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I learnt that $\mathbb{P}^1 \times \mathbb{P}^1$ is rigid, but can be deformed to a non-rigid Hirzebruch surface $S$. Suppose $\pi: M \to B$ is such deformation such that $\mathbb{P}^1 \times \mathbb{P}^1 \cong M_{t_0}$ and $S \cong M_{t_1}$.

I want to understand the meaning of "rigidity". Does this mean that: (1) there exists an open set $t_0 \in U \subseteq B$, such that for all $t\in U$, $M_t \cong \mathbb{P}^1 \times \mathbb{P}^1$? (Or maybe more strongly, $M_U \cong U \times (\mathbb{P}^1 \times \mathbb{P}^1)$?); or (2) only the first order deformation of $\mathbb{P}^1 \times \mathbb{P}^1$ is trivial (because $H^1(\mathbb{P}^1 \times \mathbb{P}^1, \Theta)$ = 0), and it could well happen that no matter how $t \in B$ closes to $t_0$, $M_t$ may not isomorphism to $\mathbb{P}^1 \times \mathbb{P}^1$?

If (1) is the meaning of "rigidity", then I feel it is strange that for some $t$, $M_t$ "suddenly" becomes NOT the same as $\mathbb{P}^1 \times \mathbb{P}^1$. Moreover, I think the construction of aforementioned deformation is by considering extension of vector bundles $V, W$, and ${\rm Ext}^1(W,V)$ is the base $B$. So for $t \neq t_0$, the extension will never be trivial, and hence the corresponding variety should not be the same as $\mathbb{P}^1 \times \mathbb{P}^1$.

If (2) is the meaning of "rigidity", and suppose $\mathfrak{M} \to \mathfrak{B}$ is the Kuranishi family of $\mathbb{P}^1 \times \mathbb{P}^1$. Then I guess $T_{\mathfrak B, t_0} \cong H^1(\mathbb{P}^1 \times \mathbb{P}^1, \Theta) = 0$. But according to the previous discussion, $\mathfrak B$ is not of dimensional $0$. Hence $t_0$ must be singular at $\mathfrak B$. (By the way, is there a way to compute the dimension of Kuranishi space?)

My knowedge of deformation theory is floppy. Any comment/reference is very well appreciated!

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  • $\begingroup$ The point is that in this case you do not have a separate moduli space, so it can happen (and, in fact, it happens) that the variety is locally rigid but not globally rigid. The phenomenon you are talking about and the explicit description for the Kuranishi family of any Hirzebruch surface can be found in Catanese's paper moduli of algebraic surfaces, Theory of Moduli, Montecatini Terme, 1985, Springer LNM 1337 (1988). $\endgroup$
    – Francesco Polizzi
    Commented May 20, 2016 at 14:43
  • $\begingroup$ Which moduli space are you talking about? $\endgroup$
    – Li Yutong
    Commented May 20, 2016 at 14:54
  • $\begingroup$ The variety $\mathbb{P}^1 \times \mathbb{P}^1$ does not have a separate (=Hausdorff) moduli space. This allows the existence of a global deformation $\mathcal{X} \to \Delta$ such that $X_t \cong \mathbb{P}^1 \times \mathbb{P}^1$ for $t \neq 0$ and $X_0 \cong \mathbb{F}_2$. In other words, the rigidity in this example is only local, not in the large. $\endgroup$
    – Francesco Polizzi
    Commented May 20, 2016 at 15:22
  • $\begingroup$ Note that local rigidity is actually sufficient to imply that the Kuranishi family of $\mathbb{P}^1 \times \mathbb{P}^1$ has $0$-dimensional basis. This gives in turn $$H^1(\Theta_{\mathbb{P}^1 \times \mathbb{P}^1})=0$$ (this last condition is called infinitesimal rigidity). $\endgroup$
    – Francesco Polizzi
    Commented May 20, 2016 at 15:29
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    $\begingroup$ Such deformation family is easy to construct, by considering a pencil of quadrics in $P^3$. An open subset in the base of this family corresponds to smooth quadrics (each isomorphic to $P^1\times P^1$), while finitely many points correspond to quadratic cones. If you consider a small resolution of singularities of the total family, the special fibers will turn into the Hirzebruch surface $F_2$. $\endgroup$
    – Sasha
    Commented May 20, 2016 at 15:32

1 Answer 1

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Rigidity means, if you have a family $F$ of surfaces such that one $f\in F$ of them is $\mathbb P^1 \times \mathbb P^1$, then there is an open set $U\subseteq F, U \ni f$ each of whom is $\mathbb P^1 \times \mathbb P^1$.

The source of confusion, I think, is in mixing up "family of varieties" with "moduli space of complex varieties with a fixed underlying real manifold" (or somesuch). In the latter, we want all the varieties to be nonisomorphic. Most families one meets are not of this type.

For a simpler example than yours, consider the conics $xy=t z^2$ in $\mathbb P^2$, $t\in \mathbb A^1$. They're all isomorphic except for $t=0$. I hope it seems less strange that $M_t$ is suddenly different there.

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  • $\begingroup$ Thank you very much for you explanation! I have two questions according to your answer: (1) Does "moduli space of complex varieties with a fixed underlying real manifold" make sense in general? (2) Is this rigidity the same as "there exists a $U$ such that $F_U \cong U \times (\mathbf{P}^1 \times \mathbf{P}^1)$"? Or weaker than this locally triviality? $\endgroup$
    – Li Yutong
    Commented May 21, 2016 at 0:54
  • $\begingroup$ (1) Let me change it to: smooth proper families $F \to S$ with nonisomorphic fibers. In the complex case, all the fibers will be diffeomorphic. (2) I'm pretty sure I'd only expect etale-locally trivial (again, with $\mathbb P^1$-fiber for simplicity) but I don't have an example off the top of my head. $\endgroup$
    – Allen Knutson
    Commented May 22, 2016 at 18:25

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