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Let $X$ be a smooth complex projective variety and $L$ be some ample divisor. For a holomorphic map $u:\Sigma \to X$, we define its degree to be $deg(u^*L)$.

Question: For a given positive integer $M$, is there a positive integer $N$ such that for a generic smooth hypesurface $D$ in the linear system $|NL|$, there are no nontrivial holomorphic maps of both degree and genus less than $M$ into $D$?

Remark1: for a given degree $d$ and genus $g$, virtual dimension of moduli spaces of genus $g$ degree $d$ maps into a hypersurface $D\in |NL|$ is $$ c_1^D(d)+(dim(D)-3)(1-g) = c_1^X(d)-Nd +(1-g)(dim(D)-3).$$ If $g$ and $d$ are bounded, then for $N$ big enough, virtual dimension will be negative.

Remark2: It seems that the answer to some similar question in symplectic setting is positive.

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  • $\begingroup$ I think you left out the bound on the degree in the formulation of your question. $\endgroup$ Commented Feb 5, 2014 at 21:53
  • $\begingroup$ "degree and genus less than M". I am adding a "both" now. $\endgroup$ Commented Feb 5, 2014 at 22:23

2 Answers 2

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Without loss of generality $L$ is very ample. This gives an embedding $X \to \mathbb P^n$. The genus $g\leq M$ degree $d\leq M$ curves in $X$ live in finitely many finite-dimensional moduli spaces (it's a subspace of finitely many components of the Hilbert scheme of $\mathbb P^n$). For each curve $C$, the codimension of the space of hypersurfaces of degree $N$ containing $C$ in the space of hypersurfaces of degree $N$ is an increasing function of $N$ - in fact for $N$ sufficiently large it is is $H^0(C, \mathcal O_C(N))$. So for $N$ large enough, the codimension will be greater than the dimension of the moduli space. This will lead to a generic hypersurface not containing any curve.

or: we can view the virtual dimension formula, after adding the dimension of the space of hypersurfaces, as a formula for the dimension of the space of pairs of a hypersurface and a curve on it. Then we can easily check that the formula is accurate (possibly up to a finite error). When the original virtual dimension is negative, the dimension of the space of hypersurfaces containing appropriate curves is less than the dimension of the whole space of hypersurfaces, and so generic hypersurfaces are not in that space.

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  • $\begingroup$ The argument after "or" is more plausible, i.e. looking at the space of pairs. In the first argument you are doing it curve by curve which is not then obvious what happens in families. $\endgroup$ Commented Feb 5, 2014 at 23:33
  • $\begingroup$ It's supposed to be two parts of on argument. By fibering over the moduli space of curves, computing the dimension curve by curve is sufficient to compute the total dimension. $\endgroup$
    – Will Sawin
    Commented Feb 6, 2014 at 14:32
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For $D$ a surface, there is work of Bogomolov. The results are weaker for higher dimensional varieties, but they do rule out genus $0$ and genus $1$ curves, cf. the work of G. Xu, Lawrence Ein, Claire Voisin and Gianluca Pacienza.

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  • $\begingroup$ What goes wrong with Will Sawin's proposal. $\endgroup$ Commented Feb 6, 2014 at 15:44
  • $\begingroup$ @MohammadF.Tehrani: Sorry, the results of Bogomolov, Su, Ein, Voisin and Pacienza are for bounded genus with no restriction on the degree (the issue I raised in my comment above). $\endgroup$ Commented Feb 6, 2014 at 15:57
  • $\begingroup$ It is still nice to hear that such strong statement is true. Thanks for recalling that. $\endgroup$ Commented Feb 6, 2014 at 22:12

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