The Solution of the Kadison-Singer Problem Adam Marcus (Crisply PDF
Preview The Solution of the Kadison-Singer Problem Adam Marcus (Crisply
The Solution of the Kadison-Singer Problem Adam Marcus (Crisply, Yale) Daniel Spielman (Yale) Nikhil Srivastava (MSR India) Outline Disclaimer The Kadison-Singer Problem, defined. Restricted Invertibility, a simple proof. Break Kadison-Singer, outline of proof. The Kadison-Singer Problem (‘59) A posi’ve solu’on is equivalent to: Anderson’s Paving Conjectures (‘79, ‘81) Bourgain-‐Tzafriri Conjecture (‘91) Feich’nger Conjecture (‘05) Many others Implied by: Akemann and Anderson’s Paving Conjecture (‘91) Weaver’s KS Conjecture 2 The Kadison-Singer Problem (‘59) A posi’ve solu’on is equivalent to: Anderson’s Paving Conjectures (‘79, ‘81) Bourgain-‐Tzafriri Conjecture (‘91) Feich’nger Conjecture (‘05) Many others Implied by: Akemann and Anderson’s Paving Conjecture (‘91) Weaver’s KS Conjecture 2 The Kadison-Singer Problem (‘59) A posi’ve solu’on is equivalent to: Anderson’s Paving Conjectures (‘79, ‘81) Bourgain-‐Tzafriri Conjecture (‘91) Feich’nger Conjecture (‘05) Many others Implied by: Akemann and Anderson’s Paving Conjecture (‘91) Weaver’s KS Conjecture 2 The Kadison-Singer Problem (‘59) 2 (` ( )) L et be a maximal Abelian subalgebra of N , A B 2 the algebra of bounded linear operators on ` ( ) N Let ⇢ : C be a pure state. A ! Is the extension of ⇢ to ( ` 2 ( ) ) unique? N B See Nick Harvey’s Survey or Terry Tao’s Blog Anderson’s Paving Conjecture ‘79 For all ✏ > 0 there is a k so that for every A n-by-n symmetric matrix with zero diagonals, 1, ..., n S , ..., S there is a partition of into 1 k { } A(S , S ) ✏ A for j = 1, . . . , k j j k k k k A = max Ax Recall k k k k x =1 k k Anderson’s Paving Conjecture ‘79 For all ✏ > 0 there is a k so that for every A n-by-n symmetric matrix with zero diagonals, 1, ..., n S , ..., S there is a partition of into 1 k { } A(S , S ) ✏ A for j = 1, . . . , k j j k k k k A = max Ax Recall k k k k x =1 k k 0 1 @ A Anderson’s Paving Conjecture ‘79 For all ✏ > 0 there is a k so that for every ` self-adjoint bounded linear operator A on , 2 S , ..., S there is a partition of N into 1 k A(S , S ) ✏ A for j = 1, . . . , k j j k k k k A = sup Ax k k k k x =1 k k Anderson’s Paving Conjecture ‘79 For all ✏ > 0 there is a k so that for every A n-by-n symmetric matrix with zero diagonals, 1, ..., n S , ..., S there is a partition of into 1 k { } A(S , S ) ✏ A for j = 1, . . . , k j j k k k k Is equivalent if restrict to projection matrices. [Casazza, Edidin, Kalra, Paulsen ‘07]
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