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..
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$\begingroup$ 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? $\endgroup$– user5117Commented Mar 31, 2011 at 12:04
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$\begingroup$ 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? $\endgroup$– Francesco PolizziCommented Mar 31, 2011 at 12:50
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$\begingroup$ Dear Francesco, I think the point of the question (which was not immediately clear to me) is that f should have some proper fibres. $\endgroup$– user5117Commented Mar 31, 2011 at 13:03
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$\begingroup$ @Francesco: ivor probably means that there exists an open set $V\subset Y$ such that $f$ is regular (i.e., defined) on $f^{-1}(V)$. $\endgroup$– Sándor KovácsCommented Mar 31, 2011 at 14:49
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$\begingroup$ @Sandor: Right, that's what I meant (but expressed badly). In his answer he clarifies that this is indeed what is meant. $\endgroup$– user5117Commented Mar 31, 2011 at 15:52
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2 Answers
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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.
answered Mar 31, 2011 at 17:08
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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.
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$\begingroup$ 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. $\endgroup$– user5117Commented Apr 1, 2011 at 9:54
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$\begingroup$ By the way, you should ask the moderators to merge your two accounts. $\endgroup$– user5117Commented Apr 1, 2011 at 9:56