Home Nilpotent perturbations of m-isometric and m-symmetric tensor products of commuting d-tuples of operators
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Nilpotent perturbations of m-isometric and m-symmetric tensor products of commuting d-tuples of operators

  • Bhagwati Prashad Duggal and In Hyoun Kim EMAIL logo
Published/Copyright: November 19, 2024
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Abstract

If T 1 and T 2 are commuting d -tuples of Hilbert space operators in B ( ) d such that ( T 1 * I + I T 2 * , T 1 I + I T 2 ) is strictly m -isometric (resp., m -symmetric) for some positive integer m , then there exist a scalar d -tuple λ and positive integers m i , 1 i 2 , such that m = m 1 + 2 m 2 2 , ( T 1 * + λ ¯ , T 1 + λ ) is m 1 isometric, and T 2 λ is m 2 -nilpotent (resp., m = m 1 + m 2 1 , ( T 1 * + λ ¯ , T 1 + λ ) is m 1 -symmetric and ( T 2 * λ ¯ , T 2 λ ) is m 2 symmetric).

MSC 2010: 47A05; 47A55; 47A11; 47B47

1 Introduction

Let B ( X ) d (resp., B ( ) d ) denote the product of d -copies of the algebra of operators, i.e., bounded linear transformations, on an infinite dimensional complex Banach space X into itself (resp., on an infinite dimensional complex Hilbert space into itself). A d -tuple A = ( A 1 , , A d ) B ( X ) d is a commuting d -tuple if [ A i , A j ] = A i A j A j A i = 0 for all 1 i , j d . Recall from Gleeson and Richter [1] that a commuting d -tuple A B ( ) d is m -isometric if

A * , A m ( I ) = j = 0 m ( 1 ) j m j β = j j ! β ! A * β A β = 0 ,

where β = ( β 1 , , β d ) , β i are non-negative integers for all 1 i d , β = i = 1 d β i , β ! = i = 1 d β i ! , A β = i = 1 d A i β i , and A * β = i = 1 d A i * β i . Generalizing this to commuting d -tuple pairs ( A , B ) of Banach space operators, the pair ( A , B ) is said to be m -isometric if

A , B m ( I ) = j = 0 m ( 1 ) j m j β = j j ! β ! A β B β = 0

[13]. Analogously [3], the commuting d -tuple ( A , B ) is said to be m -symmetric if

δ A , B m ( I ) = j = 0 m ( 1 ) j m j i = 1 d A i m j i = 1 d B i j = 0 .

The classes consisting of m -isometric and m -symmetric commuting d -tuples have been considered by a number of authors in the recent past (see [16] for further references), and it is known that these classes share a number of, though not all, properties with their single operator (i.e., 1-tuple) kin. For example, if A , B , S and T are commuting d -tuples in B ( X ) d , A commutes with S , B commutes with T , and A , B m ( I ) = S , T n ( I ) = 0 (resp., δ A , B m ( I ) = δ S , T n ( I ) = 0 ) for some positive integers m and n , then A S , B T m + n 1 ( I ) = 0 (resp., δ A S , B T m + n 1 ( I ) = 0 ); see [3, Corollary 4.2] for the general case and [7] for the 1-tuple case. The converse fails, i.e., given commuting d -tuples A , B , S , and T such that A commutes with S , B commutes with T , and A S , B T m + n 1 ( I ) = 0 (resp., δ A S , B T m + n 1 ( I ) = 0 ) it does not follow that A , B and S , T (or, scalar multiples thereof) satisfy A , B m ( I ) = S , T n ( I ) = 0 (resp., δ A , B m ( I ) = δ S , T n ( I ) = 0 ). A case where there is a positive answer is that of the tensor product pairs ( A S , B T ) : if A , B , S , T are commuting d -tuples such that ( A S , B T ) is m -isometric (resp., m -symmetric), then there exist integers m i > 0 and a non-zero scalar c such that m = m 1 + m 2 1 , ( A , c B ) is strict m 1 -isometric, and ( S , 1 c T ) is strict m 2 -isometric (resp., ( A , c B ) is strict m 1 -symmetric, and ( S , 1 c T ) is strict m 2 -symmetric). Recall here that the pair ( A , B ) is strict m -isometric (resp., strict m -symmetric) if A , B m ( I ) = 0 and A , B m 1 ( I ) 0 (resp., δ A , B m ( I ) = 0 and δ A , B m 1 ( I ) 0 ). This article considers tensor sum pairs ( S I + I Q , T I ) , S , T , and Q commuting d -tuples in B ( X ) d , and ( T 1 * I + I T 2 * , T 1 I + I T 2 ) , T 1 and T 2 commuting d -tuples in B ( ) d . We prove that if ( S I + I Q , T I ) is strictly m -isometric, then there exist positive integers m i and a scalar d -tuple λ = ( λ 1 , , λ d ) such that m = m 1 + m 2 1 , ( S + λ , T ) is m 1 -isometric and Q λ is m 2 -nilpotent; if ( T 1 * I + I T 2 * , T 1 I + I T 2 ) is strict m -symmetric (resp., strict m -isometric), then there exist positive integers m i and a scalar d -tuple λ = ( λ 1 , , λ d ) such that m = m 1 + m 2 1 , ( T 1 * + λ ¯ , T 1 + λ ) is m 1 -symmetric and ( T 2 * λ ¯ , T 2 λ ) is m 2 -symmetric (resp., m = m 1 + 2 m 2 2 , ( T 1 * + λ ¯ , T 1 + λ ) is m 1 -isometric and T 2 λ is m 2 -nilpotent). This generalizes [8, Theorem 22] and [9, Theorem 1.3] to the case of commuting d -tuples of operators. The argument that we use to prove these results is, in spirit, similar to the argument used in proving related results in [5,7] and depends upon reducing the d -tuple problem to a 1-tuple case.

The plan of the article is as follows. We introduce all relevant terminology, notation, and complementary results in Section 2. The main results are proved in Section 3, and the final section lists the references.

2 Preliminaries

Given operators A , B B ( X ) , let L A and R B B ( B ( X ) ) denote, respectively, the operators L A ( X ) = A X and R B ( X ) = X B of left multiplication by A and right multiplication by B . For commuting d -tuples A = ( A 1 , , A d ) , B = ( B 1 , , B d ) in B ( X ) d and a scalar d -tuple α = ( α 1 , , α d ) such that α i is a non-negative integer for all 1 i d , let L A α , R B α , the convolution ( L A * R B ) j ( X ) and the multiplication ( L A × R B ) ( X ) , X B ( X ) , be defined by

L A α = i = 1 d L A i α i , R B α = i = 1 d R B i α i , ( L A R B ) j ( X ) = α = j j ! α ! L A α R B α ( X ) = i = 1 d L A i R B i j ( X ) ( all integers j 0 , α ! = α 1 ! α d ! ) , α = i = 1 d α i and ( L A × R B ) ( X ) = i = 1 d L A i i = 1 d R B i ( X ) .

Define the operator i = 1 d A i X B i n by

i = 1 d A i X B i n = i = 1 d A i i = 1 d A i X B i n 1 B i for all positive integers n .

(Thus, A n = A A n 1 I = I A n 1 A = = A n , A B n = A A B n 1 B = = A n B n , and i = 1 d A i B i n = i = 1 d A i i = 1 d A i B i n 1 B i .) The commuting d -tuples A and B commute, [ A , B ] = 0 , if

[ A i , B j ] = A i B j B j A i = 0 for all 1 i , j d .

It is clear that

[ L A , R B ] = 0

and if [ A , B ] = 0 , then

[ L A , L B ] = [ R A , R B ] = 0 .

A pair ( A , B ) of commuting d -tuples of B ( X ) d operators is m -isometric if

0 = A , B m ( I ) = j = 0 m ( 1 ) j m j β = j j ! β ! A β B β = j = 0 m ( 1 ) j m j i = 1 d A i B i j = j = 0 m ( 1 ) j m j i = 1 d L A i R B i j ( I ) = j = 0 m ( 1 ) j m j ( L A R B ) j ( I ) = ( I L A R B ) m ( I )

and the pair ( A , B ) is m -symmetric if

0 = δ A , B m ( I ) = j = 0 m ( 1 ) j m j i = 1 d A i m j i = 1 d B i j = j = 0 m ( 1 ) j m j i = 1 d L A i m j i = 1 d R B i j ( I ) = j = 0 m ( 1 ) j m j L A m j × R B j ( I ) = ( L A R B ) m ( I ) .

Commuting tuples of m -isometric, similarly n -symmetric, operators are known to share a large number of properties with their single operator counterparts: thus, just as for the 1-tuple case, A , B m ( I ) = 0 implies A , B t ( I ) = 0 , similarly δ A , B m ( I ) = 0 implies δ A , B t ( I ) = 0 , for integers t m . However, there are instances where a property holds for the single operator version but fails for the d -tuple version: for example, whereas

A , B m ( I ) = 0 A 1 , B 1 m ( I ) = 0 for all invertible A and B

(similarly, for n -symmetric ( A , B ) ), these properties fail for commuting d -tuples (see [3]).

If ( A , B ) ( X , m ) -isometric (i.e., A , B m ( X ) = ( I L A R B ) m ( X ) = 0 , X B ( X ) ), then

A , B m ( X ) = 0 ( I L A R B ) ( A , B m 1 ( X ) ) = 0 ( L A R B ) A , B m 1 ( X ) = A , B m 1 ( X ) ( L A R B ) t A , B m 1 ( X ) = A , B m 1 ( X ) ,

and if ( A , B ) ( X , n ) -symmetric (i.e., δ A , B n ( X ) = ( L A R B ) n ( X ) = 0 , X B ( X ) ), then

δ A , B n ( X ) = 0 ( L A R B ) ( δ A , B n 1 ( X ) ) = 0 L A ( δ A , B n 1 ( X ) ) = R B ( δ A , B n 1 ( X ) )

L A t ( δ A , B n 1 ( X ) ) = R B t ( δ A , B n 1 ( X ) )

for all integers t 0 , where

L A ( δ A , B n 1 ( X ) ) = L A j = 0 n 1 ( 1 ) j n 1 j L A n 1 j × R B j ( X ) = j = 0 n 1 ( 1 ) j n 1 j L A n j × R B j ( X )

and

R B ( δ A , B n 1 ( X ) ) = j = 0 n 1 ( 1 ) j n 1 j L A n 1 j × R B j + 1 ( X ) .

Let X X denote the completion, endowed with a reasonable cross norm, of the algebraic tensor product of X with X has overline. Let S T denote the tensor product of S , T B ( X ) . The tensor product of the d -tuples A = ( A 1 , , A d ) and B = ( B 1 , , B d ) is the d 2 -tuple

A B = ( A 1 B 1 , , A 1 B d , , A d B 1 , , A d B d ) .

(Thus, A I = ( A 1 I , , A d I ) .) A commuting d -tuple Q = ( Q 1 , , Q d ) is n -nilpotent for some positive integer n if Q n = i = 1 d Q i α i = 0 for all d -tuples α = ( α i , , α d ) of non-negative integers α i such that α = i = 1 d α i = n and Q α is non-zero for at least one α with α n 1 . Clearly, Q is n -nilpotent if and only if the tensor product operators Q I and I Q are n -nilpotent. Furthermore, ( A , B ) is m -isometric (resp., m -symmetric) if and only if ( A I , B I ) and ( I A , I B ) are m -isometric (resp., m -symmetric).

Given a commuting d -tuple A = ( A 1 , , A d ) and a scalar λ = ( λ 1 , , λ d ) , we abbreviate ( A λ I ) = ( A 1 λ 1 I , , A d λ d I ) to A λ .

3 Results

We start with a technical lemma. Recall that a pair ( A , B ) of commuting d -tuples in B ( X ) d is strict m -isometric if A , B m ( I ) = 0 and A , B m 1 ( I ) 0 . Since A , B m ( I ) = 0 implies A , B t ( I ) = 0 for integers t m , ( A , B ) strict m -isometric implies the linear independence of the sequence { A , B r ( I ) } r = 0 m 1 . A similar statement holds for strict n -symmetric operators, i.e., if δ A , B n ( I ) = 0 and δ A , B n 1 ( I ) 0 , then the sequence { δ A , B r ( I ) } r = 0 n 1 is linearly independent.

Lemma 3.1

Given commuting d-tuples A , B B ( X ) d , if ( A , B ) is strict m -isometriic, then the sequence

{ A , B r ( I ) B β } r = 0 m 1 , β = r ,

is linearly independent.

Proof

The proof is by contradiction. Suppose that there exist scalars a j , not all zero, such that

M = j = 0 m 1 a j A , B j ( I ) B β = 0 , β = j .

The hypothesis ( A , B ) is m -isometric implies

( L A R B ) A , B m 1 ( I ) = A , B m 1 ( I ) ( L A * R B ) t A , B m 1 ( I ) = A , B m 1 ( I ) β = t t ! β ! A β A , B m 1 ( I ) B β = A , B m 1 ( I )

for all positive integers t . We have

A , B m 1 ( M ) = j = 0 m 1 a j A , B m 1 + j ( I ) B β = 0 a 0 A , B m 1 ( I ) = 0 a 0 = 0 , A , B m 2 ( M ) = j = 1 m 1 a j A , B m 2 + j ( I ) B β = 0 a 1 A , B m 1 ( I ) B β = 0 , β = 1 a 1 β = 1 A β A , B m 1 ( I ) B β = 0 a 1 A , B m 1 ( I ) = 0 a 1 = 0 .

Repeating the argument, we eventually have

a 0 = a 1 = = a m 2 = 0 , M = a m 1 A , B m 1 ( I ) B β , β = m 1 a m 1 β = m 1 ( m 1 ) ! β ! A β A , B m 1 ( I ) B β = 0 a m 1 A , B m 1 ( I ) = 0 a m 1 = 0 .

This being a contradiction, the lemma is proved.□

If, for commuting d -tuples S , T , and Q , the pair ( S , T ) is m 1 -isometric and Q is m 2 -nilpotent, then ( S I , T I ) is m 1 -isometric, I Q is m 2 -nilpotent, and ( S I + I Q , T I ) is ( m 1 + m 2 1 ) -isometric [3]. The converse has a solution, as the following theorem, proved for the single linear operators case in [8] (see also [9]), proves.

Theorem 3.2

If S , T , and Q are commuting d-tuples in B ( X ) d such that ( S I + I Q , T I ) is strict m-isometric for some positive integer m, then there exist positive integers m i , 1 i 2 , and a scalar d-tuple λ = ( λ 1 , , λ d ) such that m = m 1 + m 2 1 , ( S + λ , T ) is m 1 -isometric and Q λ is m 2 -nilpotent.

Proof

The m -isometric property of ( S I + I Q , T I ) implies

S I + I Q , T I m ( I I ) = { ( I L S I * R T I ) L I Q * R T I } m ( I I ) = j = 0 m ( 1 ) j m j S I , T I m j ( L I Q * R T I ) j ( I I ) .

Since

S I , T I m j ( I I ) = k = 0 m j ( 1 ) k m j k ( L S I * R T I ) k ( I I ) = k = 0 m j ( 1 ) k m j k i = 1 d L S i I R T i I k ( I I ) = k = 0 m j ( 1 ) k m j k i = 1 d L S i R T i k ( I ) I = S , T m j ( I ) I

and

( L I Q R T I ) j ( I I ) = i = 1 d L I Q i R T i I j ( I I ) = i = 1 d ( I Q i ) ( T i I ) j = i = 1 d T i Q i j = β = j j ! β ! T β Q β ,

where

T β = Π i = 1 d T i β i and Q β = Π i = 1 d Q i β i .

Since σ a ( Q ) is non-empty, there exists a λ = ( λ 1 , , λ d ) σ a ( Q ) . Let { y p } X be a sequence of unit vectors such that

lim p ( Q i λ i ) y p = 0 all 1 i d .

Let x X be a unit vector, and let x * , y p * X * be such that

x * ( x ) = x , x * = 1 = y p , y p * = y p * ( y p ) .

Then

lim p Q β y p , y p * = λ β

and

0 = S I + I Q , T I m ( I I ) = j = 0 m ( 1 ) j m j β = j j ! β ! ( S , T m j ( T β ) Q β )

imply

0 = lim p S I + I Q , T I m ( I I ) x y p , x * y p * = lim p j = 0 m ( 1 ) j m j β = j j ! β ! ( S , T m j ( T β ) Q β ) x y p , x * y p * .

Consequently

j = 0 m ( 1 ) j m j β = j j ! β ! S , T m j ( T β ) x , x * lim p Q β y p , y p * = 0 j = 0 m ( 1 ) j m j β = j j ! β ! S , T m j ( T β ) x , x * λ β = 0 j = 0 m ( 1 ) j m j β = j j ! β ! S , T m j ( λ β T β ) x , x * = 0 j = 0 m ( 1 ) j m j S , T m j ( λ T ) j = 0 S + λ , T m ( I ) = 0 .

Thus, ( S + λ , T ) is m -isometric, and hence, there exists a positive integer m 1 m such that ( S + λ , T ) is strict m 1 -isometric. Let m 2 = m m 1 + 1 . Then,

0 = S I + I Q , T I m ( I I ) = ( S + λ ) I + I ( Q λ ) , T I m ( I I ) = j = 0 m ( 1 ) j m j ( L I ( Q λ ) R T I ) j ( ( S + λ ) I , T I m j ( I I ) ) ( since [ L I ( Q λ ) , L ( S + λ ) I ] = 0 ) = j = 0 m ( 1 ) j m j ( L I ( Q λ ) R T I ) j ( ( S + λ ) , T m j ( I ) I ) ( since ( S + λ ) I , T I m j ( I I ) = ( S + λ ) , T m j ( I ) I ) = j = 0 m ( 1 ) j m j i = 1 d L I ( Q i λ i ) R T i I j ( ( S λ ) , T m j ( I ) I ) = j = 0 m ( 1 ) j m j β = j j ! β ! ( I ( Q λ ) β ) ( ( S λ ) , T m j ( I ) I ) ( T β I ) = j = 0 m ( 1 ) j m j β = j j ! β ! ( ( S λ ) , T m j ( I ) T β ( Q λ ) β ) = j = m 2 m ( 1 ) j m j β = j j ! β ! ( ( S λ ) , T m j ( I ) T β ( Q λ ) β ) ,

since S + λ , T t ( I ) = 0 for all t m 1 . This, in view of the fact that the sequence { S + λ , T m j ( I ) T β } j = m 2 m is linearly independent, implies ( Q λ ) m 2 = 0 . Evidently, ( Q λ ) m 2 1 0 (for the reason that if it were so, then ( ( S + λ ) I + I ( Q λ ) , T I ) , hence ( S I + I × Q , T I ) would be ( m 1 + m 2 2 ) -isometric). This completes the proof.□

A similar argument works for m -symmetric operator pairs ( S I + I Q , T I ) . Thus, if

δ S I + I Q , T I m ( I I ) = ( δ S I , T I + L I Q ) m ( I I ) = j = 0 m m j δ S I , T I m j L I Q j ( I I ) = 0

and if λ σ a ( Q ) , then

δ S I + I Q , T I m ( I I ) = 0 j = 0 m m j δ S I , T I m j L I λ j ( I I ) = 0 j = 0 m m j δ S I , T I m j L λ I j ( I I ) = 0 δ ( S + λ ) I , T I m ( I I ) = 0 δ S + λ , T m ( I ) = 0 .

Assuming ( S + λ , T ) to be strict m 1 -symmetric, m 1 m , and setting m 1 + m 2 1 = m , it then follows that

0 = δ S I + I Q , T I m ( I I ) = δ ( S + λ ) I + I ( Q λ ) , T I m ( I I ) = j = 0 m m j δ ( S + λ ) I , T I m j L I ( Q λ ) j ( I I )

= j = m 2 m m j L I ( Q λ ) j ( δ ( S + λ ) I , T I m j ( I I ) ) ( since δ ( S + λ ) I , T I t ( I I ) = 0 for all t > m 1 , [ δ ( S + λ ) I , T I , L I ( Q λ ) ] = 0 ) = j = m 2 m m j I i = 1 d L Q i λ i j ( δ S + λ , T m j ( I ) I ) = j = m 2 m m j ( δ S + λ , T m j ( I ) L Q λ j ( I ) ) .

The sequence { δ S + λ , T m j ( I ) } m 2 m being linearly independent,

L Q λ m 2 ( I ) = 0 Q λ is m 2 -nilpotent .

A more general result, at least for Hilbert space d -tuples, is possible, as the following analogue of a result of Paul and Gu [9] (see also Gu [8]) shows.

Theorem 3.3

If T i = ( T i 1 , , T i d ) , 1 i 2 , are commuting d-tuples in B ( ) d such that ( T 1 * I + I T 2 * , T 1 I + I T 2 ) is strict m-symmetric for some positive integer m, then there exist positive integers m i and a scalar β = ( β 1 , , β d ) such that m = m 1 + m 2 1 , ( T 1 * + β ¯ , T 1 + β ) is m 1 -symmetric and ( T 2 * β ¯ , T 2 β ) is m 2 -symmetric.

Proof

For convenience, let

δ 1 = δ T 1 * I , T 1 I , δ 2 = δ I T 2 * , I T 2 and δ 12 = δ T 1 * I + I T 2 * , T 1 I + I T 2 .

Then

δ 12 = L T 1 * I + I T 2 * R T 1 I + i T 2 = ( L T 1 * I R T 1 I ) + ( L I T 2 * R I T 2 ) = δ 1 + δ 2 ,

where [ δ 1 , δ 2 ] = 0 and

δ 12 m ( I I ) = k = 0 m m k δ 1 m k δ 2 k ( I I ) = k = 0 m m k j = 0 m k ( 1 ) j m k j ( L T 1 * I m k j × R T 1 I j ) × t = 0 k ( 1 ) t k t ( L I T 2 * k t × R I T 2 t ) ( I I ) = k = 0 m m k j = 0 m k ( 1 ) j m k j i = 1 d T 1 i * I m k j × i = 1 d T 1 i I j t = 0 k ( 1 ) t k t i = 1 d I T 2 i * k t i = 1 d I T 2 i t = k = 0 m m k j = 0 m k ( 1 ) j m k j t = 0 k ( 1 ) t k t × i = 1 d T 1 i * m k j i = 1 d T 2 i * k t i = 1 d T 1 i j i = 1 d T 2 i t .

In the following, we start by proving the existence of a complex scalar β = ( β 1 , , β d ) , β ¯ = β , such that

δ 12 = δ 1 ( β ¯ ) + δ 2 ( β ) , δ 1 ( β ¯ ) = δ ( T 1 * + β ) I , ( T 1 + β ¯ ) I , δ 2 ( β ) = δ I ( T 2 * + β ¯ ) , I ( T 2 + β ) .

The proof is then completed by proving the existence of an integer m 2 m such that ( I ( T 2 * + β ¯ ) , I ( T 2 + β ) ) is strictly m 2 -symmetric and ( ( T 1 * + β ) I , ( T 1 + β ¯ ) I ) is m 1 ( = m m 2 + 1 ) symmetric.

For this, let T i = ( T i 1 , , T i d ) , 1 i 2 , λ = ( λ 1 , , λ d ) σ a ( T 1 ) and μ = ( μ 1 , , μ d ) σ a ( T 2 ) . Then, there exist sequences { x p } and { y p } of unit vectors in such that

lim p ( T 1 i λ i ) x p = 0 = lim p ( T 2 i μ i ) y p , 1 i d ,

and

lim p i = 1 d T 1 i * m k j i = 1 d ( T 2 i * ) k t i = 1 d T 1 i j i = 1 d T 2 i t x p y p , x p y p = lim p i = 1 d T 1 i j x p , i = 1 d T 1 i m k j x p i = 1 d T 2 i t y p , i = 1 d T 2 i k t y p = i = 1 d λ i j i = 1 d λ i ¯ m k j i = 1 d μ i t i = 1 d μ i ¯ k t .

Since

j = 0 m k ( 1 ) j m k j i = 1 d λ i j i = 1 d λ i ¯ m k j = i = 1 d ( λ i ¯ λ i ) m k ,

similarly,

t = 0 k ( 1 ) t k t i = 1 d μ i t i = 1 d μ i ¯ k t = i = 1 d ( μ i ¯ μ i ) k ,

δ 12 m ( I I ) x p y p , x p y p = k = 0 m m k i = 1 d ( λ i ¯ λ i ) m k i = 1 d ( μ i ¯ μ i ) k = i = 1 d ( ( λ i ¯ λ i ) + ( μ i ¯ μ i ) ) m = 0 .

Let

λ i = α i , i ( λ i ) = β i , α = ( α 1 , , α d ) , β = ( β 1 , , β d ) and [ β ] = i = 1 d β i .

Then,

λ = α + β = ( α 1 + β 1 , , α d + β d ) and [ β ] = 1 2 i = 1 d ( μ i ¯ μ i ) .

For a given complex d -tuple τ = ( τ 1 , , τ d ) , define δ i ( τ ) , 1 i 2 , by

δ 1 ( τ ) = δ ( T 1 * + τ ¯ ) I , ( T 1 + τ ) I = δ 1 + ( L τ ¯ I R τ I )

and

δ 2 ( τ ) = δ I ( T 2 * + τ ¯ ) , I ( T 2 + τ ) = δ 2 + ( L I τ ¯ R I τ ) .

Then, since β ¯ = β ,

δ 12 = δ 1 + δ 2 = δ 1 ( β ¯ ) + δ 2 ( β ) + ( ( L β I + R β ¯ I ) + ( L I β ¯ + R I β ) ) = δ 1 ( β ¯ ) + δ 2 ( β ) + { ( L I β L β I ) + ( R I β R β I ) } = δ + ( L + R ) ,

where we have set

δ 1 + δ 2 = δ , L I β L β I = L and R I β R β I = R .

Evidently, [ L , R ] = 0 ( = [ L I β , L β I ] = [ R I β , R β I ] ) and, for all positive integers t ,

δ 12 t ( I I ) = s = 0 t ( 1 ) s t s δ t s ( L + R ) s ( I I ) .

Since

( L + R ) s ( I I ) = k = 0 s s k L s k R k ( I I ) = k = 0 s ( 1 ) k s k L s k j = 0 k ( 1 ) j k j i = 1 d I β i k j i = 1 d β i I j = k = 0 s ( 1 ) k s k L s k j = 0 k ( 1 ) j k j [ β ] k ( I I ) ,

( L + R ) s ( I I ) = 0 for all k > 0 and if k = 0 , then it equals L s ( I I ) . A similar argument shows that L s ( I I ) = 0 for all s > 0 ; hence,

δ 12 t ( I I ) = ( δ 1 + δ 2 ) t ( I I ) = ( δ 1 ( β ¯ ) + δ 2 ( β ) ) t ( I I )

for all non-negative integers t . Again, considering

δ 1 ( λ ) = δ 1 + ( L λ ¯ I ) R λ I = δ 1 ( β ¯ ) ( L α I R α I ) ,

δ 1 ( λ ) t ( I I ) = { δ 1 ( β ¯ ) + ( L α I R α I ) } t ( I I ) = k = 0 t t k δ 1 ( β ¯ ) t k { ( L α I R α I ) } t ( I I ) = k = 0 t t k δ 1 ( β ¯ ) t k j = 0 k ( 1 ) j k j i = 1 d α i I k j i = 1 d α i I j = k = 0 t t k δ 1 ( β ¯ ) t k j = 0 k ( 1 ) j k j i = 1 d α i I k .

Hence, δ 1 ( λ ) t ( I I ) = 0 for all k > 0 , and if k = 0 , then δ 1 ( λ ) t ( I I ) = δ 1 ( β ¯ ) t ( I I ) .

Let y be a unit vector in and let { x p } be a sequence of unit vectors such that

lim p ( T 1 i λ i ) x p = 0 ; 1 i d .

Then,

δ 1 ( β ¯ ) m k ( I I ) = δ 1 ( λ ¯ ) m k ( I I ) = j = 0 m k ( 1 ) j m k j i = 1 d ( T 1 i * λ i ¯ ) I m k j i = 1 d ( T 1 i λ i ) I j lim p δ 1 ( β ¯ ) m k ( I I ) x p y , x p y

= j = 0 m k ( 1 ) j m k j lim p i = 1 d ( T 1 i λ i ) j x p , i = 1 d ( T 1 i λ i ) m k j x p y , y = 0

for all m k > 0 , and if m k = 0 , then

0 = δ 12 m ( I I ) = k = 0 m m k δ 1 ( β ¯ ) m k δ 2 ( β ) k ( I I ) = k = 0 m m k δ 2 ( β ) k ( δ 1 ( β ¯ ) m k ) ( I I ) , since [ δ 1 ( β ¯ ) , δ 2 ( β ) ] = 0 = δ 2 ( β ) m ( I I ) .

Let m 2 m be the least positive integer such that δ 2 ( β ) m 2 ( I I ) = 0 and δ 2 ( β ) m 2 1 ( I I ) 0 . Since

δ 2 ( β ) t ( I I ) = j = 0 t ( 1 ) j t j i = 1 d I ( T 2 i * β i ¯ ) t j i = 1 d I ( T 2 i + β i ) j = j = 0 t ( 1 ) j t j I i = 1 d ( T 2 i * β i ¯ ) t j I i = 1 d ( T 2 i β i ) j = I δ T 2 * + β ¯ , T 2 + β t ( I )

for all positive integers t , the assumption ( T 2 * + β ¯ , T 2 + β ) is strict m 2 -symmetric and implies that the sequence { δ T 2 * + β ¯ , T 2 + β r ( I ) } r = 0 m 2 1 is linearly independent.

To complete the proof, we observe that

0 = δ 12 m ( I I ) = k = 0 m m k δ 1 ( β ¯ ) m k δ 2 ( β ) k ( I I ) = k = 0 m m k δ 1 ( β ¯ ) m k ( I δ T 2 * + β ¯ , T 2 + β ( I ) ) = k = 0 m m k ( δ T 1 * + β , T 1 + β ¯ m k ( I ) I ) ( I δ T 2 * + β ¯ , T 2 + β k ( I ) ) = k = 0 m m k ( δ T 1 * + β , T 1 + β ¯ m k ( I ) δ T 2 * + β ¯ , T 2 + β k ( I ) ) .

Since δ T 2 * + β ¯ , T 2 + β t ( I ) = 0 for all t m 2 , the linear independence of the sequence { δ T 2 * + β ¯ , T 2 + β r ( I ) } r = 0 m 2 1 implies that δ T 1 * + β , T 1 + β ¯ t ( I ) = 0 for all t m 1 = m m 2 + 1 . In particular, ( T 1 * + β , T 1 + β ¯ ) is m 1 -symmetric.□

Theorem 3.3 begs the question of whether an anlogous result holds for m -isometric sums ( T 1 * I + I T 2 * , T 1 I + I T 2 ) . The following theorem is a result in this direction.

Theorem 3.4

If T i , 1 i 2 , are commuting d-tuples in B ( ) d such that ( T 1 * I + I T 2 * , T 1 I + I T 2 ) is strict m-isometric for some positive integer m, then there exist a scalar λ = ( λ 1 , , λ d ) and positive integers m i m such that T 2 μ is m 2 -nilpotent and ( T 1 * + λ ¯ , T 1 λ ) is m 1 -isometric.

Proof

The proof below depends in an essential way upon Theorem 3.2. The spectrum of T 2 being non-empty, there exists a scalar λ = ( λ 1 , , λ d ) σ a ( T 2 ) . Let { x p } be a sequence of unit vectors in such that lim p ( T 2 i λ i ) x p = 0 ; 1 i d . For such a scalar λ ,

T 1 * I + I T 2 * , T 1 I + I T 2 m ( I I ) = T 1 * I + I T 2 * , ( T 1 + λ ) I + I ( T 2 λ ) m ( I I ) = ( ( I L T 1 * I + I T 2 * R I ( T 2 λ ) ) + ( L T 1 * I R I ( T 2 λ ) + L I T 2 * R I ( T 2 λ ) ) ) m ( I I ) = j = 0 m m j m j ( M 1 + M 2 ) j ( I I ) ,

where we have set

( I L T 1 * I + I T 2 * R I ( T 2 λ ) ) m j = T 1 * I + I T 2 * , I ( T 2 λ ) m j = m j , L T 1 * I R I ( T 2 λ ) = M 1 , L I T 2 * R I ( T 2 λ ) = M 2 .

Let y be a unit vector and let { x p } be a sequence of unit vectors such that

lim p ( T 2 i λ i ) x p = 0 .

Then,

( M 1 + M 2 ) j ( I I ) = k = 0 j j k ( L T 1 * I R I ( T 2 λ ) ) j k ( L I T 2 * R I ( T 2 * λ ) ) k ( I I ) = k = 0 j j k ( L T 1 * I R I ( T 2 λ ) ) j k β = k k ! β ! T 2 * β ( T 2 λ ) β

implies

lim p ( M 1 + M 2 ) j y x p , y x p = k = 0 j j k ( L T 1 * I R I ( T 2 λ ) ) j k y , y β = k k ! β ! lim p ( T 2 λ ) β x p , T 2 β x p .

Hence, lim p ( M 1 + M 2 ) j y x p , y x p = 0 for all k > 0 , and if k = 0 , then

lim p ( M 1 + M 2 ) j y x p , y x p = lim p ( L T 2 * I R I ( T 2 λ ) ) j y x p , y x p = β = j j ! β ! y , T 2 y lim p ( T 2 λ ) β x p , x p = 0

for all j > 0 (and if j = 0 then it equals I I ). Conclusion:

T 1 * I + I T 2 * , T 1 I + I T 2 m ( I I ) = m ( I I ) = T 1 * I + I T 2 * , ( T 1 + λ ) I m ( I I ) = 0 .

Let n m be the least positive integer such that T 1 * I + I T 2 * , ( T 1 + λ ) I n ( I I ) = 0 . Then, an application of Theorem 3.2 proves the existence of a scalar μ ¯ = ( μ 1 ¯ , , μ d ¯ ) σ a ( T 2 * ) and positve integers m i , 1 i 2 , such that n = m 1 + m 2 1 , T 2 * μ ¯ (hence, T 2 μ ) is m 2 -nilpotent and ( T 1 * + μ ¯ , T 1 + λ ) is m 1 -isometric. The conclusion that T 2 μ is m 2 -nilpotent implies σ ( T 2 μ ) = σ a ( T 2 μ ) = σ a ( T 2 ) { μ } = { 0 } ; hence, λ σ a ( T 2 ) if and only if λ = μ . Consequently, ( T 1 * + λ ¯ , T 1 + λ ) is strict m 1 -isometric, T 2 λ is m 2 -nilpotent, and ( T 1 * + λ ¯ ) I + I ( T 2 * λ ¯ ) , ( T 1 + λ ) I + I ( T 2 λ ) , equivalently ( T 1 * I + I T 2 * , T 1 I + I T 2 ) , is m 1 + 2 m 2 2 = n + m 2 1 = m isometric.□

  1. Funding information: The second named author was supported by Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education (NRF-2019R1F1A1057574).

  2. Author contributions: All authors contributed equally to the writing of this article. All authors have accepted responsibility for entire content of the manuscript and approved its submission.

  3. Conflict of interest: The authors declare that they have no conflict of interest.

  4. Data availability statement: This article has no associated data.

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Received: 2023-06-27
Accepted: 2023-12-21
Published Online: 2024-11-19

© 2024 the author(s), published by De Gruyter

This work is licensed under the Creative Commons Attribution 4.0 International License.

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