Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

For $K$ a compact Lie-group with maximal torus $T$, I'd like to know the cohomology $\text{H}^{\ast}(K/T)$ of the flag variety $K/T$.

If I'm not mistaken, this should be isomorphic to the algebra of coinvariants of the associated root system, according to the "classical Borel picture", as it is often called in the literature (sadly, often without a reference). Unfortunately, Borels original paper is quite long and so I have the following

Question: Does anybody know a short proof for the theorem?

For example, can it be proved using a direct computation of the Lie-algebra cohomology $\text{H}^{\ast}({\mathfrak k},{\mathfrak t})$?

As a side-question: Is it correct that for a complex semisimple Lie-group $G$ with Borel $B$, compact real form $K$ and maximal torus $T$ the map $K/T\to\ G/B$ is a homotopy equivalence? What is a good reference for these things?

I'm sorry if this is too elementary for MO, but apart from Borel's original paper I couldn't find good sources.

share|improve this question
add comment

3 Answers 3

up vote 14 down vote accepted

Borel's lengthy 1953 Annals paper is essentially his 1952 Paris thesis. It was followed by work of Bott, Samelson, Kostant, and others, which eventually answers your side question affirmatively. For a readable modern account in the setting of complex algebraic groups rather than compact groups, try to locate a copy of the lecture notes: MR649068 (83h:14045) 14M15 (14D25 20F38 57N99 57T15) Hiller, Howard, Geometry of Coxeter groups. Research Notes in Mathematics, 54. Pitman (Advanced Publishing Program), Boston, Mass.-London, 1982. iv+213 pp. ISBN 0-273-08517-4. (This was based on his 1980 course at Yale. Eventually he left mathematics to work for Citibank.) The identification of the cohomology ring with the coinvariant algebra of the Weyl group has continued to be important for algebraic and geometric questions, for instance in the work of Beilinson-Ginzburg-Soergel. While Hiller's notes are not entirely self-contained, they are helpfully written. (But note that his short treatment of Coxeter groups has a major logical gap.)

ADDED: In Hiller's notes, Chapter III (Geometry of Grassmannians) is most relevant. For connections with Lie algebra cohomology, the classical paper is: MR0142697 (26 #266) 22.60 (17.30) Kostant, Bertram, Lie algebra cohomology and generalized Schubert cells. Ann. of Math. (2) 77 1963 72–144. Nothing in this rich circle of ideas can be made quick and easy; a lot depends on what you already know.

P.S. Keep in mind that Hiller tends to give explicit details just for the general linear group and grassmannians, but he also points out how the main results work in general, with references. I don't know a more modern textbook reference for this relatively old work. But the intuitive connection between the Borel picture and the Bott/Kostant cohomology picture is roughly this: The Lie subalgebra spanned by negative root vectors plays the role of tangent space to the flag manifold/variety. In the Lie algebra cohomology approach you get an explicit graded picture for each degree in terms of number of elements in the Weyl group of a fixed length, whereas the Borel description in terms of Weyl group coinvariants makes the algebra structure of cohomology more transparent. (What I don't know is whether a simpler proof of Borel's theorem can be derived using Lie algebra cohomology.)

Concerning the relationship between $K/T$ and $G/B$, this goes back to the work around 1950 on topology of Lie groups (Iwasawa, Bott, Samelson): all the topology of a connected, simply connected Lie group comes from a maximal compact subgroup. So the two versions of the flag manifold are homeomorphic. In later times, emphasis has often shifted to treating $G$ as a complex algebraic group, so that $G/B$ is a projective variety. For me the literature is hard to compactify.

One more reference, which treats the Borel theorem in a semi-expository style: MR1365844 (96j:57051) 57T10 Reeder, Mark (1-OK), On the cohomology of compact Lie groups. Enseign. Math. (2) 41 (1995), no. 3-4, 181–200.

share|improve this answer
    
Thank you, these links were very helpful to me. –  B. Bischof Apr 17 '10 at 19:23
1  
Another important paper in the direction of Kostant's paper is that by Bernstein, Gel'fand, Gel'fand, "Schubert cells and Cohomology of the spaces G/P" –  B. Bischof Apr 19 '10 at 14:37
add comment

Note that your statement is only true in rational cohomology. For example, $H^\ast(SO(5)/T)$ is not generated in degree $2$ (though it is rationally).

The easiest proof I know starts from equivariant cohomology:

$ H^\ast_T(K/T) = H^\ast_{T\times T}(K) = H^\ast_{T\times K\times T}(K\times K) = H^\ast_K(K/T \times K/T) $

So far this uses $H^\ast_F(X) = H^\ast(X/F)$ for free actions. Now use the equivariant Künneth formula:

$... = H^\ast_K(K/T) \otimes_{H^\ast_K} H_K(K/T) = H^\ast_T \otimes_{H^\ast_K} H^\ast_T$

Rationally, the base ring $H^\ast_K$ is $(H^\ast_T)^W$, the invariants. Since you didn't want equivariant but ordinary cohomology, kill the left factor, leaving ${\mathbb Q} \otimes_{(H^\ast_T)^W} H^\ast_T$, which is your desired ring of coinvariants.

(I'm having a bunch of trouble with $H$ vs. $H^\ast$ in typesetting here, sorry!)

share|improve this answer
    
Having had the same trouble myself, I fixed the typesetting by using \ast instead of * for the asterisks. (Backticks also work.) –  Tim Perutz Apr 17 '10 at 16:59
3  
I am not sure I agree that equivariant cohomology is supreme as a candidate for the quickest way to go if one wants an algebro-topological proof. One can use the Eilenberg-Moore spectral sequence which has the form (with a field $k$ as coefficient ring say) $$ \mathrm{Tor}^*_{H^*(K)}(H^*(T)),k) \implies H^*(K/T)$$ and as $H^*(T)$ is free as $H^*(K)$-module (when $k$ has characteristic $0$) this degenerates to give what one wants. In any case if either of these proofs qualifies as "short" depends on whether your definition of length is recursive or not.... –  Torsten Ekedahl Apr 17 '10 at 18:10
    
Comment to my own comment: When I wrote $H^*(K)$ and $H^*(T)$ I meant $H^*(BK)$ and $H^*(BT)$. –  Torsten Ekedahl Apr 19 '10 at 5:26
    
Maybe even faster: $H_K(K/T)=H_T$, so $H(K/T)=\mathbb Q \otimes_{H_K} H_T=\mathbb Q \otimes_{H_T^W} H_T$ the ring of coinvariants. –  Jan Weidner Oct 25 '11 at 12:06
    
add comment

Concerning the side question, the map K/T → G/B is actually an isomorphism of real manifolds, not just a homotopy equivalence. Not sure about the references, but this is essentially textbook material (unfortunately I forgot where I learned about this). Considering the case of U(n) ⊂ GL(n) acting on the flags in Cn is instructive.

share|improve this answer
    
Oddly, in the infinite-dimensional case the most relevant map is the other way, and is only a homotopy equivalence, as one can read about in [Pressley-Segal] "Loop Groups". –  Allen Knutson Apr 19 '10 at 1:11
1  
One way of getting this isomorphism is that it is a special case of the Iwasawa decomposition for a general semi-simple (real) Lie group. This allows us to get many specific references, Helgason for instance. –  Torsten Ekedahl Apr 19 '10 at 5:34
    
Oh, so this is called the Iwasawa decomposition. I used to know this, but it was a long time ago. –  Leonid Positselski Apr 19 '10 at 20:03
add comment

Your Answer

 
discard

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.