Segre classes vs Chern classes

For a smooth projective variety X and a coherent sheaf S on it, you can consider the projective cone $\pi:\mathbb{P}S \rightarrow X$. You can consider $\mathcal{O}(1)$ on $\mathbb{P}S$ and define the Segre series a la Fulton:

$$s(S,t) = \pi_*\left(\frac {1}{t-c_1(\mathcal{O}(1))}\right) \in A^*(X)$$ I can also take a finite projective resolution of $S$ by vector bundles, and ask how to express $s(S,t)$ in terms of the Chern classes of these bundles.

• It appears that the fraction in your displayed formula is improperly formed; that;s why the formula doesn't parse properly. – Thierry Zell Sep 22 '11 at 3:44
• I've edited the LaTeX, please check if I have ruined what you meant to write. – David Roberts Sep 22 '11 at 4:36

For a vector bundle $E$ its Segre class is defined as $$s(E) = \prod(1+x_i)^{-1},$$ where $x_i$ are Chern roots of $E$. Because of this it is clear that Segre class is multiplicative in short exact sequences. Therefore for a coherent sheaf $S$ if $$0 \to E_n \to \dots \to E_2 \to E_1 \to E_0 \to S \to 0$$ is a locally free resolution then $$s(S) = s(E_0)s(E_1)^{-1}s(E_2)\cdots(E_n)^{(-1)^n}.$$
In this case (proj dim of $\mathcal{S} \geq 2$), one should not expect that $s(\mathcal{S}) = c^{-1}(\mathcal{S})$. In fact, the example I care about is such that the coherent sheaf is supported on a codimension 2 subvariety of $X$ and one has:
$$s(\mathcal{S},t) = c^{-1}(\mathcal{S},t) - 1 + t\cdot c_1(\mathcal{S})$$ I did this computation in mathematica, via equivariant localization, for the very particular case I had in mind. But I do not know how to prove it, or what does it generalize to.