The Kadison-Singer and Paulsen Problems in Finite Frame Theory

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2 Peter G. CasazzaThe Kadison-Singer Problem and the Paulsen Problem. We warn the readerthat because we are restricting ourselves to finite dimensional Hilbert spaces,a significant body of literature on the infinite dimensional versions of theseproblems does not appear here.1.2 The Kadison-Singer ProblemFor over 50 years the Kadison-Singer Problem [36] has defied the best effortsof some of the most talented mathematicians of our time.Kadison-Singer Problem 1 (KS) Does every pure state on the (abelian)von Neumann algebra D of bounded diagonal operators on l 2 , the Hilbert spaceof square summable sequences on the integers, have a unique extension to a(pure) state on B(l 2 ), i.e., the von Neumann algebra of all bounded linearoperators on the Hilbert space l 2 ?A state of a von Neumann algebra R is a linear functional f on R for whichf(I) = 1 and f(T ) ≥ 0 whenever T ≥ 0 (whenever T is a positive operator).The set of states of R is a convex subset of the dual space of R which iscompact in the ω ∗ -topology. By the Krein-Milman theorem, this convex setis the closed convex hull of its extreme points. The extremal elements in thespace of states are called the pure states (of R).This problem arose from the very productive collaboration of Kadison andSinger in the 1950s when they were studying Dirac’s Quantum Mechanicsbook [30] which culminated in their seminal work on triangular operatoralgebras.It is now known that the 1959 Kadison-Singer Problem is equivalent tofundamental unsolved problems in a dozen areas of research in pure mathematics,applied mathematics and engineering (See [1, 2, 3, 4, 17, 26, 27, 37]and their references). We will not develop this topic in detail here since it isfundamentally an infinite dimensional problem and we are concentrating onfinite dimensional frame theory. In this chapter we will look at a number ofthese finite dimensional problems which are equivalent to KS and which areimpacted by finite frame theory. Most people today seem to agree with theoriginal statement of Kadison and Singer [36] that KS will have a negativeanswer and so all the equivalent forms will have negative answers also.1.2.1 The Paving ConjectureA significant advance on KS was made by Anderson [2] in 1979 when hereformulated KS into what is now known as the Paving Conjecture (Seealso [3, 4]). Lemma 5 of [36] shows a connection between KS and Paving. For
1 The Kadison-Singer and Paulsen Problems in Finite Frame Theory 3notation, if J ⊂ {1, 2, . . . , n}, the diagonal projection Q J is the matrixwhose entries are all zero except for the (i, i) entries for i ∈ J which are allone. For a matrix A = (a ij ) N i,j=1 let δ(A) = max 1≤i≤N |a ii |.Definition 1. An operator T ∈ B(l N 2 ) is said to have an (r, ɛ)-paving if thereis a partition {A j } r j=1 of {1, 2, . . . , N} so that‖Q Aj T Q Aj ‖ ≤ ɛ‖T ‖.Paving Conjecture 1 (PC) For every 0 < ɛ < 1, there is a natural numberr so that for every natural number N and every linear operator T on l N 2whose matrix has zero diagonal, T has an (r, ɛ)-paving.It is important that r not depend on N in PC. We will say that an arbitraryoperator T satisfies PC if T − D(T ) satisfies PC where D(T ) is the diagonalof T .The only large classes of operators which have been shown to be pavableare “diagonally dominant” matrices [6, 7, 11, 31], l 1 -localized operators [24],matrices with all entries real and positive [32], matrices with small coefficientsin comparison with the dimension [13] (See [40] for a paving into blocks ofconstant size), and Toeplitz operators over Riemann integrable functions (Seealso [33]). Also, in [9] there is an analysis of the paving problem for certainSchatten C p -norms.Theorem 1. The Paving Conjecture has a positive solution if any one of thefollowing classes satisfies the Paving Conjecture:1. Unitary operators. [27]2. Orthogonal projections. [27]3. Orthogonal projections with constant diagonal 1/2. [18]4. Positive operators. [27]5. Self-adjoint operators. [27]6. Gram matrices (〈ϕ i , ϕ j 〉) i,j∈I where T : l 2 (I) → l 2 (I) is a bounded linearoperator, and T e i = ϕ i , ‖T e i ‖ = 1 for all i ∈ I. [27]7. Invertible operators (or invertible operators with zero diagonal). [27]8. Triangular operators [38]Recently, Weaver [43] provided important insight into KS by giving anequivalent problem to PC in terms of projections.Conjecture 1 (Weaver). There exist universal constants 0 < δ, ɛ < 1 andr ∈ N so that for all N and all orthogonal projections P on l N 2 with δ(P ) ≤ δ,there is a paving {A j } r j=1 of {1, 2, . . . , N} so that ‖Q A jP Q Aj ‖ ≤ 1 − ɛ, forall j = 1, 2, . . . , r.This needs some explanation since there is nothing in [43] that looks anythinglike Conjecture 1. Weaver observes that the fact that Conjecture 1
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