Why does a homologically trivial cycle have trivial projections? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T17:40:51Z http://mathoverflow.net/feeds/question/111546 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/111546/why-does-a-homologically-trivial-cycle-have-trivial-projections Why does a homologically trivial cycle have trivial projections? Johan Commelin 2012-11-05T11:07:55Z 2012-11-05T15:42:49Z <p>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.</p> <p>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.</p> <p>In [Zhang] there is more context, but I think that I stated the relevant input for the claim.</p> <h2>Reference</h2> <p>[Zhang] Shou-Wu Zhang. <em>“Gross–Schoen Cycles and Dualising Sheaves”</em>. arXiv: <a href="http://arxiv.org/abs/0812.0371" rel="nofollow">http://arxiv.org/abs/0812.0371</a> .</p> <hr> <p><em>(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.)</em></p> http://mathoverflow.net/questions/111546/why-does-a-homologically-trivial-cycle-have-trivial-projections/111547#111547 Answer by Angelo for Why does a homologically trivial cycle have trivial projections? Angelo 2012-11-05T11:20:15Z 2012-11-05T15:42:49Z <p>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.</p> <p>[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.</p>