Given a smooth scheme $X$ over a field $k$, we can consider its motive $M(X)$. It is an object in Voevodsky's triangulated category of motives $DM(k,\Lambda)$ where $\Lambda$ is the ring of coefficients. It is constructed from the presheaf with transfers represented by $X$ by forcing $A^1$-invariances and invertibility of the Lefschetz motive. Now, if $X$ is acted on by a finite group in such a way that the categorical quotient $X/G$ exists in the category of smooth schemes, I can consider the motive $M(X/G)$. On the other hand, without any assumption on the action, I can construct a motive $M(X)_G$ by taking the homotopy quotient of $M(X)$ by $G$ in $DM(k,\Lambda)$. There is a slight abuse here as one needs to work at the point set level (using model categories or infinity-categories) in order to define this homotopy quotient. As far as I understand $M(X)_G$ is also the motive of the quotient stack $[X/G]$ Clearly there is a map $$M(X)_G\to M(X/G)$$ and I am wondering whether this map is an isomorphism. I think that it is the case when we work in $DM_{et}(k,\Lambda)$, the variant of $DM(k,\Lambda)$ that uses the étale topology instead of the Nisnevich topology because in that case the map $X\to X/G$ is an étale cover. Is it also true in $DM(k,\Lambda)$ ? And if it is not true in general are there conditions under which it is known to be true ?
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4$\begingroup$ I'd guess not unless $\Lambda$ contains $\mathbb Q$, because in the case $X$ is a point, $X/G$ is a point, but applying some cohomology realization functor to $M(X)_G$, I'd guess we obtain the cohomology of $BG$, which does not equal the cohomology of the point. On the other hand with rational coefficients it should be OK at least on the level of cohomology realization functors. $\endgroup$– Will SawinCommented Apr 30, 2018 at 17:29
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1$\begingroup$ Well, with rational coefficients "Nisnevich motives" are isomorphic to etale ones.:) $\endgroup$– Mikhail BondarkoCommented Apr 30, 2018 at 17:44
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2$\begingroup$ I suspect that it suffices to take $k=\mathbb{R}$, $X=\operatorname{Spec} \mathbb{C}$, and $G=\mathbb{Z}/2\mathbb{Z}$ to obtain an example. Probably, the equality you want is fulfilled only if $G$ acts on $X$ by permuting connected components. $\endgroup$– Mikhail BondarkoCommented Apr 30, 2018 at 19:52
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