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I have two (not unrelated) questions. Let me first give a short introduction.

Introduction

For a general overview of the setup I refer to the introduction (§1) of [Zhang].

Let $k$ be a number field or a function field of a curve. Let $X/k$ be a curve, and assume that $X$ has a semistable model. Let $Y$ be the triple product $X^{3}$.

In [§5.1 (pp. 77v.), Zhang] he defines a motive $M$ as the kernel of the map $$ \bigwedge^{3} H^{1}(X)(2) \to H^{1}(X)(1) \colon a \wedge b \wedge c \mapsto a(b \smile c) + b(a \smile c) + c(a \smile b). $$ I think I have figured out what he means, and why this is a Chow motive.

A little further, he defines $\text{Ch}(M)$ as a subgroup of the kernel of the cycle map $\text{Ch}^{2}(Y) \to H^{4}(Y)$. It is the subgroup of elements $z$ satisfying

  • $z$ is stable under the action of the symmetric group $S_{3}$ on $Y$;
  • the push-forward $\text{pr}_{12,*} z$ equals $0$, where $\text{pr}_{12}$ is the projection on the first two coordinates;
  • if $i$ denotes the embedding $X^{2} \to Y \colon (x,y) \mapsto (x,x,y)$, and $\text{pr}_{2} \colon X^{2} \to X$ the projection on the second coordinate, then $\text{pr}_{2,*}i^{*} z = 0$.

Questions

Q1. Why this definition?

Q2.a. The notation suggests to me that there is a more general definition of $\text{Ch}(M)$, for Chow motives $M$. Is this true?

Q2.b. If so, can someone give me pointers (reference request) to a definition of $\text{Ch}(M)$, since all my Googles for "Chow group motive" give me results that use Chow groups to define motives (and I am actually not surprised to get those results.

I think that if $M$ is the motive corresponding to a smooth projective variety $X$, then $\text{Ch}(M)$ should equal $\text{Ch}(X)$. But what about the other motives?

References

[Zhang] Shou-Wu Zhang. “Gross–Schoen Cycles and Dualising Sheaves”. url: http://arxiv.org/0812.0371.pdf.

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  • $\begingroup$ Is there a unique exact functor from the category of motives to abelian groups such that it takes a smooth projective variety to its Chow group and a morphism of smooth projective varieties to the pullback morphism on Chow groups? $\endgroup$
    – Will Sawin
    Commented Oct 19, 2012 at 1:05
  • $\begingroup$ @Will I think that is what I am looking for. But I have no idea whether such a functor exists. Uniqueness would be a second issue. $\endgroup$
    – jmc
    Commented Oct 19, 2012 at 1:30
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    $\begingroup$ I mean, given a variety $X$ and an idempotent correspondence $e$ on it, $e$ acts by intersection on the Chow group as an idempotent, so the kernel of that correspondence should be the kernel of that action on the Chow group, and the cokernel should be the cokernel. I think that takes care of uniqueness. $\endgroup$
    – Will Sawin
    Commented Oct 19, 2012 at 1:48
  • $\begingroup$ Hmm, that seems to make sense. I am not yet very familiar with all this. So I will try to write it out for myself in a minute. Do you by chance have anything to say about Q1? Are you `surprised' by this definition, or are there other places in literature where I can read about this? $\endgroup$
    – jmc
    Commented Oct 19, 2012 at 2:55
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    $\begingroup$ I haven't had time to read the actual question carefully, but here is an answer to Will's comment: if $X$ is a smooth projective variety and $h(X)$ its motive, then there are functorial isomorphisms $\mathrm{Hom}(h(X),\mathbb L^s) = \mathrm{CH}^s(X)$ and $\mathrm{Hom}(\mathbb L^s, h(X)) = \mathrm{CH}_s(X)$. Here $\mathbb L$ is the Tate motive $H^2(\mathbf P^1)$. $\endgroup$
    – Dan Petersen
    Commented Oct 19, 2012 at 6:08

1 Answer 1

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One can define the Chow groups of a Chow motive $M$ by the formulas $$ \newcommand{\Hom}{\mathrm{Hom}}\newcommand{\L}{\mathbb{L}} \newcommand{\Ch}{\mathrm{CH}} \Ch^s(M) = \Hom(M,\L^s) $$ and $$ \Ch_s(M) = \Hom(\L^s,M).$$ Here $\L = H^2(\mathbf P^1).$ A reference for this definition is e.g. Scholl's text in the first Motives volume. When $M$ is the motive of a smooth projective variety $X$, then $\Ch^s X = \Ch^s M$ etc. But I must confess that I don't see why the definition of $\mathrm{Ch}(M)$ given in the question is consistent with this.

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  • $\begingroup$ @Dan Thank you very much for the answer, and the reference. I will accept this answer, and try to figure out the answer to Q1 myself. $\endgroup$
    – jmc
    Commented Oct 19, 2012 at 12:31
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    $\begingroup$ Shouldn't a morphism from the motive of a variety X to the motive of a variety Y be a codimension dimX cycle, i.e. CH^n(X)=Hom(h(X),L^{n-dimX}) ...? $\endgroup$
    – Konrad Voelkel
    Commented Dec 4, 2012 at 10:40

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