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For the matrix lie group $SO(p,q)$, what is in the center? Is there anything other than $I_n$ (or $-I_n$ in the case $p+q=n$ is even)?

Also where can I find a reference?

Michelle
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  • I could not find anything about the center on that wikipedia article. – Michelle Apr 02 '15 at 19:54
  • Sorry. I meant to say that I didnt find anything about the center of SO(p,q) on that page. – Michelle Apr 02 '15 at 20:06
  • This link here might be useful. – Dietrich Burde Apr 02 '15 at 20:31
  • possible duplicate of Center of the Orthogonal Group and Special Orthogonal Group – hjhjhj57 Apr 02 '15 at 21:32
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    @hjhjhj57 the question you linked does not contain an answer to this question. This is the special orthogonal group of a nondegenerate bilinear form that is not necessarily positive definite. – Matt Samuel Apr 02 '15 at 22:15
  • @MattSamuel wouldn't the second answer work with some little changes? – hjhjhj57 Apr 02 '15 at 22:17
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    @hjhjhj57 I don't know enough about the structure of the group to answer that question, but if you see a way to make it work this case perhaps you should answer the question. It's not close enough to qualify as a duplicate in my opinion. – Matt Samuel Apr 02 '15 at 22:21
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    Keep in mind that $\text{SO}(2) \cong S^1$ and $\text{SO}(1, 1) \cong \mathbb{R}^{\times}$ are both abelian, so in these cases the center is the whole thing. – Qiaochu Yuan Apr 03 '15 at 07:53
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    While reading Shlomo Sternberg's third section of "On charge conjugation" I came across the affirmation that $$ \begin{pmatrix}\pm I_{p\times p} & 0 \ 0 & \pm I_{q\times q} \end{pmatrix}$$ is the center of $O(p,q)$. Of course this isn't a proof, but it may be useful. Let me know if you want the complete reference. – hjhjhj57 Apr 11 '15 at 00:46
  • Thanks hjhjhj577. This answers my question. However, the Sternberg paper doesn't have an explanation nor a reference to why it is the case. – Michelle Apr 17 '15 at 23:44

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