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Connectedness of a suborbital graph for congruence subgroups

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.

1 Introduction

Let Q ˆ =Q{} be the extended rationals and Γ=PSL(2,Z) be the modular group acting on Q ˆ as with the upper half-plane H={zC:Imz>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 adbc=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)={gΓ:c0(modn)} 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)={gΓ:ad1(modn),c0(modn)}, 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 gΓ. 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 (ab if and only if a=b) and the universal relation (ab for all a,b Q ˆ ).

From the above, we can easily verify that

r s x y if and only if xr(modn),ys(modn).

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 γδ 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 , n0, (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 x1(modn) and y0(modn)}. 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.
    xur(modn)

    , yus(modn) and rysx=n or

  2. 2.
    xur(modn)

    , yus(modn) and rysx=n.

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

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

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

Figure 1
figure1

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.
    xur(modn)

    , rysx=n, or

  2. 2.
    xur(modn)

    , rysx=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 rysx=1 or

  2. 2.
    u=1

    and rysx=n or

  3. 3.
    u=n1

    and rysx=n.

Proof Let r s x y be an edge in H u , n . Since r s and x y are in [], x,r1(modn). Therefore, according to Theorem 2, we have 1u(modn), rysx=n or 1u(modn), rysx=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=n1, which gives (3).

For the converse, it is enough to verify (3) only. For this, let u=n1 and rysx=n. Then xr(n1)0(modn). 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 gΓ. Since v, g(v)g(), that is, wg(). 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 mn, then ϕ(v)= n v m is a homomorphism from H u , n to H u , m .

  2. 2.

    Let mn and mn; 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(ab)=ba, 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 mn and mn, m+11(modn). 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), rysx=n. So, sxry=n. That is, using Theorem 3(3), x y n r s n is an edge in H n 1 , n . □

Corollary 1 If mn, 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 vnv, 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 mN, m2, 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 12 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 n2.

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 Rez 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 Rez= 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 Rez=1. These conclude that H 1 , 3 is not connected.

Figure 2
figure2

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 n3.

Proof Since H 1 , n is periodic with period 1, we can, again, work in the strip 1 n Rez 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 nN. 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
figure3

H 1 , n .

Suppose that an edge crosses Rez= 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 Rez=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 n3. □

Theorem 8 H u , n is connected if and only if n2.

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=n1. 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,yZ. If x n < y n , then xy=1. Since x n and y n [], xy0(modn). So, n=1. If x n > y n , then xy=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 n2, 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 n3. 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. □

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Acknowledgements

Dedicated to Professor Hari M Srivastava.

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Kesicioğlu, Y., Akbaş, M. & Beşenk, M. Connectedness of a suborbital graph for congruence subgroups. J Inequal Appl 2013, 117 (2013). https://doi.org/10.1186/1029-242X-2013-117

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Keywords

  • modular groups
  • congruence subgroups
  • suborbital graphs