This is a basic question about Grothendieck's conjectural category $M_k$ of pure motives (over a field $k$). This construction first produces a category (the "false category of motives") which need not be Tannakian; the category undergoes a modification of the commutativity constraints which produces a category (the "true" category) which could be Tannakian (this is spelled out on page 451 Jannsen's great paper).

The modification requires that the Kunneth components of the diagonal be algebraic (part of the standard motivic conjectures).

Question: Is there an alternate (geometric) construction (modulo standard conjectures) which produces the "true" category bypassing this artificial commutativity modification?

Does this happen in Voevodsky's setup?



I don't know any alternate geometric construction of the commutativity contraint, but there is a natural way to explain this, as a simple fact of homological algebra (and assuming very optimistic conjectures about mixed motives).

Let $DM_{gm}(k)$ be Voevodsky's triangulated category of geometric (=constructible) motives over $k$, with rationnal coefficients. For any (smooth) $k$-scheme, we have its homological motive $M(X)$ in $DM_{gm}(k)$ (this construction being functorial in $X$). Voevodsky proved that the full pseudo-abelian subcategory of $DM_{gm}(k)$, closed under arbitrary Tate twists, spanned by objects of shape $M(X)^\wedge=\underline{Hom}(M(X),\mathbf{Q})$ for $X$ smooth and projective is equivalent to the category $Chow(k)$ of Chow motives (the cohomological version, constructed from rational equivalence on cycles).

Now comes the conjectural part. Assume that there exists a motivic $t$-structure on $DM_{gm}(k)$; see this paper of Beilinson for what this means and for the link with the standard conjectures. Let us denote by $MM(k)$ the heart of this $t$-structure. The objects of the category $MM(k)$ are, by definition, mixed motives. This category is tannakian: for any prime number $\ell$, the $\ell$-adic realization functor defines a conservative $t$-exact and symmetric monoidal functor from $DM_{gm}(k)$ to the bounded derived category of finite dimensional $\mathbf{Q}_\ell$-vector spaces (this is part of what it means to be a motivic $t$-structure).

Call a mixed motive $M$ pure of weight $n$ if $M[-n]$ is pure of weight zero in the sense of Bondarko (see this paper for the link between $t$-structures and weight structures): this simply means that $M[-n]$ belongs to $Chow(k)$, seen as a subcategory of $DM_{gm}(k)$. For $X$ smooth and projective over $k$, as $M(X)^\wedge$ is pure of weight zero, for any integer $n$, the cohomology object $H^n(X)=\tau^{\leq n}\tau^{\geq n}(M(X)^\wedge)[n]$ is pure of weight $n$. This implies that we have a (non canonical) isomorphism $M(X)^\wedge\simeq \oplus_n H^n(X)[-n]$ (the so called `Chow-Künneth decomposition'.

Denote by $M(k)$ the full subcategory of $MM(k)$ which consists of finite sums of objects $M$ of $MM(k)$ which are pure of weight $n$ for some integer $n$. We then have a natural functor $$X\mapsto \oplus_n H^n(X)$$ from the category of smooth and projective $k$-schemes to $M(k)$. This functor factors (by construction) through the category $Chow(k)$: $$Chow(k)\to M(k)$$ and factors through numerical equivalence (just because the motivic $t$-structure is compatible with the monoidal structure, using Poincaré duality in $DM_{gm}(k)$), inducing an equivalence of categories $$Chow(k)/\text{numerical equivalence} \simeq M(k)$$ Now, the point is that $M(k)$ is a tannakian subcategory of $MM(k)$, and the modified commutativity constraint of the tensor product on the category of pure motives up to numerical equivalence is the one inherited from the natural monoidal structure of $M(k)$ through the equivalence above.

| cite | improve this answer | |
  • 2
    $\begingroup$ Many thanks for a wonderful clear answer! Your explanations were exactly to the point! And thanks for answering both my questions! I have also learned a lot from your other answers on MO! $\endgroup$ – SGP Apr 17 '11 at 0:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.