Let $X$ be a smooth projective variety of dimension $n$. Take the bundle $TX \oplus Sym^2(TX)$ over $X$ where $Sym^2(TX)$ is the second symmetric product of the tangent space. The Grassmannian bundle $Gr(n,TX \oplus Sym^2(TX))$ has a canonical section, namely $TX$.

My question is: what is the Poincare dual of this section in the cohomology ring of the Grassmannian bundle?

The cohomology ring is

$H^*(Gr(n,TX \oplus Sym^2(TX)))=H(X)[c_1,\ldots, c_n,d_1,\ldots, d_{{n+1 \choose 2}}]/$

$(1+c_1+\ldots +c_n)(1+d_1+\ldots +d_{{n+1 \choose 2}})=c(TX \oplus Sym^2(TX))$

This is probably a trivial question I am a bit confused about it now.

  • $\begingroup$ Hi Gergely! This should be not extremely difficult but for sure not completely trivial... I'll think about it later, if no one answers you meanwhile! All the best! $\endgroup$ – diverietti Jan 26 '11 at 11:57
  • $\begingroup$ My guess: it is the top Chern class of $Hom(TX,Sym^2(TX))$, that is $\prod(\beta_i-\alpha_j)$ where $\beta_i$s and $\alpha_j$s are the Chern roots of the $d$`s and $c$'s respectively, i.e $\prod(1+\beta_i)=1+d_1+\ldots$ $\endgroup$ – Gergely Berczi Jan 26 '11 at 12:18

Let $U$ denote the tautological subbundle. I guess that $c_i = c_i(U^*)$ in your notation. Consider the composition map $$ U \to p^*(TX + S^2TX) \to p^*S^2TX, $$ where $p:Gr \to X$ is the projection. Then your section is the zero locus of this map. In other words, it is the zero locus of a global section of the vector bundle $U^*\otimes p^* S^2TX$. The section is regular, so the class of the zero locus equals the top Chern class of the bundle. Thus the answer is $$ c_{n^2(n+1)/2}(U^*\otimes p^*S^2TX). $$ The rest is a straightforward computation.


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