The Kadison-Singer Problem: An Overview and Test Problems Vern Paulsen LarsonFest July 16, 2012 Vern Paulsen Kadison-Singer
Kadison-Singer Problem Let G be a countably infinite, discrete set, e.g., G = N , let ℓ 2 ( G ) denote the Hilbert space of square-summable functions on G , with canonical orthonormal basis { e g : g ∈ G } and let B ( ℓ 2 ( G )) denote the bounded operators on ℓ 2 ( G ) . We identify X ∈ B ( ℓ 2 ( G )) with a matrix X = ( x g , h ) where x g , h = � Xe h , e g � . Let ℓ ∞ ( G ) denote the bounded sequences on G and regard ℓ ∞ ( G ) ⊆ B ( ℓ 2 ( G )) by identifying a bounded sequence with the corresponding diagonal operator. Let E : B ( ℓ 2 ( G )) → ℓ ∞ ( G ) , be defined by letting E ( X ) be the diagonal matrix with entries x g , g . Vern Paulsen Kadison-Singer
If we let β G denote the Stone-Cech compactification of G , then we also may identify ℓ ∞ ( G ) = C ( β G ) . For each point w ∈ β G there is a homomorphism s w : ℓ ∞ ( G ) → C given by evaluation at w . These are the pure states. The map s w ◦ E : B ( ℓ 2 ( G )) → C gives a canonical way to extend this map to a state on B ( ℓ 2 ( G )) . The Kadison-Singer Problem For each w ∈ β G , is s w ◦ E the only way to extend s w to a state, i.e., a positive linear functional, on B ( ℓ 2 ( G )) ? Shorthand: I’ll write “KSP” for “the Kadison-Singer Problem”. Vern Paulsen Kadison-Singer
Outline ◮ A Brief History ◮ Anderson’s Paving: Progress and Problems ◮ Syndetic Sets and Paving ◮ Casazza’s Approach: Frames and the Feichtinger Conjecture ◮ Bessel Sequences Indexed by a Group ◮ Feichtinger Conjecture and Fourier Frames Lecture II: ◮ Reproducing Kernels and the Nikolski Problem ◮ Dynamical Systems and Ultrafilters Vern Paulsen Kadison-Singer
A Brief History Kadison-Singer: 1) Proved that KSP is equivalent to the “discrete” case of a “fact” stated in Dirac’s book Quantum Mechanics. 2) When the discrete diagonal ℓ ∞ ( G ) is replaced by a “continuous diagonal” like L ∞ [0 , 1] ⊆ B ( L 2 [0 , 1]) , then Dirac’s statement is false. 3) Showed problem equivalent to convergence of certain “pavers”. J. Anderson: 4) Refined paving idea into a set of “paving conjectures” all of which were equivalent to KSP. Anderson’s work motivated a great deal of research on these paving conjectures in 1970–1980’s. Including work by Birman-Halperin-Kaftal-Weiss and Bourgain-Tzafriri. Vern Paulsen Kadison-Singer
P. Casazza, et al: 5) About 2004, discovered that some questions in signal processing about Gabor frames due to Feichtinger would be implied by KSP. Proved that a generalization of Feichtinger’s problem was equivalent to KSP. Today: 6) My work replaced N by an arbitrary countable, discrete group G . Used results about the behaviour of the action of G on β G to seek potential ultrafilters for which extension is unique or non-unique. Yielded syndetic sets results, we will see relations to paving. Tomorrow: 7) Nikolskii’s approach–frames in reproducing kernel Hilbert spaces. 8) Careful look at groups and ultrafilters, source of syndetic sets results. Vern Paulsen Kadison-Singer
Anderson’s Paving Problems Given T ∈ B ( ℓ 2 ( G )) a ( K , ǫ ) -paving of T is a partition of G into K disjoint sets A 1 , . . . , A K such that � P A j [ T − E ( T )] P A j � ≤ ǫ � T − E ( T ) � , where P A j is the diagonal projection onto the subspace spanned by { e g : g ∈ A j } . Theorem (Anderson) The following are equivalent: ◮ KSP is true, ◮ (paving) for each T ∈ B ( ℓ 2 ( G )) there is ǫ T < 1 and K T such that T has a ( K T , ǫ T ) -paving, ◮ (strong paving) for each ǫ < 1 , there exists K = K ( ǫ ) such that every T ∈ B ( ℓ 2 ( G )) has a ( K , ǫ ) -paving. Vern Paulsen Kadison-Singer
Solution of an Old Paving Problem Let G = N or Z , then T ∈ B ( ℓ 2 ( G )) is lower triangular provided t i , j = 0 when i < j . Recall that not every operator can be written as T 1 + T ∗ 2 with T i lower triangular. Theorem (Raghupathi-P) Every operator can be paved iff every lower triangular operator can be paved. Every Toeplitz operator(respectively, Laurent operator) can be paved iff every Toeplitz operator(resp, Laurent) with analytic symbol can be paved. Proof used uniqueness of state extensions instead of paving and lemma about states. Vern Paulsen Kadison-Singer
Open Paving Test Problems Problem (Weiss, et al) Let G = N or Z . Can the Toeplitz or Laurent operators be paved? Let G be any countable discrete group, let λ : G → B ( ℓ 2 ( G )) denote the left regular representation and let L ( G ) = λ ( G ) ′′ ⊆ B ( ℓ 2 ( G )) denote the group von Neumann algebra. Problem Is there any G for which the elements of L ( G ) are pavable? Answer not known for any countable discrete group! Vern Paulsen Kadison-Singer
Syndetic Sets and Paving A set S ⊂ G is syndetic if there exists a finite set { g 1 , ..., g m } such that G = ∪ m i =1 g i · S . A subset S ⊆ Z n is syndetic iff it has bounded gap size, i.e., iff there exists M > 0 , so that every n -cube with side M contains at least one element of S . Theorem Let G be a countable group and let T ∈ L ( G ) . Then T can be ( K , ǫ ) -paved iff T can be ( L , ǫ ) -paved by syndetic sets, L ≤ K . Consequently, if a Laurent operator can be paved, it can be paved with sets of positive lower Beurling density. So the hope is that this result could be used to prove that there exists G and T that cannot be paved and hence show that KSP is false. Vern Paulsen Kadison-Singer
The Feichtinger Conjecture A set of vectors { f g : g ∈ G } in a Hilbert space H is a Bessel sequence, provided that there exists B such that |� x , f g �| 2 ≤ B � x � 2 . � g A frame sequence provided that there exists B , A > 0 such that A � x � 2 ≤ |� x , f g �| 2 ≤ B � x � 2 . � g A Bessel sequence is bounded (below) if inf {� f g � : g ∈ G } > 0 . A set of vectors { f g : g ∈ G } is a Riesz basic sequence provided that there exist constants A , B > 0 , such that α g f g � 2 ≤ B ( � � � | α g | 2 ) ≤ � | α g | 2 ) A ( g g g for all α g . Vern Paulsen Kadison-Singer
The Feichtinger Conjecture(s) Every bounded Bessel(respectively, frame) sequence can be partitioned into finitely many Riesz basic sequences. Theorem (Casazza-Tremain) The following are equivalent: ◮ KSP is true, ◮ every bounded Bessel sequence can be partitioned into finitely many Riesz basic sequences, ◮ every bounded frame sequence can be partitioned into finitely many Riesz basic sequences. Vern Paulsen Kadison-Singer
KSP and Frame Redundancy A frame { f n } is called a Parseval frame if A = B = 1 , which is equivalent to: � h � 2 = |� h , f n �| 2 and h = � � � h , f n � f n and uniform if � f n � is constant. If { f 1 , ..., f n } is a uniform, Parseval frame for C k , then necessarily, n 1 k = � f j � 2 . For this reason, given a uniform Parseval frame for a Hilbert space 1 we call � f n � 2 the redundancy of the frame. Vern Paulsen Kadison-Singer
Theorem (Casazza-Kalra-P, Bodmann-Casazza-P-Speegle) The following are equivalent: 1. KSP is true, 2. every uniform Parseval frame of redundancy 2 can be partitioned into finitely many Riesz basic sequences, 3. there is a constant K such that every uniform Parseval frame of redundancy 2 can be partitioned into K Riesz basic sequences. 4. for a fixed r > 1 every uniform Parseval frame of redundancy r can be partitioned into finitely many Riesz basic sequences, 5. for every r there is K = K ( r ) such that every uniform Parseval frame of redundancy r can be partitioned into K Riesz basic sequences. Vern Paulsen Kadison-Singer
Theorem (Bodmann-Casazza-P-Speegle) Let F = { f 1 , .., f n } be a uniform Parseval frame for C k and write n = dk + q . Then F can be partitioned into d linearly independent and spanning sets (of k vectors each) and a “left over” set of q linearly independent vectors. Hence, in finite dimensions the strong version (5) holds with K ( r ) = ⌈ r ⌉ . Problem Find a uniform Parseval frame of redundancy 2 that cannot be partitioned into 3 Riesz basic sequences. Vern Paulsen Kadison-Singer
Invariant Bessel Sequences Let G be a countable (discrete) group. We will call a Bessel sequence { f g : g ∈ G } G-invariant provided that for all g , h , k ∈ G , we have � f g , f h � = � f gk , f hk � . Proposition Let G be a countable discrete group and let { f g : g ∈ G } be a Bessel sequence whose closed linear span is H . Then { f g : g ∈ G } is G-invariant iff there is a unitary representation π : G → B ( H ) such that π ( k − 1 ) f g = f gk . Theorem Let B = { f g : g ∈ G } be a bounded Bessel sequence that is G-invariant. If B can be partitioned into K Riesz basic sequences, then B can be partitioned into L ≤ K syndetic sets G = S 1 ∪ · · · ∪ S L such that { f g : g ∈ S j } is a Riesz basic sequence for all j . Vern Paulsen Kadison-Singer
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