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 $$\tag4 \phi(ax)=\phi(a)\phi(x),\qquad a,x\in A. $$