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Are there (complex, projective) varieties $X_0$ with the following property?

Every (flat, say) family $\mathfrak X\to T$ over a reasonable and reasonably big class of bases with a fiber over a closed point equal to $X_0$ is a trivial family, in the sense that $\mathfrak X$ is $T\times X_0$.

If one asks this with $T$ restricted to local Artinian rings, say, one gets the class of (infinitesimally) rigid varieties, but I wonder if there are "globally rigid" examples.

Later: As Francesco and unknowngoogle observe, it is quite unreasonable to ask for such strong rigidity... :)

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  • $\begingroup$ Maybe the requirement $\mathfrak{X}\cong T\times X_0$ is too strong? I think when $X_0$ has a nontrivial automorphism one can often construct a nontrivial family (i.e. not $\cong$ to the product family) in which every fiber (over closed points) is $\cong$ to $X_0$. $\endgroup$
    – Qfwfq
    Commented Aug 3, 2011 at 17:28
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    $\begingroup$ @unknowngoogle: not any nontrivial automorphism will do. You need one which is "not homotopically trivial" in some sense. Your observation shows that $Aut(X_0)$ must be rationally connected. Given any automorphism $\phi$, you can form a family over the nodal cubic obtained by taking the trivial family over $\mathbb P^1$ and gluing two fibers together using $\phi$. A trivialization of this family produces a rational curve in $Aut(X)$ connecting $\phi$ to the identity. $\endgroup$
    – Anton Geraschenko
    Commented Aug 3, 2011 at 18:49

3 Answers 3

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If $X_0$ is projectively embedded, we can use a Gr\"obner basis for its defining ideal to make a Rees family over ${\mathbb A}^1$ most of whose fibers are $X_0$, but whose central fiber is a monomial scheme. So $X_0$ must already be a monomial scheme. Since you said it's a variety, it must be a projective space. If you then allow me to reembed (by a Veronese), I could break that projective space, too, unless it's a point. So my answer: $X_0$ must be a point.

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If $X_0$ is a smooth projective variety over $\mathbb{C}$ such that $H^1(X_0, T_{X_0})=0$, then any deformation over a disk $D_{\epsilon}:=\{z \in \mathbb{C} | |z| < \epsilon \}$ is trivial for $\epsilon$ small enough.

Notice that, for a general scheme, $H^1(X_0, T_{X_0})=0$ only ensures that there are no infinitesimal deformations (i.e. deformations over Artin rings); however, since $X_0$ is projective one shows that nearby fibers in any flat family of projective schemes are isomorphic to $X_0$ [see Hartshorne, Deformation theory, p. 39].

For instance, the complex projective space $\mathbb{P}^n$ is globally rigid.

Another class of example is given by Fake Projective Planes: since they are $2$-dimensional ball quotients, they are globally rigid by Mostow Rigidity Theorem. In fact, their moduli space consists of $100$ distinct isolated points.

Let me also mention Beauville surfaces: they are surfaces $S$ of general type with $p_g=q=0$, $K_S^2=8$ (i.e. they are fake quadrics) obtained as $$S=(C_1 \times C_2)/G,$$ where $G$ is a finite group acting faitfully on the smooth curves $C_i$ and freely on their product, in such a way that $C_i/G \cong \mathbb{P}^1$ and the two maps $ C_i \longrightarrow C_i/G$ are both branched at three points.

Since three points on the projective line have no moduli, by general results of Catanese it follows that such surfaces are globally rigid, and in fact their moduli space is given again by a finite collection of reduced points.

Finally, notice that the $T=D_{\epsilon}$ is probably one of the most raisonable choices for checking global rigidity. In fact, if one choose as $T$ a compact complex curve, then even $X_0=\mathbb{P}^n$ fails to work: it is well known that there exist projective bundles over compact curves that are not trivial, think of Hirzebruch surfaces.

However, by using [Hartshorne, Deformation Theory, ex. 24.7 p. 163] one still have the following result:

Let $X_0$ be a projective scheme such that $H^1(X_0, T_{X_0})=0$ and let $\mathfrak{X} \to T$ be a flat family of projective schemes over any nonsingular curve such that the central fibre is isomorphic to $X_0$. Then there exist a dominant base change $T' \longrightarrow T$ such that the family $\mathfrak{X}'=\mathfrak{X} \times_T T'$ is trivial over $T'$.

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    $\begingroup$ I agree that it is probably unreasonable to ask for really global rigidity over more global things than your $D_\varepsilon$ :) As you say, even the very rigid $\mathbb P^n$ fails to work. Are your examples (very nice ones, by the way: I was not aware of the existence of neither family, thanks!) really global or only over small balls? $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Aug 3, 2011 at 19:26
  • $\begingroup$ I do not know. One of the problems is that it is difficult to give conditions ensuring the triviality of a flat family $\mathfrak{X} \to T$ of proj. varieties over a projective variety $T$. One of the few results I'm aware of is contained in the paper by L. Migliorini "A smooth family of minimal surfaces of general type over a curve of genus at most one is trivial", J. Alg. Geom. 4 (1995). But, of course, this answers your question only for a very small class of projective bases $T$... $\endgroup$
    – Francesco Polizzi
    Commented Aug 3, 2011 at 19:50
  • $\begingroup$ For instance, looking at Migliorini's result, one could ask the following question: Let $\mathfrak{X} \to T$ be a smooth family of Fake Projective Planes, with $T$ a projective curve of genus at least $2$. Then by Mostow Rigidity Theorem it follows that it is isotrivial (i.e. all the fibres are isomorphic). Is it also trivial? Notice that by the last result I mentioned in my answer, it becomes trivial after a base change $T' \to T$. $\endgroup$
    – Francesco Polizzi
    Commented Aug 3, 2011 at 20:13
  • $\begingroup$ @FP: in which cases is the property "étal-locally trivial" the same as "all the fibers are isomorphic"? $\endgroup$
    – Qfwfq
    Commented Aug 3, 2011 at 23:49
  • $\begingroup$ @FP: (other question) In the last paragraph, is the nonsingular base curve also supposed to be projective? $\endgroup$
    – Qfwfq
    Commented Aug 3, 2011 at 23:56
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There is a notion of "deformation in the large", at least for smooth projective varieties. A deformation in the large is any projective, $\textit{smooth}$ morphism over a connected base. Your question does make sense for deformations in the large. There are a number of varieties which are known to be rigid for deformations in the large, i.e., the morphism is a product étale locally over the base. The first example is projective space (which follows, e.g., from Mori's solution of the Hartshorne conjecture). I believe Siu first proved that hyperquadrics are rigid for deformations in the large. And there has recently been phenomenal work of Jun-Muk Hwang and Ngaiming Mok establishing the same result for any irreducible Hermitian symmetric space, e.g., Grassmannians.

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    $\begingroup$ I think the OP is asking more than rigidity in the large in the usual sense (any deformation of $X_0$ over a connected basis is still $X_0$): he really wants that any flat family with central fiber $X_0$ is trivial. $\mathbb{P}^1$ dos not satisfy this condition, think of Hirzebruch surfaces. $\endgroup$
    – Francesco Polizzi
    Commented Aug 3, 2011 at 21:25

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