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Let $X_k$ be $\mathbb{P}^2$ blown up at $k$ points (where $k$ is $0$ to $8$). Let $\beta \in H_2(X_k, \mathbb{Z}) $ be a homology class given by $$ \beta := n L + m_1 E_1 + \ldots + m_k E_k $$ where $L$ is the homology class of a line and $E_i$ are the exceptional divisors. My question is as follows:

Which homology classes are $\textit{indecomposable}$? By definition, a homology class is indecomposable if:

a) It can be represented by a non constant holomorphic map $u:\mathbb{P}^1 \longrightarrow X_k $ and

b) It cannot be written as $\beta = \beta_1 + \ldots \beta_n$ for some $n \geq 2$ such that each $\beta_i$ has a non constant holomorphic representative (as a map from $\mathbb{P}^1 $ to $X_k$).

My motivation for asking the question is as follows: I am explicitly trying to work out what is $N_{\beta}$, the number of rational curves in $X_k$ (through the right number of generic points) that represent the class $\beta$. Kontsevich and Mannin have given a recursive formula for this number in their paper (page 29)

http://www.ihes.fr/~maxim/TEXTS/WithManinCohFT.pdf

In order to actually calculate what is $N_{\beta}$, we need some initial conditions. I think the initial condition is that $N_{\beta} =1$ if $\beta$ is indecomposable.

$\textbf{Added Later:}$ Based on Mark's observation (and one further question I have about Kontsevich Mannin's paper) I have posted a separate question on mathoverflow

Are genus zero Gromov Witten Invariants on Del-Pezzo surfaces enumerative?

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  • $\begingroup$ If $k \geq 2$ the extremal rays on the cone of curves are all the classes of $(-1)$-curves (which are in particular rational), and I guess these classes should be indecomposable. Any other effective class can be written as a positive $\mathbb Q$-linear combination of these, but maybe not as an integer combination, I suppose. Do you know any examples of indecomposable classes that aren't just $(-1)$-curves? $\endgroup$
    – user47305
    May 27, 2015 at 2:11
  • $\begingroup$ I guess when $k = 8$ you have the class of a nodal cubic through the points. So there are more than just $(-1)$-curves. (I think for that class $N_\beta = 12$, though I could be wrong.) $\endgroup$
    – user47305
    May 27, 2015 at 2:17
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    $\begingroup$ @Mark: see arxiv.org/abs/math/0309111, especially Corollary 3.3. $\endgroup$
    – user5117
    May 27, 2015 at 5:12
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    $\begingroup$ @Ritwik There are no such curves if the points are general. The only way that class can be effective is if all $8$ points are collinear. $\endgroup$
    – user47305
    May 27, 2015 at 13:30
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    $\begingroup$ @Ritwik There is no curve representing the class $L-E_1 -\cdots- E_8$ unless the 8 points are collinear. That is the class of a line going through all eight points. $\endgroup$
    – user47305
    May 27, 2015 at 13:39

1 Answer 1

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Let me just summarize the comment thread in case someone more knowledgeable is willing to intervene.

Artie observes that The Cox Ring of a Del Pezzo surface by Batyrev--Popov shows that every effective class on a del Pezzo with $k \geq 2$ is a sum of $(-1)$-curve classes, with one exception. This means that no effective classes can be indecomposable in the sense above except for those of $(-1)$-curves. All the $(-1)$-curve classes are clearly indecomposable and have $N_\delta = 1$. And of course it is easy to write down these classes explicitly.

The one exception is that if $k = 8$ the anticanonical class is not a sum of two nonzero effective classes. This class is represented by (the strict transforms of) the pencil of cubics through the 8 blown up points. There are 12 singular cubics in the pencil, hence 12 that are rational, and it seems that we should have $N_\delta = 12$ by the OP's definition. The catch is that Kontsevich-Manin "expect" $N_\delta = 1$ for an indecomposable class on a del Pezzo (page 29 of the article in the question, following Claim 5.2.3).

So the question is, have we misunderstood something along the way (more likely, e.g. the definition of $N_\delta$), or did the authors forget a minor case? (A secondary question: is this already worked out somewhere?)

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  • $\begingroup$ I have asked this as a separate question. There is a further question I had about Kontsevich Mannis's paper. I combined that and your question in this thread mathoverflow.net/questions/207812/… $\endgroup$
    – Ritwik
    May 28, 2015 at 6:17

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