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.

  • $\begingroup$ Something is off. An irreducible representation cannot be decomposed as a direct sum of representations (it would have non-trivial commutant). $\endgroup$ Feb 25, 2018 at 1:56
  • $\begingroup$ 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. $\endgroup$
    – ervx
    Feb 25, 2018 at 16:17
  • 1
    $\begingroup$ 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 $A$. $\endgroup$ Feb 25, 2018 at 16:36
  • $\begingroup$ What if $A$ and $B$ and $\phi$ are all unital. Would that change things? $\endgroup$
    – ervx
    Feb 25, 2018 at 17:25
  • $\begingroup$ 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 $C[0,1]$. $\endgroup$ Feb 25, 2018 at 19:46

2 Answers 2


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.

  • $\begingroup$ I do not see how “we can not understand this as a sum of representations” and “every representation is a direct sum of irreducibles” go together. They seem to be immediate contradictions and I do not see where any conditions differ in a relevant way, e.g. $A$ is separable by Weierstrass, right? $\endgroup$ Jun 10, 2021 at 18:35
  • $\begingroup$ @Hermann: the representation $\pi\circ\phi$ is of the form $\pi_0\oplus\pi_0$, where $\pi_0:A\to\mathbb C$ is the representation $\pi_0(x)=x(0)$. The other part of the assertion (the first one) says "there is no natural way to see it as a sum of representations of $A$" (you missed the "of $A$"). The representation $\pi$ maps $(x,M)$ to $M$, and there is absolutely no reference to the structure of $A$: you can choose any $A$ whatsoever, and $\pi$ does not change. $\endgroup$ Jun 10, 2021 at 18:44
  • $\begingroup$ That is obviously true because it would be plain nonsensical to write a representation of $B$ as a sum of representations of $A$. However, I understood this comment math.stackexchange.com/questions/2665158/… and the general context to mean that you had agreed to consider the question of decomposing $\pi\circ\phi$ into a sum of irreducibles. $\endgroup$ Jun 10, 2021 at 18:54
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    $\begingroup$ Not every representation is the direct sum of irreducible representations (if that was true, every normal operator would be diagonal). The cited Corollary is about approximate unitary equivalence. $\endgroup$
    – Orr Shalit
    Dec 13, 2022 at 13:48

The answer to your question is no.

Consider the case where $A = B$ and $\phi$ is the identity. In this case one can see that the answer to your question is no, for the reason that not every *-representation is the direct sum of irreducible representations. For an example, consider the identity representation of the $C^*$-algebra generated by the multiplication operator $M_x$ on $L^2[0,1]$; this representation has no irreducible summand, and so in particular it is not a direct sum of irreducible representations.

  • $\begingroup$ As it’s currently written, your answer is unclear. Please edit to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. $\endgroup$
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    Dec 13, 2022 at 16:59

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