THE GROWTH DEGREE OF VERTEX REPLACEMENT RULES

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1 THE GROWTH DEGREE OF VERTEX REPLACEMENT RULES JOSEPH P. PREVITE AND MICHELLE PREVITE 1. Introduction. 2. Vertex Replacement Rules. 3. Marked Graphs Definition 3.1. A marked graph (G, p) is a graph with one marked vertex p G. The set G of isometry classes of all marked graphs is a metric space with dist G ((G 1, p 1 ), (G 2, p 2 )) = 2 M, where M is the largest integer such that the balls of radius M centered at p i G i are isometric. For a marked graph (G, p) where p is replaceable, we let R(G, p) be the set of marked graphs (R(G), q), where q is any vertex inside the copy of H i that replaced p. Definition 3.2. A sequence satisfying (G i, p i ) R(G i 1, p i 1 ) for all i 1 is called an R orbit of (G 0, p 0 ). What follows is prop 6.1 pg 26 of Joe s paper. Proposition 3.3 (J. Previte). Every R orbit has limit points in G, and the space R of the limit points of all R orbits is compact. Moreover, all of R can be realized as the limit points of the set of marked graphs (H j, p), where the H j s are replacement graphs. We need to check that this prop doesn t depend upon R being an expanding replacement rule. Definition 3.4. A vertex replacement rule R is called invertible if for any ( G, p) R, there is a unique (G, p) R such that ( G, p) R(G, p). Date: May 27,

2 2 For R invertible, let π be the map on R which undoes replacement, i.e., π[( G, p)] := (G, p) for all ( G, p) R(G, p). Since R is invertible, then π is well-defined and continuous on R. 4. The Growth Degree Definition 4.1. The growth function of a locally finite graph X with respect to a vertex x V (X) is given by f X (x, n) = {y V (X) : d(x, y) n, 0 n}, where d(x, y) denotes the distance between x and y. Definition 4.2. We say that a locally finite graph X has exponential growth if there is a vertex x V (X) and a constant c > 1 such that f X (x, n) c n for all n N. Otherwise, X has non-exponential growth. In particular, X has polynomial growth if for each x V (X), there are constants c and d such that f X (x, n) cn d for all n N. Furthermore, we say that X has intermediate growth if it has non-exponential growth but grows faster than any polynomial. Definition 4.3. The growth degree at x of a graph X with polynomial growth is inf {d : f X(x, n) c} d R n d for some vertex x V (X) and some constant c. Lemma 4.4. Let X be locally finite with x, y, V (X) and d(x, y) <. (i) If X has exponential growth at x then it does at y. (ii) If X has polynomial growth at x then it does at y, and the growth degrees are the same. Proof. Suppose that X has exponential growth at x. Let k = dist(x, y). For k > 1, f X (y, n) f X (x, n k) c n k, which implies that X has exponential growth at y. Next, suppose X has polynomial growth at x with growth degree d. Then, f X (y, n) f ( ) d X(x, n + k) n + k c. n d n d n The right most term is bounded, since it converges to c as n, hence X has polynomial growth at y with growth degree at most d. If X had a growth degree at y of d < d then a symmetric argument to

3 that above would show that the growth degree at x would be at most d. Hence, the growth degrees must agree. Consider the graphs R n (v) R n (G). Define the set R n (v) to be all vertices w R n (v) that are adjacent to v. Define the following two functions: and a(n) = dist R n (v)(u, u ), where u, u R n (v) for u u b(n) = sup {dist R n (v)(u, z) : u R n (v)}. z R n (v) Lemma 4.5. Let R = {H} be a replacement rule and suppose σ is a simple boundary connecting path in H. For all n, m N, there exist positive constants κ 1 and κ 2 such that κ 1 a(n) b(n) κ 2, (1) and κ 1 N(σ) m κ 1 N(σ) m b(n) b(n + m) κ 2 N(σ) m, (2) a(n) b(n + m) κ 2 N(σ) m. (3) I couldn t find this explicitly stated in Joes paper. It is Lemma 3.6 in my Hausdorff paper pg 12. I need to check that there aren t any conditions on H since we worked with primitive replacement rules in Lemma 3.6. The following lemma is Proposition 3.3 from Joe s paper. Lemma 4.6. J. Previte There is a number K (depending on H) such that for n > K, a(n) = L(σ) + N(σ)a(n 1). In particular, if N(σ) > 1, then ( ) L(σ) a(n) = N(σ) n K N(σ) 1 + a(k) L(σ) N(σ) 1. For any edge e and any replaceable vertex v in any finite graph G, let Q(e, v) be the number of (different) edges adjacent to v that can be connected to e by a nonreplaceable path in G. Let Q(e) = v V Q(e, v), where V is the set of replaceable vertices in G. Let Q(G) = max e E(G) Q(e). if there exists a nonreplaceable path in H (or in R(H)) containing a boundary vertex b in H (or in R(H)) in its interior and whose endpoints are replaceable. 3

4 4 The following is an adaptation of Lemma 4.1 in Joe s paper (pg. 21): Lemma 4.7. Let H define a vertex replacement rule R which is nonexponential with N(σ) 2. Then for all n 0, we have that Q(R n (G)) = Q(R(G)). Proof. If the boundary vertices of H are replaceable, then we have that Q(R(e)) = Q(e) for R(e) R(G). So Q(R n (G)) Q(G). Suppose the boundary vertices are not replaceable. Let e be an edge in G and let e 1,...e n be the distinct edges that are adjacent to particular replaceable vertex v G. After replacement, R(e 1 ),..R(e n ) will be attached to n distinct boundary vertices b 1,..., b n in a copy of H = R(v). By the nonexponential assumption, each b i has a unique replaceable vertex w i R(v) to which it can only be connected via a nonreplaceable path in H. Note, a priori, some of the w i can be equal. Now assume that w i is adjacent to two (or more) edges E 1 and E 2 from which b i can access w i via a nonreplaceable path in R(v). Then after replacement, R(w i ) will have two new boundary vertices B 1 and B 2. Again, these new boundary vertices will each be associated with a unique replaceable vertex W 1 and W 2. Now if W 1 W 2, then the replacement rule is exponential. Hence W 1 = W 2. This implies that N(σ) = 1. This shows that each w i is actually accessible by exactly one edge from b i. In particular, Q(e) = Q(R(e)). Now suppose that there is a boundary vertex b H such that its corresponding replaceable vertex v b H has at least two distinct adjacent edges e 1 and e 2 so that a nonreplaceable path in H from b to a replaceable vertex w H may either first pass through e 1 and then v b or it may first pass through e 2 and then v b. Call such nonreplaceable paths from b through e 1 and e 2 to v b γ 1 and γ 2, respectively. START FIXING HERE that Since R is nonexponential, then every nonreplaceable path in R(H) between replaceable vertices contains no interior boundary vertices in R(H). Figure 1. An initial graph Γ 0 and a replacement rule R = {K}. Example 4.8. Let Γ 0 be as in Figure 1. Let R = {K}, where K consists of a graph with total length 2l having 2(l 1) nonreplaceable and 2 replaceable vertices (at the ends) with a single nonreplaceable boundary vertex in the middle (see Figure 1). Then any limit graph (X, x) of R has exponential growth degree.

5 Proof. There are two cases. First assume that (X, x) is a limit graph with dist(x, V ) =, where V is the set of replaceable vertices of X. Then X is realized by the limit of an R-orbit of marked graphs so that π n (x) π n (X) is not replaceable for all n. In this case, x is at the base of an infinite binary tree with branch length l. (Note that if dist(x, G) <, then the infinite tree is in the opposite direction of G). Thus for n 1, f (X,x) (x, nl) 2 n 1 l. Thus, f (X,x) (x, n) 2 n/l l = l 2 [21/l ] n for all n N. So (X, x) has exponential growth. Next, assume (X, x) is a limit graph obtained from an R-orbit of marked graphs with π n (x) π n (X) replaceable for some n N. Then for any limit graph (X, x) of such an R-orbit, the marked point x is a finite distance from a degree one vertex (see Figure 2). Then for v a replaceable vertex, f (X,x) (v, nl) = 3l2 n 1 2. To see this, note that in Figure 2, if one takes the midpoint of the horizontal path with endpoint v of length nl (n even) then, one will 2 arrive at the root of three different binary trees each having branch length l and total length at least nl. 2 Thus, f (X,x) (v, n) 2 n/l l = l 2 [21/l ] n for all n N. So (X, x) has exponential growth. 5 v a path of length ln/2 for n = 10 Figure 2. A limit graph (X, x) where each edge represents a path of length l.

6 6 Definition 4.9. We say that a vertex replacement rule R given by a single replacement graph H is central if for each boundary vertex b H, there exists a unique replaceable vertex v b H such that any path in H from b to a replaceable vertex w H must first pass through v b. Any replacement rule with replaceable boundary vertices is central. Any replacement rule with exactly one replaceable vertex is central. If a vertex replacement rule has two or more replaceable vertices and is non-central, then there must exist a path in H containing no interior replaceable vertices and a boundary vertex b in its interior, with its two ends vertices v 1 and v 2 being replaceable. Theorem Let (G 0, p 0 ) be a finite initial marked graph with at least one replaceable vertex, and let R = {H} be a replacement rule with N(H) 2 but R not central. Then each graph in the set of limit graphs R (G 0, p 0 ) of the R orbits of (G 0, p 0 ) has exponential growth. Proof. By hypothesis, there exist replaceable vertices v 1, v 2 H, a boundary vertex b H, and a nonreplaceable simple path γ in H between v 1 and v 2 and containing b. For i = 1, 2, let γ i be the subpath of γ between the replaceable vertex v i and the boundary vertex b. Then there are essentially two cases, but both involve showing that (X, x) contains a binary tree as in Example (4.8) of branch length l = L(γ 1 ) + L(γ 2 ). First assume that (X, x) is a limit graph with π n (x) π n (X) not replaceable for any n. Then (X, x) contains an infinite binary tree of branch length l (see Figure). Similarly, if (X, x) is a limit graph with π n (x) π n (X) replaceable for some n, then x is a finite distance from a replaceable vertex and so (X, x) contains the end of an binary tree of length l similar to the second case of Example (4.8). In either case, (X, x) has exponential growth. Theorem Let (G 0, p 0 ) be an initial graph with at least one replaceable vertex, and let R = {H} be a central replacement rule. Then any limit graph in R (G 0, p 0 ) has polynomial growth with growth degree N(H)/N(σ). Proof. We will show that every limit graph in R (G 0, p 0 ) has polynomial growth with growth degree N(H)/N(σ). We first consider the case when G 0 is a finite graph. Let (X, x) be a limit graph in R (G 0, p 0 ) and let p be a vertex in (X, x). We want to compute the growth function f (X,x) (p, m). Let n be such that a(n 1) m a(n). Then all paths of length m can

7 intersect at most Q copies of R n (w), where w is a replaceable vertex. Thus, f (X,x) (p, m) Q[N(H)] n. By Lemma 4.7, Q is bounded for all vertices p (X, x). To do this, argue that for all p (X, x), we have that π n (p) π n (X, x) has at most Q nonreplaceable paths in π n (X, x) to replaceable vertices. To show that the growth degree of (X, x) is N(H)/N(σ), we will show that f (X,x) (p, m) m d (4) is bounded when d = N(H)/N(σ) and is unbounded when d < N(H)/N(σ). We proved this part the same way as Mary Tryon did. First, we show that Expression 4 is bounded when d = N(H)/N(σ). We need a lower bound on m. Recall that a(n 1) m, and by Lemma 4.6, there is a constant C > 0 such that for all n. So a(n 1) CN(σ) n 1 m d = m N(H)/N(σ) a(n 1) N(H)/N(σ) [CN(σ) n 1 ] N(H)/N(σ) = C N(H)/N(σ) N(H) (n 1)N(σ)/N(σ) (by Mary T s lemma) = C N(H)/N(σ) N(H) n 1. 7 So when d = N(H)/N(σ), we have f (X,x) (p, m) m d QN(H) n C N(H)/N(σ) N(H) n 1 = QN(H) C N(H)/N(σ) for all m. Next, we need to show that when d < N(H)/N(σ), we have that f (X,x) (p, m) m d is unbounded. For this part, we need an upper bound on m and a lower bound on f (X,x) (p, m). Recall that m a(n). So for some ɛ > 0, we

8 8 have m d < m N(H)/N(σ) ɛ a(n) N(H)/N(σ) ɛ [CN(σ) n ] N(H)/N(σ) ɛ since Lemma 4.6 implies there exists a constant C > 0 such that a(n) CN(σ) n for all = CN(H)/N(σ) ɛ N(σ) nn(h)/n(σ) N(σ) nɛ = CN(H)/N(σ) ɛ N(H) nn(σ)/n(σ) N(σ) nɛ = CN(H)/N(σ) ɛ N(H) n N(σ) nɛ. (by Mary T s lemma) We also need a lower bound on f (X,x) (p, m). Since m a(n 1), then f (X,x) (p, m) N(H) n 1. Thus for d < N(H)/N(σ), we have f (X,x) (p, m) m d N(H)n 1 N(σ) nɛ C N(H)/N(σ) ɛ N(H) n = which is unbounded as n m since N(σ) > Examples. References N(σ) nɛ C N(H)/N(σ) ɛ N(H) [1] A. Berman and R. J. Plemmons, Nonnegative Matrices in the Mathematical Sciences, SIAM, New York [2] K. Falconer, The Geometry of Fractal Sets, Cambridge University Press, Cambridge, [3] M. Gromov, Groups of polynomial growth and expanding maps (with an appendix by Jacques Tits), IHES 53 (1981) [4] J. Munkres, Topology, 2nd ed., Prentice Hall, Upper Saddle River, NJ, [5] J. Previte, Graph substitutions, Ergodic Theory Dynam. Systems 18 (1998) [6] J. Previte, M. Previte, and M. Vanderschoot, Limits of vertex replacement rules, Rocky Mountain Journal of Mathematics (to appear). [7] M. Previte, The dimensions of limits of vertex replacements, Illinois J. of Math. (to appear). [8] M. Previte and M. Vanderschoot, The topological dimension of limits of graph substitutions, Forum Math. 15 (2003) [9] M. Previte and S.-H. Yang, The topological dimension of limits of vertex replacements, Topology Appl. 153 (2006)