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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.

ervx
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  • 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 01:56
  • 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
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    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
  • Thank you very much for the explanation. I am thinking what the author intended is that $\pi\circ\phi$ decomposes into a direct sum of a unique set of irreducible representations of $A$. Is this possible? – ervx Feb 25 '18 at 19:55
  • Yes, for sure. The map $\pi\circ\phi$ is a representation of $A$, and so you just need that classic result that any representation decomposes as a direct sum of irreps. – Martin Argerami Feb 25 '18 at 20:04
  • Thank you very much! I couldn't find this result in Murphy's book. Do you know of another textbook that has a proof of this? Also, if you post your answer, I am happy to accept it. – ervx Feb 25 '18 at 20:06
  • I wrote an answer, including the reference. – Martin Argerami Feb 25 '18 at 21:02

2 Answers2

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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.

Martin Argerami
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  • 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? – Hermann Döppes Jun 10 '21 at 18:35
  • @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. – Martin Argerami Jun 10 '21 at 18:44
  • 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 https://math.stackexchange.com/questions/2665158/decomposing-a-representation-of-a-c-algebra-into-a-direct-sum-of-irreducib/2666554?noredirect=1#comment5507027_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. – Hermann Döppes Jun 10 '21 at 18:54
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    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. – Orr Shalit Dec 13 '22 at 13:48
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Not every representation can be decomposed into a direct sum of irreducible representations. In particular, in the setting of the question, not every irreducible representation of B decomposes as the direct sum of irreducible representations of A.

Consider the case where $H = L^2[0,1]$, $B = B(H)$, $A = C([0,1])$, and $\phi : A \to B(H)$ is given by $\phi(f) = M_f$ --- the multiplication operator $M_f : h \mapsto fh$ on $L^2$. Let $\pi$ be the identity representation of $B$ on $H$. In this case one can see that the $\pi$ is an irreducible $*$-representation of $B$. However, $\pi \circ \phi = \phi$ is not a direct sum of irreducible representations. Indeed, every irreducible representation of a commutative C*-algebra is one dimensional, but $\phi$ cannot have one dimensional irreducible subrepresentations, because this would imply the existence of an eigenvector for the operator $M_x$.

Orr Shalit
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  • 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. – Community Dec 13 '22 at 16:59