Let's consider a vector bundle $E$ of rank $n$ over a compact manifold $X$. Consider the associated Grassmannian bundle $G$ for some $k < n$, obtained by replacing each fiber $E_x$ by $Gr(k,E_x)$.

Let's suppose that there is a full flag of subbunldes $F_1 \subset F_2 \dots \subset F_n \subset E$. I think that in this case we are able to define relative Schubert cycles on G which restrict to usual Schubert cycles on each fiber so that we can apply Leray-Hirsh theorem to deduce that $H^*(G) = H^*(X) \otimes H^*(Gr(k,n))$.

  1. Is the reasoning above correct?
  2. Can we still compute $H^*(G)$ in the case when the full flag of subbundles doesn't exist?

EDIT: I meant complex vector bundles and complex Grassmannians. Also the bundle can be assumed holomorphic or algebraic if it makes a difference.

EDIT: Ben in his answer mentions Serre's spectral sequence that can be used in this case. Is there a reason why it will degenerate to leave $H^*(X) \otimes H^*(Gr(k, n))$ as a result?


Re 1.: the reasoning is correct additively, but not multiplicatively (did you check the case $k=1$?).

Re 2.: given any complex bundle $E$ over $X$ of rank $n$, the cohomology of the associated Grassmannian bundle of $k$-planes is $$H^{\bullet}(X,\mathbf{Z})\otimes\mathbf{Z}[c_1,\ldots,c_k,c_1',\ldots,c_{n-k}']/(1+c_1+\cdots +c_k)(1+c_1'+\cdots +c_{n-k}')=c(E)$$ where $\deg c_i=\deg c'_i=2i$ and $c(E)$ is the total Chern class of $E$.

This a modification of the Grothendieck trick which computes the cohomology of the projectivized vector bundle.

upd: here is a sketch of a proof of the above formula. On the total space $G$ of the Grassmannian vector bundle there is the tautological $k$-plane bundle $S$, which is a subbundle of the pullback $E'$ of $E$ under the bundle projection $p:G\to X$. Let $Q$ be the quotient bundle. We have $c(S)c(Q)=c(E')$. So we get an algebra map from the above algebra to the cohomology of $G$ (taking an element of $H^{\bullet}(X,\mathbf{Z})$ to its pullback, $c_i$ to $c_i(S)$ and $c_i'$ to $c_i(Q)$) and we have to show that it is an isomorphism.

Surjectivity: let us pick a $\mathbf{Z}$-basis of the cohomology of the Grassmannian, express each element as a polynomial in the Chern classes of the tautological bundle and take the resulting polynomials in $c_i(S)$'s. When restricted to any fiber these form a $\mathbf{Z}$-basis of the cohomology of the fiber, so by the Leray-Hirsch principle, $H^{\bullet}(G,\mathbf{Z})$ is a free $H^{\bullet}(X,\mathbf{Z})$-module spanned by these classes.

Injectivity: the above algebra is also a free module over $H^{\bullet}(X,\mathbf{Z})$ and the the above map from it to $H^{\bullet}(G,\mathbf{Z})$ takes a basis to a basis.

  • $\begingroup$ sorry, hit the wrong button the first time around $\endgroup$ – algori Jan 13 '10 at 5:48
  • $\begingroup$ Is it correct then that the additive structure doesn't depend on c(E)? $\endgroup$ – Evgeny Shinder Jan 13 '10 at 15:47
  • 1
    $\begingroup$ Evgeny -- yes, for a simple reason: over the (total space of the) Grassmannian bundle there is the tautological vector bundle. The Chern classes of this bundle restricted to any fiber of the Grassmannian bundle generate multiplicatively the cohomology of the fiber, which is why we can apply Leray-Hirsch. $\endgroup$ – algori Jan 13 '10 at 17:43

That sounds reasonable, though remember, asking for a full flag like that is a VERY restrictive condition; in the smooth category, it's equivalent to asking your vector bundle to be a sum of line bundles.

For the general case, there is a spectral sequence of which you've written the $E^2$-term. In cases where the base is nice, you might have some control over what's going on (for example, in the complex case, if the base has no odd cohomology, you're set). The most helpful general statement I know is the "Hirsch lemma." See 3.1 in this paper of Deligne, Griffiths, Sullivan and Morgan.


algori's answer is completely correct. In case you want a reference, this material is in Chapter 14 of Fulton's Intersection Theory. (Except that Fulton is writing about Chow rings and you asked the question in cohomology.) In particular, the additive structure is Proposition 14.6.5 and the ring structure is 14.6.6.

I'm not sure what the best references are for the analogous statements in cohomology, but you could try tracing Fulton's references and see if they help.

  • $\begingroup$ Example 19.1.11 would then provide the answer in some situations - but not in general it seems? $\endgroup$ – mdeland Jan 13 '10 at 22:23

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