# The Self-Commutator of a Subnormal Operator

Doctoral Thesis
written under the direction of John B. Conway
University of Tennessee, 1997

This work is directed toward a study of the self-commutator of a subnormal operator. If T is any operator, then the self-commutator of T is [T*,T] = T*T-TT*. For a given operator T, there are some natural self-adjoint operators associated with it: the real and imaginary parts of T, the absolute value of T, and the self-commutator of T. If one claims to understand an operator, or a class of operators, then one should understand these naturally associated operators.

In addition to being naturally associated with an operator, the self-commutator is part of a complete unitary invariant for pure subnormal operators. As such it deserves special consideration.

Several topics were investigated in my thesis, all have some connection with the self-commutator of a subnormal operator.

Topics:

## Trace Estimates

### The Berger-Shaw Theorem for Cyclic Subnormal Operators

One of the most fundamental results on self commutators is the Berger-Shaw Theorem. It implies that for any rationally cyclic hyponormal operator T, [T*,T] is trace class and the trace is at most (1/pi)Area[spectrum(T)]. We give a sharp form of this for cyclic subnormal operators.

Theorem If S is a cyclic subnormal operator, then [S*,S] is trace class and

tr[S*,S] = (1/pi) Area[ spectrum(S) - essential spectrum(S) ].

A version of this theorem was also obtained for operators that commute with a cyclic subnormal operator. Namely, if S is a pure cyclic subnormal operator and T commutes with S, then T = f(S) for some bounded (analytic) function f and the trace, tr[T*,T], is computed as the Dirichlet integral of f over the set, spectrum(S) - essential spectrum(S), of analytic bounded point evaluations.

It is known that if T is any hyponormal operator, K is its essential spectrum and Area[K] = 0, then [T*,T] is trace class if and only if the index function of T is (area) integrable on the complement of K. Unfortunately, this does not apply to the above theorem, or to operators of the form multiplication by z on a Hilbert space of functions analytic in a region G, simply because the essential spectrum is contained in the boundary of G (often equal to the boundary) and the boundary need not have area zero. Now, the proper way to measure smallness of a compact set K for subnormal operators is by requiring R(K) = C(K). Hence it is natural to try to extend the above mentioned result on hyponormal operators to subnormal operators S satisfying R(K) = C(K) where K is the essential spectrum of S. Unfortunately this does not work.

Example There exists a pure subnormal operator S such that R(K) = C(K), where K is the essential spectrum of S, and the index function of S is integrable off K, yet [S*,S] is not trace class.

## Compactness Criteria

### An Essential Version of Putnam's Inequality

Putnam's Inequality estimates the norm of the self commutator of any hyponormal operator T saying that it is at most (1/pi)Area[spectrum(T)]. We have proven a version for the essential norm that applies to not only to hyponormal operators, but even to the larger class of essentially hyponormal operators.

If P is a C*-property, then we say that an operator has property essentially P if the image of the operator in the Calkin algebra has property P in the Calkin algebra. Thus we may speak of essentially normal operators, essentially hyponormal operators and even essentially subnormal operators. An operator T is essentially hyponormal if [T*,T] = A + K where A is positive and K is compact. The essentially subnormal operators are characterized below (see Essentially Subnormal Operators), but one easily sees that any operator of the form S+K where S is subnormal and K is compact is essentially subnormal.

Theorem Let S be an operator and let K be its essential spectrum.

1. If S is essentially hyponormal, then the essential norm of [S*,S] is at most (1/pi)Area(K).
2. If S is essentially subnormal, then the essential norm of [S*,S] is at most dist[z-bar to R(K)]^2.

### Characterization of the Compactness of [S*,S]

Theorem Let S be a subnormal operator and let f be the Cauchy transform of the characteristic function of the essential spectrum of S. If T is the Toeplitz operator (constructed from S) with symbol f, then [S*,S] is compact if and only if [T,S] is compact.

In several cases one can check that T and S actually commute, hence [T,S] = TS - ST = 0. For example, if the area of the essential spectrum is zero, then f = 0, hence T = 0. Thus [S*,S] is compact. On the other hand, notice that f is continuous on the whole complex plane and analytic off of the essential spectrum. So, if S is multiplication by z on a Hilbert space H of functions analytic in a region G and if the essential spectrum of S is contained in the boundary of G, then f is in A(G), the disk algebra over G. Thus, if A(G) is contained in H, then T and S commute, hence S has compact self commutator.

Corollary Let S be a subnormal operator, K its essential spectrum and let G = spectrum(S) - essential spectrum(S). If A(G) is contained in the restriction algebra for S, then S has compact self commutator.

## Diagonalizability of the Self Commutator

Knowing that an operator S has compact self commutator has several advantages, for example, it implies that the index function of S is always finite and that S may be written as a direct sum of irreducible operators. For many operators, these two properties are easily seen directly. However, it is a nontrivial result that rationally cyclic subnormal operators may be written as direct sums of irreducibles; and this depends on the fact that the self commutator is compact (by the Berger - Shaw Theorem).

Even though many operators that one comes across regularly have compact self commutator, not all subnormal operators do - try the unilateral shift of infinite multiplicity. In fact, the self commutator can even have an absolutely continuous spectral measure (example: try tensor products). Thus it is natural to ask about the diagonalizability of [S*,S].

One natural way that operators without compact self commutator arise is by restricting a "nice" operator S to an invariant subspace (with infinite index). For example, if S is the Bergman shift, then there are invariant subspaces M for S such that S(M) has infinite codimension in M, thus S restricted to M will not have compact self commutator (because its index function is infinite). However, the next theorem implies that the self-commutator is diagonalizable.

Theorem If S is a pure subnormal operator, G is the interior of the Sarason Hull of S, the essential spectrum of S equals the boundary of G, and T is S restricted to any invariant subspace, then [T*,T] is diagonalizable. Furthermore, if [T*,T] is not compact, then the essential spectrum of [T*,T] is either finite or a sequence converging to zero and it depends only on G.

If S is (subnormal and) of the form multiplication by z on a Hilbert space H of functions analytic in a region G, then the above hypothesis is that (G is simply connected and) the polynomials are weak* dense in the bounded analytic functions on G. Even though this seems like a strong hypothesis, there are no shortage of non-trivial examples to which the above theorem applies.

It is an amazing fact that if [T*,T] is not compact, then the essential spectrum of [T*,T] depends only on G - not on the invariant subspace that S was restricted to and not even on the operator S itself! In fact, it is shown that if M is the (subnormal) operator of multiplication by z on H^2(G) and S is any operator satisfying the hypothesis of the theorem, then the essential spectrum of [T*,T] is the same as the spectrum of [M*,M].

## Compact Toeplitz Operators with Continuous Symbols

If S is a subnormal operator on H and N is the minimal normal extension of S, then given any bounded Borel function f on the spectrum of N, f(N) is well defined. So for such a function f, the Toeplitz operator S_{f} is the compression to H of f(N); that is, S_{f} is Pf(N) restricted to H, where P is the projection onto H.

It is known that if S has compact self commutator and f is a continuous function, then S_{f} is compact if and only if f = 0 on the essential spectrum of S. What if [S*,S] is not compact?

If S is a subnormal operator, then we define the essential normal spectrum of S as the normal spectrum of S in the Calkin algebra. It is known that if x is any subnormal element of any C* algebra, then its normal spectrum in that algebra is well defined. Thus, if S is a subnormal operator, then its image in the Calkin algebra is a subnormal element of the Calkin algebra; as such its normal spectrum in the Calkin algebra is well defined - we shall call it the essential normal spectrum.

One easily sees that if S has compact self commutator, then the essential normal spectrum is exactly the essential spectrum. However, in general the essential normal spectrum is only a subset of the essential spectrum.

Several results about this new subset are obtained. For example, the essential normal spectrum lies between the approximate point spectrum and the essential spectrum; furthermore, the essential spectrum is obtained from the essential normal spectrum by filling in holes.

Example If K is any compact subset of the closed unit disk that contains the unit circle, then there exists a pure subnormal operator S such that the approximate point spectrum of S is the unit circle, the essential spectrum of S is the closed unit disk and the essential normal spectrum of S is K.

Theorem If S is any subnormal operator and f is a continuous function on the spectrum of S, then the Toeplitz operator S_{f} is compact if and only if f = 0 on the essential normal spectrum of S.

## A Characterization of Essentially Subnormal Operators

An operator S is an essentially subnormal operator, if its image in the Calkin algebra is a subnormal element of the Calkin algebra. Note that since subnormality may be defined in any C* algebra, essentially subnormal operators are well defined. The following result characterizes these operators in a very manageable way, and makes it possible to begin studying this new class of operators.

TheoremIf S is any operator, then the following are equivalent:

1. S is essentially subnormal;
2. S has an essentially normal extension;
3. S has an extension of the form "Normal plus Compact".

In proving the above theorem, we develop a new characterization of subnormality - one that is easy to use in a C*-algebra setting.

### A New Characterization of Subnormal Operators

Theorem An operator S on a Hilbert space H is subnormal if and only if there exists a compact set K in the complex plane and a positive linear map p:C(K) into B(H) such that p(1) = I, p(z) = S and p(z-bar z) = S*S. Furthermore, if p exists, then the normal spectrum of S is contained in K.

This characterization of subnormality is useful when studying C*-algebras generated by subnormal operators. Several results on the C*-algebra generated by a subnormal operator were obtained. For example, the above Theorem shows that subnormality is preserved not only under *-homomorphisms but also under certain positive linear maps.

## The Algebraic Equivalence of Operators

Given two operators S,T when does there exists a *-isomorphism p:C*(S) onto C*(T) with p(S) = T? In this case one says that S and T are algebraically equivalent. Let S^(\infty) denote the infinite inflation of S; that is, the direct sum of an infinite number of copies of S. The following nice result is obtained.

### Algebraic Equivalence for Essentially Normal Operators

Theorem If S and T are pure essentially normal operators, then S and T are algebraically equivalent if and only if S^(\infty) is unitarily equivalent to T^(\infty).

This Theorem is quite nice in that it shows that several things are preserved under algebraic equivalence for pure operators. In particular, it implies that for pure cyclic subnormal operators, algebraic equivalence is the same as unitary equivalence!

### Algebraic Equivalence for General Operators

Of course, the above theorem is not true if purity is dropped. For one knows that two normal operators are algebraically equivalent precisely when they have the same spectrum. However, a general result does hold that naturally generalizes the above last Theorem for any pair of operators.

Two operators S, T are said to be approximately equivalent when there exists a sequence of unitary operators { U_{n} } such that U*_{n}SU_{n} converges in norm to T.

Theorem If S and T are any two operators, then S and T are algebraically equivalent if and only if S^(\infty) is approximately equivalent to T^(\infty).

Recently, both hyponormal and subnormal operators with finite rank self commutator have received considerable attention and have made connections with quadrature domains. Even before the appearance of Thomson's Theorem (on the existence of analytic bounded point evaluations), the pure cyclic subnormal operators with finite rank self commutator were characterized by Olin, Thomson, and Trent; they showed, in particular, that the analytic bounded point evaluations of such an operator form a quadrature domain.

More recently, McCarthy and Yang characterized the pure rationally cyclic subnormal operators with finite rank self commutator. They also showed that the set G of analytic bounded point evaluations for such an operator is not only nonempty, but that G is a quadrature domain. Furthermore the spectral measure is of the form harmonic measure for G plus a finite sum of point masses in G.

In requiring certain "smallness" conditions on the self commutator, the next natural step after considering those operators S such that [S*,S] has a finite dimensional range, is to consider those operators S such that the range of [S*,S] is not dense. This is equivalent to requiring that zero is an eigenvalue for the self commutator. It is amazing that this much larger class of operators still makes a connection with another very special (& new) class of domains - generalized quadrature domains.

A quadrature domain is characterized by the existence of a Schwarz function; that is, a meromorphic function S in G that is continuous up to the boundary of G, has only finitely many poles in G, and it equals z-bar on the boundary of G. An example is the unit disk; its Schwarz function is 1/z.
The author defines a generalized quadrature domain to be a domain that has a generalized Schwarz function ; that is, a Nevanlinna class function on G that has boundary values almost everywhere on the boundary of G with respect to harmonic measure and these boundary values agree with z-bar a.e. on the boundary (w.r.t. harmonic measure).

One can define what it means for a Nevanlinna class function (a quotient of bounded analytic functions) to have "boundary values" on any bounded region - but not all such functions have boundary values (think of the slit disk, the nontangential limits must agree from above and below along the slit). However, for simplicity here, lets just consider regions bounded by rectifiable Jordan curves. In this case, all Nevanlinna functions have boundary values - given as nontangential limits.

The connection with operator theory begins with the next result.

Theorem If G is a bounded region and S is multiplication by z on the Hardy space over G, then [S*,S] does not have dense range if and only if G is a generalized quadrature domain.
So, which regions are generalized quadrature domains? The next result shows that such domains not only exist but are very rigid. Recall that a simply connected domain G is a quadrature domain if and only if the Riemann map from the unit disk onto G is a rational function.

Theorem If G is a simply connected region and f is the Riemann map from the unit disk only G, then the following hold:
(a) G is a generalized quadrature domain if and only if f has a pseudocontinuation to the exterior of the unit disk.
(b) If f is analytic across the unit circle, then G is a generalized quadrature domain if and only if f is a rational function.
(c) There are simply connected regions G bounded by smooth Jordan curves such that G is a generalized quadrature domain and yet not a quadrature domain.

In proving part (c) above, a general method is given for constructing bounded univalent functions that are smooth up to the boundary, have pseudocontinuations to the exterior of the unit disk and are not rational functions.

The next step is to consider irreducible cyclic subnormal operators S whose self commutators do not have dense range; recall that this is the same as ker[S*,S] is non-zero. For these operators, we have the following partial result.

Theorem (a) If S is an irreducible cyclic subnormal operator and there is a non-zero bounded function in ker[S*,S], then the set of analytic bounded point evaluations for S form a generalized quadrature domain.
(b) If the set G of abpe's is a quadrature domain, then there is a non-zero bounded function in ker[S*,S] if and only if the spectral measure for S is harmonic measure for G plus a sum of point masses in G where the points form an H^{\infty}(G) zero set.

Conjecture: If S is an irreducible cyclic subnormal operator, then [S*,S] does not have dense range if and only if the set G of analytic bounded point evaluations forms a generalized quadrature domain and the spectral measure is equivalent to harmonic measure for G plus a sum of point masses in G where the points form an H^{\infty|}(G) zero set.

The above results strongly support this conjecture, however it still remains open.