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Let $G$ be a finite group. Atiyah proved that the $K$-cohomology of $BG$ vanishes in odd degrees and in even degrees is the completion of the representation ring of $G$ at the augmentation ideal.

What is the $K$-homology of $BG$? I mean the stable homotopy groups of the complex $K$-theory spectrum smash $BG$.

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In "$K$-homology of universal spaces and local cohomology of the representation ring" Greenlees proves that $$ K_i(BG_+) \simeq H_J^i(R(G)) $$ for $i=0,1$, where $J$ is the augmentation ideal and $H_J^i(R(G))$ refers to local cohomology. In fact Greenlees proves something more general using equivariant homotopy theory (for which the above is a special case), and this more general object is also studied in Chapter 18 of Greenlees and May's "Generalized Tate Cohomology"

Equivalently, one can use Anderson's universal coefficient theorem for $K$-theory (see here). Anderson shows that if $G$ is even a compact connected Lie group that $KU_1(BG) = 0$ and $$KU_0BG = \operatorname{lim}_n\operatorname{Hom}(R(G)/J^n,\mathbb{Z}).$$

There is a discussion of the relationship between the two results in Remark 2.8 of "Topological $K$-(co)homology of classifying spaces of discrete groups", by Joachim and Lück (which also contains more results along these lines).

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  • $\begingroup$ Thanks Drew. I thought K_i(BG+) depends only on the parity of i. Is that also true for H^i_J(RG)? $\endgroup$
    – KHBG
    Commented Jul 25, 2014 at 6:13
  • $\begingroup$ The $K$-homology groups are two-periodic. I think Greenlees' formula only holds for $i=0,1$, but then you can just use the periodicity. I'll edit the answer. $\endgroup$
    – Drew Heard
    Commented Jul 25, 2014 at 6:22
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    $\begingroup$ Thanks again. Does H^2_J(R(G)) and higher vanish? $\endgroup$
    – KHBG
    Commented Jul 25, 2014 at 15:00
  • $\begingroup$ I believe not. By the Joachim and Lück paper the $n$-th local cohomology groups can identified with a colimit of $\text{Ext}^n$'s $\endgroup$
    – Drew Heard
    Commented Jul 26, 2014 at 0:47
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    $\begingroup$ @KHBG: yes, if $G$ is finite then $R(G)$ has Krull dimension one and so $H^k_J(M)=0$ for any $k>1$ and any $R(G)$-module $M$; see Chapter 6 of Brodmann and Sharp, for example. $\endgroup$
    – Neil Strickland
    Commented Jul 27, 2014 at 7:10
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For finite-type torsion spectra $X$ there is a natural isomorphism $$ KU_{n-1}(X) \simeq \text{Hom}_c(KU^n(X),\mathbb{Q}/\mathbb{Z}). $$ Here $\text{Hom}_c$ denotes the group of homomorphisms that are continuous with respect to the skeletal topology on $KU^*(X)$ and the discrete topology on $\mathbb{Q}/\mathbb{Z}$. The isomorphism can be obtained in a fairly straightforward way using $X\wedge S\mathbb{Q}=0$ (so $X=\Sigma^{-1}X\wedge S(\mathbb{Q}/\mathbb{Z})$) and standard exactness arguments.

If $G$ is a finite group then we can take $X$ to be $\Sigma^\infty BG$ (without disjoint basepoint) to get reduced $K$-homology groups $KU_{2n}(BG)=0$ and
$$ KU_{2n-1}(BG) = \text{Hom}_c(\widehat{J}(G),\mathbb{Q}/\mathbb{Z}). $$ Here $J(G)$ is the augmentation ideal in the representation ring, and $\widehat{J}(G)$ is the completion of $J(G)$ with respect to itself. If $G$ is a $p$-group then we just have $\widehat{J}(G)=J(G)\otimes\mathbb{Z}_p$.

The above answer is of course the same as you get via local cohomology, but this argument is more elementary.

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