# Chern class of direct image sheaf

Let $$C$$ be a smooth projective curve in a surface $$S$$. Suppose $$E$$ is a vector bundle of rank $$r$$ on $$C$$. Then what is the total Chern class of the sheaf $$i_*E$$, where $$i$$ is an embedding of $$C$$ in $$S$$ ?

• Fulton, intersection theory, Theorem 15.3.
– ssx
Commented Aug 7, 2019 at 11:15
• Heard of Grothendieck-Riemann-Roch?
– abx
Commented Aug 7, 2019 at 12:12
• yes...i tried to calculate using that...but could not conclude about $c_2$. Commented Aug 7, 2019 at 13:24
• I find $c_2= 3 C^2-\deg(E)$.
– abx
Commented Aug 7, 2019 at 15:53
• Suppose $E$ is a globally generated line bundle of degree $d$, the elementary transformation $E$ of $H^0(L) \otimes O_S --->i_*L$ has $c_2 = d$. But we have $c_t(i_*L)c_t(E)=1$, which implies $(1 +tC +t^2c_2(L))(1+t(-C) +t^2.d)= 1$, which gives $c_2(L) = C^2-d.$. Please correct me if i am wrong. Commented Aug 8, 2019 at 5:08

The following is the way I like to do this computation. It is entirely equivalent to using GRR, and ends up being longer, but is a little more elementary. The final answer will be $$c_1(i_*E)=rC$$ and $$c_2(i_*E)=\frac{1}{2}r(r+1)C^2 -d$$ where $$r=rk(E)$$ and $$d=deg(E)$$.
We may assume for this sort of computation that there is a vector bundle $$F\to S$$ such that $$F|_C = E$$. This might not be literally true, but it is sort of like the spitting principle, you can assume it for cohomological computations. Then tensoring the exact sequence $$0\to \mathcal{O}_S(-C) \to \mathcal{O}_S \to \mathcal{O}_C \to 0$$ by $$F$$ and taking Chern characters we get \begin{align*} ch(i_*E) &= ch(F) - ch(F)ch(\mathcal{O}(-C)) \\ &=(r,c_1(F),ch_2(F))\cdot (0,C,-\frac{1}{2}C^2)\\ &=(0,rC, -\frac{r}{2}C^2 +c_1(F)\cdot C)\\ &=(0,rC,-\frac{r}{2}C^2 +d) \end{align*} So we have that $$ch_2(i_*E) = -\frac{r}{2}C^2 +d = \frac{1}{2}c_1^2(i_*E)-c_2(i_*E) = \frac{1}{2}(rC)^2-c_2(i_*E)$$ Solving for $$c_2(i_*E)$$ we find $$c_2(i_*E)=\frac{1}{2}r(r+1)C^2 -d$$.