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projective unitary group

Definition

Definition

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

$$PU(\mathscr{H}):=U(\mathscr{H})/U(1).$$P U(\mathcal{H}) := U(\mathcal{H})/U(1) \,.

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

Properties

Homotopy type

Prop

If $\mathscr{H}$ 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 $U(\mathscr{H})$ 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 $U(\mathscr{H})\to PU(\mathscr{H})$ is a model for the circle group-universal principal bundle and in particular the topological space underlying $PU(\mathscr{H})$ is equivalent to the classifying space $BU(1)\simeq B^2\mathbb{Z}\simeq K(\mathbb{Z},2)$.

Prop

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

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

Proof

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

$$BPU(\mathscr{H})\simeq BBU(1)\simeq B^3\mathbb{Z}\simeq K(\mathbb{Z},3).$$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 $PU(\mathscr{H})$-principal bundle to a $U(\mathscr{H})$-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 $BPU(\mathscr{H})\to B^{2}U(1)$ that is part of the long fiber sequence

$$BU(1)\to BU(\mathscr{H})\to BPU(\mathscr{H})\to B^{2}U(1).$$\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 $U(\mathscr{H})$ is contractible, on the underlying topological spaces this is

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

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

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

For more on this see also twisted K-theory.

Action on Fredholm operators

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

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

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

Related concepts

• unitary group

  • special unitary group, projective unitary group

• orthogonal group

  special orthogonal group