It is known that for any ring $R$, $$K_{1}(R)=H_{1}(GL_{\infty}(R), \mathbb{Z})$$ $$ K_{2}(R)= H_{2}(E_{\infty}(R),\mathbb{Z})$$ $$ K_{3}(R)= H_{3}(St_{\infty}(R),\mathbb{Z})$$ where $GL_{\infty}= lim GL_{n}$, $E_{\infty}=lim E_{n}$ (elementary matrices) , $St_{\infty}=lim St_{n}$ (Steinberg groups).


Is there a same description of higher groups of algebraic K-theory. i.e. for each $n>0$ there exist a group $M^{n}=lim M_{i}$ such that $$ K_{n}(R)= H_{n}(M^{n}(R),\mathbb{Z})$$ where every thing is functorial. in our case $M^{1}= GL_{\infty}$, $M^{2}=E_{\infty}$ and $M^{3}=St_{\infty}$.

  • 4
    $\begingroup$ At the least, there is a (perfect) group $M^n(R)$ which achieves this, but I'm not sure that it's a direct limit of nontrivial stuff. Quillen defines $K_n(R)$ as $\pi_n$ of $BGL(R)^+$, and we can find an $(n-1)$-connected cover $X_n\to BGL(R)^+$ which is an isomorphism on $\pi_n$, and now we can invoke the Kan-Thurston theorem to find $M^n(R)$ such that $\pi_n(BX_n^+)\cong\pi_n(BM^n(R)^+)\cong H_n(M^n(R))$. $\endgroup$ Jul 8, 2017 at 17:26
  • $\begingroup$ Chris Gerig, this construction is functorial in R ? $\endgroup$
    – Ofra
    Jul 8, 2017 at 17:50
  • $\begingroup$ I don't know (I should've clarified my phrase "achieves this" was referring to the homology statement alone). $\endgroup$ Jul 8, 2017 at 17:53
  • $\begingroup$ Yes. The non-stable Volodin groups. See link.springer.com/chapter/10.1007/BFb0062181 $\endgroup$
    – user40276
    Jul 13, 2017 at 23:19
  • $\begingroup$ @ChrisGerig If I understand the comments, it seems that you have answered the question. If this is correct, perhaps you can leave a (slightly expanded?) answer? $\endgroup$ Sep 12, 2017 at 0:53


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.