$\DeclareMathOperator\Var{Var}\DeclareMathOperator\Motives{Motives}$Let us assume for the moment that we have a "nice" category of motives, that is for fields $k$ we have a contravariant functor $$\Var(k)\to \Motives(k).$$ Now for any field extension $l/k$, we have a natural forgetful functor $$\Var(l)\to\Var(k).$$ Would we expect to have an extension of this functor to motives, that is a corresponding functor $$\Motives(l)\to \Motives(k)$$ compatible with the forgetful functor of schemes? I've only found references for the "other direction", that is restriction of scalars.
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1$\begingroup$ I'm always surprised by the great answers that even sometimes underspecified questions can get, but I'm a little puzzled here by what it means to ask about the conjectural properties of a conjectural object. $\endgroup$– LSpiceNov 26, 2022 at 22:17
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1$\begingroup$ @LSpice there are actual categories of motives that people use. Depending on which one you choose, certain of the expected properties hold by definition or are conjectural, but the categories exist unconditionally, and formal properties like this one can often be easily verified. $\endgroup$– R. van Dobben de BruynNov 26, 2022 at 23:24
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2$\begingroup$ The forgetful functor only exists for finite extensions $l/k$: if $l/k$ is infinite, a variety over $l$ does not become a variety over $k$ by forgetting $l$. $\endgroup$– nafNov 27, 2022 at 1:52
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2 Answers
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This would be a motivic analogue of the induced representation functor, and should be adjoint to the restriction functor, at least for finite field extensions.
This is just because the cohomology of a variety obtained by the forgetful functor is the induced representation of the cohomology.
It should be possible to construct this straightforwardly in most existing theories of motives.
For infinite field extensions it may depend on exactly which category of motives you mean and which finiteness conditions it might have - it certainly doesn't exist for Chow motives, say.
answered Nov 26, 2022 at 22:20
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For Voevodsky-type motivic categories over various schemes and a morphism $f$ of base schemes there exist the following functors: $f_*$ exists unconditionally and $f_{\#}$ (which is closely related to the functor $f_!$) exists if $f$ is smooth (thus, it should be finite and separable if you consider motives over fields). You may read https://link.springer.com/book/10.1007/978-3-030-33242-6 http://dml.mathdoc.fr/item/hal-01077507/ or http://user.math.uzh.ch/ayoub/PDF-Files/THESE.PDF on these matters.
answered Nov 30, 2022 at 8:40