Trees and Markov convexity James R. Lee Institute for Advanced Study [ with Assaf Naor and Yuval Peres ] RdRd x y

Distortion: Smallest number C ¸ 1 such that: the Euclidean distortion problem Given a metric space (X,d), determine how well X embeds into a Euclidean space. Why study this kind of geometry (in CS)? - Applicability of low-distortion Euclidean embeddings - Understanding semi-definite programs - Optimization, harmonic analysis, hardness of approximation, cuts and flows, Markov chains, expansion, randomness… Euclidean embedding: An injective map f : X ! R k (or L 2 )

Distortion: Smallest number C ¸ 1 such that: the Euclidean distortion problem Given a metric space (X,d), determine how well X embeds into a Euclidean space. Euclidean embedding: An injective map f : X ! R k (or L 2 ) Actually, then the distortion is A ¢ B.

the problem for trees One of the simplest families of metric spaces are the tree metrics. graph-theoretic tree T = (V,E) + edge lengths len : E ! R + len(e) x y d(x,y) = length of shortest geodesic

the problem for trees [Bourgain 86]: The complete binary tree B k of height k has Euclidean distortion [Matousek 99]: Every n-point tree metric embeds with distortion at most [Gupta-Krauthgamer-L 03]: A tree metric T embeds with constant distortion into a finite-dimensional Euclidean space if and only if T is doubling. When does does a tree embed into some Euclidean space (arbitrary dimension) with bounded distortion?

the problem for trees [Bourgain 86]: The complete binary tree B k of height k has Euclidean distortion [Matousek 99]: Every n-point tree metric embeds with distortion at most [Gupta-Krauthgamer-L 03]: A tree metric T embeds with constant distortion into a finite-dimensional Euclidean space if and only if T is doubling. e1e1 e2e2 e3e3

why don’t trees embed in Hilbert space? O NE A NSWER : (EQUILATERAL) FORKS If both of these paths of length 2 are embedded isometrically in a Euclidean space, then A and B must conincide! A B Quantitative version holds: If both 2-paths are embedded with distortion 1 + , then

uniform convexity A B W Z paralellogram identity: for any pair of vectors a,b 2 R 2, f(W)=0 a=f(A) b=f(B) 4 ± O(  ) O(  )

forks in complete binary trees RdRd Ramsey style proof: If B k is embedded into L 2 with distortion then there exists some almost-isometric fork. [Matousek] C ONTRADICTION!

on forks Natural question: Are forks the only obstruction? The problem isn’t forking; it’s forking, and forking, and forking… T HEOREM: For a tree metric T, the following conditions are equivalent. -- T embeds in a Euclidean space with bounded distortion -- The family of complete binary trees {B k } do not embed into T with bounded distortion. In other words, a tree embeds into a Euclidean space if and only if it does not “contain” arbitrarily large binary trees!

quantitative version T HEOREM: Let c 2 (T) be a tree’s Euclidean distortion, then (up to constants), D EFINITION: For a metric space (X,d), i.e. the height of the largest complete binary tree that embeds into T with distortion at most 2. Let’s prove this…

monotone edge colorings If T=(V,E) is a (rooted) tree, then an edge-coloring of T is a map The coloring is monotone if every color class is a monontone path in T (monotone path = continguous subset of root-leaf path) A coloring  is  -good if, for every u,v 2 T, at least an  -fraction of the u-v path is monochromatic. u v

monotone edge colorings If T=(V,E) is a (rooted) tree, then an edge-coloring of T is a map The coloring is monotone if every color class is a monontone path in T (monotone path = continguous subset of root-leaf path) A coloring  is  -good if, for every u,v 2 T, at least an  -fraction of the u-v path is monochromatic. u v

good colorings ) good embeddings We associate to every color class j 2 {1, 2, …, C}, a unit vector  j 2 R C. Given a vertex x whose path from the root uses edges e 1, e 2, …, e k, we define our embedding f : T ! R C by e1e1 e2e2 e3e3 e4e4 x f(x) = [len(e 1 )+len(e 2 )]  1 + len(e 3 )  2 + len(e 4 )  3

good colorings ) good embeddings We associate to every color class j 2 {1, 2, …, C}, a unit vector  j 2 R C. Given a vertex x whose path from the root uses edges e 1, e 2, …, e k, we define our embedding f : T ! R C by e1e1 e2e2 e3e3 e4e4 x Claim: f is non-expansive, i.e. (triangle inequality)

good colorings ) good embeddings We associate to every color class j 2 {1, 2, …, C}, a unit vector  j 2 R C. Given a vertex x whose path from the root uses edges e 1, e 2, …, e k, we define our embedding f : T ! R C by e1e1 e2e2 e3e3 e4e4 x Claim: For every x,y 2 T, lca(x,y) y x Monotonicity ) disjoint colors lca(x,y) x

good colorings ) good embeddings L EMMA: If T admits an  -good coloring, then the Euclidean distortion of T is at most 2/ . The hard part comes next… T HEOREM: If  * is the biggest  for which T admits an  -good coloring, then

good colorings ) good embeddings L EMMA: If T admits an  -good coloring, then the Euclidean distortion of T is at most 2/ . The hard part comes next… T HEOREM: If  * is the biggest  for which T admits an  -good coloring, then

good colorings ) good embeddings L EMMA: If T admits an  -good coloring, then the Euclidean distortion of T is at most 2/ . The hard part comes next… T HEOREM: If  * is the biggest  for which T admits an  -good coloring, then C OROLLARY: [ stronger embedding technique gives ]

good colorings ) good embeddings L EMMA: If T admits an  -good coloring, then the Euclidean distortion of T is at most 2/ . The hard part comes next… T HEOREM: If  * is the biggest  for which T admits an  -good coloring, then Proof outline:1. Give some procedure for coloring the edges of T. 2. If the procedure fails to construct an  -good coloring, find a complete binary tree of height O( 1 /  ) embedded inside T.

constructing a good coloring First, we define a family of trees {M k }: These are just {B k } with an extra “incoming” edge… M k = BkBk M0M0 M1M1 M2M2 Given a tree T, we say that T admits a copy of M k at scale j if… 1. M k embeds into T with distortion at most The root of M k maps to the root of T. 3. The edges of M k have length ¼ 4 j.

constructing a good coloring Now, suppose we have a “scale selector” function g : T ! Z which assigns a “scale” to every vertex in T. We produce a coloring as follows… T1T1 T2T2 T3T3 T4T4 v How to continue a coloring: Continue toward the T i which admits the largest copy of M k at scale g(v)… (break ties arbitrarily) j = g(v) 4j4j

constructing a good coloring Suppose we failed to produce an  -good coloring… u v D ·  D [ assume ¼  D ¼ 4 j ] Assume that g(w) = j for every breakpoint w on the u-v path. w In this manner, we construct a complete binary tree of height ¼ 1 /  inside T. But what about our assumptions on g(w)?

constructing a good coloring Suppose we failed to produce an  -good coloring… u v D Can define g so that every sufficiently dense set of breakpoints contains a large subset with the “right” g-values using hierarchical nets. j j+2 j+1 j+3 Points with g(w) ¸ k form a 4 k -net. At most a ¼ fraction of the 4 k -net points have label higher than k (geometric sum). Now reconstruct a complete binary tree of height  1 /  ) just using the green nodes.

cantor trees So we have these bounds: this upper bound is tight There exists a family of trees {C k } for which [ so the “branching” lower bound only gives ]

cantor trees Spherically symmetric trees (SST): Every path with marked vertices yields a binary SST.

cantor trees The Cantor trees are binary SSTs based on inductively defined paths… P 0 = P2P2 P2P2 P k+1 = PkPk PkPk length 2 k+1 P 1 = P 2 = P 3 = len(P k ) = 2 len(P k-1 ) + 2 k = k ¢ 2 k log log |C k | ~ k br(C k ) ~ k Claim: c 2 (C k ) ~ √k

strong edge colorings A monotone edge coloring  is  -strong if, for every u,v 2 T, at least half of the u-v path is colored by classes of length at least  ¢ d(u,v). T HEOREM: If  * is the biggest  for which T admits a  -strong coloring, then Proof sketch: 1. Show that  -strong colorings yield good embeddings. 2. Give some procedure to construct a monotone coloring. 3. If the coloring fails to be  -strong, show that T must contain a Cantor-like subtree. 4. Show that every Cantor-like subtree requires large distortion to embed in a Euclidean space.

cantor trees The Cantor trees do not have (good) strong colorings… P 0 = P2P2 P2P2 P k+1 = PkPk PkPk length 2 k+1 P 1 = P 2 = P 3 = 1/k 1/2k 1/4k ) best coloring is 2 -k/2 strong!

Markov convexity Idea: Look at Markov chains wandering in a Euclidean space; must satisfy special properties, e.g. symmetric random walk on Z, Z 2, … t=0t=0 t=kt=k

Markov convexity Idea: Look at Markov chains wandering in a Euclidean space; must satisfy special properties, e.g. symmetric random walk on Z, Z 2, … A metric space (M,d) is Markov 2-convex if, for every Markov chain {X t } taking values in M, and every number m 2 N, we have for some constant C ¸ 0.

Markov convexity T HEOREM: Every Euclidean space is Markov 2-convex. (with some universal constant C) A metric space (M,d) is Markov 2-convex if, for every Markov chain {X t } taking values in M, and every number m 2 N, we have for some constant C ¸ 0.

discrepancy with Euclidean space ) distortion ~ √m ~ √ log k lower bounds from Markov convexity If {X t } is the downward random walk on B k, then… 2m2m 1)1) m¢2m)m¢2m) (with the leaves as absorbing states)

lower bounds from Markov convexity P2P2 P2P2 P k+1 = PkPk PkPk length 2 k+1 P 3 = Let {X t } be the downward random walk on C k.

lower bounds from Markov convexity P 3 = Let {X t } be the downward random walk on C k. Key fact: At least a j/k fraction of P k is covered by segments whose length is at most 2 j.

conclusion M AIN T HEOREM: For every tree T, we have and -- Markov convexity is a notion for general metric spaces (X,d). Can we relate non-trivial Markov convexity to the non-containment of arbitrarily large complete binary trees? -- What about other Markov-style lower bounds for Hilbert space? -- Can we use reversible Markov chains to construct NEG metrics? -- Are these techniques useful for studying the bandwidth of trees? Q UESTIONS?