Subject: Re: question abt Grassmanian Date: Thu, 12 Jul 2001 03:26:41 +0000 From: Tom Goodwillie To: Don Davis Here's another way to do it. Let's generalize the statement to the flag manifold U(k_1+...+k_r)/U(k_1)x...xU(k_r) and then use the added generality to make a proof. Call the above manifold F(k_1,...,k_r) for short. Over it are bundles E_1, ... E_r. I claim that an additive basis for the cohomology is given by all monomials in the Chern classes c_j(E_i), i=1,...,r-1, j=1,...,k_i such that for each i the degree w.r.t. the c_j(E_i) is at most k_{i+1}+...+k_r. This statement in the Grassmanian case implies the general case. For example F(k_1,k_2,k_3) fibers over F(k_1,k_2+k_3) with fiber F(k_2,k_3). Use the Serre SS. You have cohomology bases for base and fiber; combining them you get a basis for E_2; but E_2=E_infty because all these spaces have their cohomology in even degrees. The same line of argument lets you start with the very well-known case of F(1,n-1) and deduce the case of F(1,1,...,1,n-k). But from this last case you can deduce the Grassmanian case (and I suppose the general case): The cohomology of F(k,n-k) maps into that of F(1,...,1,n-k). For i=1,...,k let c_i be the ith Chern class of the rank k bundle on F(k,n-k) and let t_i be the first Chern class of the ith line bundle on F(1,...,1,n-k). Then c_i is mapping to the ith elementary symmetric polynomial in the t_i's. We have P --> cohomology of F(k,n-k) | | v v Q --> cohomology of F(1,...,1,n-k) where P is the group of all polynomials of degree n-k or less in the indeterminates c_i and Q is the group of all polynomials in the indeterminates t_i having degree n-k or less **in each variable separately** The left vertical arrow (which is defined because each elementary symmetric polynomial has degree at most 1 in each variable) is split injective The lower horizontal arrow is split injective because our result holds in the case F(1,...,1,n-k). Therefore the upper horizontal arrow is also split injective. Tom Goodwillie