# How to calculate the Chern class of the tensor product of a torsion free sheaf with a line bundle

I'am try to work with Chern class of the coherent sheaves, in this sense. If I have a vector bundle $E$ of rank $r$ and $L$ a line bundle we have the Chern class property

$$c_{r}(E\otimes L) = \sum_{i = 0}^{r}c_{i}(E)c_{1}(L)^{r-i}.$$

I need a similar result for coherent sheaf.

If I have a coherent sheaf $\mathcal{F}$ over a smooth projective variety X we consider a locally free resolution

$$0 \longrightarrow E_{n} \longrightarrow \cdots \longrightarrow E_{0} \longrightarrow \mathcal{F} \longrightarrow 0.$$

We can define the total Chern class of $\mathcal{F}$ by

$$c(\mathcal{F}) = \displaystyle\Pi_{i = 0}^{n} c(E_{i})^{(-1)^{i}}.$$

My question: Is true that $c_{n}(\mathcal{F}\otimes L) = \sum_{i = 0}^{n}c_{i}(\mathcal{F})c_{1}(L)^{n-i},$ for the coherent sheaf $\mathcal{F}$? If this does not exist, we can have the similar relationship?

Thank you !!

## 1 Answer

No, this is false. Take the simple example $\mathcal{F}=\mathcal{O}_Y$, where $Y$ is a hypersurface in $X$. Using the resolution $0\rightarrow L(-Y)\rightarrow L\rightarrow L\otimes \mathcal{O}_Y\rightarrow 0$, one gets $c_1(L\otimes \mathcal{O}_{Y})=c_1(L)-c_1(L(-Y))= [Y]$, so it is independent of $L$, while your formula gives $c_1(L\otimes \mathcal{O}_{Y})=c_1(L)+ c_1(\mathcal{O}_{Y})=c_1(L)+[Y]$.

To get a general formula it is much easier to use the Chern character: we have $\operatorname{ch}(\mathcal{F}\otimes L)=\operatorname{ch}(\mathcal{F})\operatorname{ch}(L)$. Indeed the formula is classical if $\mathcal{F}$ is a vector bundle, and it extends readily to coherent sheaves using a finite locally free resolution.

• abx, thank you! It's great, very well. Thus if I need to calculate the Chern class of a coherent sheaf the most coerent is to use a locally free resolution. – Student85 Jun 19 '18 at 23:20