On the behavior of growth of polygons in semi-regular hyperbolic tessellations

In this work we consider tessellations (or tilings) of the hyperbolic plane by copies of a semi-regular polygon with alternating angles and we study the behavior of the growth of the polygons, edges, and vertices when the distance increase from a fixed initial polygon. Subjects: Science; Mathematics & Statistics; Advanced Mathematics; Geometry


Introduction
It is a basic fact of hyperbolic geometry that there exists a tessellation (tiling) of the hyperbolic plane by a regular polygon with p sides and with q other p-gons meeting in each vertex if, and only if, 1∕p + 1∕q < 1∕2. Beside the regular case, the study of non regular tilings, which includes archimedian e quasiregular tilings, is an active research topic, see Dolbilin and Schattschneider (1997), Durand (1999), Lucic and Molnár (1990), Martinez (1986), and Sadoc (1990) (with applications in crystallography).
Fixed a p-gon P 0 , with alternating inner angles ∕m and ∕n, let F 1 be the family of copies of P 0 obtained either by reflections (if n = m) or translations of it on its edges. Let F 2 obtained from F 1 in the same way, and so on. Giving the family F k , subset of a tessellation {p, m, n}, how is the behavior of vertices, edges, and polygons when F k grows? If k is really great, from a fixed external vertex v of

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From simple geometric arguments, we may prove that the Euclidean plane may be paved with copies of a same regular polygon using only squares, hexagons, and equilateral triangles. Is the same property true in the Hyperbolic plane? Roughly speaking, hyperbolic geometry is a geometry which does not satisfy Euclide's fifth Axiom: in a plane, given a line and a point not on it, at most one line parallel to the given line can be drawn through the given point. It may proved that there are infinite options of regular polygons to pave the Hyperbolic plane.
On the other hand, when we are interested to pave a plane, either Euclidean or Hyperbolic using non regular polygons, we are faced with a much more complicated problem. In this work, we give conditions for the existence of tilings, using semiregular polygons which have alternating inner angles. F k , is it possible that the tiles which origin from the edges of v will intercept, creating a closed "hole" in some next set F j , j >> k?
This question is meaningful. For instance, if  is a torus and {P j } is a tiling of  by squares, fixing a square P 0 and considering the sets F k as described above, for k great enough, the external tiles of F k will meet, giving origin to a "hole". This may happen even for a noncompact surface. If  is a infinite circular cylinder, considering a square tiling of  we will also get a "hole" for F k with k great enough.
In a more technical setting, in this work we are interested in tessellations (or tilings) of the hyperbolic plane by copies of a semi-regular polygon, which has edges with same length and alternating angles, and the distance in consideration is the natural distance within the usual setting of group theory. Given two of such tiles, we first take an interior point in each of them and consider continuous paths connecting those points avoiding the vertices of the polygons in the tessellation. Then, the distance between the tiles will be the minimal number of edges crossed by such a path. Fixed a polygon in a such tessellation called initial polygon, we study the behavior of the growth of the tessellation from this initial point and we prove several geometric results, including there is not holes when the tessellation grows.
We also call the attention that hyperbolic geometry has become an important mathematical tool in communication theory (Silva, Costa, Firer, & Palazzo, 2006;Silva, Firer & Palazzo, 2003) and quantum error correction codes (Albuquerque, Palazzo, & Silva, 2009), where the tilings play a fundamental role.

The semi-regular hyperbolic tesselations {p, m, n}
Let us consider in the hyperbolic plane 2 a polygon  p, m, n such that it has p equal-length edges and p alternating internal angles , and in this case we have a regular polygon denoted by  p,q . If m ≠ n, then p must be even, and the following theorem (see Beardon, 1983, Th. 7.16.2) guarantees its existence.
Theorem 1 Let m, n and r be integers such that m, n > 2 and r ≥ 2. Then, there exists a polygon P with p = 2r equal-length edges and internal angles alternating between 2 ∕m and 2 ∕n, if and only if Besides, P has incenter.
It is possible to tile 2 with  p, q , for all 1∕p + 1∕q < 1∕2, in the sense there is a family { n |n ∈ } of isometric copies of  p,q such that 2 = ∪ n∈  n and in the case when  n ∩  m ≠ �, this intersection is either a common edge or a common vertex. We say that the family  p, q = { n |n ∈ } defines a tiling (or tessellation) and each  n is called a tile. An edge (vertex) of the tilling is an edge (vertex) of a tile. We denote, respectively, by  p, q and  p,q the set of edges and vertices of  p, q . For some polygons  p, m, n , it is also possible to tile 2 , which is given for the following result.
Theorem 2 Let m, n and l be integers such that m, n, l ≥ 3 and satisfying 1 m + 1 n = 2 l . Then, there exists a family  p, m, n = { j |j ∈ } of polygons  j =  j p,m,n with p = 2k, k ≥ 2, which tile 2 .
Proof Initially, we consider a more general construction. Let k, l be integers, k ≥ 2 and l ≥ 3. Construct a hyperbolic triangle with angles of ∕k, , and such that + = ∕l. Apply reflections in opposite sides to and (also for images of the triangle). As a result, we obtain an equilateral semiregular 2k-gon with alternating angles equal to 2 and 2 . The symmetry group of the 2k-gon is the dihedral group D k , which has the rotation of order k and k reflections in straight lines.
Take the mid-point of each side of the 2k-gon as the center of half-rotation and rotate the polygon around each center by angle of . In this way, we obtain neighbors along all 2k sides of the polygon. Apply the rotation around each vertex of the 2k-gon by angle of ∕l. Repeating the rotations at each vertex of new polygons yields a tiling of the hyperbolic plane. The fundamental domain of this tiling is the first constructed triangle. To obtain desired tilings of theorem, we take the angles and be equal to ∕m and ∕n, respectively. We find such natural numbers m, n, and l by solving the Diophantine equation 1∕m + 1∕n = 2∕l, where m ≥ 3, n ≥ 3, l ≥ 3, m, n, and l are integers. ✷
Of course, if m = n then q = m and we recover the regular case, for any p, q such that 1∕p + 1∕q < 1∕2.
Let us denote by Δ 0 the initial tile of our tessellation and let p 0 be its incenter, and Γ p, m, n denotes the group of isometries which generate the tiling such that every tile Δ n ∈  p, m, n is the image f (Δ 0 ) for a unique f ∈ Γ p, m, n . If we consider on Γ p, m, n the Cayley metric d  (⋅, ⋅), we find that , ∀g, h ∈ Γ p, m, n , we may consider metric constructions in  p, m, n or in Γ p, m, n without any distinction.
Let B k denotes the closed ball in Γ p, m, n with center in the identity and with radius k, and let C k : = B k ⧵B k−1 be the circumference of radius k, which we call the k-th stage (level) of the tessellation: From the correspondence between the elements of the group and the polygons of the tessellation, we denote by P k the set of polygons of tessellation corresponding to the ball B k in the group, and by NP k : = P k ⧵P k−1 the set corresponding to the circumference C k , where the prefix N stands for new: Remark 4 Naturally, we have that where |P k |, the cardinality of P k , is the k-th coefficient of the growth series of the group Γ p, m, n . This suggests, what is actually true, that the techniques used in this work may be used to determine the growth series of the group Γ p, m, n , studied in Bartholdi and Ceccherini-Silberstein (2002) and Cannon (1984), for m = n and for (Floyd & Plotnick, 1994) for some particular cases of m ≠ n.
In this work, we study the growth of the tessellations  p, m, n and hence the growth of the groups Γ p, m, n , considering the topological shape of this growth, in the senses that, for every k ≥ 0, the set P k has a trivial topology.

The growth of the tessellation  p, m, n
Considering the polygons in P k−1 , let us observe that ∈ E k−1 is an edge of either one or two polygons in P k−1 : if is an edge of only one polygon Δ ∈ P k−1 , we have (Δ) (the isometry giving the neighbor polygon in the next level) is a new polygon in NP k ; if is an edge of two polygons in P k−1 , then the polygon gives rise to none new element in P k . If an edge in E k−1 is edge of only one polygon, then ∈ NE k−1 .
We define the function t k :V k → {2, … , q} which for each vertex v ∈ V k gives the number of edges of E k which have v as vertex. We say t k (v) is the k-type of the vertex v, and the notation v w k is used to say that w = t k (v). We also need the following notations: where k is the stage of the tessellation and w Considering an edge ∈ E k , we characterize it using the k-type of both its vertices: we say that has k-type (t k ( ( ), t k ( ( )))), where ( ) and ( ) are the initial and final vertices, respectively, which determine . It is assumed that t k ( ( )) ≤ t k ( ( )) and, without ambiguity, we may assume the nota- In a similar way as done for the vertices, we use the following notation for the set of edges in a given level k: Let us note that the vertices and edges of the tessellation were defined in accordance with the tessellation level and this information will be used to distinguish vertices NV k from V k ∖NV k .
We want to determine the types of vertices and edges which generate new polygons in the next stage and the type of these new vertices and edges.
An edge is k-outer (external) if it is an edge of a unique polygon in the stage k, and we say it is inner (internal) on the contrary case. When an edge : = i j , j k, j has j = q edges adjacent to its terminal vertex, we say that the edges having ( ) as a vertex is a closed cycle of edges if there are also q polygons in P k having ( ) as a vertex. We note that j = q does not imply that we have a closed cycle of edges at j , since there may be an outer edge among those q edges. If i j , j k, j is such an outer edge, then the reflection i j , j k, j closes the cycle of the edges, since q is the maximum number of polygons in the tilling having a common vertex.
k, j } be the set of edges of E k which have v j k, t as a vertex, considering the edges numbered anticlockwise. Since we are interested only in the case of vertices which in the stage in question have some external edge (because otherwise all polygons having it as a vertex had already been counted), we may assume  has one, and consequently, at least, two exterior edges. Thus, we assume the ordinance of the edges of  is done in such a way that i 1 , 1 k, 1 and i j , j k, j , the first and the last edge, are k-exterior edges. Such an ordering of  depends only on the choice of the first edge and, as we will see later (following Proposition 6), there is a canonical way to do it.
We begin with some definitions. Two edges with a common vertex are consecutive if the angle between them (at the common vertex) is either In Proposition 1 will be proved the edges of any given level, having a common vertex is always a set of consecutive edges and this will remove any ambiguity in the numbering of To prove this proposition, we will need some results from the theory of Coxeter groups. To avoid the introduction of new terms, we enunciate those results in the context in question (groups Γ p, q ), replacing the original terms "chambers" and "walls" by "polygons" and "edges," respectively. All of these concepts and results may be find in Humphreys (1990).
Given a set C ⊂ 2 , its (topological) interior is denoted by (C) • and its closure by C. Given a tessellation  with a set of vertices , we say that c:I → 2 is a regular path if c(t ) is a continuous function and c(t) ∉  p, m, n , ∀t ∈ I = [0, 1].
Hence, a support geodesic may be characterized by the existence of two polygons Δ 1 and Δ 2 contained in each one of these two disjoint half-spaces determined by and satisfying that Δ 1 ∩ Δ 2 is an edge contained in and, in this case, we say that supports the polygons Δ 1 , Δ 2 , as well as the edge Δ 1 ∩ Δ 2 .
Let be a support geodesic of the tessellation ( ∈ Γ p, m, n ). We say separates the tiles g(Δ 0 ) and h(Δ 0 ) if those tiles belong to different connected components of 2 ∖ , that is, every continuous path connecting these polygons intercepts .
Let be an edge of the tessellation with initial vertex v and its support geodesic. We denote by Given a support geodesic of the tessellation ∈ S and Δ i , Δ j ∈ P k , we are considering the following notation: Finally, it may be proved that the tessellation {p, q} does not generate, in any stage, a set of discontinuous edges with a common vertex.
Proposition 6 Given a vertex v l k ∈ V k , the set of edges with terminal vertex v l k is a set of consecutive edges.
Proof The proof will be made by absurd, assuming that some vertex does not possess a consecutive set of edges and thus contradicting the minimality of the distance in some level of the tessellation. We keep the notation introduced above for the group Γ p, m, n and the base tile Δ 0 . Whenever there is no ambiguity about the level of the tessellation in consideration, it will be omitted unnecessary indices of the notation.
Let us consider a vertex v = v l k with k-type l and l < q (in the case l = q the edges are consecutive because the cycle of edges is closed). The number of edges is counted module q, since the angle between two consecutive edges is either 2 m or 2 m , and we may have at most q edges with v as a vertex.
Let us suppose by absurd that the edges of v may be separated into two disjoint sets of consecutive edges. We order these edges in the counterclockwise direction, |c| g, h = |{crossed edges by c (t), counted with multiplicities}|.
Δ i | Δ j : = separates the polygons Δ i e Δ j , Δ i ∤ Δ j : = does not separate the polygons Δ i and Δ j .
with r > 1. Let + i be the geodesic rays starting at v, which support the edges of i with i ∈ {1, 2, … , s, s + r, … , l} and − i be the their opposite ray (both + i and − i are contained in the same geodesic i , with initial point v but opposite direction). Let us consider the tiles that have edges in  1 and in  2 , ordered in the counterclockwise direction, such that i and i+1 are the two consecutive edges of Δ i with common vertex v. By the hypothesis, this is a set with discontinuity of the edges and it follows that 1 is not a consecutive edge with l and s is not a consecutive edge with s+r , i.e. the angle 1 between the geodesic rays + 1 and + l which contains 1 and l , respectively, has measure | 1 | > max 2 m , 2 n . In the same way, the angle 2 between the geodesic rays + s and + s+r , which contains s and s+r , respectively, has measure | 2 | > max 2 m , 2 n .
The geodesics 1 , s , s+r and l are not necessarily distinct, but they may not be all equal, since both 1 and 2 are greater than max 2 m , 2 n . Hence, they determine at least two pairs of distinct hyperplanes and Δ 0 is contained in atleast one of them.
We claim that at least one of the exterior tiles Δ 1 , Δ s−1 , Δ s+r , or Δ l is separated from Δ 0 by one of these geodesics. This fact is obvious if it happens that Δ 0 is contained in more than one of these hyperplanes. Also, if Δ 0 is contained in only one of those hyperplanes, we must have 1 = s+r and s = l and also in this case the claim follows easily. Let us suppose, losing no generality, that 1 separates Δ 0 and Δ 1 . Then, every path connecting Δ 0 to Δ 1 must cross the geodesic 1 . Moreover, considering the isometry 1 which copies Δ 1 to the next level, since we are assuming that Δ 1 is a k-external tile, we have that d(Δ 0 , 1 (Δ 1 )) = d(Δ 0 , Δ 1 ) + 1. Let c 0 be a minimal path connecting Δ 0 to Δ 1 , c 1 a minimal path connecting Δ 1 to 1 (Δ 1 ), with the initial point of c 1 coinciding with the final point of c 0 . Thus, the path c obtained by the concatenation of c 0 and c 1 is a path connecting Δ 0 to 1 (Δ 1 ). By the construction, one has |c| = |c 0 | + |c 1 | = k + 1 and since 1 (Δ 1 ) ∈ NP k+1 , c is a minimal path. However, both c 0 and c 1 cross the geodesic 1 . But a minimal path may not cross a reflection geodesic twice, as assured by well-known result concerning to Coxeter groups (as stated in Ronan, 2009, Proposition 2.6). ✷ The following result, tell us about the type of the vertices which appear when the tessellation grows.
Proposition 7 Let  p, m, n be a tessellation of 2 . If q = 2mn∕(m + n) is even, then the k-type of a vertex is always even.
Proof The proof will be done by induction over the stage k of the tessellation, and we remember the maximum number of edges with a common vertex is q = 2s.
Now, let us assume that in the stage k all vertices have even k-type. We prove here the vertex in the stage k + 1 has even (k + 1)-type.
To obtain the tiles in NP k+1 , we need to reflect tiles of NP k in the k-outer edges of the tessellation. By Proposition 6, each vertex with a cycle that is not completed has exactly two k-outer edges. Each one of these edges will contribute in the next stage with a new adjacent edge to this vertex (we may have none if the cycle of edges is completed in the stage k). Since q is even, the two edges generated by the outer edges of the previous stage cannot be the same.
Thus, by the induction hypothesis, the (k + 1)-type of the vertices of NV k+1 is even. Since k-type of vertices is even, the same holds for the edges of the tessellation. ✷ In the next proposition, it is given more precise information about the types of the new vertices in the tessellation.
Proposition 8 Consider a  p, m, n tessellation and let v i k ∈ V k . If q = 2mn∕(m + n) is even, then v i k ∈ NV k if and only if i = 2. If q is odd, then v i k ∈ NV k if and only if either i = 2 or i = 3 and if i = 3 there is an edge with v = i( ) and t k ( ( )) = q.
To prove this proposition, we will need some results and new definitions.
Lemma 9 (duality of consecutive vertices). Let Δ ∈ NP k , where its vertices are numbered in anticlockwise 1 , 2 , … , p . The indices of the vertices may be decomposed into two disjoint sets, I = {i, i + 1, … , i + r} and J = {i + r + 1, i + r + 2, … , i + p − 1}, (where the indices are taken modulo p) in such a way that l ∈ NE k if and only if l ∈ I.
Proof Let q = 2mn∕(m + n) integer. The dual tessellation of  p, m, n is a tessellation  * q constructed as follows: (1) The vertices of  * q are the incenters of the polygons in  p, m, n .
(2) Two vertices v, v ′ in  * q determined by polygons Δ and Δ � in  p, m, n are connected by an edge iff Δ and Δ � have a common edge.
(3) The polygons of  * q are determined by the edges of  * q that intersect any edge of  p,m,n which has a given vertex of  p, m, n as initial or final point.
Given an edge ∈  p, m, n , let us denote by ̃ the unique edge of the dual tessellation  * q that intersects .
Since the vertices of Δ are labeled modulo p, it may assume that 1 ∈ NE k and 1 ∉ NE k−1 . To suppose that the indices of vertices of Δ may not be decomposed as stated, it is equivalent to assume the existence of 1 ≤ r < s ≤ t < p such that The dual edges ̃1,̃r,̃s,̃t,̃p are edges belonging to the k-stage of the dual tessellation which have the following property: But all those edges in  * q share a common vertex, and hence it constitutes a set of non-consecutive vertices, thus contradicting Proposition 6. Thus, the proof is complete. ✷ Definition 10 A k-hole in a tessellation is an open set D ⊂ P k ⊂ 2 with a non trivial fundamental group ( 1 (D) ≠ 0 ) such that its boundary D may be described as a union of edges in NE k .
We remark that if a tessellation has a k-hole, then this is equivalent to state that the fundamental group of P k is nontrivial, i.e. 1 (P k ) ≠ 0 . The next proposition ensures it does not happen.
Proof Suppose that D is a k-hole. Let us assume that k is the minimal stage with such property and let 1 , … , l ∈ NP k be the edges that determine the boundary D, the edges being labeled anticlockwise.
We may have more than one such edge belonging to the same polygon D s , but in this case the previous proposition ensures there are i s and j s ≥ 0 such that i s , i s +1 , … , i s +j s ⊆ D s and these are all such vertices.
To simplify the notation, we denote Thus, each ̂ s is a polygonal line with � s ⊂ D s , D s ∈ NP k and D s ≠ D t for s ≠ t.
and ̂ i and � ⊥ i intersect only at its common initial and final vertices. By Lemma 9, if we write we find that those constitute a set of consecutive edges and � ⊥ i, j ∈ E k−1 , for all j = 1, … , p − j.
Consider the vertices of � ⊥ i labeled in the following way: v 0 , v 1 are the vertices of We observe that for v l (l ≠ 0, p − j i ) one has v l ∈ V k−1 . If we consider the paths ̂ i (obtained by concatenation of the external edges of Δ i ) and � ⊥ i (obtained by concatenation of the inner edges of Δ i ), since their union represents the border of Δ i , a polygon, then ̂ i and � ⊥ i are homotopic in P k . Thus, the paths (obtained by concatenation of the polygonal segments ̂1, … ,̂m) and ⊥ (obtained by concatenation of the polygonal segments ̂⊥ 1 , … ,̂⊥ m ) are homotopic in P k .
We are assuming that D is a k-hole, so that its boundary D is a curve not homotopic to a constant in P k . But D is just the polygonal line , so that is not homotopic to a constant in P k . Since and ⊥ are homotopic in P k , we find that also ⊥ is not homotopic to a constant in P k , and also so in P k−1 , since P k−1 ⊂ P k . But ⊥ ⊂ P k−1 , and this contradicts the minimality of k. ✷

Proof of Proposition 8.
Proof We first prove the proposition for q even, since the proof in the odd case follows the same steps.
(⇒) Let v i k, l ∈ V k with i = 2. We suppose v i k, l ∉ NV k , i.e. v i k, l ∈ NV t for some t < k. By the Proposition 7, the k-type of a vertex is always even, then the t-type of v i k, l in the stage t is at least i = 2, so in the stage k should have type min{2 + 2(k − t), q} ≥ 2, what is an absurd.
We recall that 2 = Δ 1 ∩ Δ 2 (since the edges are consecutive). The edge 1 is the first segment of a line connecting the common vertex {v} = ∩ i j=1 j to P k−1 along the boundary of Δ 1 . There is also a line connecting v to P k−1 such that it has l as initial segment. This gives rise to a k-hole, unless 1 = i . But, in this case, we changed the parity of vertices v when passing from stage k − 1 to stage k, contradicting Proposition 7, and we conclude the proof. ✷ As a consequence, we have that an edge is in NE k if it has a vertex with k-type 2 and the polygon containing the edge = 2, k+1 is a new polygon in the tessellation. Thus, each edge with k-type 2 will give rise to a new polygon in the stage k + 1.
Proof Let v = ( ). Since v has k-type 2 there is another edge � ∈ V k such that ( � ) = v. It follows that and ′ are edges of the same polygon Δ ∈ P k hence n k ( ) = 1. Since both ( ) and ( � ) have k-type 2, we have that , � ∈ NE k and both will give rise to new polygons in the next stage, so that n k+1 ( ) = 3, and concluding the proof. ✷ If 2, k is a k-outer edge, then it contributes with a new polygon in the level k + 1. If < q then each isometry acting on the polygons sharing a same vertex will generate different tiles. However, if = q this tile will be counted twice, since there is another k-outer edge that generates this same polygon. It follows that the vertex v = ( 2, k ) will have k + 1-type 4. In the special case that p = 3 and m = n, this will imply the existence of an edge � ∈ NE k+1 of k = 1-type (4, 4) and we have proved the following: Corollary 13 If = i, k, l ∈ NV k then (i, ) = (2, ) except for the case = q, p = 3 and m = n, when it may happen that (i, ) = (4, 4).