*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 is a representation of a C*-algebra , we say is a *cyclic vector for *if is cyclic for the C*-algebra (This means that cyclic vector is a vector such that the closure of the linear span of equals ). If 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 be a C*-algebra and . Then there is a unique vector such that*

*, for* .

*Moreover, is a unit cyclic vector for and*

,* for *.

*Proof: *The function

, ,

is well-defined, linear and norm-decreasing (well-defined: suppose that such that . Then we have that and hence . Now by Theorem 3.2.2:

;

linear: ; norm-decreasing: ), so we can extend it to a norm-decreasing linear functional on . By the Riesz representation theorem, there is a unique vector such that for all . Thus, is the unique element of such that , for all .

Let . Then

,

and since this holds for all , we have (the first equality above is got by choosing a representative such that . Now

).

Hence, is dense in , since it is the space . Therefore, is cyclic for . Consequently, acts non-degenerately on (Suppose is dense in and suppose that for all . Then, for all and :

But is also dense in . Therefore we have that and hence acts non-degenerately (there is *something *in the family sending every non-zero to something non-zero).). If is an approximate unit for , then is one for , and therefore it converges strongly to (firstly it is obvious that if is unital then so is , and I’m happy that is an approximate unit. This is then an element of as is complete.). Hence

,

so is a unit vector

We call in Theorem 5.1.1 the *canonical cyclic vector for .*

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

## Theorem 5.1.2

*Let be a state and a positive linear functional on a C*-algebra , and suppose that . Then there is a unique operator such that*

.

Moreover, commutes with all () and .

*Proof *: Define a sesquilinear form on by setting

.

This is well-defined as (Let . Now

, and

,

so we want to prove that if which we have already seen implies that . However is positive so we can say that

,

where the last two implications used Theorems 3.2.2 and 3.2.7 respectively. That is and we are done. I’m sure there’s no problem that we fixed .). Observe that as

.

We can therefore extend to a bounded sesquilinear form (also denoted ) on which also has a norm not greater than . Hence, there is an operator on such that for all , and ().

We have that is positive because . Also

.

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

If , then

.

Hence, for all , so . Also,

,

so if is an approximate unit for , then , and therefore in the limit .

To see uniqueness, suppose that and for all . Then

,

and therefore

for all . Hence,

It is an easy exercise to show that we can go in the opposite direction also; that is, if and , then the equation defines a positive linear function on such that .

A representation of a C*-algebra is *non-degenerate *if the C*-algebra acts non-degeneratively on .

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

If is a non-degenerate representation for and an approximate unit of , then is an approximate unit of , so the net converges strongly to .

Let be an arbitrary representation of . If is a closed vector subspace of invariant for , then the map

, ,

is a *-homomorphism, so the pair is a representation of also. If , then is invariant for and the representation is non-degenerate (I might have to come back to this as Murphy has ). Moreover, for all . We shall often use this device to reduce to the case of non-degenerate representation.

## Theorem 5.1.3

*Let be a non-degenerate representation of a C*-algebra . Then it is a direct sum of cyclic representations of .*

*Proof *: For each , set . An easy application of Zorn’s Lemma shows that there is a maximal set of non-zero elements of such that the spaces are pairwise orthogonal for . If , then for all we have , so , and therefore the spaces are orthogonal. Observe that since is non-degenerate, . It follows from the maximality of that . Therefore, is the orthogonal direct sum of the family of Hilbert spaces . Obviously, these spaces are invariant for , and the restriction representation

, ,

has as a cycle vector. Since is the direct sum of the representations the theorem is proved

Two representations and of a C*-algebra are *unitarily equivalent *if there is a unitary such that for all . It is readily verified that unitary equivalent is indeed an equivalence relation.

## Theorem 5.1.4

*Suppose that and are representations of a C*-algebra with cyclic vectors and , respectively. Then there is a unitary such that and for all if and only if for all .*

*Proof *: The forward implication is obvious ().

Suppose, therefore, that we have for all . Define a linear map by setting . That this is well-defined and isometric follows from the equations

.

We extend to an isometric linear map , and since , is unitary ().

If , then

.

Therefore, for all . Now

,

so . By non-degeneracy of , therefore,

A representation of a C*-algebra is *irreducible *if the algebra acts irreducibly on . If two representations are unitarily equivalent, then irreducibility of one implies the irreducibility of the other. If 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 be a non-zero representation of a C*-algebra . *

*is irreducible if and only if .**If is irreducible, then every non-zero vector of is cyclic for .*

*Proof *: Condition 1. is immediate from Theorem 4.1.12.

Suppose that is irreducible, and that is a non-zero vector of . The space is invariant for , and therefore is equal to or . Because is non-zero, there is some element and some element such that . Hence , so is non-degenerate. It follows that is not the zero space, so ; that is, is a cyclic vector for

We say a state on a C*-algebra is *pure *if it has the property that whenever is a positive linear functional on such that , necessarily there is a number such that .

The set of pure states on is denoted by .

## Theorem 5.1.6

*Let be a state on a C*-algebra .*

*is pure if and only if is irreducible**If is Abelian, then is pure if and only if it is a character on .*

*Proof *: Suppose that is a pure state. Let be an element of such that . Then the function

, ,

is a positive linear functional on such that . Hence, there exists such that , and therefore for all (from Theorem 5.1.1 we know that . We have and obviously ). Consequently (the second equality comes from the fact that by assumption — so we can have ),

for all . Therefore , since is dense in . It follows that , so is irreducible by Theorem 5.1.5.

Now suppose conversely that is irreducible, and let be a positive linear functional on such that . By Theorem 4.1.2 there is a unique operator in such that and for all . But , again by Theorem 5.1.5, so for some . Hence, , so is pure. This proves the equivalence in condition 1.

Assume now that is abelian.

If is pure, then . But , so consists of scalars, and therefore (I don’t agree with this — however, if is pure, then is irreducible so that has no invariant subspace of . We want to prove that — i.e. that is one dimensional. Suppose on the contrary that is *not *one dimensional — in which case it has a two dimensional subspace with basis vectors . However as all of the are scalars and this is a contradiction as acts irreducibly). Hence, . Therefore, if they are scalars and (using the fact that is a unit vector and looking at ) it is clear and therefore a character on .

Now suppose conversely that is a character on , and let be a positive linear functional on such that . If , then , so . Since , therefore (via an application of Cauchy-Schwarz). Hence, , and it follows from elementary linear algebra that there is a scalar such that

Choose a . Now for all , we have

.

Now we can write

;

that is with . Suppose . Now choose an . Now any is of the form , with and as above:

.

If is non-zero we therefore have that . Now

.

In other words, .

Choose such that . Then , so , and therefore . This shows that is pure and the equivalence in condition 2. is proved

It follows from Theorem 5.1.6 that for an arbitrary abelian C*-algebra , . The only thing not obvious is that a character on must have norm . To see this, let be an approximate unit for . Then is also an approximate unit. Hence

by Theorem 3.3.3, so , and therefore .

## Theorem 5.1.7

*Let be a representation of a C*-algebra , and let be a unit cyclic vector for . Then the function*

,

*is a state of and is unitarily equivalent to . Moreover, if is irreducible, then is pure.*

*Proof *: Clearly is a positive linear functional on (this follows by defining a sesquilinear form and the fact that *-homomorphisms are positive as .). If is an approximate unit for , then because is non-degenerate (as it admits a cyclic vector) the net is strongly convergent to (). Hence

,

s0 . For all ,

,

so and are unitarily equivalent by Theorem 5.1.4.

If is irreducible, so is , so by Theorem 5.1.6, is pure

### Example

Let be a non-zero Hilbert space, and let . We are going to determine the pure states of . If , then the functional

, ,

is positive (), and if is a unit vector, is a state (it is straightforward to show that and if we choose then we have equality.).

The pure states of are precisely the states where is a unit vector of .

To prove this, suppose first that is a unit vector of , and let be the inclusion map. The representation is irreducible, since (this is a result from Section 4.4 which I did not cover — seems perfectly reasonable). Hence, is a cyclic vector for ( is dense in ? Let be a basis of . As is a unit vector, there exists an such that . Hence define a family of compact linear maps

Then has the property that is dense in so that is cylic.), and it follows from Theorem 5.1.7 that the representations and are unitarily equivalent and is pure.

Now suppose conversely that is a pure state of . By Theorem 4.2.1, there is a trace-class operator on such that for all . For any unit vector , the operator is a projection and therefore positive, so

.

This shows that the operator is positive. Since is a compact normal operator (), it is diagonalisable by Theorem 2.4.4; that is, there is an orthonormal basis and there is a family of scalars such that . Choose . If ,

.

Thus the pure state majorises the positive linear functional , so there exists such that . Since both and are of norm one, , so ; that is is of the required form

An interesting consequence of our characterisation of the pure states of is that every non-zero irreducible representation is unitarily equivalent to the identity representation of . To see this, let be a unit vector in . The function

, ,

is a pure state on , and is unitarily equivalent to by Theorem 5.1.7. Hence, there exists a unit vector such that . Thus, is unitarily equivalent to , and we have already seen above that is unitarily equivalent to .

## Theorem 5.1.8

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

*Proof *: It is easy to check that is weak* closed in the closed unit ball of (, 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 (). Convexity of is clear ().

Let be the set of extreme points of .

First we show . Suppose that , where and . If , then

Therefore on , and therefore on (by Theorem 3.3.2), so .

Next we show that . Suppose that is a pure state of , and that , where and the . Then is majorised by , so there exists such that , because is pure. Since (with equality in the triangle inequality as they’re all linearly dependent)

,

we have . It follows that

,

so . Hence, , so .

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

,

and , we have , because . If is a non-zero positive linear functional on majorised by, and not equal to, then for , we have

,

since . Hence, , since is an extreme point of . Therefore, . This proves that is a pure state on

## Corollary 5.1.9

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

*Proof *: Apply the Krein-Milman theorem

## Corollary 5.1.10

*If is a unital C*-algebra, then is the weak* closed convex hull of the pure states of .*

*Proof *: The set 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 is a face of , where is as in Theorem 5.1.8, so by that theorem the extreme points of are the pure states of

## Theorem 5.1.11

*Let be a positive element of a non-zero C*-algebra . Then there is a pure state of such that .*

*Proof *: We may suppose that . The function

, ,

is weak* continuous and linear, and by Theorem 3.3.6 , where is the weak* compact convex set of all norm-decreasing positive linear functionals on . The set is a weak* compact face of by Lemma A.13 (*Let be a non-empty convex compact set in a locally convex space and suppose that is a continuous linear functional on . Let be the supremum of all where ranges over . Then the set of all such such that is a compact face of *. — I think this is after identifying with the linear functional .), and therefore has an extreme point by the Krein-Milman theorem. Since is a face of , the functional is an extreme point of also. Now , since and . Therefore, is a pure state by Theorem 5.1.8

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

## Theorem 5.1.12

*Let be a C*-algebra, and . Then there is an irreducible representation of such that .*

*Proof *: By the preceding theorem, there is a pure state of such that . By Theorem 5.1.6, the representation is irreducible. Since

,

therefore

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

## Theorem 5.1.13

*Let be a C*-subalgebra of a C*-algebra , and let be a pure state on . Then there is a pure state on extending .*

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

## Theorem 5.1.14

*Let be a unital C*-algebra. Suppose that is a subset of such that if a hermitian element satisfies the condition for all , then necessarily . Then the weak* closed convex hull of is and the weak* closure of contains .*

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

Suppose that and we shall obtain a contradiction. Since clearly holds, there exists a such that . By Theorem 3.32 there is a weak* continuous linear functional and there is a real number such that

for all . There is an element such that . If , then

for all . Since () for all , the hypothesis implies that . Therefore, , so . But , a contradiction

If is a unit vector in a Hilbert space , we denote by the state

, .

## Theorem 5.1.15

*Let be a C*-algebra and suppose that is a family of representations of . Suppose also that is a pure state of such that *

.

*Then belongs to the weak* closure in of the set*

.

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

## 1 comment

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December 14, 2014 at 7:49 pm

momumomuin Theorem 5.1.1 I have several questions:

1. is complete. What we use?

2.therefore it converges strongly to How?

3. why?

Thank in advance.