# The Weyl group of $SL(2, F)$

Let $G= SL(2, F)$, given a torus $T$, the Weyl group with respect to $T$ is defined to be $W=N(T)/Z(T)$, the quotient of the normalizer $N(T)$ of the torus by the centralizer $Z(T)$ of the torus.

My question might be basic and elementary for some of you, but I would like to be in a complete understanding.

Why Weyl group of $SL(2, F)$ is the group $\mathbb{Z}/2\mathbb{Z}$?

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It's not made explicit here what field $F$ is given, though the tag p-adic-groups is suggestive. Usually "the" Weyl group is defined for an algebraic group over an arbitrary algebraically closed field, relative to any given maximal torus (all such tori being conjugate). It would help to know what kind of textbook source you are starting with, since the question is certainly "basic and elementary" once the context is clear. But more precision is needed if you start to look at fields of definition which are not algebraically closed. –  Jim Humphreys May 6 '12 at 13:53
The field $F$ is non-Arch. local field and the tag for p-adic group because am dealing with $SL(2,F)$. –  Dragon May 6 '12 at 15:52

Two people (so far) have tried to provide some sensible information, but I think the question itself is too loosely formulated to have a real answer. What you are looking at is the group of rational points of an algebraic group of rank 1 over a certain type of infinite field, which could be of characteristic 0 or not. This group itself is just an abstract group, which doesn't immediately have a definite "Weyl group" attached. It makes more sense to start with the algebraic group $G=\mathrm{SL}_2$, which is defined and split over any prime field. (The language of group schemes is appropriate here.) Over any field $F$, the group of rational points is just the usual group of $2 \times 2$ matrices of determinant 1 over $F$. It makes no difference for your question what $F$ is.

In $G$ there is a single well-defined (up to isomorphism) Weyl group $N_G(T)/T$ which just has order 2, as described concretely by Mrc Plm in the split case. All maximal tori of $G$ are conjugate as well as self-centralizing, so the choice doesn't matter.

However, if $F$ fails to be algebraically closed, the resulting abstract group $\mathrm{SL}_2(F)$ is more challenging because $G$ typically has more than one class of maximal tori defined over $F$. In your rank 1 case it's not so bad because there are just two possibilities, the "split" and "anisotropic" (or "compact") types of maximal tori. The structure of the normalizer modulo centralizer of each type has to be looked at case-by-case in terms of the "absolute" Weyl group. In higher ranks the situation gets more complicated. Anyway you have to take account of the variation of structure in the points of $G$ over an arbitrary field.

Again I'd emphasize that the basic theory coming from study of algebraic groups doesn't depend on which field of definition $F$ (even finite) you are ultimately interested in, though of course the end results about rational points depend heavily on that field. There are somewhat different expositions of all this in standard textbooks, but the idea is always the same in your limited situation.

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Let $T$ be a split torus over a field $F$.

The Weyl group is by definition a subgroup of the group of outer morphisms (defined over $F$).

There are only two automorphism (the inverse and the identity) of $T$ (defined over F), so the group of outer automorphism is isomorphic to $\mathbb{Z}/2$.

Both can realized by the conugation via the non-trivial element $\begin{pmatrix} 0 & -1 \newline 1 & 0 \end{pmatrix}$ and by the identity elment respectively.

So the Weyl group has at least two elements, hence exactly two elements.

Similiar you can argue for the maximal split torus in $SL(n)$ or $GL(n)$.

For non-split tori, the question is more difficult to answer. I know only the $GL(n)$-situation. In $GL(2)$, the tori is isomorphic to a multiplicative group of a quadratic field extensions, and the outer automorphisms of the tori are isomorphic to the Galois group. Now these can be realized by conjugation of elements $GL(2)$, but they can not be realized by elements of determinant $1$.

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For your last sentence, note that $O_2(\mathbb{R})$ does not embed in $SL_2(\mathbb{R})$. –  S. Carnahan May 18 '12 at 8:19
Yes. I guess it is clear that every thing should be defined uniquely up to conjugation, and surely the determinant will be preserved, so I win my bet;) –  Marc Palm May 18 '12 at 11:20

One can approach this from different point of views:

On a matrix level, the subgroup consisting of the diagonal matrices is an example of a torus $T$. Its centralizer is $T$ itself, and its normalizer $N(T)$ the group of monomial matrices in $G$. One then verifies that $N(T) / Z(T)$ is the cyclic group on two elements.

One can also use the projective line $\mathbb{P}^1(F)$. Then the centralizer is the subgroup fixing the `apartment', which in this case is a pair of points of the projective line, pointwise. The normalizer is the subgroup of $G$ stabilizing this apartment. The Weyl group (the quotient) is then the group of faithful automorphisms of the apartment. So the Weyl group here consists of the identity and an involution switching the two points of the apartment.

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