# Fibrations with isomorphic fibers, but not Zariski locally trivial

(I posted this same question on MSE. Sorry if it is too elementary.)

I am looking for examples of fibrations $f:X\to Y$ where the fibers are all isomorphic, but $f$ is not Zariski locally trivial. In particular, I am interested in understanding how much such examples are "rare". (I believe they are not that rare)

First of all, by fibration I mean a proper flat surjective morphism of (complex) varieties. But I am not sure this is the correct definition of fibration in Algebraic Geometry; in that case, any correction is much appreciated.

By $f:X\to Y$ being Zariski locally trivial, I mean there is a variety $F$ such that every point in the base $Y$ has a Zariski open neighborhood $U$ such that $f^{-1}(U)\to U$ is isomorphic to the projection $F\times U\to U$. Here $F$ is called the fiber of $f$ (in particular, Zariski locally trivial fibrations do have isomorphic fibers).

One example I came up with is that of an étale cover of curves: the fibers are discrete of the same size, hence isomorphic, but it is not Zariski locally trivial in general.

Remark. Sometimes a fibration is required to have connected fibers; if this was the correct definition of a fibration, my example would not be an example.

Probably there are many important examples that I am missing here. I would very much appreciate if you could help me to fill in this picture!

Thank you.

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Do you mean all the fibres or just the fibres over closed points ? –  Damian Rössler Feb 23 at 21:23
I mean over closed points. But just because I think most of the times it seems to be somehow implicit. Isn't the generic fiber defined over another (bigger) field than $\mathbb C$? –  Brenin Feb 23 at 23:21

They are not rare. A general construction goes as follows. Start with a finite (étale) group (scheme) $G$ which acts on varieties $F$ and $\tilde Y$, where the second action has no fixed points. Let $Y =\tilde Y/G$. Then $(F\times \tilde Y)/G\to Y$ has all its fibres isomorphic to $F$. It is locally trivial in the étale topology, but not usually in the Zariski topology. This includes the above examples, and many others.

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An interesting example is the case where the fiber is $\mathbb{P}^n$. Then $X$ is called a Severi-Brauer variety over $Y$; such fibrations, modulo those which are Zariski locally trivial, are classified by an important invariant of $Y$, the Brauer group. Many varieties have a nontrivial Brauer group and therefore admit $\mathbb{P}^n$-fibrations which are not Zariski locally trivial.

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Very nice example! –  Peter Crooks Feb 23 at 17:10
Dear @abx, thanks for your answer! I was browsing to know more about these Severi-Brauer varieties and I found this MO post: mathoverflow.net/questions/95009/…. According to the first answer, point ($0$), these fibrations seem to be projectivizations of vector bundles (if we try to define them algebraically, which was implicit in my question). What am I missing here? –  Brenin Feb 23 at 20:50
Oh, is that what the Brauer group is? –  Ryan Reich Feb 23 at 21:49
@Brenin: No, in this post, there is a "if": if the fibration is a projectivization, it is not interesting -- in fact its class in the Brauer group is trivial. See here for more information. –  abx Feb 24 at 6:01

There lots of isotrivial elliptic surfaces that are not topologically trivial. E.g., they may have nontrivial monodromy; such a finration will not become trivial (even topologically) after removing a few points.

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The fibres of an elliptic surface are not isomorphic to each other in general. –  Damian Rössler Feb 23 at 21:21
The key word is isotrivial. They are easily constructed as quotients of $(\text{disk})\times(\text{elliptic curve})$. –  Alex Degtyarev Feb 23 at 21:22
@AlexDegtyarev: Just to be sure: is "isotrivial" a synonym of "having isomorphic fibers", isn't it? This also means, for the family, to be birational to the product family (I guess). –  Brenin Feb 25 at 15:09
@Brenin Yes, constant $j$-invariant. No, this is not birational to product. If the fiber has nontrivial automorphism, the monodromy $\pi_1(\text{base})\to\operatorname{Aut}(\text{fiber})$ may be nontrivial. I don't see how this can be undone birationally. –  Alex Degtyarev Feb 25 at 17:56