Pure states on commutative C* algebra are exactly the characters - elementary proof

Question

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.

Answer 1

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

Answer 2

Here is an argument that uses the Gelfand theorem, namely that abelian unital $C^*$-algebras are of the form $C(X)$ where $X$ is compact and Hausdorff and the Riesz representation theorem (and of course the definition of a pure state, which for me is going to be "an extreme point of the state space" (at least for the unital case). By the Gelfand theorem it suffices to prove that pure states on $C(X)$ are multiplicative, where $X$ is compact and Hausdorff.

Recall that, from the Riesz representation theorem we have an affine bijection from the set of regular Borel probability measures on $X$ namely $\text{Prob}(X)$, to the state space of $C(X)$, given by $$\text{Prob}(X)\ni\mu\mapsto\int_X\cdot d\mu\in S(C(X))$$

Note that this affine bijection sends extreme points to extreme points; so, in order to identify a pure state on $S(C(X))$, which are by definition the extreme points of $S(C(X))$, we need to identify precisely the extreme points of $\text{Prob}(X)$.

Let $\mu$ be an extreme point of $\text{Prob}(X)$; this yields that, for any Borel set $E\subset X$, $\mu(E)=0$ or $\mu(E)=1$. Indeed, assume that there exists a Borel set $E\subset X$ with $0<\mu(E)<1$. Then setting $\mu_1(F)=\frac{1}{\mu(E)}\mu(F\cap E)$ and $\mu_2(F)=\frac{1}{1-\mu(E)}\mu(F\cap(X\setminus E))$ we get $\mu=\mu(E)\cdot\mu_1+(1-\mu(E))\cdot\mu_2$. But then, since $\mu$ is extreme we get $\mu_1=\mu_2=\mu$, which is a contradiction, since $\mu_1(E)=1$ and $\mu_2(E)=0$.

Now consider the set $$S:=\{x\in X:\mu(U)=1\text{ for every open }U\subset X\text{ with }x\in U\}$$ Assume that $S$ is empty to reach a contradiction. If $S=\emptyset$, then for each $x\in X$ we can find an open set $U_x$ with $x\in U_x$ and $\mu(U_x)=0$. Consider the open cover $\{U_x\}_{x\in X}$ of your space and by compactness reduce it to a finite one, say $\{U_{x_i}\}_{i=1}^n$. Then $\mu(X)=\mu(\bigcup_{i=1}^nU_{x_i})\le\sum_{i=1}^n\mu(U_{x_i})=0$, a contradiction. So $S$ is non-empty.

This is a closed set: if $(x_\lambda)\subset S$ is a net with $x_\lambda\to x$, then let $U$ be a neighborhood of $x$. Find $\lambda$ large so that $x_\lambda\in x$. Then $U$ is also a neighborhood of $x_\lambda$ which is an element of $S$, so $\mu(U)=1$. This shows that $\mu(U)=1$ for any neighborhood $U$ of $x$, i.e. $x\in S$.

Now since $S$ is closed it is Borel and we can compute $\mu(S)$: by regularity of $\mu$, we have $$\mu(S)=\inf\{U\subset X: U\text{ is open and }S\subset U\}$$ But if $U\subset X$ is open and $S\subset X$, then $x_0\in U$ for any point $x_0\in S$, and thus $\mu(U)=1$. Thus $\mu(S)=1$. Now assume that $x_1,x_2\in S$ are two distinct points. Since the space is Hausdorff, we can find disjoint open sets $V,W$ with $x_1\in V$ and $x_2\in W$. But then $\mu(V\cup W)=\mu(V)+\mu(W)=1+1=2$, a contradiction to the fact that $\mu(X)=1$. Thus $S$ is a singleton.

Therefore $\mu$ is precisely the dirac mass at some point $x_0\in X$. But then the corresponding functional, let's call it $\rho$, which is integration with respect to this measure gives us $$\rho(fg)=\int_Xfgd\mu=fg(x_0)=f(x_0)g(x_0)=\int_Xfd\mu\cdot\int_Xgd\mu=\rho(f)\rho(g)$$ for all $f,g\in C(X)$, i.e. $\rho$ is multiplicative, as we wanted.