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$U(p,q)$ is a indefinite unitary group with $pq\not=0$.

Is it compact? If not, what's the maximal compact subgroup?

Is it connected? How much connnected components?

What's the homotopy group of its connected component?

346699
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  • What are your thoughts on this? Have you tried to answer any of these questions on your own? – Ben Grossmann Nov 30 '16 at 00:19
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    Note: wikipedia has a page on the indefinite orthogonal group, and some of the ideas can be applied to the unitary group as well. – Ben Grossmann Nov 30 '16 at 00:25
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    @Omnomnomnom Yes, I think $O(p,q,R)\subset U(p,q)$, so it is noncompact. I guess that the maximal comact subgroup is $U(p)\times U(q)$. But I cannot answer whether it is connected. And I guess the homotopy group is $\pi_1(U(p))\times \pi_1(U(q))$ – 346699 Nov 30 '16 at 00:27
  • @user34669, Did you find an answer? – PPR Jan 30 '17 at 14:43
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    @user34669, I'd wager it is not connected at least for the case p=q. Using the following argument: the identity matrix and diag(1,-1) are both in the group. Using the \det map we find a non-constant continuous function into {+-1}. – PPR Feb 01 '17 at 20:55
  • @PPR This argument does not work because $\mathrm{det}$ takes values on the unit circle, which is connected. For example $U(1,1)$ is diffeomorphic to $S^1 \times S^1 \times D$, where $D$ is the unit disc. – user420261 May 07 '17 at 07:05
  • @user420261, right, I take it back. I guess for $U(n, n)$ one could use that this group is unitarily isomorphic to the Hermitian symplectic group. Then use the same argument as for the usual symplectic group to show that it is connected (this for example: https://math.stackexchange.com/questions/389610/the-symplectic-group-is-connected) or use an SVD-like decomposition of that group to path-connect any given matrix to the identity? – PPR May 07 '17 at 21:29

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