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The Solution of the Kadison-Singer Problem Adam Marcus (Crisply, Yale) Daniel Spielman (Yale) Nikhil Srivastava (MSR/Berkeley) IAS, Nov 5, 2014

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The Kadison-Singer Problem (‘59) A positive solution is equivalent to: Anderson’s Paving Conjectures (‘79, ‘81) Bourgain-Tzafriri Conjecture (‘91) Feichtinger Conjecture (‘05) Many others Implied by: Akemann and Anderson’s Paving Conjecture (‘91) Weaver’s KS 2 Conjecture

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A positive solution is equivalent to: Anderson’s Paving Conjectures (‘79, ‘81) Bourgain-Tzafriri Conjecture (‘91) Feichtinger Conjecture (‘05) Many others Implied by: Akemann and Anderson’s Paving Conjecture (‘91) Weaver’s KS 2 Conjecture The Kadison-Singer Problem (‘59)

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A positive solution is equivalent to: Anderson’s Paving Conjectures (‘79, ‘81) Bourgain-Tzafriri Conjecture (‘91) Feichtinger Conjecture (‘05) Many others Implied by: Akemann and Anderson’s Paving Conjecture (‘91) Weaver’s KS 2 Conjecture The Kadison-Singer Problem (‘59)

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Let be a maximal Abelian subalgebra of, the algebra of bounded linear operators on Let be a pure state. Is the extension of to unique? See Nick Harvey’s Survey or Terry Tao’s Blog The Kadison-Singer Problem (‘59)

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Anderson’s Paving Conjecture ‘79 For all there is a k so that for every n-by-n symmetric matrix A with zero diagonals, there is a partition of into Recall

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Anderson’s Paving Conjecture ‘79 For all there is a k so that for every self-adjoint bounded linear operator A on, there is a partition of into

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Anderson’s Paving Conjecture ‘79 For all there is a k so that for every n-by-n symmetric matrix A with zero diagonals, there is a partition of into Is unchanged if restrict to projection matrices. [Casazza, Edidin, Kalra, Paulsen ‘07]

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Anderson’s Paving Conjecture ‘79 There exist an and a k so that for such that and then exists a partition of into k parts s.t. Equivalent to [Harvey ‘13]:

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Moments of Vectors The moment of vectors in the direction of a unit vector is 2.51 4

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Moments of Vectors The moment of vectors in the direction of a unit vector is

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Vectors with Spherical Moments For every unit vector

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Vectors with Spherical Moments For every unit vector Also called isotropic position

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Partition into Approximately ½-Spherical Sets

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because

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Partition into Approximately ½-Spherical Sets because

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Big vectors make this difficult

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Weaver’s Conjecture KS 2 There exist positive constants and so that if all then exists a partition into S 1 and S 2 with

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Weaver’s Conjecture KS 2 There exist positive constants and so that if all then exists a partition into S 1 and S 2 with Implies Akemann-Anderson Paving Conjecture, which implies Kadison-Singer

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Main Theorem For all if all then exists a partition into S 1 and S 2 with Implies Akemann-Anderson Paving Conjecture, which implies Kadison-Singer

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A Random Partition? Works with high probability if all (by Tropp ‘11, variant of Matrix Chernoff, Rudelson)

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A Random Partition? Works with high probability if all (by Tropp ‘11, variant of Matrix Chernoff, Rudelson) Troublesome case: each is a scaled axis vector are of each

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A Random Partition? Works with high probability if all (by Tropp ‘11, variant of Matrix Chernoff, Rudelson) Troublesome case: each is a scaled axis vector are of each chance that all in one direction land in same set is

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A Random Partition? Works with high probability if all (by Tropp ‘11, variant of Matrix Chernoff, Rudelson) Troublesome case: each is a scaled axis vector are of each chance that all in one direction land in same set is Chance there exists a direction in which all land in same set is

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The Graphical Case From a graph G = (V,E) with |V| = n and |E| = m Create m vectors in n dimensions: If G is a good d-regular expander, all eigs close to d very close to spherical

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Partitioning Expanders Can partition the edges of a good expander to obtain two expanders. Broder-Frieze-Upfal ‘94: construct random partition guaranteeing degree at least d/4, some expansion Frieze-Molloy ‘99: Lovász Local Lemma, good expander Probability is works is low, but can prove non-zero

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We want

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Consider expected polynomial with a random partition.

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Our approach 1.Prove expected characteristic polynomial has real roots 2.Prove its largest root is at most 3.Prove is an interlacing family, so exists a partition whose polynomial has largest root at most

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The Expected Polynomial Indicate choices by :

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The Expected Polynomial

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Mixed Characteristic Polynomials For independently chosen random vectors is their mixed characteristic polynomial. Theorem: only depends on and, is real-rooted

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Mixed Characteristic Polynomials For independently chosen random vectors is their mixed characteristic polynomial. Theorem: only depends on and, is real-rooted

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Mixed Characteristic Polynomials For independently chosen random vectors is their mixed characteristic polynomial. Theorem: only depends on and, is real-rooted

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Mixed Characteristic Polynomials For independently chosen random vectors is their mixed characteristic polynomial. The constant term is the mixed discriminant of

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The constant term When diagonal and, is a matrix permanent.

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The constant term is minimized when Proved by Egorychev and Falikman ‘81. Simpler proof by Gurvits (see Laurent-Schrijver) If and When diagonal and, is a matrix permanent. Van der Waerden’s Conjecture becomes

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The constant term is minimized when This was a conjecture of Bapat. If and For Hermitian matrices, is the mixed discriminant Gurvits proved a lower bound on :

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Real Stable Polynomials A multivariate generalization of real rootedness. Complex roots of come in conjugate pairs. So, real rooted iff no roots with positive complex part.

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Real Stable Polynomials Isomorphic to Gårding’s hyperbolic polynomials Used by Gurvits (in his second proof) is real stable if it has no roots in the upper half-plane for all i implies

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Real Stable Polynomials Isomorphic to Gårding’s hyperbolic polynomials Used by Gurvits (in his second proof) is real stable if it has no roots in the upper half-plane for all i implies See surveys of Pemantle and Wagner

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Real Stable Polynomials Borcea-Brändén ‘08: For PSD matrices is real stable

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Real Stable Polynomials implies is real rooted real stable implies is real stable real stable (Lieb Sokal ‘81)

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Real Roots So, every mixed characteristic polynomial is real rooted.

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Interlacing Polynomial interlaces if

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Common Interlacing and have a common interlacing if can partition the line into intervals so that each contains one root from each polynomial ) ) ) ) ( (( (

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Common Interlacing ) ) ) ) ( (( ( If p 1 and p 2 have a common interlacing, for some i. Largest root of average

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Common Interlacing ) ) ) ) ( (( ( If p 1 and p 2 have a common interlacing, for some i. Largest root of average

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Without a common interlacing

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If p 1 and p 2 have a common interlacing, for some i. Common Interlacing ) ) ) ) ( (( ( for some i. Largest root of average

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and have a common interlacing iff is real rooted for all ) ) ) ) ( (( ( Common Interlacing

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is an interlacing family Interlacing Family of Polynomials if its members can be placed on the leaves of a tree so that when every node is labeled with the average of leaves below, siblings have common interlacings

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is an interlacing family Interlacing Family of Polynomials if its members can be placed on the leaves of a tree so that when every node is labeled with the average of leaves below, siblings have common interlacings

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Interlacing Family of Polynomials Theorem: There is a so that

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Our Interlacing Family Indicate choices by :

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and have a common interlacing iff is real rooted for all We need to show that is real rooted. Common Interlacing

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and have a common interlacing iff is real rooted for all We need to show that is real rooted. It is a mixed characteristic polynomial, so is real-rooted. Set with probability Keep uniform for Common Interlacing

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An upper bound on the roots Theorem: If and then

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An upper bound on the roots Theorem: If and then An upper bound of 2 is trivial (in our special case). Need any constant strictly less than 2.

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An upper bound on the roots Theorem: If and then

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An upper bound on the roots Define: is an upper bound on the roots of if for

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An upper bound on the roots Define: is an upper bound on the roots of if for Eventually set so want

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Action of the operators

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The roots of Define: Theorem (Batson-S-Srivastava): If is real rooted and

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The roots of Theorem (Batson-S-Srivastava): If is real rooted and Gives a sharp upper bound on the roots of associated Laguerre polynomials.

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The roots of Theorem (Batson-S-Srivastava): If is real rooted and Proof: Define Set, so Algebra + is convex and monotone decreasing

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An upper bound on the roots Theorem: If and then

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A robust upper bound Define: is an -upper bound on if it is an -max, in each, and Theorem: If is an -upper bound on, then is an -upper bound on, for

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A robust upper bound Theorem: If is an -upper bound on, and is an -upper bound on, Proof: Same as before, but need to know that is decreasing and convex in, above the roots

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A robust upper bound Proof: Same as before, but need to know that is decreasing and convex in, above the roots Follows from a theorem of Helton and Vinnikov ’07: Every bivariate real stable polynomial can be written

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A robust upper bound Proof: Same as before, but need to know that is decreasing and convex in, above the roots Or, as pointed out by Renegar, from a theorem Bauschke, Güler, Lewis, and Sendov ’01 Or, see Terry Tao’s blog for a (mostly) self-contained proof

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Getting Started

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at

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An upper bound on the roots Theorem: If and then

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A probabilistic interpretation For independently chosen random vectors with finite support such that and then

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Main Theorem For all if all then exists a partition into S 1 and S 2 with Implies Akemann-Anderson Paving Conjecture, which implies Kadison-Singer

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Anderson’s Paving Conjecture ‘79 Reduction by Casazza-Edidin-Kalra-Paulsen ’07 and Harvey ’13: There exist an and a k so that if all and then exists a partition of into k parts s.t. Can prove using the same technique

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A conjecture For all there exists an so that if all then exists a partition into S 1 and S 2 with

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Another conjecture If and then is largest when

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Questions How do operations that preserve real rootedness move the roots and the Stieltjes transform? Can the partition be found in polynomial time? What else can one construct this way?

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Parikshit Gopalan Georgia Institute of Technology Atlanta, Georgia, USA.

Parikshit Gopalan Georgia Institute of Technology Atlanta, Georgia, USA.

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