0
$\begingroup$

Given a variety $X$, a coherent sheaf $F$ which may not have finite locally free resolution, and a curve(integral) $C$ in $X$. How can we define intersection $F.C$ properly?(I mean 'properly' by having good behavior on short exact sequence and have projection formula)

Here is my attempt:

  1. One can define $F.C$ as degree of $F|_C$ on $C$, suggest by Riemann-Roch, namely $\chi(F|C)-rank(F|C)\chi(\mathcal{O}_C)$. However, in this setting, projection formula only hold for locally free sheaf。

  2. we take locally free resolution of $F$, and then pull them back(in fact we get $Li^\ast(F)$, where $i:C\rightarrow X$ is the inclusion) to $C$ and define intersection of $Li^\ast(F)$ with $C$. Where we have problem that $Li^\ast(F)$ may be unbounded.

So both attempt fail.

$\endgroup$
1
  • $\begingroup$ I don't think you can. This is reflected in the fact that Chow homology in the sense of Baum-Fulton-MacPherson does not have a ring structure. Notice also that the Grothendieck group of coherent sheaves does not have a ring structure in general. $\endgroup$ Commented Apr 2, 2013 at 8:53

1 Answer 1

2
$\begingroup$

The problem boils down to the intersection of a curve and a Weil divisor (which is not necessarily Cartier divisor). Indeed, for any sheaf $F$ its first Chern class is well defined as a Weil divisor, and you can define $F\cdot C := c_1(F)\cdot C$.

So the question is how one can define the intersection $D\cdot C$. Sometimes it is easy. For example, if $X$ is $Q$-factorial then for some $n$ the divisor $nD$ is Cartier, so the intersection $(nD)\cdot C$ is well defined and you can put $D\cdot C := ((nD)\cdot C)/n$. Note that this also works if $X$ is $Q$-factorial only along $C$.

If $X$ is not $Q$-factorial, probably you can take its $Q$-factorial modification and take the intersection of the proper preimages of $D$ and $C$ on it, but I am not sure that this is well defined.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .