This should be an easy question about centralizers in reductive lie groups, but I wonder if it is already available from the literature.

Consider $G$ a connected non-compact semi-simple Lie group, with a Cartan involution $\sigma$, and $H$ a reductive subgroup, stablized by $\sigma$. Then the fixed part of $\sigma$ in $H$ is a maximal compact subgroup $K_H$ in $H$. Compare the two centralizers $Z(K_H,G)$ and $Z(H,G)$. Are they equal, or at least, their derived parts equal to each other?


  • $\begingroup$ Seems not. Take $G = {\rm{SL}}_2(\mathbf{R})$ and $\sigma(g) = (g^t)^{-1}$, so the corresponding maximal compact subgroup $G^{\sigma}$ is ${\rm{SO}}_2(\mathbf{R})$. Then the diagonal subgroup $H \simeq \mathbf{R}^{\times}$ is reductive (in the sense of being $\mathbf{R}$-points of a connected reductive $\mathbf{R}$-subgroup of ${\rm{SL}}_2$) and stable under $\sigma$, with maximal compact subgroup $K_H := H^{\sigma} = \langle -1 \rangle$ equal to the center of $G$. So $Z(K_H,G) = G$ whereas $Z(H,G) = H$. $\endgroup$ – BCnrd Oct 4 '10 at 13:47
  • $\begingroup$ thanks for the counter example. Does anything change if $H$ is required to be conencted and semi-simple? $\endgroup$ – turtle Oct 4 '10 at 16:33
  • $\begingroup$ Heh-heh, I guessed that you would follow up with that further restriction on $H$. So let's instead take $H = G$! Then $Z(H,G)$ is the center of $G$ (I am assuming that the notation $Z(H,G)$ means centralizer of $H$ in $G$, by the way), which is some finite guy, whereas $Z(K_H,G)$ is not finite if $K_H$ is not semisimple (i.e., has nontrivial central torus). For instance, $H = G = {\rm{SL}}_2(\mathbf{R})$. Please contemplate subgroups of ${\rm{SL}}_2$ and ${\rm{SL}}_3$ before the next round, if we try again. $\endgroup$ – BCnrd Oct 5 '10 at 2:57
  • $\begingroup$ @BCnrd: thanks so much for the reply and comments. I'm really sorry about overlooking these starting examples. It seems that reductive subgroups not of maximal ranks are complicated, as I have also found from similar discussions in mathoverflow. Thanks again, and I'd call a stop for this question. $\endgroup$ – turtle Oct 5 '10 at 12:08
  • $\begingroup$ If $G$ and $H$ are complex points of algebraic groups, then the answer is obviously yes, since $K_H$ is Zariski dense in $H$. $\endgroup$ – Venkataramana Dec 8 '12 at 22:25

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