A Sketch of Menshikov s Theorem

Transcription

1 A Sketch of Menshikov s Theorem Thomas Bao March 14, 2010 Abstract Let Λ be an infinite, locally finite oriente multi-graph with C Λ finite an strongly connecte, an let p<p s H( Λ). Menshikov s theorem states that there exists an α> 0 s.t. P s p(x n ) exp( αn/(log n) 2 ) for all sites x an n 2. In this paper, we begin backwars, explaining the importance of this theorem an then giving a sketch of the proof, emphasizing the intuition an motivations behin the methos use. Unfortunately, ue to constraints on space, we ll sometimes simply summarize a statement an efer its rigorous proof to the text, when the proof is relatively short or simple or when the proof is too technical for an article. 1 Introuction Physicists have long believe that P T = P H. However, only until the 1986 paper by Menshikov has this been proven. In fact Menshikov s theorem is proven only for site percolation on a locally finite oriente graph. However, as Theorem 1 of section 4.2 in [1] reveals, site percolation on finite oriente graphs is nearly equivalent to that of bon percolation. Similarly, we can take an unoriente graph an create an equivalent oriente graph by replacing any bon unirecte bon xy, with two irecte ones xy an xy, so that the probability measures are the same. Nonetheless, there is still one more snag to get aroun. The way Menshikov s theorem implies P T = P H requires a very mil aitional assumption on the number of elements exactly n away from the center of the cluster (see Theorem 7 on page 100 of [1]). But, as before, further work extens Menshikov s work on the almost exponential ecay of the raius to purely exponential ecay (see section 4.4 of [1]). Finally, this can be generalize even further to exponential ecay of the volume (see section 4.5 of [1]). 2 Preliminaries We will try to follow the book s notation as much as possible, so that one can flip as easily as possible between the book an this paper. Also, in orer to save space, rather than writing out all of the notation, simply consult the textbook s list of notation on page

2 That being sai, anything not explaine in the textbook s list of notation, I will explain as they come up. In the event that I accientally ommitte a efinition, it can be foun in section 4.3. Finally, I will be skipping the first part of section 4.3, which rigorously shows that the orere set of pivot points {b 1,b 2,..., b r } for a path P from x to S + n (x) remains in the same orer regarless of which path P we choose. Now, before we begin, I highly suggest taking a quick look at Theorem 7 of section 4.3, to see how exactly Menshikov s theorem implies P T = P H. The proof is quite simple an short. Anyways, let s begin. 3 Menshikov s theorem & some intuition behin it Menshikov s Theorem: Let Λ be an infinite, locally finite oriente multi-graph with C Λ finite an strongly connecte, an let p<p s H ( Λ). Then there exists an α> 0 s.t. P s p (x n ) exp( αn/(log n) 2 ) for all sites x an n 2. Now before we ive right in let s think about how we might get to this theorem. Notice the theorem is a statement about rates, so it is natural to look towars the Margulis-Russo formula. This naturally leas us to think about the expecte number of pivots. Now, remember that Menshikov s theorem is ultimately use to prove that P H = P T. That is, when we take p = P H ɛ, we want to show E p ( C x ) < in contrast with E PH ( C x )=. So naturally, we shoul hope that the rate of change is large (otherwise E p ( C x ) remains infinite, rather than finite). So, we want the expecte number of pivots to be large. In aition, Menshikov is also a statement only for p < p s H ( Λ), that is when θ(p) =lim n ρ n (x, p) = 0, where ρ n (x, p) =P s p (R n(x)). Thus, we shoul make use of our choice of p an that ρ n (x, p) 0. So we seek an inequality which gives us a lower boun for the expecte number of pivots (which by the Margulis Russo formula is the rate of change) in terms of ρ n (x, p). Lemma 5: Let x be a site of an oriente graph Λ, an let n an k be positive integers. Then E p ((N(R n (x)) R n (x)) n/k (1 sup y ρ k (y, p)) n/k where y is any site an N(E) enotes the number of pivotal sites for an event E. Lemma 5 will inee turn out to be crucial, but it actually follows quite nicely from: Lemma 4: Let x be a site of an oriente graph Λ, an let n, r, an i, 1 i < r, be positive integers. For 1 k n r 1 i=1 i, we have where E is the event P s p (D r >k E) sup{ρ k (y, p) :y Λ }, E = R n (x) {D 1 = 1,D 2 = 2,..., D r 1 = r 1 } where D i is the istance of the shortest open path from b i 1 to b i, where b i is the ith pivot point (as mentione in the preliminaries). 2

3 Let s first partition our event E, so we have a little bit more information to start with. Since we primarily care about D r, then let s partition it by the location of the r 1th pivot, call it y, an I, an interior cluster consistent with E {b r 1 = y}. In other wors: E y,i = R n (x) {y = b r 1 } { I = I}. Now let s say E y,i hols. Then so oes F y,i, the event that all sites in I {y} are open, an I\{y}, the set of outneighbors of I not incluing {y}, are close. (Otherwise, if p I\{y} was open, p I, since it s an open site that is ajacent to an element in I.) Further notice, that E y,i hols if an only if F y,i hols an there is an open path P from a neighbor of y to S n + (x) isjoint from I I. (Note that y I.) Let us write G y,i for the event that there is such a path P. Let s X =(I I) c, an let P x p enote the prouct probbility measure in which each site of X is open inepenently with probability p. Now take a look at figure 10 page 94. Notice that we mae sure that I\{y} was close. This allows us to look at F y,i without peeking at G y,i. In fact, we can easily efine such an exploration algorithm, where we look at the open neighbors of x, then the new open neighbors of those neighbors, an so on, until we either have no new open neighbors or we hit y (we first chose y = b r 1 when we create the partition E y,i so we know its location without looking at G y,i ). So G y,i an F y,i occur inepenently, an by our earlier if an only if statement (i) P p (E y,i )=P p (F y,i )P X p (G y,i ). Now suppose again that E y,i hols an that D r > k. Since y = b r 1, the next pivot, b r, will be greater than k away from y. That is, b r lies outsie of the sphere S k (y). Since b r is a pivot, then there must be two open paths from y to b r, completely isjoint from one another except at y an b r. Since b r lies outsie the sphere S k (y), there must be a path from a neighbor of y to S + k (y) that is completely isjoint from G y,i. Thus, letting H y,i enote the event that X contains an open path from a neighbour of y to S + k (y), G y,i H y,i hols. Since E y,i {D r >k} is a subset of F y,i an we ve just shown that when it oes hol G y,i H y,i also hols, (ii) P p (E y,i {D r >k}) P p (F y,i )P X p (G y,i H y,i ). Now by the van en Berg-Kesten inequality Combining the three relations above Since P X p (H y,i ) P p (R k (y)) = ρ k (y, p), then P X p (G y,i H y,i ) P X p (G y,i )P X p (H y,i ). P p (E y,i {D r >k}) P p (E y,i )P X p (H y,i). P p (E y,i {D r >k}) P p (E y,i )ρ k (y, p) P p (D r >k) E y,i ) ρ k (y, p). Since the events E y,i partition E, Lemma 4 follows. So to summarize, we first partitione the event E by the location of y = b r 1. Then we note that F y,i an G y,i occur iff E y,i hols, an use the fact that we can look at F y,i without peeking at G y,i to get (i). Then we ae the aitional conition that D r > k, which implie b r lies outsie of S k (y). Since there must be two isjointpaths from a neighbor of y to a neighbor 3

4 of b r / S k (y) (otherwise b r wouln t be the next pivot), G y,i H y,i hols. Using this observation, we got (ii), an the rest followe from van en Berg-Kesten an stacking our inequalities. As mentione in the abstract, ue to constraints on space please see page 95 for the proof of Lemma 5 from Lemma 4. (It s relatively short an follows almost immeiately) We are now reay for: Menshikov s Theorem: Let Λ be an infinite, locally finite oriente multi-graph with C Λ finite an strongly connecte, an let p<p s H ( Λ). Then there is an α> 0 s.t. P s p(x n ) exp( αn/(log n) 2 ) for all site x an integers n 2. Fix p<p H ( Λ). For any up-event E, by Margulis Russo, we have: p P p(e) =E p (N(E)) E p (N(E)1 E )=E p (N(E) E)P p (E), where 1 E is the characteristic function for the set E. Notice that E coul not occur but still have 1 or more pivots, so that the inequality above is not just a stylistic choice. Thus, p logp p(e) = 1 P p (E) p P p(e) E p (N(E) E) Now let ρ n (p) =sup x Λ ρ n (x, p). Taking E = R n (x) an applying Lemma 5 to the inequality above p log ρ n(x, p) n/k (1 ρ k (p)) n/k Fix p <p +. Since ρ k (p) monotonically increases with p, for any p [p,p + ], p log ρ n(x, p) n/k (1 ρ k (p)) n/k n/k (1 ρ k (p + )) n/k. ˆ p+ ˆ p+ T hus : log ρ n (x, p + )=logρ n (x, p )+ log ρ n (x, p)p log ρ n (x, p )+ n/k (1 ρ k (p + )) n/k p p p log ρ n(x, p ) ρ n (x, p + ) (p + p ) n/k (1 ρ k (p + )) n/k because, e x is monotonic : Finally, as the upper boun oes not epen on x, ρ n (x, p ) ρ n (x, p + ) exp( (p + p ) n/k (1 ρ k (p + )) n/k ). ( ) ρ n (p ) ρ n (p + )exp( (p + p ) n/k (1 ρ k (p + )) n/k ). At this point, we reach the toughest part of the proof. How o we go from the above inequality to Menshikov s theorem, which epens soley on n an not the value of ρ n for some p? You woul think we require some aitional work on how fast ρ n (p + ) 0, but we on t. Instea, we will pass through to a sequence that allows us to create a constant polynomial boun an then refine this to get the esire result. 4

5 Fix p, p 0, s.t. p<p 0 <p H. Since ρ n (p 0 ) 0. Writing ρ 0 = ρ n0 (p 0 ) we may fix n 0, sufficiently large s.t. ρ 0 1/100 an ρ 0 log(1/ρ 0 ) (p 0 p)/6 (Letting y =1/x, we see lim x 0 x log(1/x) = log y lim y y =0by l hopital s). Now, let us inuctively efine two sequences, n 0 n 1..., an p 0 p 1...: (i) n i+1 = n i 1/ρ i an p i+1 = p i ρ i log(1/ρ i ). Amazingly this sequence has the following key properties: (ii) ρ i+1 := ρ ni+1 (p i+1 ) ρ (4/3) i an p i >p := (p 0 + p)/2 i. The proof of the above properties relies on two facts, ( ) an ρ n (p 0 ) 0, as n. Because the proof is fairly technical though short, we will skip ahea. (The proof is on page 98.) Now let s i = 1/ρ i, so n i = n 0 j<i s j, an s 0 = 1/ρ Since s i s 0 100, we will simply ignore the rouning. By (ii), s i+1 = giving (iii) s j = j<i so by (i), Let n n 0 be arbitrary. Then i, s.t. Thus, 1 j i 1/ρ 4/3 i s 4/3 i. Thus, for 1 j i, s i j s (3/4)j i, j 1 s i j s (3/4)j i = s 3 i, (iv) n i+1 = n i s i = n 0 ( j<i s j )s i n 0 s 4 i. (v) n i n n i+1 n 0 s 4 i. ρ n (p ) ρ ni (p ) ρ ni (p i )=ρ i 1/s i (n 0 /n) 1/4 =(n/n 0 ) 1/4, with the secon inequality following from (ii) an the final inequality following from (v). Since n 0 is fixe, we can take c =1/n 1/4 0 ( ) ρ n (p ) cn 1/4. To summarize, we cleverly efine s i to make use of the boun in (ii). Then we ingeniously erive (iii) by noting the symmetry to get the first equality. After that, (v) naturally followe, allowing us to boun ρ n (p ) in terms of n. The key was to somehow boun s i, or 1/s i, in terms of n an n 0, which all starte with (ii). We re still far from the theorem, but we re close. All that remains is to apply ( ) repeately to ( ). In the interest of keeping things in line with the book, we ll realx ( ) to get ρ n (p ) cn 1/5 ( ). Fix p with p < p <p. for n 1, let k 1 = k 1 (n) = n 5/6. Again, let us ignore rouning, so n/k 1 n 1/6, an, from ( ), ρ k1 (p ) cn 1/6. Then 1 ρ k1 (p ) 1 cn 1/6, so (1 ρ k1 (p )) n/k1 (1 cn 1/6 ) 1/6 e c 5

6 as n. Since convergent sequences are boune, it follows that (1 ρ k1 (p )) n/k1 is boune away from 0, by some constant, Ω. Applying ( ) with k = k 1 (n),p + = p an p = p, we get ρ n (p ) ρ n (p )exp( (p + p )n 1/6 Ω), since ρ n (p ) < 1, an p + p is constant an can be absorbe by Ω, then ρ n (p ) exp( Ωn 1/6 ) Now, fix p with p < p <p, an let k 2 = k 2 (n) = (log n) 7 (for n 2). By the inequality above, ρ k2 (p ) exp( Ω(log n) 7/6 )=o(n 1 ) for some constant c, since exp( Ω(log n) 7/6 ) goes to 0 much faster than n 1. (f = o(g), if f/g 0 as n.) Thus, (1 ρ k2 (p )) n/k2 1, since we ve just shown ρ k2 (p )=o(n 1 ). Applying ( ) with k = k 2 (n), an using the convergence towars 1, we get ρ n (p ) exp( Ω(n/(log n) 7 ). (We again ignore ρ n (p + ) since it is less than 1, so the inequality still hols. Iterating again, k 3 = log n(log log n) 8, it follows that ρ k3 (p )=o(n 1 ), an applying ( ) again, as we ve one, we fin ρ n (p) =exp( Ω(n/(log n(log log n) 8 ))), an the theorem follows. References [1] B. Bollobás an O. Rioran (2006), P ercolation, Cambrige U. Press, Cambrige UK. 6