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Let $M$ be a two dimensional compact complex manifold and $A \in H_2(M, \mathbb{Z})$ a fixed homology class. Define a rational curve in $M$ to be $\textit{1-cuspidal}$ if the singularities of the curve consist of nodes and exactly $1$ cusp. Let $C_A$ denote the number of 1-cuspidal rational curves in $M$ representing the homology class $A$ and passing through the right number of generic points. In the case when $M:= \mathbb{CP}^2$ and $A$ is $d$ times the homology class of a line, there is an explicit formula for $C_A$. This formula is obtained by R.Pandharipande in this paper (page 1503, Propn 1):

http://www.ams.org/journals/tran/1999-351-04/S0002-9947-99-01909-1/S0002-9947-99-01909-1.pdf

My question is now the following: is $C_A$ known for any surface other than $\mathbb{CP}^2$? For instance $\mathbb{CP}^1 \times \mathbb{CP}^1$? It seems to me that the method applied by the author should go through in many other cases.

Note that for $\mathbb{CP}^1 \times \mathbb{CP}^1$, there is an explicit formula known for the number of rational curves representing a homology class of type $(d_1, d_2)$. It can be found in this paper by Di-Francesco and Itzykson (page 31, Propn 7)

http://arxiv.org/abs/hep-th/9412175

So, in particular should one expect a formula for cuspidal curves if the corresponding formula for just rational curves (with no cusps) is known?

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Given a collection of isolated plane curve singularity type $\alpha = (\alpha_1,\ldots,\alpha_l)$, J. Li and Y.-J. Tzeng proved the existence of a polynomial $T_\alpha$ such that for any sufficiently ample line bundle $L$ on a complex projective surface $S$, a general sub-linear system of $|L|$ of right codimension contains exactly $T_\alpha(L^2,L\cdot K_S,c_1(S)^2,c_2(S))$ curves with singular type $\alpha$. (See http://arxiv.org/abs/1203.3180). The strategy is outlined in the introduction of their article for cuspidal curves.

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  • $\begingroup$ @Lin: Is the polynomial $T_\alpha$ known explicitly? $\endgroup$
    – Ritwik
    Commented Feb 12, 2015 at 6:32
  • $\begingroup$ @Lin: In fact I noticed on page 2 of the paper, that their proof is not constructive (see the last paragarph). $\endgroup$
    – Ritwik
    Commented Feb 12, 2015 at 6:34

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