Assume k=mathbfC and W homogeneous. Let X=Proj(k[x1,cdots,xn]/(W)). X is then a smooth hypersuraface in mathbbPn−1.
Assume n=2d is even. Corollary 3.10 of the paper you quoted says that theta=0 for all pairs iff the homological Chow group CHd−1hom modulo [h]d−1 is not torsion (here [h] is the class of the hyperplane section). So your question, in this case, is equivalent to
(l=d−1):
Examples of smooth hypersurfaces of dimension 2l such that CHlhom/([h]l) is not torsion ?
(By the way, I think if you phrased your question this way, it probably would become more popular, consider how many geometry-inclined people visit this site! So if you want more and better answers, consider changing the title.)
Now, a cheap way to get examples you want is to take W=x1xd+1+cdots+xdx2d. Then the cycle defined by (x1,...,xd) will not be a multiple of a power of the hyperplane section. Why? Because, I am waving my hand a bit here, if it is then the intersection with the cycle (xd+1,cdots,x2d) would be positive. But they are disjointed in X!
The same trick works for generalized quadrics, i.e. if W=f1g1+cdots+fdgd.
EDIT: Let me give more details here. In this situation you can easily make W non-homogeneous as you desire. But the trouble is you can't use my argument above as there is no longer a projective variety X. But one can get around this. Let S=k[x1,cdots,x2d]m
here m is the irrelevant ideal. Suppose W=f1g1+cdots+fdgd and assume that (f1,cdots,fd,g1,cdots,gd) is a full system of parameters in S. Let R=S/(W), P=(f1,cdots,fd) and Q=(g1,cdots,gd). I claim that thetaR(R/P,R/Q)neq0.
The reason is that thetaR(R/P,R/Q)=chiS(S/P,S/Q), the Serre's intersection multiplicity (see Hochster's original paper). Because dimS/P+dimS/Q=d+d=dimS, we must have chiS(S/P,S/Q)>0 by Positivity, which is known in this case.
More exotic examples should be abound, and I am sure people who know more intersection theory can provide some, once they are aware of what this question is about. I would be interested in hearing more answers along that line.
No comments:
Post a Comment