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.

mightbe odd, so you could refine the question.) $\endgroup$ – M P Apr 26 '13 at 9:09