Representations of C*-Algebras: Irreducible Representations and Pure States

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 $A$ .

$(H,\varphi)$ is irreducible if and only if $\varphi(A)'=\mathbb{C}1_H$ .
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 $A$ .

$\rho$ is pure if and only if $(H_\rho,\varphi_\rho)$ is irreducible
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

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

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 $\bullet$

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December 14, 2014 at 7:49 pm

momumomu

in Theorem 5.1.1 I have several questions:
1. $B(H_\rho)$ is complete. What we use?
2.therefore it converges strongly to $I_{H_\rho}$ How?
3. $\lim_\lambda \rho(u_\lambda)=\|\rho\|$ why?
Thank in advance.

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Representations of C*-Algebras: Irreducible Representations and Pure States

Theorem 5.1.1

Theorem 5.1.2

Theorem 5.1.3

Theorem 5.1.4

Theorem 5.1.5

Theorem 5.1.6

Theorem 5.1.7

Example

Theorem 5.1.8

Corollary 5.1.9

Corollary 5.1.10

Theorem 5.1.11

Theorem 5.1.12

Theorem 5.1.13

Theorem 5.1.14

Theorem 5.1.15

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Representations of C*-Algebras: Irreducible Representations and Pure States

Theorem 5.1.1

Theorem 5.1.2

Theorem 5.1.3

Theorem 5.1.4

Theorem 5.1.5

Theorem 5.1.6

Theorem 5.1.7

Example

Theorem 5.1.8

Corollary 5.1.9

Corollary 5.1.10

Theorem 5.1.11

Theorem 5.1.12

Theorem 5.1.13

Theorem 5.1.14

Theorem 5.1.15

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