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Let's consider a algebraic contact structure $P$ on $\mathbb CP^3$ and a algebraic curve $C$ degree $d$ and genus $g$. Let's assume that contact structure has degree $p$ (see Polynomial contact structures on $RP^3$ about algebraic contact structure).

It seems that there is some constant $f(d,g,p)$ (maybe, even polynomial!) such that if $C$ is tangent to $P$ at $f$ points then $C$ is tangent to $P$ everywhere.

For example, it is easy to prove that $f(d,0,0)$ equals $2d-1$.

Somebody can expect that this question is about some homological conditions : generic curve is tangent to $P$ at $a$ points, so, if it is tangent to $P$ at $a+1$ points then it is tangent everywhere. It seems that it is true because pull-back of contact form to tangent bundle of $C$ is a holomorphic form, so it has some degree..

So, my questions are:

1) How can we prove that $f(d,g,p)$ exists? At least for some values $d,g,p$? It seems that the case $p=0$ is the most easy.

2) Is it true that $f$ is a polynomial? (Thom polynomial of something...) I'm sure that it is known or similar constructions are already examined.

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3 Answers 3

up vote 3 down vote accepted

Let $i:C \to \mathbb P^3$ be the normalization of an irreducible curve $C_0\subset \mathbb P^3$ of degree $d$ and geometric genus $g$.

If $\mathcal D$ is a distribution on $\mathbb P^3$ of degree $p$ then it is defined by a section $\omega$ of $\Omega^1_{\mathbb P^3} \otimes \mathcal O_{\mathbb P^3}(p+2)$. To compute the tangencies between $C_0$ and $\mathcal D$ we pull-back $\omega$ to $C$ using $i$. What we get is a section of $\Omega^1_C \otimes \mathcal O_C(p+2)$. Notice that $\mathcal O_C(p+2) = i^* \mathcal O_{\mathbb P^3}(p+2)$ is a line-bundle of degree $d(p+2)$ over $C$.

If $i^* \omega$ vanishes identically then $C_0$ is everywhere tangent to $\mathcal D$. Otherwise, it is a section of a line-bundle of degree equal to $\deg(\Omega^1_C) + \deg(\mathcal O_C(p+2)$. As such it has exactly $2g -2 + d(p+2)$ zeros counted with multiplicities. Therefore $$ f(d,g,p) = 2g -2 + d(p+2) +1 $$ is a polynomial. If one recalls that the genus of degree $d$ irreducible curve is bounded by $(d-1)(d-2)/2$ then one sees that $$ f(d,g,p) \le d( p + d -1) + 1 . $$ This is in accordance with Serge R. claim that the bound can be taken independent of $g$.

The argument above works equally well for integrable and non-integrable distributions, and also works for codimension one distributions on $\mathbb P^n$, $n\ge 2$.

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how do you compute number of zeros for a generic curve? Geometrically I understand, let's consider rational curve and than resolve each node, resolution gives us two points where curve is tangent to the distribution, but how should these consideration be formalized? –  Nikita Kalinin Oct 23 '11 at 11:37
You look at the normalization (=desingularization) of your curve and use it to pull-back the $1$-form defining the distribution. In the case of a nodal curve you will not have necessarily tangencies at the nodes, only those branches which are indeed tangent to the distribution will contribute. If instead you have a cusp singularity, it will contribute since the normalization will have a critical point over it. –  jvp Oct 23 '11 at 11:53
thank you very much! –  Nikita Kalinin Oct 25 '11 at 9:17

I think that the answer to 1) is positive. $f$ could even be chosen not to depend on $g$ and the distribution would not need to be taken contact. This would be a consequence of the fact that the complex field has a so-called strongly minimal theory and that the tangency condition is closed.

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I've tried to google "strongly minimal theory" and found something about logic. Could you give me a true reference? –  Nikita Kalinin Oct 23 '11 at 11:41
I don't have my books now but Poizat "Cours de théorie des modèles" should contain some about it (and is translated in many languages, some translations being freely available on the web). Marker's "Model theory: an introduction" should do too (chapter 3 and 8 ?). –  Serge R. Oct 24 '11 at 7:46
The proof would be similar to jvp's: as the parameters that describe the distribution and the curve vary, the set of points on the curve at which the distribution is tangent gives a family of sets each of dimension 1 or 0. When the dimension is 1, we get all the curve to be tangent. Else we get a finite collection of points; but by strong minimality, there should be an uniform bound for the number of these finitely many points. –  Serge R. Oct 24 '11 at 7:50
to Serge R.: jvp's answer depends on $g$. –  Nikita Kalinin Oct 24 '11 at 15:39
Indeed we can avoid the dependance on g, as the genus is bounded by the degree. –  jvp Oct 24 '11 at 20:04

there is an article of Quo-Shin Chi "The dimension of the moduli space of superminimal surfaces of a fixed degree and conformal structure in the 4-sphere" where the dimension of contact curves moduli space is computed.

It lays between $2d-4g+4$ and $2d-g+4$ for a fixed complex structure on a curve

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