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Conjecture((generalized Bloch conjecture)): Suppose that $\text{H}^{k,0}(X)=0$ for all $k>0$. Then $\text{CH}_0(X)=\mathbb{Z}.$

Is generalized Bloch conjecture known for complete intersections of sufficiently large degrees? In which case generalized Bloch conjecture is known?

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    $\begingroup$ Maybe you should say what you call the generalized Bloch conjecture ... $\endgroup$
    – abx
    Commented Oct 21, 2013 at 14:10
  • $\begingroup$ I gave the statement of conjecture. $\endgroup$
    – mwZhang
    Commented Oct 21, 2013 at 15:05
  • $\begingroup$ What do you mean by "complete intersections of sufficiently large degrees"? For a fixed ambient variety $Y$, for a Cartier divisor $X$, if $\mathcal{O}_Y(X)$ is "sufficiently ample", then $h^0(X,\omega_X)$ will be nonzero. It seems to me it is better to consider complete intersections of sufficiently small degree so that every $h^{k,0}(X)$ equals $0$. $\endgroup$
    – Jason Starr
    Commented Oct 21, 2013 at 15:06

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Are you considering complete intersections in projective space? In that case, for a complete intersection $X$ in $\mathbb{P}^n$ of type $(d_1,\dots,d_c)$, then $\omega_X$ is isomorphic to $$\mathcal{O}_{\mathbb{P}^n}(-n-1+d_1 + \dots + d_c)$$. Thus $h^0(X,\omega_X)$ is nonzero unless $d_1+\dots+d_c \leq n$. In this case, $X$ is a Fano manifold. Thus, by work of Kollár-Miyaoka-Mori and Campana, $X$ is rationally chain connected (I think you can usually prove this directly for complete intersections; though definitely not for general Fano manifolds). Thus $\text{CH}_0(X)$ is just $\mathbb{Z}$ since every pair of points is connected by a chain of rational curves in $X$.

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  • $\begingroup$ What does "general complete intersections" mean in Voisin's paper "The generalized Hodge and Bloch conjectures are equivalent for general complete intersections". $\endgroup$
    – mwZhang
    Commented Oct 22, 2013 at 3:15
  • $\begingroup$ "General complete intersection" is what you think, but it is your version of the generalized Bloch conjecture which is different (and much weaker) from hers. Read the introduction of the paper, it is quite clear. $\endgroup$
    – abx
    Commented Oct 24, 2013 at 16:44

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