Algebraic K-theory question

1. May 10, 2007

sparkster

The classical definitions of K_0, K_1, and K_2 for a ring R are

K_0(R)= Grothendieck completion of the set of isomorphism classes of finitely generated projective R-modules.

K_1(R)=GL(R)/E(R)=[GL(R)]^{ab}

For K_2, Milnor used the Steinburg St(R) group which maps onto E(R) and defined K_2(R) to be the kernel of this map. He proved that this is isomorphic to H_2(E(R),Z), which has a nice representation in terms of generators and relators of E(R) by Hopf's formula.

For the higher groups, Quillen used his plus construction and defined K(R)=K_0(R) x (BGL(R)+) and defined K_i(R)=pi_i(K(R)). So for i>0, K_i(R)=pi_i(BGL(R)+) since K_0 has no homotopy. For i=0, the definitions agree since K_0(R) is discrete and BGL(R)+ is path connected. For For i=1, the plus construction kills the homotopy of E(R), so the definitions agree.

I'm having trouble seeing how they agree for i=2. If GL(R) is discrete, I think there is a fibration that will give me the result, but if GL(R) is topological or a manifold, I don't how to show they are the same.

Last edited: May 10, 2007
2. May 10, 2007

mathwonk

do you know they are the same?

3. May 10, 2007

Hurkyl

Staff Emeritus
If you believe wikipedia.

4. May 10, 2007

sparkster

They're rigged so that they agree with the classical definitions. At the end of the wikipedia article they give a link to an online book. I've seen the book and they prove it via spectral sequences. I don't understand spectral sequences yet, so I was looking for an easier proof. Knudson's book The Homology of Linear Groups has an appendix with a quick sketch of algebraic k-theory. He comments that it's easy to see the two definitions agree, so I was hoping someone here had a more basic argument (ie, one that doesn't appeal to spectral sequences).

5. May 12, 2007

mathwonk

you might ask steve gersten, at utah.

6. May 12, 2007

sparkster

I think I'll just some time in the library next week looking.