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 $S$ ?
 A: 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 $c_2(i_*E)=\frac{1}{2}r(r+1)C^2 -d$.
