Non trivial commutant implies proper projection

Question

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

Answer 1

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.

Answer 2

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?