The Kadison-Singer Problem: An Overview and Test Problems

Transcription

1 The Problem: An Overview and Test Problems LarsonFest July 16, 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 If we let βg denote the Stone-Cech compactification of G, then we also may identify l (G) = C(βG). For each point w βg 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 Problem For each w βg, 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 Problem.

4 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

5 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 (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 s. Including work by Birman-Halperin-Kaftal-Weiss and Bourgain-Tzafriri.

6 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 βg 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.

7 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.

8 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.

9 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!

10 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, ɛ)-paved iff T can be (L, ɛ)-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.

11 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.

12 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.

13 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

14 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.

15 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.

16 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.

17 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?

18 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.

19 Problem What about quasi-periodic analogues of the Halperin-Kaftal-Weiss result? Theorem (unpubl) 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. Moreover, there exists a syndetic set S and n 1,..., n L Z N such that n i + S S i with S i \(n i + S) either empty or syndetic. Problem If a Fourier frame can be partitioned into Riesz basic sequences, then can a partition always be taken of the form (n i + S) for some syndetic set S? Can an example of a set be found that fails this stronger form of paving?

20 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 every set with m(e) = 1/2 can be partitioned into finitely many Riesz basic sequences, then can every Fourier frame be partitioned into finitely many Riesz basic sequences? 3. (Hadwin-P) Let E [0, 1] with m(e (a, b)) > 0 and m(e c (a, b)) > 0, for every (a, b). Can the Fourier frame for E be partitioned into finitely many Riesz basic sequences? The answer is yes for some such sets, but believe that there exist sets of this type that are counterexamples. 4. 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?

21 Lecture II Nikolski s Approach and RKHS Groups and Ultrafilters

22 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 in H, or more shortly, that NP holds in H.

23 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(and others): The NP holds for the Segal-Bargmann space, K(z, w) = e z w and the Dirichlet space. Problem Does the NP hold for the Drury-Arveson space?

24 My student Sneh Lata obtained 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.

25 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)?

26 Ultrafilters and βg The Problem For each w βg, 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))? So this is really a question about uniqueness of extension for individual points in βg. One way to represent points in βg is as ultrafilters. So to prove or disprove KSP we could be studying ultrafilters! But the paving approach somehow sweeps this under the rug! Consequently, very little work done on trying to determine uniqueness/non-uniqueness of extension for particular ultrafilters. Notable exceptions are Reid who proved that the extension is unique for rare ultrafilters. A nicer proof of this fact was given by Anderson. Whole thing a bit odd, because existence of rare ultrafilters depends on using continuum hypothesis.

27 Recent work of Bice gives first examples of free ultrafilters, i.e, points in βg\g for which the extension is unique and that exist without using additional set theoretic hypotheses. My work uses dynamical systems classification of points in βg. If KSP true, then get uniqueness of extension for special ultrafilters. This is what leads to the syndetic sets results.

28 Groups: The Basic Construct Given a countable discrete group G, there are two Arens-like associative multiplications on βg. These make βg a compact semi-group! One is defined as follows: let s = lim λ g λ and t = lim µ h µ two points in βg, and set s t = lim λ [lim µ g λ h µ ]. For s = g G this product defines a left action of G on βg. This allows us to look at dynamical properties of points in βg and try to relate to KSP. For a fixed t βg define π t : C(βG) C(βG) via π t (f )(s) = f (s t). This is covariant and extends to a *-homomorphism, π t λ : C(βG) r G C(βG) r G.

29 When we identify C(βG) = l (G) B(l 2 (G)) with the diagonal operators, then C(βG) r G is exactly the C*-algebra generated by l (G) B(l 2 (G)) and λ(g) B(l 2 (G)), which is often called the Roe algebra. So I ll denote this concrete C*-subalgebra by Roe(G) B(l 2 (G)). Thus, π t λ : Roe(G) Roe(G) is defined on finite sums by π t λ( D g λ(g)) = π t (D g )λ(g).

30 On the other hand every element X B(l 2 (G)) can be written as a formal series X g G D g λ(g) and we could try to define Φ t : B(l 2 (G)) B(l 2 (G)) by Φ t (X ) g G π t (D g )λ(g).

31 Theorem Φ t is a well-defined completely positive map and Φ t (Y ) = Y for Y L(G), the state s t extends uniquely to B(l 2 (G))(respectively, L(G)) iff Φ t is the unique cp extension of π t λ from Roe(G) to B(l 2 (G))(respectively, L(G)), Φ t (AXB) = π t λ(a)φ t (X )π t λ(b) for A, B Roe(G), and same for any cp extension, Φ s Φ t = Φ s t. Note that π t λ vanishes on the compacts whenever t βg\g, so Φ t is very singular. In fact Φ t vanishes on any operator whose diagonals are all c 0.

32 This theorem allows one to find many C*-algebra and W*-algebra connections with KSP that could potentially settle the problem. Theorem If KSP is true, then for every countable discrete group G, and every point t βg the injective envelope of π t λ(roe(g)) contains L(G) as a C*-subalgebra. Also, the Φ t maps are L(G)-bimodule maps and can be used to show that some conjectures about behaviour of bimodule maps over W*-algebras are stronger than KSP. See work of Neufang. I used these maps to settle a conjecture of Solel in the negative.

33 Problem: Can one use dynamical properties of t to get information about uniqueness or non-uniqueness of extensions? Definition We say that X is in the uniqueness domain for the state s t : l (G) C provided that every extension of s t to B(l 2 (G)) has the same value when applied to X. Theorem (unpubl) Let t be a minimal idempotent and let X L(G). Then X is in the uniqueness domain for t iff X is in the uniqueness domain for every w βg iff X has a (K, ɛ)-paving for all ɛ > 0. Problem If for a minimal idempotent t the state s t has a unique extension, then is KSP true? Syndetic results also come from studying the maps Φ t when t is a minimal idempotent point.

34 Proposition Let t be a minimal idempotent, A G, then π t (P A ) = P S with S G either empty or syndetic. Now suppose that T L(G) and T has a (K, ɛ)-paving, G = A 1 A K, then P Ai (T E(T ))P Ai < ɛ Hence, Φ t (P Ai (T E(T ))P Ai ) < ɛ. But Φ t (P Ai (T E(T ))P Ai ) = π t (P Ai )Φ t (T E(T ))π t (P Ai ) = P Si (T E(T ))P Si. Since I = P A1 + + P AK we have I = P S1 + + P SK.

35 Further Results Caution: If t 1 βg 1 and there exists f : G 1 G 2 one-to-one, onto with f (t 1 ) = t 2 then uniqueness/non-uniqueness of extension for t 1 is equivalent for t 2. But this Rudin-Kiesler equivalence messes with most dynamical properties. The set G = βg\g is often called the corona. This set is invariant under the action of G. We will call a point w G strongly wandering if it has an open neighborhood w U G such that g 1 U g 2 U =, for all g 1 g 2. Rare ultrafilters have the property that they are strongly wandering no matter what the group is!

36 Problem 1. Are Bice s ultrafilters universally strongly wandering in this sense? 2. If a point w G is strongly wandering then is the extension unique? Theorem (unpubl) Every w G is Rudin-Kiesler equivalent to a point t G that is strongly wandering. If there exists a group G such that the state extension is unique for every strongly wandering point in G then KSP true. So enough to study strongly wandering points to determine truth/falsity of KSP. But syndetic results came from idempotent points, which are very strongly recurrent! So once again KSP is being the clever trickster!