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I would like that someone helps me to find an article on the net treating the following decomposition : $ \mathrm{CH}^k (X)_{ \mathbb{Q} } = \displaystyle\bigoplus_{ i + j = k } \mathrm{CH}^{i,j} (X) $. I don't remember exactly where i found that article on the net. Can anyone help me please ?

Thanks a lot to all of you.

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    $\begingroup$ I don't think it is reasonable to expect such a decomposition for all projective varieties; what is expected is a filtration (the conjectural Bloch-Beilinson filtration). A decomposition exists for some particular varieties, e.g. abelian varieties. $\endgroup$ – abx Nov 14 '15 at 17:13
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abx has basically answered your question in a comment, but let me add some detail. There is no natural such direct sum decomposition; you are probably thinking of the Bloch-Beilinson filtration on Chow groups, whose existence in general is only conjectural. A very nice reference is a paper by Jannsen in one of the "Motives" volumes.

There are some examples where the conjectural filtration is expected to split, so that one really gets a direct sum decomposition. If $X$ is an abelian variety, then there is the Beauville decomposition of the Chow ring, which is defined unconditionally using the Fourier transform. It gives rise to a graduation (as opposed to a filtration) on Chow groups, satisfying all expected properties of the BB filtration except a conjectural vanishing property.

A conjecture of Beauville predicts that for any hyperkähler variety $X$, the Bloch-Beilinson filtration on the Chow groups of $X$ splits. Here a direct sum decomposition satisfying all expected properties was constructed by Beauville and Voisin in the case of a K3 surface.

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  • $\begingroup$ Sorry; usually these filtrations are defined on $Q$-linear Chow groups only, aren't they? Then what's the point of saying that the filtration splits? Are you speaking about some canonical splitting here? Moreover, in the general case there quite natural reasons for this filtration to split (though not quite canonically). $\endgroup$ – Mikhail Bondarko Nov 16 '15 at 11:28
  • $\begingroup$ I guess I should've said "Chow ring" rather than "Chow group" - I meant splitting compatibly with the intersection product. $\endgroup$ – Dan Petersen Nov 16 '15 at 14:32

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