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Let $\mathcal{H}$ be a unitary universe for some group $G$. As $\mathcal{H}$ is a faithful representation the representation map is an injection $G \to \mathcal{U(H)}$, so there's a free $G$ action on $\mathcal{U(H)}$. By Kuiper's theorem $\mathcal{U(H)}$ is also non-equivariantly contractible. So therefore it's a model for $EG$.

I am unsure as to the choice of topology of $\mathcal{U(H)}$ to make the following work (all the maps and actions are continuous). We can have $\mathcal{U(H)}$ act on the Fredholm operators $\mathcal{F(H)}$ via conjugation. Thus the standard action of $G$ on $\mathcal{F(H)}$ factors though $\mathcal{U(H)}$.

I know that the standard action of $G$ on $\mathcal{F(H)}$ is not continuous, but that in the appendix of "Twisted K-theory" by Atiyah and Segal they produce a replacement set of objects $\mathcal{F} _{G-\text{cont.}} (\mathcal {H})$ and $\mathcal{U} _{G-\text{cont.}} (\mathcal {H})$ where the conjugation action by $G$ is continuous. As my action of $G$ on $\mathcal{U(H)}$ is basically composition (different from theirs) is it a continuous action?

EDIT: By the paper that @DavidRoberts linked in the comments, we have $\mathcal{U(H)}$ has a free action of $G$ and it is contractable. It however does not have an obvious $G$-CW structure. For it to be a model for $EG$, it needs to be at least homotopic to a $G$-CW complex. I'm not entirely sure about this point, there are notes out there that suggest that this isn't needed but that could be a little bit incorrect.

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  • $\begingroup$ In "A conjecture concerning positive Ricci curvature and the Witten genus" (in the proof of 5.1 on page 795) Stolz actually used this very construction for $G=U(1)$ to produce $PU(H)$ as a model for $K(Z,2)$. $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Apr 16, 2015 at 7:50
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    $\begingroup$ @მამუკაჯიბლაძე I think the question is more along the lines of using the space of Fredholms, which is not what the title is saying, unfortunately. It should be noted that the paper by Atiyah and Segal is a little bit mistaken about the strong topology on U(H); they say it is not a topological group, when in fact it is (see arxiv.org/abs/1309.5891). They are correct, though, that GL(H) is not a topological group, IIRC. $\endgroup$
    – David Roberts ♦
    Commented Apr 16, 2015 at 11:42
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    $\begingroup$ I've changed the title. The paper you linked seems to be very helpful at first glance. $\endgroup$
    – Sven Cattell
    Commented Apr 16, 2015 at 14:37
  • $\begingroup$ What are the assumptions on $G$, exactly? And what do you mean by "a/the unitary universe"? Is it the universal representation (so that $G$ is secretly locally compact)? $\endgroup$
    – Alain Valette
    Commented Apr 16, 2015 at 19:25
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    $\begingroup$ @DavidRoberts, it works out, under the strong topology $\mathcal{U(H)}$ is a contractible topological group with an injection of $G$. The action of $G$ on $\mathcal{F(U)}$ also factors through $\mathcal{U(H)}$. $\endgroup$
    – Sven Cattell
    Commented Apr 23, 2015 at 21:14

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