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. $K^M_n(R)\cong \pi_n(K(F(R))$?

  • 1
    $\begingroup$ Interesting question. I would also like to know this. $\endgroup$
    – user717
    Dec 18, 2009 at 22:57

3 Answers 3


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".)

  • 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. $\endgroup$
    – user2146
    Dec 22, 2009 at 17:38
  • $\begingroup$ Oops, for n=q of course. So I mean $K^{M}_{q}(k)=[\mathrm{spec}(k)_{+},\tilde{U}_{tr}\tilde{\mathbb{Z}}_{tr}(S^q_t)]_{sPre_*}$ $\endgroup$
    – user2146
    Dec 22, 2009 at 17:48
  • $\begingroup$ There is a definition for Milnor K-groups for local rings: math.uiuc.edu/K-theory/0865 math.uiuc.edu/K-theory/0791 $\endgroup$
    – user19475
    Jan 18, 2010 at 20:00

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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.