Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let X, Y be smooth projective (over IC), let f:X..>Y be a rational map. Assume Y is not uniruled. Is it true that f will be regular over a non-empty open subset of Y? Funny..

share|improve this question
Dear Ivor: it might be useful to give some motivation for the question. For instance, do you have a reason to think it is true? –  user5117 Mar 31 '11 at 12:04
If $f\colon X \to Y$ is dominant, then there are always non-empty, open subsets $U \subset X$ and $V \subset Y$ such that the restriction $f \colon U \to V$ is a regular map. Why are you requiring $Y$ non-uniruled? –  Francesco Polizzi Mar 31 '11 at 12:50
Dear Francesco, I think the point of the question (which was not immediately clear to me) is that f should have some proper fibres. –  user5117 Mar 31 '11 at 13:03
@Francesco: ivor probably means that there exists an open set $V\subset Y$ such that $f$ is regular (i.e., defined) on $f^{-1}(V)$. –  Sándor Kovács Mar 31 '11 at 14:49
@Sandor: Right, that's what I meant (but expressed badly). In his answer he clarifies that this is indeed what is meant. –  user5117 Mar 31 '11 at 15:52

2 Answers 2

I think you have already answered this yourself, for the most part. We have a birational morphism $X' \to X$ with some exceptional locus $E$ inside $X'$, which is some finite union of ruled components $\mathbb{P}^1 \times Z_i$. We also have the (proper) map $f': X' \to Y$, and the restriction of the map to any $\mathbb{P}^1 \times Z_i \to Y$ does not factor through $\mathbb{P}^1 \times Z_i \to Z_i$, so the image inside $Y$ will be uniruled. As such, the image of $E$ inside $Y$ is some closed set $V$ which by hypothesis cannot be all of $Y$. Restricting to the complement $U \subset Y$, the map $f'^{-1}(U) \to U$ has domain away from the exceptional locus inside $X'$, which is what you wanted, yes?

Also, the analysis you gave in your answer almost works, I think, except there is a slight problem in that you can only take $p$ to be very general (in the complement of a countable union of closed proper subsets) rather than merely general. For example, a very general K3 surface has infinitely many rational curves despite not being uniruled, so the locus in $Y$ to which your analysis applies would be precisely the complement of these infinitely many rational curves.

share|improve this answer
Thank you Arnav, I think that settles it. –  ivor Apr 1 '11 at 6:54

By 'regular over U' I meant f is regular everywhere over U. Sometimes this is called 'almost holomorphic'. For example, a standard projection IP_2->IP_1 is not almost holomorphic.

OK, some motivation: After blowing up X we obtain X' and a holomorphic map X'->Y. Positive dimensional fibers of X'->X are rationally chain connected, i.e., rational curves connect any two points. On the other hand, Y is not covered by rational curves by assumption.

Let p be a general point in Y and denote by F_p the fiber of X'->Y over p. The above remarks should imply: if F_p meets a positive dimensional fiber of X'->X, then it already contains this fiber (p general!). This should mean X->Y is a fibration over some non-empty open subset of Y.

It is either false or well known, I don't know.

share|improve this answer
I looked around a little but I couldn't find a reference for this: however, a result with a very similar flavour is Corollary 1.44 in Debarre's book, which says that if f: X ---> Y is a rational map, where Y is proper and contains no rational curve, then f is everywhere defined. –  user5117 Apr 1 '11 at 9:54
By the way, you should ask the moderators to merge your two accounts. –  user5117 Apr 1 '11 at 9:56

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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