Let $X$ be an irreducible smooth projective variety of dimension $d$. Do there exist irreducible smooth projective curves $C_1, C_2,\ldots, C_d$, an open subset $U\subset C_1\times C_2\times\ldots\times C_d$ and a dominant morphism $f:U\to X$.

  • 8
    $\begingroup$ This is not true. For example, this does not hold for sufficiently general hypersurfaces of large degree (and dimension $> 1$) by results of C. Schoen "Varieties dominated by product varieties." Internat. J. Math. 7 (1996), no. 4, 541–571. $\endgroup$ – ulrich Jun 4 '12 at 14:30
  • $\begingroup$ Tony Scholl's comment at mathoverflow.net/questions/33665/… looks relevant. $\endgroup$ – David E Speyer Jun 4 '12 at 18:51
  • 1
    $\begingroup$ @Ulrich: please add that as an answer. $\endgroup$ – Steven Gubkin Jun 4 '12 at 19:55
  • $\begingroup$ @David: Thanks for pointing out that comment. $\endgroup$ – Rex Jun 5 '12 at 18:51

Ulrich observes in the comments that C. Schoen provides counterexamples in his paper "Varieties dominated by product varieties". Beyond Schoen's work, this problem has some interesting history, which I learned from this note of Oort.

Grothendieck, in attempting to prove the Weil conjectures, had hoped that every variety was rationally dominated by a product of curves. He asked Serre if this was true; Serre showed that a sufficiently general surface contained in an explicit Abelian variety of dimension $5$ is a counterexample. (See p. 145 of the Grothendieck-Serre Correspondence, which is a really amazing book.) The counterexample is quite beautiful and very simple, and of a rather different nature than Schoen's.

Essentially Serre observes that of $S\subset A$ is a smooth surface passing through the origin and satisfying the following property:

$(*)$ If $C, C'$ are curves contained in $S$, then $C+C'$ is not contained in $S$

then $S$ cannot be rationally dominated by a product of curves. This is because the rational map must extend to a morphism (as it is a map into an Abelian variety) given by adding two maps $C\to A, C'\to A$. So it suffices to find a surface $S$ satisfying $(*)$. Serre does this by writing down an explicit analytic germ at the origin satisfying $(*)$ (not too hard) and then approximating this germ by an honest surface (which one may take to be a complete intersection, for example).

Oort notes that Schoen seems not to have been aware of Serre's counterexample.

  • $\begingroup$ The link to the Oort note is broken. The title is Did earlier thoughts inspire Grothendieck?, and it is now available at this link. $\endgroup$ – R. van Dobben de Bruyn Feb 9 '17 at 17:21

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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