TITLE: Equivariant Fredholm operators classify equivariant K-theory
QUESTION [15 upvotes]: Let $\mathcal{F}$ be the space of Fredholm operators on a separable Hilbert space $H$ with the topology induced by the operator norm. 
If $X$ is compact,
Atiyah-Jänich proved that
$$[X,\mathcal{F}]\simeq K^0(X). $$
Here $[\cdot,\cdot]$ denotes the set of homotopy classes.
For the equivariant version, for a compact Lie group $G$, let $\mathcal{F}(G)$ be the space of Fredholm operators on $L^2(G,H)$. The group $G$ acts on $\mathcal{F}(G)$ in a natural way. For compact $G$-space $X$, one has 
$$[X,\mathcal{F}(G)]_G\simeq K_G^0(X). $$
Here $[\cdot,\cdot]_G$ denotes the set of $G$-homotopy classes of $G$-maps.
(Matumoto, T.,
Equivariant K-theory and Fredholm operators.
J. Fac. Sci. Univ. Tokyo Sect. I A Math. 18 1971 109–125.) 
On the other hand, let $\hat{\mathcal{F}}_*$ be the space of self-adjoint Fredholm operators on $H$ such that its elements have infinite positive as well as infinite negative spectrum. Atiyah proved that
$$[X,\hat{\mathcal{F}}_*]\simeq K^1(X). $$
My question:
is there a equvariant isomorphism for $K_G^1$ as in $K_G^0$?
That is,
$$[X,\hat{\mathcal{F}}(G)_*]_G\simeq K_G^1(X)? $$
As the proof of Atiyah, I think it turns to prove $\hat{\mathcal{F}}(G)_*\rightarrow \Omega \mathcal{F}(G)$ is a $G$-homotopy equivalence.

REPLY [5 votes]: The isomorphism for $K_G^1$ in the question is right.
Freed, Daniel S.; Hopkins, Michael J.; Teleman, Constantin,  Loop groups and twisted K-theory I. J. Topol. 4 (2011), no. 4, 737–798. 
Section 3.5.4 and Appendix A.5