User ben williams - MathOverflow

Ostensibly different products on Ext-groups

2011-03-18T02:44:09Z

The following is presumably not the greatest generality in which this question makes sense.

Given a ring $k$, graded-commutative if it helps, and a Hopf-algebra $A$ over $k$, there is a Yoneda product making $\textrm{Ext}_A^*(k, k)$ into a ring (since $k$ is graded, this actually is a bigraded object, but I suspect the grading on $k$ is immaterial so I have suppressed it). We have from this construction an
There is also a product defined the following way $\newcommand{\E}{\textrm{Ext}}\E_A^*(k,k) \otimes_A \E^*_A(k,k) \to \E_{A \otimes_A A}^*(k\otimes_A k, k\otimes_A k) \to \E_A^*(k,k)$ using the external product on $\textrm{Ext}$.
One has a development of this idea: $\newcommand{\E}{\textrm{Ext}}\E_A^*(k,k) \otimes_k \E^*_A(k,k) \to \E_{A \otimes_k A}^*(k\otimes_k k, k\otimes_k k) \to \E_A^*(k,k)$ using the coproduct structure $A \to A \otimes_k A$ to change the base-ring. This appears to give a $k$-algebra rather than an $A$-algebra structure on $\textrm{Ext}$.

Should we expect these products to coincide? If they don't always coincide, are there conditions that ensure they do?

Answer by Ben Williams for Kunneth formula for motivic cohomology

2010-09-22T20:52:44Z

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).

Answer by Ben Williams for Homotopy theory of schemes examples

2009-10-27T02:35:09Z

If X is not smooth, then it is possible for the Chow groups and the A^1-represented motivic cohomology theory to disagree.

For instance, if we take X to be two copies of A^1 identified at a point then CH^0(X) has rank 2, but the sheaf represented by X in the A^1 category is contractible, so H^0(X,Z(0)) has rank 1.

To see this last point, we consider X as the colimit of a diagram A^1 <- * -> A^1. Since the maps in this diagrams are monomorphisms of schemes, they are cofibrations. The colimit of the diagram is therefore equivalent to the homotopy colimit, which is invariant under pointwise equivalences of diagrams. Since A^1 is contractible, we are left with the colimit of * <- * -> *, a point.

Morally, A^1 is contractible, so we can shrink down the A^1s without changing the A^1 homotopy type.

Comment by Ben Williams

2012-04-10T04:56:13Z

That's my guess. I think you understood it fine, and it's an erratum in the book.

Comment by Ben Williams

2012-04-10T00:25:36Z

My guess is that the numbers in example 1.E were changed to (7,1), (3,0) and (0,2) from (10,1), (4,0) and (0,3), perhaps to fit the page better, but the discussion a page later was not changed.

Comment by Ben Williams

2011-03-18T06:52:01Z

I suppose I should really try to understand the exterior product construction as a construction using the derived tensor product in the derived category of $A$-modules.

Comment by Ben Williams

2011-03-18T06:47:04Z

Thanks especially for the preprint, which I think meets my needs. I have qualms about the concrete construction though, because neither the functor $\otimes_A$ or $\otimes_k$ is exact (I assume $\otimes_k$ is what is meant by $\otimes$ in the explicit calculation).

Comment by Ben Williams

2011-03-18T03:53:59Z

Thanks for the answer, which is both helpful and promising. I am afraid I don't understand it, being inexperienced with this sort of thing. What is $E \otimes E'$? I think if I understood this I would understand everything. Thanks again.