(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.

  • $\begingroup$ Do you mean all the fibres or just the fibres over closed points ? $\endgroup$ Commented Feb 23, 2014 at 21:23
  • $\begingroup$ 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 $\mathbb C$? $\endgroup$
    – Brenin
    Commented Feb 23, 2014 at 23:21

3 Answers 3


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.


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.

  • $\begingroup$ Very nice example! $\endgroup$ Commented Feb 23, 2014 at 17:10
  • $\begingroup$ 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? $\endgroup$
    – Brenin
    Commented Feb 23, 2014 at 20:50
  • $\begingroup$ Oh, is that what the Brauer group is? $\endgroup$
    – Ryan Reich
    Commented Feb 23, 2014 at 21:49
  • 2
    $\begingroup$ @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. $\endgroup$
    – abx
    Commented Feb 24, 2014 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.

  • $\begingroup$ The fibres of an elliptic surface are not isomorphic to each other in general. $\endgroup$ Commented Feb 23, 2014 at 21:21
  • 2
    $\begingroup$ The key word is isotrivial. They are easily constructed as quotients of $(\text{disk})\times(\text{elliptic curve})$. $\endgroup$ Commented Feb 23, 2014 at 21:22
  • $\begingroup$ @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). $\endgroup$
    – Brenin
    Commented Feb 25, 2014 at 15:09
  • $\begingroup$ @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. $\endgroup$ Commented Feb 25, 2014 at 17:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.