16
$\begingroup$

Let $X$ be a smooth proper toric variety, $Z\subseteq X$ a smooth subvariety.

Is the fundamental class $[Z] \in H_\ast(X) = A_\ast(X)$ nonzero?

Background

If $X$ is a Kaehler variety, this is of course true, since the intersection of $Z$ with some power of the Kaehler class is positive. But our $X$ is not Kaehler if it is not projective (the Hodge structure is trivial + Kodaira embedding theorem).

I don't think that the statement is true for general complex algebraic varieties, for example the "Hironaka twist" (an example of a $3$-dimensional non-projective smooth proper algebraic variety) has two disjoint smooth curves $M_1$, $M_2$ with the property that $M_1 + M_2$ is numerically zero. So the union of these curves should give a zero class in homology. EDIT. This wouldn't be possible if $M_1$ and $M_2$ intersected - see Dustin Cartwright's and ulrich's comments below.

Chow rings (or homology) of smooth proper toric varieties is very well understood: it is generated by the boundary divisors, which are smooth toric varieties themselves, intersecting transversely. Relations are obvious from the fan description on $X$. In particular, we need to prove that $Z$ has non-zero intersection with some intersection of the boundary divisors. This could allow for an induction-on-dimension argument, if only $Z$ intersected the boundary nicely. We cannot hope for that, but maybe again there is a "perturbation" argument that may ensure this.

EDIT (continued). Dustin Cartwright's comment below shows that homology classes of irreducible curves in smooth proper varieties are nonzero. Therefore I am tempted to ask more than in the original question:

Let $X$ be a smooth proper variety, $Z\subseteq X$ a smooth subvariety. Is the fundamental class $[Z] \in H_*(X, \mathbb{Q})$ nonzero?

Improve this question
$\endgroup$
5
  • 4
    $\begingroup$ To answer your wondering about the Hironaka example: there is no such perturbation of $M_1 + M_2$. Any irreducible curve in a smooth proper algebraic variety has non-zero homology class. In affine coordinates, you can find a divisor intersecting the curve but not containing it. The closure of this divisor intersects the curve positively, so by the compatibility of intersection numbers and the cup product, this means that the curve must have non-zero homology class. $\endgroup$
    – Dustin Cartwright
    Commented Sep 17, 2012 at 3:03
  • $\begingroup$ @Dustin Cartwright: Thank you! I tried to use this argument for the general case, but then it's difficult to control what happens at the boundary after you take the closure - I didn't notice that it works in the curve case. I edited the question. $\endgroup$
    – Piotr Achinger
    Commented Sep 17, 2012 at 3:58
  • 4
    $\begingroup$ In Hironaka's example the curves are disjoint. The union of two irreducible curves meeting in at least one point cannot have zero homology class for the same reason as that mentioned by Dustin Cartwright: take an affine neighbourhood of a point of intersection! $\endgroup$
    – naf
    Commented Sep 17, 2012 at 5:38
  • $\begingroup$ @ulrich: thanks! I forgot that blowing up makes them disjoint. Another edit... $\endgroup$
    – Piotr Achinger
    Commented Sep 17, 2012 at 18:02
  • $\begingroup$ Piotr: for your question at the end, wouldn't $M_1\cup M_2$ in the Hironaka example would give you a smooth subvariety with a zero homology class? Or did you mean "irreducible"? $\endgroup$
    – Sándor Kovács
    Commented Jan 19, 2013 at 23:57

0

Reset to default

You must log in to answer this question.