Noncontractible loop in $\operatorname{SL}(2,\mathbb{C})$?

Question

It is well-known that $\operatorname{SL}(2,\mathbb{C})$ is a simply connected Lie group. However I seem to be able to think of a loop that is not contractible. Where could my mistake be?

Let $g(t) = \operatorname{diag}(e^{it},e^{-it})$ be defined from $0$ to $2\pi$. It is a loop based at identity, and its eigenvalues trace two loops around the punctured complex plane $\mathbb{C}\setminus\{0\}$. Deforming this loop to the constant loop at identity would induce those eigenvalue loops to contract to the constant loop at $1$. But this is impossible due to the puncture in the complex plane. (The eigenvalues can never be zero because the deformation must stay in $\operatorname{SL}(2,\mathbb{C})$.)

Edit: For an answer, please state an explicit contraction as that is the only way to clear my doubts completely.

Answer 1

Write

$$ D(t)=\begin{bmatrix} \exp(it) & 0 \\ 0 & \exp(-it)\end{bmatrix}, \quad C(s)=\begin{bmatrix} \cos(s) & -\sin(s) \\ \sin(s) & \phantom{-}\cos(s) \end{bmatrix}. $$

Define

$$ H(t,s)=\begin{cases} D(t) & 0\le t\le \pi \\ C(s)D(t)C(s)^{-1} & \pi\le t\le 2\pi.\end{cases} $$

for $0\le s\le\pi/2$. Observe that $H(t,0)=D(t)$ for all $t$ whereas $H(t,\pi/2)$ traces out a semicircle for $0\le t\le \pi$ and then backtracks for $\pi\le t\le 2\pi$. Thus we have homotoped $D(t)$ to a more obviously contractible loop.

Note that conjugation doesn't affect the spectrum, so nothing happens to that along the way. As the others have said, the way to interpret the spectrum is as two eigenvalues moving away from $1$ on the unit circle in opposite directions, then instead of passing over each other interpret them as bouncing off of each other at $-1$. This loop (in configuration space) is clearly contractible.

If you want to understand where I got the example from, I can explain that too, but it requires some background in quaternions. (Although all of the background can be gleaned from past posts of mine that I can link if need be.)

This also has a visual interpretation. If we interpret $\mathrm{SU}(2)$ as $\mathrm{Spin}(3)$ (the double cover of $\mathrm{SO}(3)$), then this copy of $S^1$ in $\mathrm{Spin}(3)$ double covers a copy of $S^1$ in $\mathrm{SO}(3)$. Visualize this as two rotations around an axis in three dimensions. We can slowly "edit" this "animation" by rotating the axis itself during the second revolution until the axis is upside down - rotating around the axis upside down goes backwards. This is essentially equivalent to the plate trick.

Answer 2

I suspect that your error has to do with implicitly thinking of the eigenvalues as an "ordered pair" instead of as an unordered pair of complex numbers.

In fact, the space of unordered pairs of complex numbers which multiply to 1 is in natural correspondence with the space of quadratic polynomials of the form $x^2+cx+1$, for $c \in \mathbb C$, which is just homeomorphic to $\mathbb C$ itself.

Of course, that's just a space which $\operatorname{SL}_2(\mathbb C)$ surjects onto, so it doesn't yield a direct answer to how to contract your loop. What I think it does do is explain why your intuition is faulty.

Answer 3

To expand on Dustan Levenstein's answer, note that because the eigenvalues are unordered, we don't have to treat the eigenvalues as forming two loops around the origin. At $t=0$, the eigenvalues start at 1. Then the eigenvalues move in opposite directions (one clockwise and one counterclockwise) until, at $t=\pi$, they are both at -1. Now, because the eigenvalues are unordered, then instead of completing two loops, both eigenvalues can go back to 1 using the path they started. Now there are no loops, and the contradiction is averted.