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Given a (flat) family of complex algebraic varieties $X_t$ (say parametrized by $\mathbb{C}$) and a specific $t_0$, how does one proceed to check if $X_{t_0}$ is a 'generic element'?

More precisely, I have the following:

  1. flat morphism $\mathcal{X} \to C$ (where $C$ is the affine line) such that every fiber is a normal (rational) projective surface, and

  2. a divisor $D$ on $\mathcal{X}$ such that $D_t := D|_{X_t}$ is a flat family of (reduced, possibly singular) irreducible curves.

I have some information about $X_{t_0}$ and $D_{t_0}$ for a specific $t_0$, and I am guessing (or, hoping) that the following hold in my case:

(A) for generic $t$, the surfaces $X_t$ (and the curves $D_t$) are homeomorphic,

(B) $X_{t_0}$ (resp. $D_{t_0}$) is homeomorphic to generic $X_t$ (resp. $D_t$).

How should I proceed to check if my guesses are correct? In particular, what are general criteria which guarantee (A) and/or (B)? (E.g. does any knowledge about cohomologies of the sheaf of sections of $D_t$ help?)

Any reference would be appreciated.

Edit 1: to be explicitly clear regarding (B): I know (B) does not hold in general, i.e. for specific values of $t$, $X_t$ and $D_t$ may not be generic. But I am hoping for some criteria which guarantee that they are generic, e.g. something like: if such and such cohomology groups vanish, then they will be generic.

Edit 2: An additional (related) question in view of Qfwfq's comment:

(C) Say you know $X_t$ and $X_{t_0}$ are homeo-/diffeomorphic. Then under what conditions on $D$ is $D_t$ homeo-/diffeomorphic to $D_{t_0}$? Note that I know all $D_t$'s are singular (curves) - so Ehresmann's theorem does not apply directly. (I am thinking of playing with the flows in the proof of Ehresmann's theorem - but there have to be some general results, no?)

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I think A true, B false. But for now I have no time to detail this. I'll come back soon! –  IMeasy Jun 3 '13 at 21:38
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I'm guessing the best cohomological way to detect genericity is going to be the vanishing cycle sheaf. You want a statement like: If there are no vanishing cycles, the fiber is homeomorphic to generic $X_t$. –  Will Sawin Jun 3 '13 at 23:03
    
I'm not sure which conditions are needed to make this true. –  Will Sawin Jun 3 '13 at 23:03
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If $\mathcal{X}\to C$ is a smooth morphism, then under your hypothesis (projectiveness) a theorem of Ehresmann assures the fibers are all diffeomorphic. –  Qfwfq Jun 3 '13 at 23:25
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1 Answer

For (A) your guess is correct. In fact it is true under much more general conditions by Verdier, Stratification de Whitney et theoreme de Bertini-Sard, Inventiones 1976, cor 5.1.

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Thanks! In Verdier's article there is a reference to a paper of Varchenko - which proves the results in the algebraic setting, and seems easier to read (e.g. I found the English translation :) - so I will try to read that first. –  auniket Jun 4 '13 at 13:47
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