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

Let me ask you a question about $Z/pZ$ - acyclic affine toric varieties (for some $p$) i.e. toric varieties $X$ such that homologies $H_{j}(X,Z/pZ) = 0$ for $j > 0$ and $H_{0}(X,Z/pZ) = Z/pZ$. I am interested how big such class of $Z/pZ$ - acyclic affine varieties is? For example, it is known that the affine spaces are the examples of such $Z/pZ$ - acyclic varieties. On the other hand, are $(C^{*})^r \times (C^{+})^{n-r}$ $Z/pZ$ - acyclic varieties? So, again:

Question: how big class of $Z/pZ$ - acyclic affine toric varieties is?

I would be interested in any reference about this question.

Thanks for you your answer!

share|improve this question
add comment

1 Answer

up vote 0 down vote accepted

It turns out not to be the same as every other notion of acyclic toric varieties.

If the associated cone of affine toric variety has nonempty interior, equivalently, if the toric variety contains a torus-invariant point, then it will be acyclic, because it will be homotopic to a point. We just choose a one-parameter subgroup of the torus that converges to the point and act on everything in the toric variety with that one-parametr subgroup, so they all converge to the point, forming a contraction.

If the associated cone has empty interior, equivalently, if there are no torus-invariant points, then there will be a subgroup of the torus which the stabilizer of every point is contained in. Taking the quotient by this subgroup will give a map from your algebraic variety to the quotient torus with connected fibers. One can pull back a nontrivial cohomology class in $H^1$ along this map, and it will remain nontrivial.

share|improve this answer
    
Many thanks for your answer! –  Andriy Regeta Jul 24 '13 at 18:37
add comment

Your Answer

 
discard

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.