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!



Deformation of $\mathbb{P}^1 \times \mathbb{P}^1$
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.


