Trees and Markov convexity

James R. LeeInstitute for Advanced Study

[ with Assaf Naor and Yuval Peres ]Rd

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 ! Rk (or L2)

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 ! Rk (or L2)

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 treeT = (V,E)

+edge lengthslen : E ! R+

len(e)

x

y

d(x,y) = length of shortest geodesic

the problem for trees

[Bourgain 86]: The complete binary tree Bk 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 Bk 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.

e1

e2

e3

why don’t trees embed in Hilbert space?

ONE ANSWER: (EQUILATERAL) FORKS

If both of these paths of length 2 areembedded isometrically in a Euclideanspace, then A and B must conincide!

A BQuantitative 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 R2,

f(W)=0

a=f(A) b=f(B)

4 ± O()4 ± O() O()

forks in complete binary trees

Rd

Ramsey style proof: If Bk is embedded into L2 with distortion then there exists some almost-isometric fork. [Matousek]

CONTRADICTION!

on forks

Natural question: Are forks the only obstruction? The problem isn’t forking; it’s forking, and forking, and forking…

THEOREM: 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 {Bk} do not embed into T with bounded distortion.

In other words, a tree embeds into a Euclidean space if and only if it doesnot “contain” arbitrarily large binary trees!

quantitative version

THEOREM: Let c2(T) be a tree’s Euclidean distortion, then (up to constants),

DEFINITION: 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, atleast 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, atleast 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 RC. Given a vertex x whose path from the root uses edges e1, e2, …, ek, we define our embedding f : T ! RC by

e1

e2

e3

e4

x

f(x) = [len(e1)+len(e2)]1 + len(e3) 2 + len(e4) 3

good colorings ) good embeddings

We associate to every color class j 2 {1, 2, …, C}, a unit vector j 2 RC. Given a vertex x whose path from the root uses edges e1, e2, …, ek, we define our embedding f : T ! RC by

e1

e2

e3

e4

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 RC. Given a vertex x whose path from the root uses edges e1, e2, …, ek, we define our embedding f : T ! RC by

e1

e2

e3

e4

xClaim: For every x,y 2 T,

lca(x,y)

y

x

Monotonicity)

disjoint colors

lca(x,y)

x

good colorings ) good embeddings

LEMMA: If T admits an -good coloring, then the Euclidean distortion of T is at most 2/.

The hard part comes next…

THEOREM: If * is the biggest for which T admits an -good coloring, then

good colorings ) good embeddings

LEMMA: If T admits an -good coloring, then the Euclidean distortion of T is at most 2/.

The hard part comes next…

THEOREM: If * is the biggest for which T admits an -good coloring, then

good colorings ) good embeddings

LEMMA: If T admits an -good coloring, then the Euclidean distortion of T is at most 2/.

The hard part comes next…

THEOREM: If * is the biggest for which T admits an -good coloring, then

COROLLARY:

[ stronger embedding technique gives ]

good colorings ) good embeddings

LEMMA: If T admits an -good coloring, then the Euclidean distortion of T is at most 2/.

The hard part comes next…

THEOREM: 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 {Mk}: These are just {Bk} with an extra “incoming” edge…

Mk =Bk

M0

M1

M2

Given a tree T, we say that T admits a copy of Mk

at scale j if… 1. Mk embeds into T with distortion at most 4. 2. The root of Mk maps to the root of T. 3. The edges of Mk have length ¼ 4j.

constructing a good coloring

Now, suppose we have a “scale selector” function g : T ! Z which assignsa “scale” to every vertex in T. We produce a coloring as follows…

T1 T2

T3

T4

vHow to continue a coloring: Continue toward the Ti which admits the largest copy of Mk at scale g(v)… (break ties arbitrarily)

j = g(v)

4j

constructing a good coloring

Suppose we failed to produce an -good coloring…

u

v

D

· D [ assume ¼ D ¼ 4j ]Assume that g(w) = j for every breakpoint w on the u-v path.

w

In this manner, we construct a completebinary 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 ofbreakpoints contains a large subset with the“right” g-values using hierarchical nets.

jj+2

j+1

j+3Points with g(w) ¸ k form a 4k-net. At mosta ¼ fraction of the 4k-net points have labelhigher than k (geometric sum).

Now reconstruct a complete binary tree of height1/) just using the green nodes.

cantor trees

So we have these bounds:

this upper bound is tight

There exists a family of trees {Ck} 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…

P0 =

P2 P2

Pk+1 = Pk Pk

length 2k+1

P1 =

P2 =

P3 =

len(Pk) = 2 len(Pk-1) + 2k = k¢2k

log log |Ck| ~ k br(Ck) ~ k

Claim: c2(Ck

) ~ √k

strong edge colorings

A monotone edge coloring is -strong if, for every u,v 2 T, at leasthalf of the u-v path is colored by classes of length at least ¢d(u,v).

THEOREM: 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…

P0 =

P2 P2

Pk+1 = Pk Pk

length 2k+1

P1 =

P2 =

P3 =1/k

1/2k1/2k

1/4k 1/4k 1/4k 1/4k

) best coloring is 2-k/2 strong!

Markov convexity

Idea: Look at Markov chains wandering in a Euclidean space; mustsatisfy special properties, e.g. symmetric random walk on Z, Z2, …

t=0 t=k

Markov convexity

Idea: Look at Markov chains wandering in a Euclidean space; mustsatisfy special properties, e.g. symmetric random walk on Z, Z2, …

A metric space (M,d) is Markov 2-convex if, for every Markovchain {Xt} taking values in M, and every number m 2 N, we have

for some constant C ¸ 0.

Markov convexity

THEOREM: Every Euclidean space is Markov 2-convex. (with some universal constant C)

A metric space (M,d) is Markov 2-convex if, for every Markovchain {Xt} 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 {Xt} is the downward random walk on Bk, then…

2m

1)

m¢2m)

(with the leaves as absorbing states)

lower bounds from Markov convexity

P2 P2

Pk+1 = Pk Pk

length 2k+1

P3 =

Let {Xt} be the downward random walk on Ck.

lower bounds from Markov convexity

P3 =

Let {Xt} be the downward random walk on Ck.

Key fact: At least a j/k fraction of Pk is covered by segments whose length is at most 2j.

conclusion

MAIN THEOREM: 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?

QUESTIONS?