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Recall that a C*-algebra is subhomogeneous if there is some $n$ such that the dimension of $H$ is at most $n$ for every irreducible representation $\pi \colon A \to \mathcal{B}(H)$ on some Hilbert space $H$.

For instance, abelian C*-algebras are subhomogeneous (just take $n=1$).

Question: is there a reference for the structure of a subhomogeneous C*-algebra?

By 'structure' I'm thinking about some explicit description, such as

$A \cong p_0 (C_0(X_0) \otimes M_{m_0}) p_0 \oplus \dots \oplus p_n (C_0(X_n) \otimes M_{m_n}) p_n$

for some locally compact Hausdorff spaces $X_0, \dots, X_n$, some $m_0, \dots, m_n \in \mathbb{N}$ and some projections $p_i \in C_0(X_i) \otimes M_{n_i}$. In my mind each $X_i$ would be the closed component of the primitive ideal of $A$ corresponding to representations of dimension $i$, for instance.

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    $\begingroup$ The structure you're describing is of homogeneous C*-algebras (which I think is the definition of such). The structure of a subhomogeneous C*-algebra is usually that it embeds in $M_n(C_0(X))$ for some $X$ and $n$. This is immediate by considering the inclusion into the bidual and using the structure theorem of type I von Neumann algebras (found for instance in Takesaki's first book). $\endgroup$
    – Jamie Gabe
    Commented Oct 4 at 19:31
  • $\begingroup$ @JamieGabe thanks! I think homogeneous algebras are as above with only 1 possible dimension for $H$ though. I agree that subhomogeneous do embed into $M_n(C_0(X))$, but I was hoping for a more explicit description. Thanks in any case! $\endgroup$
    – Diego Martinez
    Commented Oct 4 at 20:27
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    $\begingroup$ There might not be a more explicit description than the definition. For example prime dimension drop algebras are unital and have no non-trivial projections and therefore resist the sort of description that you are hoping for. $\endgroup$
    – Caleb Eckhardt
    Commented Oct 4 at 22:45
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    $\begingroup$ In support of @CalebEckhardt's comment: if $G$ is a virtually abelian group then $C^\ast(G)$ is subhomogeneous, but if $G$ is also torsion-free then $C^\ast(G)$ has no non-trivial projections. $\endgroup$
    – Yemon Choi
    Commented Oct 5 at 0:52
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    $\begingroup$ Also: consider the subalgebra of $M_2(C([0,1]))$ consisting of all continuous functions $f:[0,1]\to M_2({\mathbb C})$ such that $f(0)$ and $f(1)$ are diagonal. This algebra $A$ does have projections, but they do not decompose $A$ in the way that you are hoping for. $\endgroup$
    – Yemon Choi
    Commented Oct 5 at 1:32

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