The Kadison-Singer Problem: An Overview and Test Problems Vern - - PowerPoint PPT Presentation

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 ℓ2(G) denote the Hilbert space of square-summable functions on G, with canonical orthonormal basis {eg : g ∈ G} and let B(ℓ2(G)) denote the bounded operators on ℓ2(G). We identify X ∈ B(ℓ2(G)) with a matrix X = (xg,h) where xg,h = Xeh, eg. Let ℓ∞(G) denote the bounded sequences on G and regard ℓ∞(G) ⊆ B(ℓ2(G)) by identifying a bounded sequence with the corresponding diagonal operator. Let E : B(ℓ2(G)) → ℓ∞(G), be defined by letting E(X) be the diagonal matrix with entries xg,g.

Vern Paulsen Kadison-Singer

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 sw : ℓ∞(G) → C given by evaluation at w. These are the pure states. The map sw ◦ E : B(ℓ2(G)) → C gives a canonical way to extend this map to a state on B(ℓ2(G)). The Kadison-Singer Problem For each w ∈ βG, is sw ◦ E the only way to extend sw to a state, i.e., a positive linear functional, on B(ℓ2(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(L2[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 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.

Vern Paulsen Kadison-Singer

Anderson’s Paving Problems

Given T ∈ B(ℓ2(G)) a (K, ǫ)-paving of T is a partition of G into K disjoint sets A1, . . . , AK such that PAj[T − E(T)]PAj ≤ ǫT − E(T), where PAj is the diagonal projection onto the subspace spanned by {eg : g ∈ Aj}.

Theorem (Anderson)

The following are equivalent:

◮ KSP is true, ◮ (paving) for each T ∈ B(ℓ2(G)) there is ǫT < 1 and KT such

that T has a (KT, ǫT)-paving,

◮ (strong paving) for each ǫ < 1, there exists K = K(ǫ) such

that every T ∈ B(ℓ2(G)) has a (K, ǫ)-paving.

Vern Paulsen Kadison-Singer

Solution of an Old Paving Problem

Let G = N or Z, then T ∈ B(ℓ2(G)) is lower triangular provided ti,j = 0 when i < j. Recall that not every operator can be written as T1 + T ∗

2 with Ti 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(ℓ2(G)) denote the left regular representation and let L(G) = λ(G)′′ ⊆ B(ℓ2(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 {g1, ..., gm} such that G = ∪m

i=1gi · S.

A subset S ⊆ Zn 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 {fg : g ∈ G} in a Hilbert space H is a Bessel sequence, provided that there exists B such that

• g

|x, fg|2 ≤ Bx2. A frame sequence provided that there exists B, A > 0 such that Ax2 ≤

• g

|x, fg|2 ≤ Bx2. A Bessel sequence is bounded(below) if inf{fg : g ∈ G} > 0. A set of vectors {fg : g ∈ G} is a Riesz basic sequence provided that there exist constants A, B > 0, such that A(

• g

|αg|2) ≤

• g

αgfg2 ≤ B(

• g

|αg|2) for all αg.

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 {fn} is called a Parseval frame if A = B = 1, which is equivalent to: h2 =

• |h, fn|2 and h =
• h, fnfn

and uniform if fn is constant. If {f1, ..., fn} is a uniform, Parseval frame for Ck, then necessarily, n k = 1 fj2 . For this reason, given a uniform Parseval frame for a Hilbert space we call

1 fn2 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
• f 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 = {f1, .., fn} be a uniform Parseval frame for Ck 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 {fg : g ∈ G} G-invariant provided that for all g, h, k ∈ G, we have fg, fh = fgk, fhk.

Proposition

Let G be a countable discrete group and let {fg : g ∈ G} be a Bessel sequence whose closed linear span is H. Then {fg : g ∈ G} is G-invariant iff there is a unitary representation π : G → B(H) such that π(k−1)fg = fgk.

Theorem

Let B = {fg : 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 = S1 ∪ · · · ∪ SL such that {fg : g ∈ Sj} is a Riesz basic sequence for all j.

Vern Paulsen Kadison-Singer

Application: Fourier Frames

Let N be a natural number. Given n = (n1, ..., nN) ∈ ZN and t = (t1, ..., tN) ∈ [0, 1]N, we set en(t) = exp(2πi(n1t1 + · · · + nNtN)). Then {en(t) : n ∈ ZN} is an orthonormal basis for L2([0, 1]N). If E ⊆ [0, 1]N is a set of positive Lebesgue measure, then {en(t)χE : n ∈ ZN} is a uniform Parseval frame for L2(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?

Vern Paulsen Kadison-Singer

Corollary

Let E ⊆ [0, 1]N have positive Lebesgue measure. If the Fourier frame for L2(E) can be partitioned into K Riesz basic sequences, then ZN can be partitioned into L ≤ K syndetic sets S1, ..., SL such that {en(t)χE : n ∈ Sj} is a Riesz basic sequence for all 1 ≤ j ≤ L. Proof: The Fourier frame is ZN-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 χE 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

• ur result those partitions can be taken syndetic.

Vern Paulsen Kadison-Singer

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 L2(E) can be partitioned into K Riesz basic sequences, then ZN can be partitioned into L ≤ K syndetic sets S1, ..., SL such that {en(t)χE : n ∈ Sj} is a Riesz basic sequence for all 1 ≤ j ≤ L. Moreover, there exists a syndetic set S and n1, ..., nL ∈ ZN such that ni + S ⊆ Si with Si\(ni + 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 (ni + S) for some syndetic set S ? Can an example of a set be found that fails this stronger form of paving?

Vern Paulsen Kadison-Singer

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?

Vern Paulsen Kadison-Singer

Lecture II

◮ Nikolski’s Approach and RKHS ◮ Groups and Ultrafilters

Vern Paulsen Kadison-Singer

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 kx ∈ H, such that f (x) = f , kx. The two variable function K(x, y) = ky(x) is called the reproducing kernel. We let hx =

kx kx denote the so-called

“normalized” kernel. The Nikolskii Problem Given a RKHS H on X and a countable set of points Y such that {hx : 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.

Vern Paulsen Kadison-Singer

This is motivated by the following:

◮ Nikolskii: The NP holds for the Hardy space H2(D). ◮ Siep: The NP holds for the Bergman space B2(D). ◮ Baronov-Dyakonov: The NP holds for some Sz.-Nagy-Foias

“model” spaces, i.e., spaces of the form H2(D) ⊖ φH2(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) = ez ¯

w and the Dirichlet space.

Problem

Does the NP hold for the Drury-Arveson space?

Vern Paulsen Kadison-Singer

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 ⊆ H2(D) and that for f ∈ H, f H2 ≤ f H. For example, the Bergman space B2(D) is not, but the Sz.-Nagy-Foias model spaces and the Segal-Bargmann space are.

Vern Paulsen Kadison-Singer

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’ H2

β space?

• 2. Does NP hold for Shields’ H2

β spaces?

• 3. Let φ be a singular inner function. Does the FP hold for

H2(D) ⊖ φH2(D)?

Vern Paulsen Kadison-Singer

Ultrafilters and βG

The Kadison-Singer Problem For each w ∈ βG, is sw ◦ E the only way to extend sw to a state, i.e., a positive linear functional, on B(ℓ2(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.

Vern Paulsen Kadison-Singer

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.

Vern Paulsen Kadison-Singer

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 = limλ gλ and t = limµ hµ two points in βG, and set s · t = lim

λ [lim µ gλhµ].

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 πt : C(βG) → C(βG) via πt(f )(s) = f (s · t). This is covariant and extends to a *-homomorphism, πt × λ : C(βG) ×r G → C(βG) ×r G.

Vern Paulsen Kadison-Singer

When we identify C(βG) = ℓ∞(G) ⊂ B(ℓ2(G)) with the diagonal

• perators, then C(βG) ×r G is exactly the C*-algebra generated by

ℓ∞(G) ⊆ B(ℓ2(G)) and λ(G) ⊂ B(ℓ2(G)), which is often called the “Roe algebra.” So I’ll denote this concrete C*-subalgebra by Roe(G) ⊂ B(ℓ2(G)). Thus, πt × λ : Roe(G) → Roe(G) is defined on finite sums by πt × λ(

• Dgλ(g)) =
• πt(Dg)λ(g).

Vern Paulsen Kadison-Singer

On the other hand every element X ∈ B(ℓ2(G)) can be written as a formal series X ∼

• g∈G

Dgλ(g) and we could try to define Φt : B(ℓ2(G)) → B(ℓ2(G)) by Φt(X) ∼

• g∈G

πt(Dg)λ(g).

Vern Paulsen Kadison-Singer

Theorem

◮ Φt is a well-defined completely positive map and Φt(Y ) = Y

for Y ∈ L(G),

◮ the state st extends uniquely to B(ℓ2(G))(respectively, L(G))

iff Φt is the unique cp extension of πt × λ from Roe(G) to B(ℓ2(G))(respectively, L(G)),

◮ Φt(AXB) = πt × λ(A)Φt(X)πt × λ(B) for A, B ∈ Roe(G),

and same for any cp extension,

◮ Φs ◦ Φt = Φs·t.

Note that πt × λ vanishes on the compacts whenever t ∈ βG\G, so Φt is very singular. In fact Φt vanishes on any operator whose diagonals are all c0.

Vern Paulsen Kadison-Singer

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 πt × λ(Roe(G)) contains L(G) as a C*-subalgebra. Also, the Φt maps are L(G)-bimodule maps and can be used to show that some conjectures about behaviour of bimodule maps

• ver W*-algebras are stronger than KSP. See work of Neufang. I

used these maps to settle a conjecture of Solel in the negative.

Vern Paulsen Kadison-Singer

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 st : ℓ∞(G) → C provided that every extension of st to B(ℓ2(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, ǫ)-paving for all ǫ > 0.

Problem

If for a minimal idempotent t the state st has a unique extension, then is KSP true? “Syndetic” results also come from studying the maps Φt when t is a minimal idempotent point.

Vern Paulsen Kadison-Singer

Proposition

Let t be a minimal idempotent, A ⊆ G, then πt(PA) = PS with S ⊆ G either empty or syndetic. Now suppose that T ∈ L(G) and T has a (K, ǫ)-paving, G = A1 ∪ · · · ∪ AK, then PAi(T − E(T))PAi < ǫ Hence, Φt(PAi(T − E(T))PAi) < ǫ. But Φt(PAi(T − E(T))PAi) = πt(PAi)Φt(T − E(T))πt(PAi) = PSi(T − E(T))PSi. Since I = PA1 + · · · + PAK we have I = PS1 + · · · + PSK .

Vern Paulsen Kadison-Singer

Further Results

Caution: If t1 ∈ βG1 and there exists f : G1 → G2 one-to-one,

• nto with f (t1) = t2 then uniqueness/non-uniqueness of extension

for t1 is equivalent for t2. 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 g1 · U ∩ g2 · U = ∅, for all g1 = g2. Rare ultrafilters have the property that they are strongly wandering no matter what the group is!

Vern Paulsen Kadison-Singer

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!

Vern Paulsen Kadison-Singer