What do higher Chow groups mean? - MathOverflow

What do higher Chow groups mean?

2009-11-10T02:09:06Z

Let $z^i(X, m)$ be the free abelian group generated by all codimension $i$ subvarieties on $X \times \Delta^m$ which intersect all faces $X \times \Delta^j$ properly for all j < m. Then, for each i, these groups assemble to give, with the restriction maps to these faces, a simplicial group whose homotopy groups are the higher Chow groups CH^i(X,m) (m=0 gives the classical ones).

Does anyone have an intuition to share about these higher Chow groups? What do they measure/mean? If I pass from the simplicial group to a chain complex, what does it mean to be in the kernel/image of the differential?

Could one say that the higher Chow groups keep track of in how many ways two cycles can be rationally equivalent (and which of these different ways are then equivalent etc.)?

Finally: I don't see any reason why the definition shouldn't make sense over the integers or worse base schemes. Is this true? Does it maybe still make sense but lose its intended meaning?

Answer by Benjamin Antieau for What do higher Chow groups mean?

2009-11-15T16:47:50Z

This may not be particularly helpful, but when $X=\mathrm{spec} k$, for a field $k$, then $CH^{2n}(X,n)=K_n^M(k)$, the Milnor $K$-theory of $k$. I do not know if there are any other useful characterizations. Very little is known, except for the formal properties like $A^1$-invariance and such.

I believe that one can certainly define the higher Chow groups over other base schemes. For a reference, I would check Levine's book on motivic cohomology.

Answer by Jinhyun Park for What do higher Chow groups mean?

2010-01-23T19:49:27Z

Benjamin Antineau's answer needs a minor correction. For $X= spec (k)$, we have $CH^n (X, n) = K^M _n (k)$, not $CH^{2n} (X, n)$. Indeed not too many is known, but it is a very interesting subject (at least for me) to pursue. A good start would be Burt Totaro's paper 'Milnor K-theory is the simplest part of K-theory' or something similarly titled, where you can find the cubical version of it. Using $\mathbb{A}^1$-invariance with some spectral sequence arguments, one can prove that the above "simplicial version" and "cubical version" are isomorphic, thus equivalent.

Going back to Peter Arndt's question about 'intuition', the easiest one would be to look it as an algebro-geometric version of singular homology theory.

For instance, when $X$ is a topological space, a singular $n$-simplex is given by a continuous map $s: \Delta ^n \to X$. We collect their formal finite sums over the integers, and apply some simplicial formalisms. That's how we get the singular complex.

When $X$ is a variety, the problem is bad, even if we take $\Delta^n$ to be the algebraic n-simplex. One problem would be that there aren't enough morphisms of varieties $s: \Delta^n \to X$ to begin with. So, a way out is to take all "correspondences", i.e. closed subvarieties in the product space $\Delta^n \times X$. One problem that still persists here is that, to be able to apply the simplicial formalism, one has to have a good intersection property of correspondences with the faces of $\Delta^n$, but by taking all algebraic cycles, one may not get it. Consequently, we put conditions such as proper intersection with all faces.

That's why we define things in this way.

Regarding the question of what kernel/image does: it is difficult to explain everything, but the easiest case might worth paying attention: for instance, $z^i (X, 0)$ is the codimension i algebraic cycles on $X$, and the boundary map $z^i (X, 1) \to z^i (X, 0)$ by definition gives the rational equivalence of cycles on $X$. In this way, from the cokernel for instance, we recover the Chow group.

Answer by Steven Landsburg for What do higher Chow groups mean?

2011-08-07T15:15:03Z

I believe Bloch's original insight was something like the following:

First, if $X$ is a regular scheme, you can filter $K_0$ by ``codimension of support''; that is, view $K_0(X)$ as the Grothendieck group of the category of all finitely generated modules and let $F^iK_0(X)$ be the part generated by modules with codimension of support greater than or equal to $i$.

Next, suppose you want to mimic this construction for $K_m$ instead of $K_0$. The first step is to notice that if you patch two copies of $\Delta^m_X$ together along their "boundary" (i.e. the union of the images of the various copies of $\Delta^{m-1}_X$) and call the result $S^m_X$, then Karoubi-Villamayor theory tells you that $K_m(X)$ is a direct summand of $K_0(S^m_X)$. (The complementary direct summand is $K_0(X)$.)
So it suffices to find a "filtration by codimension of support" on $K_0(S^m_X)$.

The usual constructions don't work because $S^m_X$ is not regular (so that in particular, not all modules correspond to $K$-theory classes.)

But: a cycle in $z^i(X,m)$ has a positive part $z_+$ and a negative part $z_-$ which, (if it is homologically a cycle) must agree on the boundary. Therefore you can imagine taking $\Delta^m_X$-modules $M_+$ and $M_-$ supported on these positive and negative parts and patching them along the boundary to get a module on $S^m_X$. If this module has finite projective dimension (which it ``ought'' to because of all the proper-meeting conditions, and as long as it has no bad imbedded components), then it gives a class in $K_0(S^m_X)$, hence a class in $K_m(X)$, and we can take the $i-th$ part of the filtration to be generated by the classes that arise in this way.

The Bloch-Lichtenbaum work largely bypasses this intuition, but this was (I think) the original intuition for why it ought to work.