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 $M\cong N$.


1 Answer 1


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.

  • $\begingroup$ Thanks, that's interesting and more general. What does effective mean in this case, e.g. in the Chow ring? $ch_d(Q)$ is a sum of codimension d subvarieties, and all coefficient are non negative? And how can i see that $ch_d(Q)$ has to be effective in this case? – $\endgroup$
    – TonyS
    Oct 29, 2010 at 15:21
  • 1
    $\begingroup$ Yes, effective means it is a non-zero sum of classes of subvarieties. You can deduce this from Grothendieck-Riemann-Roch - the key point is that the Todd class always starts with 1 in the top degree. However, a better argument is to use Hirzebruch-Riemann-Roch: if the Chern character is trivial, then the Hilbert polynomial is trivial by HRR, hence the sheaf is trivial. $\endgroup$ Oct 30, 2010 at 15:06

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