Random Walks on Hyperbolic Groups I - University of...

Random Walks on Hyperbolic Groups I

Steve Lalley

University of Chicago

January, 2014

Random Walks on GroupsRandom Walk on Integer LatticeRandom Walk on Free Group

Speed, Entropy, Spectral RadiusSubadditivityCriterion for Positive SpeedCarne-Varopoulos Theorem

Nonamenability: Kesten’s TheoremNonamenable GroupsKesten’s TheoremLamplighter RWBougerol’s TheoremCartwright’s ExampleProof of Kesten’s Theorem⇐=

What is a Random Walk?

Definition: A right random walk on a group Γ is a sequence (Xn)n≥0 ofΓ−valued random variables such that Xn+1 = Xnωn+1 whereω1, ω2, . . . are IID random variables with common distribution µ, i.e.,

P(n⋂

i=1

{ωi ∈ Bi}) =n∏

i=1

µ(Bi )



P(n⋂

i=1

{ωi ∈ Bi}) =n∏

i=1

µ(Bi )

Notation: Use Px to denote probabilities computed for the randomwalk with initial state x , and for any n ≥ 0 and x , y ∈ Γ,

pn(x , y) = Px{Xn = y} = µ∗n(x−1y)



P(n⋂

i=1

{ωi ∈ Bi}) =n∏

i=1

µ(Bi )

Standing Assumptions:I Symmetry: µ(x) = µ(x−1) and hence pn(x , y) = pn(y , x).I Bounded Steps: µ has finite support. Also called FRRW.I Irreducibility: support(µ) generates Γ (as semigroup).

Cayley Graph

If Γ is finitely generated with symmetric generating set A = {a±i } thenthe Cayley graph C(Γ; A) is the homogeneous graph with verticesx ∈ Γ and edges connecting x to xa±i . Edges (x , xapm

i ) are assignedcolors i . The natural metric on C(Γ; A) is the word metric.

A symmetric random walk on Γ with step distribution µ supported byA can be viewed as a homogeneous, reversible Markov Chain onC(Γ; A). When µ is the uniform distribution on A the random walk iscalled simple random walk.

Example: Cayley Graph of SL(2,Z)

Vertices of Cayley graph are the tiles of the tessellation. Two verticesare connected by an edge if and only if tiles share a side.

Example: Simple Random Walk on Z2

Out[25]=-60 -40 -20 20 40 60

-40

-20

20

40

Simple Random Walk on Zd

Central Limit Theorem: Let Sn be the position of simple random walkon Zd after n steps. Then

Sn√n

=⇒ Gaussion(0, I)

Local Central Limit Theorem: Uniformly for |x |/√

n in any compact setand such that x has correct parity,

P{Sn = x} ∼ exp{−|x |2/2n}(2πn)d/2

Simple Random Walk on Zd

Observe: For simple random walk on the d−dimensional integerlattice, return probabilities decay polynomially in the number of steps:

Local Limit Theorem: P0{X2n = 0} ∼ (4πn)−d/2

Proof: By Fourier analysis. Let µ =uniform distribution on the 2d unitvectors; then

µ̂(θ) =

∫exp{i〈θ, x〉}dµ(x) =

1d

d∑j=1

cos θj

and the F-transform of µ∗2n is the (2n)th power of µ̂. Hence, byFourier Inversion,

P0{X2n = 0} = (2π)−d∫

[−π,π]d

1d

d∑j=1

cos θj

2n

dθ1dθ2 · · · dθd

Now use Laplace’s method of asymptotic expansion. SeeSpitzer,Principles of Random Walk, ch. 2, for details.

Laplace’s Method

Theorem: Let g : [−A,A]→ R be a C2 function with unique maximumg(0) = 0 at 0, and such that g′′(0) = −B < 0. Then as n→∞,∫ A

−Aeng(x) dx ∼

√2πnB

.

Proof: First check that the limits of integration can be changed to ±εwhere ε > 0 is small. Then use two-term Taylor series to get∫ ε

−εeng(x) dx ∼

∫ ε

−εe−nBx2/2 dx

Example: Simple Random Walk on Free Group

Free Group FK : Given S = {a1,a2, . . . ,aK}, the free group FK on kgenerators is the set of all reduced finite words in the symbols a±1

j . Aword is reduced if no generators occurs next to its inverse.Multiplication is by concatenation followed by reduction.

Example: Free Group on 2 Generators

Example: Simple Random Walk on Free Group

Simple Random Walk on FK : Step distribution is the uniformdistribution on the generating set A = S±. This distribution is radiallysymmetric =⇒ Yn := distance(Xn,1) is reflecting p − q random walkon Z+ with p = (2K − 1)/2K .

Consequences:

(A) Positive Speed: ` = lim Yn/n = (2K − 2)/K > 0(B) Exponential Decay of Return Probabilities:

lim P1{X2n = 1}1/2n = % where % < 1.(C) Boundary Convergence: Xn converges to a an “end” of the

Cayley Graph.

Subadditivity

Let Xn be a FRRW on a discrete group Γ. Then each of the followingsequences is subadditive:

(A) − log P1{X2n = 1}(B) −E log wn(Xn) where wn(x) = P1{Xn = x}(C) Ed(Xn,1) where d is any metric on Γ

Fekete’s Lemma: If an+m ≤ an + am for all m,n then

limx→∞

an/n = infn≥1

an/n.

Note: Unless otherwise specified assume that the metric d is theword metric with respect to a specified set of generators.

Subadditivity

Hence each of the following limits exists:

% = lim P1{X2n = 1}1/2n

h = lim−E log wn(Xn)/n where wn(x) = P1{Xn = x}` = lim Ed(Xn,1)/n where d is a metric on Γ.

These limits are the spectral radius, entropy, and speed.

The limits %,h, ` are of basic interest:

I How are they related?I How do they depend on the group?I How do they depend on the step distribution?

Kingman’s Theorem

Kingman’s Subadditive Ergodic Theorem: Let wn : Γn → R bebounded, measurable functions such that

wm+n(x1, . . . , xm+n) ≤ wm(x1, . . . , xm) + wn(xm+1, . . . , xm+n)

If ξ1, ξ2, . . . are IID random variables valued in Γ then

limn→∞

n−1wn(ξ1, . . . , ξn) = inf En−1wn(ξ1, . . . , ξn) almost surely.

Kingman’s Theorem implies that

h = lim− log wn(Xn)/n almost surely and` = lim d(Xn,1)/n almost surely

Elementary Relation

Proposition: h ≥ − 12 log %.

Proof: By symmetry of the RW

P1{X2n = 1} =∑x∈Γ

wn(x)2.

Since the distribution of Xn is concentrated on those x for whichwn(x) ≈ e−nh it follows that for any ε > 0 and all large n

P1{X2n = 1} ≥ e−2n(h+ε).

The proposition follows.

Positive Speed

Theorem: Let Xn be a symmetric FRRW on a finitely generated groupΓ. Then the following are equivalent:

(A) Positive Entropy: h > 0(B) Positive Speed: ` > 0(C) Failure of Liouville Property: ∃ nonconstant, bounded harmonic

functions.

Definition: u : Γ→ R is harmonic if u(x) = Exu(X1) for every x ∈ Γ.

Corollary: Exponential decay of return probability (% < 1) impliespositive speed.

Note: The nontrivial equivalence (A)⇐⇒(C) is due independently toDerriennic and to Kaimanovich & Vershik.

Positive Entropy =⇒ Positive Speed

Exercise: Prove that for any FRWW on a finitely generated group,positive entropy =⇒ positive speed.

Hint: Fekete’s Lemma implies that the cardinality of the ball of radiusn grows exponentially with n.

Proof that Positive Speed =⇒ Positive Entropy

Carne-Varopoulos Theorem: Let Xn be a symmetric, FRRW on Γwhose step distribution is supported by a symmetric generating set A.Let d be the word metric on Γ relative to A (i.e., the usual metric onthe Cayley graph C(Γ,A)). Let % = lim P1{X2n = 1}1/2n. Then ∀ x ∈ Γ,

P1{Xn = x} ≤ 2%n exp{−d(1, x)2/2n}

Corollary: Positive speed =⇒ positive entropy.

Proof of Carne-Varopoulos

Markov Operator: For any random walk Xn (or more generally anyMarkov chain) the associated Markov operator T : `2(Γ)→ `2(Γ) isdefined by be the Markov operator for a symmetric random walk Xnon Γ, defined by

Tf (x) = Ex f (X1) =⇒T nf (x) = Ex f (Xn)




If the random walk is symmetric then the Markov operator T isself-adjoint, and so by the Spectral Theorem there are spectralmeasures νf such that

〈f ,Tf 〉 =

∫ ‖T‖−‖T‖

t dνf (t) =⇒

〈f ,T nf 〉 =

∫ ‖T‖−‖T‖

tn dνf (t)




Consequently, the transition probabilities are given by

pn(x , y) = 〈δx ,T nδy 〉 =

∫tn dνx,y where

νx,y =12

(νδx +δy − νδx − νδy )


Chebyshev Polynomials:

Qn(cos θ) = cos nθ =⇒|Qn(s)| ≤ 1 for all |s| ≤ 1

Inversion Formula: Let {Sn}n≥0 be simple random walk on Z. Then

sn =∑k∈Z

P{Sn = k}Qk (s)

Corollary: For any self-adjoint linear operator T with norm % <∞,

T n =∑k∈Z

P{Sn = k}Qk (T )





sn =∑k∈Z

P{Sn = k}Qk (s)

The inversion formula is simply a reformulation of the fact that theFourier transform of µ∗n is the nth power of the Fourier transform ofµ, because for µ = (δ−1 + δ1)/2 the Fourier transform is cos θ.


T n =∑k∈Z

P{Sn = k}Qk (T )





sn =∑k∈Z

P{Sn = k}Qk (s)


T n =∑k∈Z

P{Sn = k}Qk (T )


For the Markov operator T of a symmetric random walk the spectralradius % is the norm of T . Consequently, since Qk is a polynomial ofdegree k , the Inversion Formula

T n/%n =∑k∈Z

P{Sn = k}Qk (T/%)

implies

Px{Xn = y} = %n∑k∈Z

P{Sn = k}〈δx ,Qk (T/%)δy 〉

≤ 2%n∑

k≥d(x,y)

P{Sn = k}


For the Markov operator T of a symmetric random walk the spectralradius % is the norm of T . Consequently, since Qk is a polynomial ofdegree k , the Inversion Formula

T n/%n =∑k∈Z

P{Sn = k}Qk (T/%)

implies

Px{Xn = y} = %n∑k∈Z

P{Sn = k}〈δx ,Qk (T/%)δy 〉

≤ 2%n∑

k≥d(x,y)

P{Sn = k}

The result now follows from the Hoeffding Inequality for simplerandom walk on Z:

P{|Sn| ≥ α} ≤ 2 exp{−2α2/n}.

Nonamenability

Definitions: (1) The isoperimetric constant κ of a graph G = (V , E) is

inf|∂F ||F |

.

(2) A finitely generated discrete group Γ is nonamenable if for some(every) finite generating set A the Cayley graph G(Γ; A) has positiveisoperimetric constant.

Folner’s Criterion: Γ is amenable if and only if for every ε > 0 andevery finite set K ⊂ Γ there is a finite set A ⊂ Γ such that

|Ax∆A| < ε|A| ∀ x ∈ K .

Sufficient Condition for Nonamenability

Proposition: A finitely generated countable group Γ is amenable onlyif for every Γ−action on a compact metric space Y there is aΓ−invariant Borel probability measure on Y .



Proof: A probability measure ν on Y is Γ−invariant if and only if it isg−invariant for every g in a symmetric generating set A. Amenability=⇒ for each m ≥ 1 there exists finite set Fm ⊂ Γ such that

|Fm∆gFm| < m−1|Fm| ∀ g ∈ A

Fix y ∈ Y ; let νm =uniform distribution on Fmy . Then‖νm − νm ◦ g−1‖ < 2/m for each g ∈ A. Compactness of Y implies νmhas weak−∗ convergent subsequence whose limit is a probabilitymeasure ν. This measure is A−invariant.



Example: The free group Fd on d ≥ 2 generators is nonamenable.

Proof: Use the criterion with the natural action of Fd on the boundary(space of ends) of the 2d−regular tree T2d



Example: Any co-compact discrete subgroup Γ of PSL(2,R) isnonamenable.

Proof: Use the criterion with the natural action of Γ on the boundaryof the hyperbolic plane.

There are many known equivalent conditions for amenability.

Examples

I Zd is amenable.I Fd (free group) is nonamenable if d ≥ 2.I SL(n,Z) is nonamenable if n ≥ 2.I Fundamental group of a genus ≥ 2 surface is nonamenable.I Hyperbolic groups are nonamenable.I Lattices of connected, noncompact, semi-simple Lie groups are

nonamenable.

Final Score: Nonamenable 5, Amenable 1.

Kesten’s Theorem.

Theorem: Let Xn be a symmetric FRRW on Γ such that every point ofΓ is accessible from 1. Then % < 1 if and only if Γ is nonamenable.

Reminder: % = limn→∞ p2n(1,1)1/2n.

Corollary: Every symmetric FRRW on a nonamenable group haspositive speed and positive entropy, and hence is transient.

Problem: What is the precise asymptotic behavior of P1{X2n = 1}?

Example: The Lamplighter Random Walk

At each site x ∈ Zd is a lamp, which can be either on or off. A randomwalker performs a simple random walk on Zd , and at each stepadjusts the state of the lamp at his/her current site by a fair coin toss.

This is a random walk on the wreath product Gd = Zd o ZZd

2 . Thegroup Gd is amenable but has exponential growth.

Theorem: (Kaimanovich-Vershik) If d ≥ 3 then the lamplighterrandom walk has nonconstant bounded harmonic functions, hencepositive entropy and positive speed. If d = 1,2 then the speed andentropy are 0.

Theorem: (Revelle) For the simple lamplighter random walk on G1,

Pe{X2n = e} ∼ C1n−1/6 exp{−C2n1/3}.

Local Limit Theorems: Semisimple Lie Groups

Theorem: (Bougerol 1981) Let Xn = ξ1ξ2 · · · ξn be a symmetric, rightrandom walk on a connected, semisimple Lie group G with finitecenter whose step distribution has a density with respect to Haarmeasure and rapidly decaying tails. Then the n−step transitionprobability densities pn(x , y) satisfy

pn(x , y) ∼Cx,y

Rnn−p/2

for some R > 1 depending on the step distribution. The exponentp ∈ N depends only the rank and number of positive roots of the (Liealgebra of) G. For G = SL(2,R), the exponent is p = 3.

Conjecture: For any lattice Γ of a connected, semisimple Lie groupwith finite center, a symmetric FRRW on Γ obeys a local limit theoremwith the same exponent p:

P1{X2n = 1} ∼ CR−2n(2n)−p/2.

Cartwright’s ExampleFree Product: If Γ1, Γ2 are two discrete groups the free product Γ1 ∗ Γ2is the group whose elements are finite words

a0b1a2b2 · · ·

where each ai ∈ Γ1, each bi ∈ Γ2, and for i ≥ 1 the elements ai and biare distinct from the identities in Γ1 and Γ2.

I Multiplication is by concatenation and reduction.I Identity is the word a0 = 1.

Example: Z2 ∗ Z2 ∗ Z2 has Cayley graph T3

Cartwright’s Example

Theorem: Consider the random walk on Zd ∗ Zd with step distribution(αν) ∗ (1− α)ν where ν is the uniform distribution on the 2d unitvectors of Zd . If d ≥ 5 then

I For α = 1/2,p2n(1,1) ∼ C%−2nn−d/2

I For some α ∈ (0,1),

p2n(1,1) ∼ Cα%−2nα n−3/2

Thus, the form of a local limit theorem for a nearest neighbor,symmetric random walk is not uniquely determined by the group.

Proof of Kesten’s Theorem

Objective: Show that Γ nonamenable =⇒ spectral radius of T is < 1.

Lemma: (Isoperimetric Inequality) If Γ is nonamenable then for anysymmetric, irreducible random walk ∃ α > 0 and n ≥ 1 such that forevery finite set A ⊂ Γ, ∑

x∈A

∑y 6∈A

pn(x , y) ≥ α|A|




x∈A

∑y 6∈A

pn(x , y) ≥ α|A|

Remark: In proving Kesten’ theorem we may assume

(a) random walk is irreducible (replace µ by (µ+ µ∗2)/2);(b) isoperimetric inequality holds with n = 1 (replace µ by µ∗n).




x∈A

∑y 6∈A

pn(x , y) ≥ α|A|

Proof: Nonamenability (Folner criterion) =⇒ there are finitely manygi ∈ Γ and ε > 0 such that for every finite A ⊂ Γ,

|A∆Agi | ≥ ε|A| for some i

Irreducibility of the random walk =⇒ for some n

µ∗n(gi ) > 0 for every i .

Sobolev Inequality

Sobolev Norm: For any function f : Γ→ R define

S(f ) =12

∑x,y∈Γ

|f (y)− f (x)|p(x , y)

Proposition: Isoperimetric inequality (with n = 1) implies Sobolevnorm inequality

S(f ) ≥ α‖f‖1.

Proof: Without loss of generality f ≥ 0.

S(f ) =∑

x,y : f (x)<f (y)

|f (y)− f (x)|p(x , y)

=

∫ ∞0

∑x,y

1{f (x) < t < f (y)}p(y , x) dt

≥∫ ∞

0α|{y : f (y) > t}|dt = α‖f‖1.

Sobolev Inequality Implies Spectral Gap

Objective: Show that Sobolev inequality implies ‖T‖ < 1. Since T isself-adjoint, it suffices (exercise!) to show

〈(I − T )f , f 〉 ≥ ε‖f‖22

Dirichlet Form: D(f , f ) = 12

∑∑(f (y)− f (x))2p(x , y)

Lemma: D(f , f ) = 〈(I − T )f , f 〉.

Conclusion: Relate D(f , f ) to S(f 2). Sobolev inequality implies

α‖f‖22 = α‖f 2‖1 ≤ S(f 2)

=12

∑x,y

|f (x)2 − f (y)2|p(x , y)

=12

∑x,y

|f (x)− f (y)||f (x) + f (y)|p(x , y)

≤ C√D(f , f )‖f‖2

Sobolev Inequality Implies Spectral GapObjective: Show that Sobolev inequality implies ‖T‖ < 1. Since T isself-adjoint, it suffices (exercise!) to show

〈(I − T )f , f 〉 ≥ ε‖f‖22


∑∑(f (y)− f (x))2p(x , y)

Lemma: D(f , f ) = 〈(I − T )f , f 〉.

Proof: Since p(x , y) = p(y , x),

D(f , f ) =12

∑∑(f (x)2 − 2f (x)f (y) + f (y)2)p(x , y)

=∑∑

(f (x)2 − f (x)f (y))p(x , y)

= 〈(I − T )f , f 〉.


α‖f‖22 = α‖f 2‖1 ≤ S(f 2)

=12

∑x,y

|f (x)2 − f (y)2|p(x , y)

=12

∑x,y

|f (x)− f (y)||f (x) + f (y)|p(x , y)

≤ C√D(f , f )‖f‖2

Sobolev Inequality Implies Spectral Gap

Objective: Show that Sobolev inequality implies ‖T‖ < 1. Since T isself-adjoint, it suffices (exercise!) to show

〈(I − T )f , f 〉 ≥ ε‖f‖22


∑∑(f (y)− f (x))2p(x , y)

Lemma: D(f , f ) = 〈(I − T )f , f 〉.


α‖f‖22 = α‖f 2‖1 ≤ S(f 2)

=12

∑x,y

|f (x)2 − f (y)2|p(x , y)

=12

∑x,y

|f (x)− f (y)||f (x) + f (y)|p(x , y)

≤ C√D(f , f )‖f‖2

Random Walks on Hyperbolic Groups I - University of...

Documents

Transcript of Random Walks on Hyperbolic Groups I - University of...