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Let $G$ be a connected reductive Lie group over $\mathbb{C}$, $K$ a maximal compact subgroup of $G$ and $T$ a maximal torus of $K$.

Question: For the normalizers of $T$, is it true that the normalizer $N_G(T)=N_K(T)C_G(T)$? Any "elementary" proof?

Here $C_G(T)$ denotes the centralizer of $T$ in $G$.

Thank you in advance.

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  • $\begingroup$ It might be helpful to explain what you mean by "reductive Lie group", or else to use a more common name lkke "semisimple". The term "reductive" comes from the theory of linear algebraic groups; it suggests complete reducibility of finite dimensional representations, which is only true in characteristic 0, etc. $\endgroup$
    – Jim Humphreys
    Commented Sep 28, 2019 at 2:13
  • $\begingroup$ @JimHumphreys. I think the term "reductive" is quite clear also in the context of Lie groups. I take it to mean that the Lie algebra is a product of a semisimple and an abelian one. If there are problems with the concept, let me know. $\endgroup$
    – Sebastian Goette
    Commented Sep 29, 2019 at 7:06
  • $\begingroup$ @Sebastian: I think the ambiguity comes into the proviso that the Lie algebra involves an abelian part. Does the additive Lie group qualify as "reductive", or the product of this group with a simple Lie group? This is pretty far from the Borel-Tits notion of reductive algebraic group. $\endgroup$
    – Jim Humphreys
    Commented Sep 29, 2019 at 21:31
  • $\begingroup$ @JimHumphreys I see your point. But for the purpose of this question, I think we may assume that noncompact abelian groups are allowed as factors of $G$, and see if we can still get an answer. For motivation, in the category of Lie groups one can (maybe should) always ask which results extend to the universal covering. $\endgroup$
    – Sebastian Goette
    Commented Sep 30, 2019 at 8:31
  • $\begingroup$ @Sebastian: "Abelian Lie group" is too broad here. Even Wallach's book Real Reductive Groups I gives an algebraic defini.tion of "real reductive Lie group". The problem is that a vector group is not a reductive Lie group (say over $\mathbb{C}$). $\endgroup$
    – Jim Humphreys
    Commented Oct 4, 2019 at 14:51

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