Cayley Graphs on Billiard Surfaces
CCayley Graphs on Billiard Surfaces
Joanna GrzegrzolkaLee University [email protected]
Jaime Lynne McCartneyDalton State College [email protected]
Jason SchmurrLee University [email protected]
October 16, 2019
In this article we discuss a connection between two famous constructions in mathematics: a Cayleygraph of a group and a (rational) billiard surface. For each rational billiard surface, there is a naturalway to draw a Cayley graph of a dihedral group on that surface. Both of these objects have theconcept of “genus” attached to them. For the Cayley graph, the genus is defined to be the lowestgenus amongst all surfaces that the graph can be drawn on without edge crossings. We prove thatthe genus of the Cayley graph associated to a billiard surface arising from a triangular billiard tableis always zero or one. One reason this is interesting is that there exist triangular billiard surfaces ofarbitrarily high genus [AI88], so the genus of the associated graph is usually much lower than the genusof the billiard surface.
Let Γ be group. We say that a subset S = { g , g , ..., g n } of Γ is a generating set for Γ if every elementof Γ can be expressed as a product of elements of S . We do not require S to be minimal – that is,we do not exclude the possibility that a proper subset of S may also be a generating set of Γ. The Cayley graph
Cay ( S, Γ) of a group Γ with generating set { g , g , . . . , g n } is a graph whose vertices arethe elements of Γ, and whose edges represent multiplication by an element of the generating set. Wedraw an edge from x to y if y = g i x for some generator g i . For example, in Figure 1 we see a drawingof the Cayley graph Cay ( { a, b, c } , D ), where D is the dihedral group with 6 elements, and { a, b, c } is the set of the three reflection elements in D . In graph theory, this graph is known as the completebipartite graph K , . A graph is bipartite if its vertex set can be partitioned into two subsets V and V such that each element of V is adjacent only to elements of V . Note that in general a Cayley graphis a directed graph. However, because the generating elements we use in this paper all have order 2,we replace the pairs of oppositely directed edges with single undirected edges to form an undirectedgraph. Definition 1
The genus of a graph is the smallest non-negative integer g such that the graph can bedrawn on a surface of genus g . A graph of genus zero is called planar . A classic theorem which can aid in computing graph genusis Kuratowksi’s Theorem.
Theorem 1 (Kuratowski) A graph G is not planar if and only if G contains a subgraph that is asubdivision of either the complete graph K or of the complete bipartite graph K , . a r X i v : . [ m a t h . GN ] O c t ababa(ba) a(ba) a Figure 1: The Cayley graph for the (1,1,1) triangle is isomorphic to K , .A subdivision of a graph G is obtained from that graph by adding vertices of order 2 to the edgesof G . See Figure 6 for an example of a graph which is a subdivision of K , .Another tool for computing the genus of a graph is the concept of a graph rotation . The followingdefinitions are from [HR03]. A rotation of a vertex is an ordered cyclic listing of the vertices adjacentto that vertex. A graph rotation consists of rotations of each vertex of the graph. This term is not tobe confused with the concept of a Euclidean rotation of the plane.A circuit of a graph is a sequence of vertices v i and edges E i of the form v E v E v . . . E n v n suchthat v = v n and such that for each i , E i connects v i − to v i . We say that such a circuit has length n . We may also choose to represent a circuit by listing only the subsequence consisting of the vertices,and omitting the final vertex since it is equal to the first: v . . . v n − .Observe that for any graph G = Cay ( { a, b, c } , D n ) where a , b , and c represent reflections:1. Every circuit v E v E v . . . E n v n corresponds to a relation E E . . . E n = R , where R is theidentity element of D n .2. It follows from the previous observation that since each E i is a Euclidean reflection and R is aEuclidean rotation, circuits in G are always of even length. From a graph theoretic perspective,this is true because G is bipartite: the reflections and rotations form the two bipartite sets.A graph rotation of a graph G induces a set of circuits on G such that each edge is traveled oncein each direction. The circuits are obtained in the following way: the circuit which contains ...v i v j continues as ...v i v j v k , where v k is the vertex directly following v i in the rotation of v j . For example,consider the following rotation of K , : v .v v v v .v v v v .v v v v .v v v v .v v v v .v v v w.v i v j v k means that the vertex w is adjacent to exactly the vertices v i , v j , v k , andthat the cyclic listing of these vertices is v i v j v k .The rotation in our example induces three circuits. This includes two circuits of length four: v v v v and v v v v ; and one circuit of length eight: v v v v v v v v v v .Let r ( ρ ) denote the number of circuits induced by a rotation ρ of a graph G . We call ρ a maximalrotation of G if r ( ρ ) = max ρ i { r ( ρ i } where the ρ i vary over all possible rotations of G .The following formula provides the connection between graph genus and graph rotations.It is relatedto Euler’s characteristic formula. Theorem 2 [HR03] Let G be a connected graph, and let ρ be a maximal rotation of G. Then the genusof G is g , where p − q + r ( ρ ) = 2 − g . The rational polygonal billiard surface is a famous construction in topological dynamics. See [MT06]for an excellent survey. Briefly, a rational polygonal billiard surface is a Riemann surface with aflat metric constructed from a polygon with angles that are rational multiples of π . It arises fromconsideration of the dynamical system of billiards.Although billiard surfaces have served both as motivation for and examples of recent advances insophisticated mathematics such as the work of Mirzhakani on moduli spaces, they have an intuitiveconstruction. When following the path of a point mass bouncing around inside a polygon, when thepoint strikes a wall of the polygon we continue its path in a straight line in a reflected copy of thepolygonal table. Copies of the table that correspond to identical directions of the billiard ball areidentified. This is known as “unfolding” the path. In this way, billiard paths can be viewed as straightlines on a compact surface instead of as a collection of line segments of various slopes.Here is a more precise formulation. Let R be a polygonal region whose interior angles are rationalmultiples of π . Suppose that the internal angles of R are p πn , p πn , . . . p m πn , where n, p , . . . , p m ∈ Z ,and suppose that n is the “least common denominator” in the sense that gcd( n, p , . . . , p m ) = 1. Itcan be shown that the dihedral group generated by (the derivatives of) Euclidean reflections in thesides of R has order 2 n . Suppose a particle moves within this region at constant speed and withinitial direction vector v , changing directions only when it reflects off the sides of R , with the angle ofincidence equaling the angle of reflection. Every subsequent direction vector for the particle is of theform δ · v , for some element δ ∈ D n , where D n acts on R via Euclidean reflections.The rational billiards construction consists of a flat surface corresponding to this physical system.Consider the set D n · R of 2 n copies of R transformed by the elements of D n . For each edge e of R ,we consider the corresponding element ρ e ∈ D n which represents reflection across e . For each δ ∈ D n ,we glue ρ e δ · R and δ · R together along their copies of e . The result is a closed Riemann surfacewith flat structure induced by the tiling by 2 n copies of R . See Figure 2 for a diagram of the surfacearising from billiards in the triangle with angles 3 π
10 , 3 π
10 , and 4 π
10 . The numerical labels indicate sideidentifications. This construction has been described as far back as 1936 by Fox and Kershner in[FK36].
There is a close relationship between a billiard surface X and its associated dihedral group. Whenthe billiard table R is a rational triangle, we can view D n as being generated by the set of threereflections { a, b, c } determined by the sides of R . The triangulation of X by copies of R defines adrawing of a graph on X , called a map in graph theory. See Figure 2. The dual of this map obtainedby exchanging vertices and faces is exactly the Cayley graph Cay ( { a, b, c } , D n ). See Figure 3. To see3
31 34 5 5111 24 6 67 788 99 101011
Figure 2: The billiard surface X (3 , , Cay ( { a, b, c } , D n ), consider some copy of R in the triangulation of X ; call it Rα , where α is an element of D n acting on R on the right. There is another copy Rβ whichshares an edge with Rα and which is obtained by reflecting Rα across an edge congruent to the edge of R corresponding to a . Observe that Rβ = Rα ( α − aα ) = Raα . That is, if we use right-action notation,then the edges of the dual of the triangulation of X do in fact correspond to left-multiplication by theelements { a, b, c } corresponding to reflections in the sides of R .
231 34 5 51 24 6 67 788 99 1010
Figure 3: The graph G (3 , ,
4) arising from X (3 , , Let T ( p , p , p ) denote a rational triangle with internal angles α i = p i πn , where p + p + p = n andgcd( p , p , p ) = 1. Let X ( p , p , p ) denote the billiard surface of T . Write G = G ( p , p , p ) for theCayley graph with generators a , b , c which are reflections across lines through the origin parallel tothe sides of T opposite α , α , α respectively. See Figure 4. Write Γ( p , p , p ) for Cay ( { a, b, c } , D n ).We write Rot ( θ ) to denote Euclidean rotation about the origin by θ and Ref ( θ ) to denote reflectionacross the line through the origin that makes an angle of θ with the positive horizontal axis. Then thefollowing formulas hold: Ref ( θ ) Ref ( θ ) = Rot (2[ θ − θ ])4 c b α α α α Figure 4: The reflections that generate
Cay ( { a, b, c } , D n ). Ref ( θ ) Rot ( θ ) = Ref (cid:18) θ − θ (cid:19) Rot ( θ ) Ref ( θ ) = Ref (cid:18) θ + 12 θ (cid:19) Remark 3
Suppose that our triangle is as in Figure 4, oriented so that the base of the triangle isparallel to the horizontal axis. It follows from the elementary properties of Euclidean reflections androtations listed above that:1. ab is rotation by α ac is rotation by − α aba is reflection across a line making angle α with the positive horizontal axis. Remark 4
There is a correspondence between closed billiard paths on T ( p , p , p ) , cylinders on X ( p , p , p ) and circuits in G ( p , p , p ) . Specifically, every infinite family of parallel equal-lengthclosed billiard paths in T ( p , p , p ) unfolds to a cylinder on X ( p , p , p ) . The sequence of sides of T struck by each path in this family determines a sequence of reflections in Γ( p , p , p ) , which form aclosed circuit in G ( p , p , p ) . For example, every right triangle has an infinite family of billiard paths consisting of striking fourwalls each; this corresponds to the closed circuit abcb in the corresponding Cayley graph (see Lemma1). Similarly, the length six circuit identified in Theorem 5 corresponds to the family of doubles of thewell-known “Fagnano orbit” that exists as a closed billiard path in any acute triangle.See Figure 2 and Figure 3 for an example of a billiard surface and its accompanying Cayley graph.Although the genus of X (3 , ,
4) is 4, we shall show that the genus of G (3 , ,
4) is zero.
The goal of this section is Theorem 5, which states that the genus of G ( p , p , p ) is always zero orone. Lemma 1 G = G ( p , p , p ) has a length four circuit if and only if T = T ( p , p , p ) is either isoscelesor a right triangle. a baaba(ba) a(ba) (ba) a(ba) (ba) a(ba) (ba) a(ba) (ba) a(ba) e Figure 5: A drawing of
Cay ( { a, b, c } , D ). Proof.
Suppose that G ( p , p , p ) has a length four circuit. Then without loss of generality we maysay that the circuit is either abab or abac , using the notation of Figure 4. If abab is a circuit in G then abab is the identity in Γ = Γ( p , p , p ). Since ab ∈ Γ is the Euclidean rotation by 2 α , we then havethat 4 α = 2 π , so that α = π T is a right triangle.If abac is a circuit in G then abac is the identity in Γ. Since ac ∈ Γ is the Euclidean rotation by − α , this implies that 2 α − α = 0; hence α = α and T is isosceles.Conversely, suppose that T ( p , p , p ) is isosceles. Without loss of generality say that p = p .Then we see that abac is the identity in Γ, so abac is a length 4 circuit in G . Finally, suppose insteadthat T ( p , p , p ) is a right triangle. Say α = π abab ∈ Γ is the identity andhence abab is a circuit in G . Lemma 2
For isosceles and right triangles, two of the three reflections generate D n . Proof.
Suppose that T = T ( p , p , p ), with n = p + P + p . We have that ab = Rot (cid:18) p πn (cid:19) . Theorder of ab is the smallest positive integer k such that:( ab ) k = Rot (0)
Rot (cid:18) p kπn (cid:19) = Rot (0) p k ≡ n ). This implies that if gcd( p , n ) = 1 then the order of ab is n . In turn, if ab has order n then itgenerates an index 2 subgroup of D n , and it follows that a and b together generate all of D n .Hence to prove the lemma it suffices to show that one of the p i is relatively prime to n . Firstsuppose that T is isosceles. Without loss of generality say that p = p . Let k = gcd( n, p ). Thensince p = n − p we see that k divides p . But gcd( p , p , p ) = 1, so k = 1. Therefore p and n arerelatively prime.Now suppose instead that T is a right triangle. Without loss of generality, say that α is the rightangle, so that p = n p + p = p . Note that p and p cannot both be even, because if theywere then p would also be even, but gcd( p , p , p ) = 1. So let p be odd and write k = gcd( n, p ).Since p is odd, k is odd. Note that 2 p = n − p , so since k is odd, k divides p by Euclid’s Lemma.6 a baaba(ba) a(ba) (ba) a(ba) (ba) a(ba) (ba) a(ba) (ba) a(ba) e Figure 6: A subgraph of the Cayley graph for an isosceles triangle when n = 7, pictured as a subdivisionof K , . The dashed edges form a subdivision of a single edge of K , , as do the dotted edges. a baaba(ba) a(ba) e(ba) (ba) a(ba) a(ba) (ba) a(ba) Figure 7: The Cayley graph for an isosceles triangle when n = 6.But then k also divides p = p + p , and hence k = 1 since gcd( p , p , p ) = 1. Thus p is relativelyprime to n .Note that it is not the case for all rational triangles that at least one of the p i is relatively primeto n . For example, consider T (5 , , Lemma 3
Let G = G ( p , p , p ) be the Cayley graph for an isosceles triangle with interior angles p πn , p πn , p πn . Then the genus of G is if n is odd and if n is even. Proof.
Suppose that n is odd. We will show that the corresponding graph is not planar. Without lossof generality, we label our isosceles triangle as in Figure 4 so that α = α . Then c = aba and we havea Cayley graph such as the one depicted in Figure 5. Note that Lemma 2 guarantees that the subgraph Cay ( { a, b } , D n ) consists of a single cycle. We choose two sets of three vertices: V = { e, ba, ( ba ) } and V = { a, aba, a ( ba ) } to be our bipartite sets. We already have edges connecting ba to each elementof V ; we also have edges connecting e to a and aba ; and we have edges connecting ( ba ) to aba and a ( ba ) . This leaves two edges to complete a subdivision of K , . As exemplified in Figure 6, we canconnect a to ( ba ) via the sequence a, ( ba ) n − , a ( ba ) n − , ( ba ) n − , a ( ba ) n − , . . . , a ( ba ) , ( ba ) . Similarly,7e can connect e to a ( ba ) via the sequence e, a ( ba ) n − , a ( ba ) n − , a ( ba ) n − , ( ba ) n − , . . . , ( ba ) , a ( ba ) .Hence by Kuratowski’s Theorem, the graph is not planar; that is, its genus is not zero.Now suppose that n is even. We will show that the graph is planar. Since a and b generate D n , wecan draw a loop in the plane with all vertices of D n on it, using all the a and b edges. It remains todraw all the c edges without creating any edge crossing. As exemplified in Figure 7, this can be doneby drawing all the c edges connecting ( ba ) k to a ( ba ) k +1 inside the loop and drawing all the c edgesconnecting ( ba ) k +1 to a ( ba ) k +2 outside the loop. Thus, the graph is planar; that is, its genus is zero.As an example of Lemma 3, consider Figure 8, which illustrates the Cayley graph for the (1,1,5)triangle, drawn on a torus. a ba aba(ba) a(ba) (ba) a(ba) (ba) a(ba) (ba) a(ba) (ba) a(ba) e A AA BB
Figure 8: The Cayley graph for the (1,1,5) triangle, drawn on a torus.
Lemma 4 If T ( p , p , p ) is a right triangle then G = G ( p , p , p ) is not planar. Proof.
As in the previous lemma we will demonstrate a subgraph of G = G ( p , p , p ) isomorphicto a subdivision of K , . Since T is a right triangle, let α = π p = p + p , and so n = p + p + p = 2( p + p ) is even. Since gcd( p , p , p ) = 1, at least one of p and p is odd. Hencewe may let p be odd. It follows that gcd( p , n ) = 1. Therefore k = n kp ≡ n )2 kp ≡ n (mod n ) kp ≡ n n ) Ref (cid:18) kp πn (cid:19) = Ref (cid:18) ( n/ πn (cid:19) c = a ( ba ) k Again, Lemma 2 guarantees that the subgraph
Cay ( { a, b } , D n ) consists of a single cycle. Con-sider Figure 9, which demonstrates that for n > V = { e, ( ba ) , ( ba ) n/ } and V = { aba, a ( ba ) n/ , a ( ba ) n/ } can be used as the bipartite sets for a subgraph of G that is a subdivisionof K , . This diagram suffices for n >
4; if n = 4 then T = T (2 , ,
1) is isosceles so the claim followsfrom Lemma 3.
Theorem 5
The genus of G = G ( p , p , p ) is always 0 or 1. In particular, the genus is zero if andonly if T ( p , p , p ) is isosceles and n = p + p + p is even. a ba aba(ba) a(ba) n/2 (ba) n/2+1 a(ba) n/2+2 Figure 9: A subgraph of the Cayley graph for a right triangle which is isomorphic to a subdivision of K , . The three blue vertices form a bipartite set; the three green blue vertices form the other. Proof.
Since D n has 2 n elements, G has 2 n vertices. Since G is 3-regular, it follows that G has (2 n ) = 3 n edges. Let ρ be a maximal rotation of G . By Theorem 2, 2 n − n + r ( ρ ) = 2 − g , so g = 1 + n − r ( ρ )2 .Now consider the graph rotation ρ defined on G in the following way. Each vertex x of G isadjacent to three other vertices v x,a , v x,b , v x,c via edges with labels a , b , and c , respectively. Using thisnotation, for each x , define the rotation ρ at x to be x.v x,a v x,b v x,c . Then any walk whose first edge is a will begin with the edge string abcabc . Since the element abc ∈ D n is a reflection, we know that it isits own inverse. Hence abcabc is the identity in D n , from which it follows that abcabc always describes acircuit in the directed graph. Indeed, any directed edge in G is a part of such a circuit. Since G has 6 n directed edges, it follows that ρ induces exactly r ( ρ ) = 6 n n circuits. Thus g ≤ n − r ( ρ )2 = 1.That is, the genus of G is at most 1.Next we determine when the genus of G ( p , p , p ) is 0. Suppose that g = 0, and that ρ is amaximal rotation of G . Since all circuits of G have even length of at least 4, it follows that ρ mustinduce a circuit of length 4. By Lemma 1, this can only occur if T ( p , p , p ) is isosceles. Lemma 3finishes the proof. 9 ibliography [FK36] R. Fox and R. Kershner. “Concerning the transitive properties of geodesics on a rationalpolyhedron”. In: Duke Math. J.
J. Geom. Phys.
Pearls in Graph Theory, A Comprehensive Introduction .Dover, 2003. isbn : 9780486432328.[MT06] Howard Masur and Serge Tabachnikov.