Some overlap with PU(ℋ). Needs to be disentangled.
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
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.
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 $K(\mathbb{Z}, 2)$.
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 $B U(1) \simeq B^2 \mathbb{Z} \simeq K(\mathbb{Z},2)$.
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 $X$
By prop. we have that the classifying space of $P U(\mathcal{H})$ itself is an Eilenberg-MacLane space
This is the classifying space for degree-3 integral cohomology (see Eilenberg-MacLane spectrum for more on this).
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.
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
Since the topological space underlying $U(\mathcal{H})$ is contractible, on the underlying topological spaces this is
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).
For more on this see also twisted K-theory.
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: $(g, F) \mapsto g F g^{-1}$.
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):