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 $S$ ?



Chern class of direct image sheaf
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 $c\_2(i\_\*E)=\frac{1}{2}r(r+1)C^2 -d$.


