MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $B \cap B^- = T$.

share|cite|improve this question
up vote 8 down vote accepted

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 $\mathfrak b$.

share|cite|improve this answer
Thank you ! I was trying to do the same thing but on the group-level, meaning that I was trying to look at the $U_\alpha$ (whose Lie algebras are the $\mathfrak{g}_\alpha$). However, I was trying to explicitely determine the image $w_0(\alpha)$ and I couldn't figure it out... I'll look at the global image of the positive roots then. – Sep 12 '11 at 19:21
@Faisal: What you outline over an algebraically closed field of characteristic 0 for the Lie algebra is the right blueprint for passing to the algebraic (or Lie) group using the classical dictionary. But in prime characteristic the Lie algebra becomes too cumbersome to work with, as Chevalley found, even though his end results are remarkably close to the classical ones. – Jim Humphreys Sep 13 '11 at 22:45
@Jim: My outline does betray the fact that I'm much more accustomed to working over an alg. closed field of char. 0. However, I did check my references to make sure what I wrote still held true in the case of positive characteristic. Did I overlook something? I would appreciate any comments on the matter, as I am ignorant of this aspect of the theory. In any case, as you mention in your answer, the outline does rely on some nontrivial structure theory, but that is to be expected. – Faisal Sep 13 '11 at 23:04
In prime characteristic, your description of the Lie algebra of a reductive group depends on knowing virtually all of the Borel-Chevalley structure theory for the groups. But the bigger complication is using information about subalgebras of the Lie algebra to get back to the group. There is no machine that does this directly, only a lot of partial results which often depend on the prime or Lie type. Chevalley abandoned the Lie algebra approach to get uniform results including classification of semisimple groups. But finding the root system (etc.) in the group is not easy. – Jim Humphreys Sep 14 '11 at 13:25

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.

share|cite|improve this answer
Indeed, I worked with those 3 references and especially with yours, and I was a wondering if I missed something about this longest element or not. Though you are right, with 26.3(b) I thought I was really next to a proof, but still, as I commented above, one has to examine precisely the action of $w_0$, and the fact that $w_0$ send $\phi^+$ on $-\phi^+$ does not seem obvious to me, without using the action on Weyl chambers. – Sep 12 '11 at 20:21

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.

share|cite|improve this answer
Thank you, indeed, I've seen a proof that a Weyl chamber is sent to the opposite under the action of the longest element using combinatorics, but that wasn't really my setting... I should have been more precise. – Sep 12 '11 at 19:16

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.