Let $f: X \to S$ be an arithmetic surface, where $S=\operatorname{Spec } O_K$ for a number field $K$. It is well known that if we want to introduce a reasonable intersection theory on $X$ we have to deal with Arakelov geometry. In this question I want to stick to the usual theory of schemes, so no fibres at infinity are involved.

I'd like to know if there exist a unique pairing:

$$(,):\operatorname{Div} X\times\operatorname{Div} X\to \operatorname{Div} S$$

Satisfying the following properties:

1) it is bilinear and symmetric

2) It descends to the so called Deligne pairing:

$$\operatorname{Pic} X\times\operatorname{Pic} X\to \operatorname{Pic}S$$

3) If $D,E$ are two smooth divisors on $X$ meeting transversally, then:

$$(D,E)=\sum_{x\in D\cap E} [k(x): k(f(x))] f(x)$$

My idea is that one can follow the classical proof of the existence of an intersection pairing for algenbraic surfaces (see Hartshorne chap. 5 for example) and just modify things for arithmetic surfaces. What do you think?

Remark: Keep in mind that I'm not asking about the existence of an intersection pairing, but just about this "cup product" at the level of divisors. For example, if we composed such a map with the degree of divisors on $S$, this wouldn't give an intersection pairing because the degree is not $0$ on principal divisors.

  • 2
    $\begingroup$ In Deligne's famous letter, "Le d'eterminant de la cohomologie", he defines the Deligne pairing on divisor classes by first defining a multilinear, symmetric pairing for ordered tuples of sufficiently transverse divisors (not divisor classes). Then he uses a moving lemma (basically the trick of writing each divisor class as a difference of very ample divisor classes) to prove that this gives a well-defined pairing on divisor classes. $\endgroup$ Jul 27 '18 at 15:46
  • 2
    $\begingroup$ I have a correction to my previous comment. I just checked Deligne's letter, specifically Section 6, pp. 141--142. His pairing has domain $\text{Div}\ X \times \text{Div}\ X$, but the target is $\text{Pic}\ S$. His proof of symmetry of the pairing uses computations with invertible sheaves on $S$, not divisors. $\endgroup$ Jul 27 '18 at 16:07
  • $\begingroup$ Yes it seems that the target is the Picard group of $S$. My problem is to define something with image in the group of divisors. $\endgroup$
    – manifold
    Jul 27 '18 at 16:12

As mentioned in the second comment, the "usual" construction requires that the target is the divisor class group, rather than the group of divisors. In the geometric case, this is clear: how do you define the pairing of $D$ and $E$ in the nontransverse case, i.e., when $E$ equals $D$? As a divisor class, there are several ways to defining the pairing of $D$ with itself, e.g., form the pushforward to $S$ of the divisor class of the restriction to $D$ of the invertible sheaf $\mathcal{O}(D)$ of $D$. If $D$ surjects to $S$, I see no way to naturally choose a divisor in this divisor class. You can "arbitrarily" choose a divisor in that class, but then the construction is unlikely to be multilinear.


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.