Informally, my question is the following:

Is there an "inverse theorem" for the first cohomology group $H^1$ of (the projective completion of) an algebraic surface $S$? Namely, can we give a description of those surfaces for which $H^1(S)$ is non-trivial?

The motivation for my question comes from surfaces over finite fields ${\bf F}_q$, although the question could be posed over any other field of course, such as the complex numbers; one could also consider higher dimensional varieties than surfaces, but the surface case appears to be the key one. One of course has to be precise as to what cohomology one wishes to use here (etale, Cech, singular, etc.), but I think the answer should not be too sensitive to this choice.

More concretely, if $S \subset {\bf A}^3$ is an absolutely irreducible surface in affine 3-space defined over a finite field ${\bf F}_q$, then in the regime where the degree of $S$ is bounded and the characteristic of ${\bf F}_q$ is large, the Lang-Weil inequality tells us that the number $|S({\bf F}_q)|$ of ${\bf F}_q$-points of $S$ is given by the approximate formula

$$ |S({\bf F}_q)| = q^2 + O( q^{3/2} ). \qquad (1)$$

However, a simple probabilistic argument shows that if one chooses a "random" such surface (e.g. by taking the zero set of a degree $d$ polynomial with all coefficients chosen uniformly and independently from ${\bf F}_q$, and verifying that this gives an absolutely irreducible surface with high probability), one should "generically" be able to obtain the improvement

$$ |S({\bf F}_q)| = q^2 + O( q ). \qquad (2)$$

If one plugs in the Grothendieck-Lefschetz fixed point formula and the results of Deligne's Weil II paper (together with some known bounds on Betti numbers), one sees (I think) that this improvement will occur when the ell-adic first cohomology group $H^1_c(\overline{S}, {\bf Q}_\ell)$ with compact supports of a projective completion $\overline{S}$ of $S$ is trivial. So this suggests that "most" surfaces should have trivial $H^1$ (and thus obey (2)), and the ones which do not should be special and have some additional algebraic structure.

One likely obstruction to having vanishing $H^1$ is if there is a dominant map $\phi: S \to C$ from the surface $S$ to a positive genus curve $C$, since this creates a map from the non-trivial group $H^1(C)$ to $H^1(S)$ (and, using the counting rational points interpretation, the existence of $\phi$ suggests that $|S(F_q)| \approx q |C(F_q)|$ and $|C(F_q)| = q + O(q^{1/2})$, suggesting that the Lang-Weil bound (1) is optimal in this case). A little bit more generally, if $S$ (or some Zariski open dense subset thereof) has some finite cover $\tilde S$ that maps onto a positive genus curve $C$, then this also suggests (though does not quite force) $H^1(S)$ to be non-trivial.

I can't think of any further obstruction, so I (somewhat naively) conjecture that any irreducible algebraic surface $S$ with non-trivial $H^1$ must have a generically finite cover that maps onto a positive genus curve (assuming that the characteristic is sufficiently large depending on the degree, or alternatively one could phrase the question over ${\bf C}$). But perhaps there is an obvious counterexample to such a claim.

The work of Deligne does show that $H^1$ vanishes when the projective completion of $S$ is smooth; there are variants of this fact due to Hooley, but they don't seem to say anything stronger in the case of surfaces in ${\bf A}^3$ (although they do suggest that the analogous problem for higher-dimensional varieties would reduce to the surface case). So $\overline{S}$ has to be singular somewhere if $H^1$ is to be non-vanishing, but this doesn't seem to come close to a sufficient condition.

Using abstract nonsense one can rephrase non-vanishing $H^1$ in other ways (e.g. non-vanishing first Betti number, non-trivial abelianisation of the fundamental group, existence of non-trivial principal ${\bf Q}_\ell$-bundles, etc.), but I wasn't able to get too far with any of these reformulations.