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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.

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  • $\begingroup$ Something is off. An irreducible representation cannot be decomposed as a direct sum of representations (it would have non-trivial commutant). $\endgroup$
    – Martin Argerami
    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
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    $\begingroup$ 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 $A$. $\endgroup$
    – Martin Argerami
    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 $C[0,1]$. $\endgroup$
    – Martin Argerami
    Feb 25, 2018 at 19:46

2 Answers 2

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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.

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  • $\begingroup$ 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? $\endgroup$
    – Hermann Döppes
    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. $\endgroup$
    – Martin Argerami
    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. $\endgroup$
    – Hermann Döppes
    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
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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.

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    – Community Bot
    Dec 13, 2022 at 16:59

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