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Oct 2, 2010 at 9:05 comment added Hanno Ok, thanks alot for your help, Ben!
Oct 2, 2010 at 8:07 comment added Ben Webster 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.
Oct 2, 2010 at 7:30 vote accept Hanno
Oct 2, 2010 at 7:17 comment added Hanno 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?
Oct 1, 2010 at 21:25 comment added Ben Webster Roughly, yes, though you can state everything more canonically in terms of Chern roots.
Oct 1, 2010 at 21:20 history edited Ben Webster CC BY-SA 2.5
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Oct 1, 2010 at 19:44 comment added Hanno 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?
Oct 1, 2010 at 19:34 comment added Hanno @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.
Oct 1, 2010 at 18:47 history answered Ben Webster CC BY-SA 2.5