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.

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    $\begingroup$ Besides mapping onto a curve of positive genus, $S$ could also get nontrivial $H^1$ by being birational to an abelian surface, or more generally to a subvariety of some abelian variety. $\endgroup$ Apr 27, 2014 at 17:41
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    $\begingroup$ And indeed by considering the Albanese, one can see that a smooth projective surface $S$ has nonzero $H^1$ iff it maps onto either a curve of positive genus, or another surface embedded in an abelian variety. $\endgroup$ Apr 27, 2014 at 18:23
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    $\begingroup$ Noam and Donu: thanks, this makes a lot of sense! From the counting-rational-points perspective, what appears to be happening is that if S has nontrivial $H^1$, then it has nontrivial image in the Albanese, and then all the 1D slices of S have Jacobian that maps into that Albanese, but from the rigidity of maps between abelian varieties, the images of these Jacobians are basically the same up to homomorphisms and translations, so the cardinality of each of the 1D slices of S is roughly constant, so one doesn't get the additional cancellation of (2) when averaging over slices. $\endgroup$
    – Terry Tao
    Apr 28, 2014 at 1:17

1 Answer 1


Let $\overline{S}\subset \mathbb{P}^2$ be a singular surface. Then by the Lefschetz hyperplane theorem we have that $h^1(\overline{S})=0$, but $h^3(\overline{S})$ can be nonzero, and surfaces with this property are a likely candidate for not satisfying the bound (2). (The surfaces with $h^3(\overline{S})=0$ satisfy bound (2).)

The probability that a surface has nonisolated singularities is very small. To ease the presentation I will ignore them and assume that $\overline{S}$ has isolated singularities.

Let $\Sigma$ be the singular locus of $\overline{S}$. Then $H^3(\overline{S})$ equals the cokernel of $H^2(\overline{S}\setminus \Sigma)\to H^3_{\Sigma}(\overline{S})$ (with the latter group I intend cohomology with support in $\Sigma$). So a necessary condition to have $h^3(\overline{S})$ nonzero is that $h^3_{\Sigma}(S)$ is nonzero.

If $S$ has only ADE singularities (and the characteristic is sufficiently large) then $H^3_{\Sigma}(\overline{S})$ vanishes. Hence the singularities of $S$ need to be sufficiently bad.

Moreover, it is my experience that $H^3_{\Sigma}(\overline{S})$ needs to be large in order to have a nonzero cokernel, and the position of the singularities is very special. Most results in this direction are for threefolds. For a threefold $T\subset \mathbb{P}^4$ with isolated singularities the "mysterious" cohomology group is $H^4(T)$ and the rank of this group equals one (the hyperplane class) plus the dimension of the cokernel of $H^3(T\setminus \Sigma)\to H^4_{\Sigma}(T)$.

In the threefold case we have that even for a node $p\in T$ the group $H^4_{p}(T)$ is nonzero. It turns out that over the complex numbers one has that a nodal threefold $T$ in $\mathbb{P}^4$ of degree $d$ with less than $(d-1)^2$ nodes satisfies $h^4(T)=1$ (the minimum possible), but a general threefold of degree $d$ containing a plane has precisely $(d-1)^2$ nodes, all on the plane and $h^4(T)$ equals $2$.

If $h^3(\overline{S})$ is nonzero then every resolution of singularities $\tilde{S}$ of $\overline{S}$ has nonzero $h^3$ and by Poincare duality has a nonzero $h^1$. In characteristic zero this implies that the Albenese of $\tilde{S}$ is nonzero, and $\tilde{S}$ would admit a morphism to an abelian variety. If one of the isogeny factors of this abelian variety is a curve then you have the desired map. In characteristic zero Libgober has a series of examples where if you put sufficient restricts on the singularities of $\overline{S}$ then the Albenese of $\tilde{S}$ is isogenous to a product of elliptic curves.


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