# How can I see that when $X$ is a trivial $G$-space, $K(X) \otimes R(G) \rightarrow K_G(X)$ is an isomorphism?

Here, $K$ is complex K-theory, $R(G)$ is the complex representation ring of $G$ (which -- for now, though it shouldn't matter -- is a finite group), and $K_G$ is $G$-equivariant complex K-theory.

In fact, I don't even see how the map is supposed to run. If $E\rightarrow X$ is a vector bundle and $G \rightarrow \mbox{Aut}(V)$ is a $G$-representation, then I should be able to get a vector bundle out of the pair $(E,V)$. (If so, I can extend in the obvious way to formal differences in both slots.) If $\mbox{rk }E = \dim V =r$, then we can just convert $E$ into a principal $U(r)$-bundle $\mathcal{F}(E)$ and apply the Borel construction $\mathcal{F}(E) \times_{U(r)} V$; up to messing with which side things are acting on, this admits a fiberwise $G$-action and we're good to go. I'll denote this composite construction by $E \cdot V$ for brevity. If $E$ and $V$ don't have the same rank, then presumably we should stabilize until they do and then proceed as before, but I'm having trouble working this out. If $\mbox{rk }E=n$ and $\dim V=n+d$, then we could carry this out as $E\cdot V = (E\oplus\underline{\mathbb{R}^d})\cdot V - \underline{\mathbb{R}^d} \cdot V$ ... but now we're stuck trying to define the subtrahend, which is no better than before. Ditto for when the ranks are switched.

Then, why is this an isomorphism if $G$ acts trivially on $X$? I've seen this stated breezily in two different sources, with no indication of why it should be true. To describe a backwards map, we might take a $G$-equivariant bundle $E \rightarrow X$, choose a point $x\in X$ and then decompose $E \stackrel{\sim}{\leftarrow} \mathcal{F}_{E_x}(E) \times_{U(r)} E_x$ (where by $\mathcal{F}_{E_x}(E)$ I mean the frames modeled on the complex vector space $E_x$). But I'm having trouble convincing myself that this is indeed $G$-equivariant. I suspect I'm missing something basic here; I'm not sure I have a single equivariant bone in my whole body.

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I feel for you - I too, have no equivariant bones! But have you checked 'Segal - Equivariant K-theory'? The map $K(X) \otimes R(G) \to K_G(X)$ seems to be fairly easy to construct - there is always a map from $X$ to a point, and so this gives a map $R(G) \to K_G(X)$. For a trivial action we also have a homomorphism $K(X) \to K_G(X)$ and the combination is the map in question. Segal then explicitly constructs an inverse to this map –  Juan S Sep 2 '12 at 22:49
@JuanS: Ah, I've looked at it but I'll check again! This was one of the two sources I was referring to in my last paragraph. (The other is Rezk's notes on power operations.) IIRC it's on the very first page that Segal asserts this isomorphism as a self-evident fact. Maybe I should have read further. –  Aaron Mazel-Gee Sep 3 '12 at 0:12
See Proposition 2.2 –  Juan S Sep 3 '12 at 2:08
Not having looked at the paper in a long time, I think the point should just be that you can consider your G-representation as a vector space (with G-action) and cross with X to get a trivial bundle (with fiberwise G-action) and then tensor that (G-)vector bundle with your bundle $E\rightarrow X$. This is the same as what Juan said, essentially; I'm just saying it less functorially. –  Dan Ramras Sep 3 '12 at 6:39
@DanRamras: Yeah, that's essentially what I was trying to do in my final paragraph. I'll expand in a CW answer. –  Aaron Mazel-Gee Sep 3 '12 at 10:56

Here is one (maybe not to enlightning) way to see that at least these two groups are abstractly isomorphic. Please let me know if I got something wrong.

My proof will use that topological $K$-theory is a special case of bivariant $K$-theory. Moreover it is important that the group $G$ is compact (or discrete and finite). Let me also assume for simplification that $X$ is compact. So here we go:

$K^G(X) \cong K_0^G(C(X))$ due to an equivariant version of the Theorem of Swan. Now we use bivariant $K$-theory to get $K_0^G(C(X)) \cong KK_0^G(\mathbb{C},C(X))$.

Now there is a Theorem called the Theorem of Green and Julg, saying that $KK^G_0(\mathbb{C},C(X)) \cong KK_0(\mathbb{C},C(X)\rtimes G)$ where $C(X)\rtimes G$ denotes the socalled crossed product $C^*$-algebra of the $G$-$C^*$-algebra $C(X)$. This is only true if $G$ is compact. For a proof see for example the book "basic bundle theory and $K$-cohomology invariants" partly written by Husemöller. There is a proof of the Green-Julg Theorem in there by S. Echterhoff using $KK$-theory.

Now it is a fact from $C^*$-algebra theory that if $G$ acts trivially on an algebra like $C(X)$ in your case, that $C(X)\rtimes G \cong C(X)\otimes C^*G$. In the case where $G$ is discrete this is just $C(X)\otimes \mathbb{C}G$.

Now using the Künneth-type sequence of Claude Schochet (both algebras are nuclear, (at least if $G$ is finite and discrete) and therefore in the bootstrap category).

$0 \to K_*(A)\otimes K_*(B) \to K_*(A\otimes B) \to \mathrm{Tor}(K_*(A),K_*(B)) \to 0$

for the case $A=C(X)$ and $B= \mathbb{C}G$ and using the fact that $K_0(\mathbb{C}G) \cong R_{\mathbb{C}}(G)$ and $K_1(\mathbb{C}G) = 0$ we see that the Tor term vanishes and that

$K^G_0(C(X)) \cong KK_0(\mathbb{C},C(X)\rtimes G) \cong K_0(C(X)\otimes \mathbb{C}G) \cong K_0(C(X))\otimes R_{\mathbb{C}}(G)$

Note that we also use that $KK_*(\mathbb{C},C(X)) \cong K_0(C(X))$.

Actually we could have avoided $KK$-theory, but I do not know a reference of the Green-Julg isomorphism that does not use $KK$-theory.

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Haha. This is cool, but you're right that it's not particularly enlightening (at least for those of us who aren't $C^*$-algebraists!). In any case, it's good to know that this result is a special case of a much more general theory. –  Aaron Mazel-Gee Sep 3 '12 at 13:12
so I just checked, also in the case where $G$ is non-discrete but compact, it still is true that $G$ is amenable, hence the full groups $C^*$-algebra $C^*G$ is nuclear. In particular your isomorphism should also hold for compact groups if we identify or replace $R(G)$ by $K_0(C^*G)$. :) .. thanks that you still like the approach, although it is kind of abstract nonsense.. –  mland Sep 3 '12 at 14:34

So it turns out that I actually got the map wrong, or at least I was being too complicated by trying to use frame bundles; rather than define it on arbitrary monomials, one just uses the universal property of the tensor product. Specifically, as Juan S points out in the comments, there's a "trivial $G$-action" map $K(X) \rightarrow K_G(X)$, and there's the map $R(G)=K_G(\mbox{pt}) \rightarrow K_G(X)$, induced by the terminal map $X \rightarrow \mbox{pt}$ (which is a map of trivial $G$-spaces).

To prove that the induced map $\varphi: K(X) \otimes R(G) \rightarrow K_G(X)$ is an isomorphism, we construct an explicit inverse $\psi$ (as I attempted to do above). This assumes that $G$ is compact (so that its set $\mbox{simp}(G)$ of isomorphism classes of simple representations is finite, maybe?). Our explicit inverse $\psi:K_G(X) \rightarrow K(X) \otimes R(G)$ is given on a $G$-vector bundle $E$ by $$\psi(E) = \sum_{V \in \mbox{simp}(G)} \mbox{Hom}^G(\underline{V},E) \otimes [V].$$ On the one hand, $\varphi \circ \psi = \mbox{id}_{K_G(X)}$ since $$\sum_{V \in \mbox{simp}(G)} \mbox{Hom}^G(\underline{V},E) \otimes [V] \mapsto \bigoplus_{V \in \mbox{simp}(G)} \mbox{Hom}^G(\underline{V},E) \otimes \underline{V} \stackrel{\sim}{\rightarrow} E$$ (as this is true fiberwise). On the other hand, we can check factorwise that $\psi \circ \varphi = \mbox{id}_{K(X) \otimes R(G)}$. First, $$E \otimes 1 \mapsto E \mapsto \sum_{V \in \mbox{simp}(G)}\mbox{Hom}^G(\underline{V},E) \otimes [V] = \mbox{Hom}^G( \underline{1},E) \cong E$$ since any nontrivial $V\in \mbox{simp}(G)$ induces a $G$-bundle $\underline{V}$ which has no nontrivial maps to $E$. Second, $$1 \otimes [V_0] \mapsto \underline{V_0} \mapsto \sum_{V\in \mbox{simp}(G)} \mbox{Hom}^G(\underline{V},\underline{V_0}) \otimes [V] = \mbox{Hom}^G(\underline{V_0},\underline{V_0}) \otimes [V_0] = 1 \otimes [V_0]$$ since nonisomorphic simple representations don't have any nontrivial maps between them.

QED. Note that we're using (twice) the fact that $\mbox{Hom}^G (\underline{V}, \underline{V})=\underline{1}$ (i.e. the endomorphisms of a simple representation are only the scalars). In general, the endomorphisms of a simple representation form an associative division algebra over the base field; as this must be finite-dimensional (since it's contained in the endomorphism ring of the underlying vector space) and $\mathbb{C}$ is algebraically closed, it follows that the endomorphisms in the complex case are just the complex scalars.

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$Hom_G (V,V) = 1$ always holds for complex representations. This is a consequence of Schur's lemma (which says that every map of irreps is either 0 or an isomorphism). For take $T \in Hom_G(V,V)$. Since this is over $\mathbb C$, $T$ has an eigenvalue $c$. Now $T-cId \in Hom_G(V,V)$ and is not invertible so it must be 0 by Schur. Thus $T = c Id$. –  Eric O. Korman Sep 3 '12 at 13:50
@Eric: Thanks, I just discovered this myself a few minutes ago; I must have been editing the final paragraph while you were commenting. –  Aaron Mazel-Gee Sep 3 '12 at 13:55