3
$\begingroup$

I've been reading Kenneth Davidson's book on C*-Algebras and found the following assertion in the section of representation of C*-algebras: "Given a C*-algebra $\mathfrak{A}$ and a representation $\pi$ on a Hilbert space $\mathcal{H}$, if $\pi(\mathfrak{A})'$ (the commutator of $\pi(\mathfrak{A}))$ isn't equal to $\mathbb{C}I$ then there exists a proper projection in $\pi(\mathfrak{A})'$." He states that it's due to a theorem relating convex hull of set of projections but I didn't understood well that method. In other book I saw that it could be obtained using the polar decomposition and constructing the projection that way, I just don't know why it should be proper. Basically I'd like to know why there's an operator $T\in \pi(\mathfrak{A})$ such that $\overline{T(\mathcal{H})}\lneq \mathcal{H}$.

$\endgroup$
3
$\begingroup$

The C$^*$-algebra $\pi(\mathfrak A)'$ is a von Neumann algebra (i.e., it is weak-operator closed). This allows you to perform Borel functional calculus on normal operators. So, take any nontrivial positive operator $T\in\pi(\mathfrak A)'$, and write its spectrum $\sigma(T)=A\cup B$, with $A,B$ disjoint Borel sets. Then $P=1_A(T),Q=1_B(T)$ are nonzero pairwise orthogonal projections in $\pi(\mathfrak A)'$, and so they are nontrivial.

$\endgroup$
  • $\begingroup$ I haven't studied much about Borel functional calculus, can I just take something like a partition of unity on the spectrum of said positive operator? $\endgroup$ – Julio Cáceres Jul 11 '17 at 21:32
  • $\begingroup$ I don't see why. A partition of the unity (in the Real Analysis sense) will not give you projections. $\endgroup$ – Martin Argerami Jul 12 '17 at 15:10
  • $\begingroup$ oh, that's true $\endgroup$ – Julio Cáceres Jul 12 '17 at 20:47
  • 1
    $\begingroup$ The point is that $\pi(\mathfrak A)'$ is not trivial. That means precisely that there exists positive $T$ with non-singleton spectrum. $\endgroup$ – Martin Argerami Jul 13 '17 at 3:34
  • 1
    $\begingroup$ @C.Ding: No. First, $T$ is already a nontrivial projection, which was the point. Second, $Q=I-T$. $\endgroup$ – Martin Argerami Jul 14 '17 at 13:06
1
$\begingroup$

The method of "convex hull" is an easy way:

Since $\pi(\mathfrak A)'$ is a von Neumann algebra, it is the closed linear span of its projections. Therefore if all the projections in $\pi(\mathfrak A)'$ are trival, then $\pi(\mathfrak A)'=\mathbb{C}1.$

Where are you confused?

$\endgroup$

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.