I'm trying to come up with an 'elementary proof' that if $A$ is a commutative $C^*$ algebra, then the pure states of $A$ are exactly its characters.
I have already come up with a proof which uses properties of the GNS representation in order to show this. One can also identify the algebra with $C(X)$ and it's probably easier to prove this on this specific algebra.
But I was wondering if the is a more 'algebraic' proof which uses relatively elementary results and does not require much more than the basic definitions and properties of pure states.
I can show by simple arguments that all characters are pure states. But I'm having difficulty with the other direction – showing that in the unital and commutative case, pure states must be multiplicative.
Does anyone have an idea?
Thanks in advance.
Best Answer
There is no need for machinery. Properly, one needs a tiny bit of ad-hoc functional calculus to use square roots; concretely, if $0\leq x\leq 1\ $ then one can define $$ x^{1/2}=1-\sum_{n=1}^\infty \frac{2(2n-2)!}{4^nn!(n-1)!}\,(1-x)^n $$ and properties of series show that $x^{1/2}≥0$ and $(x^{1/2})^2=x$.
Suppose that $\phi$ is pure. Fix $a\in A_+$ with $a\leq1$. If $\varphi(a)\ne0$, $\varphi(1-a)\ne0$, define states $$ \phi_a(x)=\frac1{\phi(a)}\,\phi(ax),\qquad \phi_{1-a}(x)=\frac1{\phi(1-a)}\,\phi((1-a)x). $$ The fact that $A$ is commutative guarantees that $\phi_a$ is a state, because if $x\geq0$ then $$ \phi_a(x)=\frac1{\phi(a)}\,\phi(a^{1/2}xa^{1/2})\geq0. $$ Since $$ \phi(x)=\phi(a)\,\phi_a(x)+\phi(1-a)\,\phi_{1-a}(x), $$ the fact that $\phi$ is pure gives us $\phi_a=\phi$. This is $$\tag1 \phi(ax)=\phi(a)\phi(x). $$ If $\phi(a)=0$, then for $x\geq0$ $$ \phi(ax)=\phi(a^{1/2}xa^{1/2})\leq \|x\|\,\phi(a)=0. $$ So again $$\tag2 \phi(ax)=\phi(a)\phi(x),\qquad x\geq0. $$ If $\phi(1-a)=0$ instead, the same argument applied to $1-a$ gives us $\phi((1-a)x)=\phi(1-a)\phi(x)$, which is again $(2)$.
As the positive elements span the algebra, $(2)$ becomes $$\tag3 \phi(ax)=\phi(a)\phi(x),\qquad x\in A. $$ Now for arbitrary $a\geq0$ we apply the above to $a/\|a\|$, and then use again that positive elements span $A$ to get $$\tag4 \phi(ax)=\phi(a)\phi(x),\qquad a,x\in A. $$