# Decomposing a representation of a C$^{*}$-algebra into a direct sum of irreducible ones

Suppose we have C$^{*}$-algebras $A$ and $B$ and a $*$-homomorphism $\phi\colon A\to B$. I came across a paper, where it was claimed that every irreducible representation of $B$ decomposes into a direct sum of a unique set of irreducible representations of $A$.

I am confused as to how we actually do this. If $\pi$ is a representation of $B$, then, clearly, $\pi\circ\phi$ is a representation of $A$. I am aware that every non-degenerate representation decomposes as a direct sum of cyclic representations, but this is not the same thing.

Any help would be much appreciated.

• Something is off. An irreducible representation cannot be decomposed as a direct sum of representations (it would have non-trivial commutant). – Martin Argerami Feb 25 '18 at 1:56
• Thank you for your answer. My understanding is that the representation $\pi\circ\phi$ may not be irreducible, but that it may be broken up into irreducible ones. – ervx Feb 25 '18 at 16:17
• You are right. But I'm failing to see what your question is, then. Uniqueness? Also, if you don't put conditions on $\phi$, nothing prevents it from being zero, say. In that case, there is no relationship (via $\phi$) between the irreps of $B$ and those of $A$. – Martin Argerami Feb 25 '18 at 16:36
• What if $A$ and $B$ and $\phi$ are all unital. Would that change things? – ervx Feb 25 '18 at 17:25
• It doesn't look like it to me. Consider $A=C[0,1]$, $B=C[0,1]\oplus M_2(\mathbb C)$, with $\phi(x)=(x,x(0)I)$, $\pi(x,M)=M$. You have $\pi\circ\phi(x)=x(0)I$. I fail to see in which sense one can say that $\pi$ is a direct sum of evaluations of $C[0,1]$. – Martin Argerami Feb 25 '18 at 19:46

As stated, I don't think the assertion makes sense. Consider for instance $A=C[0,1]$, $B=C[0,1]\oplus M_2(\mathbb C)$, and $$\phi(x)=(x,x(0)I),\ \ \ \ \pi(x,M)=M.$$ Then $\pi$ is irreducible, but there is no natural way to see it as a sum of representations of $A$ (which are all point evaluations).
What does hold is that $\pi\circ\phi$ is a representation of $A$ and, at least in the separable case, every representation is a direct sum of irreducibles. This is a nontrivial result, see for instance Corollary II.5.9 in Davidson's book. I think that the general version follows from work by Hadwin, but I don't have a reference.