Ergodic behaviour of "signed voter models"
aa r X i v : . [ m a t h . P R ] D ec Ergodic behaviour of “signed voter models”
E. Andjel G. Maillard T.S. Mountford Abstract
We consider some questions raised by the recent paper of Gantert, L¨owe and Steif(2005) concerning “signed” voter models on locally finite graphs. These are voter modellike processes with the difference that the edges are considered to be either positive ornegative. If an edge between a site x and a site y is negative (respectively positive) thesite y will contribute towards the flip rate of x if and only if the two current spin valuesare equal (respectively opposed). MSC
Key words and phrases.
Particle system, voter model, random walk, coupling.
Acknowledgment.
The research of EA was partially supported by the European ScienceFoundation programme: Phase Transitions and Fluctuation Phenomena. The researchof GM and TSM is partially supported by the SNSF, grants − − / CMI, Universit´e de Provence, 13453 Marseille cedex 13, France. [email protected] Institut de Math´ematiques, ´Ecole Polytechnique F´ed´erale, Station 8, 1015 Lausanne, Switzerland. gregory.maillard@epfl.ch, thomas.mountford@epfl.ch Introduction
This work arises from questions raised in the recent article by Gantert, L¨owe and Steif, [4].In this paper we consider voter model like processes called “signed” voter models. For sucha process we suppose given a locally finite graph G = ( V, E ) and a function s : E → {− , } .Our model ( η ( t ) : t ≥
0) will simply be a spin system on {− , } V with operatorΩ f ( η ) = X x ∈ V (cid:0) f ( η x ) − f ( η ) (cid:1) d ( x ) X y : { x,y }∈ E { η ( x ) η ( y ) = s ( { x, y } ) } . (1.1)Here the usual spins, 0 and 1 are replaced by − d ( x ) is the degree of vertex x and configuration η x is simply the elementof {− , } V with spins equal to those of η except at site x . From now on we will abusenotation and write s ( x, y ) for s ( { x, y } ); we will call this the sign of edge { x, y } . This can beseen as a generalization of the classical voter model (see e.g. [6], [1]) in that if the function s is identically 1 (or equivalently if all signs are positive) then the corresponding process is thevoter model. Definition 1.1
A nearest neighbour path ( γ ( s ) : 0 ≤ s ≤ t ) having finitely many jumps attimes ≤ t ≤ t ≤ · · · t n ≤ t is said to be even or positive if the number of ≤ i ≤ n sothat s ( γ ( t i − ) , γ ( t i )) = − is even. Otherwise the path is said to be odd or negative . If it ispositive we write sgn( γ ) = 1 otherwise sgn( γ ) = − . As with the voter model the easiest and most natural way to realize the voter model isvia a Harris construction: we introduce for each ordered pair ( x, y ) with an edge betweenthem a Poisson process, N x,y , of rate 1 /d ( x ) with all Poisson processes being independent.The process is built by stipulating that at times t ∈ N x,y , the spin at x becomes equalto s ( x, y ) η t ( y ). A.s. no two distinct Poisson processes have common points so the rule isunambiguous. It can easily be checked that with probability one this rule specifies η t ( x ) forall t and x just as in the classical voter model (see [1]). The Markovian nature is simplyinherited from that of the system of Poisson processes. It is then easily seen that this isindeed the desired process. As with the voter model, duality plays the dominant role inunderstanding the “signed” voter model. For fixed t ≥ x ∈ V we define the randomwalk on G , X x,t = ( X x,t ( s ) : 0 ≤ s ≤ t ) by the recipe: X x,t (0) = x , the random walk jumpsfrom y to z at time s ∈ [0 , t ] if immediately before time s it was at site y and t − s ∈ N y,z .As in [1], we recover η t ( x ) via the identity η t ( x ) = η (cid:0) X x,t ( t ) (cid:1) sgn (cid:0) X x,t (cid:1) . (1.2)It should be noted that for fixed t the random walks X x,t ( · ) and X y,t ( · ) are coalescing. If thetwo paths meet for the first time at s o ∈ [0 , t ], then irrespective of η we have η t ( x ) η t ( y ) = sgn (cid:0) γ x,y,s o (cid:1) , (1.3)where γ x,y,s o : [0 , s o ] → V is the concatenation of the path ( X x,t ( s ) : 0 ≤ s ≤ s o ) with thepath ( X y,t ( s o − s ) : 0 ≤ s ≤ s o ). For more discussion of the dual see the next section.As written above this article is written to address questions raised by [4]; it also followsfor instance the article of [9] which addresses signed voter models on the integer lattice wherethe signs are assigned to the edges in i.i.d. fashion. See [4] for a fuller bibliography.A major preoccupation of [4] was with unsatisfied cycles that are defined as follows.2 efinition 1.2 Unsatisfied cycles are nearest neighbour cycles in G whose sign is negative. Such cycles are important since in their absence the vertices can be divided into a “positive”set, V + and a “negative” set, V − so that the process ( η ′ t : t ≥
0) for η ′ t ( x ) = η t ( x ) for x ∈ V + , η ′ t ( x ) = − η t ( x ) for x ∈ V − is a classical voter model. Equally, the presence of unsatisfiedcycles precludes the existence of fixed configurations η for which the total flip rate is zero (see[4], Section 2 for details). For the classical voter model the configurations 1 of all 1s and − −
1s are fixed in this sense and so the voter model is never ergodic in the sense of [6],i.e., there exists a unique equilibrium µ and for every initial η , η t converges in distributionto µ as t tends to infinity. In the case of “signed” voter models ergodicity in this sense is areal possibility. A simple criterion for ergodicity was the existence of unsatisfied cycles andthe recurrence of the associated simple random walk, see Theorem 1.1 of [4]. The question ofwhether, for these processes, if there existed a unique equilibrium the process must necessarilybe ergodic was raised in [4]. In fact this holds and can be seen to be a consequence of Matloff’slemma (Lemma 3.1 in [7]), see also Lemma V.1.26 of [6]. Theorem 1.3
If the “signed” voter model has a unique equilibrium, then the “signed” votermodel is ergodic.
Another question we are fully able to resolve is the second open question listed in [4]:
Definition 1.4
For a path γ = ( γ ( s ) : 0 ≤ s ≤ t ) and ≤ s ≤ t ≤ t , γ s ,t signifies the path ( γ ( s ) : s ≤ s ≤ t ) . If s = 0 we write γ t instead of γ s ,t . For a path γ = ( γ ( s ) : s ≥ on V , we say that γ traverses infinitely many unsatisfied cycles if there exists sequences ( s i ) i ≥ and ( t i ) i ≥ tending to infinity so that γ s i ,t i are unsatisfied cycles. Theorem 1.5
For the graph Z with usual edge set and any sign assignation, s , either theprocess is not ergodic or a random walk must a.s. traverse infinitely many unsatisfied cycles. By Proposition 1.9 below the two statements in Theorem 1.5 are exclusive. The peculiarityof this result is highlighted by the next result
Theorem 1.6
For the graph Z d , d ≥ there are sign functions s on the edge set so that theassociated voter model is ergodic but the random walk must a.s. traverse only finitely manyunsatisfied cycles. The Theorem 1.1 of [4] shows that in dimensions 1 and 2, if there is an unsatisfied cyclethen necessarily the associated “signed” voter model is ergodic so the above results are in asense definitive. We finally consider another raised question ([4], question one). Proposition1.2 of this work gives a useful robust criterion for there to exist multiple equilibria for a signedvoter model: there exists a subset W ⊂ V that satisfies firstly that with positive probability arandom walk (starting from an appropriate site) will never leave W and secondly that W , withinherited edge set, has no unsatisfied cycles. The question raised was whether this criterionwas in fact necessary as well as sufficient. Proposition 1.7
In general if there are multiple equilibria, it does not follow that we canfind a region W ⊂ V on which the inherited graph has no unsatisfied cycle and for which therandom walk will with strictly positive probability never leave. Proposition 1.8
If the graph G = ( V, E ) is of bounded degree and the sign function is suchthat there are multiple equilibria, then we can find a region W ⊂ V on which the inherited graphhas no unsatisfied cycle and for which the random walk will with strictly positive probabilitynever leave. Finally in the last section we show
Proposition 1.9
If the random walk on G = ( V, E ) , ( X ( t ) : t ≥ satisfies with probability1, X ( · ) traverses infinitely many unsatisfied cycles then the signed voter model is ergodic. An important tool we will use is the fact that for two Markov chains on a state space S where the jump rates satisfy sup x ∈ S q ( x, x ) < ∞ , (1.4)has a “time shift” coupling. By this we mean that Lemma 1.10
Under condition (1.4), and given
T < ∞ and ǫ > , there exists a finite t so that for any s ∈ [0 , T ] and any x ∈ S , two realizations of the Markov chain starting at x , ( X ( t ) : t ≥ and ( X ′ ( t ) : t ≥ may be coupled so that with probability at least − ǫ (a) for all t ≥ t , X ( t ) = X ′ ( t + s ) and(b) the sequence of sites visited (allowing repeat visits) by the process X ( · ) up to time t isequal to that for X ′ ( · ) up to time t + s . (Remark in particular that sgn(( X ′ ) t + s ) = sgn( X t ) ∀ t ≥ t .)The rest of the paper is organized as follows: Sections 2, 3 and 4 are respectively devotedto the proofs of Theorem 1.3, 1.5 and 1.6, and Sections 5 and 6 to the proofs of Propositions1.7, 1.8 and 1.9. The following proof for Theorem 1.3 is really just a transcription of Lemma V.1.26 of [6]. Itis included for completeness. It rests on a property of the dual for the signed voter model,which we now describe in detail.We suppose, as usual, a given Harris system for generating signed voter models ( η t : t ≥
0) from a given initial configuration η . That is a collection of independent Poisson pro-cesses N x,y of rate d ( x ) for ordered neighbour pairs ( x, y ). Given an initial configuration η , a time t ≥
0, an integer r and r points in vertex set V , x , x , · · · , x r , the values of( η t ( x ) , η t ( x ) , · · · , η t ( x r )) are determined by the dual process X t ( u ) = (cid:16)(cid:0) X t, ( u ) , i t, ( u ) (cid:1) , (cid:0) X t, ( u ) , i t, ( u ) (cid:1) , · · · , (cid:0) X t,r ( u ) , i t,r ( u ) (cid:1)(cid:17) , (2.1)where X t,j ( u ) ∈ V , i t,j ( u ) ∈ {− , } for all u ∈ [0 , t ]. The process (piecewise constant)evolves as follows: X t ( · ) jumps at time u ∈ [0 , t ] if and only if there exists j ≤ r so that t − u ∈ N X t,j ( u − ) ,z for some z neighbouring X t,j ( u − ). This being the case4i) for every index k so that X t,k ( u − ) = X t,j ( u − ), there will be no change: X t,k ( u ) = X t,k ( u − ) and i t,k ( u ) = i t,k ( u − ),(ii) for every index k so that X t,k ( u − ) = X t,j ( u − ), we will have X t,k ( u ) = z and i t,k ( u ) = i t,k ( u − ) s ( X t,j ( u − ) , z ) with s defined as in Definition 1.1.Given this dual one recovers the values η t ( x k ) by η t ( x k ) = η (cid:0) X t,k ( t ) (cid:1) i t,k ( t ) . (2.2)The key point for the proof is that over the interval [0 , t ] the process X t will evolve as a Markovchain whose jump rates are bounded and which does not depend on t so that the couplingresult mentioned at the end of the introduction may be applied. That is given integer r < ∞ and ǫ >
0, uniformly over all x , x , · · · , x r there exists t so that (cid:13)(cid:13)(cid:13) X t ( t ) − X t + u ( t + u ) (cid:13)(cid:13)(cid:13) TV = (cid:13)(cid:13)(cid:13) X t + u ( t ) − X t + u ( t + u ) (cid:13)(cid:13)(cid:13) TV < ǫ (2.3)for all t ≥ t and u ∈ [0 , t ], where by abuse of notation we identify the random variables withtheir law.We may now turn directly to the proof of Theorem 1.3. We consider η as fixed. Itis sufficient to show that all limit points of the distribution of η t as t tends to infinity areequilibria. We suppose that for sequence { t n } n ≥ tending to infinity η t n → ν in law. (2.4)Let h be a cylinder function depending on, say, the spin values at x , x , · · · , x r , i.e., h ( η ) = g ( η ( x ) , η ( x ) , · · · , η ( x r )). We have that < ν, h > = lim n →∞ E η (cid:2) h ( η t n ) (cid:3) = lim n →∞ E (cid:2) h ′ (cid:0) η , X t n ( t n ) (cid:1)(cid:3) , (2.5)where by abuse of notation we have h ′ (cid:0) η , X t ( t ) (cid:1) = g (cid:16) η (cid:0) X t, ( t ) (cid:1) i t, ( t ) , η (cid:0) X t, ( t ) (cid:1) i t, ( t ) , · · · , η (cid:0) X t,r ( t ) (cid:1) i t,r ( t ) (cid:17) . (2.6)But equally for any fixed t we have (our signed voter model is easily seen to be a Feller process) < ν, P t h > = lim n →∞ E η (cid:2) P t h ( η t n ) (cid:3) , (2.7)where as usual ( P t ) t ≥ denotes the Markov semigroup of our signed voter model. The quantityinside the limit in the r.h.s. of (2.7) can be rewritten as E η [ h ( η t n + t )] which in the notationintroduced in (2.6) is equal to E (cid:2) h ′ (cid:0) η , X t n + t ( t n + t ) (cid:1)(cid:3) . (2.8)But, as already noted, as t n tends to infinity k X t n ( t n ) − X t n + t ( t n + t ) k TV tends to zero andso lim n →∞ (cid:16) E (cid:2) h ′ (cid:0) η , X t n + t ( t n + t ) (cid:1)(cid:3) − E (cid:2) h ′ (cid:0) η , X t n ( t n ) (cid:1)(cid:3)(cid:17) = 0 (2.9)which implies that < ν, h > = < ν, P t h > . By the arbitrariness of t and h we must concludethat measure ν is an equilibrium but, given our hypotheses that there is a unique equilibrium,we have established that any limit point ν must equal this equilibrium. That is we haveestablished ergodicity. 5 The integer lattice in three dimensions
In this section we consider the signed voter model on Z with simple random walk motion.We address the question of whether the existence of a single equilibrium implies that thesimple random walk must a.s. run infinitely many unsatisfied cycles. Given the possibility ofadapting the example of the preceding section to three dimensions we interpret the randomwalk “running infinitely many unsatisfied cycles” to mean: there exist s i , t i ↑ ∞ with s i < t i for all i ≥ B ( s i ) = B ( t i ) for all i ≥ B ( s ) : s i ≤ s ≤ t i ) := B s i ,t i is odd . (3.1)We do not require that the path B s i ,t i visits each site in the range exactly once, with theexception of B ( s i ) = B ( t i ).Our approach uses the following simple properties of simple random walks found in e.g.Lawler, [5].(A) There exists k ∈ (0 , ∞ ) so that for a random walk ( X ( t ) : t ≥
0) starting at X (0) = 0and any x ∈ ∂B (0 , n ) 1 kn d − ≤ P (cid:16) X (cid:0) T ∂B (0 ,n ) (cid:1) = x (cid:17) ≤ kn d − (3.2)(see [5], Lemma 1.7.4).(B) Harnack principle: for all α < k < ∞ so that1 k ≤ P z (cid:16) X (cid:0) T ∂B (0 ,n ) (cid:1) = x (cid:17) P (cid:16) X (cid:0) T ∂B (0 ,n ) (cid:1) = x (cid:17) ≤ k (3.3)uniformly over z ∈ B (0 , αn ) and n (see [5], Theorem 1.7.6.).Let C r = ∂B (0 , r ), the external boundary, and B r = B (0 , r ). Consider the quantity H ( z ) = ∞ X n =1 X x ∈ Cny ∈ Cn +1 P (cid:0) X ( T C n ) = x | X (0) = z (cid:1) P (cid:0) X ( T C n +1 ) = y | X (0) = x (cid:1) N x,yn , (3.4)where for all x ∈ C n and y ∈ C n +1 N x,yn = min n P x (cid:16) path X T Cn +1 is even (cid:12)(cid:12) X ( T C n +1 ) = y (cid:17) ,P x (cid:16) path X T Cn +1 is odd (cid:12)(cid:12) X ( T C n +1 ) = y (cid:17)o . (3.5)Then, by (3.2) and (3.3), the following are clear:(i) H ( · ) ≡ ∞ or H ( z ) < ∞ ∀ z ;(ii) H ( z ) < ∞ if and only if I < ∞ with I = ∞ X n =1 X x ∈ Cny ∈ Cn +1 n +2 N x,yn . (3.6)6urthermore,(iii) I = ∞ implies that for all random walks a.s. P (cid:16) path X T Cn is even (cid:12)(cid:12) X ( T C n ) (cid:17) → / Proposition 3.1 If I = ∞ then a.s. the random walk runs infinitely many unsatisfied cycles. Proposition 3.2 If I < ∞ then a.s. the signed voter model has multiple equilibria.Proof of Proposition 3.1 . If I = ∞ then one of X n = i mod 6 X x ∈ Cny ∈ Cn +1 n +2 N x,yn = ∞ (3.8)for i = 0 , , , , ,
5. Without loss of generality we suppose the first. The “mixing” propertiesof Brownian motion ensure that then a.s. X n =0 mod 6 N X ( T Cn ) ,X ( T Cn +1 ) n = ∞ (3.9)for any random walk ( X ( s ) : s ≥ D n as P (2 n , , (cid:16) X ′ hits X T Cn − ,T Cn − before C n +2 (cid:17) ≥ C (3.10)for X ′ an independent random walk, where C > n large P ( D n ) > / . (3.11)Define D ′ n the event X T Cn +1 ,T Cn +2 \ X T Cn − ,T Cn − = ∅ . (3.12)By (3.3) and (3.11), if D n occurs then P (cid:16) D ′ n (cid:12)(cid:12) F T Cn − (cid:17) > C ′ , (3.13)for some universal C ′ not depending on n , where {F t } t ≥ is the natural filtration for randomwalk X ( · ). Now (3.2) ensures that X n =0 mod 6 I D ′ n N X ( T Cn ) ,X ( T Cn +1 ) n = ∞ a.s. (3.14)under conditions given. We now introduce the discrete filtration J ′ n = F T C n +2 and G ′ n = σ (cid:16) J ′ n , X T C n +1 ,T C n +2 (cid:17) (3.15)7nd consider the filtration (over indices n = 0 mod 6) G ′ , J ′ , G ′ , · · · , G ′ n , J ′ n , G ′ n +1 , · · · . (3.16)Note that on D ′ n ∈ G ′ n we can define measurably t n ∈ [ T C n − , T C n − ], s n ∈ [ T C n +1 , T C n +2 ]so that X ( t n ) = X ( s n ). Note that P (cid:16) X t n ,s n is odd (cid:12)(cid:12) G ′ n (cid:17) ≥ N X ( T C n ) ,X ( T C n +1 )6 n (3.17)So by (3.14) and L´evy 0-1 law (see e.g. [2]) we have a.s. infinitely many unsatisfied cycles. Proof of Proposition 3.2 . As before, we denote by C r the external boundary of B (0 , r ), theEuclidean ball centered at the origin of radius 2 r . For x ∈ B r +1 = B (0 , r +1 ), v ∈ C r +1 ,the law P x,v,r is the law of the random walk started at x conditioned to exit B (0 , r +1 ) at v . Now for α < − α ≪ ≤ /
4) and for x ∈ C r there are twocomplementary sets: S ( x, r ) = (cid:8) v ∈ C r +1 : N x,vr < − α (cid:9) and U ( x, r ) = (cid:8) v ∈ C r +1 : N x,vr ≥ − α (cid:9) . (3.18)For v ∈ S ( x, r ) one can speak of a sign of v with respect to x : v is even or positive withrespect to x if P x,v,r (path from x to v is even) ≥ / v is odd or negative withrespect to x . If v is positive with respect to x at level r , we write sgn( x, v, r ) = 1. We writesgn( x, v, r ) = − v ∈ S ( x, r ) (for α = 3 /
4) but v is not positive with respect to x at level r . For v ∈ U ( x, r ) there is (at precision level 1 − α ) a reasonable chance of a path from x to v being either even or odd, we write sgn( x, v, r ) = 0. Therefore, for u ∈ C r and z ∈ C r +1 ,sgn( u, z, r ) = P u,z,r ( X is even) > / − P u,z,r ( X is odd) > / . (3.19)We first have Lemma 3.3
For a random walk ( X ( t ) : t ≥ on Z and for any α < , under condition I < ∞ a.s. X ( T C r +1 ) ∈ S ( X ( T C r ) , r ) (3.20) for all r sufficiently large. Lemma 3.4
For any x ∈ C r , w ∈ C r +1 with w ∈ S ( x, r ) , the P x,w,r probability that the path X ( · ) satisfies for all t ≤ T C r +1 sgn( X t ) sgn( X ( t ) , w, r ) = sgn( x, w, r ) (3.21) is at least − α ) .Proof . Suppose without loss of generality that sgn( x, w, r ) = 1. Then the P x,w,r probabilityof event A = n path X T Cr +1 is odd o (3.22)is less than 1 − α . Consider, with respect to the natural filtration, the c`adl`ag martingale M t = E (1 A | F t ). By Doob’s optional sampling theorem (see e.g. [2]) the probability that thisvalue ever gets above 1 / − α ). This gives the resultThe following is a simple consequence of (3.3).8 emma 3.5 There exists a universal c > so that for any x ∈ C r and w ∈ C r +1 , P x,w,r (cid:16) capacity (cid:0) X τ r ,σ r (cid:1) > r c (cid:17) > c , (3.23) where τ r = inf (cid:26) t : | X ( t ) | ≥ × r − (cid:27) and σ r = inf (cid:26) t > τ r : | X ( t ) | ≥ × r − or ≤ × r − (cid:27) . (3.24) Corollary 3.6
There exists strictly positive c so that for any x, y ∈ C r and w, v ∈ C r +1 , if X is a P x,w,r motion and X ′ is a P y,v,r motion, then with probability c the conditional probabilitygiven X ′ that X τ r ,σ r intersects ( X ′ ) τ ′ r ,σ ′ r is at least c , where τ r , σ r (resp. τ ′ r , σ ′ r ) are associatedto X (resp. X ′ ). Definition 3.7
We say { x, y, v, w } with x, y ∈ C r and v, w ∈ C r +1 are - compatible if sgn( x, v, r ) sgn( x, w, r ) sgn( y, v, r ) sgn( y, w, r ) = 1 . (3.25)In the following we assume that α has been fixed so large that 240 K (1 − α ) < c for c theconstant of Corollary 3.6 and K the constant defined in (3.29–3.30) below. Lemma 3.8
Suppose that { x, y, v, w } with x, y ∈ C n and v, w ∈ C n +1 are not -compatibleand that for each u ∈ { v, w } , N x,un < − α then for at least one u ∈ { v, w } , there existssome universal constant K > so that N y,un > c/ (128 K ) , where c is the constant defined inCorollary 3.6.Proof . We suppose without loss of generality that v and w are both positive with respect to x but that while w is positive with respect to y , v is not. By our assumption on the largenessof α we have by Lemma 3.4 and Corollary 3.6, that there exists a nearest neighbour path γ ( · )from x to w on which for all times s ,sgn( γ s ) sgn( γ ( s ) , w, r ) = 1 . (3.26)Furthermore for τ ′ n , σ ′ n defined for path γ , we have P z,u,n (cid:16) X τ n ,σ n hits γ τ ′ n ,σ ′ n (cid:17) > c, (3.27)for each ( z, u ) ∈ { ( x, v ) , ( x, w ) , ( y, v ) , ( y, w ) } . We consider two processes, ( Z x ( t ) : t ≥
0) and( Z y ( t ) : t ≥
0) starting respectively in x and y , running until C n +1 is hit and so that for u ∈ { x, y } the process ( Z u ( t ) : t ≥
0) has law 1 / P u,v,n + 1 / P u,w,n . Then we define themeasures µ u ( z ) by µ u ( { z } ) = P (cid:0) Z u ( T γ ) = z, τ ′ n < T γ < σ ′ n (cid:1) ∀ u ∈ { x, y } , z ∈ γ τ ′ n ,σ ′ n . (3.28)From facts (3.2–3.3), we have that there exists universal K so that1 K µ y ( { z } ) ≤ µ x ( { z } ) ≤ K µ y ( { z } ) ∀ z ∈ γ τ ′ n ,σ ′ n (3.29)9nd for either u , P (cid:0) Z u ( T C n +1 ) = v | Z u ( T γ ) = z, τ ′ n < T γ < σ ′ n (cid:1) ∈ (1 /K, − /K ) ∀ z ∈ γ τ ′ n ,σ ′ n . (3.30)We classify the points in γ of size between 5 × n − and 7 × n − into five sets: A ++ = n z : P z,w,n (cid:16) X T Cn +1 is even (cid:17) ≥ / , P z,v,n (cid:16) X T Cn +1 is even (cid:17) ≥ / o A + − = n z : P z,w,n (cid:16) X T Cn +1 is even (cid:17) ≥ / , P z,v,n (cid:16) X T Cn +1 is odd (cid:17) ≥ / o A − + = n z : P z,w,n (cid:16) X T Cn +1 is odd (cid:17) ≥ / , P z,v,n (cid:16) X T Cn +1 is even (cid:17) ≥ / o A −− = n z : P z,w,n (cid:16) X T Cn +1 is odd (cid:17) ≥ / , P z,v,n (cid:16) X T Cn +1 is odd (cid:17) ≥ / o D = n z : P z,v,n (cid:16) X T Cn +1 is odd (cid:17) ∈ (1 / , / o . (3.31)We have by the optimal stopping time reasoning of proof of Lemma 3.4 and our assumptionson x and v that µ x ( D ) < − α ) . (3.32)By (3.29), this implies that µ y ( D ) < − α ) K. (3.33)We claim that µ x ( A + − ) < ǫ = 40 K (1 − α ) . (3.34)To see this suppose the contrary, then we must have either P (cid:16) Z x ( T γ ) ∈ A + − , τ ′ n < T γ < σ ′ n , ( Z x ) T γ is even (cid:17) ≥ ǫ/ P (cid:16) Z x ( T γ ) ∈ A + − , τ ′ n < T γ < σ ′ n , ( Z x ) T γ is odd (cid:17) ≥ ǫ/ . (3.36)In the former case we have via our choice of KP x,v,n (cid:16) X ( T γ ) ∈ A + − , τ ′ n < T γ < σ ′ n , X T γ is even (cid:17) ≥ ǫ/ (2 K ) (3.37)and so by the Markov property P x,v,n (cid:16) X T Cn +1 is odd (cid:17) ≥ ǫ/ (4 K ) > − α, (3.38)which contradicts our hypothesis on x and v . Similarly in the other case we are forced toconclude that P x,v,n ( X T Cn +1 is odd) > − α . Arguing similarly with set A + − replaced by A − + , we are able to deduce that µ x ( A − + ) < ǫ. (3.39)The Harnack principle (see (3.3)) now permits us to conclude that µ y ( A − + ∪ A + − ) < Kǫ .We thus conclude that either µ y ( A ++ ) ≥ ( c − Kǫ ) / µ y ( A −− ) ≥ ( c − Kǫ ) /
2. Withoutloss of generality we suppose the former. Note that our assumptions on the closeness of α to1 ensures that ( c − Kǫ ) / > c/
3. Then for identical reasons, either P (cid:16) Z y ( T γ ) ∈ A ++ , τ ′ n < T γ < σ ′ n , ( Z x ) T γ is even (cid:17) ≥ ( c − Kǫ ) / > c/ P (cid:16) Z y ( T γ ) ∈ A ++ , τ ′ n < T γ < σ ′ n , ( Z x ) T γ is even (cid:17) ≥ ( c − Kǫ ) / > c/ . (3.41)Again without loss of generality we suppose the former. In this case we have P y,w (cid:16) X ( T γ ) ∈ A ++ , τ ′ n < T γ < σ ′ n , X T γ is even (cid:17) ≥ c/ (12 K ) (3.42)and so P y,w (cid:16) X T γ is even (cid:17) ≥ c/ (32 K ) . (3.43)Consider two independent random walks X ( t ) and Y ( t ) then for any α < r sufficiently large N X ( T Cr ) ,X ( T Cr +1 ) r < − α and N Y ( T Cr ) ,Y ( T Cr +1 ) r < − α, (3.44)that is { X ( T XC r ) , Y ( T YC r ) , X ( T XC r +1 ) , Y ( T YC r +1 ) } are 1-compatible. Definition 3.9
We say { x, y, z, w } with x ∈ C r − , y, z ∈ C r and w ∈ C r +1 are - compatible if sgn( x, y, r −
1) sgn( x, z, r −
1) sgn( y, w, r ) sgn( z, w, r ) = 1 . (3.45) Lemma 3.10
Under the hypothesis that
I < ∞ , for any two independent random walks ( X ( t ) : t ≥ and ( Y ( t ) : t ≥ with probability one { X ( T C r − ) , X ( T C r ) , Y ( T C r ) , X ( T C r +1 ) } are -compatible for all r large. Here, as before T C r as an argument denotes the stopping time appropriate to the process. Proof . We will show that with probability one { X ( T C r − ) , X ( T C r ) , Y ( T C r ) , X ( T C r +1 ) } are 2-compatible for all r large and even. The proof for r odd is entirely analogous. We first observethat under the condition I < ∞ , we have a.s. X r even N X ( T Cr − ) ,X ( T Cr +1 ) r + < ∞ , (3.46)where for all x ∈ C r − and y ∈ C r +1 N x,yr + = min n P x (cid:16) path X T Cr +1 is even (cid:12)(cid:12) X ( T C r +1 ) = y (cid:17) ,P x (cid:16) path X T Cr +1 is odd (cid:12)(cid:12) X ( T C r +1 ) = y (cid:17)o . (3.47)Given (3.2), we have easily that there exists universal constant K so that the probability that N X ( T Cr − ) ,vr − or N v,X ( T Cr +1 ) r > /
100 (3.48)or sgn( X ( T C r − ) , v, r −
1) sgn( v, X ( T C r +1 ) , r ) = sgn( X ( T C r − ) , X ( T C r +1 ) , r +) (3.49)is bounded by KN X ( T Cr − ) ,X ( T Cr +1 ) r + , where sgn( X ( T C r − ) , X ( T C r +1 ) , r +) is given its obviousmeaning. The result now follows from (3.2) again and L´evy’s 0-1 law.11iven Lemmas 3.8 and 3.10 we can find a path realization X = X ( t, ω ) so that for a.s.every random walk path Y = Y ( t, ω ) the conclusion of the lemmas hold (here we use thenotation X ( t, ω ) to underline the fact that we consider a fixed path of the random walk( X ( s ) : s ≥
0) at time t ). That is let us pick and fix a “good” path X so that for a.s.path Y we have that for r large, { X ( T C r − ) , X ( T C r ) , Y ( T C r ) , X ( T C r +1 ) } are 2-compatible and { X ( T C r ) , Y ( T C r ) , Y ( T C r +1 ) , X ( T C r +1 ) } are 1-compatible and also such that for any α < N X ( T Cr ) ,X ( T Cr +1 ) r < − α . We will use this path to designate sites in C r as positiveof negative: we say that X C is a positive site, subsequently if P X ( T Cr − ) ,X ( T Cr ) ,r (cid:16) X T r is odd (cid:17) ≤ / , (3.50)then X ( T C r ) has the same sign as X ( T C r − ). Given this assignation we now assign signs toarbitrary y ∈ C r . If P X ( T Cr − ) ,y,r (cid:16) X T r is odd (cid:17) ≤ / , (3.51)then y ∈ C r has the same sign as X ( T C r − ), otherwise it is the opposite. Lemma 3.11
With probability one there exists a finite random r so that either ∀ r ≥ r , sgn (cid:16) Y T Cr (cid:17) sgn (cid:0) Y ( T C r ) (cid:1) = 1 (3.52) or ∀ r ≥ r , sgn (cid:16) Y T Cr (cid:17) sgn (cid:0) Y ( T C r ) (cid:1) = − . (3.53) Proof . We first observe that for r large enough all the terms N X ( T Cr − ) ,X ( T Cr ) r − , N X ( T Cr − ) ,Y ( T Cr ) r − are less than, say, 1 / r large, 1- and 2-compatibility givesgn (cid:16) X ( T C r − ) , X ( T C r ) , r − (cid:17) sgn (cid:16) X ( T C r − ) , Y ( T C r ) , r − (cid:17) × sgn (cid:16) X ( T C r ) , X ( T C r +1 ) , r (cid:17) sgn (cid:16) Y ( T C r ) , X ( T C r +1 ) , r (cid:17) = 1 (3.54)and sgn (cid:16) X ( T C r ) , Y ( T C r +1 ) , r (cid:17) sgn (cid:16) Y ( T C r ) , Y ( T C r +1 ) , r (cid:17) × sgn (cid:16) X ( T C r ) , X ( T C r +1 ) , r (cid:17) sgn (cid:16) Y ( T C r ) , X ( T C r +1 ) , r (cid:17) = 1 . (3.55)Therefore their productsgn (cid:16) X ( T C r − ) , X ( T C r ) , r − (cid:17) sgn (cid:16) X ( T C r − ) , Y ( T C r ) , r − (cid:17) × sgn (cid:16) X ( T C r ) , Y ( T C r +1 ) , r (cid:17) sgn (cid:16) Y ( T C r ) , Y ( T C r +1 ) , r (cid:17) = 1 . (3.56)12sing our assumptions, we havesgn (cid:0) Y ( T C r +1 ) (cid:1) = sgn (cid:0) X ( T C r ) (cid:1) sgn (cid:0) X ( T C r ) , Y ( T C r +1 ) , r (cid:1) = sgn (cid:0) X ( T C r − ) (cid:1) sgn (cid:0) X ( T C r − ) , X ( T C r ) , r − (cid:1) sgn (cid:0) X ( T C r ) , Y ( T C r +1 ) , r (cid:1) = sgn (cid:0) X ( T C r − ) (cid:1) sgn (cid:0) X ( T C r − ) , X ( T C r ) , r − (cid:1) sgn (cid:0) X ( T C r ) , Y ( T C r +1 ) , r (cid:1) × sgn (cid:0) X ( T C r − ) , Y ( T C r ) , r − (cid:1) = sgn (cid:0) X ( T C r − ) (cid:1) sgn (cid:0) X ( T C r − ) , Y ( T C r ) , r − (cid:1) sgn (cid:0) X ( T C r − ) , X ( T C r ) , r − (cid:1) × sgn (cid:0) X ( T C r ) , Y ( T C r +1 ) , r (cid:1) sgn (cid:0) X ( T C r − ) , Y ( T C r ) , r − (cid:1) = sgn (cid:0) Y ( T C r ) (cid:1) sgn (cid:0) X ( T C r − ) , X ( T C r ) , r − (cid:1) sgn (cid:0) X ( T C r ) , Y ( T C r +1 ) , r (cid:1) × sgn (cid:0) X ( T C r − ) , Y ( T C r ) , r − (cid:1) . (3.57)Therefore, combining (3.56) and (3.57), we get for all r largesgn( Y ( T C r +1 )) = sgn( Y ( T C r )) sgn( Y ( T C r ) , Y ( T C r +1 ) , r ) . (3.58)Now, conditional upon to Y ( T C r ) , Y ( T C r +1 ), the probability thatsgn (cid:0) Y T Cr (cid:1) sgn (cid:0) Y ( T C r ) (cid:1) = sgn (cid:0) Y T Cr (cid:1) sgn (cid:0) Y ( T C r +1 ) (cid:1) (3.59)is simply N X ( T Cr ) ,X ( T Cr +1 ) r . Hence the result follows by Lemma 3.3.Define the function h ( x ) = P x (cid:16) for all r large sgn (cid:16) Y T Cr (cid:17) sgn (cid:0) Y ( T C r ) (cid:1) = 1 (cid:17) − P x (cid:16) for all r large sgn (cid:16) Y T Cr (cid:17) sgn (cid:0) Y ( T C r ) (cid:1) = − (cid:17) (3.60)and the product measures µ ± by µ + ( { η : η ( x ) = 1 } ) = (1 + h ( x )) / µ − ( { η : η ( x ) = 1 } ) =(1 − h ( x )) /
2. We have by L´evy’s 0-1 law and the Markov property that with probability 1lim t →∞ | h ( Y ( t )) | exists and equals 1. So there exists x ∈ Z for which | h ( x ) | is arbitrarilyclose to 1 and in particular for which h ( x ) = 0. But in this case we have for all t by dualityand the Markov property that P t µ ± (cid:0) { η : η ( x ) = 1 } (cid:1) = (1 ± h ( x )) / . (3.61)Then using a similar argument as in [4] (Section 7, Proof of Proposition 1.2), this impliesnon-uniqueness of equilibria. We show Theorem 1.6 in this section. For notational convenience we give the proof for fourdimensions but the proof is easily seen to hold in all dimensions.Our purpose is to choose a sequence of integer scales R n so that R n +1 /R n tends to infinitysufficiently rapidly. Then we will give sign +1 to all edges except those of the form ( x, x + e )for e = (1 , , ,
0) and x ∈ { R n } × [ − R n , R n ] . The basic idea is to consider a random walkstarting at a site in [ − R n / , R n / , say, and run until it hits ∂ [ − R n , R n ] . Even given theinitial and final points uncertainty as to the sign of the random walk will be introduced.13n the first part of this section we argue from invariance principle considerations that if R n +1 /R n ≥ n + 1) /K n +2 for constants ( K n ) n ≥ small then almost surely a random walkdoes not run though infinitely many unsatisfied cycles. Then we argue that if we increase therequirement to R n +1 /R n ≥ n + 1) /K n +2 , (4.1)then we will have ergodicty.We now undertake the first part of the program. Consider a Brownian motion in 4 di-mensions, ( B ( t ) : t ≥ V ir , r ≥ i = 3 , − r, r ] i and given a process( Y ( t ) : t ≥ T n = inf { t : Y ( t ) leaves V n } . It follows from the a.s. nonexistence of doublepoints for 4-dim Brownian motion (see [3]) (and the fact that two dimensional subspaces of ∂V are polar) that, with probability 1, there does not exist t , t ≤ T n so that t < t and( B ( t ) , B ( t ))or( B ( t ) , B ( t )) ∈ (cid:0) { } × V (cid:1) × (cid:0) ∂V \ ( { } × V ) (cid:1) (4.2)and B ( t ) = B ( t ) for t ≤ t ≤ t ≤ t . (4.3)Bearing in mind that B does not hit the intersections of the faces of ∂V , there exists K n > − / n K n ≤ inf | B ( t ) − B ( t ) | (4.4)for t , t , t , t as above. Now (possibly reducing K n ) we can also have that this is so forBrownian motion starting in V K n uniformly over the initial point. Now let us inductivelydefine R n as follows: R is such that for a 4-dimensional random walk (starting at 0) X , theprobability that(i) there exist t < t ≤ T R (recall that T R is the leaving time of V R ) so that( X ( t ) , X ( t ))or( X ( t ) , X ( t )) ∈ (cid:16) { R } × V R (cid:17) × (cid:16) ∂V R \ (cid:0) { R } × V R (cid:1)(cid:17) ;(ii) there exists t ≤ t ≤ t ≤ t so that t ≤ T R and | X ( t ) − X ( t ) | ≤ K R is less that ≤ /
4. Such an R exists by the invariance principle, see e.g. [2]. Now, given R j − take R j ≥ j R j − /K j +1 , so that for any random walk X ( · ) starting in V K j +1 R j , theprobability that(i) there exists t < t < T j +1) R j so that( X ( t ) , X ( t ))or( X ( t ) , X ( t )) ∈ (cid:16) { R j } × V R j (cid:17) × (cid:16) ∂V R j \ (cid:0) { R j } × V R j (cid:1)(cid:17) ;(ii) there exists t ≤ t ≤ t ≤ t so that t ≤ T j +1) R j and | X ( t ) − X ( t ) | ≤ K j +1 R j
14s bounded by 1 / j + 1) . Now take the configuration of ± Z as follows: allbonds are +1 except bonds ( x, x + e ) for x ∈ { R j } × V R j . (4.5)Then by Borel-Cantelli there exists j < ∞ such that for all j ≥ j (i) If we consider the random walk between hitting V R j until hitting V jR j there is no t < t so that ( X ( t ) , X ( t ))or( X ( t ) , X ( t )) ∈ (cid:16) { R j } × V R j (cid:17) × (cid:16) ∂V R j \ (cid:0) { R j } × V R j (cid:1)(cid:17) ;there exists t ≤ t ≤ t ≤ t such that X ( t ) = X ( t ).(ii) The random walk does not return to V R j after hitting V jR j .This easily implies that X does not run thought infinitely many unsatisfied cycles.It is easily seen that the Harnack principle (property (B) in section 3) yields: Lemma 4.1
Let π r ( w, · ) be the harmonic measure for a random walk starting at w , at theboundary of the ball B (0 , r ) . Then lim m →∞ lim sup r →∞ sup x,y ∈ B (0 ,r ) z ∈ ∂B (0 ,mr ) π mr ( x, z ) π mr ( y, z ) = 1 . (4.6)Let ( R n ) n ≥ satisfying (4.1) and consider C n = ∂B (0 , n ) and S n = nR n . Lemma 4.2
There exists k ∈ (0 , / so that for all n large enough and all x ∈ C S n , y ∈ C S n +1 P x (cid:16) X T CSn +1 is odd (cid:12)(cid:12) X (cid:0) T C Sn +1 (cid:1) = y (cid:17) > k (4.7) and P x (cid:16) X T CSn +1 is even (cid:12)(cid:12) X (cid:0) T C Sn +1 (cid:1) = y (cid:17) > k . (4.8) Proof . By the invariance principle we have that if n is large, uniformly for each x ∈ C S n the probability of leaving the box V R n +1 for the first time through { R n +1 } × V R n +1 / , thenpassing to ∂V R n +1 without leaving [2 R n +1 / , + ∞ ) × V R n +1 / is greater than k ∈ (0 , k . From here, uniformly over the random hitting point of ∂V R n +1 , theconditional probability of hitting ∂B (0 , ( n + 1) R n +1 /
2) before hitting V R n +1 will be greaterthan k ∈ (0 ,
1) provided n is large. This follows from the invariance principle and the classicalhitting estimates of Lawler (see properties (A) and (B) of Section 3). From property (A) ofSection 3 we have the existence of a constant k so that1 k S n +1 ≤ P w (cid:16) X (cid:0) T C Sn +1 (cid:1) = z (cid:17) ≤ k S n +1 . (4.9)So using P w (cid:16) X (cid:0) T C Sn +1 (cid:1) = z, T V Rn +1 > T C Sn +1 (cid:17) ≥ P w (cid:16) X (cid:0) T C Sn +1 (cid:1) = z (cid:17) − sup z ∈ V Rn +1 P z (cid:16) X (cid:0) T C Sn +1 (cid:1) = z (cid:17) P w (cid:16) T C Sn +1 > T V Rn +1 (cid:17) , (4.10)15e obtain P w (cid:16) X (cid:0) T C Sn +1 (cid:1) = z, T V Rn +1 > T C Sn +1 (cid:17) ≥ k S n +1 (4.11)for n large uniformly over w ∈ ∂B (0 , ( n + 1) R n +1 / x ∈ C S n and y ∈ C S n +1 P x (cid:16) X T CSn +1 is odd (cid:12)(cid:12) X (cid:0) T C Sn +1 (cid:1) = y (cid:17) ≥ k k k S n +1 . (4.12)This given, P x (cid:16) X (cid:0) T C Sn +1 (cid:1) = y (cid:17) ≤ k S n +1 (4.13)gives P x (cid:16) X T CSn +1 is odd (cid:12)(cid:12) X (cid:0) T C Sn +1 (cid:1) = y (cid:17) ≥ k k k . (4.14)We argue similarly for the second part.The following is a simple consequence of Lemmas 4.1 and 4.2. Corollary 4.3
There exists N so that for all r ∈ Z + and all x ∈ C S N , y ∈ C S N r (cid:12)(cid:12)(cid:12)(cid:12) P x (cid:18) X T CSN r is odd (cid:12)(cid:12)(cid:12) X (cid:16) T C SN r (cid:17) = y (cid:19) − (cid:12)(cid:12)(cid:12)(cid:12) ≤ (1 − k ) r (4.15)We are now ready to complete the proof of Theorem 1.6. Proposition 4.4
For R n and sign functions as previously described, the signed voter modelis ergodic.Proof . We need only show that as t tends to infinity the difference in absolute variation of themeasures µ x,t, − and µ x,t, + tends to zero for each x ∈ Z , where µ x,t, ± is defined on Z by µ x,t, + ( y ) = P x (cid:0) X ( t ) = y, X t is even (cid:1) and µ x,t, − ( y ) = P x (cid:0) X ( t ) = y, X t is odd (cid:1) . (4.16)However we have by our basic coupling that for any y ∈ Z and any T ≥ t →∞ sup s ∈ [0 ,T ] (cid:16)(cid:13)(cid:13) µ y,t, + − µ y,t − s, + (cid:13)(cid:13) TV + (cid:13)(cid:13) µ y,t, − − µ y,t − s, − (cid:13)(cid:13) TV (cid:17) = 0 (4.17)(see Lemma 6.1 for a statement and a proof in a more general setting). We consider x ∈ Z , r ≥ T < t fixed. Let ν r ( ds, y ) be the joint law (under P x ) of ( T C Sr , X ( T C Sr )). Then,by stong Markov property, µ x,t, + = Z T X y ∈ C Sr (cid:20) P x (cid:16) X T CSr is odd (cid:12)(cid:12) T C Sr = s, X (cid:0) T C Sr (cid:1) = y (cid:17) µ y,t − s, − + P x (cid:16) X T CSr is even (cid:12)(cid:12) T C Sr = s, X (cid:0) T C Sr (cid:1) = y (cid:17) µ y,t − s, + (cid:21) ν r ( ds, y ) + µ rx,t, + , (4.18)where µ rx,t, + ( z ) = P x (cid:16) X ( t ) = y, T C Sr > T, X t is even (cid:17) , (4.19)16nd similarly for µ rx,t, − . But as t → ∞ sup y ∈ C Sr +1 sup ≤ s ≤ T (cid:13)(cid:13) µ y,t − s, − − µ y,t, − (cid:13)(cid:13) TV −→ µ x,t − s, + . Thus as t → ∞ µ x,t, + = Z T X y ∈ C Sr (cid:20) P x (cid:16) X T CSr is odd , T C Sr ≤ T (cid:12)(cid:12) X (cid:0) T C Sr (cid:1) = y (cid:17) µ y,t, − + P x (cid:16) X T CSr is even , T C Sr ≤ T (cid:12)(cid:12) X (cid:0) T C Sr (cid:1) = y (cid:17) µ y,t, + (cid:21) ν r ( ds, y ) + µ rx,t, + + o (1)(4.21)and similarly for µ x,t, − . Thus (cid:13)(cid:13) µ x,t, + − µ x,t, − (cid:13)(cid:13) TV = X y ∈ C Sr (cid:20) P x (cid:16) X T CSr is odd , T C Sr ≤ T (cid:12)(cid:12) X (cid:0) T C Sr (cid:1) = y (cid:17) µ y,t, − + P x (cid:16) X T CSr is even , T C Sr ≤ T (cid:12)(cid:12) X (cid:0) T C Sr (cid:1) = y (cid:17) µ y,t, + (cid:21) ν r ([0 , T ] , y )+ (cid:13)(cid:13) µ Tx,t, + − µ Tx,t, − (cid:13)(cid:13) TV + o (1) . (4.22)Now let t → ∞ and then T → ∞ to getlim T →∞ lim t →∞ (cid:13)(cid:13) µ x,t, + − µ x,t, − (cid:13)(cid:13) TV = X y ∈ C Sr (cid:12)(cid:12)(cid:12)(cid:12) P x (cid:16) X T CSr is odd (cid:12)(cid:12) X (cid:0) T C Sr (cid:1) = y (cid:17) − P x (cid:16) X T CSr is even (cid:12)(cid:12) X (cid:0) T C Sr (cid:1) = y (cid:17)(cid:12)(cid:12)(cid:12)(cid:12) . (4.23)Now letting r → ∞ and using Corollary 4.3, we obtain desired the result. Gantert et al. [4] ask whether the converse of Proposition 1.2 of their article held. This statedthat if the graph G = ( V, E ) had the property that there existed W ⊂ V, x ∈ V so that(i) P x ( T W = ∞ ) > T W = inf { t : X ( t ) ∈ W c } and(ii) W with its inherited edge set contained no unsatisfied cycles,then “necessarily” the signed voter model could not be ergodic. The question was raised atthe end of the paper as to whether a converse existed: can it be that whenever a signed votermodel is non ergodic such a W can be found? We first show this is not the case, but thenshow that with the additional hypothesis that the graph is of bounded degree, it is indeedtrue. We first state without proof (it follows from [4], Proposition 1.2). Proposition 5.1
If a.s. for all random walk X = ( X ( t ) : t ≥ on the graph G , there existsrandom T so that on [ T, ∞ ) , X does not traverse a negative edge then the signed voter modelhas multiple equilibria.
17e will build our counterexample out of a rooted tree with only positive edges by adding anumber of negative edges whose density is so small that the property of multiple equilibria isunchanged. Consider a rooted tree so that each i th generation has n i “children” where n i → ∞ as i → ∞ and is always even. We now amend T as follows. We pick strictly increasing V n ↑ ∞ so that n V n ≥ n . At the V th n generation we pair up the vertices so that each vertex of the V th n generation is paired with a member having the same father. We add the corresponding edges.For the resulting graph all original edges are fixed positive and the extra “within generation”edges negative. Though this new graph has cycles, we retain use of the words descendantsinherited from the original rooted tree. By the Borel-Cantelli lemma and Proposition 5.1, thesigned voter model has multiple equilibria. Let W be a subject of V with the property that,with initial point suitably chosen, the probability of a random walk on G ever leaving W isstrictly positive. Then by L´evy’s 0-1 law (see e.g. [2]), on the event that the random walk( X ( t ) : t ≥
0) never leaves W we must have with probability tending to 1 as t tends to infinity X (cid:0) T r ( t ) (cid:1) is a descendant of X ( t ) (5.1)and both X (cid:0) T r ( t ) (cid:1) and its pair belong to W, (5.2)where r ( t ) is the next V n level below the current level of X ( t ) and T r ( t ) is the hitting timeof this generation. But this must mean that with probability tending to one as t tends toinfinity, the cycle of length 3 involving the point X ( T r ( t ) ), its pair and their (common) fatheris unsatisfied.This counterexample is somewhat cheap, the “real” question is whether the converse toProposition 1.2 holds for graphs of bounded degree. We now show Proposition 1.8.In the following let M = sup x ∈ V d ( x ). We first consider that there exists an equilibrium µ so that for some x ∈ V , µ ( { η : η ( x ) = 1 } ) = 1 /
2. Let α = sup x ∈ V (cid:12)(cid:12) µ ( { η : η ( x ) = 1 } ) − µ ( { η : η ( x ) = − } ) (cid:12)(cid:12) . (5.3)Without loss of generality we have α = sup x ∈ V (cid:0) µ ( { η : η ( x ) = 1 } ) − µ ( { η : η ( x ) = − } ) (cid:1) . (5.4)Now we have (see e.g. [6] or [7]) for any x ∈ V and t ≥ h ( x ) := µ ( { η : η ( x ) = 1 } ) − µ ( { η : η ( x ) = − } ) = E x (cid:2) h ( X ( t )) sgn( X t ) (cid:3) . (5.5)Now fix ǫ > ǫ ≪ x ∈ { y : h ( y ) > α − ǫ } . For 0 ≤ t ≤ T , where T is fixed, let M t = E (cid:2) η T ( x ) = 1 | F t (cid:3) − E (cid:2) η T ( x ) = − | F t (cid:3) = h (cid:16) X x,Tt (cid:17) sgn (cid:16)(cid:0) X x,T (cid:1) t (cid:17) . (5.6)(Here F t = Harris system on interval [ T − t, T ].) Note that | M t | ≤ α for all 0 ≤ t ≤ T . Let σ = inf { t ≥ | M t | ≤ α − ǫ } then by optional sampling theorem we have α − ǫ < h ( x ) = E [ M σ ∧ T ] ≤ ( α − ǫ ) P ( σ ≤ T ) + αP ( σ > T ) , (5.7)from which we deduce P ( σ ≤ T ) ≤ /
10. Now this (and the arbitrariness of T ) implies thatif W is the component of { y : | h ( y ) | ≥ α − ǫ } containing x , then P x ( T W = ∞ ) ≥ /
10. We18ow show that, provided ǫ is sufficiently small W has no unsatisfied cycles: Suppose not andlet x , x , · · · , x r be an unsatisfied cycle in W . The point is that for all i ∈ { , · · · , r } h ( x i ) = X y ∼ x i h ( y ) s ( x i , y ) d ( x i ) , (5.8)thus h ( x i ) = h ( x i +1 ) M s ( x i , x i +1 ) + R i (cid:18) M − M (cid:19) , (5.9)where | R i | ≤ α . From which we have for h ( x i ) > α − ǫ ) ≤ h ( x i +1 ) s ( x i , x i +1 ) M + M − M α. (5.10)That is h ( x i +1 ) s ( x i , x i +1 ) ≥ α − M ǫ > ǫ is sufficiently small. Similarly if h ( x i ) < h ( x i +1 ) s ( x i , x i +1 ) < − ( α − M ǫ ) < ǫ sufficiently small. This gives a contradiction.In the following a signed random walk (on graph G ) shall be a process (( X ( t ) , i ( t )) : t ≥ X ( t ) : t ≥
0) is a random walk and the process ( i ( t ) : t ≥
0) takes values on {− , } ,starts at value 1 and only changes when X ( · ) changes. At a jump time t for X ( · ), we have i ( t ) = i ( t − ) s ( X ( t − ) , X ( t )). We now suppose that there exists multiple equilibria but thateach equilibria has h ( x ) = µ ( { η : η ( x ) = 1 } ) − µ ( { η : η ( x ) = − } ) ≡ , (5.11)that is for all x ∈ V , µ ( { η : η ( x ) = 1 } ) = 1 /
2. Let us denote by µ the canonical equilibriumwhere under µ the spins η ( x ) , · · · , η ( x n ) can be obtained byA) running coalescing signed random walks ( X x j ( · ) , i j ( · )), 1 ≤ j ≤ n for “ ∞ ” to obtaincoalesced classes C = n x i (1) , · · · , x i ( r ) o , C = n x i (1) , · · · , x i ( r ) o , · · · , C n = n x i n (1) , · · · , x i n ( r n ) o . (5.12)B) assigning signs to x , · · · , x n so that η ( x i ) must be compatible with η ( x j ) if x i , x j belongto same cluster but are independent and equiprobable if they belong to distinct clusters.Note: automatically we have µ = µ if two independent random walks on G = ( V, E ) mustalmost surely meet.Thus, summarizing the foregoing, it will be enough to show that if every equilibrium µ satisfies µ ( { η : η ( x ) = 1 } ) = 1 / x ∈ V , then there is a unique equilibrium, the canonicalmeasure µ . Thus we consider the evolution of the dual (( X ,t ( · ) , i ,t ( · )) , · · · , ( X n,t ( · ) , i n,t ( · )))for x , · · · , x n fixed but t variable (and ultimately tending to ∞ ). For any fixed n andcoalescing random walks X i ( · ) starting at x i ∈ V for i ∈ { , · · · , n } let event A ( T ) be definedby A ( T ) = n ∃ s > T, j, k ∈ { , · · · , n } : X j ( T ) = X k ( T ) but X j ( s ) = X k ( s ) o . (5.13)We then have for all ǫ >
0, there exists T so that P x , ··· ,x n ( A ( T )) < ǫ . From this we see that toshow our result it is sufficient to show for all r and for all sequence of r -tuples ( y n , y n , · · · , y nr )so that lim n →∞ P y n , ··· y nr ( A (0)) = 0, we have for all ( d , · · · , d r ) ∈ {− , +1 } r lim n →∞ µ (cid:16)n η ( y nj ) = d j , j = 1 , · · · , r o(cid:17) = 2 − r . (5.14)19emark that even though we have supposed that with positive probability two independentrandom walks may avoid each other for ever, nothing prevents the existence of an integer r so that for all distinct y , y , · · · , y r , P y ,y , ··· ,y r ( A (0)) = 0.To make our claim we will argue by induction. The result for r = 1 is simply our hy-pothesis on the equilibria of our signed voter model. Suppose now that the result holds for r − r -tuples ( y n , y n , · · · , y nr ) and ( d , d , · · · , d r ) ∈ {− , } r .As a building block we consider the following measure γ t,y n ,y n , ··· ,y nr on {− , } V given by γ t,y n ,y n , ··· ,y nr ( A ) = µ ( P t A | L t ), where as usual ( P t ) t ≥ denotes the semigroup for the signedvoter model and L t is event that independent signed random walks random walks (also inde-pendent of the voter model) (( X j ( s ) , i j ( s )) : s ≥
0) beginning at y nj with 2 ≤ j ≤ r , satisfy η ( X j ( t )) i j ( t ) = d j . We have by induction that as n tends to infinity, the probability of event L t tends to 2 − ( r − . It follows, just as in the proof of Theorem 1.3 (given time stretchingproperties of the duals) that any limit point of duals γ t,y n ,y n , ··· ,y nr as t → ∞ is an equilibrium.In particular we have lim t →∞ γ t,y n ,y n , ··· ,y nr (cid:0)(cid:8) η : η ( y n ) = d (cid:9)(cid:1) = 1 / . (5.15)This implies that for ǫ > n sufficiently large independent signed random walks randomwalks (also independent of the η ) (( X j ( s ) , i j ( s )) : s ≥
0) beginning at y nj with j ∈ { , · · · , r } ,we have (cid:12)(cid:12) µ (cid:0)(cid:8) η : η ( X jt ) i j ( t ) = d j , j = 1 , · · · , r (cid:9)(cid:1) − − r (cid:12)(cid:12) < ǫ (5.16)for t large enough. But, given our assumptions on the sequence ( y n , y n , · · · , y nr ), implies thatfor n large and then t large enough for coalescing signed random walks (( Y j ( s ) , i j ( s )) : u ≥ y nj with j ∈ { , · · · , r } , we have (cid:12)(cid:12) µ (cid:0)(cid:8) η : η ( Y jt ) i j ( t ) = d j , j = 1 , · · · , r (cid:9)(cid:1) − − r (cid:12)(cid:12) < ǫ. (5.17)But µ is an equilibrium and this means (cid:12)(cid:12) µ (cid:0)(cid:8) η : η ( y nj ) = d j , j = 1 , · · · , r (cid:9)(cid:1) − − r (cid:12)(cid:12) < ǫ. (5.18)The result follows from the arbitrariness of ǫ . To show Proposition 1.9 we will need the following result. Let, for x ∈ V , t ≥
0, the measures µ x,t, ± on V be defined by µ x,t, + ( y ) = P x (cid:0) X ( t ) = y, X t is even (cid:1) , (6.1) µ x,t, − ( y ) = P x (cid:0) X ( t ) = y, X t is odd (cid:1) . (6.2) Lemma 6.1
For fixed T ∈ (0 , ∞ ) , and ǫ > , there exists T < ∞ so that uniformly over s ∈ [0 , T ] , x ∈ V and t ≥ T k µ x,t, + − µ x,t − s, + k TV + k µ x,t, − − µ x,t − s, − k TV < ǫ. (6.3)20he proof relies on using the coupling of [8] for two continuous time random walks on V starting at x , ( X ( r ) : r ≥
0) and ( X ′ ( r ) : r ≥ V starting from x is chosen to be the same for the two continuous time processes. Tocomplete the realizations of the continuous time processes it is then just a question of addingthe associated i.i.d. exponential random variables giving the resting times at each site: { e i } i ≥ for process ( X ( r ) : r ≥
0) and { e ′ i } i ≥ for process ( X ′ ( r ) : r ≥ n large P ni =1 e i = P ni =1 e ′ i + s . The time for this to occur does notdepend on the initial x and is tight over s in compact intervals. Proof of Proposition 1.9 . We have always the existence of the equilibrium which is the limitof the distribution ( η t : t ≥
0) for ( η ( x )) x ∈ V i.i.d. Bernoulli (1 /
2) with associated distribution µ . So we must show that for any initial r and x , x , · · · , x r ∈ V the distribution( η t ( x ) , · · · , η t ( x r )) converges to that of µ. (6.4)That is for any η the joint law of η ( X x ,t ( t )) sgn (cid:0) ( X x ,t ) t (cid:1) , η ( X x ,t ( t )) sgn (cid:0) ( X x ,t ) t (cid:1) , · · · , η ( X x r ,t ( t )) sgn (cid:0) ( X x r ,t ) t (cid:1) (6.5)converges to that of η ( x ) , η ( x ) , · · · , η ( x r ) under µ . Now the ( X x i ,t ( s ) : 0 ≤ s ≤ t ) arecoalescing random walks. But for fixed x , x , · · · , x r the probability of any further coalescenceof the random walks on interval [ T, t ] converges to zero as T → ∞ , uniformly in t > T . Fromthis we see that to show the desired ergodicity it is enough to show for y , · · · , y n fixed in V and ( Z y i ( s ) : s ≥
0) independent random walks on G , { η ( Z y i ( t )) sgn(( Z y i ) t ) : 1 ≤ i ≤ n } converges in law as t → ∞ to that of independent Bernoulli (1 / n . We suppose the desired convergence holds for integer n − n = 1). It is enough to show that as t → ∞ , the conditionalprobability that η ( Z y ( t )) sgn(( Z y ) t ) = 1 given η ( Z y i ( t )) sgn(( Z y i ) t ) ∀ ≤ i ≤ n ) convergesto 1 / α > /
2. Fix ǫ >
0, a small strictly positive constant whichwill be more fully specified later. Fix T ≫ P (cid:0) Z y ( s ) has not traversed an unsatisfied cycle for 0 ≤ s ≤ T (cid:1) < ǫ/ . (6.6)We suppose that t ≥ T + T for T given by Lemma 6.1 for this ǫ and T . Consider themartingale M s = P (cid:0) η ( Z y ( t )) sgn(( Z y ) t ) = 1 | Z y ( s ) , sgn (cid:0) ( Z y ) s (cid:1) , η ( Z y i ( t )) sgn(( Z y i ) t ) ∀ ≤ i ≤ n (cid:1) . (6.7)On { M > α } , we have, conditional on this initial value, by the optional sampling theoremfrom [2] P ( T α < T ) ≤ − α )3 − α (6.8)for T α = inf (cid:26) s : M s < / α (cid:27) . (6.9)Thus if ǫ is sufficiently small then with strictly positive probability(i) for all 0 ≤ s ≤ T, M s > / α and(ii) there exists 0 ≤ s ≤ s ≤ T so that ( X ( s ) : s ≤ s ≤ s ) traverses an unsatisfied cycle.21ut by Lemma 6.1 and our assumption on t we have that (cid:13)(cid:13) µ X ( s ) ,t − s , + − µ X ( s ) ,t − s , + (cid:13)(cid:13) TV + (cid:13)(cid:13) µ X ( s ) ,t − s , − − µ X ( s ) ,t − s , − (cid:13)(cid:13) TV < ǫ. This and the fact that M s > (1 / α ) / M s < − (1 / α ) / ǫ . But if ǫ is cho-sen sufficiently small then this will contradict (i) above. Thus we have that in fact for α > / η ( Z y ( t )) sgn(( Z y ) t ) = 1 given η ( Z y i ( t )) sgn(( Z y i ) t ∀ ≤ i ≤ n ) is less than α for t large. We similarly have that it must equally be greater than 1 − α and we are done. References [1]
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