47

Let's define intersection numbers as follows.

Consider a collection $f_1,\dots, f_n$ of holomorphic functions on some neighborhood of zero in $\mathbb C^N$ cutting out divisors $D_1$, all of which vanish at $0$. Define $$\omega(f_1,\dots, f_n)=\frac{df_1}{f_1}\wedge\dots \wedge \frac{df_n}{f_n}.$$ We say the local intersection number is defined by $$(D_1,\dots, D_n)=\operatorname{Res}_{\{0\}}\omega(f_1,\dots, f_n).$$

Clearly this is a local definition and the obvious modifications give a definition of intersection number for hypersurfaces intersecting at nonzero points.

I'm looking for a complex-analytic proof that this definition of intersection satisfies the general version of Bézout's theorem for $n$ hypersurfaces in $\mathbb P^n$ (as opposed to one that shows this is equivalent to the algebraic definition in terms of local rings and then uses a standard algebraic proof). Surely this is written down somewhere. Where can I find it?

Potato
  • 41,411
  • 3
    Maybe Griffiths/Harris, Ch. 5, specifically p. 670 is what you are looking for. – Takumi Murayama Sep 10 '15 at 13:45
  • @TakumiMurayama That idea (using the global residue theorem) looks right, but I'd have to work out the details. Note that unfortunately they don't provide an explicit proof. (And where is the global residue theorem proved?) – Potato Sep 10 '15 at 17:40
  • @TakumiMurayama So the global residue theorem is on p. 656, I think. But now I'm confused about something simple: the residue theorem says a certain sum is zero, but we want a sum that equals $\Pi d_i$. In particular, if we take the form to be $\frac{df_1}{f_1}\wedge\dots\wedge \frac{df_i}{f_i}$, doesn't the global residue theorem tells us the sum of the residues is $0$ and not the product of the degrees, as we desire? – Potato Nov 11 '15 at 05:12
  • Could this be relevant? – Lonidard Nov 11 '15 at 07:27
  • 1
    @TakumiMurayama Actually I think I'm being really stupid. Consider for example $f_1 = x, f_2 = y$. in $\mathbb P^2$. We have the right residue at $0$, and then I think we pick up the extra stuff when we switch charts and consider the line at infinity, so it all works out. The sum is zero, but one of the terms is exactly what we want, so we can just rearrange and get Bezout. – Potato Nov 11 '15 at 15:06
  • @Potato I thought about this for a while, but I realized I didn't know where Griffiths/Harris made the connection between degree of a divisor and the degree of the polynomials defining each divisor, if anywhere. Maybe you know? And what you said about the extra stuff happening at zero seems correct—this is basically the content of the calculation done on pp. 670–671. – Takumi Murayama Nov 16 '15 at 22:51
  • I don't know if there is still any interest in this question but these residues compute the localization of the top Chern class of $\mathcal{O}{\mathbb{P}^n}(d_1)\oplus \cdots \oplus \mathcal{O}{\mathbb{P}^n}(d_n)$. Hence their sum is the product $d_1\cdots d_n$. – Alan Muniz Mar 05 '20 at 01:30

1 Answers1

1

(Posting this in hopes of getting this question out of the Unanswered category.)

This question appears to be answered in the comments, by the OP:

Consider for example $f_1=x,f_2=y$ in $\mathbb P^2$. We have the right residue at $0$, and then I think we pick up the extra stuff when we switch charts and consider the line at infinity, so it all works out. The sum is zero, but one of the terms is exactly what we want, so we can just rearrange and get Bezout.

brainjam
  • 9,172