### Topics in Operator Theory, Lecture 2: Dilations of contractions

#### by Orr Shalit

In this lecture we will study the first chapter in the theory of dilations of contractions. To proceed in our study of operator spaces and operator algebras, the material we will cover is not strictly needed. However, this is where I want to begin, for several reasons:

- The objects and theorems here motivate (and have motivated historically) the development of the general theory, and help understand it better and appreciate it more.
- We will reach very quickly a nontrivial application of operator theory to function theory, which is quite different from what you all saw in your studies, probably.
- I am stalling, so that the students who need to fill in some prerequisites (like commutative C*-algebras and the spectral theorem, will have time to do so).
- I love this stuff!

Okay, enough explaining, let us begin.

#### 1. Overview

**Definition: **An operator is said to be

if ,**selfadjoint****normal****unitary****isometric****an****isometry**) if (equivalently, if for all ,**coisometric****a****coisometry**) if (i.e., if is an isometry),**contractive****a****contraction**) if .

Normal (and hence, selfadjoint and unitary) operators are well understood. In this lecture our goal will be to understand first isometries (and hence also coisometries) and then contractions a little bit better. Before we begin with this, I will now repeat some material from the first lecture in my von Neumann algebras notes to remind ourselves of the nice structure theorem for normal operators.

The spectral theorem is the basic structure theorem for normal operators. It tells us how a general normal operator looks like. Recall that if is a normal operator acting on a finite dimensional space , then is unitarily equivalent to a diagonal operator, that is, there exists a unitary operator such that

,

where (some points in are possibly repeated).

Moreover, if is a compact normal operator on a Hilbert space, then unitarily equivalent to a diagonal operator (an infinite diagonal matrix, acting by multiplication on ), the diagonal of which corresponds to the eigenvalues of , which form a sequence converging to :

.

If is unitarily equivalent to a diagonal operator where the diagonal elements form a bounded sequence of numbers (not necessarily converging to ), then is a bounded normal operator (which is not necessarily compact). However, a general bounded normal operator need not be unitarily equivalent to a diagonal operator.

**Example: **The operator given by is a selfadjoint bounded operator, and it is an easy exercise to see that this operator has no eigenvalues (so it cannot be unitarily equivalent to a diagonal operator). However, the operator in this example is rather well understood, and it is “sort of” diagonal. The general case is not significantly more complicated than this.

To understand general normal operators, one needs to recall the notions of measure space and of spaces. Let be a measure space and consider the Hilbert space . Every defines a (normal) bounded operator on .

**Exercise A: **In case you never have, prove the following facts (or look them up; Kadison-Ringrose have a nice treatment relevant to our setting). Let be a -finite measure space and .

- (where is the
of , which is defined to be ).**essential supremum** - .
- If and defines a bounded operator on , then is essentially bounded: .
- If , then and .
- is selfadjoint if and only if is real valued almost everywhere.

The algebra is an abstract C*-algebra with the usual algebraic operations, the -operation , and norm . The map

is a -representation (i.e., and algebraic homomorphism that preserves the adjoint ), which is isometric (), so omitting we can think of as a C*-subalgebra of . Since , the operator is always normal. It is selfadjoint if and only is a.e. real valued, and it is unitary if and only if a.e., which happens if and only if it is isometric. The operator , where , is called a * multiplication operator. *Multiplication operators form a rich collection of examples of normal operators. The spectral theorem says that this collection exhausts all selfadjoint operators: every normal operator is unitarily equivalent to a multiplication operator.

**Theorem 1 (the spectral theorem):** *Let be a normal operator on a Hilbert space . Then is unitarily equivalent to a multiplication operator, that is, there exists a measure space , a unitary operator , and a complex valued , such that*

*. *

*When is separable, can be taken to be a locally compact Hausdorff space, and a regular Borel probability measure. *

The spectral theorem allows, in principle, to “solve all problems about normal operators”. Okay, that’s maybe an exaggeration; the spectral theorem reduces any problem about normal operators to problems about multiplication operators. As an example, we prove

**Proposition 2 (von Neumann’s inequality for normal operators): ***Let be a normal contraction. Then for any polynomial , *

.

**Proof:** Since unitary equivalence preserves everything, we may assume that . Now, is a contraction so a.e. So we find

,

and since a.e., we have a.e., so

,

as required. (The second equality simply follows from the maximum modulus principle, which, by the way, can be proved using the methods we will use in this lecture).

#### 2. Isometries

Unitary operators are isometric, that’s one basic example. An isometry that is not a unitary is called a ** proper isometry**.

**Example (the unilateral shift): **Let be a Hilbert space, suppose . Let be the Hilbert space consisting of square summable sequences with values in :

.

has an inner product

.

On we define **the****unilateral shift*** *(of multiplicity ) to be given by

.

A simple computation reveals and so

.

**Definition:** Let be an isometry. A subspace is called a **wandering subspace*** *for if

.

(Equivalently for all ). We define

(*) .

We have that , and this means that .

**Example: **Let be a subspace. Fix , and let be given by

and for all .

Then is a wandering subspace for . In particular,

is a wandering subspace.

**Definition: **An isometry is said to be ** a unilateral shift** (of multiplicity ) if it has a wandering subspace (with ) for such that .

Note that is determined uniquely by (*) as

(**) .

Alright: we have defined THE unilateral shift and the notion of A unilateral shift, what gives?

**Exercise A: **Let be a unilateral shift. Show that

- is unitarily equivalent to THE unilateral shift on , where is given by (**).
- A unilateral shift is unitarily equivalent to if and only if .
- (in other words, ).

**Definition: **Let . A subspace is said to be

if ,**invariant**if (equivalent, if is invariant for ).**coinvariant**if it is invariant for both and (equivalently, if and are both invariant for ).**reducing**

Sometimes, to clarify the operator with respect to which the subspace should satisfy the properties, the terminology **invariant*** for * (etc.) is used.

**Theorem 3 (Wold decomposition):** *Let be an isometry. Then there exists a decomposition of to two reducing subspaces such that is unitary on and is a unilateral shift. Moreover, this decomposition is determined uniquely by and , where . *

**Proof: **If we define , then is wandering for because for all we have by definition .

Let us define , and . Then is clearly invariant, and is a unilateral shift.

We will soon show that . Assuming that for the moment, we have that , so is invariant, and thus are reducing. Moreover, is a surjective isometry, so it is a unitary (showing that a surjective isometry is a unitary is a basic exercise).

To show that we take and note that if and only if for every . But

where we cancelled out all the terms in the above “telescopic sum”. Now, for all is the same as for all , and we have established that .

We leave the uniqueness as an exercise.

**Definition: **Let , and suppose that is a Hilbert space containing . An operator is said to be an ** extension **of if .

**Theorem 4 (unitary extension of an isometry):** *Every isometry has a unitary extension. *

**Proof: **Let be an isometry and let be its Wold decomposition on . By Exercise A, up to unitary equivalence, , the unilateral shift of some multiplicity on . Letting be the bilateral shift on , that is, is defined by

.

Clearly, is an extension of , and so is an extension of .

**Corollary: ***Von Neumann’s inequality holds for isometries. *

**Proof:** Let be an isometry and let be a unitary extension. From follows and so .

#### 3. Contractions

In the previous section we showed that every isometry has a unitary extension, and this immediately led to the application that isometries satisfy von Neumann’s inequality. Of course, we cannot hope to show that any operator other than an isometry has a unitary extension, so we need another trick.

**Definition:** Let , and suppose that is a Hilbert space containing . An operator is said to be an ** dilation **of if for all .

Sometimes, the term * dilation* refers to the relation , and then the above notion is called a

**power dilation***.*There are also situations when is only assumed for all for some , and then is said to be an

*of . It is usually not hard to understand from context what notion is the one intended.*

**N-dilation****Examples: **If is an extension of , then is a dilation of . Note that in this case

.

Likewise, if is a **co-extension**** **of (meaning that is a dilation of ) then it is a dilation, and in this case we have the picture:

,

or (simply changing basis)

.

Both of the above situations are a special case of the following picture

(#)

in which is easily seen to be a dilation of :

.

It turns out that (#) is the most general form of a dilation.

**Proposition 5 (Sarason’s Lemma):** *Let be a Hilbert spaces, and suppose that is a (power) dilation of . Then there exist two subspaces invariant for such that . *

**Remark: **The case where corresponds to the case where is invariant and is an extension of ; the case where correspond to the case where is co-invariant and is a co-extension of .

**Remark:** In the situation of the above proposition, we say that is a * semi-invariant subspace *for . More generally, if is an operator algebra, then a subspace is said to be

*if the map is a homomorphism. Clearly, is semi-invariant for if and only it is semi-invariant for .*

**semi-invariant for****Proof of Sarason’s Lemma: (maybe I should leave this as an exercise)**

Let us write as in the above remark, and let (this notation means the closed linear subspace spanned by where and ). Then is clearly invariant, and . For this to work out we really have no choice but to define . The only thing that remains to do is to prove that is invariant for .

If then the definition of semi-invariant means that , and so for every we have that . By taking sums and limits, we find that for all .

To show that is invariant, let , , and consider . Since , to show that it is in we need to show that it is orthogonal to ; equivalently, we need to show that . But by the previous paragraph , because . This completes the proof.

**Theorem 6 (Sz.-Nagy’s isometric dilation theorem):** *Let be a contraction on a Hilbert space . Then there exists a Hilbert space and an isometry such that *

*is a co-extension of .**is the smallest subspace of invariant for that contains .*

*Moreover, the pair is unique, in the sense that if is another such pair satisfying the above requirements, then there exists a unitary such that for all , and such that . *

**Remark:** The isometry (or more precisely, the pair ) is known as * the minimal isometric dilation of *. The theorem can be reformulated (and sometimes is) by saying that every contraction has a minimal coisometric extension.

**Proof: **Define the * defect operator* . Note that is equivalent to

**,**so the square root makes sense. The operator measures how much fails to be an isometry. We now define (the

*) and*

**defect****space**.

On we define an operator by

or, in matrix form

(all the empty slots are understood to be zero).

We leave it as an exercise for the reader to complete the proof.

**Exercise B:** Prove that is an isometry, a co-extension of , and that it satisfies the minimality requirement. While you are at it, please also show that is unique.

The uniqueness of the minimal dilation can be strengthened as follows.

**Exercise C: **Prove that if is an isometric co-extension of , the there exist two reducing subspaces , such that and , and such that is unitary equivalent to the minimal isometric dilation of . Thus, breaks up as that , where is essentially irrelevant.

Note that if we didn’t insist on creating a minimal isometric dilation, we could have taken

and to be given by the same formula:

**Exercise D: **If are contractions, then there exists a Hilbert space and isometries on such that

for all and all .

**Theorem 7 (Sz.-Nagy’s unitary dilation theorem):** *Let be a contraction on a Hilbert space . Then there exists a Hilbert space and a unitary such that *

*is a dilation of .**is the smallest subspace of reducing for that contains .*

*Moreover, the pair is unique, in the sense that if is another such pair satisfying the above requirements, then there exists a unitary such that for all , and such that . *

**Remark: **The unitary (or more precisely, the pair ) is called * the minimal unitary dilation of *.

**Proof: **Let be the minimal isometric dilation of , and let be the unitary extension of constructed in Theorem 4. Then is a dilation of (“my dilation’s dilation is a dilation of mine”). If is not minimal by some chance, then restrict to (the notation should be self-explanatory), and then is a minimal unitary dilation. Uniqueness is left as an exercise.

**Corollary (von Neumann’s inequality): ***Let be a contraction on a Hilbert space. For every polynomial , *

.

**Proof: **

.

[…] of , given by Ando’s theorem. Then is an isometric co-extension of , so by Exercise C in the previous post, and , where and is the minimal isometric co-extension of . We can therefore write, with respect […]