8
$\begingroup$

Let $\mathbb F$ be an algebraic closure of the field of order $p$. Let $S=\textrm{Spec}(\mathbb F[[z]])$ with special point $s$ and generic point $\eta$. I'm looking for an example of a smooth projective morphism: $X\rightarrow S$ and integers $i,j$ so that $$h^i(X_{\eta},\Omega^j_{X_\eta})<h^i(X_{s},\Omega^j_{X_s}).$$

Note that this can't happen in characteristic $0$ (see e.g. Hodge numbers in a family).

I haven't been able to find such an example referenced in Deligne-lllusie.

$\endgroup$
3
  • 2
    $\begingroup$ The answer by @gdb is correct, explicit, and relates this problem to moduli of finite flat group schemes (something else that is special to characteristic $p$). If you want to see more, I recommend that you read about Enriques surfaces in characteristic $2$. The moduli space is connected (though reducible). $\endgroup$ Feb 20 '19 at 10:55
  • 3
    $\begingroup$ Here is a useful reference for Hodge numbers of Enriques surfaces in characteristic two mentioned by Jason Starr: L. Illusie "Complexe de de Rham-Witt et cohomologie cristalline", section II 7.1. $\endgroup$ Feb 20 '19 at 17:52
  • 1
    $\begingroup$ Dear @JasonStarr, do you think the following is ridiculous: given such an example $X\rightarrow S$, there exists a finite flat map $X'\rightarrow X$ such that the Hodge numbers of $X'$ don't jump? $\endgroup$
    – Raju
    Feb 21 '19 at 5:33
17
$\begingroup$

Update: The details of this construction are now available in my blog post with Sean Cotner on Thuses.

I was recently interested in exactly the same question. But I failed to find any reference where this issue is discussed. So I came up with an example (hopefully correct) of such a family on myself. Presumably, this example is well-known to experts, I followed the footnotes of Grothendieck (page 10) in his paper "On the de Rham cohomology of algebraic varieties" and Serre's example of a projective smooth variety with a failure of Hodge symmetry. Here is an idea of the construction.

Fix a base $S:=\operatorname{Spec} \bar{\mathbf{F}}_p[T]_{(T)}$. Then there is a smooth affine group scheme $G' \to S$ such that its generic fibre is isomorphic to $\mathbf G_m$ and its special fibre is isomorphic to $\mathbf G_a$ (it is an example at the end of SGA 3, XVI, paragraph 3). Since $G$ is smooth of relative pure dimension $1$, we know that the Frobenius homomorphism $Fr_{G/S}: G' \to G'^{(p)}$ is finite flat of degree $p$. Consider its kernel $G:=\operatorname{Ker} (Fr_{G/S})$ -- a finite flat $S$-group scheme of order $p$. We know by the construction that its generic fibre $G_{\eta}$ is isomorphic to $\mu_p$ and the special fibre $G_{s}$ is isomorphic to $\alpha_p$.

Now the idea is to mimic a construction of Serre (Propostion 16 in Serre's lecture here). We want to find a complete intersection $S$-scheme $X \subset \mathbf P^m_S$ of relative dimension at least $3$, such that $X$ admits a free $G$-action and such that $Y:=X/G$ is smooth. Then $Y$ will be an example of a family, where hodge numbers jump. Observe that it is projective as a quotient of projective $S$-scheme by a finite flat $S$-group scheme.

Assume for a moment that we have such $X$ and denote the quotient morphism by $h: X \to Y$. Then it is rather easy to show (using Lefschetz hyperplane section for Picard groups, which is explained in detail in SGA 2, Exp. XII) that $\mathbf{Pic}^{\tau}_{X/S}$ is a trivial group scheme over $S$. A general descent type argument guarantees that $$ \mathbf{Pic}^{\tau}_{Y/S}=\operatorname{Ker} (h^*: \mathbf{Pic}^{\tau}_{Y/S} \to \mathbf{Pic}^{\tau}_{X/S})=\operatorname{Hom}_{S-gp}(G, \mathbf G_{m,S})=G^D. $$

So, we come up with an example of a smooth projective $S$-scheme such that the torsion component of its Picard functor is isomorphic to $\mathbf Z/p =\mu_p^{D}$ on a generic fibre and isomorphic to $\alpha_p=\alpha_p^D$ on a special fibre.

Recall that the Torsion component of the Picard scheme is open in the Picard scheme for any locally projective flat family with geometrically integral fibres (as explained in FGA Explained) and the Lie algebra of the Picard scheme (or equivalently of its torsion component) is given by the first cohomology group of the structure sheaf. Thus $$ \mathrm{H}^1(Y_{\eta}, \mathcal O_{Y_{\eta}})=\operatorname{Lie}(\mathbf{Pic}_{Y_{\eta}/k(\eta)})=\operatorname{Lie}(\mathbf{Pic}^{\tau}_{Y_{\eta}/k(\eta)})=\operatorname{Lie}(\mathbf Z/p)=0 $$ and

$$ \mathrm{H}^1(Y_s, \mathcal O_{Y_s})=\operatorname{Lie}(\mathbf{Pic}_{Y_{s}/k(s)})=\operatorname{Lie}(\mathbf{Pic}^{\tau}_{Y_{s}/k(s)})=\operatorname{Lie}(\alpha_p)=\bar{\mathbf{F}}_p. $$ So, indeed, $h^{0,1}$ jumps in this family.

The last thing that we should address is the construction of such $X$. The idea is to adapt the proof of Serre's Proposition 15 for any Noetherian local ring. But before doing this we need to find at least one action of $G$ on some projective space such that a closed locus of non-free action has codimension at least $3$ in the special fibre.

To construct such an action we consider a regular representation of $G$. Namely $G$ acts on $\mathcal O_G(G)$ and this action is "algebraic", meaning that it defines an action on an affine scheme $V:=\operatorname{Spec_S}(Sym^{\bullet}\mathcal O_G(G)^{\vee})$. This induces an action of $G$ on the projective space $\mathbf P_S(V)$. And I claim that the special fibre has exactly one point, where the action is not free (in a schematic sense). This point corresponds to a line of constants in a space of regular functions of $G$. Once we veriefied this we see that this action satisfies the condition of Serre's Lemma on special fibre for any $p\geq 5$. If $p\leq 3$ then one can take $\mathcal O_G(G) \otimes \mathcal O_G(G)$ insead and this will work.

Finally, there is a way to generalize Proposition 16 from Serre's paper in our context. But It will take too much space here. The basic idea is that you do the same things on a special fibre and then lift hyperplanes in an arbitrary way over the whole $S$. This is possible since $S$ is local. I wrote notes for myself a week ago. I can show it to you, if you write me a personal message, but I don't want to make them public (they are not polished at all).

The last thing to say here is that if you want a smooth projective family of finite type $\bar{\mathbf F}_p$-schemes, where hodge numbers jump, you need to use standard spreading out techniques to spread this family over an open subscheme of $\mathbf A^1_{\bar{\mathbf F}_p}$.

$\endgroup$
2
  • 1
    $\begingroup$ Dear GDB, Thanks! This is beautiful. I guess the same thing happens for BG, the classifying stack over S (though not "sm projective" as asked for in bed question.) $\endgroup$
    – Raju
    Feb 20 '19 at 14:55
  • $\begingroup$ Hi GDB, I don't know how to write private messages, but I'd be delighted to see your notes. My gmail address is krishnamoorthy. $\endgroup$
    – Raju
    Feb 21 '19 at 3:08
6
$\begingroup$

This is a comment on gdb's gorgeous answer, for people like me who aren't so comfortable with the Picard functor. Let $X$ be a space with a free action by one of the groups schemes $\mathbb{Z}/p$, $\mu_p$ or $\alpha_p$ and let $X \to Y$ be the quotient map. Assume also that $H^0(X, \mathcal{O}) = k$ (our ground field).

Here is a concrete check that $H^1(Y, \mathcal{O}) \to H^1(X, \mathcal{O})$ is injective in the $\mu_p$ case, but has kernel for $\alpha_p$ and $\mathbb{Z}/p$.

The $\mu_p$ case A $\mu_p$ action on a ring is equivalent to a $(\mathbb{Z}/p)$-grading on that ring, and this passes to localization, so a $\mu_p$ action on $X$ is a $(\mathbb{Z}/p)$-grading $\mathcal{O}_X = \bigoplus \mathcal{O}^j_X$ of the sheaf of rings $\mathcal{O}_X$. The ringed space $Y$ is the same topological space with structure sheaf $\mathcal{O}_Y = \mathcal{O}_X^0$. So $H^1(Y, \mathcal{O}_Y)$ is a direct summand of $H^1(X, \mathcal{O}_X) = \bigoplus H^1(X, \mathcal{O}_X^j)$.

The $\alpha_p$ case An $\alpha_p$ action on a ring is a $p$-nilpotent derivation, and again this passes to localizations, so an $\alpha_p$ action on $X$ is a derivation $D : \mathcal{O}_X \to \mathcal{O}_X$ with $D^p=0$. We have $\mathcal{O}_Y \cong \mathrm{Ker}(D)$. Choose an affine cover $V_i = \mathrm{Spec}(B_i)$ of $Y$ and let $U_i =\mathrm{Spec}(A_i)$ be the preimage cover of $X$. Using that the $\alpha_p$ action is free, for each $i$, choose $x_i \in A_i$ with $D x_i = 1$. Then $D(x_i-x_j)=0$, so $x_i-x_j$ descends to $V_i \cap V_j$ and is a Cech cocycle in $H^1(Y, \mathcal{O})$ which is a coboundary in $H^1(X, \mathcal{O})$. I claim that it is not a coboundary in $H^1(Y, \mathcal{O})$. If it were, so we had $x_i - x_j = f_i - f_j$ with $f_i \in B_i$, then $x_i - f_i$ would glue to a global section $s$ of $\mathcal{O}_X$ with $Ds=1$, contradicting that the only global sections of $\mathcal{O}_X$ are constants.

The $\mathbb{Z}/p$ case (used in Serre's original paper, though not in gdb's answer): A $\mathbb{Z}/p$-action on a ring is an automorphism $\phi$ with $\phi^p = \mathrm{Id}$. So we have such an automorphism of $\mathcal{O}_X$. Put $\Delta(a) = \phi(a) - a$, so $\Delta^p=0$. Again, the freeness of the action lets us find $x_i \in A_i$ with $\Delta(x_i)=1$ (in other words, write $A_i$ as an Artin-Schreier cover $A_i = B_i[x_i]/(x_i^p-x_i-y_i)$ for some $y_i \in B_i$, normalized so that $\phi(x_i) = x_i +1$). Then $\Delta(x_i-x_j) = 0$ so $x_i-x_j$ descends to a cocycle on $Y$, and the argument goes just as in the previous case (replacing $Ds=1$ with $\Delta(s)=1$).

$\endgroup$
3
  • 2
    $\begingroup$ Thanks David -- I didn't know this description of $\alpha_p$-action. One "concrete" way of seeing this $\alpha_p$ thing in action: let $E/\mathbb F$ be supersingular $\alpha_p$. Then $\alpha_p$ may be realized as the Frobenius kernel; on the other hand, the pullback of absolute Frobenius induces the 0-map on $H^1(E^{(p)},\mathcal O)$. $\endgroup$
    – Raju
    Feb 21 '19 at 5:23
  • $\begingroup$ You're welcome! The more common statement in the literature is that a $\mathbb{G}_a$ action is a nilpotent derivation, so the statement I gave here just comes to figuring out when it factors through $\alpha_p$. $\endgroup$ Feb 21 '19 at 11:22
  • $\begingroup$ Oh, grr, a $\mathbb{G}_a$ action in characteristic $0$ is a derivation. In characteristic $p$ is a list of commuting $p$-nilpotent derivations, all but finitely many of which are $0$. (Let the pull back of the action be $u \mapsto \sum D_k(u) t^k$. The condition that this is an action translates to $D_i D_j = \binom{i+j}{j} D_{i+j}$. In char. $0$, this means that $D_k = \tfrac{D^k}{k!}$. In characteristic $p$, we deduce that $D_{p^j}$ is $p$-nilpotent and all other $D_k$ are monomials in the $D_{p^j}$. Also, in char. $0$, $D_1$ is a derivation; in char. $p$, all the $D_{p^j}$ are.) $\endgroup$ Feb 21 '19 at 14:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.