Denote by $U(H)$ and $PU(H)$ the unitary and projective unitary groups on an infinite-dimensional Hilbert space $H$. Recall that $U(H)$ is contractible by Kuiper's theorem and that $PU(H)$ is a $K(Z,2)$.

In Section 6 of https://arxiv.org/pdf/hep-th/9702147v2.pdf it was shown that there is a principal $PU(H)$-bundle \begin{equation} PU(H) \rightarrow U(T) \rightarrow U(T)/PU(H) \end{equation} where $T$ is the Hilbert space of Hilbert-Schmidt operators on $H$. This comes from the closed embedding of Banach Lie groups \begin{eqnarray} i: PU(H) &\rightarrow& U(T) \\ [a] &\mapsto& Ad(a) \end{eqnarray} where $Ad(a)$ sends $t\in T$ to $ata^*$. It follows that $U(T)/PU(H)$ is a $K(Z,3)$.

I'm interested the possible generalization to general linear groups. We can consider the map \begin{eqnarray} i: PGL(H) &\rightarrow& GL(T) \\ [a] &\mapsto& Ad(a) \end{eqnarray} where $Ad(a)$ sends $t\in T$ to $ata^{-1}$. I tried to imitate the proof in the paper and had to prove the following:

1) $i$ is well-defined

2) $i$ is injective

3) $i$ is continuous

4) $i(PGL(H))$ is closed

5) $i$ is a homeomorphism onto its image

I was able to prove 1)-3) so far.

My questions:

Are statements 4) and 5) above true?

Regardless, do we anyway have a fiber bundle $PGL(H) \rightarrow GL(T) \rightarrow GL(T)/PGL(H)$?

In any case, is $GL(T)/PGL(H)$ a model for $K(Z,3)$?

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    $\begingroup$ ($U(H)$,norm topology) is contractible by Kuiper's theorem. But the norm topology sucks. ($U(H)$,strong operator topology) is a much better group to work with. See the review paper arxiv.org/pdf/1309.5891.pdf for more info. ($U(H)$,strong operator topology) is also contractible, and the proof is much, much simpler than Kuiper's theorem. Here's the proof: write $H$ as $L^2([0,1])$, and use the nullhomotopy $u\mapsto \{u_t\}$, where $u_t$ acts like $u$ on $L^2([t,1])$, and acts like $id$ on $L^2([0,t])$. $\endgroup$ – André Henriques Apr 19 '18 at 21:20
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    $\begingroup$ Just in case anybody stumbles across this question while searching for models of $K(\mathbb Z, 3)$ more generally (which seems likely given the title), let me link to this old gem. $\endgroup$ – Tim Campion Apr 21 '18 at 21:29

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