the solution of the kadison -singer problem adam marcus (crisply, yale) daniel spielman (yale)
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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. The Kadison -Singer Problem (‘59). A positive solution is equivalent to: Anderson’s Paving Conjectures (‘79, ‘81) - PowerPoint PPT PresentationTRANSCRIPT
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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 KS2 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 KS2 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 KS2 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 everyn-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 everyself-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 everyn-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 VectorsThe moment of vectors in the direction of a unit vector is
2.51 4
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Moments of VectorsThe 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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Partition into Approximately ½-Spherical Sets
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Partition into Approximately ½-Spherical Sets
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Partition into Approximately ½-Spherical Sets
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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Big vectors make this difficult
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Weaver’s Conjecture KS2
There exist positive constants and so that
if all
then exists a partition into S1 and S2 with
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Weaver’s Conjecture KS2
There exist positive constants and so that
if all
then exists a partition into S1 and S2 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 S1 and S2 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 whichall land in same set is
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The Graphical Case
From a graph G = (V,E) with |V| = n and |E| = mCreate m vectors in n dimensions:
If G is a good d-regular expander, all eigs close to dvery 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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We want
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We want
Consider expected polynomial with a random partition.
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Our approach
1. Prove expected characteristic polynomialhas 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 PolynomialIndicate 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 termWhen 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 discriminantGurvits 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 iimplies
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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 iimplies
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 ifcan partition the line into intervals so that each contains one root from each polynomial
)))) ( ( ((
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Common Interlacing
)))) ( ( ((
If p1 and p2 have a common interlacing,
for some i.Largest rootof average
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Common Interlacing
)))) ( ( ((
If p1 and p2 have a common interlacing,
for some i.Largest rootof average
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Without a common interlacing
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Without a common interlacing
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Without a common interlacing
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Without a common interlacing
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If p1 and p2 have a common interlacing,
for some i.
Common Interlacing
)))) ( ( ((
for some i.Largest rootof 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 PolynomialsTheorem: There is a so that
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Our Interlacing FamilyIndicate 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 probabilityKeep 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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Action of the operators
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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 boundTheorem: 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 boundProof: 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 boundProof: 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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Getting Started
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 vectorswith finite support
such that and
then
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Main Theorem
For all
if all
then exists a partition into S1 and S2 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 S1 and S2 with
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Another conjecture
If and then
is largest when
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Questions
How do operations that preserve real rootednessmove 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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