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I was wondering when the Kunneth formula holds for motivic cohomology: $$ H^p(X,A(\alpha)) = \bigoplus_{i+j=p;\beta+\gamma = \alpha} H^j(X,A(\beta)) \otimes H^i(X,A(\gamma)) $$ where $H^p(X,A(\alpha))$ is defined as you wish: by higher Chow groups, Hom groups in $DM(X)$, etc... The case I'm most interested in is $A= \mathbb{Q}$ and $M(X) \in DM(\mathbb{Q})_{\mathbb{Q}}$ in the thick sub-triangagulated category generated by the $\mathbb{Q}(n)$, $n\in \mathbb{Z}$.

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up vote 8 down vote accepted

I now remember a nice argument, why there's no Kunneth formula for Chow groups of $X \times X$ unless $X$ has a Tate motive. Let $X$ be smooth projective of dimension $d$. We start with a decomposition of a diagonal: $$ [\Delta] = \sum_{i,j} \alpha^i_j \beta^{d-i}_j \in \oplus_i CH^i(X) \otimes CH^{d-i}(X) $$ We can assume $\alpha^i_j$ are linearly independent. In this case we can show that $\alpha^i_j$ form a basis of Chow groups and $\beta^{d-i}_j$ is the dual basis.

Indeed, as a correspondence $[\Delta]$ acts as identity on Chow groups, so for any class c, $$c = [\Delta]c = \sum_{i,j} \alpha^i_j deg(\beta^{d-i}_j \cup c),$$ and the claim follows if we substitute $c = \alpha^i_j$.

Now $CH_i(X) = Hom(\mathbb Z(i)[2i], M(X))$ and we can consider the set of $\alpha^i_j$ as a morphism of motives $$\oplus_{i,j}\mathbb Z(i)[2i] \to M(X).$$ A simple computation shows that it is an isomorphism with the inverse given by $\beta^i_j$.

And of course, on the other hand, if $X$ has a Tate motive, then Kunneth formula for Chow groups follows (it doesn't answer the question, since I only consider smooth projective varieties).

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That's really nice thanks. – AFK Jan 26 '10 at 1:22

I don't think so; for example if X is Spec(Q) then this doesn't seem to be true. If you define, for any $Y$ in this thick subcategory, $\mathbb{H}(Y)$ to be the bigraded ring $\oplus H^s(Y, \mathbb{Q}(t))$, then I'd expect there to be a spectral sequence of the form $$ {\rm Tor}^{{\mathbb H}({\rm Spec} \mathbb{Q}))}(\mathbb{H}(X),\mathbb{H}(X)) \Rightarrow \mathbb{H}(X \times X) $$ instead, i.e. you should take the module structure over the motivic cohomology of a point into account on the righthand side of the formula you were proposing.

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Correct me if I'm wrong. I think here is how it goes. For any $X$ and any Tate motive $M$, there is an isomorphism of modules over $H^{\*,\*}(Spec F)$: $$H^{\*,\*}(M(X) \otimes M) = H^{\*,\*}(X) \otimes_{H^{\*,\*}(Spec F)} H^{\*,\*}(M).$$


  1. Right hand side makes sense, since $H^{\*,\*}(M)$ is indeed a module over $H^{\*,\*}(Spec F)$

  2. When $M = M(\mathbf P^n)$, this is the the statement of the projective bundle theorem.

  3. It's unclear to me how to write down the individual terms $H^{p,q}(M(X) \otimes M)$ it terms of cohomology $M(X)$ and $M$.

To prove the theorem in this form it is sufficient to consider the case $M = \mathbf Q(n)$. Note that $H^{\*,\*}(\mathbf Q(n))$ is $H^{\*,\*}(Spec F)$ with bidegree shifted by (0,n), therefore tensoring with this module is the same as shifting the bidegree by (0,n). And the same thing is in the left-hand side, according to Voevodsky's Cancellation Theorem.

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There is a paper by Dugger & Isaksen called Motivic Cell Structures in which they establish a Künneth formula for various cohomology theories that are represented in the $\mathbb{A}^1$-homotopy category, provided the object $X$ satisfies some sort of cellularity condition which is similar to the requirement of having a Tate motive.

Of course, as Tyler suggested this Künneth formula is of the form of a spectral sequence over the motivic cohomology of the ground field, to wit $\mathrm{Tor}_{H(\mathrm{spec}\; k)} (H(X) , H(Y)) \Rightarrow H(X \times Y)$.

In general, this spectral sequence fails, as can probably be seen by all the counterexamples already given, and certainly can be seen by considering $\mathrm{spec}\; \mathbb{C} \times_\mathbb{R} \mathrm{spec} \; \mathbb{C}$ with $\mathbb{Z}/2$-coefficients, where the motivic cohomology rings are known in their entirities (These calculations also appear in papers of Dugger & Isaksen on the motivic Adams spectral sequence).

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Very briefly, I believe the following is true: Motivic cohomology does not satisfy a Kunneth formula on the level of cohomology groups, but it does satisfy a kind of Kunneth formula on the level of some suitable derived category of sheaves. This should hold true in general for any Bloch-Ogus cohomology theory, I think.

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Did you mean $H^p(X \times X, A(\alpha) = \dots$?

I 'm pretty sure Kunneth formula doesn't hold for motivic cohomology in general. I think you'll get the wrong statement even when you specialize to the case of Picard group of a product of two curves of genus > 0.

However, for Tate motives over $\mathbf Q$ it probaby does hold. Seems like it's sufficient to check the case of $\mathbf{P^n}$ for which motivic cohomology "coincide" with the usual cohomology.

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This is exactly what I had in mind. In the case of a mixed Tate motives, it feels like this is equivalent to Ext^2(A(0),A(n)) = 0 for the base field but I don't know how to turn this intuition into a proof. Any thoughts? – AFK Nov 25 '09 at 21:09
I had a wrong impression, that $H^{p,q}(Spec(F), \mathbf Q)$ is nonzero only for $p=q=0$. However, since it is not the case I agree with Tyler Lawson, that the formula must have tensor product over $H^{*,*}(Spec F)$ or something like that. Otherwise even H^{,}(X \times Spec(F)) doesn't satisfy the decomposition. – Evgeny Shinder Nov 25 '09 at 21:39

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