Given two *-probability spaces and , we want to consider maps . For example, we can look homomorphisms, which preserve the *-algebra operations, and can also consider restricting to state-preserving maps satisfying . In algebraic probability theory, however, it is often necessary to include a continuity condition, leading to the idea of *normal* maps, which I look at in this post. In fact, as we will see, all *-homomorphisms between *commutative* probability spaces which preserve the state are normal, so this concept is most important in the noncommutative setting.

In contrast to the previous few posts on algebraic probability, the current post is a bit of a gear-change. We are still concerned with with the basic concepts of *-algebras and states. However, the main theorem stated below, which reduces to the Radon-Nikodym theorem in the commutative case, is deeper and much more difficult to prove than the relatively simple results with which I have been concerned with so far.

The weak topology on a *-probability space is the weakest topology making the maps

continuous, for all . This definition means that a net converges to a limit in the weak topology if and only if for all . In fact, the weak topology on the whole of can be a bit too weak, so we will restrict to bounded sets. The seminorm is defined on any *-probability space , and elements are said to be bounded if is finite. The unit ball in with respect to this seminorm is,

I will only define normal maps for *bounded* *-probability spaces and, although the notion of normality can be extended to more general linear maps, I concentrate on homomorphisms in this post. We say that is bounded if every is bounded.

Definition 1Let and be bounded *-probability spaces. Then, a *-homomorphism isnormaliff it is weakly continuous on .

We previously showed that, in many cases, homomorphisms between *-probability spaces are -continuous. This also holds for normal maps.

Lemma 2Let and be bounded *-probability spaces and be a normal *-homomorphism. Then, is -isometric, so that for all . In particular,

(1)

*Proof:* By lemma 2 of the post on homomorphisms, it is sufficient to show that (1) holds. Suppose, on the contrary, that this is false. Then, there exists such that , but . So,

as . So, in the topology and, hence, also in the weak topology. Then, there exists such that and . As

is increasing in ,

does not tend to zero, contradicting weak continuity. ⬜

A consequence of this is that compositions of normal maps are normal. That is, if and are normal homomorphisms and, respectively, , then the composition between and is also normal. This follows from the fact that compositions of continuous maps are continuous.

Corollary 3Compositions of normal homomorphisms are normal.

Normal homomorphisms have various characterisations. For example, we will show the following result (see theorem 8 below), which is related to the existence of conditional expectations in classical probability theory.

Theorem 4Let and be bounded *-probability spaces. Then, a *-homomorphism is normal if and only if, for every , there exists sequences satisfying

(2) such that

(3)

for all .

Note that condition (2) is sufficient for the sum (3) to be absolutely convergent.

Before going into detail on normal homomorphisms, we note that the concept applies more generally than just for *-probability spaces. In fact, normal maps are usually studied in the context of von Neumann algebras. In order to include such generalisations, I will not restrict solely to *-probability spaces below. As in the previous post, we call a pair a **-algebra representation*, where is a *-algebra, is a semi-inner product space, and acts on by left multiplication. The operator seminorm on is denoted by , and the representation will be called bounded if is finite for all . As previously, denotes the sequences in such that is finite, which has the semi-inner product . An element acts diagonally on by , so that is a *-algebra representation.

A *-probability space defines a *-algebra representation with semi-inner product and with acting on itself by left-multiplication.

#### The Main Theorem

The following result, applying to linear functionals on , is the main result of this post. A linear map is called *normal* if it satisfies the equivalent conditions of theorem 5. This relates the weak, strong, ultraweak and ultrastrong operator topologies.

Theorem 5Let be a bounded *-algebra representation, and be a linear function. Then, the following are equivalent.

- is weakly continuous on .
- is strongly continuous on .
- is ultraweakly continuous.
- is ultrastrongly continuous.
- There exists such that for all .

*Proof:* The implications *1 ⇒ 2* and *3 ⇒ 4* follow immediately from the fact that the strong and ultrastrong topologies are stronger than the weak and ultraweak topologies respectively. The implications *3 ⇒ 1* and *4 ⇒ 1* follow from lemma 7 of the operator topology post. The implication *5 ⇒ 3* is immediate from the definition of the ultraweak topology.

The result now follows once we establish the implication *2 ⇒ 5*. However, this is by far the most involved part of the theorem, so the proof is left until later. ⬜

It is important to note that, for a bounded *-probability space, the state is always normal. This is an immediate consequence of lemma 10 of the operator topology post.

Lemma 6Let be a bounded *-probability space. Then, is normal.

For classical probability spaces, theorem 5 is effectively the Radon-Nikodym theorem.

Lemma 7Let be a probability space. With respect to the *-probability space , the following are equivalent for any linear map ,

- is continuous in probability on the unit ball of .
- for some .
- is normal.

*Proof:* The result follows from theorem 5 since, by lemma 12 of the operator topology post, the first statement of the lemma is equivalent to statement 2 of the theorem the second is equivalent to statement 5 of the theorem. ⬜

#### Normal Homomorphisms

Applying theorem 5 to homomorphisms between algebras gives the result below, and any *-homomorphism satisfying the equivalent conditions is called *normal*. I note that theorem 4 is a consequence of the equivalence between statements 1 and 5.

Theorem 8Let and be bounded *-algebra representations, and be a *-homomorphism. Then, the following are equivalent.

- is weakly continuous on .
- is strongly continuous on .
- is ultraweakly continuous.
- is ultrastrongly continuous.
- is normal for all normal linear maps .

Furthermore, if these conditions hold then is norm-bounded such that for all or, equivalently, .

*Proof:* *1 ⇒ 2:* If tends to strongly to zero, then tends weakly to zero, by lemma 2 on operator topologies. Hence, tends weakly to zero and, again by lemma 2, strongly.

*3 ⇒ 4:* If tends to zero ultrastrongly then ultraweakly, by lemma 5 on operator topologies. Hence, tends ultraweakly to zero and, again by lemma 5, ultrastrongly.

*3 ⇒ 1:* By lemma 9 on operator topologies, the weak and ultraweak topologies coincide on , so is continuous with respect to the weak topology on and the ultraweak topology on . As the ultraweak topology is stronger than the weak topology, this means that is continuous with respect to the weak topologies on and .

*4 ⇒ 2:* By lemma 9 on operator topologies, the strong and ultrastrong topologies coincide on , so is continuous with respect to the strong topology on and the ultrastrong topology on . As the ultrastrong topology is stronger than the strong topology, this means that is continuous with respect to the strong topologies on and .

*5 ⇒ 3:* Let be a net tending ultraweakly to zero. Then, for any normal linear map , is normal. Hence, tends to zero, showing that ultraweakly.

*2 ⇒ 5:* Let us first show that . If not, then there exists with and . Then, as . In particular, strongly. Setting , choose with and . Also, define by . This is a positive linear map satisfying . Hence, by lemma 7 of the post on states, is increasing in . So,

contradicting the condition that strongly. Hence, as required.

Now, suppose that is a normal linear map. Then, is strongly continuous on and, hence, is strongly continuous on , so is normal. ⬜

Recall that a homomorphism of *-probability spaces and is a *-homomorphism preserving the state, . Any such map is automatically an -isometry,

For commutative *-algebras, this is sufficient to imply that that it is normal. More generally, it is enough to know that is tracial so that for all .

Lemma 9Let be a homomorphism between bounded *-probability spaces and . If is commutative or, more generally, if is tracial, then is normal.

*Proof:* As explained above, we know that is -continuous. However, the strong topology coincides with ultrastrong on , which is stronger then convergence and, as is tracial, convergence is stronger than strong convergence on . So, is strongly continuous on and, by theorem 8, is normal. ⬜

For noncommutative algebras, and non-tracial states, homomorphisms of *-probability spaces need not be normal. The following result can be useful, in some situations, to ensure normality.

Lemma 10Let and be bounded *-probability spaces and be a *-homomorphism. Each of the following statements implies the next one:

- is ultrastrongly dense in ,
- is -dense in ,
and, if preserves the state,

- is normal.

*Proof:* *1 ⇒ 2:* By lemma 11 on operator topologies, the ultrastrong topology is stronger than the topology, so is -dense in .

*2 ⇒ 3:* For any then, if preserves the state,

As is -dense in ,

for all and, hence, .

Now suppose that tends strongly zero. Then, for any and ,

so, taking the limit,

As is -dense in , can be chosen to make the right-hand-side as small as we like and, hence, strongly. So, is normal. ⬜

As an example of theorem 8 applied to classical probability spaces, it immediately implies the existence of conditional expectations.

Corollary 11Let be a probability space and be a sub-sigma-algebra of . Then, for any , there exists satisfying for all .

*Proof:* Let , , and be the inclusion map. As is commutative, lemma 9 says that is normal. Now, for any , lemma 12 of the operator topology post applies, so that is a normal map on and, by theorem 8 above, is normal. Applying lemma 12 on operator topologies another time, there exists satisfying

for all . ⬜

Finally, I note that homomorphisms of *-probability spaces are not always normal. Example 2 of the post on homomorphisms gives an example of a homomorphism which is not -continuous, so cannot be normal. However, even if is -continuous, it need not be normal.

Example 1An -continuous homomorphism between bounded *-probability spaces and which is not normal.

In this example, let be the Hilbert space of square-integrable functions with inner product

and let be the *-algebra of bounded linear maps . If , then we define the state

on . For bounded measurable , define to be multiplication by , . Let be the subalgebra of consisting of the elements for continuous functions which are constant on each of the intervals and . We then let be the restriction of to , and be inclusion. This is clearly a homomorphism between the *-probability spaces and .

The -norm of with respect to is the supremum of and, with respect to it is the supremum on the unit interval. As is constant on the intervals and , these are the same, so is -isometric,

Next, consider for , for , and for . Then, and we compute,

As is commutative, this says that strongly in . However, the unitary defined by satisfies , where is equal to on the unit interval. Hence,

for all , showing that does not tend weakly or strongly to zero in . So, is not normal.

#### Proof of Theorem 5

Throughout this section, I take to be a bounded *-algebra representation, and use to denote the operator norm of . For any any , define the linear map

(4) |

Denote the space of all such linear maps by . The remainder of this post is devoted to the following result, which will complete the proof of theorem 5.

Theorem 12If is a linear map and is strongly continuous on , then .

This statement is, by far, the most mathematically involved result of this post, so we prove it step-by-step. It will also require a couple of powerful, but standard, theorems of operator algebras and functional analysis. Namely, we will rely on the *Krein–Smulian* and *Kaplansky density* theorems. As the Krein–Smulian theorem gives a condition for a linear functional on the *dual* of a Banach space to be weakly continuous, the method we will use to prove theorem 4 will start by showing that is a Banach space and that the relevant completion of is its Banach dual. The completion that we use will be the von Neumann algebra generated by , although I will not make use of the theory of von Neumann algebras here.

For now, note that the inequality

shows that every is bounded, with operator norm . As elements of can be expressed in the form in many different ways, define a norm by taking the infimum over all such representations,

In particular, for and , the following inequality holds,

(5) |

As yet, we have not even shown that is closed under linear combinations, so is a vector space, nor have we shown that is a norm. In fact, not only are these statements true, but is a Banach space.

Lemma 13is a Banach space under the norm .

*Proof:* We need to show that is a vector space, and that is a norm with respect to which is complete.

Is is straightforward that is positive definite since, if , then (5) gives . Now suppose that and . Then, for some . So, is in and,

Taking the infimum over all such and gives .

Next, consider a sequence with finite. Using the fact that

we can define by the absolutely convergent sum

We show that is in .

Fixing , choose such that and . By scaling we can assume, without loss of generality, that . Decomposing and then,

Letting be a bijection from to then,

where we have defined by and . So, and,

So,

As is arbitrary, we obtain

(6) |

In particular, for any , taking for in the argument above shows that is in with . Hence, is a vector space and is a norm.

It only remains to prove completeness. So, consider a Cauchy sequence . By (5), is Cauchy for any and, hence, we can define by . Choosing a subsequence satisfying whenever , by the argument above and inequality (6), is in with

Letting increase to infinity,

shows that is norm convergent to . ⬜

The following is a standard result for a *-algebra acting on a *Hilbert* space.

Lemma 14Let be a bounded *-algebra representation, with a Hilbert space. Then, for a linear map , the following are equivalent:

- is weakly continuous.
- is strongly continuous.
- There exists finite sequences () such that

(7)

*Proof:* *1 ⇒ 2:* This is immediate, as the strong convergence is stronger than weak.

*2 ⇒ 3:* As is a strong neighbourhood of 0, there exist a finite sequence in such that every satisfying for all is in . Define in . Letting act diagonally on , so that for , we see that whenever . So, .

Letting , then this shows that for a bounded linear map . As is a Hilbert space, the Riesz representation theorem states that there is an in the closure of such that , which is equivalent to (7).

*3 ⇒ 1:* By definition, the maps are weakly continuous, so is weakly continuous. ⬜

Locally convex topologies on a vector space which generate the same collection of linear functionals are called *compatible*. It is known that topologies are compatible if and only if they possess the same collection of closed convex sets, which is a consequence of the fact that the closure of a convex set can be described in terms of the continuous linear functionals.

Lemma 15Let be a locally convex topological vector space over and be convex. Then, is in the closure of if and only if is in the closure of for all continuous linear .

*Proof:* Let denote the closure of . First if then, for each continuous functional , is in the closure of by continuity. Conversely, suppose that . By the Hahn–Banach theorem, there exists a continuous functional and satisfying for all . Hence, is not in the closure of . ⬜

I will use for the closure of a set under, respectively, the weak, strong, ultraweak and ultrastrong topologies.

Corollary 16If is convex then .

*Proof:* Lemma 14 states that a linear functional on is weakly continuous if and only if it is strongly continuous. So, the result follows from lemma 15. ⬜

Due to lemma 14, things are made easier if we work with Hilbert spaces rather than general semi-inner product spaces. So, for a bounded *-representation , consider the completion of ,

That is, is a Hilbert space and is an isometry with dense image. By continuous linear extension, every has a unique bounded linear action on satisfying

for all , and this action has the same operator norm as the action on . We now have two different *-algebra representations, and . The construction of given above can be performed with respect to either of these, although they give the same result. I also use the notation (4) for .

Lemma 17A function is in iff for some . Then, .

*Proof:* If then, by definition, for some . Then, and are in , and as required.

Conversely, suppose that for . As has dense image, there exists sequences such that and . Then . Writing

we obtain,

as . Lemma 13 says that is a Banach space so that is in and,

as required. ⬜

The `ultra’ topologies remain unchanged when we pass to the Hilbert space completion.

Lemma 18The ultraweak and, respectively, the ultrastrong topology on are the same whether defined with respect to or .

*Proof:* Let be the natural homomorphism. The ultraweak topologies are generated by the linear functions for, respectively, and . Lemma 17 implies that these two collections of functions are the same, so the ultraweak topologies coincide.

A net tends to zero in the ultrastrong topology iff in the -ultraweak topology. By what we have just shown, this is equivalent to weakly in . Hence weakly wrt . Then, in the ultrastrong topology on . ⬜

Lemma 14 can also be applied the ultraweak and ultrastrong topologies on .

Lemma 19If is a linear map then, the following are equivalent.

- is ultraweakly continuous.
- is ultrastrongly continuous.
- .

*Proof:* By lemmas 17 and 18, each of the statements of the lemma correspond to the statements of lemma 14 applied to the representation . ⬜

Corollary 20If is convex then .

*Proof:* Lemma 19 states that a linear functional on is ultraweakly continuous if and only if it is ultrastrongly continuous. So, the result follows from lemma 15. ⬜

Next, as the elements of act as bounded linear operators on the Hilbert space , this defines a unique *-homomorphism from to the *-algebra of bounded linear operators on , given by . The image, , is a *-subalgebra of , the completion of which will be denoted by . It does not matter which of the operator topologies is used, as the weak, strong, ultraweak and ultrastrong completions all give the same result, which is the von Neumann algebra generated by . We only require the ultraweak and ultrastrong topologies, and do not even make use of the fact that is an algebra.

Theorem 21Under the pairing

is the Banach dual of .

*Proof:* It needs to be shown that, for every , the linear map given by is bounded with operator norm equal to and, conversely, for every bounded linear map there exists a unique satisfying .

First, for ,

so we see that is bounded with operator norm .

Now, let be a bounded linear map. To complete the proof, we need to show that there exists a unique such that for all , and that .

Fixing , consider the linear map given by (where the overline denotes complex conjugation). This satisfies

so has operator norm bounded by . By the Riesz representation theorem, there exists a unique satisfying

for all , and then . As depends linearly on the choice of , we write , so that is linear with and,

As , it remains to show that and that for all .

For any , we use that fact that converges in the norm so that,

Finally, for all ultraweakly continuous linear , lemma 19 says that for some . If vanishes on then . If does not vanish on then, by scaling, it maps onto , so . Hence, by lemma 15, . ⬜

I now state the Kaplansky density theorem, which is the first of the two `big’ results that we need. This states that if is a Hilbert space, is a *-subalgebra of , and is both in the strong closure of and in the unit ball , then it is in the strong closure of . By scaling, this means that if is in the strong closure of , so that there is a net tending strongly to , then the net can be chosen such that for all .

Theorem 22 (Kaplansky)Let be a Hilbert space and be a *-subalgebra of . If is in the strong closure of , then it is in the strong closure of .

The usual proofs of the Kaplansky density theorem make use of the continuous functional calculus to replace a net converging strongly to by , for a continuous bounded function . I simply state this theorem here, with reference to the Wikipedia entry, as it is standard. As in equation (6) from the post on operator topologies, I express the theorem in the form of a string of useful identities. As usual, is a bounded *-algebra representation.

Lemma 23If is a *-subalgebra of then

*Proof:* The first equality is given by corollary 20. Setting , which is a convex subset of , the equalities and hold as the weak and strong topologies coincides with their ultra counterparts on the unit ball (lemma 9 of the post on operator topologies). Corollary 20 above gives . Only the equality remains.

As above, let be the Hilbert space completion of , and define the *-homomorphism by . The Kaplansky density theorem applied to the action of on gives,

As is an -isometry, and the ultrastrong topology for defined with respect to its action on and for its action on coincide (lemma 18), this gives as required. ⬜

The next big result that we require is the Krein–Smulian theorem (see here for a proof). This is pure functional analysis, involving Banach spaces. Given a Banach space , its dual is defined to be the space of continuous linear maps which, under the operator norm, is itself a Banach space. If, for and , we use to denote the application of to , then is bilinear. The weak-* topology on is defined to be the weakest topology making the maps , , continuous for each . The Krein–Smulian theorem shows that a convex set is weak-* closed if and only if is weak-* closed for all . In the case where is a subspace then, by scaling, it is sufficient to show that is weak-* closed.

Theorem 24 (Krein–Smulian)Let be a Banach space with dual . Then, a convex subset is weak-* closed iff is weak-* closed for each . In particular, if is a subspace, then in it weak-* closed iff is weak-* closed.

In our case, we showed in lemma 13 that is a Banach space and then, in theorem 21, that its Banach dual is the closure of in . The weak-* topology then coincides with the ultraweak topology, so the Krein–Smulian theorem gives a necessary and sufficient condition for convex subsets of to be ultraweakly closed. I conclude this post with the proof of theorem 12.

*Proof of Theorem 12:* Let be linear, and strongly continuous on . It needs to be shown that or, by lemma 19, that is ultraweakly continuous. As above, we define to be the Hilbert space completion, and the *-homomorphism by . Setting , which is a *-subalgebra of , we will consider the linear map given by . To see that this is well-defined, consider the case where . Then, for all , meaning that is in the strong closure of the single point so, by strong continuity, as required. As the ultrastrong topology is the same for the action of on as for its action on (lemma 18), it follows that is ultrastrongly continuous on norm-bounded subsets of .

Now, let be the ultraweak closure of . Using to denote the closed -ball, for each positive real , lemma 23 gives

As is ultrastrongly continuous on , it has a unique ultrastrong extension to . By uniqueness, the extensions to and must agree on the intersection. As , we can define a map such that it agrees with on and is ultrastrongly continuous on each .

In summary, we have defined a linear map which is ultrastrongly continuous on and satisfies . As is ultraweakly continuous, it only remains to show that is ultraweakly continuous. Equivalently, is ultraweakly closed, for which it is sufficient to show that is closed under the ultraweak topology. Since theorem 21 states that is the Banach dual of , and the weak-* topology corresponds to ultraweak convergence, the Krein–Smulian theorem says that it is sufficient to show that is ultraweakly closed. The fact that is ultrastrongly continuous on immediately gives that is ultrastrongly closed, and finally is ultraweakly closed by corollary 20 above.

⬜

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