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](https://math.stackexchange.com/questions/1656851/realizing-a-reflection-group-as-a-weyl-group) of this question has already appeared on MSE.



If $N_G(S)/Z_G(S)$ is a reflection group, is it a Weyl group?
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.


