Equality of chern classes and isomorphism

2010-10-29T14:58:42Z

Given two torsion free coherent sheaves $M$ and $N$ wit $rk(M)=rk(N)=r$ on an smooth projective surface $S$, by definition $det(M):=\Lambda^r(M)^{**}$.

Is the following criterion correct?

$M\cong N$ $\Leftrightarrow$ $M \hookrightarrow N$ and $c_i(M)=c_i(N)$ for $i=0,1,2$

One only has to look at "$\Leftarrow$":

So we have $0\rightarrow M\rightarrow N\rightarrow Q \rightarrow0$. Because of $c_i(M)=c_i(N)$ we see that $c_1(Q)=0$ and $c_2(Q)=0$. Since $M$ and $N$ have the same rank $codim(supp(Q))\geq 1$. We also have an induced map $det(M)\hookrightarrow det(N)$ of line bundles, i.e. $det(N)\cong det(M)\otimes O_S(D)$ for some effective divisor $D$. Now $det(Q)\cong det(N)\otimes det(M)^{-1}\cong O_S(D)$.

So by definition $c_1(Q)=c_1(det(Q))=D$, but $c_1(Q)=0$, so $D$ is effective and 0, i.e. $Q$ has no support in codimension 1, so $codim(supp(Q))\geq 2$.

So $Q$ is an Artinian sheaf, and for those one has $c_2(Q)=-dim(H^0(S,Q))$. Since $c_2(Q)=0$ we have $H^0(S,Q)=0$, but $H^0(S,Q)=\bigoplus\limits_{s\in supp(Q)} Q_s$. So $Q_s=0$ for all $s\in supp(Q)$, i.e. $Q=0$. So we have $M\cong N$.

Answer by ABayer for Equality of chern classes and isomorphism

2010-10-29T15:07:09Z

That's correct. A slightly shorter argument is: if $\mathcal{Q}$ has support in codimension $d$, then its Chern character $\mathrm{ch}_d(\mathcal{Q})$ is non-zero and effective. So a sheaf is trivial if and onlfy if $\mathrm{ch} = 0$, which is true if and only if the rank and the Chern classes vanish. In particular, the result holds for any dimension.

Equality of chern classes and isomorphism - MathOverflow

Equality of chern classes and isomorphism

Answer by ABayer for Equality of chern classes and isomorphism