Understanding Algebraic K-theory through Quillen's Plus Construction

  • Thread starter Thread starter sparkster
  • Start date Start date
Click For Summary
The discussion focuses on the classical definitions of algebraic K-theory, specifically K_0, K_1, and K_2 for a ring R, and how they relate to Quillen's plus construction. K_0 is defined through the Grothendieck completion of finitely generated projective R-modules, while K_1 involves the abelianization of the general linear group GL(R). K_2 is linked to the Steinburg group and its kernel, showing isomorphism with H_2(E(R),Z). The participants express confusion regarding the agreement of definitions for K_2, particularly when considering the topology of GL(R). They seek simpler proofs or explanations, avoiding complex concepts like spectral sequences, and suggest consulting additional resources or experts for clarity.
sparkster
Messages
153
Reaction score
0
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:
Physics news on Phys.org
do you know they are the same?
 
mathwonk said:
do you know they are the same?
If you believe wikipedia.
 
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).
 
you might ask steve gersten, at utah.
 
I think I'll just some time in the library next week looking.
 
Thread 'How to define a vector field?'
Hello! In one book I saw that function ##V## of 3 variables ##V_x, V_y, V_z## (vector field in 3D) can be decomposed in a Taylor series without higher-order terms (partial derivative of second power and higher) at point ##(0,0,0)## such way: I think so: higher-order terms can be neglected because partial derivative of second power and higher are equal to 0. Is this true? And how to define vector field correctly for this case? (In the book I found nothing and my attempt was wrong...

Similar threads

  • · Replies 26 ·
Replies
26
Views
770
  • · Replies 3 ·
Replies
3
Views
802
Replies
2
Views
2K
  • · Replies 0 ·
Replies
0
Views
760
  • · Replies 0 ·
Replies
0
Views
2K
  • · Replies 80 ·
3
Replies
80
Views
9K
  • · Replies 3 ·
Replies
3
Views
5K
  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 3 ·
Replies
3
Views
1K