Using Hodge theory (and the ill-defined Lefschetz principle), one can show that in characteristic 0, given a proper smooth family $X \rightarrow B$, the cohomology groups of the structure sheaf of the fibers are locally constant (as a function on $B$). I'm aware of the existence of a number of counterexamples to the corresponding statement in positive characteristic. Given the success of this question, I want to ask:

What are the simplest examples of a proper smooth family exhibiting jumping of some cohomology group of the structure sheaf?

By the above discussion, such an example will necessarily be in positive characteristic.

By "simplest", I mean by one of the following measures.

(best) An example whose proof is as elementary as possible, and ideally short.

An example with a simple conceptual underpinning. (Well, the best answer would do well by both of the first two measures.)

A known example that is simple to state, but may have a complicated proof. (Ideally there should be a reference.)

An expected, folklore, or conjectured example.

  • 2
    $\begingroup$ Dear Ravi, This answer may be relevant: math.stackexchange.com/questions/51146/… Junecue Suh's thesis contains examples (I think), but I can't find it online. Cheers, Matt $\endgroup$
    – Emerton
    Commented Jul 21, 2011 at 23:50
  • $\begingroup$ Thanks Matt! Junecue should post his thesis on the arXiv... $\endgroup$
    – Ravi Vakil
    Commented Jul 23, 2011 at 14:10
  • $\begingroup$ The introduction to the question is mildly confusing. Does "ill-defined" mean that the Lefschetz principle is false? How can you show something if you also know counterexamples to it? (Perhaps you meant to say "I'm aware of a number of counterexamples in positive characteristic," since this is what you say later on, but again the introduction is mildly confusing as it stands to a spectator like me.) $\endgroup$ Commented Jul 25, 2011 at 13:11
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    $\begingroup$ There is no need for arXiv: gradworks.umi.com/32/56/3256586.html The first of three sections is also journals.cambridge.org/article_S0010437X08003527 $\endgroup$
    – Junkie
    Commented Jul 26, 2011 at 2:34
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    $\begingroup$ @Junkie: thanks! Gradworks charges money, so I won't get the thesis. My library has access to Compositio, so at least I have access to that. I still would argue for posting on the arXiv, which would allow everyone with internet access the opportunity to see Junecue Suh's work. $\endgroup$
    – Ravi Vakil
    Commented Jul 26, 2011 at 16:58

4 Answers 4


One example is given by Enriques surfaces in characteristic $2$. There are three types depending on the value of $\mathrm{Pic}^\tau$ (as a group scheme) which can be either $\mathbb Z/2$, $\mu_2$ or $\alpha_2$. In the first case $\omega_X$ is the generator so in particular it is non-trivial and $H^2(X,\mathcal O_X)=0$ (using of course Serre duality). In the two other cases $\omega_X$ is trivial (as it is numerically trivial and all numerically trivial line bundles are trivial) so that $h^2(X,\mathcal O_X)=1$. Now, $\alpha_2$ can be deformed to both $\mathbb Z/2$ and $\mu_2$ and such deformations can be lifted to deformations of Enriques surfaces (in fact $\mathrm{Pic}^\tau$ is flat in families of Enriques surfaces and the functor from deformations of the surfaces to those of $\mathrm{Pic}^\tau$ is formally smooth - Liedtke: arXiv:1007.0787, Ekedahl-Shepherd-Barron: unpublished). If we pick a connected family of Enriques surfaces with some special value being an $\alpha_2$-surface and generically $\mathbb Z/2$, then we get an example.

Such an example can be constructed (very) explicitly without deformation theory. Here is a semi-explicit construction which works in any positive characteristic. Fixa a a group scheme $A$ of order $p$ over $\mathbb A^1$ localised at $0$ (say) which is $\alpha_p$ at $0$ and $\mathbb Z/p$ elsewhere. By the Godeaux construction (which Raynaud - Prop. 4.2.3, p-torsion du schema de Picard, Astérisque 64 - showed works for such families) there is a free action of $A$ on a flat complete intersection $Y$ (of any dimension, which we assume is $\ge 2$) such that $X=Y/A$ is smooth (note that contrary to the case of an étale group scheme $Y$ will not be smooth). As $Y$ is a complete intersection we have that $\mathrm{Pic}^\tau_Y=0$ and from that it follows that $\mathrm{Pic}^\tau_X=A$. Now, $H^1(-,\mathcal O_-)$ is the tangent space of $\mathrm{Pic}^\tau$ so it is zero outside of $0$ and $1$-dimensional at $0$. (By being careful one can get Enriques surfaces for $p=2$, this I guess was the inspiration for the Godeaux construction).

  • $\begingroup$ Very nice! Christian Liedtke had of course told me about his work and your work with Shepherd-Barron, and reminded me that they answered this question. I very much appreciate the explicit construction in the second paragraph --- it highlights the geometry of what is going on in this example. $\endgroup$
    – Ravi Vakil
    Commented Jul 23, 2011 at 14:13

This is an attempt to realise Sándor's program of getting an example based on Kodaira vanishing or non-vanishing varying in a family. It will be done by keeping the surface fixed but varying the line bundle.

I shall start not with the examples of Raynaud but rather a variation given by Raynad-Szpiro (for details on it see Szpiro's article in Astérisque 64 or Flexors exposé in Astérisque 86). Recall that a Tango-Raynaud structure on a relative smooth and proper curve $f\colon X\to C$ consists of a line bundle $\mathcal L$ on $X^{(p)}$ together with a map $\mathcal L\to B^1_{X/C}$, where $B^1_{X/C}\subseteq \alpha_\ast\Omega^1_{X^{(p)}/C}$ is the image of $d\colon\alpha_\ast\mathcal O_{X}\rightarrow\alpha_\ast\Omega^1_{X/C}$ (and $\alpha\colon X\to X^{(p)}$ is the relative Frobenius), such that the adjoint map $\alpha^\ast\mathcal L\rightarrow \Omega^1_{X/C}$ is an isomorpism. If $C$ is a proper smooth curve and the fibration is non-isotrivial, then such an $\mathcal L$ is ample as $\Omega^1_{X/C}$ is by Szpiro (and $\alpha$ is finite) and on the other hand, by looking at the exact sequence $0\rightarrow\mathcal O_X\rightarrow\alpha_\ast\mathcal O_{X^{(p)}}\rightarrow B^1_{X/C}\rightarrow0$ tensored with $\mathcal L^{-1}$ we get an embedding $H^0(X,B^1\bigotimes\mathcal L^{-1})\hookrightarrow H^1(X^{(p)},\mathcal L^{-1})$ and thus the given map $\mathcal L\to B^1_{X/C}$ gives a non-zero element of $H^1(X^{(p)},\mathcal L^{-1})$. The map $X\rightarrow C$ is constructed by the Kodaira procedure (using that if the second starting curve $D$ is Tango-Raynaud, then so is $f$).

The aim is now to show that there exists an $\mathcal{M}\in\mathrm{Pic}^0(X^{(p)})$ such that $H^1(X^{(p)},\mathcal L\bigotimes \mathcal{M})=0$. Of course, for every $\mathcal{M}$, $\mathcal L\bigotimes \mathcal{M}$ is ample and $\mathrm{Pic}^0(X^{(p)}$ is connected so we would be finished if we could do this. Note that $\mathrm{Pic}^0(X^{(p)})$ is positive dimensional (as it contains a subgroup isogenous to $\mathrm{Pic}^0(C)\times \mathrm{Pic}^0(D)$) and I shall in fact show that the vanishing is true for all but a finite number of $\mathcal{M}$'s.

I shall use $\mathcal L'$ to denote a general element of the form $\mathcal L\bigotimes\mathcal{M}$, where $\mathcal{M}\in\mathrm{Pic}^0(X^{(p)})$.

  • We have that $H^1(X^{(p)},\mathcal L'^{-p^n})=0$ for $n>0$. This is proven by descending induction on $n$, the statement being true for large $n$ by Serre and ampleness of $\mathcal L'$. It is therefore enough to show that the $p$'th power map $H^1(X^{(p)},\mathcal L'^{-p^n})\rightarrow H^1(X^{(p)},\mathcal L'^{-p^{n+1}})$ is injective. This is the map induced by adjunction $\mathcal L''\rightarrow F_\ast F^\ast\mathcal L''$, where $F\colon X\rightarrow X$ is the absolute Frobenius (on both $X$ and $X^{(p)}$) and $\mathcal L'':=\mathcal L'^{-p^n}$. We prove this by factoring $F$ as $\alpha\circ\beta$, where $\beta\colon X^{(p)}\rightarrow X$ is the base change of the Frobenius on $C$. Hence it will be enough to show that $H^1(X^{(p)},\mathcal L'')\rightarrow H^1(X,\alpha^\ast\mathcal L'')$ and $H^1(X,\alpha^\ast\mathcal L'')\rightarrow H^1(X,\beta^\ast\alpha^\ast\mathcal L'')$ are both injective. For the first we have that its kernel is equal to $\mathrm{Hom}(\mathcal L'',B^1_{X/C})$ and an element of it gives (by inclusion and adjunction as above) rise to a map $\alpha^\ast\mathcal L''\rightarrow \Omega^1_{X/C}\cong \alpha^\ast\mathcal L$. Such a map is zero as because of the ampleness of $\mathcal L$, $\mathcal L''$ is more positive than $\mathcal L$. As for the injectivity of $H^1(X,\alpha^\ast\mathcal L'')\rightarrow H^1(X,\beta^\ast\alpha^\ast\mathcal L'')$ we get similarly that the kernel is equal to $\mathrm{Hom}(\alpha^\ast\mathcal L'',f^{\ast}B^1_{C})$. However, $\alpha^\ast\mathcal L''$ has strictly positive degree on the fibres of $f$ so all such maps are zero.
  • Hence to show that $H^1(X^{(p)},\mathcal L'^{-1})=0$ for general $\mathcal{M}$ it would be enough to show that for such $\mathcal{M}$ the $p$'th power map $H^1(X^{(p)},\mathcal L'^{-1})\rightarrow H^1(X^{(p)},\mathcal L'^{-p})$ is injective. Again we can factor it and the second part of the argument works as before so it will be enough to show that there are only a finite number of $\mathcal{M}$'s for which $\mathrm{Hom}(\mathcal L',B^1_{X/C})$ could be non-zero. As for $\mathcal L$ a non-zero such map would give a non-zero map $\alpha^\ast\mathcal L'\rightarrow \Omega^1_{X/C}\cong \alpha^\ast\mathcal L$. However, $\mathcal L'$ is numerically equivalent to $\mathcal L$ and hence the map would have to be an isomorphism. This implies that $\mathcal{M}$ would have to lie in the kernel of $\alpha^\ast$ which is finite as $\alpha$ is finite flat (and surjective).
  • $\begingroup$ Very nice! For a moment I thought about varying only the line bundle, but this is not my arena. Thanks for working this out. It's beautiful. $\endgroup$ Commented Jul 25, 2011 at 8:14
  • $\begingroup$ I agree with S'andor. It is beautiful, and highly motivated. $\endgroup$
    – Ravi Vakil
    Commented Jul 25, 2011 at 16:51
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    $\begingroup$ Thanks for the appreciation but let's not forget the beauty of Sándor's idea which was the reason I was motivated to spend time on making things work. $\endgroup$ Commented Jul 25, 2011 at 18:38

Here is an idea of how one (let's say I) might try to construct such an example. I think this idea, if it works, might qualify under rule #2. Also, it is possible that the answer Matt gave on MSE help fill the gap in this. Even though this is not a complete answer I think it still might be useful.

So, let's say that $X$ is a smooth projective variety and $\mathscr L$ an ample line bundle on $X$ such that $H^1(X,\mathscr L^{-1})\neq 0$. This can obviously happen only in positive characteristic by Kodaira vanishing, but it can happen there. I think Raynaud was the first to give examples of this (in 1978?) and Ekedahl showed that this can even happen with $\mathscr L=\omega_X$ in characteristic $2$. Others gave various examples for this to happen, I think maybe including Mukai.

Anyway, suppose that such an $X$ comes in a polarized family (that is, there is a relatively ample line bundle on the total space that restricts to $\mathscr L$ on $X$) where the general fiber does not have satisfy this (for any power of the line bundle that's the deformation of $\mathscr L$). I am not sure if this is an outrageous expectation, but I am kind of thinking that the failure of Kodaira vanishing is special and so a general deformation will not be a counter-example to Kodaira vanishing, in other words Kodaira vanishing holds on it. One potential problem I see with this is that it is possible that such an $\mathscr L$ would perhaps not deform, so I could not reasonably take a "general" deformation. Well, actually if one uses $\mathscr L=\omega_X$ as in Torsten's example, then it definitely deforms! I am sure that Torsten will read this and tell me why I am wrong. :)

In any case, what I need is just that $\dim H^1(X,\mathscr L^{-1})$ would jump and this still seems an easier task than the original since $\mathscr L$ is undetermined, so one has a little more freedom.

So, if we have that, then we're in business. Take a general section of a high power of the global line bundle that restricts to $\mathscr L$ and take the finite cover it determines. This finite cover has the property that the cohomology of $\mathscr O$ of the fibers of this cover over the original base is the sum of the cohomologies of the negative powers (up to the degree minus one) of the deformations of $\mathscr L$. The assumption implies that either the original family gave you an example or there will be a jump in this one.

  • $\begingroup$ This definitely is a helpful example of type #2 --- it would be great to see this worked out in more detail. I like that this example bootstraps on another fundamental (counter)example, the failure of Kodaira vanishing. $\endgroup$
    – Ravi Vakil
    Commented Jul 23, 2011 at 14:15
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    $\begingroup$ My example seems generic in the sense that I think that my construction of examples probably gives a component in the moduli space of surfaces so I don't think it will work. The examples based on (generalisations of) Raynaud's construction looks more promising. $\endgroup$ Commented Jul 24, 2011 at 7:07
  • $\begingroup$ Torsten, thanks! This was just an amateur idea. Cheers! $\endgroup$ Commented Jul 24, 2011 at 14:26
  • $\begingroup$ On the other hand the Raynaud method seems to work, see separate answer. $\endgroup$ Commented Jul 25, 2011 at 7:39

By the way, for Enriques this more than just an example - it reflects their classification: the moduli space of Enriques surfaces is connected in any characteristic. All such surfaces arise as desingularizations of quotients $Y/G$, where Y is a (possibly singular) complete intersection of $3$ quadrics in $\mathbb{P}^5$, and $G$ is a finite flat group scheme of length $2$.

In characteristic $p\neq2$, all Enriques surfaces satisfy $h^{01}=h^{10}=0$, and their moduli space is irreducible. Also, over an algebraically closed field of characteristic $p\neq2$, the only possibility for $G$ is $\mathbb{Z}/2\mathbb{Z}$, which is isomorphic to $\mu_2$.

In characteristic $p=2$, we have $h^{01}=0$, $h^{10}=1$ if the surface arises as a quotient by $G=\mu_2$ ("classical"), we have $h^{10}=h^{01}=1$ if $G=\alpha_2$ ("supersingular"), and finally, we have $h^{01}=1, h^{10}=0$ if $G=\mathbb{Z}/2\mathbb{Z}$ ("singular"). The moduli space is connected, but has two irreducible components: one corresponds to $\mu_2$-quotients, the other to $\mathbb{Z}/2\mathbb{Z}$-quotients, and their intersection to $\alpha_2$-quotients.

Thus, the Hodge numbers reflect the position in the moduli space, and a jumping corresponds to a change in type.


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