Survey
* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project
Ramanujan Graphs
from Finite Free
Convolutions
Nikhil Srivastava
UC Berkeley
joint with Adam Marcus (Princeton) and
Daniel Spielman (Yale)
Expander Graphs
An expander is a sparse well-connected graph.
Every set of vertices has many neighbors.
Random walks mix quickly.
Large spectral gap
No bi-Lipschitz embedding into normed spaces.
Etc.
Expanders
Non-Expanders
Spectral Expanders
Let G be a graph and A be its adjacency matrix
a
b
e
c
d
0
1
0
0
1
1
0
1
0
1
eigenvalues π1 β₯ π2 β₯ β― ππ
0
1
0
1
0
0
0
1
0
1
1
1
0
1
0
Spectral Expanders
Let G be a graph and A be its adjacency matrix
a
b
e
c
d
0
1
0
0
1
1
0
1
0
1
0
1
0
1
0
0
0
1
0
1
1
1
0
1
0
eigenvalues π1 β₯ π2 β₯ β― ππ
If d-regular, then π΄π = ππ so
If bipartite then eigs are symmetric
about zero so
π1 = π
ππ = βπ
βtrivialβ
Spectral Expanders
Let G be a graph and A be its adjacency matrix
a
0
b
1
G is connected iff π02
e
c
0
1
d
1 0 0
0 1 0
π΄1 <0 π 1
0 1 0
1 0 1
1 interesting
1
0
1
0
eigenvalues π1 β₯ π2 β₯ β― ππ
If d-regular, then π΄π = ππ so
If bipartite then eigs are symmetric
about zero so
π1 = π
ππ = βπ
βtrivialβ
Spectral Expanders
Definition: G is a good expander
if all non-trivial eigenvalues βͺ π
[
-d
0
]
d
(this implies all of the other connectivity and
pseudorandomness properties)
Spectral Expanders
Definition: G is a good expander
if all non-trivial eigenvalues βͺ π
[
]
0
d
-d
e.g. πΎπ and πΎπ,π have all nontrivial eigs -1,0.
Spectral Expanders
Definition: G is a good expander
if all non-trivial eigenvalues βͺ π
[
]
0
d
-d
e.g. πΎπ and πΎπ,π have all nontrivial eigs 0.
Goal: construct infinite families, π fixed, π β β.
Spectral Expanders
Definition: G is a good expander
if all non-trivial eigenvalues βͺ π
[
]
0
d
-d
e.g. πΎπ and πΎπ,π have all nontrivial eigs 0.
Goal: construct infinite families, π fixed, π β β.
Question: What are the best expanders, for fixed π?
Spectral Expanders
Definition: G is a good expander
if all non-trivial eigenvalues βͺ π
[
]
0
d
-d
e.g. πΎπ and πΎπ,π have all nontrivial eigs 0.
Goal: construct infinite families, π fixed, π β β.
Alon-Boppanaβ86: Canβt beat
[β2 π β 1, 2 π β 1]
The meaning of 2 π β 1
The infinite d-ary tree
π π΄ π = [β2 π β 1, 2 π β 1]
(Cayley graph of the free group with π/2 generators)
The meaning of 2 π β 1
The infinite d-ary tree
π π΄ π = [β2 π β 1, 2 π β 1]
Alon-Boppanaβ86: This is the best possible spectral expander.
Ramanujan Graphs:
Definition: G is Ramanujan if all non-trivial eigs
have absolute value at most 2 π β 1
[
-d
[
-2 π β 1
0
]
2 πβ1
]
d
Ramanujan Graphs:
Definition: G is Ramanujan if all non-trivial eigs
have absolute value at most 2 π β 1
[
-d
[
-2 π β 1
0
]
2 πβ1
]
d
Margulis, Lubotzky-Phillips-Sarnakβ88: Infinite
sequences of Ramanujan graphs exist for π = π + 1
(Cayley graphs, analysis based on number theory)
Ramanujan Graphs:
Definition: G is Ramanujan if all non-trivial eigs
have absolute value at most 2 π β 1
[
-d
[
-2 π β 1
0
]
2 πβ1
]
d
Margulis, Lubotzky-Phillips-Sarnakβ88: Infinite
sequences of Ramanujan graphs exist for π = π + 1
Friedmanβ08: A random d-regular graph is almost
Ramanujan : 2 π β 1 + π(1)
More Recently
[MSSβ13]. Infinite families of bipartite Ramanujan
graphs exist for every π β₯ 3.
New Theorem
[MSSβ13]. Infinite families of bipartite Ramanujan
graphs exist for every π β₯ 3.
[MSSβ15]. For every even π and π β₯ 3, there is a
d-regular bipartite Ramanujan (multi)graph.
New Theorem
[MSSβ13]. Infinite families of bipartite Ramanujan
graphs exist for every π β₯ 3.
[MSSβ15]. For every even π and π β₯ 3, there is a
d-regular bipartite Ramanujan (multi)graph.
This Talk: Suppose G is a union of d random perfect
matchings on n vertices. Then, with nonzero probability
π2 π΄πΊ β€ 2 π β 1
Random Graph Model
Let
π΄ = π1 ππ1π + π2 ππ2π + β― + ππ ππππ
For random permutations π1 , β¦ , ππ β ππ¦ππ .
π = adjacency matrix of a fixed perfect
matching on π vertices.
Random Graph Model
Let
π΄ = π1 ππ1π + π2 ππ2π + β― + ππ ππππ
For random permutations π1 , β¦ , ππ β ππ¦ππ .
π = adjacency matrix of a fixed perfect
matching on π vertices.
π΄=
π1 ππ1π +
π2 ππ2π
Random Graph Model
Let
π΄ = π1 ππ1π + π2 ππ2π + β― + ππ ππππ
For random permutations π1 , β¦ , ππ β ππ¦ππ .
π = adjacency matrix of a fixed perfect
matching on π vertices.
Theorem: With nonzero probability:
π2 π΄ < 2 π β 1
Eigs of a random 5-regular graph
Eigs of a random 5-regular graph
Eigs of a random 5-regular graph
Limiting Spectral Distribution
Kesten-McKay Law
[Mckayβ81] Let π΄π be a sequence of random π βregular
graphs. Then the spectral distributions of the π΄π converge
weakly to the spectrum of ππ :
ππΎπ π₯ =
π 4 πβ1 βπ₯ 2
2π(π 2 βπ₯ 2)
on
[β2 π β 1, 2 π β 1]
infinite
"π΄β "
2 πβ1
[Mckayβ81]
π΄π
finite
infinite
"π΄β "
2 πβ1
[Mckayβ81]
π΄π
?
finite
infinite
"π΄β "
2 πβ1
[Mckayβ81]
π΄π
No control on π2:
convergence in distribution
oblivious to extreme eigs.
finite
infinite
"π΄β "
2 πβ1
πΌπ π΄π = πΌ det π₯πΌ β π΄π
π΄π
expected
characteristic
polynomial
finite
Random Graph Model
Let
π΄ = π1 ππ1π + π2 ππ2π + β― + ππ ππππ
For random permutations π1 , β¦ , ππ β ππ¦ππ .
π = adjacency matrix of a fixed perfect
matching on π vertices.
Random Graph Model
Let
π΄ = π1 ππ1π + π2 ππ2π + β― + ππ ππππ
For random permutations π1 , β¦ , ππ β ππ¦ππ .
π = adjacency matrix of a fixed perfect
matching on π vertices.
Traditional Approaches:
1. Moments of Eigenvalues πΌππ(π΄π ).
2. Quadratic form πΌ sup π₯ π π΄π₯
Random Graph Model
Let
π΄ = π1 ππ1π + π2 ππ2π + β― + ππ ππππ
For random permutations π1 , β¦ , ππ β ππ¦ππ .
π = adjacency matrix of a fixed perfect
matching on π vertices.
Let π π΄ = det(π₯πΌ β π΄) be the characteristic poly.
We are interested in π2 π΄ = π2 (π π΄ ).
Random Graph Model
Let
π΄ = π1 ππ1π + π2 ππ2π + β― + ππ ππππ
For random permutations π1 , β¦ , ππ β ππ¦ππ .
π = adjacency matrix of a fixed perfect
matching on π vertices.
Let π π΄ = det(π₯πΌ β π΄) be the characteristic poly.
We are interested in π2 π΄ = π2 (π π΄ ).
Consider:
β1 π π₯ πβπ πΌππ (π΄)
πΌπ π΄ = πΌ det π₯πΌ β π΄ =
π
Outline of the Proof
1. Prove that πΌπ(π΄) has real roots.
2. Prove that the roots are Ramanujan:
π2 πΌπ π΄ β€ 2 π β 1
3. Prove π2 π΄ β€ π2 (πΌπ π΄ ) with nonzero probability.
1. Prove that πΌπ(π΄) has real roots.
1. Prove that πΌπ(π΄) has real roots.
Quadrature Theorem. For any doubly stochastic π:
π
π
ππ ππππ = πΌπ π
πΌπ π
π=1
ππ ππππ
π=1
where π1 , β¦ , ππ are independent random (Haar)
orthogonal matrices conditioned on ππ π = π.
1. Prove that πΌπ(π΄) has real roots.
Quadrature Theorem. For any doubly stochastic π:
π
π
ππ ππππ = πΌπ π
πΌπ π
π=1
ππ ππππ
π=1
where π1 , β¦ , ππ are independent random (Haar)
orthogonal matrices conditioned on ππ π = π.
(determinant is a low degree degree poly
in entries of Q)
1. Prove that πΌπ(π΄) has real roots.
Quadrature Theorem. For any doubly stochastic π:
π
π
ππ ππππ = πΌπ π
πΌπ π
π=1
ππ ππππ
π=1
where π1 , β¦ , ππ are independent random (Haar)
orthogonal matrices conditioned on ππ π = π.
For a polynomial π of degree n
1. Prove that πΌπ(π΄) has real roots.
Quadrature Theorem. For any doubly stochastic π:
π
π
ππ ππππ = πΌπ π
πΌπ π
π=1
ππ ππππ
π=1
where π1 , β¦ , ππ are independent random (Haar)
orthogonal matrices conditioned on ππ π = π.
1. Prove that πΌπ(π΄) has real roots.
Quadrature Theorem. For any doubly stochastic π:
π
π
ππ ππππ = (π₯ β π)πΌπ π
πΌπ π
π=1
ππ ππππ
π=1
where π1 , β¦ , ππ are independent random (Haar)
orthogonal matrices conditioned on ππ π = π.
π = π orthogonal to 1
1. Prove that πΌπ(π΄) has real roots.
Quadrature Theorem. For any doubly stochastic π:
π
π
ππ ππππ = πΌπ π
πΌπ π
π=1
ππ ππππ
π=1
where π1 , β¦ , ππ are independent random (Haar)
orthogonal matrices conditioned on ππ π = π.
For the rest of the talk I will ignore the trivial
eigenspace π and the trivial root/eigenvalue π.
So ππ is now the largest root.
1. Prove that πΌπ(π΄) has real roots.
Quadrature Theorem.
π
π
ππ ππππ = πΌπ π
πΌπ π
π=1
ππ ππππ
π=1
1. Prove that πΌπ(π΄) has real roots.
Quadrature Theorem.
π
π
ππ ππππ = πΌπ π
πΌπ π
π=1
ππ ππππ
π=1
this only depends on π(π)!
1. Prove that πΌπ(π΄) has real roots.
Quadrature Theorem.
π
π
ππ ππππ = πΌπ π
πΌπ π
π=1
ππ ππππ
π=1
<definition on board>
1. Prove that πΌπ(π΄) has real roots.
Quadrature Theorem.
π
π
ππ ππππ = πΌπ π
πΌπ π
π=1
ππ ππππ
π=1
Linearization Formula:
π
ππ ππππ = π π βπ π π βπ β¦ βπ π(π)
πΌπ π
π=1
where βπ
is the Finite Free Convolution.
1. Prove that πΌπ(π΄) has real roots.
Quadrature Theorem.
π
π π π= π₯ β 1
ππ ππππ = πΌπ π
πΌπ π
π=1
π
β1
2
π₯+1
π
2
ππ ππππ
π=1
Linearization Formula:
π
ππ ππππ = π π βπ π π βπ β¦ βπ π(π)
πΌπ π
π=1
where βπ
is the Finite Free Convolution.
1. Prove that πΌπ(π΄) has real roots.
Quadrature Theorem.
π
π π π= π₯ β 1
ππ ππππ = πΌπ π
πΌπ π
π=1
π
β1
2
π₯+1
π
2
ππ ππππ
π=1
Linearization Formula:
π
ππ ππππ = π π βπ π π βπ β¦ βπ π(π)
πΌπ π
π=1
where βπ
is the Finite Free Convolution.
[Walshβ22]: The operation βπ preserves real-rootedness.
ο
1. Prove that πΌπ(π΄) has real roots.
Quadrature Theorem.
π
π π π= π₯ β 1
ππ ππππ = πΌπ π
πΌπ π
π=1
π
β1
2
π₯+1
π
2
ππ ππππ
π=1
Linearization Formula:
π
ππ ππππ = π π βπ π π βπ β¦ βπ π(π)
πΌπ π
π=1
where βπ
is the Finite Free Convolution.
[Walshβ22]: The operation βπ preserves real-rootedness.
Outline of the Proof
1. Prove that πΌπ(π΄) has real roots. Also showed:
πΌπ
π
π
π
ππ
π=1 π
π
= π π βπ β¦ βπ π π .
2. Prove that the roots are Ramanujan:
π2 πΌπ π΄ β€ 2 π β 1
3. Prove π2 π΄ β€ π2 (πΌπ π΄ ) with positive probability.
ο
Outline of the Proof
2. Prove that the roots are Ramanujan:
π2 π π βπ β¦ βπ π π β€ 2 π β 1
Outline of the Proof
2. Prove that the roots are Ramanujan:
π2 π π βπ β¦ βπ π π β€ 2 π β 1
Easy bound: π2 π βπ π β€ π2 π + π2 (π)
<proof on board>
Outline of the Proof
2. Prove that the roots are Ramanujan:
π2 π π βπ β¦ βπ π π β€ 2 π β 1
Easy bound: π2 π βπ π β€ π2 π + π2 (π)
Gives π2 π π βπ β¦ π π
β€ π,
trivial
Free Probability Detour
Sums of independent random matrices
Independent Hermitian random matrices π΄, π΅.
Q: What is the distribution of πππ π΄ + π΅ ?
No general method.
Depends on eigenvectors of π΄ and π΅.
Free Probability
Independent orthogonally invariant matrices π΄, π΅.
(π΄ and ππ΄ππ have same distribution)
Intuition: eigenvectors in `generic positionβ, so
maybe can say something about π΄ + π΅...
Free Probability
Independent orthogonally invariant matrices π΄, π΅.
Moment generating function
1
πΌ: = ππ΄ π§ = πΌππ π§πΌ β π΄
π
β1
1
=
π
π
πΌππ π΄π
π§ π+1
a.k.a. Cauchy / Stieltjes transform
Free Probability
Independent orthogonally invariant matrices π΄, π΅.
Moment generating function
1
πΌ: = ππ΄ π§ = πΌππ π§πΌ β π΄
π
β1
Cumulant generating function: functional inverse
(β1)
π§ = ππ΄ (πΌ)
Free Probability
Independent orthogonally invariant matrices π΄, π΅.
Moment generating function
1
πΌ: = ππ΄ π§ = πΌππ π§πΌ β π΄
π
β1
Cumulant generating function: functional inverse
(β1)
π§ = ππ΄ (πΌ)
[Voiculescuβ91] As π β β:
(β1)
(β1)
(β1)
ππ΄+π΅ πΌ = ππ΄
πΌ + ππ΅
πΌ β 1/πΌ
Free Probability
[Voiculescuβ91] As π β β:
(β1)
(β1)
(β1)
ππ΄+π΅ πΌ = ππ΄
πΌ + ππ΅
πΌ β 1/πΌ
Summary: Precise description of limiting spectral
distribution of sums of random matrices in generic
position. Beats triangle inequality.
Free Probability
[Voiculescuβ91] As π β β:
(β1)
(β1)
(β1)
ππ΄+π΅ πΌ = ππ΄
πΌ + ππ΅
πΌ β 1/πΌ
Summary: Precise description of limiting spectral
distribution of sums of random matrices in generic
position. Beats triangle inequality.
Can be used to show:
ππ + ππ ππ πππ
Free Probability
[Voiculescuβ91] As π β β:
(β1)
(β1)
(β1)
ππ΄+π΅ πΌ = ππ΄
πΌ + ππ΅
πΌ β 1/πΌ
Summary: Precise description of limiting spectral
distribution of sums of random matrices in generic
position. Beats triangle inequality.
Can be used to show:
ππ + ππ ππ πππ
but we want extreme eigs in finite dimensionsβ¦.
Back to the proof:
Finite R-Transform
Inequalities
Free Convolution for Polynomials
Main tool: Stieltjes/Cauchy Transform
For a real-rooted poly of degree π, consider:
1 πβ²(π₯) 1
ππ π₯ =
=
π π(π₯) π
π
1
π₯ β ππ
Define for πΌ > 0:
πΌmax π = max{π₯: ππ π₯ = πΌ}
Notice
πΌmax p > π1 (π)
Free Convolution for Polynomials
Main tool: Stieltjes/Cauchy Transform
For a real-rooted poly of degree π, consider:
1 πβ²(π₯) 1
ππ π₯ =
=
π π(π₯) π
π
1
π₯ β ππ
Define for πΌ > 0:
πΌmax π = max{π₯: ππ π₯ = πΌ}
Notice
πΌmax p > π1 (π)
A picture of πΌπππ₯
1
ππ π₯ =
π
π
1
π₯ β ππ
πΌ
πΌπππ₯
For a single matching
π π = π₯+1
So
ππ
π
π
2
π₯β1
π
2
π₯
π₯ = 2
π₯ β1
and
πΌmax π π
1 + 1 + 4πΌ 2
=
2πΌ
For d matchings
Main inequality:
πΌmax π βπ π β€ πΌmax π + πΌmax π β 1/πΌ
Meaning:
When πΌ = β same as triangle inequality.
When πΌ < β can beat the triangle
inequality when roots are spread out.
For d matchings
Main inequality:
πΌmax π βπ π β€ πΌmax π + πΌmax π β 1/πΌ
Meaning:
When πΌ = β same as triangle inequality.
When πΌ < β can beat the triangle
inequality when roots are spread out.
Proof by characterizing extremizers, relies on convexity.
For d matchings
Main inequality:
πΌmax π βπ π β€ πΌmax π + πΌmax π β 1/πΌ
Cf. Voiculescu
πΌmax π β π = πΌmax π + πΌmax π β 1/πΌ
(asymptotically, for large πΌ)
For d matchings
Main inequality:
πΌmax π βπ π β€ πΌmax π + πΌmax π β 1/πΌ
Applying π β 1 times:
πΌmax π π β β― β π π
1 + 1 + 4πΌ 2 π β 1
β€π
β
2πΌ
πΌ
For d matchings
Main inequality:
πΌmax π βπ π β€ πΌmax π + πΌmax π β 1/πΌ
Applying π β 1 times:
πΌmax π π β β― β π π
1 + 1 + 4πΌ 2 π β 1
β€π
β
2πΌ
πΌ
= 2 π β 1 for πΌ =
πβ1
πβ2
infinite
"π΄β "
Finite analogue of Free Probability.
πΌπ π΄π = πΌ det π₯πΌ β π΄π
π΄π
expected
characteristic
polynomial
finite
Outline of the Proof
1. Prove that πΌπ(π΄) has real roots. Also showed:
πΌπ
π
π
π
ππ
π=1 π
π
= π π βπ β¦ βπ π π .
2. Prove that the roots are Ramanujan:
π2 πΌπ π΄ β€ 2 π β 1
3. Prove π2 π΄ β€ π2 (πΌπ π΄ ) with positive probability.
ο
ο
infinite
"π΄β "
Finite analogue of Free Probability.
πΌπ π΄π = πΌ det π₯πΌ β π΄π
expected
characteristic
polynomial
?
π΄π
finite
Outline of the Proof
3. Prove π2 π΄ β€ π2 (πΌπ π΄ ) with positive probability.
Outline of the Proof
3. Prove: There is some setting of π1 , β¦ , ππ s.t:
π
π
ππ΄ ) with positive probability.
π
3. Proveππ2 2 π
π΄ β€ π2π(πΌπ
ππ
β€
π
πΌπ
π
ππ
π
2
π
π
π
π
π
Averaging Polynomials
Basic Question: Given
when are the roots
of the
related to roots of
?
Averaging Polynomials
Basic Question: Given
when are the roots
of the
related to roots of
?
Answer: Certainly not always
Averaging Polynomials
Basic Question: Given
when are the roots
of the
related to roots of
?
Answer: Certainly not alwaysβ¦
Averaging Polynomials
Basic Question: Given
when are the roots
of the
related to roots of
?
But sometimes roots (avg) = avg (roots):
A Sufficient Condition
Basic Question: Given
when are the roots
of the
related to roots of
?
Answer: When they have a common interlacing.
Definition.
interlaces
if
Theorem. If
interlacing,
monic have a common
Theorem. If
interlacing,
Proof.
monic have a common
Theorem. If
interlacing,
Proof.
monic have a common
Theorem. If
interlacing,
Proof.
monic have a common
Theorem. If
interlacing,
Proof.
monic have a common
Theorem. If
interlacing,
Proof.
monic have a common
Theorem. If
interlacing,
Proof.
monic have a common
Many Swaps
Generate random permutations using swaps.
Swaps generate polynomials with common
interlacings. Apply theorem inductively.
πΌπ1 πΌπ2 β¦ πΌππ π π΄ + ππ β¦ π1 π΅π1π β¦ πππ
Conclusion. There is some setting of π
π11, β¦ , π
πππ1 ,s.t:
β¦ , π1π , β¦ , πππ s.t:
π
π2 π
π
π
π
π
π
π
π
π
ππ β¦π2π1πππ1 πβ¦
ππππ
π ππ
π
π
β€ π2 πΌπ
π
π
π
π
π
π
π
πππ ππ
β¦ ππ 1 ππ1
π
π
β¦ ππ
Outline of the Proof
1. Prove that πΌπ(π΄) has real roots. Also showed:
πΌπ
π
π
π
ππ
π=1 π
π
= π π βπ β¦ βπ π π .
ο
2. Prove that the roots are Ramanujan:
π2 πΌπ π΄ β€ 2 π β 1
ο
3. Prove π2 π΄ β€ π2 (πΌπ π΄ ) with positive probability.
ο
Outline of the Proof
Quadrature + Walshβ22
1. Prove that πΌπ(π΄) has real roots. Also showed:
πΌπ
π
π
π
ππ
π=1 π
π
= π π βπ β¦ βπ π π .
Finite Analogue of
2. Prove that the roots are Ramanujan:
Voiculescuβ91
π2 πΌπ π΄ β€ 2 π β 1
Interlacing
3. Prove π2 π΄ β€ π2 (πΌπ π΄ ) with positive probability.
ο
ο
ο
infinite
"π΄β "
π 4 π β 1 β π₯2
2π(π 2 β π₯ 2 )
Finite analogue of Free Probability.
πΌπ π΄π = πΌ det π₯πΌ β π΄π
expected
characteristic
polynomial
Interlacing
? Families
π΄π
finite
Is this a coincidence?
[Marcusβ15]:
If π and π are discrete measures, each uniform on π
real points with characteristic polynomials π and π,
then
Finite
convolution
β1
lim πΆππ β ππ
ππ
πββ
πΌ =
β1
πΆπβπ (πΌ)
For all sufficiently large real πΌ.
(exact statement is a bit more technical)
Similar results for S-transform. Also: Free CLT.
Voiculescu
convolution
Is this a coincidence?
[Marcusβ15]:
If π and π are discrete measures, each uniform on π
real points with characteristic polynomials π and π,
then
Finite
convolution
β1
lim πΆππ β ππ
ππ
πββ
πΌ =
β1
πΆπβπ (πΌ)
For all sufficiently large real πΌ.
(exact statement is a bit more technical)
Similar results for S-transform. Also: Free CLT.
Voiculescu
convolution
High-level question
Why do expected characteristic polynomials give
sharp bounds?
(also: 2-Covers, Kadison-Singer, Thin Trees [Oveis-Gharan-Anariβ13])
What are expected characteristic polynomials?
High-level question
Why do expected characteristic polynomials give
sharp bounds?
(also: 2-Covers, Kadison-Singer, Thin Trees [Anari-Gharanβ15])
What are expected characteristic polynomials?
Heuristic Observation + Marcusβ15:
Finite approximations of asymptotic limits of finite
random matrices.
Open questions
β’ More finite free probability notions: entropy
β’ Nonbipartite Ramanujan graphs
β’ Other random graph models
β’ Algorithms
β’ Bounds on probability (conjectured to be
52% by Novikov, Miller, Sabelliβ06)
β’ More connections between infinite and finite
β’ Applications to free probability/vN algebras?