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

If and are vector spaces, we denote by their algebraic tensor product. This is linearly spanned by the elements (, ).

One reason why tensor products are useful is that they turn bilinear maps (a bilinear map has ) into linear maps (). More precisely, if is a bilinear map, where and are vector spaces, then there is a unique linear map such that for all and .

If are linear functionals on the vector spaces respectively, then there is a unique linear functional on such that

since the function

, ,

is bilinear.

Suppose that the finite sum , where and . If are linearly independent, then . For, in this case, there exist linear functionals such that . If is linear, we have

.

Thus for arbitrary and this shows that all the .

Similarly if the finite sum with the linearly independent, implies that all the are zero.

If and are linear maps between vector spaces, then by elementary linear algebra, there exists a unique linear map

such that for all and . The map is bilinear.

If and are normed, then there are in general many possible norms of which are related in a suitable manner to the norms on and , and indeed it is this very lack of uniqueness that creates the difficulties of the theory, as we shall see in the case that and are C*-algebras.

When the spaces are Hilbert spaces, however, matters are simple.

## Theorem 6.3.1

*Let and be Hilbert spaces. Then there is a unique inner product on such that*

### Proof

If * *are conjugate-linear maps from and , respectively, to , then there is a unique conjugate-linear map from to such that

.

(Observe that and are linear, and set .)

If is an element of a Hilbert space, let be the conjugate-linear functional defined by setting .

Let be the vector space of all conjugate-linear functionals on . The map

, ,

is bilinear (), so there is a unique linear map such that

.

The map

, ,

is a sesquilinear form on such that

That it is the unique such sesquilinear form is clear ().

If , then (finite sum always). Let be an orthonormal basis for . Then for some () and therefore,

.

Thus is positive and if , then the and . Therefore is an inner product

If and are as in Theorem 6.3.1, we shall always regard as a pre-Hilbert space with the above inner product. The Hilbert space completion of is denoted by , and called the *Hilbert space tensor product of *and . Note that ()

It is an elementary exercise to show that if and are orthonormal bases for and , respectively, then is an orthonormal basis for ().

If are closed vector subspaces of , respectively, then the inclusion map is isometric when has its canonical inner product. It follows that we may regard as a closed vector subspace .

## Lemma 6.3.2

*Let be Hilbert spaces and suppose that and . Then there is a unique such that*

.

*Moreover, .*

### Proof

The map is bilinear, so to show that is bounded, we may assume that and are unitaries, since the unitaries span the C*-algebras and . If , then we may write where the are orthogonal. Hence

.

Consequently, .

Thus, for all operators on , respectively, the linear map is bounded on and hence has an (isometric) extension to a bounded linear map on .

It is easily verified that the maps

, ,

and

, ,

are injective -homomorphisms and therefore isometric (). Hence, and , so

.

If is a sufficiently small positive number, and if , then there are unit vectors such that

and .

Hence,

,

,

so . Letting , we get

### Remark

Let and be Hilbert spaces and suppose that and . It is routine to show that

and

.

(I’m happy with the second and the first is done below I think.)

If are operators on and are linearly independent operators on such that (the finite sum) , then all of the are zero. For if , choose orthonormal vectors such that

.

Then there are scalars (which aren’t chosen uniquely — this representation is not unique.) such that for any of the . If , we have

,

so . This shows that , and therefore, by linear dependence of the , we get that the for all . It follows that so all of the are zero.

If and are algebras, there is a unique multiplication on such that

.

for all and : Let denote left multiplication by and let be the vector space of all linear maps on . If and , then , and the map

, ,

is bilinear. Hence, there is a unique map such that for all and . The bilinear map

, ,

is readily seen to be the required unique multiplication on . We call endowed with this multiplication the *algebra tensor product *of the algebras and .

If and are *-algebras, then there is a unique involution on such that for all and . The existence of such an involution is easily seen if we show that

.

If are *-algebras, then we want to make into a *-algebra. We want to define by

.

If then for some reindexing, then if , then we get that which implies that .

Suppose, therefore that , and assume without loss of generality that the are linearly independent. Now if , this implies that which means that as required.

We call with the above involution the **-algebra tensor product *of and .

If are *-subalgebras of , respectively, we may clearly regard as a *-subalgebra of .

### Remark 6.3.2

Let and be *-homomorphisms from *-algebras and into a *-algebra such that every element of commutes with every element of . Then there is a unique *-homomorphism such that

.

This follows from the observation that the map

, ,

is bilinear and so induces a linear map , which is easily seen to have the required properties (The commutativity is needed to show that ).

If are *-algebras and and are *-homomorphisms, then is a *-homomorphism. The proof is a routine exercise ().

We shall use the next result to show that there is at least one C*-norm on if and are C*-algebras.

## Theorem 6.3.3

*Suppose that and are representations of the C*-algebras and respectively. Then there exists a unique *-homomorphism such that*

.

*Moreover, if and are injective, so is .*

### Proof

The maps

, ,

and

, ,

are *-homomorphisms, and the elements of commute with the elements of (). Hence, by the previous remark, there is a unique *-homomorphism such that .

Now suppose that the representations and are faithful (that is and are injective), and let . We can write in the (finite sum) form , where the are linearly independent. Then the are linearly independent, because is injective and (). Hence, all the are zero. Since is injective, all the are zero also, so . Thus

Let and be C*-algebras with universal representations and respectively. By Theorem 6.3.3 there is a unique injective *-homomorphism such that for all and . Hence the function

, ,

is a C*-norm on , called the *spatial *C*-norm. Note that ( is an injective *-homomorphism given by . Now

,

since are injective C*-isomorphisms.).

We call the C*-completion of with respect to the *spatial tensor product *of and , and denote it by .

In general, there may be more than one C*-norm on . If is a C*-norm on , we denote the C*-completion of with respect to with respect to by .

## Lemma 6.3.4

*Let be C*-algebras and let be a C*-norm on . Then for each and the maps*

, ,

and

, ,

are continuous.

### Proof

Since is a linear map between Banach spaces, we may invoke the closed graph theorem (the domain of so we have the ‘converse’ ). Thus, to show that is continuous we need only to show that if a sequence converges to in and the sequence converges to , then (also continuity at will imply continuity as is linear). We may suppose that and are positive (by replacing by and by if necessary). Hence (). If is a positive linear functional on , then the linear functional

, ,

is positive, hence continuous. Consequently, , since . Since is an arbitrary positive linear functional on , . Therefore, is continuous. Similarly, is continuous

## Theorem 6.3.5

*Let be non-zero C*-algebras and suppose that is a C*-norm on . Let be a non-degenerate representation (, : — *basically everything isn’t sent to zero*) of . Then there exist unique *-homomorphisms and such that*

.

*Moreover, the representations and are non-degenerate.*

### Proof

Let . If , it can be written in the form

.

Suppose we have two such expressions; that is, can also be written as

,

where , and . If is an approximate unit for and , then

,

so in the limit (using Lemma 6.3.4)

.

We can therefore well-define a map by setting , if is in the ‘‘ form (). Since , it is clear that is linear (). By Lemma 6.3.4, there exists a positive number (depending on ) such that , so . Hence, is bounded. Since is dense in (because is non-degenerate), we can therefore extend uniquely to a bounded linear map on , also denoted by .

Suppose now that is an approximate unit for . Reasoning as before, for all , we have that , if is in that representation. We can therefore well-define a map by setting by . The linear map is bounded and extends uniquely to a bounded linear map on , also denoted by .

A routine verification shows that the maps

,

and

, ,

are *-homomorphisms (), and that .

Now suppose that and are another pair of *-homomorphisms such that . Suppose that is such that . Then fir all and , and therefore for all . By non-degeneracy of , we have . Thus, is non-degenerate. Similarly, is non-degenerate.

In particular, and are degenerate.

If and are approximate units as above, then the nets and converge strongly to (by non-degeneracy of and ). Hence, converges strongly to both and for all , hence . Similarly, converges strongly to both and for all so

We denote the maps and in the preceding theorem by and , respectively.

## Corollary 6.3.6

*Let and be C*-algebras and let be a C*-seminorm on . Then*

.

### Proof

Let , so is a C*-norm on . Let be the universal representation of . This is faithful and non-degenerate, so Theorem 6.3.5 applies. If and , then .

Hence

,

so

Let and be C*-algebras. Denote by the set of all C*-norms on . We define for each . If , then for any we have

by Corollary 6.3.6. Hence, .

It is readily verified that

, ,

is a C*-norm, called the *maximal *C*-norm (). We denote by the C*-completion of under this norm, and call the *maximal tensor product *of and .

If is a C*-seminorm, then . The maximal tensor product has a very useful universal property.

## Theorem 6.3.7

*Let and be C*-algebras and suppose that and are *-homomorphisms such that every element of commutes with every element of . Then there is a unique *-homomorphism such that*

.

### Proof

Uniqueness is clear (). By Remark 6.3.2 there is a *-homomorphism satisfying the equation in the statement of the theorem. The function

, ,

is a C*-seminorm. Hence for all . Therefore, is a norm-decreasing *-homomorphism, and so extends to a norm-decreasing *-homomorphism on

We say that a C*-algebra is *nuclear *if, for each C*-algebra , there is only one C*-norm on .

### Remark

If a *-algebra admits a complete C*-norm , then it is the only C*-norm on . For if is another C*-norm on , and denotes the C*-completion of with respect to , then the inclusion is an injective *-homomorphism and therefore isometric, so .

### Example

For each , the C*-algebra is nuclear. The reason is that for each C*-algebra , the *-algebra admits a complete C*-norm. This is seen by showing that the unique linear map , such that for and , is a *-isomorphism (all fine except . I had calculated this directly but there is a slicker way

Fix and . Now the linear maps and are linear maps and hence there exists a unique *linear *map .).

We are going to show that all finite-dimensional C*-algebras are nuclear and for this we shall need to determine the structure theory for such algebras.

## Theorem 6.3.8

*If is a non-zero finite-dimensional C*-algebra, it is *-isomorphic to .*

### Proof

If is simple, the result is immediate from Remark 6.2.1 (have glossed over this section). We prove the general result by induction on the dimension of . The case is obvious. Suppose the result holds for all dimensions less than . We may suppose that is not simple, and so contains a non-zero proper closed ideal , and we may take to be of minimum dimension. In this case, has no non-trivial ideals, so is *-isomorphic to for some . Hence, has a unit , so and commutes with all the elements of (). Also is a C*-subalgebra of () and the map

, ,

is a *-isomorphism. Since the algebra has dimension less than ,it is *-isomorphic to . Hence, by the Axiom of Induction, is *-isomorphic to

## Theorem 6.3.9

*A finite-dimensional C*-algebra is nuclear.*

### Proof

Let be a finite-dimensional C*-algebra, which we may suppose to be the direct sum . Let be an arbitrary C*-algebra. A routine verification shows that the unique linear map

,

such that

for all and , is a *-isomorphism. Hence, admits a complete C*-norm (), so it admits only one C*-norm. This shows that is nuclear

The next result suggests that nuclear algebras exist in abundance.

## Theorem 6.3.10

*Let be a non-empty set of C*-subalgebras of a C*-algebra which is upwards directed (if , then there exists a such that ). Suppose that is dense in and that all the algebras are nuclear. Then is nuclear.*

### Proof

Let be an arbitrary C*-algebra and suppose are C*-norms on . Set (we may regard as a *-subalgebra of for each ). Then is a *-subalgebra of and it is clear that is dense in the C*-algebras and . Now on for each , by nuclearity of , so on , and therefore the identity map on extends to a *-isomorphism . If and , then there exists a sequence in such that in , so in and in , and therefore

,

where convergence is with respect to . Therefore, . Hence for any , we have , so on . Therefore, the algebra is nuclear

### Example

If is a Hilbert space, is a nuclear C*-algebra. To see this, suppose that are orthonormal vectors in . If , then is a projection, , and the map

, \lambda|e_i\rangle\langle e_j|$,

is a *-isomorphism. We show surjectivity only: If , then

.

Since is finite-dimensional, it is nuclear (by Theorem 6.3.9).

If is an orthonormal basis for , let be the set of all finite non-empty subsets of made into an upwards-directed poset by setting if . For , let and . Each us therefore a nuclear C*-algebra, and the set is upwards-directed.

If is a finite-rank operator on , we can write for some . Therefore,

,

so , since for . From this it follows by norm-density of the finite-rank operations in that for all . Therefore . Hence, is dense in . By Theorem 6.3.10, is a nuclear .

An *AF-algebra *is a C*-algebra that contains an increasing sequence of finite-dimensional C*-subalgebras such that is dense in .

## Corollary 6.3.11

*All AF-algebras are nuclear.*

## Appendix: Non-negative Definite Kernels

Suppose that is a set. We say that is a non-negative kernel if for all , and , we have that

.

Let . Define

,

which is defined because of finite support. Then , is linear and

.

is a semi-inner product, so the Cauchy-Schwarz Inequality holds. Define . Now is a subspace by Cauchy-Schwarz, and now set

.

This defines an inner product on . Now complete to get a Hilbert space , together with a map , . Now since is a basis of , with

.

Moreover, if is any Hilbert space and a map such that m then consider the map . Since

,

there exists a unique isometry such that . In particular, if , then is unitary.

Now let , and define

.

is a non-negative definite kernel. Let and then

where is the Schur product. Moreover,

.

In other words, and — and the Schur product of positive matrices is positive (Theorem 12).

There exists a Hilbert space and a map such that if , then

,

and also .

Pick unitaries , . Define by

.

Then

.

Also , so total. Thus, by ‘uniqueness’ of , there exists a unique unitary on mapping — call this .

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September 15, 2011 at 4:32 pm

Random Walks on Finite Quantum Groups: Random Walks on Comodule Algebras « J.P. McCarthy: Math Page[…] dimensional C*-algebras are very concrete objects, namely they are multi-matrix algebras (see Theorem 6.3.8). Not every multi-matrix algebra carries a Hopf algebra structure. The simplest example is that of […]