Embeddings of the Heisenberg group and the Sparsest Cut problem - PowerPoint PPT Presentation

Embeddings of the Heisenberg group and the Sparsest Cut problem Robert Young New York University (joint work with Assaf Naor) May 2018 A.N. was supported by BSF grant 2010021, the Packard Foundation and the Simons Foundation. R.Y. was supported by NSF grant DMS 1612061, the Sloan Foundation, and the Fall 2016 program at MSRI. The research that is presented here was conducted under the auspices of the Simons Algorithms and Geometry (A&G) Think Tank.

Part 1: The Sparsest Cut problem What’s the “best” way to cut a graph into two pieces?

Part 1: The Sparsest Cut problem What’s the “best” way to cut a graph into two pieces? Problem Let G be a graph. Find | E ( S , S c ) | Φ( G ) = min | S | · | S c | . S ⊂ V ( G )

Part 1: The Sparsest Cut problem What’s the “best” way to cut a graph into two pieces? Problem Let G be a graph. Find | E ( S , S c ) | Φ( G ) = min | S | · | S c | . S ⊂ V ( G )

Sparsest Cut is a matrix problem  0 1 1 1 0 0 0  1 0 1 1 0 0 0     1 1 0 1 1 0 0     C = 1 1 1 0 0 1 0     0 0 1 0 0 1 1     0 0 0 1 1 0 1   0 0 0 0 1 1 0 If C is the adjacency matrix of G , then � i ∈ S , j ∈ S c C ij | E ( S , S c ) | Φ( G ) = min | S | · | S c | = min � i ∈ S , j ∈ S c 1 S ⊂ V ( G ) S ⊂ [ n ] (where [ n ] = { 1 , . . . , n } )

The Nonuniform Sparsest Cut problem Problem Let C (capacity) and D (demand) be symmetric n × n matrices with non-negative entries. Find: � i ∈ S , j ∈ S c C ij Φ( C , D ) = min . � i ∈ S , j ∈ S c D ij S ⊂ [ n ]

The Nonuniform Sparsest Cut problem Problem Let C (capacity) and D (demand) be symmetric n × n matrices with non-negative entries. Find: � i ∈ S , j ∈ S c C ij Φ( C , D ) = min . � i ∈ S , j ∈ S c D ij S ⊂ [ n ] This problem is NP-hard

The Nonuniform Sparsest Cut problem Problem Let C (capacity) and D (demand) be symmetric n × n matrices with non-negative entries. Find: � i ∈ S , j ∈ S c C ij Φ( C , D ) = min . � i ∈ S , j ∈ S c D ij S ⊂ [ n ] This problem is NP-hard, but there are polynomial-time algorithms to approximate Φ( C , D ) based on metric embeddings.

Relaxing the problem A cut metric is a semimetric of the form d S ( i , j ) = | 1 S ( i ) − 1 S ( j ) | where S ⊂ X .

Relaxing the problem A cut metric is a semimetric of the form d S ( i , j ) = | 1 S ( i ) − 1 S ( j ) | where S ⊂ X . Let C = { d S | S ⊂ [ n ] } ⊂ M n

Relaxing the problem A cut metric is a semimetric of the form d S ( i , j ) = | 1 S ( i ) − 1 S ( j ) | where S ⊂ X . Let C = { d S | S ⊂ [ n ] } ⊂ M n and let K ⊃ C . The relaxation Φ K of Sparsest Cut is � i , j C ij M ij Φ K ( C , D ) = min . � i , j D ij M ij M ∈K

Relaxing the problem A cut metric is a semimetric of the form d S ( i , j ) = | 1 S ( i ) − 1 S ( j ) | where S ⊂ X . Let C = { d S | S ⊂ [ n ] } ⊂ M n and let K ⊃ C . The relaxation Φ K of Sparsest Cut is � i , j C ij M ij Φ K ( C , D ) = min . � i , j D ij M ij M ∈K ◮ Then Φ K ≤ Φ C = Φ.

Relaxing the problem A cut metric is a semimetric of the form d S ( i , j ) = | 1 S ( i ) − 1 S ( j ) | where S ⊂ X . Let C = { d S | S ⊂ [ n ] } ⊂ M n and let K ⊃ C . The relaxation Φ K of Sparsest Cut is � i , j C ij M ij Φ K ( C , D ) = min . � i , j D ij M ij M ∈K ◮ Then Φ K ≤ Φ C = Φ. ◮ Is there a K such that Φ K is easy to compute and close to Φ?

Geometric relaxations If f : X → Y , let d f ∈ M n be the induced distance function d f ( i , j ) = d ( f ( i ) , f ( j )). ◮ If K 1 = { d f | f : [ n ] → L 1 } , then Φ = Φ K 1 (Linial-London-Rabinovich)

Geometric relaxations If f : X → Y , let d f ∈ M n be the induced distance function d f ( i , j ) = d ( f ( i ) , f ( j )). ◮ If K 1 = { d f | f : [ n ] → L 1 } , then Φ = Φ K 1 (Linial-London-Rabinovich) Proof: Every element of K 1 is a linear combination of cut metrics

Geometric relaxations If f : X → Y , let d f ∈ M n be the induced distance function d f ( i , j ) = d ( f ( i ) , f ( j )). ◮ If K 1 = { d f | f : [ n ] → L 1 } , then Φ = Φ K 1 (Linial-London-Rabinovich) Proof: Every element of K 1 is a linear combination of cut metrics Φ ◮ If M = { n × n distance matrices } , then log n � Φ M ≤ Φ (Linial-London-Rabinovich, Aumann-Rabani).

Geometric relaxations If f : X → Y , let d f ∈ M n be the induced distance function d f ( i , j ) = d ( f ( i ) , f ( j )). ◮ If K 1 = { d f | f : [ n ] → L 1 } , then Φ = Φ K 1 (Linial-London-Rabinovich) Proof: Every element of K 1 is a linear combination of cut metrics Φ ◮ If M = { n × n distance matrices } , then log n � Φ M ≤ Φ (Linial-London-Rabinovich, Aumann-Rabani). Proof: Every n -point metric space embeds in L 1 with log n distortion (Bourgain)

The Goemans-Linial relaxation Theorem (Goemans-Linial) Let N = { n × n distance matrices with negative type } . Then K 1 ⊂ N ⊂ M , so Φ log n � Φ M ≤ Φ N ≤ Φ . Furthermore, Φ N can be computed in polynomial time.

The Goemans-Linial question Φ( C , D ) Define the Goemans-Linial integrality gap α ( n ) = max Φ N ( C , D ) where C , D are n × n matrices.

The Goemans-Linial question Φ( C , D ) Define the Goemans-Linial integrality gap α ( n ) = max Φ N ( C , D ) where C , D are n × n matrices. 2 + o (1) (Arora-Lee-Naor). 1 ◮ α ( n ) � (log n )

The Goemans-Linial question Φ( C , D ) Define the Goemans-Linial integrality gap α ( n ) = max Φ N ( C , D ) where C , D are n × n matrices. 2 + o (1) (Arora-Lee-Naor). 1 ◮ α ( n ) � (log n ) Question (Goemans-Linial) Is α ( n ) bounded?

The Goemans-Linial question Φ( C , D ) Define the Goemans-Linial integrality gap α ( n ) = max Φ N ( C , D ) where C , D are n × n matrices. 2 + o (1) (Arora-Lee-Naor). 1 ◮ α ( n ) � (log n ) Question (Goemans-Linial) Is α ( n ) bounded? Equivalently, does every finite negative-type metric space embed in L 1 by a bilipschitz map?

The Goemans-Linial question Φ( C , D ) Define the Goemans-Linial integrality gap α ( n ) = max Φ N ( C , D ) where C , D are n × n matrices. 2 + o (1) (Arora-Lee-Naor). 1 ◮ α ( n ) � (log n ) Question (Goemans-Linial) Is α ( n ) bounded? Equivalently, does every finite negative-type metric space embed in L 1 by a bilipschitz map? But the answer is no: ◮ α ( n ) � (log log n ) c (Khot-Vishnoi)

The Goemans-Linial question Φ( C , D ) Define the Goemans-Linial integrality gap α ( n ) = max Φ N ( C , D ) where C , D are n × n matrices. 2 + o (1) (Arora-Lee-Naor). 1 ◮ α ( n ) � (log n ) Question (Goemans-Linial) Is α ( n ) bounded? Equivalently, does every finite negative-type metric space embed in L 1 by a bilipschitz map? But the answer is no: ◮ α ( n ) � (log log n ) c (Khot-Vishnoi) ◮ α ( n ) � (log n ) c ′ (with c ′ ≈ 2 − 60 ) (Cheeger-Kleiner-Naor)

The main theorem Theorem (Naor-Y.) There is an n–point subspace X (a ball in the word metric) of the Heisenberg group H 5 such that any embedding of X into L 1 has distortion at least ≍ √ log n.

The main theorem Theorem (Naor-Y.) There is an n–point subspace X (a ball in the word metric) of the Heisenberg group H 5 such that any embedding of X into L 1 has distortion at least ≍ √ log n. Corollary (Naor-Y.) � α ( n ) � log n

Part 2: The Heisenberg group Let H 2 k +1 ⊂ M k +2 be the (2 k + 1)–dimensional nilpotent Lie group  �    1 x 1 . . . x k z �    �  0 1 0 0 y 1      �      .  �  H 2 k +1 = ... . x i , y i , z ∈ R .   � 0 0 0 .   �    �  0 0 0 1 y k      �     �  0 0 0 0 1   �

Part 2: The Heisenberg group Let H 2 k +1 ⊂ M k +2 be the (2 k + 1)–dimensional nilpotent Lie group  �    1 x 1 . . . x k z �    �  0 1 0 0 y 1      �      .  �  H 2 k +1 = ... . x i , y i , z ∈ R .   � 0 0 0 .   �    �  0 0 0 1 y k      �     �  0 0 0 0 1   � This contains a lattice H Z 2 k +1 = � x 1 , . . . , x k , y 1 , . . . , y k , z | [ x i , y i ] = z , all other pairs commute � .

A lattice in H 3

A lattice in H 3 z = xyx − 1 y − 1

A lattice in H 3 z = xyx − 1 y − 1 z 4 = x 2 y 2 x − 2 y − 2

A lattice in H 3 z = xyx − 1 y − 1 z 4 = x 2 y 2 x − 2 y − 2 z n 2 = x n y n x − n y − n

From Cayley graph to sub-riemannian metric ◮ d ( u , v ) = inf { ℓ ( γ ) | γ is a horizontal curve from u to v }

From Cayley graph to sub-riemannian metric ◮ d ( u , v ) = inf { ℓ ( γ ) | γ is a horizontal curve from u to v } ◮ The map s t ( x , y , z ) = ( tx , ty , t 2 z ) scales the metric

From Cayley graph to sub-riemannian metric ◮ d ( u , v ) = inf { ℓ ( γ ) | γ is a horizontal curve from u to v } ◮ The map s t ( x , y , z ) = ( tx , ty , t 2 z ) scales the metric ◮ The ball of radius ǫ is approximately an ǫ × ǫ × ǫ 2 box.