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Let $X$ be a smooth curve over a field. Let $Y$ be the triple product $X \times X \times X$. Let $\gamma$ be a homologically trivial codimension $2$ cycle.

In the text [Zhang, p. 76] that I am currently reading it is concluded that $\pi_{i,*}(\gamma)$ is trivial (rationally). I do not see why this is true, and could not find a proof elsewhere.

In [Zhang] there is more context, but I think that I stated the relevant input for the claim.

Reference

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

(I should add a note that I am currently not in the situation to search through a library. So maybe this can be found in any book on algebraic cycles; if so, please give the reference.)

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  • $\begingroup$ Actually, I do not fully understand whay is going on from "These imply that ..." (two lines below eq. 5.1.2 on p. 76 of [Zhang]). It seems pretty dense to me. But I think the above question is the most important key to understand the proof of [Zhang, Lemma 5.1.2]. $\endgroup$
    – jmc
    Commented Nov 5, 2012 at 11:43
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    $\begingroup$ What is $\pi_i$? $\endgroup$
    – Angelo
    Commented Nov 5, 2012 at 12:56
  • $\begingroup$ $\pi_{i}$ is the projection onto the $i$-th factor. I.e., one of the factors $X$. $\endgroup$
    – jmc
    Commented Nov 5, 2012 at 21:07

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If $X$ and $Y$ are smooth complete varieties and $f\colon X \to Y$ is a morphism, there is a pushforward in cohomology, which is Poincaré dual to the pullback. This pushforward is compatible with the pushforward on cycles; that is, the cohomology class of the pushforward of a cycle is the pushforward of the cohomology class of the cycle. Hence, the pushforward of a homologically trivial cycle is homologically trivial.

[Edit:] the poster wants a proof that $\pi_{i*}\gamma$ is trivial as in the Chow ring. If $\pi_i$ is the projection onto the $i^{\rm th}$ factor, then $\pi_{i*}\gamma$ has codimension~0, and the part of the Chow ring in degree 0 is $\mathbb Z$, so any class of cycles of degree 0 that is homologically equivalent to 0 is in fact 0. The alternative is that $\pi_i$ is a projection onto a product of the other two factors, but then the statement would be false, that is, $\pi_{i*}\gamma$ could very well be not rationally equivalent to 0.

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  • $\begingroup$ Thanks for your answer. However, I interpreted Zhangs claim as if $\pi_{i,*}(\gamma)$ is rationally trivial. I should have been more clear about that in my question. I really need rational equivalence to $0$. $\endgroup$
    – jmc
    Commented Nov 5, 2012 at 11:33
  • $\begingroup$ Thanks for the edit. Do you mean that the cycle map is injective on codimension $0$ cycles? That is a fact I did not know yet. $\endgroup$
    – jmc
    Commented Nov 5, 2012 at 21:58
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    $\begingroup$ To Johan: Well, if $X$ is a smooth irreducible complete variety, the only cycles in codimension 0 are the multiples of the fundamental class of $X$. Since the top cohomology is also isomorphic to $\mathbb Z$, and is generated by the cohomology class of the fundamental class of $X$, you see that the cycle map is an isomorphism. $\endgroup$
    – Angelo
    Commented Nov 6, 2012 at 4:48
  • $\begingroup$ I see, I actually should have been able to think that up myself. Thanks for helping me! $\endgroup$
    – jmc
    Commented Nov 6, 2012 at 8:53

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