2
$\begingroup$

Given a *-representation $$ \Psi\colon C(K) \to B(H), $$ for $C(K)$ the continuous functions on a compact Hausdorff space and $B(H)$ the bounded operators on a Hilbert space $H$. We can always decompose $\Psi$ as a direct sum of cyclic representations $$ \Psi_i \colon C(K) \to B(H_i), $$ cyclic meaning that there exists $\xi_i \in H_i$ with $\overline{\big\{\Psi(f)\xi_i \colon f \in C(K)\big\}} = H_i$.

Apparently a *-representation is non-degenerate iff it can be decomposed as a direct sum of cyclic representations (e.g. Arveson, "Invitation to C*-algebras, Ex. 1.3F).

But wouldn't this mean that every *-representation of a unital commutative C*-algebra is degenerate (since every unital commutative C*-algebra is isometrically isomorphic to some $C(K)$)?

I am quite skeptical that this would be true since I have not come across the statement as in the title of my post. What am I missing?


EDIT: Turns out that I was missing the fact that all *-representations I considered were unital.

$\endgroup$

1 Answer 1

2
$\begingroup$

That's not what Arveson says. He says that a representation is non-degenerate if and only if $H$ can be decomposed as a direct sum of cyclic subspaces.

Any C$^*$-algebra has an arbitrary number of degenerate representations: given $\pi:A\to B(H)$, consider $\tilde\pi:A\to B(H\oplus H)$, where $$ \tilde\pi(a)(\xi\oplus\eta)=\pi(a)\xi\oplus 0. $$

But no one cares about them, because you can always change a representation $\pi:A\to B(H)$ into a non-degenerate one by changing $H$ by $\overline{\pi(A)H}$.

$\endgroup$
4
  • $\begingroup$ Thank you. But what about representations of the type $C(K) \to B(H)$ in particular? Are they always non-degenerate? Is it maybe because $C(K)$ has a unit? $\endgroup$
    – el_tenedor
    Commented Apr 15, 2017 at 18:05
  • $\begingroup$ Did you read my answer? Any C$^*$-algebra has degenerate representations. Take a point $k\in K$, and a fixed rank-one projection $P$ in $H$, and let $\pi:C(K)\to B(H)$ be given by $\pi(f)=f(k)P$. For any $P\ne I$, this will be degenerate. $\endgroup$ Commented Apr 15, 2017 at 18:18
  • $\begingroup$ Apparently, my claim is true if we restrict ourselves to unital representations, like this post suggests. I did not want to give you the feeling that I didn't read your answer carefully enough - I just assumed the problem was that $A$ didn't have a unit - the primary book that I work with only uses unital representations... Thanks again for your reply! $\endgroup$
    – el_tenedor
    Commented Apr 15, 2017 at 18:45
  • $\begingroup$ Of course. Any unital representation is non-degenerate. But if you are dealing with unital representations, even considering whether it is degenerate makes no sense. "Non-degenerate" is a notion that is relevant for non-unital representations. $\endgroup$ Commented Apr 15, 2017 at 18:48

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .