2
$\begingroup$

Let $E\to X$ be a rank $r$ holomorphic vector bundle on a $n$-dimensional compact complex manifold. Then, it is well known that one can recover the Segre classes of $E$ as follows. Let $\pi\colon P(E)\to X$ the projectivized bundle of lines of $E$, and $\mathcal O_E(1)\to P(E)$ the corresponding (anti)tautological line bundle.

The cohomology ring $H^\bullet(P(E),\mathbb Z)$ is then given by $$ H^\bullet(P(E),\mathbb Z)\simeq H^\bullet(X,\mathbb Z)[\xi]/(\xi^r+\pi^*c_1(E)\cdot\xi^{r-1}+\cdots+\pi^*c_r(E)), $$ where $\xi=c_1(\mathcal O_E(1))$. Therefore, since $\pi_*\xi^{r-1}=1$, and the total Chern class $c_\bullet(E)$ is the formal inverse of the total Segre class $s_\bullet(E)$, one can show that $$ \pi_*\xi^{r-1+k}=s_k(E). $$ Now, Segre classes are just particular Schur polynomials in the Chern classes of $E$, namely those corresponding to the partitions of the form $1+\cdots+1$.

Question. Is there an analogous geometric construction to obtain all the Schur polynomials in the Chern classes of $E$ as a direct image of (the appropriate self-intersection of) the first Chern class of a "tautological" line bundle?

For instance, what happens if we consider the (complete or an incomplete) flag manifold of $E$ with its tautological line bundles?

Thanks in advance.

$\endgroup$

1 Answer 1

3
$\begingroup$

The following Proposition 14.2.2, page 248 in Fulton's book Intersection Theory might interest you.

Proposition. Let $E$ be a vector bundle of rank $n$ on $X$, $d \leq n$ and let $G_d(E)$ be the Grassmann bundle of $d$-planes in $E$ with projection $\pi \colon G_d(E) \to X$.

Let $S$ be the universal subbundle of $\pi^*E$, which has rank $d$, and set $k=n-d$.

Finally, let $F$ be a vector bundle of rank $f$ on $X$. then for all $\alpha \in A_* X$ we have $$\pi_*(c_{df}(S^{\vee} \otimes \pi^*F) \cap \pi^* \alpha) = \Delta^{(d)}_{f-k}(c(F-E)) \cap \alpha.$$

$\endgroup$
1
  • $\begingroup$ Uhm... Thanks... I have to understand a little bit the notations and the meaning! I'll see! $\endgroup$
    – diverietti
    Jul 26, 2012 at 16:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.