# Question 1

*Let be normal elements of a C*-algebra , and an element of such that . Show that , using Fuglede’s theorem and the fact that the element*

*is normal in and commutes with*

*.*

*This more general result is called the Putnam-Fuglede theorem.*

## Solution

Fuglede’s theorem states that if is a normal element commuting with some , then also commutes with . Now we can show that using the normality of and . We can also show that and commute. Hence by the theorem and commute. This yields:

.

Taking conjugates:

,

as required

Question 2

*Let be a positive linear functional on .*

## (a)

*If is a closed ideal in , show that iff .*

### Solution

Suppose that is a (non-zero) closed ideal such that . Let so we have that . Consider now . Now

,

as as is an ideal. But as (as ). Hence .

Now suppose that so that .Then

,

.

Take limits to get

## (b)

*We say that is faithful if for all . Show that if is faithful, then the GNS representation is faithful.*

### Solution

To show that is faithful we must show that is injective. That is we must show that implies that . Suppose therefore that and that is a non-zero element of . Then we have that . If these are equal then we have that . That is :

.

But is faithful and is positive hence . Now by the C*-equation:

.

We can let an approximate unit play the role of so we have that

That is, and so as required

## (c)

*Suppose that is an automorphism of such that for all . Define a unitary on by setting . Show that .*

### Solution (Wills)

Consider first

Therefore has dense range in — because is an automorphism — and is an isometry so therefore it is a unitary operator.

To find the the adjoint we find a map such that . We can look at and extend to by continuity. Try :

,

and similarly .

Now let . It is now a simple calculation to show that the relation holds. Firstly . Now

# Question 3

*If is a positive linear map between C*-algebras, show that is necessarily bounded.*

## Solution

Just a rehashing of Theorem 3.3.1

## Question 4

*Suppose that is unital. Let be an automorphism of such that . Define to be the set of matrices*

*.*

*Show that is a C*-subalgebra of .*

## Solution

It is straightforward to show that is a vector subspace as it contains the zero matrix () and is closed under linear combinations (* *is linear). We need to show that it is closed under multiplication and under the involution.

Let and consider :

Now

,

(and vice versa on the other diagonal) as required.

Also

.

This is an element of with and .

That is, is a C*-subalgebra of

*Define a map by setting*

*.*

*Show that is an injective *-homomorphism.*

## Solution

To do this we must show that *, and if , then .*

Firstly,

*.*

Also

,

once we use the homomorphism property of *. *Finally the injectivity of * *is trivial

* *

*We can thus identify as a C*-subalgebra of . If we set*

*,*

*then is a self-adjoint unitary and . If is any unital C*-algebra with a self-adjoint unitary element , and is a *-homorphism such that*

*show that there is a unique *-homomorphism ** extending (in the sense that ) such that .*

### Solution (Wills)

Let . We want to choose such that on . Define a map by:

.

Linearity is easy to show and it’s clear that . To show that is a *-homomorphism we must show that it is multiplicative and respect the *-operation. Multiplicativity is a straightforward calculation where we use the fact that (the fact that is very useful here.) Similarly showing that the *-operation is respected it straightforward using this fact.

To show uniqueness, suppose that is another map with all these properties. Let . Then

.

So that . Now let and examine:

;

that is

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