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Let $H_{d_1,g_1}, H_{d_2,g_2}$ be two Hilbert schemes of curves in $\mathbb{P}^3$ with degrees $d_1, d_2$ and genus $g_1, g_2$. Denote by $H:=H_{d_1,g_1}\times H_{d_2,g_2}$ where an element in $H$ is a pair of curves $(C_1, C_2)$ with $C_i \in H_{d_i,g_i}$.

As far as I understand, a generic element of $H$ consists of pairs $(C_1,C_2)$ such that $C_1 \cap C_2 = \emptyset$.

Assume now that $g_i \not=\frac{1}{2}(d-1)(d-2)$ i.e., the Hilbert scheme do not parametrize plane curves. My question is whether we can say something similar for a degree $d$ surface in $\mathbb{P}^3$. More precisely, for a fixed $d \ge 5$, we denote by $H'$ the space parametrizing the family of all smooth degree $d$ surfaces $X$ in $\mathbb{P}^3$ such that $X$ contains at least one pair of curves $(C_1, C_2) \in H$. Then can we say that a general $X$ in $H'$ contains a non-intersection pair of curves $(C_1, C_2) \in H$. This is equivalent to saying that the dimension of an irreducible component of $H'$ is maximal (among the components of $H'$) if it parametrizes surfaces containing $(C_1, C_2) \in H$ with $C_1 \cap C_2 = \emptyset$.

Note that we need to assume $C_i$ are not plane curves because as far as I recall it gives a counter-example to the statement.

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By the Noether-Lefschetz theorem, a very general surface $X\subset \mathbb{P}^3$ of degree $d\geq 4$ has $\mathrm{Pic}(X) \cong \mathbb Z$, generated by a hyperplane section $H$. But then if $C \sim aH$ and $C'\sim bH$ on $X$, we have $C\cdot C' = abd^2 > 0$, and so $C,C'$ must intersect in at least one (and generically many) point. Thus every pair of curves on $X$ intersect each other.

(This doesn't rule out that for some choices of pairs of curve classes that what you want could be true, but it does rule out that the most general possible result is true).

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  • $\begingroup$ @Huizenga: However, the space of such surfaces in not dense in the space of degree $d$ surfaces in $\mathbb{P}^3$ (due to a result by Ciliberto, Miranda, et al.) Moreover, in most cases if the degree of the curve is somewhat lower than $d$ then the curves will not be complete intersection in the surface. $\endgroup$
    – Naga Venkata
    Commented Aug 4, 2012 at 22:24
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    $\begingroup$ Of course the space of such surfaces is dense. It doesn't contain an open dense subset (hence the quantifier "very general"), but it is dense. Basically, you can ask for exceptional cases where what you want to be true holds, but it certainly does not hold in general. $\endgroup$
    – Jack Huizenga
    Commented Aug 5, 2012 at 3:44
  • $\begingroup$ @Huizenga: I am a bit confused. The result of Ciliberto, Miranda, et. al. titled "General components of the Noether-Lefschetz locus" states that the space of degree $d$ surfaces with Picard number greater than $1$ is dense in the space of all degree $d$ surfaces (for $d \ge 4$). $\endgroup$
    – Naga Venkata
    Commented Aug 5, 2012 at 5:53
  • $\begingroup$ Certainly a set and its complement can both be dense. $\endgroup$
    – Jack Huizenga
    Commented Aug 5, 2012 at 8:14
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We can choose $d_1,g_1,d_2,g_2,d$ such that every pair of curves in $H$ contained in a single degree $d$ surface must intersect. The most well-behaved hypersurface that is not a plane is of course a quadric surface, so we choose $d=2$. A curve of degree $4$ and genus $1$ on a quadric surface must have bidegree $(2,2)$. Two such curves have intersection number $8$ and anyways must intersect. Clearly, they are not plane curves.

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  • $\begingroup$ @Sawin: The question says $d \ge 5$. Is it obvious that this example can be extended to the higher degree case? $\endgroup$
    – Naga Venkata
    Commented Aug 4, 2012 at 22:19
  • $\begingroup$ No. We want to say that the only curves whose Hilbert polynomial is the same as the Hilbert polynomial of the divisor class $nH$ are the divisor class $nH$. Thanks to the intersection theory formula for genus and degree, we can compute $D \cdot D$ and $D \cdot H$ from the Hilbert polynomial, so we know that $(D - nH)\cdot (D-nH)=0$ and $(D-nH) \cdot H=0$ for any divisor of this type. So if we know that no nonzero divisors satisfy $X \cdot X=0$ and $X \cdot H=0$, we're good. Is that dense? $\endgroup$
    – Will Sawin
    Commented Aug 4, 2012 at 23:45
  • $\begingroup$ The question is primarily intended for curves that are not of complete intersection in a surface which is to say that the divisor class $[C_i]\not=kH$ for any $k$. This is because the space of degree d surfaces in $\mathbb{P}^3$ with picard number greater than $1$ is dense in the space of all degree $d$ surfaces. So in most cases the class of the curves are not of the form $kH$ for any $k$. $\endgroup$
    – Naga Venkata
    Commented Aug 5, 2012 at 0:09
  • $\begingroup$ My idea is that if the Picard number is small we can force the curves to be of the form $kH$ using their Hilbert polynomial. Do you know whether the surfaces with indefinite Neron lattices are dense? $\endgroup$
    – Will Sawin
    Commented Aug 5, 2012 at 2:44
  • $\begingroup$ @Sawin: I do not totally understand what you mean by "force to of the form". If you mean by adding it to a multiple of the hyperplane class, that might be possible. But this does not mean that the picard number becomes of the surface is $1$. By indefinite do you mean of infinite rank? $\endgroup$
    – Naga Venkata
    Commented Aug 5, 2012 at 6:04

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