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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 $S$ ?

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    $\begingroup$ Fulton, intersection theory, Theorem 15.3. $\endgroup$
    – Samuel
    Aug 7, 2019 at 11:15
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    $\begingroup$ Heard of Grothendieck-Riemann-Roch? $\endgroup$
    – abx
    Aug 7, 2019 at 12:12
  • $\begingroup$ yes...i tried to calculate using that...but could not conclude about $c_2$. $\endgroup$
    – user130022
    Aug 7, 2019 at 13:24
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    $\begingroup$ I find $c_2= 3 C^2-\deg(E)$. $\endgroup$
    – abx
    Aug 7, 2019 at 15:53
  • $\begingroup$ 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. $\endgroup$
    – user130022
    Aug 8, 2019 at 5:08

1 Answer 1

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

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  • $\begingroup$ Thank you very much $\endgroup$
    – user130022
    Aug 10, 2019 at 9:25

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