Derived Algebraic Geometry and Chow Rings/Chow Motives

I recently heard a talk about Chow motives and also read Milne's exposition on motives. If I understand it correctly, the naive definition of the Chow ring would be that it simply consists of all algebraic cycles, but to define a multiplication one needs to impose a certain equivalence relation, either rational or numerical equivalence.

I wondered myself if an alternative definition using derived algebraic geometry would be possible. Regardless which framework of derived algebraic geometry you use, a feature should be that you can get always the correct intersection/fiber product. Therefore, one might try to define a derived Chow ring by considering 'derived algebraic cycles' (without any equivalence relation). One would probably get a space out of this instead of a set, but this wouldn't necessarily be a bad thing. Also, the associated category of 'derived Chow motives' would then be a simplicial category (or $(\infty,1)$-category).

What I would like to know is the following: Has somebody tried to build such a theory and if not, what are the problems of it or why is it perhaps a bad idea right from the start?

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 I gather this is part of the motivation behind Jacob Lurie's paper on "Structured Spaces" (which you can find on his website) though I have not understood enough of it yet to see if he carries out this particular application. – Eric Finster Aug 22 2010 at 14:40

I can't give you a complete answer apart from saying that this is definitely a hard problem!

If you have a derived scheme $X$, you can always truncate to get an every-day scheme $t_0 (X)$. On the level of rings, this corresponds to truncating a simplicial ring $A$ to $\pi_0 (A)$. The canonical morphism $A \to \pi_0(A)$ is obviously 0-connected. A result of Waldhausen then tells you that $A$ and $\pi_0(A)$ have isomorphic $K_0$ and $K_1$.

Now the Chow-Ring is more or less the same as $K_0$ (after tensoring with Q and up to rational equivalence). So I guess what this means is that on the level of cycles it is impossible to tell the difference between a derived scheme and an ordinary scheme, because $CH(X) = CH(t_0 (X)).$

I think the fancy way of saying this is that the etale topoi of $X$ and $t_0(X)$ are the same, but that is slightly over my head. This result can be found both in Luries thesis and in HAG II by Toen-Vezzosi.

The situation immediately changes though if you work with Bloch´s higher Chow groups! They see the difference starting from $K_2$, since the higher Chow groups are tied to higher $K$-groups via a Riemann-Roch map. So on the level of these cycles you could tell a difference.

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I don't know much about this stuff, so instead of answering the question, I try to formulate more precise questions, in the hope someone else will take up these questions:

One of the main reason to look for cycles is that they give realizations (their fundamental class) in all cohomology theories, which happen to have special properties (e.g., are Hodge cycles or Tate cycles), and anytime you see a Hodge (or Tate) cycle in cohomology, you expect that it comes from an algebraic cycle (the Hodge or Tate conjecture) and hence similar phenomena should occur in all cohomology theories (i.e., there is a Hodge (or Tate) cycle in all realizations).

Now, if the following were true:

1) Any 'derived algebraic cycle' gives rise to virtual fundamental classes in all cohomology theories, which again are Hodge or Tate cycles.

2) It is not clear that the virtual fundamental classes of 'derived algebraic cycles' are already fundamental classes of real algebraic cycles,

then one might formulate a 'derived' Hodge or Tate conjecture, which would have the same consequences.

Your question has another aspect, which regards a possible framework for working with these motives; I leave this aside as I understand even less about how this should work.

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