Traces and spectral properties of shift operators Contribution to - PowerPoint PPT Presentation

Traces and spectral properties of shift operators Contribution to the Aleksander Pełczyński Memorial Conference ————————————————– Albrecht Pietsch 1

K. Kuratowski: A Half Century of Polish Mathematics 2

My personal intention is to improve mathematical theories by making them simpler and more elegant. It is desirable or even necessary that such modifications yield not only all well-known results, but also new ones. The today’s lecture is a typical realization of this philosophy. I going to describe a new view on traces, which could and should have been found already 25 years ago. In the meantime, trace theory and its applications to non-commutative geometry and pseudo-differential operators have developed quickly. Since all standard books are written on the basis of the classical technique, it will be hard to accomplish any change. To ensure acceptance of mathematical concepts, they must be presented at the right moment. 3

Gorbachev whispers into Honecker’s ear, October 1989: Wer zu spät kommt, den bestraft das Leben. Life punishes those who come too late. 4

Part 1 5

Sequence ideals — > Operator ideals S ∈ L ( X , Y ) n ∈ N := { 1 , 2 , 3 , . . . } n -th approximation number: � � � S − F � : rank ( F ) < n a n ( S ) := inf . With every set z ( N 0 ) of sequences we associate the class � D z := D z ( X , Y ) X , Y of operators, whose components are defined by � � � � S ∈ L ( X , Y ) : ∈ z ( N 0 ) D z ( X , Y ) := a 2 h ( S ) , h ∈ N 0 := { 0 , 1 , 2 , . . . } . � � � � a n ( S ) is replaced by the dyadic subsequence a 2 h ( S ) . 6

Which properties of z ( N 0 ) ensure that D z is an operator ideal ? (1) z ( N 0 ) is an ideal in l ∞ ( N 0 ) , the algebra of bounded sequences. e 0 =( 1 , 0 , 0 , . . . ) ∈ z ( N 0 ) guarantees that D z contains (2) non-zero operators. (3) If z ( N 0 ) is invariant under the forward shift S + : ( α 0 , α 1 , α 2 , . . . ) → ( 0 , α 0 , α 1 , . . . ) , then we infer from S , T ∈ L a 2 h + 1 ( S + T ) ≤ a 2 h ( S ) + a 2 h ( T ) − − − − − → Y for X that S , T ∈ D z ( X , Y ) ⇒ S + T ∈ D z ( X , Y ) . Since the ideal z ( N 0 ) is solid , ( β h ) ∈ z ( N 0 ) and | α h | ≤ | β h | ⇒ ( α h ) ∈ z ( N 0 ) , it follows from A ∈ L S ∈ L B ∈ L a n ( BSA ) ≤� B � a n ( S ) � A � for X 0 − − − → X − − − → Y − − − → Y 0 that S ∈ D z ( X , Y ) ⇒ BSA ∈ D z ( X 0 , Y 0 ) . 7

Shift-monotone sequence ideals The map z ( N 0 ) �→ D z is far from being one-to-one. What condition ensures that z ( N 0 ) can be reconstructed from D z ? a =( α h ) ∈ l ∞ ( N 0 ) h , k ∈ N 0 := { 0 , 1 , 2 , . . . } k -th ordering number: | α h | . o k ( a ) := sup h ≥ k If | α 0 |≥| α 1 |≥| α 2 |≥ . . . ≥ 0, then o k ( a )= | α k | . ℓ 2 : Hilbert sequence space over N := { 1 , 2 , 3 , . . . } . D a : dyadic diagonal operator on ℓ 2 associated with the sequence 2 k terms 2 k − 1 terms � �� � � �� � d a := ( , . . . ) for a =( α h ) ∈ l ∞ ( N 0 ) . α 0 , α 1 , α 1 , . . . , α k , . . . , α k � �� � � �� � 2 k ≤ n < 2 k + 1 1 ≤ n < 2 k Lemma o k ( a ) = a 2 k ( D a ) . 8

Shift-monotone sequence ideal z ( N 0 ) : (1) z ( N 0 ) is an ideal in l ∞ ( N 0 ) . e 0 =( 1 , 0 , 0 , . . . ) ∈ z ( N 0 ) . (2) (3) z ( N 0 ) is invariant under the forward shift S + . (4) a ∈ z ( N 0 ) , b ∈ l ∞ ( N 0 ) , and o k ( b ) ≤ o k ( a ) imply b ∈ z ( N 0 ) . Theorem The map z ( N 0 ) �→ D z is one-to-one : a ∈ z ( N 0 ) ⇔ D a ∈ D z ( ℓ 2 ) . � � Proof : Substitute b = o k ( a ) in (4) . � � � � a ∈ z ( N 0 ) ⇔ o k ( a ) ∈ z ( N 0 ) ⇔ a 2 k ( D a ) ∈ z ( N 0 ) ⇔ D a ∈ D z ( ℓ 2 ) . By o k ( S − a ) ≤ o k ( a ) , z ( N 0 ) is invariant under the backward shift S − : ( α 0 , α 1 , α 2 , . . . ) → ( α 1 , α 2 , α 3 , . . . ) . Monotony (4) means that every a = ( α h ) ∈ z ( N 0 ) is dominated by a monotone sequence b =( β h ) ∈ z ( N 0 ) ; that is | α h |≤ β h and β 0 ≥ β 1 ≥ β 2 ≥ · · · ≥ 0 . 9

The S ± -invariant sequence ideal ∞ � � � l 1 ( N 0 ) : = ( α h ) : | α h | < ∞ h = 0 fails to be monotone: n 2 terms � �� � ( 1 , 0 , 0 , 0 , 1 0 , . . . , 0 , 1 n 2 , 0 , . . . ) , 4 , . . . , corresponding sequence of ordering numbers n 2 terms � �� � ( 1 , 1 4 , 1 4 , 1 4 , 1 n 2 , . . . , 1 1 1 4 , . . . , n 2 , ( n + 1 ) 2 , . . . ) . Example ∞ � � � l 1 [ 2 − h ]( N 0 ) : = 2 h | α h | < ∞ ( α h ) : h = 0 is a shift-monotone sequence ideal. 10

Example � � c 0 ( N 0 ) := ( α h ) : lim h →∞ α h = 0 is the largest shift-monotone sequence ideal � = l ∞ ( N 0 ) . Operators in D c 0 are called approximable . In what follows, z ( N 0 ) denotes a shift-monotone sequence ideal contained in c 0 ( N 0 ) . 11

Dyadic representations ∞ � S = S k , convergent with respect to the operator norm, k = 0 S ∈ L ( X , Y ) and S k ∈ L ( X , Y ) such that rank ( S k ) ≤ 2 k . Theorem An operator is approximable if and only if it admits a dyadic representation. D k : orthogonal projection from ℓ 2 onto span { e n : 2 k ≤ n < 2 k + 1 } . ∞ � D a = α k D k for a = ( α h ) ∈ l ∞ ( N 0 ) k = 0 is a dyadic representation of the dyadic diagonal operator D a , h − 1 � � � � � o h ( a ) = � D a − α k D k � = a 2 h ( D a ) . k = 0 12

z -representations ∞ � rank ( S k ) ≤ 2 k S = S k such that and k = 0 h − 1 �� � � � � � � S − ∈ z ( N 0 ) . S k ( ♦ ) � k = 0 former definition: � � � S h � ∈ z ( N 0 ) . ( � ) It follows from the S − -invariance of z ( N 0 ) that ( ♦ ) ⇒ ( � ) : h − 1 h � � � � � � � � � � � S h � ≤ � S − S k � + � S − S k � . k = 0 k = 0 aha experience: ( ♦ ) ensures the convergence of � ∞ k = 0 S k , while ( � ) does not. 13

Theorem The ideal D z consists of all operators that admit a z -representation. Proof : Given S ∈ D z ( X , Y ) , there exists F k ∈ L ( X , Y ) such that rank ( F k ) < 2 k . � S − F k � ≤ 2 a 2 k ( S ) and Note that F 0 = O . Letting S 0 := O , S 1 := O , and S k + 2 := F k + 1 − F k for k = 0 , 1 , 2 , . . . , we get h + 1 h + 1 � � � � rank ( S k ) ≤ 2 k . � � S k = F h , � S − S k � = � S − F h � ≤ 2 a 2 h ( S ) , k = 0 k = 0 Hence the S + -invariance of z ( N 0 ) yields ( ♦ ) , which means that S = � ∞ k = 0 S k is a z -representation. 14

� h − 1 � � < 2 h rank ( S k ) ≤ 2 k , we have Since rank S k whenever k = 0 h − 1 � � � � � a 2 h ( S ) ≤ � S − S k � . k = 0 � � ∈ z ( N 0 ) or S ∈ D z ( X , Y ) . So, conversely, ( ♦ ) implies a 2 h ( S ) 15

Lemma If ∞ ∞ � � S = S k and T = T k k = 0 k = 0 are z -representations of S , T ∈ D z ( X , Y ) , then ∞ � S + T = Z k with Z 0 := O and Z k + 1 := S k + T k for k = 0 , 1 , 2 , . . . k = 0 is a z -representation of their sum. h h − 1 � � � � � � � � � � � S + T − Z k = � S + T − ( S k + T k ) � � k = 0 k = 0 h − 1 h − 1 � � � � � � � � � � ≤ � S − S k � + � T − T k � . k = 0 k = 0 16

1 2 S + -invariant linear forms on z ( N 0 ) 1 2 S + -invariance: � 1 � a ∈ z ( N 0 ) . λ 2 S + a = λ ( a ) for all In the optimal case, linear forms λ on sequence ideals admit an analytic representation in the form ∞ � λ ( a ) = for a =( α h ) . λ h α h h = 0 λ ( e h )= λ h for all unit sequences e h . If λ is 1 2 S + -invariant, then λ h + 1 = λ ( e h + 1 ) = λ ( S + e h ) = 2 λ ( e h ) = 2 λ h ∞ � λ h = 2 h λ 0 2 h α h . ⇒ ⇒ λ ( a ) = λ 0 h = 0 17

Such linear forms can only live on shift-monotone sequence ideals z ( N 0 ) that are contained in ∞ � � � l 1 [ 2 − h ]( N 0 ) := 2 h | α h | < ∞ ( α h ) : . h = 0 The existence of further 1 2 S + -invariant linear forms is based on the axiom of choice . So we are mostly dealing with esoteric objects that are floating somewhere in the air. The value of λ at a sequence a =( α h ) will be denoted by λ ( α h ) , � � instead of λ ( a )= λ ( α h ) . 18

Traces The concept of a trace makes sense only for square matrices. Similarly, in the setting of operators, we must assume that the operators act on one and the same Banach space. A trace on an operator ideal A is a linear form τ , defined on all components A ( X ):= A ( X , X ) , for which the following holds: AS ∈ A ✲ X X ❅ ❅ ❅ ❅ ❅ ❅ ❅ ❅ ❅ � ✒ � ✒ � � ✒ � ✒ ✒ ❅ S ∈ A � � � � S ∈ A τ ( AS ) = τ ( SA ) whenever ❅ ❅ ❅ ❅ � ❅ ❅ ❅ ❅ ❅ ❅ A ∈ L ❘ ❅ ❅ ❘ ❅ ❅ ❘ ❘ ❅ ❘ ❘ ❅ � � � � ❘ ❅ ❅ ❘ ❅ ❘ ❘ ❅ � ✲ Y Y . SA ∈ A 19

The smallest non-trivial ideal F consists of all finite rank operators. Every trace on F is a scalar multiple of the usual trace defined by n � � x m , x ∗ m � , trace ( F ) := m = 1 where F = � n m = 1 x ∗ m ⊗ x m with x 1 , . . . , x n ∈ X and x ∗ 1 , . . . , x ∗ n ∈ X ∗ is any finite representation. We have | trace ( F ) | ≤ n � F � whenever rank ( F ) ≤ n . 20