Let <mi class="MJX-tex-caligraphic" mathvariant="script">A be a C ∗<!-

Question

Let

A

be a

C^{*}

-algebra.
(i) Let

φ

be a state on a

u \in A

-algebra

A

. Suppose that

| φ (u) | = 1

for all unitary elements u∈A. Show that φ is a pure state. [Hint:

span U (A) = A

]
(ii) Let

φ

be a multiplicative functional on a

C^{*}

-algebra

A

. Show that

φ

is a pure state on

A

.
(iii) Show that the pure and multiplicative states coincide for commutative

A

.
I managed to work out the first two problems but I have no idea about the last one. How to see from being an extreme element in the state space of a commutative

A

that the extreme element is multiplicative?

Answer 1

Suppose that

φ

is an extreme point in the state space.
Let

a \geq 0

with

0 < φ (a) < 1

(this exists unless

φ = 0

). Define

ψ_{a} (x) = \frac{φ (a x)}{φ (a)} .

Because

A

is abelian, for

x \geq 0

we have

ψ_{a} (x) = φ (a^{1 / 2} x a^{1 / 2}) \geq 0.

. So

ψ_{a}

is positive and

ψ (1) = 1

, thus a state. We have, with

t = φ (a)

,

φ (x) = t ψ_{a} (x) + (1 - t), ψ_{1 - a} (x), x \in A .

As

φ

is extreme, we have

ψ_{a} = φ

. This equality is

φ (a x) = φ (a) φ (x) .

This works for all

x \in A

and all

a \in A^{+}

with

φ (a) < 1

. As the positive elements span

A

, we get

φ (a x) = φ (a) φ (x), a, x \in A .

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