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Algebraic K-theory question

  1. May 10, 2007 #1
    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.


    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. jcsd
  3. May 10, 2007 #2


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    do you know they are the same?
  4. May 10, 2007 #3


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    If you believe wikipedia.
  5. May 10, 2007 #4
    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).
  6. May 12, 2007 #5


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    you might ask steve gersten, at utah.
  7. May 12, 2007 #6
    I think I'll just some time in the library next week looking.
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