# Equivariant Fredholm operators classify equivariant K-theory

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.

• The mathscinet link is only accessible by subscription. Could you please add text citation for the paper you're linking to? – Dan Ramras Oct 3 '15 at 21:26
• It is a paper about forty years ago. I didn't find the electronic version on the web. – Bo Liu Oct 4 '15 at 8:36
• The online version is not that difficult to find, I added a link. – Dmitri Pavlov Oct 4 '15 at 12:09

The isomorphism for $K_G^1$ in the question is right.