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?


2 Answers 2


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 $SL(2, {\mathbb R})$.

  • $\begingroup$ Thank you for the information. $\endgroup$
    – Brad
    Mar 14, 2019 at 23:07
  • 1
    $\begingroup$ Hmm, I've never seen double covers of arbitrary Lie groups called spin groups "of" $G$. Only ones for special orthogonal groups. Is this generalization to arbitrary Lie groups a convention somewhere? $\endgroup$ Mar 22, 2019 at 16:45

I just want to note that the isomorphisms you wrote down are almost correct in some sense. When talking about Lie algebras, the complexified Lorentz algebra (i.e. taking complex linear combinations of its generators) satisfies $$ \mathfrak{so}(3,1)_{\mathbb{C}}\simeq\mathfrak{su}(2)_{\mathbb{C}}\times \mathfrak{su}(2)_{\mathbb{C}}\simeq\mathfrak{so}(4)_{\mathbb{C}}$$ Though their complexifications are isomorphic, their real parts are not. Depending on how you take the real part of $\mathfrak{su}(2)_{\mathbb{C}}\times \mathfrak{su}(2)_{\mathbb{C}}$, you can get $\mathfrak{so}(3,1)$, $\mathbb{so}(4)$ or a single $\mathfrak{su}(2)_{\mathbb{C}}$ (which itself is isomorphic to $\mathfrak{sl}(2,\mathbb{C})$); for more explanations, see:

Side note: these relations are useful for building representations of the (real) Lorentz group, as it can be shown that that they correspond to the tensor product of two representations of $\mathfrak{su}(2)_{\mathbb{C}}\simeq \mathfrak{sl}(2,\mathbb{C})$. These spin representations are not independent, but rather related to each other by conjugation. See links above and also the Wikipedia page.


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