Question

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 $c(S) = \Pi_{i = 1}^n (1+ f_i t)$)?

Answer 1

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 $G(n,k)$.

Answer 2

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 Littlewood-Richardson rule depending on the circumstance), both of which can be found in Manivel's book (chapter 1) or see Wikipedia.