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Oct 5 at 1:32 comment added Yemon Choi 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.
Oct 5 at 1:28 history edited Yemon Choi CC BY-SA 4.0
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Oct 5 at 0:52 comment added Yemon Choi 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.
Oct 4 at 22:45 comment added Caleb Eckhardt 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.
Oct 4 at 20:27 comment added Diego Martinez @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!
Oct 4 at 19:31 comment added Jamie Gabe 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).
Oct 4 at 18:16 history asked Diego Martinez CC BY-SA 4.0