Can we realize Weyl group as a subgroup?

Given a semisimple Lie group G, let T be a maximal torus, W be the Weyl group defined as the quotient N(T)/C(T), where N(T) denotes the normalizer of T and C(T) denotes the centralizer.

Two questions are:

1. How many ways are there we can realize W as a subset of G?

2. Can we always realize W as a subgroup of G?

• This question (the second, more interesting question) will probably need some clarification before it can be cleanly answered. Is $G$ a semisimple real Lie group? Is $T$ a maximal split torus? – Marty Jan 27 '10 at 15:21
• It's worth noting that the questions here make equal sense and have mostly the same answers when the group is assumed to be a (connected) semisimple algebraic group over an arbitrary algebraically closed field. In any case, smaller fields need more discussion. – Jim Humphreys Apr 16 '10 at 18:45

In general it is not possible to embed the Weyl group $W$ in the group $G$: already you can see this for $SL_2(\mathbb C)$, where the Weyl group has order $2$: if the torus fixes the lines spanned by $e_1$ and $e_2$ respectively, you want to pick the linear map taking $e_1$ to $e_2$ and $e_2$ to $e_1$, but this has determinant $-1$. A lift of $W$ to $N(T)$ must be an element of order $4$ not $2$, say $e_1 \mapsto -e_2$ and $e_2 \mapsto e_1$.
In fact, Tits has shown that this is essentially the only obstruction: the Weyl group can always be lifted to a group $\tilde{W}$ inside $G$ which is an extension of $W$ by an elementary abelian $2$-group of order $2^l$ where $l$ is the number of simple roots. If I recall correctly, this lift is then unique up to conjugation.
• So is Tits' group just the $\mathbf{Z}$-points of the normalizer, after one chooses a model over $\mathbf{Z}$? – moonface Jan 27 '10 at 16:33
• One should point out that Tits' result is for G simply connected. If we use a smaller Lie group, then the rank of the $2$-group drops. I think I recall that, if G is the adjoint form of its Lie algebra, then W always lifts. – David E Speyer Jan 27 '10 at 16:37
• You're right I meant to take $G$ simply connected, but are you sure about the adjoint form case? For $E_8$ for example, the adjoint form is simply connected, so the Tits group can't get any smaller. – Kevin McGerty Jan 27 '10 at 17:03