# From Weyl groups to Weyl groupoids?

I'm trying to find a framework where the choices in the classical construction of a root system of a semi-simple lie algebra are not needed.

Let $\mathfrak{g}$ be a semisimple lie algebra.

Definition (Weyl groupoid): Let $\mathcal{W}$ be the following groupoid:

• $Obj(\mathcal{W})=\{\mathfrak{b}\subset \mathfrak{g} | \space \mathfrak{ b} \text{ is a maximal solvable algebra}\}$

• $Mor(\mathcal{W}) = \{Ad(e^x):\mathfrak{b}_1 \to \mathfrak{b}_2 \space |\space x \in \mathfrak{g}\}$

• Objects in $\mathcal{W}$ are in 1-1 correspondence with choosing: cartan subalgebra & Weyl chamber (root basis).
• Morally there's sort of an exponential map $exp: \mathcal{W} \to G/B$ (this is not meant to be a precise statement).

A couple of questions:

1. Is this the correct categorification?

2. How do I get the weyl group out of this?

3. I read somewhere that a choice of borel subalgebra is equivalent to a choice of complex structure on the complexified tangent bundle of $G/B$. Can this groupoid be interpreted in terms of a groupoid of complex structures?

• It's easier to have this discussion on the Lie group rather than Lie algebra level. There for $G$ a compact connected Lie group you can consider the groupoid whose objects are maximal tori and whose morphisms are conjugations. Every object in this groupoid is isomorphic, and all of their automorphism groups are the Weyl group. – Qiaochu Yuan Apr 30 '16 at 15:54
• I would normally choose morphisms to be automorphisms seen as elements of $GL(\mathfrak{g})$ instead of exponentials - this is not a big deal as long as you work over the real or complex numbers. To get a Weyl group, you consider a groupoid whose objects are Cartan subalgebras, together with the obvious functor from $\mathcal{W}$. Fibers are naturally W-torsors. – S. Carnahan Apr 30 '16 at 15:54
• @S.Carnahan That's great! Can I make this geometric? i.e. is there a weyl group torsor $W \to G/B \to G/T$ ? – Saal Hardali Apr 30 '16 at 16:00

Warning: This answer went through a massive editing.

This answer does not regard the specific questions asked, but the motivation behind: finding a framework where the choices in the classical construction of a root system of a semi-simple Lie algebra are not needed.

Let $\mathfrak{g}$ be a Lie algebra and $G=\text{Aut}(\mathfrak{g})^0$. Let $C^+$ be the set of pairs (Cartan,Borel) where the Cartan subalgebra is contained in the Borel subalgebra ($C^+$ is a subset of a product of two Grassmannians of $\mathfrak{g}$). Note that $G$ acts transitively on $C^+$, stabilizers being centralizers of maximal tori.

It is not hard to see that the centralizer of the $G$-action on $C^+$ is isomorphic to the Weyl group (it acts trivially on the Cartan coordinate and transitively on all Borels including a given Cartan). Let's call it "the choice free Weyl group" $W$.

There is a natural vector bundle $H$ over $C^+$, the fiber over (Cartan,Borel) being the vector space underlying the Cartan. Note that (since the stabilizer of a point acts trivially on its fiber) the vector space of all $G$-invariant sections is isomorphic to each fiber. Let's call it "the choice free Cartan" $\mathfrak{h}$. We get a natural action of $W$ on $\mathfrak{h}$.

Note that the dual space, $\mathfrak{h}^*$, could be identified with the space of invariant sections to the dual vector bundle. We can single out those invariant sections which are in the root system at each fiber and obtain "the choice free root system" $\Phi$. Note that the second coordinate in $C^+$ is a Borel. This gives a choice of positivity at each fibers, thus we can also single out "the choice free positive roots" $\Phi^+$.

For each parabolic type we can consider the space of parabolic subalgebras of this given type $P$. Note that between two such, $P,P'$, either there exists a unique $G$-equivariant map or none. If there exists such we will say $P\leq P'$. We may regard a $G$-equivariant maps $C^+\to P$ as a simplex whose faces are the simplices $C^+\to P'$ for each $P'$ such that $P\leq P'$. We will get a simplicial complex on which $W$ acts, "the choice free Coxeter complex".

The following is unrelated, just a comment regarding the question's title. The name "Weyl Groupoid" should be reserved to the following category.

• Objects: Cartan subalgebras.
• Morphisms: A morphism between two Cartan algebras is a Lie-algebra isomorphism which could be extended to a global automorphism in $G$.

This is a connected groupoid and the isotropy group of every object is isomorphic to the Weyl group.

Remark: I was tacitly assuming above that everything is over an algebraically closed field, or split. One can easily adjust the definitions to handle the general case.

• The isotropy group of a Cartan in your groupoid is not the Weyl group; you also get elements coming from diagram automorphisms. For $\mathfrak{sl}(n)$, minus transpose is an algebra automorphism preserving the usual Cartan, but doesn't come from an element of the Weyl group. – Ben Webster Apr 30 '16 at 18:53
• (Deleted a foolish comment: I misread and thought that $C$ was a particular Cartan, rather than the variety of all of them.) – LSpice Apr 30 '16 at 19:27
• @Ben Webster, thanks, definition corrected. Really I have in mind the variety of centralizer of max tori in $G$ where $G$ is connected, hence the mistake. – Uri Bader Apr 30 '16 at 19:54
• @user89334 your edit is very helpful thanks! – Saal Hardali May 1 '16 at 6:51
• @SaalHardali I made a major edition in order to make the "choice free construction" clear. I decided to neglect your wish for "categorfication" as the natural setting, I find, is simply to consider $G$-equivariant objects. – Uri Bader May 1 '16 at 19:56

It's a bit unclear to me precisely what you want, but the usual solution is take the abstract Cartan (discussed a bit here), which is defined to be $\mathbb{T}=B/[B,B]$ for any Borel $B$. This is canonical, since any two Borels are conjugate, and every Borel is self-normalizing so $\mathbb{T}$ is unique up to the map induced by conjugation by an element of $B$. The action of the Weyl group is given by looking at all the isomorphisms induced between $\mathbb{T}$ and any fixed torus $T$ by all the different Borels containing $T$ (which are all conjugate under $N(T)$). Composing one of these isomorphisms with the inverse of another gives all the elements of the Weyl group.

• As the sidebar informs, this point of view is also discussed in mathoverflow.net/questions/103751/… . – LSpice Apr 30 '16 at 19:28
• Thank you! That was helpful. Mainly what I was after is a way to remember what choice gives me what and in what sense are they equivalent. S. Carnahan gave me the idea of encoding it in a torsor $\mathcal{W} \to G/B \to G/T$ where a point in the base is a cartan subalgebra and a point in the total space is a borel subalgebra. The weyl group acts transitively on the fibers. Does this work? – Saal Hardali Apr 30 '16 at 19:29
• I must say that I fail to see how any of this is choice free. Surely, everything here is independent of the choices made, but choices are made. – Uri Bader Apr 30 '16 at 20:05
• @user89334 Fair enough. Like I said, I wasn't really sure what the user wanted. – Ben Webster May 1 '16 at 2:56
• @SaalHardali You can change what the accepted answer is. – Ben Webster May 1 '16 at 15:58