# Continous Trace C*-Algebras

## Definitions

Recall that for $A$ a $C^\ast$-algebra with Hausdorff spectrum $X=\operatorname{Spec}(A)$, each primitive quotient $A(x)$ has up to equivalence a unique irreducible representation $\pi_x$. The dimension of a subspace is preserved by a unitary operator, hence whenever $p\in A$ s.t. $p(x) = [\pi_x](p)$ is a projection, the rank of $p(x)$ is well-defined, and is the dimension of the range of $\pi_x(p(x))$.

Definition(Continous-trace $C^\ast$-algebra, version 1) . A continuous-trace $C^\ast$-algebra is a $C^\ast$-algebra $A$ with Hausdorff spectrum $X=\operatorname{Spec}(A)$ s.t. for each $x\in X$ there are neighbourhood $U$ of $x$ and $a\in A$ s.t. for all $s\in U$, $a(s)$ is a rank-one projection.

The definition seems unrelated to the notion of trace. There is another version of the definition. For any Hilbert space, there is a trace defined on positive operators $T$, $$\operatorname{tr}(T) = \sum_n (Te_n|e_n)$$ where $\{e_n\}$ is an orthonormal basis. The trace is independent of the basis chosen, so that $\operatorname{tr}(UTU^\ast)=\operatorname{T}$ and extens to a functional on the ideal of trace-class operators $$I_1(\mathcal{H}) \colon =\{T\in B(\mathcal{H})|\operatorname{tr}(|T|)<\infty\}.$$

A positive element $a\in A$ is a continuous-trace element if $x\mapsto \operatorname{tr}(a(x))$ is a finite continuous function on $\operatorname{Spec}(A)$. These elements form the postivie part of an ideal $\mathfrak{m}(A)$: $$\mathfrak{m}(A) = \{a\in A|a(x)\text{ is trace-class }, x\mapsto \mathrm{tr}(a(x))\text{ is continous on }X, \forall x\in X = \mathrm{Spec}(A)\}$$

Definition(Continous-trace $C^\ast$-algebra, version 2) . A continuous-trace $C^\ast$-algebra is a $C^\ast$-algebra $A$ s.t. $\mathfrak{m}(A)$ is dense in $A$.

The two definitions are equivalent.

Theorem(Fell's condition) . A $C^\ast$-algebra is of continous-trace (satisfies the condition for defintion version 2) iff it has a Hausdorff spectrum and there are local rank-one projections for every $x\in \operatorname{Spec}(A)$.

There is still one more definition

Definition(Continous-trace $C^\ast$-algebra, version 3) . A continuous-trace $C^\ast$-algebra is a $C^\ast$-algebra $A$ with Hausdorff spectrum $X=\operatorname{Spec}(A)$ s.t. $A$ is locally Morita equivalent to $C\_0(X)$. This means for each $x\in X$ there is a compact neighbourhood $U$ s.t. $A^U$ is Morita equivalent to $C(U)$.

A continuous-trace algebra $A$ has a charcteristic class $\delta(A)\in H^3(X,\mathbb{Z})$, stable under tensoring with $\mathcal{K}$. The classification is due to Dixmier and Douady. See Dixmier-Douady theory for details.

Here is an explanation. Suppose that $A$ with spectrum $X$ is isomorphic to the algebra $\Gamma_0(X,E)$ of sections vanishing at $\infty$ of a locally trivial bundle $E\to X$ with typical fiber $\mathcal{K}$. The structure group of $E$ is $\operatorname{Aut}(\mathcal{K})\cong PU(\mathcal{H})$ where $\operatorname{dim}(\mathcal{H})=\aleph_0$. $U(\mathcal{H})$ is contractible for infinite dimensional $\mathcal{H}$, and $U(1)$ acts freely and transitively on it, hence $PU$ has the homotopy type $BU(1)$ which is a $K(\mathbb{Z},2)$-space. Principal PU-bundles over $X$ are classified by $$[X,BPU] = [X,K(\mathbb{Z},3)] = H^3(X,\mathbb{Z}).$$ So every stable continuous-trace algebra defines a class in $H^3(X,\mathbb{Z})$ and every class in $H^3(X,\mathbb{Z})$ comes from a stable continous-trace algebra.

## Automorphisms

### The group $\operatorname{Aut}_{C_0(X)} A$

Theorem. For $A$ a continuous-trace algebra with spectrum $X$, there is a homomorphism $\xi:\operatorname{Aut}_{C_0(X)}A\to H^2(X,\mathbb{Z})$ s.t. $$0\to \operatorname{Inn}A \to \operatorname{Aut}_{C_0(X)}A\xrightarrow{\xi}H^2(X;\mathbb{Z})$$ is an exact sequence of groups. $\xi$ is surjective when $A$ is stable.

In particular, the outer automorphisms $\operatorname{Ait}_{C_0(X)} A/\operatorname{Inn}A$ form an abelian group.

### The full automorphism group

Theorem. For $A$ a continuous-trace algebra with spectrum $X$, there is an exact sequence $$0\to\operatorname{Aut}_{C_0(X)}A\to \operatorname{Aut}A\xrightarrow{\rho}\operatorname{Homeo}_{\delta(A)}X.$$ If $A$ is $\sigma$-unital and stable, $\rho$ is surjective.

## Relation with bundle gerbes

Recall that given a central extension of topological groups $$1\to U(1) \to \tilde{G} \xrightarrow{q} G \to 1$$ and a principal $G$-bundle $\pi: Y\to X$, there is a canonical bundle gerbe, called the lifting gerbe: Pull the $U(1)$-bundle $q:\tilde{G}\to G$ back by the map $g:Y^{[2]}\to G$ defined by $y_1 g(y_1,y_2) = y_2$ (note that $(y_1,y_2) \in Y^{[2]}$ means $\pi(y_1)=\pi(y_2)$, i.e. $y_1,y_2$ are indeed connected by a $h\in G$): This means $$P = \{((y_1,y_2),\tilde{g})\subset Y^{[2]}\times \tilde{G}| g(y_1,y_2)=q(\tilde{g})\},$$ or equivalently $$P = \{((y_1,y_2),\tilde{g})\subset Y^{[2]}\times \tilde{G} | y_1 q(\tilde{g}) = y_2\}.$$ This is a principal $U(1)$-bundle over $Y^{[2]}$. The gerbe product is induce by the group structure of $\tilde{G}$. Namely, $$\operatorname{pr}^\ast_{ij}P = \{(((y_i,y_j),y_k),\tilde{g})\subset Y^{[3]}\times \tilde{G}| y_i q(\tilde{g})=y_j \}$$ then the product is given by $$(((y_i,y_j),y_k),\tilde{g})\times (((y_j,y_k),y_i),\tilde{h}) \mapsto (((y_i,y_k),y_j),\tilde{g}\tilde{h}).$$ This is an isomorphism.

A lifting of the bundle $Y\to X$ to a principal $\tilde{G}$-bundle gives rise to a trivialization of this gerbe.

Associated to the central extension of groups $$1\to U(1)\to U(\mathcal{H})\to PU(\mathcal{H}) \to 1,$$ since $U(\mathcal{H})$ is contractible, every principal $U(\mathcal{H})$-bundle over $X$ is trivial. Let $Y\to X$ be a principal $PU(\mathcal{H})$-bundle over $X$. If $Y\to X$ lifts to a principal $U(\mathcal{H})$-bundle over $X$, then $Y$ is trivial, i.e. $\delta=0$ exactly when the lifting gerbe is trivial.

Moreover, the Dixmier-Douady class of the gerbe is the same as for the continous-trace algebra $CT(X,\delta)$: Every $U(1)$-bundle gerbe is the lifting gerbe of some $PU(\mathcal{H})$-principal bundle through this extension.