The Kadison-Singer Problem: An Overview and Potentially Tractable Cases

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1 The Problem: An Overview and Potentially Tractable Cases November 22, 2012

2 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.

3 Let ϕ : l (G) C be a homomorphism. (These are the pure states on l (G).) The map ϕ E : B(l 2 (G)) C gives a canonical way to extend ϕ to a state on B(l 2 (G)). The Problem For each homomorphism ϕ is the map ϕ E the unique way to extend ϕ to a state, i.e., a unital positive linear functional on B(l 2 (G))? Shorthand: I ll write KSP for the Problem.

4 If we let βg denote the Stone-Cech compactification of G, then we also may identify l (G) = C(βG). Each homomorphism ϕ is given by evaluation at a point w βg. So KSP is really a question about uniquenesss of extension for individual points in βg. Points in βg are given by ultrafilters. But very little work done on trying to determine uniqueness/non-uniqueness for particular ultrafilters. Main focus has been paving.

5 Outline A Brief History Anderson s Paving: Progress and Problems Casazza s Approach: Frames and the Feichtinger Conjecture Bessel Sequences Indexed by a Group and Syndetic Sets Feichtinger Conjecture and Fourier Frames Reproducing Kernels and the Nikolski Problem Further Results

6 A Brief History : 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 (N) 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 s. Including work by G. Weiss et al and Bourgain-Tzafriri.

7 P. Casazza, et al: 5) About 2004, discovered that a question of Feichtinger s about Gabor frames would be implied by KSP. Proved that a generalization of Feichtinger s problem was equivalent to KSP. Me: 6) Tried to focus on question for particular ultrafilters. Replaced N by an arbitrary countable, discrete group G. Structure of G yields interesting ultrafilters, leads to syndetic set results. Nikolski: 7) Studies frames in reproducing kernel Hilbert spaces.

8 Anderson s Paving Problems Given T B(l 2 (G)) a (K, ɛ)-paving of T is a partition of G into K disjoint sets A 1,..., A K such that P Aj [T E(T )]P Aj ɛ T E(T ), where P Aj 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 ɛ T < 1 and K T such that T has a (K T, ɛ T )-paving, (strong paving) for each ɛ < 1, there exists K = K(ɛ) such that every T B(l 2 (G)) has a (K, ɛ)-paving.

9 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 g x, f g 2 B x 2. 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 A( g α g 2 ) g α g f g 2 B( g α g 2 ) for all α g.

10 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. Note these are akin to weaker paving, i.e., no uniform bounds. Get uniform by looking at uniform Parseval case.

11 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 k = 1 f j 2. For this reason, given a uniform Parseval frame for a Hilbert space 1 we call f n the redundancy of the frame. 2

12 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.

13 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 = d + 1. Problem Find a uniform Parseval frame of redundancy 2 that cannot be partitioned into 3 Riesz basic sequences.

14 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.

15 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 is syndetic iff it has bounded gap size, i.e., iff there exists M > 0, so that adjacent elements are at most M apart. Note that such a set has positive lower Beurling density. Similarly, a subset S Z n is syndetic iff there exists M > 0 so that every n-cube with side M contains at least one element of S. 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. Thus, if invariant Bessel sequences can be Rieszed then each set in the partition would need to be thick. Hope is that this extra structure can be used for counterexample.

16 Application: Fourier Frames Let N be a natural number. Given n = (n 1,..., n N ) Z N and t = (t 1,..., t N ) [0, 1] N, we set e n (t) = exp(2πi(n 1 t n N t N )). Then {e n (t) : n Z N } is an orthonormal basis for L 2 ([0, 1] N ). If E [0, 1] N is a set of positive Lebesgue measure, then {e n (t)χ E : n Z N } is a uniform Parseval frame for L 2 (E) called the Fourier frame for E. Note that the redundancy of the Fourier frame is m(e) 1. Problem Does the FC hold for Fourier frames? Conjecture: No! Most attempts to find a counterexample have focused on N = 1. The key might be larger N, maybe N = 5, a la Tao s solution of Fuglede?

17 Corollary Let E [0, 1] N have positive Lebesgue measure. If the Fourier frame for L 2 (E) can be partitioned into K Riesz basic sequences, then Z N can be partitioned into L K syndetic sets S 1,..., S L such that {e n (t)χ E : n S j } is a Riesz basic sequence for all 1 j L. Proof: The Fourier frame is Z N -invariant. When N = 1, a result of Halperin-Kaftal-Weiss proves that the pavings can be taken to be congruence sets mod M for some M iff averages of χ E converge uniformly to the constant function m(e). Results of Bourgain imply that there exist sets E for which the Fourier frame can be partitioned into Riesz basic sequences, but cannot be partitioned by congruence sets. By our result those partitions can be taken syndetic.

18 Problem 1. Give an example of E with m(e) = 1/2 such that the Fourier frame cannot be partitioned into 3 Riesz basic sequences. 2. If the Fourier frame for each set with m(e) = 1/2 can be partitioned into finitely many Riesz basic sequences, then is there necessarily an upper bound on the number? Does this imply that every Fourier frame can be partitioned into finitely many Riesz basic sequences? (Yes, if KSP true.) 3. If the Fourier frame for every E [0, 1] can be partitioned into finitely many Riesz basic sequences, then can the Fourier frame for every E [0, 1] N be partitioned into finitely many Riesz basic sequences?

19 Nikolskii s Approach: Reproducing Kernel Hilbert Spaces Recall that a Hilbert space H of actual functions on a set X is called a reproducing kernel Hilbert space on X, provided that for each x X, the evaluation functional is bounded, i.e., there exists k x H, such that f (x) = f, k x. The two variable function K(x, y) = k y (x) is called the reproducing kernel. We let h x = normalized kernel. kx k x The Nikolskii Problem denote the so-called Given a RKHS H on X and a countable set of points Y such that {h x : x Y } is a Bessel sequence, one can ask if the FC is true for such sequences. When FC is true for all such sequences, we say that the Nikolskii Property holds for H, or more shortly, that NP holds for H.

20 This is motivated by the following: Nikolskii: The NP holds for the Hardy space H 2 (D). Siep: The NP holds for the Bergman space B 2 (D). Baronov-Dyakonov: The NP holds for some Sz.-Nagy-Foias model spaces, i.e., spaces of the form H 2 (D) φh 2 (D) where φ is an inner function.(not known for all inner functions). Lata: The NP holds for the Segal-Bargmann space, K(z, w) = e z w and the Dirichlet space. She also gives new proofs of the aove results. Problem Does the NP hold for the Drury-Arveson space?

21 An equivalence between the NP and the FC. First we need a definition. An RKHS H of functions on D is contractively contained in the Hardy space, provided that as a set H H 2 (D) and that for f H, f H 2 f H. For example, the Bergman space B 2 (D) is not, but the Sz.-Nagy-Foias model spaces and the Segal-Bargmann space are.

22 Theorem (Lata+P) The Feichtinger Conjecture is true iff the NP holds for every space contractively contained in the Hardy space. So the KSP is false iff the NP fails for some particular contractively contained space. Which spaces are good candidates? Problem 1. Does the NP hold for every contractively contained space iff it holds for every contractively contained Shields H 2 β space? 2. Does NP hold for Shields H 2 β spaces? 3. Let φ be a singular inner function. Does the FP hold for H 2 (D) φh 2 (D)?

23 Further Results: 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. Yields NO information on relationship between paving estimates (K, ɛ) for lower triangular operators and all operators!

24 Let : B(l 2 ) B(l 2 ) be the (unbounded) operator of upper triangular truncation. Problem Is Feichtinger s conjecture true iff every bounded below Bessel(frame) sequence such that (( f j, f i )) is bounded can be partitioned into finitely many Riesz basic sequences? Problem Give a geometric characterization of Bessel sequences such that ( ( f j, f i ) ) is bounded.

25 Group von Neumann Algebras 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!

26 Theorem Let G be a countable group and let T L(G). Then T can be (K, ɛ)-paved iff it can be (L, ɛ)-paved by syndetic sets with L K. 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.

27 There is a natural action of the group G on βg, so this makes the pair a dynamical system. Problem Are there any special dynamical properties of a point in w βg that guarantees the homomorphism of evaluation at w has a unique extension? Alternatively, can one find dynamical properties that guarantee non-uniqueness of the extension? Theorem (unpubl) If there exists any countable, discrete group G such that s w has a unique extension for every strongly wandering point in βg, then is true.

The Kadison-Singer Problem: An Overview and Test Problems

The Kadison-Singer Problem: An Overview and Test Problems The Problem: An Overview and Test Problems LarsonFest July 16, 2012 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