THE GROWTH OF LIMITS OF VERTEX REPLACEMENT RULES

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1 THE GROWTH OF LIMITS OF VERTEX REPLACEMENT RULES JOSEPH PREVITE, MICHELLE PREVITE, AND MARY VANDERSCHOOT Abstract. In this paper, we give conditions to distinguish whether a vertex replacement rule given by a single graph H yields a limit graph having polynomial or exponential growth. In the case of polynomial growth, we compute the growth degree of the limit graph. 1. Introduction. The notion of a vertex replacement rule was introduced in [4]. Roughly, a vertex replacement rule is an iterative rule which replaces certain vertices of a graph with copies of other graphs. These rules generate fractals (see [8]) when the resulting graphs are uniformly scaled, and infinite graphs when the resulting graphs remain unscaled. In [8], [7], [4] and [6], the dimensions of limits of scaled replacement rules were computed. The referee for [7] was extraordinarily helpful by suggesting the further investigation of the relationship between the Hausdorff dimension of limits of scaled vertex replacements and the growth degree (also known as the dimension) of limits of unscaled replacements. This paper is the start of that investigation. The main objective for this paper is to study the growth of limits of unscaled replacement rules given by one replacement graph H. We provide conditions to distinguish whether a vertex replacement rule yields a limit having exponential growth (Theorem 4.7) or polynomial growth (Theorem 4.10) and compute the growth degree in the case of polynomial growth (Theorem 4.10). This growth degree turns out to be the same formula as the Hausdorff dimension of limits of scaled replacements (see [6]). In Section 2, we define and provide examples of vertex replacement rules. We also define some requisite functions and state a lemma about those functions. In Section 3, we prove two results (Theorems 3.1 and 3.3) that show that the limit of an unscaled vertex replacement rule given by one replacement graph is a well-defined object. Theorem 3.1 is Date: January 16,

2 2 used again at the very end of the proof of one our main results (Theorem 4.10). Section 4 is the heart of the paper. In it we define exponential and polynomial growth of infinite graphs and prove our main results (Theorems 4.7 and 4.10). The authors wish to acknowledge and thank the referees for their helpful input and insightful suggestions which have greatly improved this manuscript. 2. Vertex Replacement Rules. In this section, we define and provide some basic examples of vertex replacement rules. Throughout this paper, we assume that all graphs are connected, locally finite, unit metric graphs (i.e., each graph is a metric space and every edge is isometric to the open interval (0, 1)). Furthermore, the distance between two points in a graph is measured by the shortest path in the graph between the two points. More precisely, each graph is treated as a length space (also called a geodesic space). Definition 2.1. A graph H with a designated set of vertices {v 1,..., v k } is called symmetric about {v 1,..., v k } if every permutation of {v 1,..., v k } can be realized by an isometry of H. The vertices in such a designated set are called boundary vertices of H, and the set {v 1,..., v k } is denoted by H. The graph H shown in Figure 1 has several possible sets of boundary vertices. For example, H could consist of any pair of vertices, say v 1 v 2 v 1 v 3 v 4 Figure 1. A graph H. and v 2. Or perhaps H consists of only the single vertex v 1. However, H cannot be the set {v 1, v 2, v 3, v 4 }, since H is not symmetric about this set. Definition 2.2. A replacement graph is a graph H together with a specified set H of boundary vertices. Definition 2.3. Let G be a graph, and let (H, H) be a replacement graph. A vertex v in a graph G is replaceable (with respect to (H, H)) if deg(v) = H, where deg(v) (the degree of v) is the number of edges adjacent to v and X denotes the number of vertices in the graph X.

3 The replacement graph (H, H) yields a vertex replacement rule R that acts on a graph G to produce a new graph R(G) by substituting for each replaceable vertex v in G a copy of the replacement graph H in such a way that the deg(v) edges previously adjacent to v in G are now adjacent to the H boundary vertices of H. Since H is symmetric about H, it is irrelevant how the edges previously adjacent to v are attached to H. Thus, vertex replacement is a well-defined procedure. 3 Figure 2. A replacement graph (H, H), where all three vertices are boundary vertices. For example, let a vertex replacement rule R be given by the replacement graph (H, H) as shown in Figure 2. The vertices designated as unfilled-in circles denote the boundary vertices, about which H is symmetric. Figure 4 illustrates the construction of R(G) for the graph G given in Figure 3. Figure 3. A graph G. Notice that R(G) in Figure 4 itself contains replaceable vertices. Thus iterating the replacement rule R produces a sequence of graphs R n (G). See Figure 5. Note that each of the replaceable vertices in R(G) was originally a degree two boundary vertex in a copy of H. Hence, we make the following definition:

4 4 R(G) Figure 4. Constructing the graph R(G). G R(G) R 2 (G) Figure 5. A graph G with R(G) and R 2 (G). Definition 2.4. We say a boundary vertex v of a replacement graph (H, H) is replaceable if and only if deg(v) = H 1. In other words, it is important to keep in mind the fact that once a copy of a replacement graph (H, H) is inserted into a graph G, an edge of G is attached to each boundary vertex and its degree increases by one. Therefore, one should not treat a replacement graph (H, H) as an initial graph G, but always view it as having already replaced some vertex of a graph G. Figure 6 provides an example of a replacement graph (H, H) whose boundary vertices are not replaceable. There is a natural pointwise map π : R(G) G which undoes replacement by crushing the inserted copies of H to the vertices that they replaced. If K G then R(K) R(G) is the subset of R(G) that corresponds to K after replacement, i.e., R(K) = π 1 (K). Note that if K is entirely nonreplaceable, then R(K) is isomorphic as a graph to

5 5 Figure 6. A replacement graph (H, H) with nonreplaceable boundary vertices (denoted as unfilled circles). K, but possibly not isometric. For example, suppose G is the graph depicted in Figure 7, and K is the subset of G consisting of the degree four vertex together with the four open edges adjacent to this vertex. When the replacement rule given by the triangle in Figure 2 is applied to G, then we obtain the graph R(G) shown in Figure 7. Let x and y be two points in K that lie on either side of one of the replaceable vertices in G and are 1/2 apart. Observe that there is an isomorphism f : K R(K), but K and R(K) are not isometric since the distance between f(x) and f(y) is 3/2. K x G y f(x) f(y) R(G) R(K) Figure 7. K is isomorphic but not isometric to R(K). If v is replaceable by H in G then we denote R(v) R(G) as the copy of H that replaced v. For each replaceable vertex v G, R(v) is isomorphic as a graph to H. That is, R(v) = π 1 (v). Similarly, denote by R n (v) R n (G) the set π n (v), where v is replaceable by H in G. Denote by R n (v) the set H of vertices w R n (v) that are adjacent to both R n (v) and R n (G)\R n (v). Let R n 1 (H) denote a copy of R n (v) R n (G), where v is replaceable by H and let R n 1 (H) = R n (v).

6 6 It will be important in our calculations to keep a tally of certain vertices in a replacement graph (H, H). Define N(H) to be the number of replaceable vertices of (H, H). For H > 1, define B(H) to be the minimum number of replaceable vertices on a path connecting distinct boundary vertices. If H = 1 and the boundary is not replaceable, we define B(H) = 0. If H = 1 and the boundary is replaceable, then H is a single vertex, and we call the replacement rule trivial since R(G) = G for all G. Similarly, if N(H) = 0 then R j (G) = R(G) for all j 1, and we again call the replacement rule trivial Prior Results. We now introduce two functions that are used to determine the rate of growth of R n (G) and a lemma about those functions. For a replacement graph (H, H) with H 2, define a(n) = dist R n (v)(u, u ), where u, u R n (v) for u u. For any replacement graph (H, H), define b(n) = sup {dist R n (v)( R n (v), z)}. z R n (v) In the above definitions we specify that dist R n (v) means that distances are computed via paths inside R n (v), where v is any replaceable vertex. Observe that a(n) is well-defined since the graph H is symmetric about H. As an example, consider the replacement rule shown in Figure 2. One can see from Figure 5 that a(2) = 3, and b(2) = 1.5 It is straightforward to show that for any replacement graph (H, H) with B(H) > 1, we have a(n) b(n). 2 The following lemma is an immediate consequence of Proposition 3.3 from [4]. Lemma 2.5. Let R = (H, H) be a vertex replacement rule with B(H) > 1. There are fixed positive constants C and C (depending on (H, H)) such that and C B(H) n a(n) C B(H) n C B(H) n b(n) C B(H) n. 3. Sequences of Replacement Graphs and Their Convergence Let G be a finite graph with at least one replaceable vertex. There are several possible ways to study the sequences {R n (G)} that result from vertex replacement rules R. One approach is to scale each graph R n (G) in the sequence {R n (G)} so that all graphs have the same diameter

7 and then analyze the limit of the resulting sequence with respect to the Gromov-Hausdorff metric (see [8]). In this paper, our approach will be to consider pointed graphs (graphs with distinguished base points). In particular, we will analyze limits of sequences of pointed metric spaces (R n (G), x n ), where {x n } is a sequence of points with each x n R n (G), where π(x n ) = x n 1, and x 0 is nonreplaceable in G. (This makes replacement a well-defined operation on a pointed graph (G, x 0 ), since there is a unique x 1 so that π(x 1 ) = x 0.) Note that x n need not be a vertex and could be a midpoint of any edge. The purpose of this section is to prove that the limit of the sequence {(R n (G), x n )} is well-defined. For any pointed metric space (X, x) and any M > 0, let B M (X, x) be the open ball of radius M centered at x. Note that the distance between points in B M (X, x) may not necessarily be realized by paths within B M (X, x) (so B M (X, x) may not be a geodesic space itself). This is illustrated in Figure 1, where although v 2 and the midpoint of the edge connecting v 3 and v 4 are in B 2 (H, v 1 ), the distance between these two points is realized by a path not in B 2 (H, v 1 ). In the main theorems of this section we show that the open balls B M (R n (G), x n ) and B M (R m (G), x m ) are isometric for sufficiently large n and m. This result is proved in Theorem 3.1 for a replacement rule (H, H) with nonreplaceable boundary vertices and in Theorem 3.3 for (H, H) with replaceable vertices. Theorem 3.1. Let R = (H, H) be a nontrivial vertex replacement rule whose boundary H is not replaceable and such that B(H) 1. Let (G, x 0 ) be a pointed graph with a replaceable vertex. Then for any integer M > 0 there exists an N M N so that for all n N M, the ball B M (R n (G), x n ) is isometric to B M (R N M (G), x NM ). Proof. Let M > 0 be given and suppose (G, x 0 ) is a pointed graph (so the base point x 0 is nonreplaceable). Consider the pointed graph (R n (G), x n ), where n 4M. Since the boundary vertices of H are not replaceable, x n must lie at least 4M away from any replaceable vertex. So B M (R n (G), x n ) contains no replaceable vertices. Thus for any n 4M the balls B M (R n (G), x n ) are isomorphic as graphs. To see that they are isometric, we claim that for all n 4M, all distance realizing paths between points of B M (R n (G), x n ) contain no replaceable vertices in their interiors. In other words, the situation illustrated in Example 7 does not occur. This implies that B M (R 4M (G), x 4M ) as a subgraph of R 4M (G) is isometric to B M (R n (G), x n ) as a subgraph of R n (G). 7

8 8 Suppose by way of contradiction that there is a replaceable vertex r on the interior of a distance realizing path connecting p to q, where p, q B M (R n (G), x n ). Without loss of generality, assume that d(p, r) d(q, r). Then d(x n, r) d(x n, p) + d(p, r) which implies that 4M d(x n, r) d(x n, p) + d(p, r) M + d(p, r) so d(p, r) 3M. But since the path is distance minimizing, d(p, r) d(p, q) d(p, x n ) + d(q, x n ) 2M, which is a contradiction. Therefore, there exists N M N, namely N M = 4M, such that the ball B M (R n (G), x n ) is isometric to B M (R N M (G), x NM ) for n N M. Definition 3.2. For any replacement rule given by (H, H) and any graph G with nonreplaceable point x G, let N (G, x) denote the set of all edges and vertices that can be connected to x via a nonreplaceable path in G. Note that if G has replaceable vertices then the resulting space N (G, x) will have open edges that are adjacent to both N (G, x) and G \ N (G, x). Also note that if (H, H) has replaceable boundary, then each N (R n (G), x n ) is graph isomorphic to N (R m (G), x m ) for any n, m 0. Theorem 3.3. Let R = (H, H) be a nontrivial vertex replacement rule such that the boundary H is replaceable. Let (G, x 0 ) be a pointed graph with a replaceable vertex. Then for any integer M > 0, there exists an N M N so that for all n N M, the ball B M (R n (G), x n ) is isometric to B M (R N M (G), x NM ). Proof. Since R is nontrivial, we have that B(H) > 1. Let M > 0 be given and suppose (G, x 0 ) is a pointed graph (so the base point x 0 is nonreplaceable). Observe that since (H, H) has replaceable boundary, the number of open edges in N (R n (G), x n ) that are adjacent to replaceable vertices is constant for all n. Call this number β. Label the β replaceable vertices in R n (G) that are adjacent to these edges as v 1,n,..., v β,n. Without loss of generality, these vertices are all distinct (otherwise consider (R(G), x 1 ) instead of (G, x 0 )). By Lemma 2.5, we can find a fixed N M so that a(n M ) > 2M. For any n N M, let X n denote the subgraph of R n (G) consisting of N (R n (G), x n ) together with R N M (v i,n Nm ) attached to the vertex v i,n. (Essentially, X n is N (R n (G), x n ) with β copies of R N M 1 (H) attached to it.) Since a(n M ) > 2M, then B M (R n (G), x n ) X n for all n N M. As stated earlier, for any n, m 0, each N (R n (G), x n ) is isomorphic as a graph to N (R m (G), x m ). Since each of the R N M (v i,j ) are

9 isomorphic as graphs, then for any n, m N M there exists a graph isomorphism g n,m from X n onto X m that maps N (R n (G), x n ) onto N (R m (G), x m ) so that g n,m (x n ) = x m and maps each R N M (v i,n NM ) onto R N M (v i,m NM ) so that g n,m (v i,n ) = v i,m for 1 i β. To show that B M (R n (G), x n ) is isometric to B M (R m (G), x m ) for all n, m N M, we will show that g n,m restricted to B M (R n (G), x n ) is an isometry onto B M (R m (G), x m ). To show that g n,m BM (R n (G),x n) maps into B M (R m (G), x m ), pick z B M (R n (G), x n ) and suppose γ is a distance minimizing path between x n and z. Since z B M (R n (G), x n ), then γ has length less than M. By definition of g n,m, g n,m (γ) is a path between x m and g n,m (z) with the same length as γ, which is less than M. Thus, g n,m (z) is in B M (R m (G), x m ). Hence, g n,m (B M (R n (G), x n )) B M (R m (G), x m ). To show that g n,m BM (R n (G),x n ) maps onto B M (R m (G), x m ), pick w B M (R m (G), x m ) and suppose γ is a distance minimizing path between x m and w. Since w B M (R m (G), x m ), then γ has length less than M. By definition of g n,m, gn,m(γ) 1 is a path between x n and the point gn,m(w) 1 having length less than M. Thus, the point gn,m(w) 1 is in B M (R n (G), x n ). Hence, g n,m (B M (R n (G), x n )) = B M (R m (G), x m ). Finally, we show that g n,m BM (R n (G),x n ) is distance preserving by first proving the following claim. Claim 3.4. For all n N M, any distance minimizing path in B M (R n (G), x n ) cannot exit the graph X n. Proof. Let a, b B M (R n (G), x n ) and suppose by way of contradiction that there is a distance minimizing path γ between a and b such that γ X n. Clearly, the length of γ equals d(a, b) and d(a, b) d(a, x n ) + d(x n, b) < 2M. Suppose c is some point of γ outside of X n. This implies that d(c, x n ) > 2M since a(n M ) > 2M. Without loss of generality, we may assume d(c, b) d(a, b)/2. Then d(x n, c) d(x n, b)+d(b, c) < M+ d(a, b)/2 < M + 2M/2 = 2M. So we have a contradiction. Therefore, any distance minimizing path in B M (R n (G), x n ) must not exit X n. Let n N M. Pick z 1, z 2 B M (R n (G), x n ), and let γ be a distance minimizing path between z 1 and z 2. By Claim 3.4, γ is in X n. By definition of g n,m, g n,m (γ) is a path between g n,m (z 1 ) and g n,m (z 2 ) having the same length as γ. Suppose by way of contradiction that g n,m (γ) is not distance minimizing. Then there is a distance minimizing path γ between g n,m (z 1 ) and g n,m (z 2 ) whose length is less than the length of γ. By Claim 3.4, γ is in X m. Thus, gn,m(γ 1 ) is a path between z 1 and z 2 that has the same length as γ. But this contradicts the fact that γ is a distance minimizing path between z 1 and z 2. 9

10 10 Therefore, g n,m (γ) is a distance minimizing path between g n,m (z 1 ) and g n,m (z 2 ). So dist(z 1, z 2 ) = dist(g n,m (z 1 ), g n,m (z 2 )), which implies that g n,m is an isometry when restricted to B M (R n (G), x n ). Thus, we have B M (R n (G), x n ) is isometric to B M (R m (G), x m ) for all n, m N M. For any replacement rule R and any M > 0, let N M be an integer for which the balls B M (R n (G), x n ) are isometric to B M (R N M (G), x NM ) for all n N M. Since each of the balls are eventually isometric for any fixed radius, we can define the limit to be the pointed graph (X, x) so that for any M > 0, B M (X, x) is isometric to B M (R N M (G), x NM ). More formally, the limit (X, x) can be defined as the direct limit obtained by the direct system given by the isometric inclusion maps i j,k : B j (R N j (G), x Nj ) B k (R N k (G), xnk ), for j k. The limit object (X, x) is an infinite pointed graph which also is a geodesic metric space. Thus, the limit of (R n (G), x n ) is this well-defined graph which we will denote as (R (G), x ). 4. The Growth of Graphs In this section we find conditions on a replacement graph (H, H) that guarantee (R (G), x ) has exponential growth (Theorem 4.7) or polynomial growth (Theorem 4.10). In the case of polynomial growth, the growth degree is computed in Theorem 4.10 to be ln(n(h))/ ln(b(h)) if H is replaceable, and 1 if H is nonreplaceable. These values are the same as the Hausdorff dimensions computed in [6] for limits of scaled replacement rules. We first recall the definitions of exponential and polynomial growth of infinite graphs as well as the growth degree of graphs with polynomial growth. Let V (X) denote the set of vertices in a graph X. Definition 4.1. The growth function of a locally finite graph X with respect to a point x is given by ( f X (x, n) = V B n (X, x)). Definition 4.2. We say that a locally finite graph X has exponential growth at a point x if there are constants K > 0 and c > 1 so that f X (x, n) Kc n for all n N. We say X has polynomial growth at x if there are constants c > 0 and d 0 such that f X (x, n) cn d for all n N. (Note that one could construct a graph X having growth function 2 [[ n]]+1 with respect to a vertex x, which is neither polynomial nor exponential. Such graphs are said to have intermediate growth.)

11 Definition 4.3. The growth degree at a point x of a graph X with polynomial growth is inf{d : c R such that n 1, f X(x, n) n d c} for some constant c. In the literature, see [2] or [3, 5] for a good summary. Lemma 4.4. Let X be locally finite with x, y V (X) (1) If X has exponential growth at x then it does at y. (2) 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 with constants K and c so that f X (x, n k) Kc n. Let k = dist(x, y). For 1 k n, f X (y, n) f X (x, n k) Kc n k, which implies that X has exponential growth at y with K = K. c k Next, suppose X has polynomial growth at x with growth degree d. Then there exists c > 0 so that f X (y, n) f X(x, n + k) n d n d ( ) d n + k c. n The right most term is bounded by some number c > 0, 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 that above would show that the growth degree at x would be at most d. Hence, the growth degrees must agree. The above Lemma can be modified to include any arbitrary points x and y that are not necessarily vertices. Example 4.5. Let G be any finite graph with a degree one vertex and let (H, {b}) be as given in Figure 8. Namely, H is homeomorphic to a closed interval having exactly one boundary vertex b and with exactly l 1 > 0 vertices on one side of b and l 2 > 0 vertices on the other side of b. Then (R (G), x ) has exponential growth. Proof. Let G have a replaceable vertex v and let x 0 G be nonreplaceable. Set K = dist(v, x 0 ) and l = l 1 + l 2. After two replacements, x 2 R 2 (G) is at most K + l away from two distinct replaceable vertices. It follows by induction that after 2n replacements, x 2n is at most K + nl away from 2 n distinct replaceable vertices. 11

12 12 Thus, l 1 l 2 b Figure 8. A replacement rule (H, {b}). f (R 2n (G),x 2n )(x 2n, K + nl) 2 n. This implies that for K + nl m K + (n + 1)l, f (R 2n (G),x 2n )(x 2n, m) f (R 2n (G),x 2n )(x 2n, K + nl) 2 n 2 m K l 1. This implies that for all N 2n Therefore, f (R N (G),x N )(x N, m) 2 m K l 1. f (R (G),x )(x, m) 2 m K l 1. So (R (G), x ) has exponential growth and contains an infinite binary tree. Definition 4.6. Let (H, H) be a replacement rule with nonreplaceable boundary. We say (H, H) is branching if there exists a nonreplaceable boundary vertex b R(H) which can be connected via nonreplaceable paths to two distinct replaceable vertices in R(H). Note that if there exists a nonreplaceable boundary vertex b H which can be connected via nonreplaceable paths to two distinct replaceable vertices in H, then clearly (H, H) is branching. The replacement rule in Figure 8 is branching. Figure 9 illustrates a branching rule that does not have this stronger property. Further note that if there exists a nonreplaceable boundary vertex b R n (H) for some n > 1 which can be connected via nonreplaceable paths to two distinct replaceable vertices in R n (H), this implies that (H, H) is branching. Observe that a branching replacement rule (H, H) must have H 2 and have H nonreplaceable.

13 13 R(b 1 ) R(b 2 ) b 1 b 2 b 3 H R(H) R(b 3 ) Figure 9. A branching replacement rule. Recall that although the boundary vertices b 1, b 2, b 3 are degree 3, they not replaceable since H is viewed as R(v) for a replaceable vertex v. Theorem 4.7. Let R = (H, H) be a branching replacement rule. Let (G, x 0 ) be any finite pointed graph that has a replaceable vertex. Then (R (G), x ) has exponential growth. Proof. Let w R(H). Since R is branching, then we may denote by l 1 and l 2 the lengths of any two paths in N (R(H), w) that connect w to two distinct replaceable vertices in (H, H). Let l = l 1 + l 2 and let K = dist(v, x 0 ), where v is some replaceable vertex in G. As in the proof of Example 4.5, after four replacements, x 4 will be at most l away from two distinct replaceable vertices. By induction, the point x 4n is at most K + ln away from 2 n distinct replaceable vertices in R 4n (G), which implies that (R (G), x ) has exponential growth and contains an infinite binary tree. Definition 4.8. Let (H, H) be a nontrivial, nonbranching replacement rule with nonreplaceable boundary and let w H. Define N (H, w) to be all edges and vertices in H that can be connected to w via nonreplaceable paths in H. Similarly, define the set N (R n (H), z), where z R n (H), to be all edges and vertices in R n (H) that can be connected to z via nonreplaceable paths in R n (H).

14 14 Since H is symmetric about H, then for the boundary vertices w i H, the sets N (H, w i ) are either distinct (and isomorphic) or yield the same subset of H. Lemma 4.9. Let (H, H) be a nontrivial, nonbranching replacement rule with nonreplaceable boundary. Then for each z R n (H), the set N (R n (H), z) is adjacent to a unique replaceable vertex in R n (H). Proof. Since (H, H) is nonbranching and nontrivial, then for any w H, the set N (H, w) is adjacent to exactly one replaceable vertex. Case 1: Suppose that for any w H, the set N (H, w) has exactly one edge adjacent to its associated unique replaceable vertex. We claim that for all n, any z R n (H) can be connected via nonreplaceable paths in R n (H) to exactly one edge adjacent to a replaceable vertex in R n (H), which would prove the Lemma in this case. We proceed by way of contradiction. Observe that the case n = 1 holds since (H, H) is nonbranching. Let n 2 be the smallest number so that z R n (H) can be connected via nonreplaceable paths in R n (H) to distinct edges e 1 and e 2, each adjacent to replaceable vertices. Since n is the smallest number so that z R n (H) can be connected via nonreplaceable paths in R n (H) to distinct edges e 1 and e 2, and since the boundary is not replaceable, π(e 1 ) and π(e 2 ) must be a single replaceable vertex v in R n 1 (H). Again, by the assumption on n, there is a unique edge e in R n 1 (H) that is adjacent to v and is accessible to π(z) R n 1 (H) via a nonreplaceable path that goes through e. Thus after replacement, we obtain two nonreplaceable paths through R(e ) to e 1 and e 2 in R(v). However, this yields two nonreplaceable paths in R(v) from the unique boundary vertex of R(v) that is adjacent to R(e ) and e 1 and e 2. The existence of these paths contradicts the original assumption in Case 1. Case 2: Suppose that for w H, there exists k 2 distinct edges in N (H, w) adjacent to the unique replaceable vertex v adjacent to N (H, w). Then after replacement, each of these k edges will be attached to k distinct boundary vertices b 1, b 2,..., b k R(v). Since (H, H) is not branching, N (R(H), R(w)) also is adjacent to a unique replaceable vertex v in R(H). Thus there are nonreplaceable paths in R(v) from b 1, b 2,..., b k to v. If b is any other vertex in R(v) then there also is a nonreplaceable path that connects it to v obtained by the symmetry of H that fixes b 1 and exchanges b 2 and b. Therefore, there is a unique replaceable vertex that is accessible to R(v) via nonreplaceable paths in R(v). Next, we claim that for all n 1, any z R n (H) can be connected via nonreplaceable paths to exactly one replaceable vertex in

15 R n (H). By way of contradiction let n be the smallest number so that z R n (H) can be connected via nonreplaceable paths to distinct replaceable vertices v 1 and v 2. By this assumption on n, the vertices π(v 1 ) and π(v 2 ) must be the same replaceable vertex v in R n 1 (H). However, this implies that R(v) has two replaceable vertices that can be accessed via nonreplaceable paths in R(v) to boundary vertices in R(v), which is a contradiction of the last sentence in the previous paragraph. Thus, for any z R n (H), we have that N (R n (H), z) is adjacent to exactly one replaceable vertex in R n (H). Our last theorem shows that a nonbranching replacement rule has polynomial growth and computes its growth degree. Recall that N(H) denotes the number of replaceable vertices in H and B(H) denotes the minimum number of replaceable vertices on a path connecting distinct boundary vertices of H. Theorem Let (G, x 0 ) be a finite pointed graph with at least one replaceable vertex, and let (H, H) be a nontrivial, nonbranching replacement rule. Then (R (G), x ) has polynomial growth with growth ln B(H) degree replaceable. if H is replaceable and growth degree 1 if H is not Proof. We first handle the case where the boundary of H is replaceable. Since the replacement rule is nontrivial and H is replaceable, then 2 B(H) N(H). We will show that for any integer M > 0, 15 f (R (G),x )(x, M) M d (1) is bounded when d = and is unbounded when d <. ln B(H) ln B(H) Recall that N (G, x 0 ) denotes the set of all edges and vertices in a graph G that can be connected to x 0 via non-replaceable vertices. Also, since H has replaceable boundary, then for all n, m 0, N (R n (G), x n ) is isomorphic to N (R m (G), x m ) and the number β of open edges in N (R n (G), x n ) that are adjacent to replaceable vertices is constant for all n. Finally, recall from the proof of Theorem 3.3 that there exists N M N so that 2M < a(n M ), and for any n N M, X n denotes the subgraph of R n (G) consisting of N (R n (G), x n ) together with β copies of R N M 1 (H) attached to it. For each M > 1a(1), we may find N 2 M so that a(n M 1) 2M < a(n M ). Since 2M < a(n M ), then for all n N M, B M (R n (G), x n ) X n.

16 16 Thus for all n N M, f (R n (G),x n )(x n, M) N (R n (G), x n ) + β R N M 1 (H) N (R n (G), x n ) + β H (1 + N(H) + N(H) N(H) N M 1 ) ( N(H) N (R n N M ) 1 (G), x n ) + β H. N(H) 1 Next, consider the denominator of Expression (1). ( ) M d = M ln B(H) 1 2 a(n ln B(H) M 1) ( ) 1 ln B(H) 2 C[B(H)]N M 1 ( ) 1 = 2 C ln B(H) N(H) N M 1. by Lemma 2.5 Therefore, f (R n (G),x n)(x n, M) M d N (Rn N(H) (G), x n ) + β H ( N M 1 C) ln B(H) N(H) N M 1 ( 1 2 N(H) 1 ), where the right side of the above inequality has a limit as N M, and so is bounded for all M. Thus, Expression (1) is bounded when d =. ln B(H) Next, we show that when d = ϵ, for ϵ > 0, then f (R (G),x )(x,m) ln B(H) M d is unbounded. By Lemma 2.5, there exist positive constants C and C such that C B(H) n b(n) C B(H) n. Recall that since H is replaceable, then for all n, the sets N (R n (G), x n ) are isometric. So N (R n (G), x n ) is constant for all n. For M large enough, there exists n M > 0 so that b(n M + 1) > N (R n (G), x n ) and 3b(n M ) + N (R n (G), x n ) < M 3b(n M + 1) + N (R n (G), x n ).

17 17 To obtain an upper bound on M d, consider M d = M ln B(H) ϵ (3b(n M + 1) + N (R n (G), x n ) ) [ 4 CB(H) ] n M +1 ln B(H) ϵ = (4 C) ln B(H) ϵ B(H) (n M +1) ln B(H) B(H) (n M +1)ϵ = (4 C) ln B(H) ϵ N(H) (n ln B(H) M +1) ln B(H) B(H) (n M +1)ϵ = (4 C) ln B(H) ϵ N(H) n M +1 B(H) (n M +1)ϵ. ln B(H) ϵ By construction of the function b(n), any vertex u in R n M 1 (H) is at most b(n M ) away from its closest boundary vertex in R n M 1 (H). So, by the triangle inequality, u is at most a(n M ) + b(n M ) away from any specified boundary vertex in R n M 1 (H). Since a(n M ) 2b(n M ) and since M 3b(n M )+ N (R n (G), x n ) a(n M )+b(n M )+ N (R n (G), x n ), we have that for n n M, x n R n (G) is within M of every vertex in each of the β copies of R n M 1 (H) attached to N (R n (G), x n ). Since each copy of R n M 1 (H) contains N(H) n M replaceable vertices, the point x n is at most M away from βn(h) n M replaceable vertices. Thus for n n M, f (R n (G),x n)(x n, M) M d βn(h)nm B(H) (nm +1)ϵ = βb(h)(n M +1)ϵ (4 C) ln B(H) ϵ N(H) n M +1 (4 C) ln B(H) ϵ N(H) which is unbounded as n M since B(H) > 1. Hence, f (R (G),x )(x, M) M d is bounded when d = and is unbounded when d <. That ln B(H) ln B(H) is, (R (G), x ) has polynomial growth with growth degree ln B(H) when (H, H) is nonbranching and H is replaceable. Now we prove that (R (G), x ) has polynomial growth with growth degree 1 when (H, H) is nonbranching and H is not replaceable. Since the H is not replaceable, for any n 4M, all paths of length M from x n cannot contain replaceable vertices. So B M (R n (G), x n ) is a subset of N (R n (G), x n ). In particular, B M (R 4M (G), x 4M ) is a subset

18 18 of N (R 4M (G), x 4M ). Furthermore, by Theorem 3.1, for all n 4M, the ball B M (R n (G), x n ) is isometric to B M (R 4M (G), x 4M ). Let S be the set of replaceable vertices in G and let β = v S R(v). By Lemma 4.9, for all n 2, the set N (R n 1 (G), x n 1 ) is adjacent to at most β replaceable vertices. Thus, for all n 2, Thus, N (R n (G), x n ) N (R n 1 (G), x n 1 ) + β N (H, w). N (R 4M (G), x 4M ) N (R(G), x 1 ) + 4Mβ N (H, w). Since for all n 4M, B M (R n (G), x n ) is isometric to B M (R 4M (G), x 4M ) N (R 4M (G), x 4M ), then f (R n (G),x n )(x n, M) N (R(G), x 1 ) + 4Mβ N (H, w). Thus, for all n 4M f (R n (G),x n )(x n, M) M is bounded which, together with the fact that f (R n (G),x n )(x n, M) M implies that (R (G), x ) has growth degree one. References [1] W. Ballmann, Lectures on spaces of nonpositive curvature, Birkhäuser Verlag DMV Seminar Vol 25 (1995) [2] W. Imrich and N. Seifter, A survey on graphs with polynomial growth, Discrete Math., 95 (1991) [3] B. Mohar and W. Woess, A survey on spectra of infinite graphs, Bull. London Math. Soc. 21 (1989) [4] J. Previte, Graph substitutions, Ergodic Theory Dynam. Systems 18 (1998) [5] J. Previte, M. Previte, and M. Vanderschoot, Limits of vertex replacement rules, Rocky Mountain Journal of Mathematics 37 (2007) [6] M. Previte, The dimensions of limits of vertex replacements, Illinois J. of Math. 51 (2007) [7] M. Previte and S.-H. Yang, The topological dimension of limits of vertex replacements, Topology Appl. 153 (2006) [8] M. Previte and S.-H. Yang, A novel way to generate fractals, Amer. Math. Monthly 115 (2008)