Longest element of a Weyl group - MathOverflow

Longest element of a Weyl group

2011-09-12T15:47:28Z

Let $G$ an algebraic (reductive) group. $T$ a maximal torus, $B$ a Borel subgroup containing $T$, and $w_0$ the longest element of the Weyl group.

I'm looking for a reference explaining why when you conjugate $B$ by $w_0$, the result is the opposite Borel subgroup $B^-$.

Is there a proof involving roots of $G$ relative to $T$ ?

I've found a proof in a book of M. Geck, but this proof doesn't involve roots at all, but only the fact that $B^-$ is uniquely defined by the relation $B \cap B^- = T$.

Answer by Koen S for Longest element of a Weyl group

2011-09-12T16:27:32Z

One way to show this is by using BN-pairs and the associated (spherical) building (see section 6.2 the book "Buildings: Theory and Applications" by Abramenko and Brown, especially section 6.2.6).

The argument is basically the following. The chambers of the associated building correspond to the left cosets of $B$. The parabolic associated to a chamber $gB$ is then given by the conjugate of $B$ by $g$. The Weyl distance between two chambers associated to $gB$ and $hB$ is given by the element of the unique element $w$ in the Weyl group such that $Bg^{-1}hB$. So for your situation this yields that the chamber associated to $B$ is opposite to the chamber associated to $w_0B$.

I hope this helps. While typing this I also realized the answer may depend on which way you define the opposite Borel subgroup.

Answer by Faisal for Longest element of a Weyl group

2011-09-12T17:52:29Z

The proof depends on how you're setting things up. In my opinion the cleanest approach is the Lie algebraic one, and it goes as follows. Your Borel subalgebra $\mathfrak b$ determines a choice of simple roots $\Delta$ and consequently a choice of positive roots $\Phi^+$ : $\mathfrak b = \mathfrak t \oplus \bigoplus_{\alpha \in \Phi^+} \mathfrak g_\alpha$ . The action of $w \in W$ takes $\mathfrak b$ to $\mathfrak b_w = \mathfrak t \oplus \bigoplus_{\alpha \in \Phi^+} \mathfrak g_{w\alpha}$ . With respect to the length function defined using $\Delta$ , the longest element $w_0$ of $W$ takes $\Phi^+$ to $-\Phi^+$ . It follows that $b_{w_0}$ is the Borel subalgebra opposite to $\mathfrak b$ .

Answer by Jim Humphreys for Longest element of a Weyl group

2011-09-12T19:45:41Z

The role of the longest element in $W$ emerges only gradually in the Chevalley structure theory. This is developed similarly but in slightly different styles in the three books with the same title Linear Algebraic Groups (by Borel, Springer, and me). It's important to understand that the desired statement about opposite Borel subgroups of a reductive group depends on a long series of steps culminating in the Bruhat decomposition. In my book, the later steps occupy Sections 26-28. There's already a proof in 26.2 of the existence of a (necessarily unique) Borel subgroup intersecting a given Borel subgroup $B$ in a specified maximal torus $T$ of $B$ . Theorem 26.3(b) almost gives the answer you want, but before enough details of the structure theory are in place. The underlying strategy is to fix $B$ and the flag variety $G/B$ , then see that the fixed points of $T$ on the latter are in natural bijection with both the elements of $W$ and the Borels containing $T$ . (I don't think this got articulated explicitly enough, however.)

Only in my Section 27 is the root system explored, followed in Section 28 by the full details of the Bruhat decomposition. There is still some work to be done on the internal structure of $W$ (including the length function and longest element) in relation to the root system. In my treatment this gets folded into the more general study of $BN$ -pairs (Tits systems), which axiomatize the Bruhat decomposition efficiently. Along the way it turns out for instance that $W$ has a natural structure of Coxeter group.

Unfortunately, the precise statement you want to see a proof of is left somewhat implicit in all these textbook treatments. But this is due partly to the fact that so much heavy theory has to be developed systematically before that kind of statement becomes obvious. It's hard to take any real shortcuts, but also unnecessary to get into the theory of buildings and such.

EDIT. The basic outline here looks deceptively simple: Start with the fixed data $B$ and $T$ and Weyl group $W$ . Then $W$ has a unique longest element $w_0$ , which has order 2 and interchanges positive and negative roots. Here the positive roots are defined by the choice of $B$ , so that $B= T U$ with $U$ the product in any order of 1-dimensional root subgroups corresponding to positive roots. Then the Borel subgroup $B^- = w_0 B w_0$ contains $T$ and has the form $B^- =T U^-$ with $U^-$ the product of root subgroups for negative roots.

But to work this out for a reductive group over an algebraically closed field of arbitrary characteristic requires virtually all of the Borel-Chevalley structure theory through the Bruhat decomposition. For instance, it's highly nontrivial to find the root subgroups and define the root system intrinsically for the group, as well as to work out basic facts about the Weyl group $N_G(T)/T$ and its length function.