# nLab projective unitary group

Contents

Some overlap with PU(ℋ). Needs to be disentangled.

### Context

#### Group Theory

group theory

Classical groups

Finite groups

Group schemes

Topological groups

Lie groups

Super-Lie groups

Higher groups

Cohomology and Extensions

Related concepts

# Contents

## Definition

###### Definition

For $\mathcal{H}$ a Hilbert space, the projective unitary group $P U(\mathcal{H})$ is the quotient of the unitary group $U(\mathcal{H})$ by its center $Z U(\mathcal{H}) \simeq U(1)$, the circle group

$P U(\mathcal{H}) \coloneqq U(\mathcal{H})/U(1) \,.$

This is naturally a topological group. For $\mathcal{H}$ of finite-dimension $n$ $P U(n) := P U(\mathcal{H})$ is also naturally a Lie group.

## Properties

### Homotopy type

###### Proposition

If $\mathcal{H}$ 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 $K(\mathbb{Z}, 2)$.

###### Proof

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

###### Proposition

For $\mathcal{H}$ an infinite-dimensional separable Hilbert space $P U(\mathcal{H})$-principal bundles over a topological space $X$ are classidied by third integral cohomology of $X$

$P U(\mathcal{H}) Bund(X) \simeq H^3(X, \mathbb{Z}) \,.$

###### Proof

By prop. we have that the classifying space of $P U(\mathcal{H})$ itself is an Eilenberg-MacLane space

$B P U(\mathcal{H}) \simeq B B U(1) \simeq B^3 \mathbb{Z} \simeq K(\mathbb{Z}, 3) \,.$

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

###### Proposition

Every circle 2-bundle/bundle gerbe on $X$ is equivalent to the lifting gerbe of some $P U(\mathcal{H})$-principal bundle to a $U(\mathcal{H})$-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 $\mathbf{B} P U(\mathcal{H}) \to \mathbf{B}^2 U(1)$ that is part of the long fiber sequence

$\mathbf{B} U(1) \to \mathbf{B} U(\mathcal{H}) \to B \mathbf{P} U(\mathcal{H}) \to \mathbf{B}^2 U(1) \,.$

Since the topological space underlying $U(\mathcal{H})$ is contractible, on the underlying topological spaces this is

$K(\mathbb{Z},2) \to * \to K(\mathbb{Z},3) \stackrel{\simeq}{\to} K(\mathbb{Z}, 3) \,.$

This means that the morphism that sends $P U(\mathcal{H})$-bundles to the twist that they induce is an isomorphism.

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

### Action on Fredholm operators

Let $\mathcal{H}$ be an infinite-dimensional separable Hilbert space.

Since by the above $PU(\mathcal{H}) \simeq B U(1)$ 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 $PU(\mathcal{H}) \times Fred \to Fred$ of $PU(\mathcal{H})$ on the space of Fredholm operators (on connected components).

This is given by letting a projective unitary act by conjugation on a Fredholm operator: $(g, F) \mapsto g F g^{-1}$.

## References

On the cohomology of the classifying spaces $B PU(n)$:

On the spaces of group homomorphisms into $PU(\mathcal{H})$ (with an eye towards the universal equivariant $PU(\mathcal{H})$-bundle):

Last revised on September 19, 2021 at 11:15:13. See the history of this page for a list of all contributions to it.