5
$\begingroup$

I have a little confusion about the definition of motivic cohomology assuming the existence of a category of (mixed) motives. I've seen it defined as either $$\text{Ext}_{\mathcal{MM}_k}^i(1,M)$$ (for instance Beilinson "Notes on absolute Hodge cohomology) and $$\text{Ext}_{\mathcal{MM}_k}^i(M,1)$$ (for instance Mazza, Voevodsky and Weibel, "Lectures in Motivic Cohomology"). This is a bit confusing to me, these two cohomologies are not even functorial in the same way. Is there a conventional choice that I'm missing here (presumably on the functor from smooth schemes to motives being co- or contravariant)?

Improve this question
$\endgroup$
1
  • 1
    $\begingroup$ We expect that taking the dual is exact and induces an equivalence from mixed motives to the opposite category of mixed motives. The two definitions are related formally the same we go from singular homology to singular cohomology. The conventions are also related to the way you construct motives associated to algebraic varieties. Beilinson thought of this as a cohomology (hence contravariant) whereas Voevodsky considered this as homology. Therefore, both definitions agree, at least if you think of motivic cohomology as a cohomology of algebraic varieties. $\endgroup$
    – D.-C. Cisinski
    Commented Feb 7, 2023 at 9:57

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .