16
$\begingroup$

It may be a easy question for experts.

The definition of the Weyl group of a complex semisimple Lie algebra $\mathfrak{g}$ is well-known: We first $\textbf{choose}$ a Cartan subalgebra $\mathfrak{h}$ and we have the root space decomposition. The Weyl group now is the group generated by the reflections according to roots.

Naively this definition depends on the choices of the Cartan subalgebra $\mathfrak{h}$. Of course we can prove that for different choices the resulting Weyl groups are isomorphic.

My question is: can we define the Weyl group intrinsically such that we don't need the do check the unambiguity.

One thought is: we have the abstract Cartan subalgebra $\mathfrak{H}:=\mathfrak{b}/[\mathfrak{b},\mathfrak{b}]$ of $\mathfrak{g}$ (which is in fact not a subalgebra of $\mathfrak{g}$). Can we define the Weyl group along this way? Again is there any references for this?

$\endgroup$
3
  • 8
    $\begingroup$ Don't you end up proving the result does not depend on the Borel you picked? $\endgroup$ Commented Aug 2, 2012 at 3:00
  • $\begingroup$ @ Mariano You are right. It can be proved that for two Borel subalgebra $\mathfrak{b}$ and $\mathfrak{b}' $ , the resulting quotient $\mathfrak{b}/[\mathfrak{b}, \mathfrak{b}]$ and $\mathfrak{b}'/[\mathfrak{b}', \mathfrak{b}']$ are canknically isomorphic. It's in Representation Theory and Complex Geometry Chapter 3 by Chriss/Ginzburg. Oop, still there are choices. $\endgroup$ Commented Aug 3, 2012 at 6:12
  • 3
    $\begingroup$ @Zhaoting: I think all the approaches suggested in the answers tend to show the impossibility of giving the desired intrinsic definition "such that we don't need to check the unambiguity". Under the surface the conjugacy theorems and other scaffolding are concealed. $\endgroup$ Commented Aug 3, 2012 at 14:18

5 Answers 5

16
$\begingroup$

Probably the earliest intrinsic definition of Weyl group occurs in section 1.2 of the groundbreaking paper "Representations of Reductive Groups Over Finite Fields" by Deligne and Lusztig (Ann. of Math. 103, 1976, available at JSTOR). This is done elegantly in the closely related but more general setting of a reductive algebraic group $G$ over an arbitrary algebraically closed field (though their interest is mainly in prime characteristic). Letting $X$ denote the set of all Borel subgroups of $G$, the set of $G$-orbits on $X \times X$ provides a natural model for a universal Weyl group of $G$ (or its Lie algebra).

[ADDED] In the algebraic group setting, this intrinsic definition depends just on knowing what a connected reductive (or semisimple) group is and what a Borel subgroup is (maximal closed connected solvable subgroup). But obviously one can't exploit the "Weyl group" without knowing more of the structure theory: conjugacy theorems, Bruhat decomposition. (Is it a group? finite?) In the easier characteristic 0 Lie algebra theory, where $X$ becomes the set of Borel subalgebras (whose definition requires some theory) with conjugation action by the adjoint group, this abstract notion of "Weyl group" similarly needs unpacking. But the Deligne-Lusztig definition is a good conceptual one for their purposes and sneaks in the underlying set $X$ of the flag variety of $G$. Any intrinsic definition of the Weyl group needs serious background in Lie theory.

In the treatment by Chriss and Ginzburg, even when one is primarily interested in the Lie algebra picture, the group in the background tends to play an important role. Indeed, in the early work of Borel and Chevalley on semisimple algebraic groups, the Weyl group appears most naturally in the guise of the finite quotient $W_G(T) :=N_G(T)/T$ for a fixed maximal torus $T$. Then one sees $W$ as generated by reflections relative to roots, etc. As in the parallel Lie algebra setting in characteristic 0, the maximal tori (or Cartan subalgebras) are all conjugate under the adjoint group action, but this falls short of giving an intrinsic definition of the sort provided by Deligne-Lusztig.

[Weyl himself gave the group an awkward name, but was mainly concerned with its use in the context of a compact Lie group. The notion basically originates earlier in the work of Cartan, but it took a while to see the root system and Weyl group as combinatorial objects including the Coxeter presentation of the group as a reflection group (carried over by Witt to Lie algebras).]

$\endgroup$
7
  • 1
    $\begingroup$ Of course, this elegant "intrinsic definition" rests on all of the usual conjugacy results, though it isn't clear what the OP is really seeking by asking for a definition which avoids the need to check "the unambiguity". $\endgroup$
    – user22479
    Commented Aug 2, 2012 at 12:34
  • $\begingroup$ @quasi (if I may call you that): See my added paragraph, where I emphasize that the definition itself uses no more than basic definitions. The price for that is having to figure out what it actually means in concrete terms; then you do need more theory. $\endgroup$ Commented Aug 2, 2012 at 17:36
  • 3
    $\begingroup$ @Jim Maybe we can look at the set of $G$ -orbits of $X \times X$ and say that "this is the Weyl group". But can we define a multiplication just on this set of $G$-orbits? If we can, then this is what I am seeking for: an intrinsic definition of Weyl group. $\endgroup$ Commented Aug 2, 2012 at 23:10
  • 2
    $\begingroup$ @Zhaoting: This is all worked out carefully by Deligne-Lusztig in their section 1.2. But I'd emphasize that it uses most of the deep structure theory (including conjugation theorems and Bruhat decomposition in the group version) to reach the intrinsic formulation. $\endgroup$ Commented Aug 3, 2012 at 14:14
  • 1
    $\begingroup$ Title of @Asaf's reference: Bader and Furman - Boundaries, Weyl groups, and superrigidity. $\endgroup$
    – LSpice
    Commented Aug 24, 2021 at 12:54
22
$\begingroup$

Let $g_r$ be the set of regular semi-simple elements of the Lie algebra, and $\tilde g_r$ be the set of these elements with a choice of Borel containing it. The Weyl group is the group of deck transformations of the cover $\tilde {g}_r\to {g}_r$.

$\endgroup$
3
  • $\begingroup$ This is a great point! Maybe the remaining problem to me is that can we relate this deck transformation with the set of $G$-orbits on $X \times X$, as Jim Humphreys pointed out in his answer. $\endgroup$ Commented Aug 2, 2012 at 23:13
  • 1
    $\begingroup$ @Zhaoting: Like the other intrinsic descriptions, this requires a lot of the structure theory to relate it to the concrete Weyl group attached to a Cartan subalgebra (or maximal torus) of a semisimple Lie algebra (or group). Here you also need regular elements (Kostant/Steinberg; cf. Bourbaki Ch. 7-8): regular semisimple elements are dense and each lies in exactly $|W|$ Borel subalgebras (or subgroups), corresponding to positive systems of roots or Weyl chambers. $\endgroup$ Commented Aug 3, 2012 at 14:07
  • $\begingroup$ @Ben: This is a very nice topological viewpoint, which I guess goes back to work on the topology of compact Lie groups (Adams, Bott, Samelson, ...)? Is there a good reference for the translation to semisimple Lie algebras and Borel subalgebras? $\endgroup$ Commented Aug 3, 2012 at 14:11
15
$\begingroup$

Sometimes when you define a group using an arbitrary choice of object and then show the choice of object doesn't matter, you could have defined a groupoid without making an arbitrary choice.

For example, to define the fundamental group $\pi_1(X,x)$ of a path-connected space $X$ we need to choose a basepoint $x \in X$, but then we can show we get isomorphic groups no matter what basepoint we choose, with an isomorphism given by a homotopy class of paths between the basepoints. To avoid this maneuver we can work with the fundamental groupoid of $X$, whose objects are points of $X$ and whose morphisms are homotopy classes of paths. If $X$ is path-connected all objects in this groupoid are isomorphic, and thus the automorphism groups of all objects are isomorphic. The automorphism group of $x$ is just $\pi_1(X,x)$. The fundamental groupoid is thus equivalent, as a category, to the one-object groupoid corresponding to the group $\pi_1(X,x)$. But the advantage of the fundamental groupoid is that we can define it without choosing a basepoint, and it makes sense and works well even when $X$ is not path-connected.

Similarly, I think we can define the Weyl groupoid of a compact semisimple Lie group $G$ in a way that gives a groupoid equivalent to the usual Weyl group, but doesn't require a choice of maximal torus. The idea should go like this. The objects of the Weyl groupoid are maximal tori. A morphism $f : T \to T'$ in the Weyl groupoid is a Lie group isomorphism of the form

$$ t \mapsto g t g^{-1} \textrm{ for all } t \in T $$

for some $g \in G$. If I did this right, the automorphism group of any object $T$ in the Weyl groupoid is the usual Weyl group

$$ W_G(T) = N_G(T) / T ,$$

that is, the normalizer of $T \subset G$ modulo the centralizer of $T \subset G$, which is $T$ itself. If this is true, the Weyl groupoid will be equivalent, as a groupoid, to the usual Weyl group $W_G(T)$ for any maximal torus $T$.

$\endgroup$
10
  • 2
    $\begingroup$ Isn't this just pushing the bump under the rug? That is, if one wants to know that there is one Weyl group associated to a given semisimple Lie group, then one has to show that the resulting groupoid is connected, which in this case is the same as showing that the maximal tori are all conjugate... right? $\endgroup$ Commented Sep 12, 2013 at 17:14
  • 2
    $\begingroup$ You need to show all maximal tori are conjugate to show this groupoid is equivalent to a group, and to get a specific group that it's equivalent to, you may need to pick a specific maximal torus... but if you're happy working with groupoids instead of groups (as I am), you can avoid this arbitrary choice - and it's this arbitrary choice that was annoying the original questioner, not the difficulty of proving all maximal tori are conjugate. $\endgroup$
    – John Baez
    Commented Sep 13, 2013 at 5:34
  • 2
    $\begingroup$ @Joshua: there is potentially more at stake here than just the issue of whether the Weyl group is uniquely determined up to isomorphism; you might also want to know something about how the Weyl group behaves in families. In this case, the Weyl group a priori depends on a choice of maximal torus $T$, so there is a family of groups over the space of maximal tori whose fiber at a maximal torus is the Weyl group $W_G(T)$. Now one can ask whether this family is isomorphic to the constant family with constant value some group or not, which is stronger than asking whether the fibers are isomorphic. $\endgroup$ Commented Dec 3, 2014 at 6:24
  • 1
    $\begingroup$ An analogous issue in thinking about fundamental groups vs. fundamental groupoids is the following. You might want to convert a fibration of spaces over a space $B$ into a fibration of groups over $B$ by taking fundamental groups fiberwise. But to do this reasonably you need a continuous choice of basepoints of each fiber: in other words, you need a section of the fibration. But not only are sections not unique, they may even fail to exist! However, what you can always do without making any choices is to convert a fibration of spaces over $B$ into a fibration of groupoids over $B$. $\endgroup$ Commented Dec 3, 2014 at 6:27
  • 1
    $\begingroup$ @JohnBaez: But even if one is okay using groupoids directly, there are lots of situations where it's useful to know what the Weyl group actually is ($S_n$, hyperoctahedral, Coxeter presentations, etc.), and as you say, for that it seems one still needs to pick a maximal torus (or, as in Jim Humphreys's comments, use a lot of theory including conjugacy of tori). I wonder how far into e.g. the representation theory of semisimple complex Lie groups one can get using the Weyl groupoid without ever determining the Weyl group, or at least the connectedness of the groupoid? $\endgroup$ Commented Dec 3, 2014 at 7:11
4
$\begingroup$

Yes: this is the approach to defining the 'abstract Weyl group' introduced in "Representation Theory and Complex Geometry" by Chriss/Ginzburg on p. 135 (2nd Edition, Birkhauser).

$\endgroup$
3
$\begingroup$

I have heard that, originally, the Weyl group was designed (and worked out e.g. by Chevalley) as some Galois group which are then intrinsic.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .