2

It wouldn't seems so to me since the universal cover is basically the space of all paths from the identity to various points, since the orthogonal group has two connected components wouldn't it be impossible to define paths from the identity to the other component of the group, making it impossible to define a universal cover? Its not that I read anywhere that the Pin group is the universal cover I just want to double check.

Marc van Leeuwen
  • 115,048
Eben Cowley
  • 637
  • 2
    It's unusual to talk about the universal cover of a space that isn't path-connected. Most people would say that (when $n \ge 3$) the spin group is the universal cover of the special orthogonal group. – Qiaochu Yuan Jun 03 '19 at 21:30

2 Answers2

3

First of all, for $n\ge 3$, $\pi_1(SO(n))\cong {\mathbb Z}_2$, see for instance this discussion. By the definition, $Spin(n)$ is the connected double cover of $SO(n)$. Hence, $Spin(n)\to SO(n)$ is the universal covering of $SO(n)$ as long as $n\ge 3$. For $n=2$ this of course fails since $SO(2)\cong S^1$ and its universal cover is ${\mathbb R}$.

Moishe Kohan
  • 97,719
1

I'm no expert on this, but my understanding is as follows. A universal covering space is usually defined over a path connected and locally path connected space, and produces a simply connected space which has the local properties of the original space.

Spin is the UCS for the special orthogonal group, which is path etc connected. For the orthogonal group, you need a covering space which has separate components mapping to the components of O(n), but within each component, the covering space is simply connected.

It is more revealing to look at how the Spin space makes SO(n) simply connected. If you take two rotations in 2-space, which turn the space through 180 degrees clockwise and anticlockwise, you get the same rotation. But if you try to generate paths in rotation space, that is to generate the two rotations continuously starting from the identity map, your paths consist of distinct sets of clockwise and anticlockwise rotations. You cannot continously deform these paths into one another: SO(2) is not simply connected.

But if you take a space in which rotation through 180 degrees clockwise does not give you the same end point as rotation through the same angle anticlockwise, you eliminate the problem, because now the endpoints of the two paths are not the same. So you need a covering space in which rotation through 360 degrees doesn't bring you back to the starting point, and this is what spinor space essentially does, if I've understood correctly. Does this help?

Rory Allen
  • 41