Spinors - Groups and Double Cover of Lorentz Group

Question

As part of a project, I keep coming across a small nit-picking area regarding the spinor group $SU(2)$. The Lorentz group can be thought of as the group of rotations in $SO(1,3)$. I am under the impression that the spinors are a double cover of this Lorentz group. I keep getting confused as to whether we have the relation:

$SU(2) \times SU(2) = SO(1,3)$ or $SU(2) \times SU(2) = SO(3)$

Which case is correct and why?

Answer 1

Both of your isomorphisms are false. For instance, $SU(2)\times SU(2)$ is a compact group, while $SO(1,3)$ is a noncompact group. But, at least the two groups share the dimension (six). In contrast, $SO(3)$ is 3-dimensional, hence, it is not isomorphic to $SU(2)\times SU(2)$ either.

There are different spin groups for different Lie groups; thus you cannot talk about the spinor group, but different spinor groups. The setup is that you have a connected Lie group $G$ and a distinguished index 2 subgroup in its fundamental group $\pi_1(G)$. This data determines a connected Lie group $\tilde{G}$ which is admits a 2-fold nontrivial covering $\tilde{G}\to G$; by a slight abuse of notation, one says that $\tilde{G}$ is the spin group of $G$. The choice of the index 2 subgroup is usually canonical, so this notion of a spin-group of $G$ is well-defined.

To your examples:

1. The group $SO_+(1,3)$ is noncompact, its fundamental group is the same as $\pi_1(SO(3))\cong {\mathbb Z}_2$ since $SO(3)$ is the maximal compact subgroup of $SO_+(1,3)$. Hence, $SO(1,3)$ has a unique nontrivial 2-fold covering group called $Spin(1,3)$. The more common way to describe this groups is as follows: $$ SO_+(1,3)\cong PSL(2, {\mathbb C}), $$ hence, the spin-cover is $SL(2, {\mathbb C})\cong Spin(1,3)$.
2. The group $SU(2)=Spin(3)$, is the spinor group of $SO(3)$.

For other examples: $G=SO_+(2,1)$ has infinite cyclic fundamental group; hence, it again has a canonical index 2 subgroup of $\pi_1(G)$; the corresponding spin group is $SL(2, {\mathbb R})$.