# Extensions of sub-homogeneous $C^*$-algebras are subhomogeneous

Definition: A $C^*$-algebra $A$ is called sub-homogeneous if there exists $n\in\mathbb{N}$ such that every irreducible representation of $A$ has dimension at most $n$.

I could not find a proof or a counter example for the following claim:

Let $A,B$ be sub-homogeneous $C^*$-algebras and let $C$ be a $C^*$-algebra s.t. there is an exact sequence: $0\to A\to C\to B\to 0$. Does it follow that $C$ is sub-homogeneous?

Thanks

• It would help if you gave the definition of sub-homogeneous, as well as a motivation for the question. Dec 22 '17 at 0:19

Yes. Since $C^{**}\cong A^{**}\oplus B^{**}$, uniform bounds on the dimension of the irreducible representations of $A$ and $B$ impose a uniform bound on the dimension of the irreducible representations of $C$
• How can one see the irreducible representation from the double dual? By extending an irreducible representation to the enveloping von-Neumann algebra? Does it also mean that the dimension of irreducible representations of $C$ is bounded by the max of the other two? Thank you very much!
• @Nam For your second question, "yes." The full details about the relationship between $A^{**}$ and the representation theory of $A$ can be found in Takesaki Volume 1, Chapter III (and many other places). For your third question: Yes, it does mean that $C$ is max{n,m}-subhomogeneous where $A$ is $n$-subhomogeneous and $B$ is $m$-subhomogeneous. Dec 22 '17 at 15:53