It is known mathematically that given a bilinear form $Q$ with signature $(p,q)$ then the group $Spin(p,q)$ is the double cover of the group $SO(p,q)$ associated to $Q$, and that $Pin(p,q)$ is the double cover of the group $O(p,q)$ associated to $Q$.

I understand that all the representations of the double cover are representations of the starting group, but not viceversa, due to the properties of projective representations and the equivalence of algebras.

This question arises since in physics we are interested to spinor-bundles, arising from $Spin(p,q)$ groups which covers the $SO(p,q)$-principal bundles.


  1. Why care at all about using the $Spin$-principal bundle instead of the usual orthonormal frame bundle?

  2. If we are interested in using the most general representation, why then restrict to the double cover and not some bigger group?


1 Answer 1

  1. In quantum mechanics, we are interested in projective representations since the Hilbert space is a projective space, not a vector space. However, all projective representations of $SO(p,q)$ ane faithful representations of $Spin(p,q)$.

  2. In general, we are interested in universal covers of groups. However, $SO(p,q)$ is doubly-connected so its universal covering IS the double-cover.

  • $\begingroup$ Oh i didn't know that $SO$ had this property. Is it true also for $Pin$ groups? $\endgroup$
    – LolloBoldo
    Sep 17, 2023 at 12:01
  • $\begingroup$ The only distinction between $SO$ and $O$ are the existence of extra disconnected components. The local structure and topology of each connected component are still the same so I suspect that this is also true for $O$ and $Pin$. I might be missing some subtlety though. $\endgroup$
    – Prahar
    Sep 17, 2023 at 12:30

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