Open Access

Connectedness of a suborbital graph for congruence subgroups

Journal of Inequalities and Applications20132013:117

https://doi.org/10.1186/1029-242X-2013-117

Received: 5 December 2012

Accepted: 1 March 2013

Published: 20 March 2013

Abstract

In this paper, we give necessary and sufficient conditions for the graph H u , n to be connected and a forest.

MSC:20H10, 20H05, 05C05, 05C20.

Keywords

modular groupscongruence subgroupssuborbital graphs

1 Introduction

Let Q ˆ = Q { } be the extended rationals and Γ = P S L ( 2 , Z ) be the modular group acting on Q ˆ as with the upper half-plane H = { z C : Im z > 0 } :
g = ( a b c d ) : z = x y a z + b c z + d = a x + b y c x + d y ,

where a, b, c, and d are rational integers and a d b c = 1 .

Jones, Singerman, and Wicks [1] used the notion of the imprimitive action [24] for a Γ-invariant equivalence relation induced on Q ˆ by the congruence subgroup Γ 0 ( n ) = { g Γ : c 0 ( mod n ) } to obtain some suborbital graphs and examined their connectedness and forest properties. They left the forest problem as a conjecture, which was settled down by the second author in [5].

In this paper we introduce a different Γ-invariant equivalence relation by using the congruence subgroup Γ 1 ( n ) instead of Γ 0 ( n ) and obtain some results for the newly constructed subgraphs H u , n . In Section 4 we will prove our main theorems on H u , n which give conditions for H u , n to be connected or to be a forest, and we work out some relations between the lengths of circuits in H u , n and the elliptic elements of the group Γ 1 ( n ) . As Γ only has finite order elements of orders 2 and 3, the same is true for Γ 1 ( n ) .

Here, it is worth noting that these concepts are very much related to the binary quadratic forms and modular forms in [6] and [7, 8] respectively.

2 Preliminaries

Let Γ 1 ( n ) = { g Γ : a d 1 ( mod n ) , c 0 ( mod n ) } , which is one of the congruence subgroups of Γ. Then Γ < Γ 1 ( n ) Γ for each n, where Γ is the stabilizer of ∞ generated by the element ( 1 1 0 1 ) , and second inclusion is strict if n > 1 .

Since, by [1], Γ acts transitively on Q ˆ , any reduced fraction r s in Q ˆ equals g ( ) for some g Γ . Hence, we get the following imprimitive Γ-invariant equivalence relation on Q ˆ by Γ 1 ( n ) :
r s x y if and only if  g 1 h Γ 1 ( n ) ,

where g = ( r s ) and h is similar.

Here, as in [1], the imprimitivity means that the above relation is different from the identity relation ( a b if and only if a = b ) and the universal relation ( a b for all a , b Q ˆ ).

From the above, we can easily verify that
r s x y if and only if  x r ( mod n ) , y s ( mod n ) .

The equivalence classes are called blocks and the block containing x y is denoted by [ x y ] .

Here we must point out that the above equivalence relation is different from the one in [1]. This is because we take the group Γ 1 ( n ) instead of Γ 0 ( n ) . The main reason of changing the equivalence relation lies in the fact that in the case of Γ 1 ( n ) , as we will see below, the elliptic elements do not necessarily correspond to circuits of the same order. It was the case in [5].

3 Subgraphs H u , n

The modular group Γ acts on Q ˆ × Q ˆ through g : ( α , β ) ( g ( α ) , g ( β ) ) . The orbits are called suborbitals. From the suborbital O ( α , β ) containing ( α , β ) we can form the suborbital graph G ( α , β ) whose vertices are the elements of Q ˆ and edges are the pairs ( γ , δ ) O ( α , β ) , which we will denote by γ δ and represent them as hyperbolic geodesics in .

Since Γ acts transitively on Q ˆ , every suborbital O ( α , β ) contains a pair ( , u n ) for u n Q ˆ , n 0 , ( u , n ) = 1 . In this case, we denote the suborbital graph by G u , n for short.

As Γ permutes the blocks transitively, all subgraphs corresponding to blocks are isomorphic. Therefore we will only consider the subgraph H u , n of G u , n whose vertices form the block [ ] = [ 1 0 ] , which is the set { x y Q ˆ x 1 ( mod n )  and  y 0 ( mod n ) } . The following two results were proved in [1].

Theorem 1 There is an edge r s x y in G u , n if and only if either
  1. 1.
    x u r ( mod n )
    , y u s ( mod n ) and r y s x = n or
     
  2. 2.
    x u r ( mod n )
    , y u s ( mod n ) and r y s x = n .
     

Lemma 1 G u , n = G v , m if and only if n = m and u v ( mod n ) .

The suborbital graph F : = G 1 , 1 is the familiar Farey graph with a b c d if and only if a d b c = ± 1 .

As it is illustrated in Figure 1, the pattern is periodic of period 1. That is, if x y is an edge, then x + 1 y + 1 is an edge as well.
Figure 1

Farey graph.

Lemma 2 No edges of F cross in .

Theorem 1 clearly gives the following.

Theorem 2 Let r s and x y be in [ ] . Then there is an edge r s x y in H u , n if and only if
  1. 1.
    x u r ( mod n )
    , r y s x = n , or
     
  2. 2.
    x u r ( mod n )
    , r y s x = n .
     
Theorem 3 Let r s and x y be in [ ] . Then there is an edge r s x y in H u , n if and only if
  1. 1.
    u = 0
    and r y s x = 1 or
     
  2. 2.
    u = 1
    and r y s x = n or
     
  3. 3.
    u = n 1
    and r y s x = n .
     

Proof Let r s x y be an edge in H u , n . Since r s and x y are in [ ] , x , r 1 ( mod n ) . Therefore, according to Theorem 2, we have 1 u ( mod n ) , r y s x = n or 1 u ( mod n ) , r y s x = n . The first implies that u = 0 and u = 1 , which proves (1). The second assures that u = 0 , n = 1 or u = 1 , n = 2 or u = n 1 , which gives (3).

For the converse, it is enough to verify (3) only. For this, let u = n 1 and r y s x = n . Then x r ( n 1 ) 0 ( mod n ) . This, by Theorem 2, completes the proof. □

Theorem 4 Γ 1 ( n ) permutes the vertices and the edges of H u , n transitively.

Proof
  1. 1.

    Let v and w be vertices in H u , n . Then w = g ( v ) for some g Γ . Since v , g ( v ) g ( ) , that is, w g ( ) . Therefore, g ( ) lies in the block [ ] and so g is in Γ 1 ( n ) .

     
  2. 2.

    The proof for edges is similar. □

     

Definition 1 Let H u , n and H v , m be two suborbital graphs. If the map ϕ is an injective function from the vertex set of H u , n to that of H v , m and sends the edges of H u , n to the edges of H v , m , then ϕ is called a suborbital graph homomorphism (homomorphism for short) and it will be denoted by ϕ : H u , n H v , m .

Theorem 5
  1. 1.

    If m n , then ϕ ( v ) = n v m is a homomorphism from H u , n to H u , m .

     
  2. 2.

    Let m n and m n ; then the homomorphism in (1) is not an isomorphism.

     
  3. 3.

    Let ϕ : H 1 , n H n 1 , n , given by ϕ ( a ) = a for all vertices and ϕ ( a b ) = b a , be an isomorphism.

     
Proof
  1. 1.

    Let r s n x y n be in H u , n . To see r s m x y m is in H u , m is an easy consequence of Theorem 2.

     
  2. 2.

    Conversely, suppose that h : H u , n H u , m , h ( v ) = n v m is an isomorphism. Then there exists a vertex v in H u , n such that h ( v ) = m + 1 m . Therefore, v = m + 1 n . But, since m n and m n , m + 1 1 ( mod n ) . That is, m + 1 n is not a vertex in H u , n . This gives the proof.

     
  3. 3.

    Since the subgraphs H 1 , n and H n 1 , n have same set of vertices, ϕ is well defined. Now suppose r s n x y n is an edge in H 1 , n . Then, by Theorem 3(2), r y s x = n . So, s x r y = n . That is, using Theorem 3(3), x y n r s n is an edge in H n 1 , n . □

     

Corollary 1 If m n , then H u , n H u , m , v n v m is an isomorphism if and only if m = n .

Corollary 2 ϕ : H u , n F , given by v n v , is a homomorphism.

Proof Since H u , 1 = F , Theorem 5(1) gives the result. □

Corollary 3 No edges of H u , n cross in .

Proof By Corollary 2 there is an isomorphism from H u , n to a subgraph of F. Also, by Lemma 2, no edges of F cross in . Therefore the result follows. □

4 Main calculations

In this final section, we state all conditions for H u , n to be connected and a forest.

Definition 2 For m N , m 2 , let v 1 , v 2 , , v m be vertices of H u , n . The configuration v 1 v 2 v m v 1 (some arrows, not all, may be reversed) is called a circuit of length m.

If m = 3 , the circuit is said to be a triangle. If m = 2 , we call the self paired edge a 2-gon.

A graph is called a forest if it contains no circuits other than 2-gons.

As in examples 1 2 is a 2-gon in H 1 , 2 and 1 2 is a triangle in H 1 , 1 and furthermore we will see below that v 1 v 2 never becomes a circuit in H 1 , n for n 2 .

We now prove the connectedness of H 1 , n separately as follows.

Theorem 6 H 1 , 2 is connected.

Proof Since the situation, only for this case, coincides with the situation in [1], it is not necessary to give a proof. □

To understand subsequent proofs better, we start by giving the following example.

Example 1 The subgraph H 1 , 3 is not connected.

Solution 1 Since 1 3 is an edge in H 1 , 3 and H 1 , 3 is periodic with period 1, we just need consider the strip 1 3 Re z 4 3 . It is clear that ∞ is adjacent to 1 3 and 4 3 in H 1 , 3 , but to no intermediate vertices. We will show that no vertices of H 1 , 3 in the interval [ 2 3 , 1 ] are adjacent to vertices of H 1 , 3 outside this interval. Of course, there is some vertex of H 1 , 3 , such as 19 27 , in [ 2 3 , 1 ] .

As in Figure 2, suppose that the edge a 3 b c 3 d in H 1 , 3 crosses Re z = 2 3 . Then Corollary 2 implies that a b c d is an edge in F and furthermore a b < 2 < c d . This proves that the edges a b c d and 2 cross in F, a contradiction. A similar argument shows that no edges of H 1 , 3 cross Re z = 1 . These conclude that H 1 , 3 is not connected.
Figure 2

H 1 , 3 .

Note 1 The graphs H 1 , 3 and H 2 , 3 have at least two connected components.

Proof Example 1 and Theorem 5(3) give the result. □

We now give the following.

Theorem 7 H 1 , n is not connected if n 3 .

Proof Since H 1 , n is periodic with period 1, we can, again, work in the strip 1 n Re z n + 1 n . Note that ∞ is adjacent to 1 n and n + 1 n in H 1 , n , but to no intermediate vertices. We will show that no vertices in H 1 , n , between 2 n and 1, are adjacent to vertices outside this interval. We note that there are vertices of H 1 , n in ( 2 n , 1 ) for n N . Indeed as in Figure 3, if n is odd, take the vertex 6 n + 1 9 n in ( 2 n , 1 ) and if n is even, take the vertex n + 1 2 n in ( 2 n , 1 ) .
Figure 3

H 1 , n .

Suppose that an edge crosses Re z = 2 n , whence that it joins v = a n b to w = c n d . By Corollary 2, nv and nw must be adjacent in F. As in Example 1, this is a contradiction. A similar argument shows that no edge crosses Re z = 1 , and since vertices between 2 n and 1 are not adjacent to ∞, it follows that H 1 , n is not connected.

Consequently, since there is no circuit like 1 n v 1 n + 1 n in H 1 , n , H 1 , n is not connected for n 3 . □

Theorem 8 H u , n is connected if and only if n 2 .

Proof If n = 1 , 2 , it follows from [1]; otherwise, it follows from Theorem 7. □

Theorem 9 H u , n contains a triangle if and only if n = 1 .

Proof Let D be a triangle in H u , n . From Theorem 3, u = 1 or u = n 1 . Using Theorem 5(3), we may only work in H 1 , n . By Theorem 4, we may suppose that D has the form v 1 v 2 or v 1 v 2 . Let us do calculations only for the first triangle. We easily see that v 1 = x n and v 2 = y n for some x , y Z . If x n < y n , then x y = 1 . Since x n and y n [ ] , x y 0 ( mod n ) . So, n = 1 . If x n > y n , then x y = 1 . Therefore, again, n = 1 .

Conversely, if n = 1 , then u = 0 or 1. But since H 0 , 1 = H 1 , 1 , we have the triangle 1 0 1 1 2 1 1 0 . □

Theorem 10 H u , n contains a 2-gon if and only if n = 1 or 2.

Proof Suppose x k n y l n x k n is a 2-gon in H u , n . Then, by Theorem 3, it is easily seen that n = 1 or 2.

Conversely, if n = 1 or 2, it is clear that 1 0 1 n 1 0 is a 2-gon. □

We now give one of our main theorems.

Theorem 11 If n 2 , then H 1 , n and H n 1 , n are forests.

Proof Let n = 2 . Assume that H 1 , 2 is not a forest. Therefore we suppose that there exists a circuit D, other than 2-gon, in H 1 , 2 . By Theorem 4 and Theorem 3, we may assume that D has the form v 1 v k , where the vertices v 1 , v 2 , , v k are all different. Here, since the pattern for the subgraph H u , n is periodic with period 1, we may choose the vertices of D, apart from ∞, in the interval [ 1 2 , 3 2 ] . By Theorem 3, v 1 = 1 2 or 3 2 . If v 1 = 1 2 , then v k = 2 a + 1 2 [ 1 2 , 3 2 ] and v 1 v k give that v k = 3 2 . Since 1 is not a vertex in H 1 , 2 , Corollary 2 implies that such a circuit D does not occur. Similarly, we can show that there is not a circuit D in the case where v 1 = 3 2 . That is, H 1 , 2 is a forest.

Now let n 3 . If H 1 , n is not a forest, then, as we will see now by Theorem 3, D must be of the form v 1 v k or v 1 v k . As above, we choose the vertices in the finite interval [ 1 n , n + 1 n ] . By Theorem 3, v 1 = 1 n or n + 1 n . If v 1 = 1 n , then, as above, v k must be n + 1 n . In this case D has the form 1 n n + 1 n . As 1 is not a vertex in H 1 , n , Corollary 2 implies that such a circuit D does not occur. Similarly, we can show that if v 1 = n + 1 n , then there does not exist a circuit D like n + 1 n 1 n . Therefore H 1 , n is a forest. Using Theorem 5(3), we see that the subgraph H n 1 , n is a forest as well. Therefore the proof is completed. □

Theorem 12 If H u , n contains a triangle, then Γ 1 ( n ) contains an elliptic element of order 3.

Proof If H u , n contains a triangle, then by Theorem 9, n = 1 . So, Γ 1 ( 1 ) = Γ and ( 1 1 3 2 ) Γ 1 ( 1 ) is an elliptic element of order 3. □

Remark 1 In general, the converse of Theorem 12 is not true. For example, the element ( 1 1 3 2 ) Γ 1 ( 3 ) is an elliptic element of order 3, but H 1 , 3 does not contain a triangle. And also, by Theorem 5(3), H 2 , 3 does not contain a triangle either.

Remark 2 In [5], it is shown that the elliptic elements in Γ 0 ( n ) correspond to circuits in the subgraph F u , n of the same order and vice versa. Here, in the case of Γ 1 ( n ) , owing to Theorem 12 triangles in the subgraph H u , n correspond to elliptic elements in Γ 1 ( n ) of order 3. But the converse is not true as shown in Remark 1.

Theorem 13 H u , n contains a 2-gon if and only if Γ 1 ( n ) contains an elliptic element of order  2.

Proof If H u , n contains a 2-gon, then by Theorem 10, n = 1 or 2. So, ( 1 1 2 1 ) is an elliptic element of order 2 in both Γ 1 ( 1 ) and Γ 1 ( 2 ) .

Conversely, assume that Γ 1 ( n ) contains an elliptic element of order 2. Then there is an element of Γ 1 ( n ) of the form ( 1 + a n b c n 1 + d n ) such that 2 + ( a + d ) n = 0 . From this we get n = 1 or 2. Hence, the proof now follows from Theorem 10. □

Declarations

Acknowledgements

Dedicated to Professor Hari M Srivastava.

Authors’ Affiliations

(1)
Department of Mathematics, Recep Tayyip Erdoğan University
(2)
Department of Mathematics, Karadeniz Technical University

References

  1. Jones GA, Singerman D, Wicks K: The modular group and generalized Farey graphs. LMS Lect. Note Ser 160. Groups St Andrews 1989 1991, 316–338.Google Scholar
  2. Biggs NL, White AT LMS Lect. Note Ser. In Permutation Groups and Combinatorial Structures. Cambridge University Press, Cambridge; 1979.View ArticleGoogle Scholar
  3. Neumann PM: Finite permutation groups and edge-coloured graphs and matrices. In Topics in Group Theory and Computation. Edited by: Curran MPJ. Academic Press, London; 1977.Google Scholar
  4. Sims CC: Graphs and finite permutation groups. Math. Z. 1967, 95: 76–86. 10.1007/BF01117534MathSciNetView ArticleGoogle Scholar
  5. Akbaş M: On suborbital graphs for the modular group. Bull. Lond. Math. Soc. 2001, 33: 647–652. 10.1112/S0024609301008311View ArticleGoogle Scholar
  6. Hurwitz A: Über die reduktion der binären quadratischen formen. Math. Ann. 1894, 45: 85–117. 10.1007/BF01447227MathSciNetView ArticleGoogle Scholar
  7. Serre JP: A Course in Arithmetic. Springer, New York; 1973.View ArticleGoogle Scholar
  8. Schoeneberg B: Elliptic Modular Functions: an Introduction. Springer, Berlin; 1974.View ArticleGoogle Scholar

Copyright

© Kesicioğlu et al.; licensee Springer 2013

This article is published under license to BioMed Central Ltd. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.