12
$\begingroup$

My question is simple, but I don't expect there are any simple answers.

Let $X$ and $Y$ be a pair of schemes, and let $X(\mathbb{C})$ and $Y(\mathbb{C})$ denote their respective spaces complex points. Suppose we are given spaces $X'\simeq X(\mathbb{C})$ and $Y'\simeq Y(\mathbb{C})$, where $\simeq$ denotes weak equivalence of spaces.

What is known about necessary or sufficient conditions for a map $f':X'\to Y'$ to be homotopy equivalent to a map coming from a morphism $f:X\to Y$ of schemes?

I am specifically interested in showing that a particular map of spaces is not homotopy equivalent to a map coming from a morphism of schemes.

$\endgroup$
  • 3
    $\begingroup$ What's the map? $\endgroup$ – leibnewtz Mar 15 at 1:58
  • 1
    $\begingroup$ for $g>1$, Riemann--Hurwitz implies that surjective holomorphic map is injective, and since $X$ is smooth, a holomorphic self-bijection is a holomorphic automorphism. Now note that any integral symplectic $2gx2g$ matrix can be realized as the map on first cohomology induced by a self-diffeomorphism of the associated smooth manifold. A self-diffeomorphism of a connected closed orientable smooth manifold is not null-homotopic; since there are finitely many holomorphic automorphisms, and there are infinitely many symplectic matrices, we are done. $\endgroup$ – Aknazar Kazhymurat Mar 15 at 3:52
  • 7
    $\begingroup$ As a general rule, there are much fewer maps between algebraic varieties than there are between the underlying topological spaces. For instance, the pullback map on cohomology induced by your $f'$ has to be compatible with the mixed Hodge structures on both sides. $\endgroup$ – dhy Mar 15 at 5:03
  • 2
    $\begingroup$ Here's a simple example first, $X=\mathbb{C}^{×} $. If I'm not mistaken a non trivial loop in $Y(\mathbb{C})$, for projective $Y$, is never realizable in this manner, since the map would extend through the (simply connected) $\mathbb{P}^{1}$ $\endgroup$ – EBz Mar 15 at 12:59
  • 5
    $\begingroup$ dhy's example of a necessary condition from Hodge structures is probably the best easy necessary condition you will find. It is even sufficient in the case of maps to abelian varieties. One can perhaps sometimes improve it by working with the mixed Hodge structure on cohomology of finite covering spaces. $\endgroup$ – Will Sawin Mar 15 at 15:45

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.

Browse other questions tagged or ask your own question.