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

Hello!

Let $n,m\geq 0$ be integers. If I understand it correctly, there is the following description of the cohomology of the complex Grassmannian $\text{Gr}(m+n;m)$: denote by $\text{Sym}(n,m)$ the subring of the polynomial ring in $n+m$ variables consisting of polynomials which are symmetric in the first $n$ and the last $m$ variables, and by $\text{Sym}(n+m)$ the ring of symmetric polynomials in $n+m$ variables. Then $\text{Sym}(n,m)$ is a free $\text{Sym}(n+m)$ module, and its graded rank (written as a $q$-polynomial) equals the Poincare polynomial of the Grassmannian $\text{Gr}(m+n;m)$.

I'd like to know if this is true for more general flag varieties: Is it true that the graded rank of $\text{Sym}(m_1,...,m_k)$ over $\text{Sym}(m_1+...+m_k)$ is the Poincare polynomial of the variety of flags $(\{0\}=U_0,U_1,...,U_k=V)$ in $V := {\mathbb C}^{m_1+...+m_k}$ such that $\text{dim}(U_{i}) - \text{dim}(U_{i-1}) = m_i$ for $i=1,...,k$? If yes: how to prove it? :-)

On the other hand, there is a presentation of the cohomology of the full flag variety of ${\mathbb C}^N$ as the quotient of ${\mathbb C}[X_1,...,X_N]$ by the ideal generated by the symmetric polynomials of positive degree. How do these two descriptions of the cohomology of flag varieties relate?

Thank you!

Hanno

share|improve this question
    
Small comment on terminology: usually "flag variety" has a narrower meaning in terms of the quotient of a reductive group (e.g., general or special linear group) by a Borel subgroup. In your situation the quotient involves more general parabolic subgroups, where the terminology "generalized flag variety" is widely used. A further suggestion is to add a broader tag such as ag.algebraic-geometry. –  Jim Humphreys Oct 1 '10 at 16:52
    
Jim: Thank you, I changed the title and the tags! –  Hanno Becker Oct 1 '10 at 17:13

2 Answers 2

up vote 4 down vote accepted

Yes, this is all true. The cohomology of the partial flag variety $F(m_1,\dots,m_k)$ is the quotient of the polynomials that are symmetric under $S_{m_1}\times \cdots \times S_{m_k}$ by the positive degree ones which are fully symmetric (a special case of this is the presentation you mention for the full flag variety). Since partially symmetric polynomials are a free module over full symmetric ones, the q-rank of the free module is the same as the q-dimension of the quotient by all positive degree fully symmetric polynomials.

The "reason" this presentation works is that $F(m_1,\dots,m_k)$ carries tautological vector bundles $U_i/U_{i-1}$ (these are subquotients of the trivial bundles of rank $m_1+\cdots m_k$). The Chern classes of these bundles are the elementary symmetric polynomials in the variables corresponding to $S_{m_i}$; thus, the fully symmetric polynomials, written in terms of these are the Chern classes of the the trivial bundle of rank $m_1+\cdots m_k$ by the Whitney sum formula (which says that if I think of $c_i(V)$ as elementary symmetric functions in one set of variables, and $c_j(W)$ as elementary symmetric functions in another, the Chern classes $c_k(V\oplus W)$ are the elementary symmetric functions in the union of the variables) and the splitting principle (which says that when I have a filtration $ V_1\subset V_2\subset \cdots\subset V_n=V$, then $c_i(V)=c_i(V/V_{n-1}\oplus \cdots \oplus V_2/V_1\oplus V_1)$) are thus 0. Admittedly, you then have to do some actual geometry to see that these Chern classes generate and that these are all the relations, but to me, this is the heart of the argument.

share|improve this answer
    
@Ben: Thanks alot, this is awesome. I didn't notice that for a homomorphism of positively graded connected k-algebras $R_{\bullet}\to S_{\bullet}$ making $S_{\bullet}$ into a free graded $R_{\bullet}$-module of finite rank, the q-rank equals the q-dimension of the quotient S_{\bullet}/⟨R+⟩. That's nice :-) Concerning the actual computation of $H^{\ast}(F(m_1,...,m_k);{\mathbb C})$`, I have some questions, see next comment. –  Hanno Becker Oct 1 '10 at 19:34
    
First, what do you mean with "The Chern classes of these bundles are the elementary symmetric polynomials in the variables corresponding to $S_{m_i}$"? Do you consider the cohomology of $F(m_1,...,m_k)$ as a subring of the cohomology of the full flag variety F via the canonical map $F\to F(m1,...,mk)$? Then what you say is a consequence of the Whitney sum formula, too, isn't it? –  Hanno Becker Oct 1 '10 at 19:44
    
Roughly, yes, though you can state everything more canonically in terms of Chern roots. –  Ben Webster Oct 1 '10 at 21:25
    
Now I figured out how to compute the cohomology of $F(m_1,...,m_k)$ wit the knowledge of $F(m_1+...+m_k)$ using Leray Hirsch for the fibre bundle $F(m_1)\times ...\times F(m_k)\to F(m_1+...+m_k)\to F(m_1,...,m_k)$. Concerning Chern roots: So you want to consider the Chern classes as the elementary symmetric polynomials in the Chern roots computed after choice of an arbitrary map splitting the bundle after pullback? That's nice! But how would you proceed to show that there are no more relations, and that the chern classes generate the cohomology? –  Hanno Becker Oct 2 '10 at 7:17
    
well, the dimension is right by Schubert decomposition, so it's enough to show injectivity or surjectivity. For surjectivity, there are a couple of arguments; you can use Eilenberg-Moore as Torsten does, or I think you could use Kirwan surjectivity. Localization in equivariant cohomology would also work. It's like that there's some more down to earth method I'm just not thinking of as well, like writing down a explicit set of Chern classes which are upper-triangular with respect to Schubert classes, but I couldn't point you to such an argument. –  Ben Webster Oct 2 '10 at 8:07

One way is to use the Eilenberg-Moore spectral sequence applied to the fibration $$ \mathrm{Fl}_{m_1,\ldots,m_k} \rightarrow BU_{m_1}\times\cdots\times BU_{m_k}\rightarrow BU_{m_1+\cdots+m_k} $$ which is a spectral sequence $$ \mathrm{Tor}_*^{\mathrm{Sym}(m_1+\ldots+m_k)}(\mathrm{Sym}(m_1,\ldots,m_k),\mathbb Z) \implies H^*(\mathrm{Fl}_{m_1,\ldots,m_k}). $$ However, as $\mathrm{Sym}(m_1,\ldots,m_k)$ is projective as module over $\mathrm{Sym}(m_1+\ldots+m_k)$ (by for instance the fact that is a finite Cohen-Macaulay module over a regular ring of the same dimension) the sequence collapses and gives $$ H^*(\mathrm{Fl}_{m_1,\ldots,m_k}) = \mathrm{Sym}(m_1,\ldots,m_k)\bigotimes{}_{\mathrm{Sym}(m_1+\ldots+m_k)}\mathbb Z $$ which gives what you want.

share|improve this answer
    
Torsten: Thank you very much! I'm not very familiar with the EMSS, so I will have to spend some time thinking about what you say before I can give a smart comment ;-) If something's unclear, I will ask, if that's ok. –  Hanno Becker Oct 1 '10 at 19:49

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.