2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 '18 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 '18 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 $A$. $\endgroup$ – Martin Argerami Feb 25 '18 at 16:36
  • $\begingroup$ What if $A$ and $B$ and $\phi$ are all unital. Would that change things? $\endgroup$ – ervx Feb 25 '18 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 '18 at 19:46
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.