# Noether-Lefschetz over finite fields

The classical Noether-Lefschetz theorem asserts the following: Over the complex numbers, a very general surface $S\subset \mathbb{P}^3$ has Picard number 1 (that is, $Pic(S)\simeq \mathbb Z$), provided that $\mbox{deg }S\ge 4$.

Over finite fields (or even countable fields), the corresponding statement does not make much sense, because here `very general' refers to the surface being chosen outside a countable union of closed proper subsets of the parameter space of surfaces. Still, I think makes sense to ask:

What evidence is there for the Noether-Lefschetz statement over countable fields? I.e., is there a sense in which the set of surfaces with low Picard number constitutes a 'large' proportion of the surfaces in $\mathbb{P}^3$ e.g., over finite fields with many elements?

• K3 surfaces over finite fields have (geometrically) even Picard number: not much hope for Noether-Lefschetz for quartic surfaces! – M P Apr 26 '13 at 9:07
• (Though let me add that the Picard number over the ground field might be odd, so you could refine the question.) – M P Apr 26 '13 at 9:09
• @dhagbert: The characteristic polynomial of Frobenius on $H^2$ has real coefficients and even degree, so an even number of real roots, which have to be $\pm 1/q$ by the Weil conjectures. It vanishes at $1/q$ because the Picard number is positive, so over $\mathbb{F}_{q^2}$ it has even Picard number. This works for any smooth surface of even degree in $\mathbb{P}^3$. – Felipe Voloch Apr 26 '13 at 10:46
• @Voloch: You need to know the Tate conjecture in order to use this argument; even though it is known for K3 surfaces, at least for characteristics $> 3$, it is far from being known for all surfaces in $\mathbb{P}^3$. – ulrich Apr 26 '13 at 11:19
• @J.C. Ottem: Hilbert's irreducibility theorem can be used to show that even over countable fields (which are not algebraic over a finite field) "most" surfaces of degree $d \geq 4$ have geometric Picard number equal to $1$. See the paper "Complete intersections with middle Picard number $1$ defined over $\mathbb{Q}$ by Terasoma. – ulrich Apr 26 '13 at 11:25

There is a Noether-Lefschetz statement, but it's not quite the same as the one that holds over large fields and global fields.

Let the Tate-Picard rank be the formula for the (geometric) Picard rank implied by the Tate conjecture.

Let $n \geq 4$ be fixed, let $\mathbb F_q$ any large finite field, and let $X$ be a random smooth surface in $\mathbb P^3$ of degree $n$. I believe ne can show that:

If $n$ is odd, then with probability $1-o(1)$, the Tate-Picard rank of $X$ is $2$.

If $n$ is even, then with probability $1/2+o(1)$, the Tate-Picard rank of $X$ is $1$, and with probability $1/2+o(1)$, the Tate-Picard rank of $X$ is $3$.

Conditional on the Tate conjecture, this gives the distribution of the Picard rank. Otherwise, this is an upper bound for the Picard rank.

Here's a sketch of a proof:

Let $\mathcal F$ be the sheaf on the moduli space of smooth surfaces of degree $n$ in $\mathbb P^3$ whose stalk at a point is the primitive $2$nd cohomology of the corresponding hypersurface. Then $\mathcal F$ is lisse of rank $n^3-4n^2+6n-3$

Step 1: We compute the monodromy group of $\mathcal F$. This is probably in the literature somewhere (I know it is over $\mathbb C$, which implies it at least for sufficiently large characteristic), but it's not hard.

By Poincare duality, it is contained in $O(n^3-4n^2+6n-3)$. By Larsen's conjecture, it contains $SO(n^3-4n^2+6n-3)$ if and only if the invariant subspace of its $8$th tensor power is $105$-dimensional. By Deligne's equidsitribution theorem, this happens as long as the average of the eight power of the number of points on the hypersurface, divided by q, minus q+1+1/q, is 105. This can be easily checked by exchanging the order of summation as soon as $n\geq 4$. Finally, one can use vanishing cycles near a nodal singularity to check that the determinant character is nontrivial.

Step 2: Compute the Tate-Picard rank in terms of the characteristic polynomial of Frobenius.

The Tate conjecture says that over $\mathbb F_{q^n}$, the cohomology classes that are generated by cycles are exactly the $q^n$-eigenspace of $\operatorname{Frob}_{q^n}$. So for $X$ defined over $\mathbb F_q$, the $\lambda$-eigenspace of $\operatorname{Frob}_{q}$ is generated by algebraic cycles defined over $\mathbb F_{q^n}$ if and only if $\lambda^n = q^n$, i.e. $\lambda$ is $q$ times an $n$th root of unity.

Thus the Tate-Picard rank is defined to be the sum over all $n$th roots of unity $\zeta$ of the multiplicity of the eigenvalue $\zeta q$ of \operatorname{Frob}_{q}$, plus$1$(as we pass from primitive cohomology to all of$H^2$). Because the characteristic polynomial of Frobenius is a polynomial of degree$n^3-4n^2+6n-3$with integer coefficients, the eigenvalue$\zeta q$cannot appear unless the degree of$\mathbb Q(\zeta)$is at most$n^3-4n^2+6n-3$, so we need only check a finite set of eigenvalues. Step 3: By Deligne's equidistribution theorem, the distribution of the number of eigenvalues of Frobenius that are$q$times a root of unity of bounded degree converges as$q$goes to$\infty$to the distribution of the number of eigenvalues of a matrix in$O(n^3-4n^2+6n-3)$that are roots of unity of bounded degree. This would follow immediately from Deligne's equidistribution theorem if the number of such eigenvalues were a semicontinuous function, which it isn't. However, it is upper semicontinuous - the set where of matrices with at least$r$eigenvalues in a given finite set is closed. Moreover, the boundary of this closed set has Haar measure$0$, hence by standard measure theory, the convergence of the measure of this set follows from the convergence of the measure of continuous functions. Step 4: Calculate this distribution for a random element of$O(N)$. If$N$is odd, then any orthogonal matrix has eigenvalues$\lambda_1, \lambda^{-1}, \lambda_2,\lambda_2^{-1},\dots, \pm 1$, with the last eigenvalue equal to the determinant. The eigenvalues$\lambda_1,\lambda_2$,etc. have probability$0$of taking each eigenvalue, so the expected number of low degree roots of unity is$1$, for a Tate-Picard rank of$2$. This happens for$n$even. If$N$is even, then an orthogonal matrix has eigenvalues$\lambda_1, \lambda^{-1}, \lambda_2,\lambda_2^{-1},\dots, \lambda_{N/2},\lambda_{N/2}^{-1}$if its determinant is$1$and$\lambda_1, \lambda^{-1}, \lambda_2,\lambda_2^{-1},\dots, 1,-1$if its determinant is$-1$. Again, the variable eigenvalues have probability$0$of attaining any specific value, so the number of root of unity eigenvalues is$0$half the time and$2$half the time, for a Tate-Picard rank of$1$half the time and$3\$ half the time.