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 \rhoif \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_2U 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 APS(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

Advertisements