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I am new here, but hopefully you can help me with a concrete problem I have. I try to compute a Self-Intersection Number of a constructed curve in an analytic surface. I know the answer by some references, but have no idea how it has been found.

First my surface:

Let $N\in \mathbb N$. We take $M_N:=\coprod\limits_{n\in\mathbb Z} (\mathbb C \times \mathbb C) / \sim $, where $"\sim"$ is the following equivalence relation:

  1. $\forall a,b \in\mathbb C \setminus \{0\} \ \forall n,k\in \mathbb Z : (a,b)_n \sim (a(ab)^{-k},b(ab)^k)_{n+k}$ , where $(a,b)_n$ denotes the pair $(a,b)$ in the $n$th-copie of $\mathbb C \times \mathbb C$.
  2. $\forall b \in\mathbb C \setminus \{0\} \ \forall n,k\in \mathbb Z: (0,b)_n \sim (b^{-1},0)_{n+1}$
  3. $\forall a,b \in\mathbb C \ \forall n\in \mathbb Z : (a,b)_n \sim (a,b)_{n+N}$

The elements of $M_N$ are equivalence-classes of Elements $(a,b)_n$ under $\sim$ and shall be denoted by $[a,b]_n$.

The curve:

For $n\in\mathbb Z$ let $S_n := \{ [0,b]_{n-1}\} \cup \{ [0,0]_n\}$. Immedeatly we see: 1. $\forall n \in \mathbb Z : S_n \cup S_{n+1} = \{[0,0]_n\}$ 2. $\forall n \in \mathbb Z : S_n \simeq \mathbb{P^1(C)}$

I am searching for the Self-Intersection Number $S_n.S_n$. I know it is -2. But why is it? I have no experience in computing Intersection Numbers. The usual way seems to intersect a sligthly moved copy of the curve with the original curve. But this won't work, if the result is negative.

The idea must lie in the fact, that for $b\not =0$ the identity $[a,b]_{n-1} = [b^{-1},ab^2]_n$ ist given. But how to use this?

I am thankful for any hint. Shall I blow up anywhere?

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To compute negative self-intersection numbers, the usual strategy is to find some Y such that (X+Y) does move, and then compute X.(X+Y). From this, you work backwards to find X.X. I'm not sure how to carry this out in your case though. –  Tony Jan 5 '13 at 19:46
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I'm surprized that such an extremely technical question by a newcomer gets six votes within three hours. Also, the tag "arithmetic geometry" is inappropriate. Be that as it may, welcome to this site Greyfox! –  Georges Elencwajg Jan 5 '13 at 19:52
    
Sorry, I have no influence on the votes and don't know if I can sell or trade them. The Tag might be wrong, but the problem came arised from an arithmetic geometry problem, namely compactifiing the Elliptic modular surface of Level N and compute theta-divisors. But this I want to try on my own first. –  Greyfox Jan 5 '13 at 22:54
    
@Tony So it would be great, if I could find some Y satisfying $S_n.Y=2$ and $S_n.(S_n+Y)=0$ or similar. Maybe I try $Y=\sum_{k\in\mathbb Z} S_k$? Then I get $S_n.Y= S_n.S_{n-1}+S_n.S_n+S_n.S_{n+1}=S_n.S_n+2$. If I could show $S_n.Y=0$, I was done. Could this be the idea? It doesn't use this special equivalence yet. –  Greyfox Jan 5 '13 at 23:57
    
Oh, I think it should be $Y:=\sum\limits_{k=1}^N S_k$. otherwise I get Multiplicities. And then Y should be the principal divisor given by the function $f:M_2\to\mathbb C$, which maps $[a,b]_n$ to $ab$. And Intersection Numbers with principal divisors are zero, right? Or am I traveling in a complete wrong direction? –  Greyfox Jan 6 '13 at 0:36

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