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]