Ergodicity of the adic transformation on the Euler graph

Transcription

1 Under consideration for publication in Math. Proc. Camb. Phil. Soc. Ergodicity of the adic transformation on the Euler graph By SARAH BAILEY Department of Mathematics, CB 250, Phillips Hall, University of North Carolina, Chapel Hill, NC USA MICHAEL KEANE Department of Mathematics, Wesleyan University, 265 Church Street, Middletown, CT 0659 USA KARL PETERSEN Department of Mathematics, CB 250, Phillips Hall, University of North Carolina, Chapel Hill, NC USA IBRAHIM A. SALAMA School of Business, North Carolina Central University, P.O. Box 907, Durham, NC USA (Received ) Abstract The Euler graph has vertices labelled (n,k) for n = 0,,2,... and k = 0,,...,n, with k + edges from (n,k) to (n +,k) and n k + edges from (n,k) to (n +,k + ). The number of paths from (0,0) to (n,k) is the Eulerian number A(n,k), the number of permutations of,2,...,n + with exactly n k falls and k rises. We prove that the adic (Bratteli-Vershik) transformation on the space of infinite paths in this graph is ergodic with respect to the symmetric measure.. The Euler Graph The Euler graph is an infinite directed graph such that at level n there are n + vertices labelled (n,0) through (n,n). The vertex (n,k) has n + 2 total edges leaving it, with k + edges connecting it to vertex (n +,k) and n k + edges connecting it to vertex (n +,k + ). Define X to be the space of infinite edge paths on the Euler graph. X is a compact metric space in a natural way: if two paths x = x 0 x x 2... and y = y 0 y y 2... agree for all n less than j and x j y j, then define d(x,y) = 2 j. The number of paths from the root vertex (0,0) to the vertex (n,k) is the Eulerian number, A(n,k), which is the number of

2 2 S. Bailey, M. Keane, K. Petersen, and I. Salama Level (n) k = 0 k = k = 2 k = 0 2 Fig.. The first three levels of the Euler graph. The numbers on the diagonals give the number of edges coming out of each vertex, and k represents the label on each vertex. permutations of 0,,...,n with exactly k rises and n k falls. These numbers satisfy the recursion A(n +,k) = (n k + 2)A(n,k ) + (k + )A(n,k). ( ) A(n, k ) A(n, k) n k + 2 k + n k + A(n +, k) Fig. 2. The Euler Graph gives rise to Equation. We put a partial order on the set of paths in X. The edges e 0 through e n+ into the fixed vertex (n,k) with 0 < k < n are completely ordered; we illustrate it so that the ordering increases from left to right. If x,y are paths in X, we say that x is less than y if there exists an N such that both x and y pass through vertex (N +,k), x n = y n for all n > N, and x N < y N with respect to the edge ordering. (2,0) (2,) e 2 e e e 0 e (,) Fig.. The order on the edges coming into vertex (,): e 0 < e < e 2 < e < e. Define k n : X {0,,...,n } by agreeing that if a path x passes through vertex (n,k), then k n (x) = k. We then say that x has a left turn at level n if k n+ (x) = k n (x) and a right turn if k n+ (x) = k n (x) +. Then define X max to be the set of paths in X such that there are no greater paths with respect to the above ordering: X max = { the path with no left turns, the path with no right turns} {x X there is a j such that x has a unique left turn at x j and for all n j, x n is the maximal edge into (n,j + )}. X min is the set of paths in X such that there are no smaller paths with respect to the above ordering: X min = { the path with no left turns, the path with no right turns} {x X there is a j such that x has a unique right turn at x j and for all n j, x n is the minimal edge into (n,n j)} Both X max and X min are countable.

3 Ergodicity of the adic transformation on the Euler graph Fig.. The dashed paths are maximal, and the dotted paths are minimal. In addition, the paths following the far left edge and the far right edge are both maximal and minimal. If x X \ X max, consider the first non-maximal edge, x j, of x and let y j be the next greatest edge with respect to the edge ordering. Then define y 0 y...y j to be the minimal path into the source of y j and let T(x) = y 0...y j x j+ x j+2...(so T(x) i = x i for all i = j +,j + 2,...). Then T : X \ X max X \ X min is the Euler adic. Fig. 5. T maps the dotted path into the dashed path. Since both X max and X min are countable, for any T-invariant, nonatomic measure µ, µ(x max ) = µ(x min ) = The Symmetric Invariant Measure A cylinder set C = [c 0 c...c n ] is {x X x i = c i for all i = 0,,...,n }. Given any T invariant Borel measure, µ, on X, define the weight w n on an edge c n connecting level n and n+ to be µ([c 0...c n ] [c 0...c n ]) for n greater than 0 and w 0 = µ([c 0 ]). Then µ[c 0...c n ] = w 0...w n, where w i is the weight on the edge c i. There are two conditions which together are necessary and sufficient to ensure that a measure on X is T-invariant. The first is that if e 0 and e have the same source vertex and the same terminal vertex, then their weights are equal. The second is the diamond law. If u is the weight associated with the edges connecting vertex (n,k) to (n +,k), u 2 is the weight associated to the edges connecting (n,k) to (n+,k+), v is the weight on the edges connecting (n+,k) to (n+2,k+), and v 2 is the weight on the edges connecting (n+,k+) to (n+2,k+), then u v = u 2 v 2. Definition. The symmetric measure, η, is determined by assigning weights /(n + 2) on each edge connecting level n to level n +.

4 S. Bailey, M. Keane, K. Petersen, and I. Salama (n, k) u u 2 (n +, k) (n +, k + ) u v = u 2v 2 v 2 v (n + 2, k + ) Fig. 6. The diamond law. This measure clearly satisfies both of the above conditions and hence is T-invariant. 2 2 Fig. 7. The Symmetric Measure. The Cutting and Stacking Representation We can also view the transformation T as a map on the unit interval defined by cutting and stacking which preserves Lebesgue measure, m. Each stage of cutting and stacking corresponds to a level in the Euler graph. At each stage n = 0,,2,... we have n + stacks S n,0,s n,,...,s n,n (corresponding to the vertices (n,k), 0 k n, of the Euler graph). Stack S n,k consists of A(n,k) subintervals of [0,]. Each subinterval corresponds to a cylinder set determined by a path of length n, terminating in vertex (n,k). The transformation T is defined by mapping each level of the stack, except the topmost one, linearly onto the one above it. This corresponds to mapping each non-maximal path of length n to its successor. To proceed to the next stage in the cutting and stacking construction, each stack S n,k is cut into n + 2 equal substacks. These are recombined into new stacks in the order prescribed by the way T maps their corresponding cylinder sets. In this way, we obtain a Lebesgue measure-preserving transformation defined almost everywhere on [0,]. Fig. 8. The Euler adic as a cutting and stacking transformation.

5 Ergodicity of the adic transformation on the Euler graph 5. Ergodicity In order to prove that the Euler adic T is ergodic with respect to the symmetric measure η, we adapt the proof in [5] of ergodicity of the B(/2,/2) measure for the Pascal adic. For previous proofs of the ergodicity of Bernoulli measures for the Pascal adic, see [],[0], [8],[6],[7] and the references that they contain. Proposition. For each x X, denote by I n (x) the cylinder set determined by x 0 x...x n. Then for each measurable A X, η(a I n (x)) η(i n (x)) χ A (x) almost everywhere. Proof. In view of the isomorphism of (X,η) and ([0,],m), this is just the Lebesgue Density Theorem. Denote by ρ the measure η η on X X. Proposition 2. For ρ-almost every (x,y) X X, there are infinitely many n such that I n (x) and I n (y) end in the same vertex of the Euler graph, equivalently (n,k n (x)) = (n,k n (y)). This is equivalent to saying that for infinitely many n the number of left turns in x...x n equals the number of left turns in y...y n, or that in the cutting and stacking representation the subintervals of [0,] corresponding to I n (x) and I n (y) are in the same stack. This happens because the symmetric measure has a central tendency: if a path is not near the center of the graph at level n, there is a greater probability that at level n + it will be closer to the center than before (and the farther from the center, the greater the probability). We defer momentarily the proof of Proposition 2 in order to show how it immediately implies the main result. Theorem. The Euler adic T is ergodic with respect to the symmetric measure, η. Proof. Suppose that A X is measurable and T-invariant and that 0 < η(a) <. By Proposition, η(a I n (x)) η(i n (x)) and η(ac I n (y)) η(i n (y)) for ρ-almost every (x,y) A A c. Hence for almost every (x,y) A A c we can pick an n 0 = n 0 (x,y) such that for all n n 0, η(a I n (x)) η(i n (x)) > 2 and η(a c I n (y)) η(i n (y)) > 2. ( ) Then, by Proposition 2, we can choose n n 0 such that I n (x) and I n (y) end in the same vertex, and hence there is j Z such that T j (I n (x)) = I n (y). Since A is T-invariant, this contradicts ( ). Then we must have η(a) = 0 or η(a) =, and so T is ergodic with respect to η. It remains to prove Proposition 2. Lemma. On (X X,ρ), for each n =,2,... let D n (x,x ) = k n (x) k n (x ), and let F = B((x,x ),...,(x n,x n)) denote the σ-algebra generated by (x,x ),...,(x n,x n).

6 6 S. Bailey, M. Keane, K. Petersen, and I. Salama Let σ(x,x ) be a stopping time with respect to (F n ) such that D σ(x,x )(x,x ) > 0. Fix M > 0 and let τ(x,x ) = inf{n > σ(x,x ) : D n {0,M}}. For n = 0,,2,..., let D σ(x,x )(x,x ) if 0 n σ(x,x ) Y n (x,x ) = D n (x,x ) if σ(x,x ) < n τ(x,x ) D τ(x,x )(x,x ) if n τ(x,x ) Then (Y n (x,x ) : n = 0,,2,...) is a supermartingale with respect to (F n ). Proof. We have to check the defining inequality for supermartingales only for the range of n where Y n = D n, since otherwise Y n (x,x ) is constant in n. If x turns to the left at stage n, then k n+ (x) = k n (x), but if x turns to the right k n+ (x) = k n (x) +. From Figure 2 we see that and η{k n+ (x) = k n (x) x...x n } = k n(x) + n + 2 ( 2) η{k n+ (x) = k n (x) + x...x n } = n k n(x) +. ( ) n + 2 Without loss of generality assume that k n (x ) > k n (x). Note that D n on the set A = {k n+ (x) = k n (x),k n+ (x ) = k n (x )} {k D n+ = n+ (x) = k n (x) +,k n+ (x ) = k n (x ) + } D n + on the set B = {k n+ (x) = k n (x),k n+ (x ) = k n (x) + } D n on the set C = {k n+ (x) = k n (x) +,k n+ (x ) = k n (x )}. From ( 2) and ( ), E ρ (D n+ D n F n ) = 0 ρ(a F n ) + ρ(b F n ) ρ(c F n ) = n + 2 [(k n(x) + )(n k n (x ) + ) (k n (x ) + )(n k n (x) )] 0. Hence E ρ (D n+ F n ) D n. Lemma 2. k n(x) n 2 in measure. Proof. Let u n (x) = 2k n (x) n for all n. We will show that u n /n 0 in measure. We begin by computing the variance of u n. Note that if k n+ (x) = k n (x) then u n+ = u n, and if k n+ (x) = k n (x) + then u n+ = u n +. Following the calculations in [9], and using ( 2) and ( ), E η (u n+ u n ) = (n + )/(n + 2)u n ; so, since u 0 = 0, E(u n ) = 0 for all n =,2,.... Similarly, ( ) ( ) E η (u 2 n+ u n ) = (u n ) 2 kn (x) + n + (u n + ) 2 kn (x) + n + 2 n + 2

7 Ergodicity of the adic transformation on the Euler graph 7 ( ) ( ) = (u n ) 2 un + n + 2 n + (u n + ) 2 un + 2 2(n + 2) 2(n + 2) Then E η (u 2 n+ u n ) = and continuing this recursively we see that V (u n+ ) = E η (u 2 n+) = = nu2 n n n ( (n )u 2 ) n + +, n + 2 n + (n + )(n + 2) = n +. Then by Chebyshev s Inequality, { u } n η ǫ c 0 as n, n nǫ2 so that u n n 0 in measure, i.e. k n(x) n 2 in measure. n (i + )(i + 2) Proof of Proposition 2. From Lemma, (D n ) is a supermartingale with respect to F n = (B((x,x ),...,(x n,x n))). Fix M > 0 and define stopping times σ(x,x ) = inf{n k n (x) k n (x )} and τ(x,x ) = inf{n > σ(x,x ) D n {0,M}}. Then E ρ (D τ ) E ρ (D σ ) =. If τ is finite almost everywhere, then i=0 E ρ (D τ ) = M(ρ{D τ = M}) + 0(ρ{D τ = 0}), so that ρ{d n 0 for any n > σ(x,x )} ρ{d τ = M} /M for all M. Letting M implies that ρ{d n 0 for any n > σ(x,x )} = 0. Hence with ρ-probability there is an n 0 for which k n0 (x) = k n0 (x ). Repeat this process with σ(x,x ) = inf{n > n 0 (x,x ) k n (x) k n (x )} to see that with ρ-probability, k n (x) = k n (x ) infinitely many times. It remains to show that τ is finite almost everywhere. We have a fixed M; fix also a large L. Fix a small enough ǫ so that if k n (x)/n, k n (x )/n are in the interval (/2 ǫ,/2 + ǫ), then k n+i (x) n + i, n k n+i(x) n + i, k n+i(x ), n k n+i(x ) n + i n + i for i = 0,,...,ML. In other words, starting from (n,k n (x)) all the probabilities of going left or right for both x and x are at least / for ML steps. Let A n = {(x,x ) X X k n (x)/n,k n (x )/n (/2 ǫ,/2 + ǫ)}, and note that ρ(a n ) as n, by the convergence in measure. Let B n = {(x,x ) X X k n+i (x) = k n (x),k n+i (x ) = k n (x )+i for all i = 0,...,M}. For every n, {x τ(x) = } A n A n B c n B c n+m Bc n+(l )M = G n, since (x,x ) in B n implies D n+i (x,x ) is either 0 or M for some i M. Conditioned on the set A n, the sets B n,b n+m,...,b n+(l )M are not independent, because at each step the probabilities of going left or right, given by sums of the weights on the edges, are changing. But since the probabilities of going left or right at each step are all near /2, so that the probability of each event we are considering is near the probability that it

8 8 S. Bailey, M. Keane, K. Petersen, and I. Salama would be assigned by a genuine symmetric random walk, we can estimate the measure of G n. For each j = 0,,...,L, abbreviate E j = Bn+jM c. Then for each pair of vertices v = ((jm,k),(jm,k )),we have ρ(e j v) ( / 2M ). Thus ρ(e j E j E 0 A n ) = ρ(e j v)ρ(v E j E 0 A n ) ( / 2M ), vertices v at level jm and iterating gives ρ(e L E 0 A n ) ( / 2M) L. Therefore ρ(τ = A n ) ( / 2M ) L for all L. Letting n and then L, we conclude that ρ{τ = } = 0. Remark. In fact k n (x)/n /2 almost everywhere. We can see this as follows. Continue to let u n (x) = 2k n (x) n as in Lemma 2. Since E((n + 2)u n+ (n + )u n ) = (n + )u n, S n = (n + )u n forms a mean-0 martingale. If X n = S n S n, then the X n are a martingale difference sequence in L 2, thus mean 0 and orthogonal. The variance of X n is E(Xn) 2 = E(Sn) 2 E(Sn ) 2 = n2 + 5n. If we let b n = n 2, then E(X 2 n)/b 2 n <, so by the extension to martingales of Kolmogrov s Criterion for the Strong Law of Large Numbers (see [, p.28]) S n /b n 0 almost everywhere, that is to say, u n /n 0 almost everywhere. Remark 2. It would be interesting to determine further dynamical properties of this system, such as weak mixing, rigidity, singularity of the spectrum, and whether the rank is infinite. So far we can show that the symmetric measure (η) is the only fully supported invariant ergodic measure [2], and that (X, T, η) is totally ergodic and loosely Bernoulli, []. We thank the referee for helpful comments. REFERENCES [] Sarah Bailey, Dyanmical properties of some non-stationary, non-simple Bratteli-Vershik systems, Ph.D. thesis, University of North Carolina, Chapel Hill, in preparation. [2] Sarah Bailey and Karl Petersen, Random permutations and unique fully supported ergodicity for the Euler adic transfromation, preprint. [] William Feller, An Introduction to Probability Theory and Its Applications, vol. II, J. Wiley and Sons, Inc., New York, 966. [] A. Hajian, Y. Ito, and S. Kakutani, Invariant measures and orbits of dissipative transformations, Adv. in Math 9 (972), [5] Michael Keane, Entropy and ergodic theory, Entropy (Andreas Greven, Gerhard Keller, and Gerald Warnecke, eds.), Princeton University Press, 200, pp [6] Xavier Méla, Dynamics of a class of nonstationary adic transfomrations, to appear in Ann. Inst. H. Poincaré Prob. and Stat. [7] Xavier Méla and Karl Petersen, Dynamical properties of the Pascal adic transformation, Erg. Th. Dyn. Sys. 25 (2005), [8] K. Petersen, Information compression and retention in dynamical processes, Dynamics and Randomness (Proceedings of the December 2000 Workshop in Santiago, Chile, A. Maass, S. Martinez, J. San Martin, eds., Kluwer Acad. Publ.), 2002, pp [9] Ibrahim A. Salama and Dana Quade, The asymptotic normality of a rank correlation statistic based on rises, Stat. and Prob. Letters 2 (997), [0] A.M. Vershik and S.V. Kerov, Locally semisimple algebras, combinatorial theory and the K 0-functor, J. Soviet Math. 8 (987), 70 7.