If there is nontrivial isotrivial elliptic fibration, what will be type of singular fiber(since we have the kodaira's table of singular fiber). Is it a contradiction since in kodaira's table, smooth fiber around singular fiber seems non isomorphic. I am a bit confused?


1 Answer 1


I think one ususally uses isotrivial in this case to mean that the smooth fibers are all isomorphic. Then it is certainly the case that there can be singular fibers.

Which singular fibers can appear depends on the $j$ invariant. The $j$ invariants at which they can appear are the same as in a non-isotrivial fibration:

$I_0^*$ - any $j$ invariant.

$III, III^*$ - only $j=1728$.

$II, II^*, IV, IV^*$ - only $j=0$.

It is easy to construct examples of these:

For $j=1728$, the family $y^2= x^3 - t^k x$ will be of type $III$ if $k=1$, $I_0^*$ if $k=2$, and $III^*$ if $k=3$.

For $j=0$, the family $y^2 = x^3 - t^k x$ will be of type $(II,IV,I_0^*, IV^*, II^*)$ respectively as $k=(1,2,3,4,5)$.

And $y^2 =x^3 - t^2 ax -t^3 b$ will always have type $I_0^*$ when $y^2=x^3-ax-b$ is smooth.

I think you can check all these by Tate's algorithm, for instance.

I think in Kodaira's table he gives an example of a family for each type of singular fiber. But these are not the only examples.

  • $\begingroup$ It seems that isotrivial elliptic fibration doesn't have monodramy. But according to Kodaira, the type of singular fiber is determined by the monodramy. $\endgroup$
    – xin fu
    Aug 26, 2016 at 15:20
  • 2
    $\begingroup$ @xinfu A trivial elliptic fibration does not have monodromy. But an isotrivial fibration does. However, the mondromy must be contained in the automorphism group of the curve. This explains the restrictions I wrote down on the fiber type. $\endgroup$
    – Will Sawin
    Aug 26, 2016 at 21:10

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