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Let $G$ be a compact, connected Lie group and $S$ a torus in $G$ not assumed maximal. Then conjugation in $G$ induces a faithful representation of $N = N_G(S)/Z_G(S)$ in the Lie algebra $\mathfrak s$ of $S$.

In all the instances I know where the image of $N \to \mathrm{GL}(\mathfrak s)$ is a reflection group, it is because there is a closed, connected subgroup $K$ of $G$ such that $S$ is its maximal torus, and $N$ is isomorphic to the Weyl group of $K$.

For instance, it happens

  1. If $S$ is a maximal torus,

  2. If $S$ is a circle (edit: this memory of mine is untrue; I had shown this only in a few cases, so the answer is negative),

  3. If $S$ is the maximal torus in some $K$ such that $H^*(G;\mathbb Q) \to H^*(K;\mathbb Q)$ is surjective,

  4. If $S$ is the maximal torus in the fixed-point set of some finite-order Lie group automorphism of $G$.

Does it always happen? If $N$ is a reflection group, is it always a $W_K$?

A less concise version of this question has already appeared on MSE.

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The answer is negative, even if $S$ is a circle, i.e., $1$-dimensional. To see that let $G$ be simple such that its Weyl group contains $-1$, i.e., $G$ is not of type $A_n$, $D_{2n+1}$, or $E_6$. Then $N_G(S)/Z_G(S)=\{\pm1\}$ is a reflection group for any $1$-dimensional subgroup. $S\subseteq T$. There are infinitely many conjugacy classes of $S$. But $G$ contains only finitely many conjugacy classes of semisimple rank-$1$-subgroups.

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  • $\begingroup$ I don't understand the answer to what is negative. When $S$ has dimension 1, to say that the image is a reflection group is either an empty statement (if one says that the trivial group is a reflection group) or means that $N_G(S)/Z_G(S)$ has order 2 (or equivalently $N_G(S)\neq Z_G(S)$. You seem to say that the latter holds. So, well, I'm just confused. $\endgroup$ – YCor Sep 24 '16 at 7:18
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    $\begingroup$ Now I am confused. In these particular examples, $N=N_G(S)/Z_G(S)$ is a reflection group of order $2$. The question was whether $N$ is then the Weyl group of an intermediate subgroup $K$ having $S$ as maximal torus. Any candidate for $K$ is either isomorphic to $SU(2)$ or $SO(3)$. There are only finiteley many conjugacy classes of those. So $K$ does not exist for most $S$. But maybe I got the question all wrong. $\endgroup$ – Friedrich Knop Sep 24 '16 at 8:49
  • $\begingroup$ I'm sorry, I did all the confusion reading the question. The question is not whether the image of the representation of $N/Z$ is a reflection group, but takes it as an an assumption. (I hope I understand correctly "reflection group" as "generated by reflections", including, for instance, the trivial group.) $\endgroup$ – YCor Sep 24 '16 at 10:33
  • $\begingroup$ "Reflection group" was intended to include the trivial group! $\endgroup$ – jdc Sep 24 '16 at 19:33
  • $\begingroup$ OK, this is actually what I'd hoped for (but not the error or confusion). I think I had proved in the $A_n$ and $D_{2n+1}$ cases that there exists a rank one simple containing subgroup, back when I asked this question on MSE, and misremembered that as having shown it in general. Is it obvious there are only finitely many conjugacy classes of simple subgroups? $\endgroup$ – jdc Sep 24 '16 at 19:36

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