# The locus of rational/elliptic curves on a special surface in $\mathbb{P}^3$

Let $P$ and $Q$ be two general polynomials of the same degree $d>5$. Consider the surface $S: z^2=P(x)Q(y)$ in $\mathbb{P}^3$ (after homogenization by the variable $w$). One can show that these kinds of surfaces are of general type. I want to show that the locus of rational/elliptic curves on this special surface in $\mathbb{P}^3$ is in the part $\{zw=0 \}$.According to a conjecture made by Lang the locus swept out by entire curves is equal to the locus swept out by images of non constant rational maps from an abelian variety to a projective variety $S$. So if we assume this conjecture it is enough to prove that the locus of non-constant holomorphic functions to the surface $S$ is contained in the part $\{zw=0 \}$. \ One idea to attack to the above problem is to solve the following problem: The above statement is equivalent to prove that the functional equation $h^2 = P(f) Q(g)$ have no non-constant solutions for holomorphic functions $f,g,h$. I couldn't prove that this functional equation has no non-constant solution. I appreciate it if anyone have any idea/comment about this problem.

• "So if we assume this conjecture it is enough to prove ..." Actually, you are only using the direction that is known (and trivial). Every nonconstant rational map from an Abelian variety induces (many) entire curves whose images sweep out the image of the Abelian variety. So the union of (nonconstant) images of Abelian varieties is contained in the union of (nonconstant) images of entire maps. – Jason Starr Sep 8 '14 at 10:40
• Yes, you are right, this is the trivial part. But do you have any idea about this problem? As I mentioned in above, how to prove that the above functional equation has no non-constant solution. Or maybe you want to think by using the algebrogeometric tools to solve the problem. – mehdi Sep 8 '14 at 12:33
• Using Riemann-Hurwitz you can show a certain proportion of the zeros of $P(f)$ and $Q(g)$ must be single roots, which means a certain proportion of the zeros of $P(f)$ must be zeros of $Q(g)$ and vice versa. Here I'm thinking of $f,g$ as just rational functions on a curve. One could try to derive a contradiction from that, but I don't immediately see how. – Will Sawin Sep 8 '14 at 19:21
• thanks Will, but by applying Riemann-Hurwitz to the map $P(f):\mathbb{P}^1 \rightarrow \mathbb{P}^1$ we have: $2g_{\mathbb{P}^1} -2=(2g_{\mathbb{P}^1}-2)deg(P(f))+\sum_{p\in\mathbb{P}^1}(e_p -1)$ so we have $-2=-2d +\sum (e_p-1)$ where $d=deg(P(f))$. How we can deduce from the above that there are simple roots? – mehdi Sep 9 '14 at 6:31
• I mean that if $P(x)=(x-a_1)\cdots(x-a_n)$ then want to see that all roots of the equation, for example,$f(z)=a_1$ are of multiplicity 1. – mehdi Sep 9 '14 at 7:29

• Dear Remke, According to the above paper by Voisin every general hypersurface in $P^n$ of degree $d\ge 2n-1$ contains no rational curve. In my case the polynomials P and Q can be taken general (e.g. their coefficients are general) can I deduce that by Voisin theorem the above hypersurface for $d\ge 5$ contains no rational curve? – mehdi Sep 10 '14 at 18:41