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Let G be a connected reductive group and K a maximal compact subgroup of G, is it true that the center of K is the intersection of the center of G with K?

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    $\begingroup$ My recollection is that all maximal compact subgroups of $SL(2,R)$ are isomorphic to the circle group $T$, and so the answer would appear to be no. $\endgroup$ – Yemon Choi Jan 14 '12 at 9:21
  • $\begingroup$ It's perhaps interesting to note that $\dim Z(K)/(K\cap Z(G)) \leq 1$, so you're not so far off. $\endgroup$ – Allen Knutson Jan 16 '12 at 22:38
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To elaborate on Yemon's comment: a noncompact, irreducible Hermitian symmetric space is of the form $G/K$, where $G$ is a noncompact simple Lie group with trivial center, and $K$ is a maximal compact subgroup with nondiscrete center; see e.g. Thm 6.1 in Chap. VIII of S. Helgason, ``Differential geometry, Lie groups and symmetric spaces'', Academic Press, 1978. So there are lots of examples where the two centers are un-related.

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