Let $(X, \omega, J)$ be a compact symplectic manifold with an almost complex structure. Fix some homology class $\beta \in H_2(X, \mathbb{Z})$. An almost complex structure $J$ is said to be $\textit{regular}$ for some $J$-holomorphic curve $u$ (that represents $\beta$) if the linearization of the $\overline{\partial}$ operator is surjective (at $u$).

$\textbf{Question}:$ Is the standard almost complex structure on a del-Pezzo surface a regular almost complex structure?

$\textbf{Remark}:$ One needs the complex structure to be regular in order to conclude that the moduli space of non multiply covered holomorphic maps has the right dimension (and to compute the Gromov Witten invariants). Since the genus zero Gromov-Witten invariants of del-Pezzo surfaces are well known (see the paper by Kontsevic-Mannin http://www.ihes.fr/~maxim/TEXTS/WithManinCohFT.pdf and the paper by Pnadharipande-Gottsche http://arxiv.org/abs/alg-geom/9611012), I would imagine this is a well known fact.

$\textbf{Added Later:}$ Suppose $u:\mathbb{P}^1 \longrightarrow X$ is a degree $\beta$ map. Consider $u^* TX \longrightarrow \mathbb{P}^1 $. By Grothendick's theorem this splits holomorphically as a sum of two line bundles $L_1 \oplus L_2 \longrightarrow \mathbb{P}^1$. If one can show that $c_1(L_1) >= -1$ and $c_1(L_2) >= -1$ then the linearization of $\overline {\partial} $ at $u$ is surjective (this is Lemma 3.3.1 in McDuff and Salamon).

Is there an immediate way to see why the two Chern numers are greater than or equal to $-1$?

  • $\begingroup$ What is the linearization of the $\overline{\partial }$ operator? I thought $\overline{\partial }$ was linear. $\endgroup$ – abx Jul 25 '15 at 13:45
  • $\begingroup$ Dear Ritwik, How is this different from your previous questions about Del Pezzo surfaces? As I always do, I direct you to the Ph.D. thesis of Damiano Testa, where he proves that the moduli spaces of non-multiple cover rational curves with specified curve class on Del Pezzo surfaces of degree $\geq 2$ have the expected dimension and that they are irreducible. $\endgroup$ – Jason Starr Jul 25 '15 at 13:45
  • $\begingroup$ @Jason: I will carefully look at Testa's thesis; but I am asking a slightly different question. I am not asking if the moduli space has the expected dimension; I am asking if the linearization of the d bar operator is surjective. If the answer to my question is yes, then it implies the moduli space has the right dimension, but I would imagine not the other way round. I gather from your remark that Testa actually proves linearization of d bar is surjective (or rather the complex structure is regular). I apologize for asking this hastily without going through the reference you gave earlier. $\endgroup$ – Ritwik Jul 25 '15 at 13:52
  • $\begingroup$ @abx: $\overline{\partial}$ is not linear; it is a section of an infinite dimensional bundle. The linearization is basically the covariant derivative of this section. $\endgroup$ – Ritwik Jul 25 '15 at 13:53
  • 1
    $\begingroup$ @Ritwik. For an immersion, the derivative map $du$ is an embedding of $T\mathbb{P}^1$ in $u^*TX$, and the cokernel is the normal bundle. Thus the degree of the normal bundle is $\text{deg}_{\mathbb{P}^1}(u^*TX) - \text{deg}_{\mathbb{P}^1}(T\mathbb{P}^1) = \text{deg}_{\mathbb{P}^1}(u^*T_X) - 2$. Since $u^*TX$ has positive degree, the degree of the normal bundle is at least $-1$. $\endgroup$ – Jason Starr Jul 25 '15 at 20:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.