Quillens higher K-groups of rings can be realized as πnK(C) - the Waldhausen K-Theory of a suitable Waldhausen category C. Is this also true for Milnor K-Theory of Rings? Is there a functor F from rings to waldhausen categories s.t. $K^M_n(R)\cong \pi_n(K(F(R))$?
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1$\begingroup$ Interesting question. I would also like to know this. $\endgroup$– user717Commented Dec 18, 2009 at 22:57
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3 Answers
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I don't know if there any evidence for this to be true. Note that Quillen K-groups are defined as homotopy groups of some space (+-construction, Q-construction, Waldhausen construction etc), whereas Milnor K-groups were defined in terms of generators and relations, which generalize generators and relations for classical K_2.
More invariantly Milnor K-groups can be constructed using homology of GL_n (paper of Suslin and Nesterenko) or as certain motivic cohomology groups of a field (Suslin-Voevodsky). However, these constructions are unrelated to any homotopy groups.
Also, I'm not sure how you define Milnor K-theory for a general ring R? (I was interpreting your question with "ring R" replaced by "field F".)
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1$\begingroup$ There is a paper on Milnor K-Theory of Rings from Elbaz-Vincent and Müller-Stach, but the definition is certainly much older (they cite a 1977 paper of Guin). At least for a field,as you say, Milnor K-Theory is on the diagonal of the motivic cohomology of that field and sice motivic cohomology is representable as homotopy classes of maps into the motivic Eilenberg-MacLane spaces one can write someting like $K^M_n(k)=[\mathrm{spec}(k),\tilde U_tr\tilde\mathbb Z_tr(S^q_t)]_{sPre_*}$. I guess one can play around a little bit and write that as homotopy groups, but at all I don't know the answer. $\endgroup$– user2146Commented Dec 22, 2009 at 17:38
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$\begingroup$ Oops, for n=q of course. So I mean $K^{M}_{q}(k)=[\mathrm{spec}(k)_{+},\tilde{U}_{tr}\tilde{\mathbb{Z}}_{tr}(S^q_t)]_{sPre_*}$ $\endgroup$– user2146Commented Dec 22, 2009 at 17:48
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$\begingroup$ There is a definition for Milnor K-groups for local rings: math.uiuc.edu/K-theory/0865 math.uiuc.edu/K-theory/0791 $\endgroup$– user19475Commented Jan 18, 2010 at 20:00
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Bob Thomason proved that there is no Milnor K-theory functor for schemes, with a reasonable map to Quillen K-theory, in:
Le principe de scindage et l'inexistence d'une $K$-theorie de Milnor globale. [The splitting principle and the nonexistence of a global Milnor $K$-theory] Topology 31 (1992), no. 3, 571--588.
- John
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The original question seems not to have been answered yet. One answer might be that it would be unnatural to expect all the Milnor K-groups of a field R to arise as the homotopy groups of a single space $K(F(R))$, because the natural way they currently arise is as homotopy groups of separate spaces, or better, of separate spectra. The spectra are the Eilenberg-MacLane spectra $\mathbb Z(n)$ associated to the chain complexes that compute motivic cohomology of $R$, namely, $K_n^M R = \pi_{-n} \mathbb Z(n)$.
