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In 1976 J. Deddens and W. Wogen asked which subnormal operators have a cyclic adjoint. A similar question was raised for hyponormal operators. Several partial results have since been obtained, although it was not known if every pure subnormal operator had a cyclic adjoint.

Recently I have answered Deddens & Wogen's question by proving that every pure subnormal operator has a cyclic adjoint. More generally, a subnormal operator (not necessarily pure) has a cyclic adjoint if and only if its normal part is cyclic.

**Two Applications:**

Two applications of the main result are to the triangularlizability of subnormal operators and to the existence of separating vectors for some non-self adjoint operator algebras.

More specifically, every pure subnormal operator has a matrix representation that is "almost" lower triangular; "almost" meaning that the only non-zero entries that may appear above the main diagonal are on the super-diagonal. This is the best result possible, because it is easy to see that not every pure subnormal operator has a lower triangular matrix representation.

Secondly, if S is a pure subnormal operator, then the commutant of S* has a separating vector. This is interesting because the commutant of a subnormal operator generally does not have a separating vector (try the unilateral shift of multiplicity two).

**History of the Problem:**

D. Sarason first proved that pure isometries have cyclic adjoints, his proof is in Halmos' problem book.

In 1955 J. Bram proved that a *-cyclic normal operator is actually cyclic. From this it follows immediately that if S is multiplication by z on an invariant subspace of L^2(mu) for some measure mu, then S has a cyclic adjoint. This covers many non-trivial examples. A "test question" that J. Deddens raised asked about direct sums of operators of this type.

In 1978 W. Wogen proved that (nonconstant) analytic Toeplitz operators on the circle and pure quasinormal operators have cyclic adjoints - either of these classes seem like a natural place to possibly find a counterexample. In fact Wogen proved an amazing result; namely that there is one function in H^2 that is a cyclic vector for any nonconstant co-analytic Toeplitz operator. Thus not only do (nonconstant) analytic Toeplitz operators have a cyclic adjoint, but their adjoints have a common cyclic vector.

In 1988 K. Chan extended Wogen's result on common cyclic vectors to multipliers of more general Hilbert spaces of analytic functions satisfying some reasonable conditions.

In 1990 P. Bourdon and J.H. Shapiro actually extended Wogen's result on common cyclic vectors to multipliers on *any* Hilbert space of analytic functions. In fact in 1991, G. Godefroy and J.H. Shapiro proved that the adjoint of a nonconstant multiplier on any Hilbert space of analytic functions in supercyclic and they characterized the (nonconstant) multipliers whose adjoints are hypercyclic as those whose range intersects the unit circle.

Also in 1978 K. Clancey and D. Rogers proved that if T is any operator such that T* has "sufficiently many" eigenvectors, then T* is cyclic. From this it easily follows that if T is a pure hyponormal operator and the approximate point spectrum of T has area zero, then T has a cyclic adjoint.

**What I actually Proved:**

Recently I proved that every pure subnormal operator has a cyclic adjoint. This was done by first giving a characterization of the subnormal operators with cyclic adjoints. Namely, a subnormal operator S on a Hilbert space H has a cyclic adjoint if and only if there is a *-cyclic normal operator N on a space K and a one-to-one bounded linear operator A from H to K that intertwines S and N, so AS = NA.

Finally I proceed to show that for any pure subnormal operator S there exists a one-to-one map that intertwines S and a *-cyclic normal operator.

Furthermore, my techniques also allow me to give a stronger version of the result of Clancey and Rogers and also to show that the result of Bourdon and Shapiro follows immediately from Wogen's work.