# Non trivial commutant implies proper projection

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 $\overline{T(\mathcal{H})}\lneq \mathcal{H}$.

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.

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

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?