Roots of invertibly weighted shifts with finite defect

Abstract

Let T T be a unilateral invertibly weighted shift; i.e., T T maps a square-summable vector sequence { x 0 , x 1 , … } \left \{ {{x_{0,}}{x_1}, \ldots } \right \} from a Hilbert space H H to the sequence { 0 , A 0 x 0 , A 1 x 1 , … } \left \{ {0,{A_0}{x_0},{A_1}{x_{1,}} \ldots } \right \} , where { A n } \left \{ {{A_n}} \right \} is a uniformly bounded sequence of invertible operators on H H . If S 0 {S_0} is the identity operator on H H , and S n = A n − 1 A n − 2 ⋯ A 0 {S_n} = {A_{n - 1}}{A_{n - 2}} \cdots {A_0} for n ⩾ 1 n \geqslant 1 , then T T is unitarily equivalent to multiplication by the variable Z Z on the space H 2 ( T ) {H^2}\left ( T \right ) consisting of formal series ∑ x n Z n \sum {x_n}{Z^n} having coefficients x n ∈ H {x_n} \in H which satisfy ∑ ‖ S n x n ‖ 2 > + ∞ \sum {\left \| {{S_n}{x_n}} \right \|^2} > + \infty . The commutant of this multiplication consists of formal series ∑ F n Z n \sum {F_n}{Z^n} which define bounded operators on H 2 ( T ) {H^2}\left ( T \right ) —where each F n {F_n} is an operator on H H , and the action of such a series on an element of H 2 ( T ) {H^2}\left ( T \right ) is given by the Cauchy product of the two series. Using these characterizations, it is shown that if H H has finite dimension m ⩾ 2 m \geqslant 2 , then T T has an n n th root only if n n divides m m . Examples are given of shifts T T with (a) m = 2 m = 2 , but T T has no square root, and (b) m = 4 m = 4 , T T has a square root, but no fourth root.