No continuous square root on GL(2,R)

Idea. On the circle of rotations $R_\theta=\cos\theta\,I+\sin\theta\,J$, the square root must be $\pm e^{i\theta/2}$, and there is no way to choose the sign continuously all the way around the circle: this is the usual non-existence of a continuous square root of $z\in S^1$.

Step 1: the commutant. Let $\theta\notin\pi\mathbb{Z}$. If $M$ commutes with $R_\theta=\cos\theta\,I+\sin\theta\,J$ then, as $\sin\theta\ne0$, $M$ commutes with $J=\begin{pmatrix}0&-1\\1&0\end{pmatrix}$. Writing $M=\begin{pmatrix}a&b\\c&d\end{pmatrix}$, the relation $MJ=JM$ gives $MJ=\begin{pmatrix}b&-a\\d&-c\end{pmatrix}$ and $JM=\begin{pmatrix}-c&-d\\a&b\end{pmatrix}$, hence $d=a$, $c=-b$. So $M=aI-bJ$ is of the form $xI+yJ$, which we identify with the complex number $x+iy$ (the map $xI+yJ\mapsto x+iy$ is a ring isomorphism since $J^2=-I$). Under this identification $R_\theta=e^{i\theta}$.

Step 2: sign choices. Suppose $S$ exists. For $\theta\in(0,\pi)\cup(\pi,2\pi)$ the matrix $S(R_\theta)$ commutes with $S(R_\theta)^2=R_\theta$, so it corresponds to a complex number $z$ with $z^2=e^{i\theta}$: $z=\varepsilon(\theta)e^{i\theta/2}$, $\varepsilon(\theta)\in\{\pm1\}$. Since $S$ is continuous and $\theta\mapsto e^{i\theta/2}$ is continuous, $\varepsilon$ is continuous with values in $\{\pm1\}$, hence constant on each of the two intervals: $\varepsilon=\varepsilon_1$ on $(0,\pi)$ and $\varepsilon=\varepsilon_2$ on $(\pi,2\pi)$.

Step 3: across $\theta=\pi$. By continuity of $S$ at $R_\pi$, the limits of $S(R_\theta)$ as $\theta\to\pi^-$ and $\theta\to\pi^+$ coincide. They are $\varepsilon_1e^{i\pi/2}=\varepsilon_1i$ and $\varepsilon_2e^{i\pi/2}=\varepsilon_2i$. So $\varepsilon_1=\varepsilon_2=:\varepsilon$.

Step 4: contradiction at $\theta=0\equiv2\pi$. As $\theta\to0^+$, $S(R_\theta)\to\varepsilon e^{0}=\varepsilon I$. As $\theta\to2\pi^-$, $S(R_\theta)\to\varepsilon e^{i\pi}=-\varepsilon I$. But $R_\theta\to I$ in both cases, so by continuity of $S$ both limits must equal $S(I)$: $\varepsilon I=-\varepsilon I$, which is absurd.

Remark. The obstruction is topological: $GL_2^+(\mathbb{R})$ deformation retracts onto $SO(2)\cong S^1$, which is not simply connected. On the simply connected set $\{A:\ A\text{ has no real negative eigenvalue}\}$ a continuous principal square root does exist (given by a convergent series).

Lesson. Non-simply connected domains obstruct continuous choices of roots and logarithms.