Taken from C*-Algebras and Operator Theory by Gerald Murphy.

This section is concerned with positive linear functionals and representations. Pure states are introduced and shown to be the extreme points of a certain convex set, and their existence is deduced from the Krein-Milman theorem. From this the existence of  irreducible representations is proved by establishing a correspondence between them and the pure states.

If $(H,\varphi)$ is a representation of a C*-algebra $A$, we say $x\in H$ is a cyclic vector for $(H,\varphi)$ if $x$ is cyclic for the C*-algebra $\varphi(A)$ (This means that cyclic vector is a vector $x\in H$ such that the closure of the linear span of $\{\varphi(a)x\,:\,a\in A\}$ equals $H$). If $(H,\varphi)$ admits a cyclic vector, then we say that it is a cyclic representation.

We now return to the GNS construction associated to a state to show that the representations involved are cyclic.

## Theorem 5.1.1

Let $A$ be a C*-algebra and $\rho\in S(A)$. Then there is a unique vector $x_\rho\in H_\rho\in H$ such that

$\rho(a)=\langle a+N_\rho,x_\rho\rangle$, for $a\in A$.

Moreover, $x_\rho$ is a unit cyclic vector for $(H_\rho,\varphi_\rho)$ and

$\varphi_\rho(a)x_\rho=a+N_\rho$, for $a\in A$.

Proof: The function

$\tau_0:A/N_\rho\rightarrow\mathbb{C}$$a+N_\rho\mapsto \rho(a)$,

is well-defined, linear and norm-decreasing (well-defined: suppose that $a,\,b\in A$ such that $a+N_\rho=b+N_\rho\in A/N_\rho$. Then we have that $a-b\equiv 0\,(\text{mod }N_\rho)$ and hence $\rho((a-b)^*(a-b))=0$. Now by Theorem 3.2.2

$|\rho(a-b)|^2\leq\|\rho\|\rho((a-b)^*(a-b))=0$

$\Rightarrow |\rho(a-b)|=0\Rightarrow \rho(a-b)=0\Rightarrow\rho(a)=\rho(b)$;

linear: $\checkmark$; norm-decreasing: $\checkmark$), so we can extend it to a norm-decreasing linear functional $\tau$ on $H_\rho$. By the Riesz representation theorem, there is a unique vector $x_\rho\in H_\rho$ such that $\tau(y)=\langle y,x_\rho\rangle$ for all $y\in H_\rho$. Thus, $x_\rho$ is the unique element of $H_\rho$ such that $\rho(a)=\langle a+N_\rho,x_\rho\rangle$, for all $a\in A$.

Let $a,\,b\in A$. Then

$\langle b+N_\rho,\varphi_\rho(a)x_\rho\rangle=\langle a^*b+N_\rho,x_\rho\rangle=\rho(a^*b)=\langle b+N_\rho,a+N_\rho\rangle$,

and since this holds for all $b$, we have $\varphi_\rho(a)x_\rho=a+N_\rho$ (the first equality above is got by choosing a representative $x_0\in A$ such that $x_\rho=x_0+N_\rho$. Now

$\langle b+N_\rho,\varphi(a)(x_0+N_\rho)\rangle=\langle b+N_\rho,ax_o+N_\rho\rangle$

$=\rho((ax_0)^*b)=\rho(x_0^*(a^*b))$).

Hence, $\varphi_\rho(a)x_\rho$ is dense in $H_\rho$, since it is the space $A/N_\rho$. Therefore, $x_\rho$ is cyclic for $(H_\rho,\varphi_\rho)$. Consequently, $\varphi_\rho(A)$ acts non-degenerately on $H_\rho$ (Suppose $\varphi_\rho(A)x_\rho$ is dense in $H_\rho$ and suppose that $\varphi_\rho(a)x=0$ for all $x\in H_\rho$. Then, for all $y\in H_\rho$ and $a\in A$:

$\langle y,\varphi_\rho(a)x\rangle=0$

$\Rightarrow\langle\varphi_\rho(a^*)y,x\rangle=0$

But $\varphi_\rho(a^*)y$ is also dense in $H_\rho$. Therefore we have that $x=0$ and hence $\varphi_\rho$ acts non-degenerately (there is something in the family $\varphi_\rho(A)$ sending every non-zero to something non-zero).). If $\{u_\lambda\}_{\lambda\in\Lambda}$ is an approximate unit for $A$, then $\{\varphi_\rho(u_\lambda)\}$ is one for $\varphi_\rho(A)$, and therefore it converges strongly to $I_{H_\rho}$ (firstly it is obvious that if $A$ is unital then so is $H_\rho$, and I’m happy that $\{\varphi_\rho(u_\lambda)\}$ is an approximate unit. This is then an element of $H_\rho$ as $B(H_\rho)$ is complete.). Hence

$\|x_\rho\|^2=\langle x_\rho,x_\rho\rangle=\lim_\lambda\langle \varphi_\rho(u_\lambda)x_\rho,x_\rho\rangle=$

$\lim_\lambda \rho(u_\lambda)=\|\rho\|=1$,

so $x_\rho$ is a unit vector $\bullet$

We call $x_\rho$ in Theorem 5.1.1 the canonical cyclic vector for $(H_\rho,\varphi_\rho)$.

If $\rho,\,\tau$ are positive linear functionals on a C*-algebra $A$, we write $\rho\geq\tau$ if $\rho-\tau$ is positive. We say that $\rho$ majorises $\tau$ or $\tau$ is majorised by $\rho$if $\rho\geq \tau$.

## Theorem 5.1.2

Let $\rho$ be a state and $\tau$ a positive linear functional on a C*-algebra $A$, and suppose that $\rho\geq\tau$. Then there is a unique operator $T\in B(H_\rho)$ such that

$\tau(a)=\langle \varphi_\rho(a)Tx_\rho,x_\rho\rangle$.

Moreover, $T$ commutes with all $\varphi_\rho(A)$ ($T\in \varphi_\rho(A)'$) and $0\leq T\leq 1$.

Proof : Define a sesquilinear form  $\sigma$ on $A/N_\rho$ by setting

$\sigma(a+N_\rho,b+N_\rho)=\tau(b^*a)$.

This is well-defined as $\rho\geq\tau$ (Let $a_1+N_\rho,\,a_2+N_\rho=a+N_\rho$. Now

$\sigma(a_1+N_\rho,b+N_\rho)=\tau(b^*a_1)$, and

$\sigma(a_2+N_\rho,b+N_\rho)=\tau(b^*a_2)$,

so we want to prove that $\tau(b^*a_1)=\tau(b^*a_2)$ if $a_1\equiv a_2\,(\text{mod }N_\rho)$ which we have already seen implies that $\rho((a_1-a_2)^*(a_1-a_2))=0$. However $(a_1-a_2)^*(a_1-a_2)$ is positive so we can say that

$(\rho-\tau)((a_1-a_2)^*(a_1-a_2))\geq0$

$\Rightarrow \rho((a_1-a_2)^*(a_1-a_2))-\tau((a_1-a_2)^*(a_1-a_2))\geq0$

$\Rightarrow \tau((a_1-a_2)^*(a_1-a_2))\leq0$

$\Rightarrow \tau((a_1-a_2)^*(a_1-a_2))=0\Rightarrow \tau(a_1-a_2)=0$

$\Rightarrow \tau(b^*(a_1-a_2))=0$

where the last two implications used Theorems 3.2.2 and 3.2.7 respectively. That is $\tau(b^*a_1)=\tau(b^*a_2)$ and we are done. I’m sure there’s no problem that we fixed $b$.). Observe that $\|\sigma\|\leq 1$ as

$|\tau(b^*a)|\leq \tau(b^*b)^{1/2}\tau(a^*a)^{1/2}\leq \rho(b^*b)^{1/2}\rho(a^*a)^{1/2}=\|b+N_\rho\|\|a+N_\rho\|$.

We can therefore extend $\sigma$ to a bounded sesquilinear form (also denoted $\sigma$) on $H_\rho$  which also has a norm not greater than $1$. Hence, there is an operator $T$ on $H_\rho$ such that $\sigma(x,y)=\langle Tx,y\rangle$ for all $x,\,y\in H_\rho$, and $\|T\|\leq 1$ ($\checkmark$).

We have that $T$ is positive because $\sigma(a+N_\rho,a+N_\rho)=\tau(a^*a)\geq 0$. Also

$\tau(b^*a)=\sigma(a+N_\rho,b+N_\rho)=\langle T(a+N_\rho),b+N_\rho\rangle=\langle T\varphi_\rho(a)x_\rho,x_\rho\rangle$.

(I have no idea why Murphy makes this particular calculation!!)

If $a,\,b,\,c\in A$, then

$\varphi_\rho(a)T(b+N_\rho,c+N_\rho)=\langle T(b+N_\rho),a^*c+N_\rho\rangle=\tau(c^*ab)$

$=\langle T(ab+N_\rho),c+N_\rho\rangle=\langle T\varphi_\rho(a)(b+N_\rho),c+N_\rho\rangle$.

Hence, $\varphi_\rho(a)T=T\varphi_\rho(a)$ for all $a\in A$, so $T\in\varphi_\rho(A)'$. Also,

$\tau(a^*b)=\langle T(b+N_\rho),a+N_\rho\rangle$

$=\langle T\varphi_\rho(b)x_\rho,\varphi_\rho(a)x_\rho\rangle=\langle T\varphi(a^*b)x_\rho,x_\rho\rangle$,

so if $\{u_\lambda\}_{\lambda\in\Lambda}$ is an approximate unit for $A$, then $\tau(u_\lambda b)=\langle T\varphi_\rho(u_\lambda b)x_\rho,x_\rho\rangle$, and therefore in the limit $\tau(b)=\langle T\varphi_\rho(b),x_\rho\rangle=\langle \varphi_\rho(b)Tx_\rho,x_\rho\rangle$.

To see uniqueness, suppose that $S\in \varphi_\rho(A)'$ and $tau(a)=\langle \varphi_\rho(a)Sx_\rho,x_\rho\rangle$ for all $a\in A$. Then

$\langle S\varphi_\rho(a^*b)x_\rho,x_\rho\rangle=\tau(a^*b)=\langle T\varphi_\rho(a^*b)x_\rho,x_\rho\rangle$,

and therefore

$\langle S(b+N_\rho),a+N_\rho\rangle=\langle S(b+N_\rho),a+N_\rho\rangle$

for all $a,\,b\in A$. Hence, $S=T$ $\bullet$

It is an easy exercise to show that we can go in the opposite direction also; that is, if $T\in \varphi_\rho(A)'$ and $\|T\|\leq 1$, then the equation $\tau(a)=\langle \varphi_\rho(a)Tx_\rho,x_\rho\rangle$ defines a positive linear function $\tau$ on $A$ such that $\rho\geq\tau$.

A representation $(H,\varphi)$ of a C*-algebra $A$ is non-degenerate if the C*-algebra $\varphi(A)$ acts non-degeneratively on $H$.

It is clear that a direct sum of non-degenerate representations is non-degenerate ($\checkmark$), and also that cyclic representations are non-degenerate ($\checkmark$). Therefore, the universal representation of $A$ is non-degenerate.

If $(H,\varphi)$ is a non-degenerate representation for $A$ and $\{u_\lambda\}_{\lambda\in\Lambda}$ an approximate unit of $A$, then $\{\varphi(u_\lambda)\}$ is an approximate unit of $\varphi(A)$, so the net $\{\varphi(u_\lambda)\}$ converges strongly to $I_H$.

Let $(H,\varphi)$ be an arbitrary representation of $A$. If $K$ is a closed vector subspace of $H$ invariant for $\varphi(A)$, then the map

$\varphi_K:A\rightarrow B(K)$$a\mapsto (\varphi(a))_K$,

is a *-homomorphism, so the pair $(K,\varphi_K)$ is a representation of $A$ also. If $K=\overline{\langle \varphi(A)H\rangle}$, then $K$ is invariant for $\varphi(A)$ and the representation $(K,\varphi_K)$ is non-degenerate (I might have to come back to this as Murphy has $K=\overline{\varphi(A)H}$). Moreover, $\|\varphi(a)\|=\|\varphi_K(a)\|$ for all $a\in A$. We shall often use this device to reduce to the case of non-degenerate representation.

## Theorem 5.1.3

Let $(H,\varphi)$ be a non-degenerate representation of a C*-algebra $A$. Then it is a direct sum of cyclic representations of $A$.

Proof : For each $x\in H$, set $H_x=\overline{\varphi(A)x}$. An easy application of Zorn’s Lemma shows that there is a maximal set $\Lambda$ of non-zero elements of $H$ such that the spaces $H_x$ are pairwise orthogonal for $x\in \Lambda$. If $y\in \left(\bigcup_{x\in\Lambda}H_x\right)^\perp$, then for all $x\in\Lambda$ we have $\langle y,\varphi(a^*b)x\rangle=0$, so $\langle \varphi(a)y,\varphi(b)x\rangle=0$, and therefore the spaces $H_x,\,H_y$ are orthogonal. Observe that since $(H,\varphi)$ is non-degenerate, $y\in H_y$. It follows from the maximality of $\Lambda$ that $y=0$. Therefore, $H$ is the orthogonal direct sum of the family of Hilbert spaces $\{H_x\}_{x\in \Lambda}$. Obviously, these spaces are invariant for $\varphi(A)$, and the restriction representation

$\varphi_x:A\rightarrow B(H_x)$$a\mapsto \varphi(a)_{H_x}$,

has $x$ as a cycle vector. Since $(H,\varphi)$ is the direct sum of the representations $\{(H_x,\varphi_x)\}_{x\in\Lambda}$ the theorem is proved $\bullet$

Two representations $(H_1,\varphi_1)$ and $(H_2,\varphi_2)$ of a C*-algebra $A$ are unitarily equivalent if there is a unitary $U:H_1\rightarrow H_2$ such that $\varphi_2(a)=U\varphi_1(a)U^*$ for all $a\in A$. It is readily verified that unitary equivalent is indeed an equivalence relation.

## Theorem 5.1.4

Suppose that $(H_1,\varphi_1)$ and $(H_2,\varphi_2)$ are representations of a C*-algebra $A$ with cyclic vectors $x_1$ and $x_2$, respectively. Then there is a unitary $U:H_1\rightarrow H_2$ such that $x_2=Ux_1$ and $\varphi_2(a)=U\varphi_1(a)U^*$ for all $a\in A$ if and only if $\langle\varphi_1(a)x_1,x_1\rangle=\langle \varphi_2(a)x_2,x_2\rangle$ for all $a\in A$.

Proof : The forward implication is obvious ($\checkmark$).

Suppose, therefore, that we have $\langle\varphi_1(a)x_1,x_1\rangle=\langle \varphi_2(a)x_2,x_2\rangle$ for all $a\in A$. Define a linear map $U_0:\varphi_1(A)\rightarrow H_2$ by setting $U_0(\varphi_1(a)x_2)=\varphi_2(a)x_2$. That this is well-defined and isometric follows from the equations

$\|\varphi_2(a)(x_2)\|^2=\langle \varphi_2(a^*a)x_2,x_2\rangle=\langle\varphi_1(a^*a)x_1,x_1\rangle=\|\varphi_1(a)x_1\|^2$.

We extend $U_0$ to an isometric linear map $U:H_1\rightarrow H_2$, and since $U(H_1)=\overline{\varphi_2(A)x_2}=H_2$$U$ is unitary ($\checkmark$).

If $a,\,b\in A$, then

$U\varphi_1(a)\varphi_1(b)x_1=\varphi_2(ab)x_2=\varphi_2(a)U\varphi_1(b)x_1$.

Therefore, $U\varphi_1(a)=\varphi_1(a)$ for all $a\in A$. Now

$\varphi_2(a)Ux_1=U\varphi_1(a)x_1=\varphi_2(a)x_2$,

so $\varphi_2(a)(Ux_1-x_2)=0$. By non-degeneracy of $\varphi_2$, therefore, $Tx_1=x_2$ $\bullet$

A representation $(H,\varphi)$ of  a C*-algebra $A$ is irreducible if the algebra $\varphi(A)$ acts irreducibly on $H$. If two representations are unitarily equivalent, then irreducibility of one implies the irreducibility of the other. If $H$ is  one dimensional Hilbert space, then the zero representation of any C*-algebra is irreducible (surely so is any one dimensional representation?).

## Theorem 5.1.5

Let $(H,\varphi)$ be a non-zero representation of a C*-algebra $A$

1. $(H,\varphi)$ is irreducible if and only if $\varphi(A)'=\mathbb{C}1_H$
2. If $(H,\varphi)$ is irreducible, then every non-zero vector of $H$ is cyclic for $(H,\varphi)$.

Proof : Condition 1. is immediate from Theorem 4.1.12.

Suppose that $(H,\varphi)$ is irreducible, and that $x$ is a non-zero vector of $H$. The space $\overline{\varphi(A)x}$ is invariant for $\varphi(A)$, and therefore is equal to $0$ or $H$. Because $\varphi$ is non-zero, there is some element $y\in H$ and some element $a\in A$ such that $\varphi(a)y\neq 0$. Hence $\overline{\varphi(A)y}=H$, so $\varphi$ is non-degenerate. It follows that $\varphi(A)x$ is not the zero space, so $\overline{\varphi(A)x}=H$; that is, $x$ is a cyclic vector for $(H,\varphi)$ $\bullet$

We say a state $\rho$ on a C*-algebra is pure if it has the property that whenever $\tau$ is a positive linear functional on $A$ such that $\rho\geq \tau$, necessarily there is a number $t\in[0,1]$ such that $\tau=t\rho$.

The set of pure states on $A$ is denoted by $PS(A)$.

## Theorem 5.1.6

Let $\rho$ be a state on a C*-algebra $A$.

1. $\rho$ is pure if and only if $(H_\rho,\varphi_\rho)$ is irreducible
2. If $A$ is Abelian, then $\rho$ is pure if and only if it is a character on $A$.

Proof : Suppose that $\rho$ is a pure state. Let $S$ be an element of $\varphi_\rho(A)'$ such that $0\leq S\leq 1$. Then the function

$\tau:A\rightarrow\mathbb{C}$$a\mapsto\langle\varphi_\rho(a)Sx_\rho,x_\rho\rangle$,

is a positive linear functional on $A$ such that $\rho\geq\tau$. Hence, there exists $t\in[0,1]$ such that $\tau=t\rho$, and therefore $\langle \varphi_\rho(a)Sx_\rho,x_\rho\rangle=\langle t\varphi_\rho(a)x_\rho,x_\rho\rangle$ for all $a\in A$ (from Theorem 5.1.1 we know that $\rho(a)=\langle\varphi_\rho(a)x_\rho,x_\rho\rangle$. We have $\tau(a)=t\rho(a)$ and obviously $t\rho(a)=\langle t\varphi_\rho(a)x_\rho,x_\rho\rangle$). Consequently (the second equality comes from the fact that $S\in \varphi_\rho(A)'$ by assumption — so we can have $\varphi_\rho(b^*)S\varphi_\rho(a)=S\varphi_\rho(b^*a)$),

$\langle S(a+N_\rho),b+N_\rho\rangle=\langle S\varphi_\rho(a)x_\rho,\varphi_\rho(b)x_\rho\rangle$

$=\langle S\varphi_\rho(b^*a)(x_\rho),x_\rho\rangle$

$=\langle t\varphi_\rho(b^*a)x_\rho,x_\rho\rangle$

$=\langle t(a+N_\rho),b+N_\rho\rangle$

for all $a,\,b\in A$. Therefore $S=t1_{H_\rho}$, since $A/N_\rho$ is dense in $H_\rho$. It follows that $\varphi_\rho(A)'=\mathbb{C}1_{H_\rho}$, so $(H_\rho,\varphi_\rho)$ is irreducible by Theorem 5.1.5.

Now suppose conversely that $(H_\rho,\varphi_\rho)$ is irreducible, and let $\tau$ be a positive linear functional on $A$ such that $\rho\geq\tau$. By Theorem 4.1.2 there is a unique operator $S$ in $\varphi_\rho(A)'$ such that $0\leq S\leq 1$ and $\tau(a)=\langle\varphi_\rho(a)Sx_\rho,x_\rho\rangle$ for all $a\in A$. But $\varphi_\rho(a)'=\mathbb{C}1_{H_\rho}$, again by Theorem 5.1.5, so $S=tI_{H_\rho}$ for some $t\in [0,1]$. Hence, $\tau=t\rho$, so $\tau$ is pure. This proves the equivalence in condition 1.

Assume now that $A$ is abelian.

If $\rho$ is pure, then $\varphi_\rho(A)'=\mathbb{C}1_{H_\rho}$. But $\varphi_\rho(A)\subset \varphi(A)'$, so $\varphi_\rho(A)$ consists of scalars, and therefore $B(H_\rho)\subset \varphi_\rho(A)'$ (I don’t agree with this — however, if $\rho$ is pure, then $(H_\rho,\varphi_\rho)$ is irreducible so that $\varphi_\rho$ has no invariant subspace of $H_\rho$. We want to prove that $B(H_\rho)=\mathbb{C}1_{H_\rho}$ — i.e. that $H_\rho$ is one dimensional. Suppose on the contrary that $H_\rho$ is not one dimensional — in which case it has a two dimensional subspace with basis vectors $\{e_1,e_2\}$. However $\varphi_\rho(A)e_i\subset e_i$ as all of the $\varphi_\rho(A)$ are scalars and this is a contradiction as $\varphi_\rho(A)$ acts irreducibly). Hence, $B(H_\rho)=\mathbb{C}I_{H_\rho}$. Therefore, if $T,\,S\in B(H_\rho)$ they are scalars and (using the fact that $x_\rho$ is a unit vector and looking at $\langle \lambda \mu x_\rho,x_\rho\rangle=\langle \lambda x_\rho,x_\rho\rangle\langle\mu x_\rho,x_\rho\rangle$) it is clear $\rho(ab)=\rho(a)\rho(b)$ and therefore a character on $A$.

Now suppose conversely that $\rho$ is a character on $A$, and let $\tau$ be a positive linear functional on $A$ such that $\tau\leq\rho$. If $\rho(a)=0$, then $\rho(a^*a)=0$, so $\tau(a^*a)=0$. Since $|\tau(a)|\leq\tau(a^*a)^{1/2}$, therefore $\tau(a)=0$ (via an application of Cauchy-Schwarz). Hence, $\text{ker }\rho\subset\text{ker }\tau$, and it follows from elementary linear algebra that there is a scalar such that $\tau=t\rho$

Choose a $a\not\in\text{ker }\rho$. Now for all $b\in A$, we have

$\rho\left(b-a\frac{\rho(b)}{\rho(a)}\right)=0$.

Now we can write

$b=\left(b-a\frac{\rho(b)}{\rho(a)}\right)+a\frac{\rho(b)}{\rho(a)}$;

that is $A=\{c+\lambda a:c\in\text{ker }\rho,\lambda\in\mathbb{C}\}$ with $\rho(c+\lambda a)=\lambda \rho(a)$. Suppose $\text{ker }\rho\subsetneq\text{ker }\tau$. Now choose an $a\in\text{ker }\tau\backslash\text{ker }\rho$. Now any $b\in A$ is of the form $b=c+\lambda a$, with $c\in\text{ker }\rho$ and $a$ as above:

$\Rightarrow \tau(b)=\tau(c)+\lambda(a)=0+\lambda 0=0$

$\Rightarrow\tau=0$.

If $\tau$ is non-zero we therefore have that $\text{ker }\rho=\text{ker }\tau$. Now

$\tau(b)=\tau(c)+\lambda \tau(a)=0+\lambda \tau(a)$

$=\lambda \rho(a)\frac{\tau(a)}{\rho(a)}=\underbrace{\frac{\tau(a)}{\rho(a)}}_{:=t} \rho(b)$.

In other words, $\text{dim }A/\text{ker }\rho=1$.

Choose $a\in A$ such that $\rho(a)=1$. Then $\rho(a^*a)=1$, so $0\leq\tau(a^*a)=t\rho(a^*a)=t\leq\rho(a^*a)=1$, and therefore $t\in[0,1]$. This shows that $\rho$ is pure and the equivalence in condition 2.  is proved $\bullet$

It follows from Theorem 5.1.6 that for an arbitrary abelian C*-algebra $A$$PS(A)=\Phi(A)$. The only thing not obvious is that a character $\rho$ on $A$ must have norm $1$. To see this, let $\{u_\lambda\}_{\lambda\in\Lambda}$ be an approximate unit for $A$. Then $\{u_\lambda^2\}$ is also an approximate unit. Hence

$\|\rho\|=\lim_\lambda\rho(u_\lambda^2)=\left(\lim_\lambda \rho(u_\lambda)\right)^2=\|\rho\|^2$

by Theorem 3.3.3, so $\|\rho\|=\|\rho\|^2$, and therefore $\|\rho\|=1$.

## Theorem 5.1.7

Let $(H,\varphi)$ be a representation of a C*-algebra $A$, and let $x$ be a unit cyclic vector for $(H,\varphi)$. Then the function

$\rho:A\rightarrow\mathbb{C}$$a\mapsto\langle \varphi(a)x,x\rangle$

is a state of $A$ and $(H,\varphi)$ is unitarily equivalent to $(H_\rho,\varphi_\rho)$. Moreover, if $(H,\varphi)$ is irreducible, then $\rho$ is pure.

Proof : Clearly $\rho$ is a positive linear functional on $A$ (this follows by defining a sesquilinear form $\sigma(x,y)=\langle\varphi(a)x,y\rangle$ and the fact that *-homomorphisms are positive as $\varphi(a^*a)=\varphi(a)^*\varphi(a)$.). If $\{u_\lambda\}_{\lambda\in\Lambda}$ is an approximate unit for $A$, then because $(H,\varphi)$ is non-degenerate (as it admits a cyclic vector) the net $\{\varphi(u_\lambda)\}$ is strongly convergent to $I_H$ ($\checkmark$). Hence

$\|\rho\|=\lim_\lambda\rho(u_\lambda)=\lim_\lambda\langle \varphi(u_\lambda)x,x\rangle=\langle x,x\rangle=1$,

s0 $\rho\in S(A)$. For all $a\in A$,

$\langle\varphi_\rho(a)x_\rho,x_\rho\rangle=\rho(a)=\langle \varphi(a)x,x\rangle$,

so $(H_\rho,\varphi_\rho)$ and $(H,\varphi)$ are unitarily equivalent by Theorem 5.1.4.

If $(H,\varphi)$ is irreducible, so is $(H_\rho,\varphi_\rho)$, so by Theorem 5.1.6, $\rho$ is pure $\bullet$

### Example

Let $H$ be a non-zero Hilbert space, and let $A=K(H)$. We are going to determine the pure states of $A$. If $x\in H$, then the functional

$\omega_x:A\rightarrow \mathbb{C}$$T\mapsto \langle Tx,x\rangle$,

is positive ($\checkmark$), and if $x$ is a unit vector, $\omega_x$ is a state (it is straightforward to show that $\|\omega_x\|\leq 1$ and if we choose $T=I_{\mathbb{C}x}\subset K(H)$ then we have equality.).

The pure states of $A$ are precisely the states $\omega_x$ where $x$ is a unit vector of $H$.

To prove this, suppose first that $x$ is a unit vector of $H$, and let $I:A\rightarrow B(H)$ be the inclusion map. The representation $(H,I)$ is irreducible, since $A'=\mathbb{C}$ (this is a result from Section 4.4 which I did not cover — seems perfectly reasonable). Hence, $x$ is a cyclic vector for $A$ ($K(H)x$ is dense in $H$? Let $E=\{e_\lambda\}_{\lambda\in\Lambda}$ be a basis of $H$. As $x$ is a unit vector, there exists an $e_{\lambda_0}\in E$ such that $\langle x,e_{\lambda_0}\rangle\neq 0$. Hence define a family of compact linear maps $L=\{T_\lambda:\lambda\in\Lambda\}$

$T_\mu(e_\lambda):=\left\{\begin{array}{cc}e_\mu &\text{ if }\lambda=\lambda_0 \\ 0 & \text{ otherwise}\end{array}\right.$

Then $\mathbb{C}L\subset K(H)$ has the property that $Lx$ is dense in $H$ so that $x$ is cylic.), and it follows from Theorem 5.1.7 that the representations $(H_{\omega_x},\varphi_{\omega_x})$ and $(H,I)$ are unitarily equivalent and $\omega_x$ is pure.

Now  suppose conversely that $\rho$ is a pure state of $A$. By Theorem 4.2.1, there is a trace-class operator $T$ on $H$ such that $\rho(S)=\text{tr}(TS)$ for all $S\in A$. For any unit vector $x\in H$, the operator $|x\rangle\langle x|$ is a projection and therefore positive, so

$0\leq \rho(|x\rangle\langle x|)=\text{tr}(T|x\rangle\langle x|)=\text{tr}(|Tx\rangle\langle x|)=\langle Tx,x\rangle$.

This shows that the operator $T$ is positive. Since $T$ is a compact normal operator ($\checkmark$),  it is diagonalisable by Theorem 2.4.4; that is, there is an orthonormal basis $E\subset H$ and there is a family of scalars $\{\lambda_e\}_{e\in E}$ such that $Te=\lambda_ee$. Choose $e_0\in E$. If $S\in A^+$,

$\rho(S)=\text{tr}(ST)=\sum_{e\in E}\langle STe,e\rangle=\sum_{e\in E}\lambda_e\langle Se,e\rangle\geq \lambda_{e_0}\omega_{e_0}(S)$.

Thus the pure state $\rho$ majorises the positive linear functional $\lambda_{e_0}\omega_{e_0}$, so there exists $t\in[0,1]$ such that $\lambda_{e_0}\omega_{e_0}=t\rho$. Since both $\omega_{e_0}$ and $\rho$ are of norm one, $\lambda_{e_0}=t$, so $\omega_{e_0}=\rho$; that is $\rho$ is of the required form

An interesting consequence of our characterisation of the pure states of $A$ is that every non-zero irreducible representation $(K,\psi)$ is unitarily equivalent to the identity representation $(H,I)$  of $A$. To see this, let $y$ be a unit vector in $K$. The function

$\tau:A\rightarrow \mathbb{C}$$T\mapsto\langle\psi(T)y,y\rangle$,

is a pure state on $A$, and $(K,\psi)$ is unitarily equivalent to $(H_\tau,\varphi_\tau)$ by Theorem 5.1.7. Hence, there exists a unit vector $x\in H$ such that $\tau=\omega_x$. Thus, $(K,\psi)$ is unitarily equivalent to $(H_{\omega_x},\varphi_{\omega_x})$, and we have already seen above that $(H_{\omega_x},\varphi_{\omega_x})$ is unitarily equivalent to $(H,I)$.

## Theorem 5.1.8

If $A$ is a C*-algebra, then the set $S$ of norm-decreasing positive linear functionals on $A$ forms a convex weak* compact set. The extreme points of $S$ are the zero functional and the pure states of $A$.

Proof : It is easy to check that $S$ is weak* closed in the closed unit ball of $A^*$ ($S=B_1[0]\cap (A^*)^+$, so if we can show that the positive linear functionals are weak*-closed then we are done. I’m just going to reference Lemma 2.1 (i)), and therefore weak* compact by the Banach-Alaoglu theorem ($\checkmark$). Convexity of $S$ is clear ($\checkmark$).

Let $\partial_e A$ be the set of extreme points of $S$.

First we show $0\in\partial_e S$. Suppose that $0=t\rho+(1-t)\tau$, where $t\in(0,1)$ and $\rho,\,\tau\in S$. If $a\in A$, then

$t\rho(a^*a)+(1-t)\tau(a^*a)=0$

Therefore $\rho=0=\tau$ on $A^+$, and therefore on $A$ (by Theorem 3.3.2), so $0\in\partial_eS$.

Next we show that $PS(A)\subset\partial_eS$. Suppose that $\rho$ is a pure state of $A$, and that $\rho=t\tau_1+(1-t)\tau_2$, where $t\in(0,1)$ and the $\tau_i\in S$. Then $t\tau_1$ is majorised by $\rho$, so there exists $s\in[0,1]$ such that $t\tau_1=s\rho$, because $\rho$ is pure. Since (with equality in the triangle inequality as they’re all linearly dependent)

$1=\|\rho\|=t\|\tau_1\|+(1-t)\|\tau_2\|$,

we have $\|\tau_1\|=\|\tau_2\|=1$. It follows that

$t=\|t\tau_1\|=\|s\rho\|=s$,

so $\tau_1=\rho$. Hence, $(1-t)\tau_2=(1-t)\rho$, so $\rho\in\partial_eS$.

Finally, we suppose that $\rho$ is a non-zero element of $\partial_eS$ and show that it is a pure state. Since

$\rho=\|\rho\|\left(\frac{\rho}{\|\rho\|}\right)+(1-\|\rho\|)0$,

and $0,\,\rho/\|\rho\|\in S$, we have $\|\rho\|=1$, because $\rho\in\partial_eS$. If $\tau$ is a non-zero positive linear functional on $A$ majorised by, and not equal to, $\rho$ then for $t=\|\tau\|$, we have

$\rho=t\left(\frac{\tau}{\|\tau\|}\right)+(1-t)\left(\frac{\rho-\tau}{\|\rho-\tau\|}\right)$,

since $1-t=\|\rho-\tau\|$. Hence, $\rho=\tau/\|\tau\|$, since $\rho$ is an extreme point of $S$. Therefore, $\tau=\|\tau\|\rho$. This proves that $\rho$ is a pure state on $A$ $\bullet$

## Corollary 5.1.9

The set $S$ is the weak* closed convex hull of $latex$ and the pure states of $A$.

Proof : Apply the Krein-Milman theorem $\bullet$

## Corollary 5.1.10

If $A$ is a unital C*-algebra, then $S(A)$ is the weak* closed convex hull of the pure states of $A$.

Proof : The set $S(A)$ is a non-empty convex weak* compact set, so by the Krien-Milman theorem it is the weak* closed convex hull of it’s extreme points. It is clear that $S(A)$ is a face of $S$, where $S$ is as in Theorem 5.1.8, so by that theorem the extreme points of $S(A)$ are the pure states of $A$ $\bullet$

## Theorem 5.1.11

Let $a$ be a positive element of a non-zero C*-algebra $A$. Then there is a pure state $\rho$ of $A$ such that $\|a\|=\rho(a)$.

Proof : We may suppose that $a\neq0$. The function

$\hat{a}:A^*\rightarrow\mathbb{C}$$\tau\mapsto\tau(a)$,

is weak* continuous and linear, and by Theorem 3.3.6 $\|a\|=\text{sup}\{\tau(a):\tau\in S\}$, where $S$ is the weak* compact convex set of all norm-decreasing positive linear functionals on $A$. The set $F=\{\tau\in S|\tau(a)=\|a\|\}$ is a weak* compact face of $S$ by Lemma A.13 (Let $C$ be a non-empty convex compact set in a locally convex space $X$ and suppose that $\tau$ is a continuous linear functional on $X$. Let $M$ be the supremum of all $\text{Re}(\tau(x))$ where $x$ ranges over $C$. Then the set $F$ of all such $x\in C$ such that $\text{Re}(\tau(x))=M$ is a compact face of $C$. — I think this is after identifying $a$ with the linear functional $\hat{a}$.), and therefore has an extreme point $\rho$ by the Krein-Milman theorem. Since $F$ is a face of $S$, the functional $\rho$ is an extreme point of $S$ also. Now $\rho\neq0$, since $\|a\|=\rho(a)$ and $a\neq0$. Therefore, $\rho$ is a pure state by Theorem 5.1.8 $\bullet$

It follows from Theorem 5.1.11 that a non-zero C*-algebra $A$ has pure states.

## Theorem 5.1.12

Let $A$ be a C*-algebra, and $a\in A$. Then there is an irreducible representation $(H,\varphi)$ of $A$ such that $\|a\|=\|\varphi(a)\|$.

Proof : By the preceding theorem, there is a pure state $\rho$ of $A$ such that $\rho(a^*a)=\|a^*a\|$. By Theorem 5.1.6, the representation $(H_\rho,\varphi_\rho)$ is irreducible. Since

$\|a\|^2=\rho(a^*a)=\langle\varphi_\rho(a^*a)x_\rho,x_\rho\rangle=\|\varphi(a)x_\rho\|^2\leq\|\varphi_\rho(a)\|^2\leq\|a\|^2$,

therefore $\|a\|=\|\varphi_\rho(a)\|$ $\bullet$

The characterisation given in Theorem 5.1.8 allows us to prove another extension theorem for positive functionals.

## Theorem 5.1.13

Let $B$ be a C*-subalgebra of a C*-algebra $A$, and let $\rho_0$ be a pure state on $B$. Then there is a pure state $\rho$ on $A$ extending $\rho_0$.

Proof : The set $F$ of all states on $A$ extending $\rho_0$ is a weak* compact face of the set $S$ of norm-decreasing positive linear functionals on $A$ (by Theorem 3.3.8, $F$ is non-empty.) By the Krein-Milman theorem, $F$ admits an extreme point, $\rho$ say. Hence, $\rho$ is an extreme point of $S$, and non-zero, so by Theorem 5.1.8 $\rho$ is a pure state of $A$ $\bullet$

## Theorem 5.1.14

Let $A$ be a unital C*-algebra. Suppose that $S$ is a subset of $S(A)$ such that if a hermitian element $a\in A$ satisfies the condition $\rho(a)\geq 0$ for all $\rho\in S$, then necessarily $a\in A^+$. Then the weak* closed convex hull of $S$ is $S(A)$ and the weak* closure of $S$ contains $PS(A)$.

Proof : Let $C$ denote the weak* closed convex hull of $S$. It follows from Theorem 5.1.8 that $PS(A)$ is the set of extreme points of $S(A)$, so by the Krein-Milman theorem if we show that $C=S(A)$, then $PS(A)$ is contained in the weak* closure of $S$.

Suppose that $C\neq S(A)$ and we shall obtain a contradiction. Since $C\subset S(A)$ clearly holds, there exists a $\rho\in S(A)$ such that $\rho\not\in C$. By Theorem 3.32 there is a weak* continuous linear functional $\theta:A^*\rightarrow\mathbb{C}$ and there is a real number $t$ such that

$\text{Re}(\theta(\rho))>t>\text{Re}(\theta(\tau))$

for all $\rho\in C$. There is an element $a\in A$ such that $\theta=\hat{a}$. If $b=\text{Re}(a)$, then

$\text{Re}(\theta(\tau))=\text{Re}(\tau(a))=\tau(b)$

for all $\tau\in S(A)$. Since $\tau(t1_A-b)\geq0$ ($\checkmark$) for all $\tau\in S$, the hypothesis implies that $t1_A-b\geq0$. Therefore, $\rho(t1_A-b)\geq0$, so $t1_A\geq\rho(b)$. But $\rho(b)=\text{Re}(\theta(\rho))>t$, a contradiction $\bullet$

If $x$ is a unit vector in a Hilbert space $H$, we denote by $\omega_x$ the state

$B(H)\rightarrow\mathbb{C}$$T\mapsto\langle Tx,x\rangle$.

## Theorem 5.1.15

Let $A$ be a C*-algebra and suppose that $\{(H_\lambda,\varphi_\lambda)\}_{\lambda\in\Lambda}$ is a family of representations of $A$. Suppose also that $\rho$ is a pure state of $A$ such that

$\bigcap_{\lambda\in\Lambda}\text{ker}(\varphi_\lambda)\subset\text{ker }\rho$.

Then $\rho$ belongs to the weak* closure in $A^*$ of the set

$S=\{\omega_x\varphi_\lambda\,|\,\lambda\in\Lambda\emph{ and }x\in H_\lambda,\,\|x\|=1\}$.

Proof : Omitted as it uses hereditary C*-subalgebras and such $\bullet$