresponds to $q=0$ and $u=1$ in $\mathcal I$. The matrix $\operatorname{diag}(1,-1)$ has $z=0$, $w=1$: it is in $\mathcal J$ with $q=0$, $u=1$ (here $|w|=\sqrt{1+0}=1$).

Remark. A rank-one idempotent is determined by its image $\ell_1$ and its kernel $\ell_2$, which are two distinct lines of $\mathbb{R}^2$. The set of pairs of distinct lines is $(\mathbb{RP}^1\times\mathbb{RP}^1)\setminus\text{diagonal}\cong S^1\times(S^1\setminus\{pt\})\cong S^1\times\mathbb{R}$, confirming the cylinder. An involution $\ne\pm I$ is determined by its $(+1)$-eigenline and $(-1)$-eigenline, again a pair of distinct lines.

Lesson. In $M_2$, conjugacy classes with prescribed trace and determinant are surfaces, and the coordinates $(z,w)$ turn the two equations into explicit formulas.