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Let $X$ be a smooth quasiprojective algebraic variety over a field $k$. Then the $K$-groups $K_m(X)$ are defined, and there are two standard filtrations on them: the "codimension filtration" given by $$\operatorname{Fil}^p_{\mathrm{cod}} K_m(X) :=\bigcup\nolimits_Y \text{Ker}(K_m(X)\to K_m(X-Y))$$ where the limit is taken over all $Y$ of codimension $\ge p$; and the "$\gamma$-filtration" $\operatorname{Fil}^\bullet_\gamma$ defined by Quillen. I gather that $\operatorname{Fil}^p_\gamma \subset \operatorname{Fil}^{p-m}_{\mathrm{cod}}$, and that for $m = 0$, this becomes an isomorphism after tensoring with $\mathbb{Q}$.

  • Are there other cases (maybe under additional assumptions on $k$ or on $X$) where the two filtrations agree rationally? Grayson's article in "Handbook of $K$-theory" suggests that they should be different when $X = \mathrm{Spec}(k)$ and $k$ has large cohomological dimension, but I confess that I don't understand the reasoning behind this.

  • The two filtrations both come from spectral sequences: the codimension filtration from the Brown--Gersten--Quillen sequence $$E_2^{pq} = H^p(X, \mathscr{K}_{-q}),$$ where $\mathscr{K}_q$ denote the $K$-theory sheaves on $X$; and the $\gamma$-filtration (if I've undstood correctly) comes from the Friedlander--Suslin motivic spectral sequence $$E_2^{pq} = \operatorname{CH}^{-q}(X,-p-q) = H^{p-q}_{\mathrm{mot}}(X, \mathbb{Z}(-q)).$$

    Both converge to $K_{-p-q}(X)$ (sorry about the convoluted indexing!). Landsburg ("Relative Chow groups", Illinois J Math (35), 1991) has constructed maps $\operatorname{CH}^{-q}(X,-p-q) \to H^p(X, \mathscr{K}_{-q})$ between the $E_2$ terms of the spectral sequences, which he shows are isomorphisms if $p + q = 0$ or $-1$.

    Are Landsburg's maps compatible with the differentials in the two spectral sequences, and are there additional hypotheses which would force them to be isomorphisms (maybe up to torsion) in all degrees?

(Apologies if this is a naive question; I'm only just beginning to learn all of this stuff.)

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2 Answers 2

up vote 6 down vote accepted

At the time when I constructed those maps, the motivic spectral sequence was not known to exist. I gave a conjectural construction for that spectral sequence in an American Journal of Math paper called "Relative Cycles and Algebraic K-Theory", but I still don't know whether the gaps in that construction can be (or have been) filled.

I stopped working on this long enough ago that I could be remembering wrong, but my best recollection is that if the conjectural construction in RCAKT can be made to work, and if the spectral sequence it yields is the same as Friedlander-Suslin's (which it certainly ought to be), then you can piece together results in those two papers to show that the maps in question are compatible with the differentials. (You might also need some results from my paper "Some Filtrations on Higher K-Theory and Related Invariants".)

There might also be a much easier argument, directly using the Friedlander-Suslin construction, which would save you from mucking around in those other papers.

I apologize for many things, including my hazy memory, my failure to state this result explicitly in the original papers, and the fact that I'm not up to date on the current state of the art.

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Regarding fields, you might already know this, but if you take $p=0$, then $Ch^{-q}(Spec(k),-q)=K_{-q}^M(k)$, whereas $H^0(Spec(k),\mathscr{K_{-q} })=K_{-q}^Q(k)$, where $K^M$ and $K^Q$ are Milnor and Quillen $K$-theory. These won't agree in general; for example, for finite fields the higher Milnor $K$-groups are all zero.

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I didn't know the isomorphism between the higher Chow group and Milnor $K$-theory -- what's a good reference for this? And by $H^0(\operatorname{Spec}(k), -q)$ do you mean what I was calling $H^0(\operatorname{Spec}(k), \mathscr{K}_{-q})$, or do you mean Voevodsky-style motivic cohomology $H^0(\operatorname{Spec}(k), \mathbb{Z}(-q))$? –  David Loeffler Sep 3 '12 at 8:15
    
Hang on, I found it: Totaro, "Milnor K-theory is the simplest part of algebraic K-theory," K-theory 6 (1992). –  David Loeffler Sep 3 '12 at 8:24
    
David: Sorry; I mistyped and put $-q$ where I meant to put ${\cal K}_[-q}$. So you and I meant the same thing --- the zeroth Zariski cohomology of the sheaf ${\cal K}_{-q}$, which is of course just $K_{-q}(k)$. At the risk of rendering your comment confusing to future readers, I'm editing the answer to correct this typo. –  Steven Landsburg Sep 3 '12 at 13:45
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