Subject: Re: question abt Lie gps
From: mjh@math.mit.edu (Michael J. Hopkins)
Date: Wed, 22 Feb 2006 16:56:00 -0500 (EST)
Lisa
There's a paper from a few years back by Brylinski (I think it is in
K-theory) computing K_G(G). His argument builds on Hodgkin's proof
that K(G) is torsion free. Both arguments are sneaky versions of the
Eilenberg-Moore spectral sequence. The technique leads to the
cohomology computation you're asking about (at least with rational
coefficients).
The trick is to write
K_G(G) = K_(G\times )(G\times G)
where G\time G is acting on itself by
(a,b)(x,y) = (a x b^(-1), b y a^(-1)).
This holds because each G\times G orbit meets G \times 1, and the
stabilizer of G\times 1 is the diagonal. This equivalence can also be
thought of as the equivalence between flat G-bundles on S1 and flat
G-bundles on an annulus. Anyway, now use the Kunneth spectral sequence
Tor^R[G\times G] (K_{G\times G} (G), K_{G\times G} (G)) ==>
K_{G\times G}(G\times G) = K_G(G).
By a similar "orbit" argument,
K_{G\times G}(G) = K_G(pt) = R[G]. So the Tor calculation is just
the Hochshild homology of the representation ring. It follows that
K_G(G) = \Omega^*_R(G) -- the deRham complex of the representation ring.
Since R(G) is smooth, this is free as an R(G)-module. One now gets, by
Kunneth again
K_G(G^n) = \Omega^*_R(G) \tensor_{R(G)} ...\tensor_R(G) \Omega^*_R(G).
For the Borel construction, replace R(G) by its completion at the
augmentation
ideal.
The same argument works in cohomology with rational coefficients.
If you work at the chain level, rather than the cohomology level, you
get more or less the same picture with any coefficients.
Mike
>>
>> Subject: question for the topology mailing list
>> From: Lisa Jeffrey
>> Date: Wed, 22 Feb 2006 15:02:56 -0500 (EST)
>>
>> Let G be a compact connected Lie group (I am particularly interested in
>> the case G=SU(2)). Let G act on itself by conjugation. Can anyone
>> tell me the homotopy type of the quotient space G^n/G, or at least its
>> cohomology? I suspect this is well known but I don't know where to look
>> for it. Can anyone supply a reference?
>>
>> Many thanks, Lisa Jeffrey
>>
>>