TITLE: How can I see that when $X$ is a trivial $G$-space, $K(X) \otimes R(G) \rightarrow K_G(X)$ is an isomorphism?
QUESTION [7 upvotes]: 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.

REPLY [3 votes]: 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.