Notes on geodesics in first-passage percolation

30 Pages·2016·0.316 MB·English

by  Michael Damron

Course notes on geodesics in first-passage percolation
Michael Damron∗
Georgia Institute of Technology
July 15, 2016
Abstract
These notes accompany courses given in University of Bath and Northwestern Uni-
versityduringtheirsummerschoolsin2016. Themaintopicisgeodesicsinfirst-passage
percolation, specifically recent developments by Hoffman and by Damron-Hanson, on
directional properties of infinite geodesics. As is done in our recent survey [6], we out-
line how many of the results from Damron-Hanson ’14 [10] can be extended from two
dimensions to general dimensions.
Contents
1 Introduction 2
1.1 Existence of finite geodesics . . . . . . . . . . . . . . . . . . . . . . . . . 2
2 Infinite geodesics: Hoffman 4
2.1 Properties of Busemann functions . . . . . . . . . . . . . . . . . . . . . . 4
2.2 Hoffman’s argument . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
3 Interlude: the shape theorem 6
4 Newman and curvature 10
4.1 Newman’s theorems and conjectures . . . . . . . . . . . . . . . . . . . . 10
4.2 Proofs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
5 Weakening assumptions: back to the present day 13
5.1 In direction e . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
1
5.2 In other directions and what is known about limits . . . . . . . . . . . . 17
6 Busemann gradient fields 18
6.1 Existence of geodesics directed in sectors . . . . . . . . . . . . . . . . . . 18
6.2 Sketch of proof . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
∗M. D. is supported by NSF grant DMS-1419230 and an NSF CAREER grant.
1
7 Bigeodesics 24
7.1 Previous results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
7.2 Sketch of proof of Theorem 7.2 . . . . . . . . . . . . . . . . . . . . . . . 25
1 Introduction
Wewillspeakaboutgeodesicsinfirst-passagepercolation(FPP),specificallydirections
and existence. All of this material is contained in the recent survey [6] by Auffinger-
Damron-Hanson. Main outline:
1. Intro and Hoffman ’05
2. Newman’s conjectures and curvature
3. Busemann functions
4. Busemann gradient fields: Damron-Hanson ’13
5. Bigeodesics: Damron-Hanson ’15
Consider Zd with nearest neighbor edges Ed. We begin with a list of definitions.
FPP consists of:
• (t ), a collection of nonnegative weights assigned to the edges,
e
(cid:80)
• T(γ) = t , where γ is a lattice path,
e∈γ e
• T(x,y) = inf T(γ) for x,y ∈ Zd, T(A,B) = inf{T(x,y) : x ∈ A,y ∈ B} for
γ:x→y
A,B ⊂ Zd, and
• for x,y ∈ Rd, T(x,y) = T([x],[y]), where x ∈ [x]+[0,1)d, y ∈ [y]+[0,1)d.
• a geodesic from x to y is a lattice path γ from x to y with T(γ) = T(x,y). [Might
not exist.]
Note the triangle inequality of T:
T(x,y) ≤ T(x,z)+T(z,y).
• So a subpath of a geodesic is a geodesic: if γ is a geodesic from x to y and γ(cid:48) is
the section from w to z, if it is not a geodesic, replace γ(cid:48) by the geodesic to get
a path of lower weight.
• A geodesics can be taken to be (vertex) self-avoiding.
1.1 Existence of finite geodesics
Definition 1.1 (Zhang [36]). Let ρ be the random variable
ρ(x) = limT(x,∂B(n)),
n
where B(n) = [−n,n]d and ∂B(n) = {x ∈ B(n) : ∃y ∈ B(n)c s.t. x ∼ y}.
(Exercise.) Show that
ρ(x) = inf{T(γ) : γ is an infinite self-avoiding path from x}.
2
Lemma 1.2. 1. If ρ(0) = ∞ then geodesics exist: for all x,y, there is a geodesic
from x to y.
2. If (t ) are i.i.d. with P(t = 0) < p , then a.s., geodesics exist (between all
e e c
vertices).
3. (Exercise.) if (t ) are i.i.d. and continuous, then a.s., geodesics are unique.
e
In item 2, we are using
p = sup{p : P (∃ infinite path of 0-edges) = 0)},
c p
in a model where P (t = 0) = p = 1−P (t = 1). (It is known that p (d) ∈ (0,1) for
p e p e c
all d and p (d) ∼ 1/(2d).)
c
Proof. 1. If ρ(0) = ∞, also ρ(x) = ∞ for all x, since
|T(x,∂B(n))−T(0,∂B(n))| ≤ T(0,x) < ∞.
So pick γ any deterministic path from x to y and choose n large enough so that
x ∈ B(n) and T(x,∂B(n)) > T(γ). Then if γ(cid:48) is a path from x to y touching
B(n)c, one has
T(γ(cid:48)) ≥ T(x,∂B(n)) > T(γ),
so it cannot be a geodesic, and the infimum is over a finite set.
2. Choose δ > 0 such that P(t < δ) < p , and there is no infinite path of edges with
e c
weight < δ. Thus ρ(0) = ∞ almost surely since every infinite path has infinite
passage time.
In fact, (follows from Kesten [24, Prop. 5.8]) a stronger version of item 2 holds. If
(t ) are i.i.d. with P(t = 0) < p , then
e e c
(cid:18) (cid:19)
T(0,x)
P liminf > 0 = 1. (1.1)
x→∞ (cid:107)x(cid:107)1
(Exercise.) Show this for P(t = 0) = 0.
e
Open question. Do geodesics exist in the i.i.d. case when P(t = 0) = p ? True
e c
in 2d by Wierman-Reh [31, Cor. 1.3]. In general dimensions if > p by Zhang [37,
c
Theorem 2].
Under which conditions is ρ(0) < ∞? In 2d, Damron-Lam-Wang ’15 [14, Cor. 1.3]
showed that ρ(0) < ∞ a.s. if and only if (cid:80) F−1(p + 1/2n) < ∞. Here F is the
n c
distribution function of a single t , and F−1 is the generalized inverse:
e
F−1(t) = inf{x : F(x) ≥ t} for t > 0.
For general dimensions, I don’t know better conditions for ρ(0) = ∞ than P(t = 0) <
e
p . (But see also [36, Theorem 8.1.7] for ρ(0) = ∞ in high dimensions for Bernoulli
c
distributions.)
3
2 Infinite geodesics: Hoffman
Definition 2.1. An infinite path γ is an infinite geodesic if each segment is a finite
geodesic. Let N = N(ω) be the maximal k such that there are k disjoint infinite
geodesics.
Note: we may remove loops from γ to ensure it is self-avoiding and still an infinite
geodesic.
Proposition 2.2. The variable N is translation-invariant, so if the distribution of (t )
e
is ergodic, N is almost surely constant. If geodesics exist in ω, then N(ω) ≥ 1.
Proof. YoucancheckN ismeasurable,sothefirststatementisclear. Infinitegeodesics
exist by a subsequence argument. Let γ be a geodesic from 0 to ne and take a
n 1
subsequential limit as follows. All γ ’s must take one of the 2d edges incident to 0
n
first. Choose f such that it is the first edge of infinitely many γ ’s (say γ ’s). Then
1 n n
k
f which is the second edge of infinitely many of the γ ’s, and so on. If γ is the path
2 n
k
following edges f ,f ,..., then each segment of γ is a segment of a finite geodesic, so
1 2
is a geodesic.
2.1 Properties of Busemann functions
The main question today: what is the value of N? For this we need Busemann func-
tions.
Definition 2.3. Let γ be an infinite geodesic with starting point x and vertices
0
x ,x ,.... The Busemann function for γ is defined as
1 2
B (x) = lim[T(x,x )−T(x ,x )].
γ n 0 n
n
This limit exists by monotonicity:
T(x,x )−T(x ,x ) = T(x,x )+T(x ,x )−T(x ,x ) ≥ T(x,x )−T(x ,x )
n 0 n n n n+1 0 n+1 n+1 0 n+1
and the bound due to the triangle inequality:
|T(x,x )−T(x ,x )| ≤ T(x,x ).
n 0 n 0
Sometimes we write
B (x,y) = B (x)−B (y).
γ γ γ
Here are some properties.
1. For m < n, B (x )−B (x ) = T(x ,x ).
γ m γ n m n
2. (Exercise.) If γ ,γ are infinite geodesics that coalesce (they have finite sym-
1 2
metric difference) and have initial points x , y , then
0 0
B (x) = B (x)−B (x ) for all x.
γ1 γ2 γ2 0
So
B (x,y) = B (x,y) for all x,y.
γ1 γ2
4
3. B (x,y) = B (x,z)+B (z,y) for all x,y,z.
γ γ γ
4. Let θ be a translation of the lattice by an integer vector. Then
B (x,y)(ω) = B (θ(x),θ(y))(θ(ω)),
γ θγ
where the translated weight configuration θ(ω) is defined as t (θ(ω)) = t (ω).
e θ−1e
Furthermore θγ is the translated geodesic ray.
2.2 Hoffman’s argument
The following was shown also by Garet-Marchand [15] (and Haggstrom-Pemantle [20]
for exponential).
Theorem 2.4 (Hoffman’05[21]). Let (t ) be i.i.d., continuous, with finite mean. Then
e
N ≥ 2.
Holds under much more general assumptions.
Proof. We know N ≥ 1. Suppose for a contradiction that N = 1. Then if γ,γ(cid:48) are
infinite geodesics, they must touch infinitely many times. By unique geodesics, they
must coalesce. So we can define the function
f(x,y) = B (x,y),
γ
where γ = γ(ω) is any infinite geodesic.
(Exercise.) Show this equals lim [T(x,ne )−T(y,ne )] when we assume N = 1, so
n 1 1
it is measurable.
Also it has finite mean and is translation covariant:
f(x,y)(ω) = f(θ(x),θ(y))(θ(ω))
for any integer translation. Additivity and the ergodic theorem give for any x,
n n
1 1 (cid:88) 1 (cid:88)
f(0,nx) = f((k−1)x,kx) = f(0,x)(θk−1ω) → Ef(0,x) a.s. and in L1,
n n n
k=1 k=1
where θ is translation by −x.
What is this limit? By translation invariance, x (cid:55)→ Ef(0,x) is additive for x ∈ Zd:
Ef(0,x+y) = Ef(0,x)+Ef(x,x+y) = Ef(0,x)+Ef(0,y),
so it is determined by its values for x = ±e . By symmetry for all i,j,
i
Ef(0,±e ) = Ef(0,±e ),
i j
and
0 = Ef(0,e )+Ef(e ,0) = Ef(0,e )+Ef(0,−e ) = 2Ef(0,e ),
1 1 1 1 1
5
giving Ef(0,x) = 0 for all x. Thus for all x,
1
f(0,nx) → 0 a.s. and in L1.
n
In fact one can show a stronger statement (by an argument similar that given in
the next section for the Shape Theorem): for each (cid:15) > 0,
P(|f(0,x)| ≥ (cid:15)(cid:107)x(cid:107) for infinitely many x ∈ Zd) = 0,
1
so that if γ is the limit of geodesics from 0 to ne , with vertices 0 = x ,x ,x ,..., then
1 0 1 2
B (0,x ) f(0,x )
γ n n
= → 0 almost surely.
(cid:107)x (cid:107) (cid:107)x (cid:107)
n 1 n 1
On the other hand, since 0,x ∈ γ, B(0,x ) = T(0,x ), so this implies
n n n
T(0,x )
n
→ 0 almost surely ,
(cid:107)x (cid:107)
n 1
which contradicts (1.1).
What else is known?
• (Hoffman ’08 [22, Theorem 1.3]) N ≥ 4 in 2d.
• (Damron-Hochman ’13 [13, Theorem 1.4]) Given K, there exist continuous dis-
tributions in 2d with N ≥ K.
• In the discrete case (no unique geodesics), one considers the graph of infection
of the origin. T is the union of all geodesics from 0 to all other vertices. Let
K be the number of ends. (A graph has ≥ k ends if removing a finite set splits
it into ≥ k components.) Then: (Auffinger-Damron ’14 [4, Theorem 2.3]) if
I = infsupp t > 0 and P(t = I) is large enough (with the distribution of t not
e e e
purely atomic), then K = ∞ almost surely.
3 Interlude: the shape theorem
(This section is adapted from tutorial lectures of Jack Hanson from the Bath summer
school.)
We will need the shape theorem for all the results beyond, and particularly for
Newman’s results. It is a type of law of large numbers for the set
B(t) = {x ∈ Rd : T(0,x) ≤ t}.
Theorem3.1(Richardson[32],Cox-Durrett[9],Kesten[24]). AssumeEmin{t ,...,t }d <
1 2d
∞, where t are i.i.d. copies of t and P(t = 0) < p . There exists a deterministic,
i e e c
convex set in Rd, symmetric about the axes and with nonempty interior, such that for
any (cid:15) > 0,
P((1−(cid:15))B ⊂ B(t)/t ⊂ (1+(cid:15))B for all large t) = 1.
6
There is also a version for ergodic distributions by Boivin [8].
(Exercise.) Show that B is the unit ball of a norm, called g, and
|T(0,x)−g(x)|
limsup = 0 almost surely.
(cid:107)x(cid:107)
x→∞ 1
This can be thought of as
T(0,x) = g(x)+o((cid:107)x(cid:107) ) as x ∈ Rd → ∞.
1
Proof of shape theorem. The idea of the proof is to first show “radial” convergence;
that is, for a fixed x ∈ Zd, to show that
T(0,nx)
g(x) := lim exists.
n n
To do this, we appeal to the subadditive ergodic theorem. Then we “patch” together
convergence in many different directions x to get a uniform convergence.
Liggett’s version [28, Theorem 1.10] of Kingman’s subadditive ergodic theorem
states:
Theorem 3.2. Let {X : 0 ≤ m < n} is an array of random variables satisfying the
m,n
following assumptions:
1. for each n, E|X | < ∞ and EX ≥ −cn for some constant c > 0,
0,n 0,n
2. X ≤ X +X for 0 < m < n,
0,n 0,m m,n
3. for each m ≥ 0, the sequence {X : k ≥ 1} is equal in distribution to
m+1,m+k+1
the sequence {X : k ≥ 1}, and
m,m+k
4. for each k ≥ 1, {X : n ≥ 1} is a stationary ergodic process.
nk,(n+1)k
Then
1 1
g := lim EX = inf EX exists,
0,n 0,n
n n n n
and
1
lim X = g a.s. and in L1.
0,n
n n
We apply this theorem for a fixed x ∈ Zd to the sequence
X = T(mx,nx).
m,n
Item 2 holds by the triangle inequality, whereas 3 and 4 hold by stationarity of the
environment under integer translations. In item 1 we can take any c > 0, since T ≥ 0
a.s. The only thing to check is that ET(0,nx) < ∞ for each n. By subadditivity,
and symmetry, it suffices to check that ET(0,e ) < ∞. To do this, we construct 2d
1
edge-disjoint deterministic paths γ ,...,γ from 0 to e and note
1 2d 1
ET(0,e ) ≤ Emin{T(γ ),...,T(γ )}.
1 1 2d
Now you can check:
7
(Exercise.) If Emin{t ,...,t } < ∞ for i.i.d. edge-weights t , then the right side
1 2d i
above is finite.
By the theorem, then, we define
T(0,nx)
g(x) = lim ,
n n
which is a.s. constant. We next extend g to Qd by taking such a rational x and letting
m ∈ N be such that mx is an integer point. Then set
T(0,mnx) 1 T(0,n(mx)) 1
g(x) = lim = lim = g(mx).
n mn m n n m
g thus defined on Qd satisfies the following properties: for x,y ∈ Qd,
1. g(x+y) ≤ g(x)+g(y),
2. g is uniformly continuous on bounded sets,
3. for q ∈ Q, g(qx) = |q|g(x),
4. g is symmetric about the axes.
Item 1 follows from the triangle inequality for T, and 4 follows from symmetries of the
edge weights. Item 3 is an easy exercise, and 2 follows from 1: for h = (h ,...,h ),
1 d
|g(z)−g(z+h)| ≤ g(h) = g(h e +···+h e ) ≤ g(e )(|h |+···+|h |) ≤ (cid:107)h(cid:107) ET(0,e ).
1 1 d d 1 1 d 1 1
Then g has a continuous extension to Rd. The above properties extend to real argu-
ments, so g is a seminorm. (A norm, except it could have g(x) = 0 for some x (cid:54)= 0.) It
is a result of Kesten [25, Theorem 6.1] that g is a norm when P(t = 0) < p .
e c
Now that we have “radial” convergence to a norm g, we need to patch together
convergence in every direction to a type of uniform convergence. Here we do this
under the assumption:
P(t ∈ [a,b]) = 1, where 0 < a < b < ∞.
e
This implies
T(0,x),g(x) ∈ [a(cid:107)x(cid:107) ,b(cid:107)x(cid:107) ] for x ∈ Zd.
1 1
So define the event
(cid:26) (cid:27)
T(0,nx)
Ω(cid:48) = {t ∈ [a,b] for all e}∩ lim = g(x) for all x ∈ Zd .
e
n n
(Exercise.) Show that on the above event we actually have
T(0,αx)
lim = g(x) for all x ∈ Zd.
α→∞ α
(Here α is real instead of just being an integer.)
Fix ω ∈ Ω(cid:48), a set which has probability one, for the rest of the argument. We will
show the equivalent statement
|T(0,x)−g(x)|
limsup = 0,
(cid:107)x(cid:107)
x→∞ 1
8
so suppose it fails: for some (cid:15) > 0, there is a sequence (x ) going to infinity such that
n
|T(0,x )−g(x )| ≥ (cid:15)(cid:107)x (cid:107) for all n.
n n n 1
We may assume by compactness that x /(cid:107)x (cid:107) → z for some z with (cid:107)z(cid:107) = 1. We will
n n 1 1
show that for some x in a nearby direction to z, one cannot have T(0,nx)/n → g(x).
Fix
(cid:18) (cid:19)
(cid:15)
δ ∈ 0,
4b+1
and choose x ∈ Zd such that
(cid:13) (cid:13)
(cid:13) x (cid:13)
(cid:13) −z(cid:13) < δ,
(cid:13)(cid:107)x(cid:107) (cid:13)
1 1
so that for n large,
(cid:13) (cid:13)
(cid:13) xn x (cid:13)
(cid:13) − (cid:13) < 2δ.
(cid:13)(cid:107)x (cid:107) (cid:107)x(cid:107) (cid:13)
n 1 1 1
We will compare the passage time from 0 to x to the passage time to the “nearby”
n
point (cid:107)x (cid:107) x/(cid:107)x(cid:107) . Then
n 1 1
(cid:12) (cid:18) (cid:19)(cid:12) (cid:12) (cid:18) (cid:19) (cid:18) (cid:19)(cid:12)
(cid:12) x (cid:12) (cid:12) x x (cid:12)
|T(0,xn)−g(xn)| ≤ (cid:12)T(0,xn)−T 0,(cid:107)xn(cid:107)1 (cid:12)+(cid:12)T 0,(cid:107)xn(cid:107)1 −g (cid:107)xn(cid:107)1 (cid:12)
(cid:12) (cid:107)x(cid:107) (cid:12) (cid:12) (cid:107)x(cid:107) (cid:107)x(cid:107) (cid:12)
1 1 1
(cid:12) (cid:18) (cid:19) (cid:12)
(cid:12) x (cid:12)
+(cid:12)g (cid:107)xn(cid:107)1 −g(xn)(cid:12).
(cid:12) (cid:107)x(cid:107) (cid:12)
1
The first and last terms are bounded by
(cid:13) (cid:13) (cid:13) (cid:13)
(cid:13) x (cid:13) (cid:13) x xn (cid:13)
b(cid:13)(cid:107)xn(cid:107)1 −xn(cid:13) = b(cid:107)xn(cid:107)1(cid:13) − (cid:13) < 2bδ(cid:107)xn(cid:107)1.
(cid:13) (cid:107)x(cid:107) (cid:13) (cid:13)(cid:107)x(cid:107) (cid:107)x (cid:107) (cid:13)
1 1 1 n 1 1
However since we have radial convergence in direction x, the above exercise gives
(cid:18) (cid:19) (cid:18) (cid:19)
x x
T 0,(cid:107)x (cid:107) = g (cid:107)x (cid:107) +o((cid:107)x (cid:107) ),
n 1 n 1 n 1
(cid:107)x(cid:107) (cid:107)x(cid:107)
1 1
so for n large, the second term is bounded by δ(cid:107)x (cid:107) . In total,
n 1
(cid:15)(cid:107)x (cid:107) ≤ |T(0,x )−g(x )| ≤ (4b+1)δ(cid:107)x (cid:107) < (cid:15)(cid:107)x (cid:107) ,
n 1 n n n 1 n 1
a contradiction.
Remarks.
1. Boivin [8] showed a shape theorem in the ergodic case (first done for bounded
marginals by Derriennic, reported in [24, (9.25)]). He assumed (a) the weight
distribution is ergodic under integer translations and (b) Etd+(cid:15) < ∞ for some
e
(cid:15) > 0. (Actually this moment condition is slightly weaker in his paper.)
9
2. The key to extending the above argument to the non-bounded cases (and ergodic
cases) is to see that the relevant property used here is that
(cid:32) (cid:33)
T(0,x)
P sup < ∞ > 0.
(cid:107)x(cid:107)
x(cid:54)=0 1
Our boundedness condition allowed us to say this supremum is at most b a.s., but
this weaker condition is enough. It is implied by Emin{t ,...,t }d < ∞ in the
1 2d
i.i.d. case or Etd+(cid:15) < ∞ in the ergodic case.
e
4 Newman and curvature
4.1 Newman’s theorems and conjectures
For this section, we will assume:
1. Eeαte < ∞ for some α > 0,
2. (t ) is i.i.d., continuous, so there is a unique geodesic G(x,y) from each x to each
e
y.
Our aim is to show that there are infinite geodesics with asymptotic directions.
Definition 4.1. We say that γ (with vertices x ,x ,x ,...) has asymptotic direction
0 1 2
θ (where θ ∈ Rd has Euclidean norm 1) if
x
n
argx := → θ.
n
(cid:107)x (cid:107)
n
Newman also makes the “uniform curvature” assumption. For illustration, we will
give a slightly more transparent condition, which is stronger, but has the same flavor.
Let us assume:
A. ∂B is differentiable (meaning each x ∈ ∂B has a unique supporting hyperplane –
a hyperplane H such that (a) x ∈ H and B does not intersect both components
x x
of Hc) and
x
B. there are uniform constants C,δ > 0 such that for all x ∈ ∂B,
g(x+u)−g(x) ≥ C(cid:107)u(cid:107)2 for u with x+u ∈ H and (cid:107)u(cid:107) < δ.
x
This assumption is a type of curvature of the boundary of B. Generally it is believed
that the limit shape has differentiable boundary and is strictly convex (even with
positive curvature). However it is not even known that it is not a polygon. For a
certain class of distributions in 2d (mentioned above in the remark about K = ∞),
Auffinger-Damron ’14 [4, Cor. 2.1] showed the limit shape is not a polygon.
Open problem. Show that if (t ) are i.i.d., continuous, and Emin{t ,...,t }d < ∞,
e 1 2d
then the limit shape cannot be a polygon.
The following is a version of [29, Theorem 2.1].
Theorem 4.2 (∼Newman). Assuming 1, 2, A, B,
10