I am interested in (Chow) motives over (algebraically closed) characteristic $p>0$ fields. For $H$ being $\mathbb{Q}_l$-adic cohomology, one can consider $Ch_l(M)=\sum (-1)^i\dim_{\mathbb{Q}_l} H^i(M)$. Does this Euler characteristic depend on the choice of $l\neq p$?
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asked Jan 19, 2011 at 1:39
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I might be missing something (seeing that I don't know what a Chow motive is), but I think the answer is yes. It is a result of G.Laumon proved in Comparaison de caractéristiques d’Euler-Poincaré en cohomologie $\ell$-adique, C. R. Acad. Sc. Paris, t. 292 (1981), Série I, 209-212 that for $X$ a separated scheme of finite type over $k$ a field of characteristic $p>0$, the Euler-Poincaré characteristic (over $\bar{k}$) is independent of $\ell≠p$ (the proof consists in showing that the Euler-Poincaré characteristic coincides with the Euler-Poincaré characteristic in compact support; the latter being the degree of the Zeta function and thus independent of $\ell$). Hope this helps.
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$\begingroup$ I am sorry; could you say more? Unfortunately, I wasn't able to access the Laumon's paper (I only read reviews for it). So, my question is: how does one use zeta-function for varieties that are not defined over finite fields? $\endgroup$ Commented Jan 19, 2011 at 14:18
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$\begingroup$ The general case reduces to the case of finite $k$. The magic words here are "spreading out". $\endgroup$– OlivierCommented Jan 19, 2011 at 15:53
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$\begingroup$ Dear Olivier, for your proof to work, one would need the Zeta function of a Chow motive over a finite field to be independent of $\ell$. Is this known? Do you have a reference? $\endgroup$– jmcCommented Nov 5, 2013 at 10:38
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$\begingroup$ Dear jmc, well first of all this is not my proof at all, it is Laumon's. But more seriously, I'm not sure I understand: the zeta function of scheme is independent of $\ell$ by definition as no $\ell$ appear in it. $\endgroup$– OlivierCommented Nov 6, 2013 at 13:26
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1$\begingroup$ Dear Olivier, I agree that for a scheme it is independent of $\ell$. For a Chow motive this is a priori not clear. After some research in the literature I found a proof. It boils down to the fact that the Lefschetz trace formula works not only for morphisms, but for arbitrary correspondences. $\endgroup$– jmcCommented Nov 14, 2013 at 7:48