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 `2 (G )
denote the Hilbert space of square-summable functions on G , with
canonical orthonormal basis {eg : g ∈ G } and let B(`2 (G )) denote
the bounded operators on `2 (G ). We identify X ∈ B(`2 (G )) with a
matrix X = (xg ,h ) where xg ,h = hXeh , eg i.
Let `∞ (G ) denote the bounded sequences on G and regard
`∞ (G ) ⊆ B(`2 (G )) by identifying a bounded sequence with the
corresponding diagonal operator.
Let E : B(`2 (G )) → `∞ (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 `∞ (G ) = C (βG ).
For each point w ∈ βG there is a homomorphism sw : `∞ (G ) → C
given by evaluation at w . These are the pure states.
The map sw ◦ E : B(`2 (G )) → C gives a canonical way to extend
this map to a state on B(`2 (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(`2 (G )) ?
Shorthand: I’ll write “KSP” for “the Kadison-Singer Problem”.
Vern Paulsen
Kadison-Singer
Outline
I
A Brief History
I
Anderson’s Paving: Progress and Problems
I
Syndetic Sets and Paving
I
Casazza’s Approach: Frames and the Feichtinger
Conjecture
I
Bessel Sequences Indexed by a Group
I
Feichtinger Conjecture and Fourier Frames
Lecture II:
I
Reproducing Kernels and the Nikolski Problem
I
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 `∞ (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(`2 (G )) a (K , )-paving of T is a partition of G into
K disjoint sets A1 , . . . , AK such that
kPAj [T − E (T )]PAj k ≤ kT − E (T )k,
where PAj is the diagonal projection onto the subspace spanned by
{eg : g ∈ Aj }.
Theorem (Anderson)
The following are equivalent:
I
KSP is true,
I
(paving) for each T ∈ B(`2 (G )) there is T < 1 and KT such
that T has a (KT , T )-paving,
I
(strong paving) for each < 1, there exists K = K () such
that every T ∈ B(`2 (G )) has a (K , )-paving.
Vern Paulsen
Kadison-Singer
Solution of an Old Paving Problem
Let G = N or Z, then T ∈ B(`2 (G )) is lower triangular provided
ti,j = 0 when i < j. Recall that not every operator can be written
as T1 + T2∗ 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(`2 (G ))
denote the left regular representation and let
L(G ) = λ(G )00 ⊆ B(`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 {g1 , ..., gm }
such that G = ∪m
i=1 gi · S.
n
A subset S ⊆ Z 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 , )-paved iff T can be (L, )-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
X
|hx, fg i|2 ≤ Bkxk2 .
g
A frame sequence provided that there exists B, A > 0 such that
X
Akxk2 ≤
|hx, fg i|2 ≤ Bkxk2 .
g
A Bessel sequence is bounded(below) if inf{kfg k : 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
X
X
X
A(
|αg |2 ) ≤ k
αg fg k2 ≤ B(
|αg |2 )
g
g
g
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:
I
KSP is true,
I
every bounded Bessel sequence can be partitioned into finitely
many Riesz basic sequences,
I
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:
X
X
khk2 =
|hh, fn i|2 and h =
hh, fn ifn
and uniform if kfn k is constant. If {f1 , ..., fn } is a uniform,
Parseval frame for Ck , then necessarily,
n
1
=
.
k
kfj k2
For this reason, given a uniform Parseval frame for a Hilbert space
we call kfn1k2 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 = {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 ) = dr e.
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 hfg , fh i = hfgk , fhk i.
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(n1 t1 + · · · + nN tN )).
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.
I
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 ).
I
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.
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
I
Nikolski’s Approach and RKHS
I
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) = hf , kx i.
The two variable function K (x, y ) = ky (x) is called the
reproducing kernel. We let hx = kkkxx k 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:
I
Nikolskii: The NP holds for the Hardy space H 2 (D).
I
Siep: The NP holds for the Bergman space B 2 (D).
I
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).
I
Lata(and others): The NP holds for the Segal-Bargmann
space, K (z, w ) = e z w̄ 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 ⊆ H 2 (D) and that for
f ∈ H, kf kH 2 ≤ kf kH .
For example, the Bergman space B 2 (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’ Hβ2 space?
2. Does NP hold for Shields’ Hβ2 spaces?
3. Let φ be a singular inner function. Does the FP hold for
H 2 (D) φH 2 (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(`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.
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[lim gλ hµ ].
λ
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 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 ) = `∞ (G ) ⊂ B(`2 (G )) with the diagonal
operators, then C (βG ) ×r G is exactly the C*-algebra generated by
`∞ (G ) ⊆ B(`2 (G )) and λ(G ) ⊂ B(`2 (G )),
which is often called the “Roe algebra.” So I’ll denote this
concrete C*-subalgebra by Roe(G ) ⊂ B(`2 (G )).
Thus, πt × λ : Roe(G ) → Roe(G ) is defined on finite sums by
X
X
πt × λ(
Dg λ(g )) =
πt (Dg )λ(g ).
Vern Paulsen
Kadison-Singer
On the other hand every element X ∈ B(`2 (G )) can be written as
a formal series
X
X ∼
Dg λ(g )
g ∈G
and we could try to define Φt : B(`2 (G )) → B(`2 (G )) by
X
Φt (X ) ∼
πt (Dg )λ(g ).
g ∈G
Vern Paulsen
Kadison-Singer
Theorem
I
Φt is a well-defined completely positive map and Φt (Y ) = Y
for Y ∈ L(G ),
I
the state st extends uniquely to B(`2 (G ))(respectively, L(G ))
iff Φt is the unique cp extension of πt × λ from Roe(G ) to
B(`2 (G ))(respectively, L(G )),
I
Φt (AXB) = πt × λ(A)Φt (X )πt × λ(B) for A, B ∈ Roe(G ),
and same for any cp extension,
I
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
over 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 : `∞ (G ) → C provided that every extension of st to B(`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 , )-paving for all > 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 , )-paving,
G = A1 ∪ · · · ∪ AK ,
then kPAi (T − E (T ))PAi k < Hence, kΦt (PAi (T − E (T ))PAi )k < .
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,
onto 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 6= 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