Presentation is loading. Please wait.

Presentation is loading. Please wait.

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

Similar presentations


Presentation on theme: "The Solution of the Kadison-Singer Problem Adam Marcus (Crisply, Yale) Daniel Spielman (Yale) Nikhil Srivastava (MSR/Berkeley) IAS, Nov 5, 2014."— Presentation transcript:

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

2 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

3 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)

4 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)

5 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)

6 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

7 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

8 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]

9 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]:

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

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

12 Vectors with Spherical Moments For every unit vector

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

14 Partition into Approximately ½-Spherical Sets

15

16

17 because

18 Partition into Approximately ½-Spherical Sets because

19 Big vectors make this difficult

20

21 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

22 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

23 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

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

25 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

26 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

27 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

28 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

29 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

30 We want

31

32 Consider expected polynomial with a random partition.

33 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

34 The Expected Polynomial Indicate choices by :

35 The Expected Polynomial

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

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

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

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

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

41 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

42 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 :

43 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.

44 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

45 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

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

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

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

49 Interlacing Polynomial interlaces if

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

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

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

53 Without a common interlacing

54

55

56

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

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

59 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

60 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

61 Interlacing Family of Polynomials Theorem: There is a so that

62 Our Interlacing Family Indicate choices by :

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

64 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

65 An upper bound on the roots Theorem: If and then

66 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.

67 An upper bound on the roots Theorem: If and then

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

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

70 Action of the operators

71

72

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

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

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

76 An upper bound on the roots Theorem: If and then

77 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

78 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

79 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

80 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

81 Getting Started

82 at

83 An upper bound on the roots Theorem: If and then

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

85 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

86 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

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

88 Another conjecture If and then is largest when

89 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?

90


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

Similar presentations


Ads by Google