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Let $f:X\rightarrow Y$ be a morphism of normal projective varieties over an algebraically closed field with connected fibers.

Assume that both $Y$ and the general fiber of $f$ admit a non-trivial first order infinitesimal deformation, that is $Ext^1(\Omega_Y,\mathcal{O}_Y)\neq 0$, and $Ext^1(\Omega_{f^{-1}(y)},\mathcal{O}_{f^{-1}(y)})\neq 0$ for $y\in Y$ general.

Does this imply that $X$ admits a non-trivial first order infinitesimal deformation as well?

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  • $\begingroup$ Why do you ask? $\endgroup$ Jul 7, 2015 at 21:07

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A candidate for a counterexample is the Cartwright-Steger surface, see http://arxiv.org/abs/1412.4137.

It is complex surface $X$ of general type with $$p_g(X)=q(X)=1, \quad K_X^2=9,$$ hence $K_X^2=9 \chi(\mathcal{O}_X)$. Since equality is attained in the Miyaoka-Bogomolov-Yau inequality, $X$ is a ball quotient, i.e. a quotient of $\mathbb{B}^2_{\mathbb{C}}$ by a cocompact subgroup of $\textrm{PSU}(2, \, 1)$.

By Mostow rigidity theorem, $X$ is biholomorphically rigid, i.e. its Kuranishi deformation space consists of exactly one point, in other words there are no deformations of $X$ in the large.

On the other hand, since $q=1$ and $X$ is of general type, the Albanese map of $X$ is a fibration $$a \colon X \to E,$$ whose base $E$ is an elliptic curve (actually, $E = \mathbb{C}/(\mathbb Z \oplus \omega \mathbb{Z})$, where $\omega$ is a third root of unity) and whose general fibre $F$ of $a$ is a smooth curve of genus at least $2$ (actually, $g(F)=19$). Therefore both $E$ and $F$ admit non-trivial first-order deformations.

Note that one should check that $H^1(X, \, T_X) =0$ in order to be sure that that $X$ has no obstructed first-order deformations.

ADDENDUM. As explained here, one actually has $H^1(X, \, T_X)=0$ by a result of Calabi and Vesentini, so $X$ really provides a counterexample.

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    $\begingroup$ Do you know whether there are any ball quotient surfaces with infinitesimal deformations? That might be an interesting MO question... $\endgroup$
    – Will Sawin
    Jul 8, 2015 at 1:31
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    $\begingroup$ This is a good point. I'm not really an expert of ball quotients, so the answer is no. It seems to me that any any such a first-order deformation should be obstructed (because of rigidity in large implied by Mostow's theorem). I will look into the literature and, if I find nothing, I will post the question on MO. $\endgroup$ Jul 8, 2015 at 5:47
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    $\begingroup$ Actually, as explained in the link in the addendum, no complex ball quotient of dimension at least $2$ can have infinitesimal deformations, since $H^1(X, T_X)=0$ by a result of Calabi and Vesentini. $\endgroup$ Jul 8, 2015 at 14:20
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Where does this problem come from? Why would you hope for something like this? I am not trying to be snarky, I am just wondering if there is a true statement that might better fit your situation.

At any rate, counterexamples are easy possible to produce via varieties with a group action such that the variety deforms, but the group action is rigid. Here is one construction. Let $Y$ be a curve of genus $g\geq 2$ with an action of a finite group, $$\mu:G\times Y \to Y, $$ such that (i) the pair $(Y,\mu)$ is rigid, and (ii) there are no points of $Y$ that are fixed by the entire group $G$. In characteristic $0$, the second condition is automatic if $G$ is noncyclic (in characteristic $p$, it is true if the quotient by the maximal quasi-$p$-group is noncyclic), and the first condition is automatic if the quotient map, $Y\to Y/G$, is a map to a rational curve branched over precisely $3$ points. The Klein quartic with its full group of automorphisms is one example.

Now let $X'$ be the product $$X' = \prod_{g\in G} Y = \text{Hom}_{\text{Sets}}(G,Y).$$ In other words, $X'$ is the projective $k$-scheme of ordered tuples $x'=(y_g)_{g\in G}$ of elements $y_g\in Y$. There is a projection $\text{pr}_1:X'\to Y$ sending $x'$ to $y_1$.

There is a diagonal embedding $$\Delta : Y \to X', \ \ y \mapsto (y)_{g\in G},$$ and the action $\mu$ also induces a second embedding $$\Gamma_\mu:Y \to X', \ \ y \mapsto (\mu(g,y))_{g\in G}.$$ By hypothesis (ii) above, the images $\Delta(Y)$ and $\Gamma_{\mu}(Y)$ are disjoint. Denote the blowing up of $X'$ along the closed subscheme $\Delta(Y)\sqcup \Gamma_{\mu}(Y)$ by $\nu$, $$\nu:X\to X'.$$ Define $f$ to be $\text{pr}_1\circ \nu$.

Every fiber is a product of smooth projective curves blown up at two points, and there are deformations of all of these. The base is just $Y$, and this deforms as well. However, the minimal model / canonical model / Albanese image of $X'$ is $X$. Every deformation of a product of curves is a product of curves, cf. the work of van Opstall. The fundamental locus of the Albanese morphism is a disjoint union of two curves. But for either curve, every projections of the curves to a curve factor is an isomorphism. So you can use one of the two fundamental curves, e.g., the curve $\Delta(Y)$, to identify all curve factors with the fundamental curve. Then the second fundamental curve gives a sequence of automorphisms of that curve. Now use that $(Y,\mu)$ is rigid to conclude that $X$ is rigid.

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  • $\begingroup$ for certain values of "easy"... $\endgroup$
    – Will Sawin
    Jul 8, 2015 at 1:30
  • $\begingroup$ @WillSawin: Maybe I should change it. $\endgroup$ Jul 8, 2015 at 2:46

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