Over the rational numbers,there is a well defined product $$\text{CH}^p(X) \otimes \text{CH}^q(X) \to \text{CH}^{p+q}(X), A \otimes B \mapsto A\cdot B$$ To an algebraic cycle $Z$ one can associate its structure sheaf $\mathcal{O}_Z$. For that map to factor through $\text{CH}$ one probably has to consider some equivalence relation on sheafs, too. I was wondering, if one could relate $\mathcal{O}_{A\cdot B}$ to some (derived?) tensor product of the structure sheafs $\mathcal{O}_A,\mathcal{O}_{B }$.

  • $\begingroup$ How is $A\cdot B$ defined ? $\endgroup$ – user18119 Feb 28 '13 at 17:36
  • $\begingroup$ Are you asking because you already know this works for surfaces? If you are unfamiliar with the surface case, here is a blog post that discusses it: amathew.wordpress.com/2013/01/28/… $\endgroup$ – Matt Feb 28 '13 at 17:57

$\def\cO{\mathcal{O}}$Literally speaking, $\cO_{A \cdot B}$ doesn't make sense, because $A \cdot B$ is an equivalence class of cycles modulo rational equivalence. However, I will take your question to be asking "can I perform some natural computation on $\cO_A$ and $\cO_B$ which allows me to find $A \cdot B$". The answer is yes.

Throughout the answer, I will take the ambient space $X$ inside which $A$ and $B$ live to be smooth.

First all, a reminder of how to turn sheaves into Chow classes. Let $\mathcal{E}$ be a coherent sheaf on $X$, supported on $\bigcup_{i=1}^r Z_i$. Here the $Z_i$ are irreducible subvarieties of $X$. Let $k(Z_i)$ denote the fraction field of $Z_i$, so $\mathrm{Spec}\ k(Z_i)$ is the generic point of $Z_i$. Let $d = \max_i(\dim Z_i)$. Then we define a class $[\mathcal{E}]$ in $CH_d(X)$ by $$\sum_{\dim Z_i = d} \left( \dim_{k(Z_i)} \mathcal{E}_{\mathrm{Spec}\ k(Z_i)} \right) \cdot [Z_i].$$

In many cases, we have $[\cO_A] \cdot [\cO_B] = [\cO_{A \cap B}]$. Here I mean scheme-theoretic intersection: $\cO_{A \cap B} = \cO_A \otimes \cO_B$. What does many cases mean?

(1) Whenever the intersection $A \cap B$ is reduced and of expected dimension. This is, of course, is the motivation for calling the product in Chow "intersection product".

(2) More generally, if $A \cap B$ is of expected dimension and either $A$ or $B$ is a hypersurface.

(3) More generally, if $A \cap B$ is of expected dimension and $A$ and $B$ are both Cohen-Macaulay.

The simplest example to demonstrate that $[\cO_A] \cdot [\cO_B]$ is not representated by $\cO_{A \cap B}$ is the following: In $\mathbb{P}^4$, with homogenous coordinates $(v:w:x:y:z)$, let $A$ be the $\mathbb{P}^2$ given by the equations $\{ u=v,\ w=x \}$. Let $B_1$ be the $\mathbb{P}^2$ given by $\{ u=w=0 \}$ and let $B_2$ be the $\mathbb{P}^2$ given by $\{ v=x=0 \}$. Let $B$ be the reduced union $B_1 \cup B_2$.

Clearly, $A \cap B_1$ and $A \cap B_2$ are each reduced points. So, in Chow, we have $[A] \cdot [B] = 2 \cdot [\mathrm{pt}]$. However, if you work it out directly, you'll see that the length of $\cO_A \otimes \cO_B$ is $3$, not $2$.

If $A \cap B$ has expected dimension, the corrected version is given by Serre's Tor formula. $$[\cO_A] \cdot [\cO_B] = \sum_{r} (-1)^r [\mathcal{T}or_r^X(\cO_A, \cO_B)].$$ For a good discussion of Serre's formula in derived algebraic geometry, see this MO thread.

If $A \cap B$ does not have expected dimension, then you need to use $K$-theory (just $K_0$) or something more sophisticated. See this MO thread and the first Section of this paper of Brion for a good intro. To emphasize what is written there, in order to turn $K_0$ classes (or elements of the derived category) into Chow classes, one needs to either take the associated graded ring of a filtered ring, or apply the Chern character map. $[A] \cdot [B]$ is the image of the derived tensor product when you apply those constructions.

$K_0$ is the de-categorification of the derived category of coherent sheaves, so everything that is written here could be described in the derived language. (To be precise, to a complex $\mathcal{E}^0 \to \mathcal{E}^1 \to \cdots \to \mathcal{E}^r$ in the derived category, assign the class $\sum (-1)^i [\mathcal{E}^i]$ in $K_0$.) However, I don't think we are actually using the derived category in any deep way here.

  • $\begingroup$ Thank you for that great answer. $\endgroup$ – orbifold Mar 5 '13 at 15:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.