group theory

# Contents

## Definition

###### Definition

For $ℋ$ a Hilbert space, the projective unitary group $PU\left(ℋ\right)$ is the quotient of the unitary group $U\left(ℋ\right)$ by its center $ZU\left(ℋ\right)\simeq U\left(1\right)$, the circle group

$PU\left(ℋ\right):=U\left(ℋ\right)/U\left(1\right)\phantom{\rule{thinmathspace}{0ex}}.$P U(\mathcal{H}) := U(\mathcal{H})/U(1) \,.

This is naturally a topological group. For $ℋ$ of finite-dimension $n$ $PU\left(n\right):=PU\left(ℋ\right)$ is also naturally a Lie group.

## Properties

### Homotopy type

###### Prop

If $ℋ$ is an infinite-dimensional separable Hilbert space the underlying topological space of its projective unitary group has the homotopy type of an Eilenberg-MacLane space $K\left(ℤ,2\right)$.

###### Proof

The unitary group $U\left(ℋ\right)$ in this case is contractible (by Kuiper's theorem) and the circle group $U\left(1\right)$ acts free and faithfully on it. Therefore the quotient map $U\left(ℋ\right)\to PU\left(ℋ\right)$ is a model for the circle group-universal principal bundle and in particular the topological space underlying $PU\left(ℋ\right)$ is equivalent to the classifying space $BU\left(1\right)\simeq {B}^{2}ℤ\simeq K\left(ℤ,2\right)$.

###### Prop

For $ℋ$ an infinite-dimensional separable Hilbert space $PU\left(ℋ\right)$-principal bundles over a topological space $X$ are classidied by third integral cohomology of $X$

$PU\left(ℋ\right)\mathrm{Bund}\left(X\right)\simeq {H}^{3}\left(X,ℤ\right)\phantom{\rule{thinmathspace}{0ex}}.$P U(\mathcal{H}) Bund(X) \simeq H^3(X, \mathbb{Z}) \,.
###### Proof

By prop. 1 we have that the classifying space of $PU\left(ℋ\right)$ itself is an Eilenberg-MacLane space

$BPU\left(ℋ\right)\simeq BBU\left(1\right)\simeq {B}^{3}ℤ\simeq K\left(ℤ,3\right)\phantom{\rule{thinmathspace}{0ex}}.$B P U(\mathcal{H}) \simeq B B U(1) \simeq B^3 \mathbb{Z} \simeq K(\mathbb{Z}, 3) \,.

This is the classifying space for degree-3 integral cohomology (see Eilenberg-MacLane spectrum for more on this).

###### Prop

Every circle 2-bundle/bundle gerbe on $X$ is equivalent to the lifting gerbe of some $PU\left(ℋ\right)$-principal bundle to a $U\left(ℋ\right)$-bundle, and the equivalence classes of these structures correspond uniquely.

###### Proof

The twisted bundles of a given bundle gerbe are given by the twisted cohomology relative to the morphism $BPU\left(ℋ\right)\to {B}^{2}U\left(1\right)$ that is part of the long fiber sequence

$BU\left(1\right)\to BU\left(ℋ\right)\to BPU\left(ℋ\right)\to {B}^{2}U\left(1\right)\phantom{\rule{thinmathspace}{0ex}}.$\mathbf{B} U(1) \to \mathbf{B} U(\mathcal{H}) \to B \mathbf{P} U(\mathcal{H}) \to \mathbf{B}^2 U(1) \,.

Since the topological space underlying $U\left(ℋ\right)$ is contractible, on the underlying topological spaces this is

$K\left(ℤ,2\right)\to *\to K\left(ℤ,3\right)\stackrel{\simeq }{\to }K\left(ℤ,3\right)\phantom{\rule{thinmathspace}{0ex}}.$K(\mathbb{Z},2) \to * \to K(\mathbb{Z},3) \stackrel{\simeq}{\to} K(\mathbb{Z}, 3) \,.

This means that the morphism that sends $PU\left(ℋ\right)$-bundles to the twist that they induce is an isomorphism.

(Somebody should force me to say this in more detail).

Let $ℋ$ be an infinite-dimensional separable Hilbert space.
Since by the above $\mathrm{PU}\left(ℋ\right)\simeq BU\left(1\right)$ and since there is a canonical action of line bundles on complex vector bundles, hence on the topological K-theory of a manifold $X$, there must also be a natural action of $\mathrm{PU}\left(ℋ\right)×\mathrm{Fred}\to \mathrm{Fred}$ of $\mathrm{PU}\left(ℋ\right)$ on the space of Fredholm operators (on connected components).
This is given by letting a projective unitary act by conjugation on a Fredholm operator: $\left(g,F\right)↦gF{g}^{-1}$.