Timeline for The normalizer a maximal compact subgroup of a semi-simple Lie group
Current License: CC BY-SA 3.0
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
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| Jan 18, 2013 at 15:17 | comment | added | YCor | @Alain: the virtually connected case is implied by the connected case. Indeed by Mostow, if $G$ is virtually connected and $K$ is compact maximal then $KG^0=G$ and $K\cap G^0$ is maximal in $G^0$ (and equal to $K^0$). So if $N$ is the normalizer of $K$ then $N=K(N\cap G^0)$. By the connected case, since $N\cap G^0$ is contained in the normalizer of $K\cap G^0$, we have $N\cap G^0=K\cap G^0$. So $N=K$. | |
| Jan 18, 2013 at 14:23 | history | edited | Hugo Chapdelaine | CC BY-SA 3.0 |
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| Jan 18, 2013 at 7:55 | comment | added | Alain Valette | At least for connected groups, this question was already discussed on MO, see e.g. mathoverflow.net/questions/83694/… | |
| Jan 18, 2013 at 3:42 | history | edited | Hugo Chapdelaine | CC BY-SA 3.0 |
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| Jan 18, 2013 at 2:19 | comment | added | user30180 | I meant to say at the start of the preceding comment that I also assume $G$ has finite center (this ensures $G^0$ is finite over the identity component of the group of $\mathbf{R}$-points of the adjoint connected semisimple $\mathbf{R}$-group $H$ with the same Lie algebra as $G$, which in turn underlies the justification of passage to the case when $G = H(\mathbf{R})$). The universal cover of ${\rm{SL}}_2(\mathbf{R})$, whose center is $\mathbf{Z}$, is a counterexample otherwise (it has trivial $K$). | |
| Jan 18, 2013 at 1:53 | comment | added | user30180 | I assume $\pi_0(G)$ is finite. We want $|N_G(K^0)/K^0|<\infty$. We can assume $G=H(\mathbf{R})$ for adjoint connected semisimple $H$, so $K^0 = \mathbf{K}(\mathbf{R})$ for anisotropic connected semisimple $\mathbf{K} \subset H$. Now $N_G(K^0) = N_H(\mathbf{K})(\mathbf{R})$ and $N_H(\mathbf{K})^0$ is reductive (!), so $N_G(K^0)/K^0 = L(\mathbf{R})$ for a reductive $\mathbf{R}$-group $L$ and $L^0(\mathbf{R})$ has no non-trivial connected compact subgroups. Thus, $L^0$ has no relative roots ($S^1\subset{\rm{SL}}_2$),so $L^0$ is anisotropic and hence $L^0(\mathbf{R})$ is compact, so $L^0=1$. QED | |
| Jan 18, 2013 at 0:43 | history | asked | Hugo Chapdelaine | CC BY-SA 3.0 |