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.


1 Answer 1


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 $$\pi_*(c_{df}(S^{\vee} \otimes \pi^*F) \cap \pi^* \alpha) = \Delta^{(d)}_{f-k}(c(F-E)) \cap \alpha.$$

  • $\begingroup$ 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

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