Question

Quillens higher K-groups of rings can be realized as π_nK(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. $K^M_n(R)\cong \pi_n(K(F(R))$?

Answer 1

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".)

Answer 2

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

Answer 3

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, $K_n^M R = \pi_{-n} \mathbb Z(n)$.