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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 $\mathcal{O}_A,\mathcal{O}_{B }$.

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How is $A\cdot B$ defined ? –  user18119 Feb 28 '13 at 17:36
    
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/… –  Matt Feb 28 '13 at 17:57

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$\def\cO{\mathcal{O}}$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.

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Thank you for that great answer. –  orbifold Mar 5 '13 at 15:04

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