Given the complex grassmannian variety G$(n, k)$, I consider the tautological bundle $S$, i.e. the $n$plane bundle whose fiber at each point of G$(n, k)$ is given by the corresponding $n$plane in $\mathbf{C}^k$. I consider now the Chern polynomial of $S$, $c(S)$. How can I explicitly compute the Chern roots of $c(S)$ (i.e. cohomology classes $f_i \in H^2(G(n, k), \mathbf{Z}) , i = 1, \dots, n$ such that $c(S) = \Pi_{i = 1}^n (1+ f_i t)$)?

$\begingroup$ I am a bit confused: one assume the existence of such $f_i$ to drive formulas for the Chern class of composite bundles. But I don't think such $f_i$ actually exist. In the case you have, $H^2(Gr(n,k))$ has single element $\sigma_1$; thus, if there are such $f_i$ they should be multiples of $\sigma_1$. This would imply, all other Chern classes are powers of $\sigma_1$, but this can't be true. For example, by the answer of Steven, $c_2(S)=\sigma_1^2\sigma_2$; and I am not sure if $\sigma_2$ is expressible in terms of $\sigma_1$. $\endgroup$– Mohammad FarajzadehTehraniCommented Dec 17, 2013 at 23:47
2 Answers
You are going to need to pass to an extension first. There is a bundle over $G(n,k)$ whose fibers are the complete flags on the vector spaces in the canonical bundle. Lets call it $q:P(n,k)\rightarrow G(n,k)$. The pullback of the conical $k$plane bundle to $P(n,k)$ now splits as a direct sum of line bundles, so by the sum formula for characteristic classes it factors as a product of linear factors as you wrote above. The problem is that the factorization is in the cohomology of the total space of $P(n,k)$. Its not that bad, because the cohomology of $P(n,k)$ is a module over the cohomology of $G(n,k)$.
However to get an explicit answer you are going to have to learn to do computations in the cohomology of $G(n,k)$.

$\begingroup$ By your last sentence, are you meaning Schubert calculus? $\endgroup$ Commented Mar 7, 2010 at 21:43

1$\begingroup$ Sorry for being such a slob. I was answering it off the cuff. The construction I gave is really standard and the fact that the bundle splits into line bundles is called the splitting principle. Indeed you compute in the cohomology of $G(n,k)$ with Schubert cells. If you read Milnor (and Stasheff's) Characteristic Classes, and work examples and talk to people, you will be able to do computations like this with elan. :) $\endgroup$ Commented Mar 7, 2010 at 22:04
Depending on what you need to do with the Chern roots, it may be cleaner to just ask for the Chern classes of $S$.
In this case, let $Q$ be the quotient bundle, i.e., there is a trivial bundle ${\bf C}^k$ which contains $S$ as a subbundle, and $Q = {\bf C}^k / S$. The $i$th Chern class of $Q$ is the cohomology class of $\sigma_i$, the special Schubert class of codimension $i$ (see Proposition 3.5.5 of Manivel's book Symmetric Functions, Schubert Polynomials, and Degeneracy Loci). Using the relation $c(S)c(Q) = 1$, and knowledge of the cohomology ring of $G(n,k)$ should be enough to perform any usual calculations.
For computing with these Schubert classes, one only needs to learn the Pieri rule (and perhaps the LittlewoodRichardson rule depending on the circumstance), both of which can be found in Manivel's book (chapter 1) or see Wikipedia.