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Connectedness of a suborbital graph for congruence subgroups
Journal of Inequalities and Applications volume 2013, Article number: 117 (2013)
In this paper, we give necessary and sufficient conditions for the graph to be connected and a forest.
MSC:20H10, 20H05, 05C05, 05C20.
Let be the extended rationals and be the modular group acting on as with the upper half-plane :
where a, b, c, and d are rational integers and .
Jones, Singerman, and Wicks  used the notion of the imprimitive action [2–4] for a Γ-invariant equivalence relation induced on by the congruence subgroup 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 .
In this paper we introduce a different Γ-invariant equivalence relation by using the congruence subgroup instead of and obtain some results for the newly constructed subgraphs . In Section 4 we will prove our main theorems on which give conditions for to be connected or to be a forest, and we work out some relations between the lengths of circuits in and the elliptic elements of the group . As Γ only has finite order elements of orders 2 and 3, the same is true for .
Let , which is one of the congruence subgroups of Γ. Then for each n, where is the stabilizer of ∞ generated by the element , and second inclusion is strict if .
Since, by , Γ acts transitively on , any reduced fraction in equals for some . Hence, we get the following imprimitive Γ-invariant equivalence relation on by :
where and h is similar.
Here, as in , the imprimitivity means that the above relation is different from the identity relation ( if and only if ) and the universal relation ( for all ).
From the above, we can easily verify that
The equivalence classes are called blocks and the block containing is denoted by .
Here we must point out that the above equivalence relation is different from the one in . This is because we take the group instead of . The main reason of changing the equivalence relation lies in the fact that in the case of , as we will see below, the elliptic elements do not necessarily correspond to circuits of the same order. It was the case in .
The modular group Γ acts on through . The orbits are called suborbitals. From the suborbital containing we can form the suborbital graph whose vertices are the elements of and edges are the pairs , which we will denote by and represent them as hyperbolic geodesics in ℋ.
Since Γ acts transitively on , every suborbital contains a pair for , , . In this case, we denote the suborbital graph by for short.
As Γ permutes the blocks transitively, all subgraphs corresponding to blocks are isomorphic. Therefore we will only consider the subgraph of whose vertices form the block , which is the set . The following two results were proved in .
Theorem 1 There is an edge in if and only if either
, and or
, and .
Lemma 1 if and only if and .
The suborbital graph is the familiar Farey graph with if and only if .
As it is illustrated in Figure 1, the pattern is periodic of period 1. That is, if is an edge, then is an edge as well.
Lemma 2 No edges of F cross in ℋ.
Theorem 1 clearly gives the following.
Theorem 2 Let and be in . Then there is an edge in if and only if
, , or
Theorem 3 Let and be in . Then there is an edge in if and only if
Proof Let be an edge in . Since and are in , . Therefore, according to Theorem 2, we have , or , . The first implies that and , which proves (1). The second assures that , or , or , which gives (3).
For the converse, it is enough to verify (3) only. For this, let and . Then . This, by Theorem 2, completes the proof. □
Theorem 4 permutes the vertices and the edges of transitively.
Let v and w be vertices in . Then for some . Since , , that is, . Therefore, lies in the block and so g is in .
The proof for edges is similar. □
Definition 1 Let and be two suborbital graphs. If the map ϕ is an injective function from the vertex set of to that of and sends the edges of to the edges of , then ϕ is called a suborbital graph homomorphism (homomorphism for short) and it will be denoted by .
If , then is a homomorphism from to .
Let and ; then the homomorphism in (1) is not an isomorphism.
Let , given by for all vertices and , be an isomorphism.
Let be in . To see is in is an easy consequence of Theorem 2.
Conversely, suppose that , is an isomorphism. Then there exists a vertex v in such that . Therefore, . But, since and , . That is, is not a vertex in . This gives the proof.
Since the subgraphs and have same set of vertices, ϕ is well defined. Now suppose is an edge in . Then, by Theorem 3(2), . So, . That is, using Theorem 3(3), is an edge in . □
Corollary 1 If , then , is an isomorphism if and only if .
Corollary 2 , given by , is a homomorphism.
Proof Since , Theorem 5(1) gives the result. □
Corollary 3 No edges of cross in ℋ.
Proof By Corollary 2 there is an isomorphism from to a subgraph of F. Also, by Lemma 2, no edges of F cross in ℋ. Therefore the result follows. □
4 Main calculations
In this final section, we state all conditions for to be connected and a forest.
Definition 2 For , , let be vertices of . The configuration (some arrows, not all, may be reversed) is called a circuit of length m.
If , the circuit is said to be a triangle. If , 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 is a 2-gon in and is a triangle in and furthermore we will see below that never becomes a circuit in for .
We now prove the connectedness of separately as follows.
Theorem 6 is connected.
Proof Since the situation, only for this case, coincides with the situation in , it is not necessary to give a proof. □
To understand subsequent proofs better, we start by giving the following example.
Example 1 The subgraph is not connected.
Solution 1 Since is an edge in and is periodic with period 1, we just need consider the strip . It is clear that ∞ is adjacent to and in , but to no intermediate vertices. We will show that no vertices of in the interval are adjacent to vertices of outside this interval. Of course, there is some vertex of , such as , in .
As in Figure 2, suppose that the edge in crosses . Then Corollary 2 implies that is an edge in F and furthermore . This proves that the edges and cross in F, a contradiction. A similar argument shows that no edges of cross . These conclude that is not connected.
Note 1 The graphs and have at least two connected components.
Proof Example 1 and Theorem 5(3) give the result. □
We now give the following.
Theorem 7 is not connected if .
Proof Since is periodic with period 1, we can, again, work in the strip . Note that ∞ is adjacent to and in , but to no intermediate vertices. We will show that no vertices in , between and 1, are adjacent to vertices outside this interval. We note that there are vertices of in for . Indeed as in Figure 3, if n is odd, take the vertex in and if n is even, take the vertex in .
Suppose that an edge crosses , whence that it joins to . 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 , and since vertices between and 1 are not adjacent to ∞, it follows that is not connected.
Consequently, since there is no circuit like in , is not connected for . □
Theorem 8 is connected if and only if .
Proof If , it follows from ; otherwise, it follows from Theorem 7. □
Theorem 9 contains a triangle if and only if .
Proof Let D be a triangle in . From Theorem 3, or . Using Theorem 5(3), we may only work in . By Theorem 4, we may suppose that D has the form or . Let us do calculations only for the first triangle. We easily see that and for some . If , then . Since and , . So, . If , then . Therefore, again, .
Conversely, if , then or 1. But since , we have the triangle . □
Theorem 10 contains a 2-gon if and only if or 2.
Proof Suppose is a 2-gon in . Then, by Theorem 3, it is easily seen that or 2.
Conversely, if or 2, it is clear that is a 2-gon. □
We now give one of our main theorems.
Theorem 11 If , then and are forests.
Proof Let . Assume that is not a forest. Therefore we suppose that there exists a circuit D, other than 2-gon, in . By Theorem 4 and Theorem 3, we may assume that D has the form , where the vertices are all different. Here, since the pattern for the subgraph is periodic with period 1, we may choose the vertices of D, apart from ∞, in the interval . By Theorem 3, or . If , then and give that . Since 1 is not a vertex in , 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 . That is, is a forest.
Now let . If is not a forest, then, as we will see now by Theorem 3, D must be of the form or . As above, we choose the vertices in the finite interval . By Theorem 3, or . If , then, as above, must be . In this case D has the form . As 1 is not a vertex in , Corollary 2 implies that such a circuit D does not occur. Similarly, we can show that if , then there does not exist a circuit D like . Therefore is a forest. Using Theorem 5(3), we see that the subgraph is a forest as well. Therefore the proof is completed. □
Theorem 12 If contains a triangle, then contains an elliptic element of order 3.
Proof If contains a triangle, then by Theorem 9, . So, and is an elliptic element of order 3. □
Remark 1 In general, the converse of Theorem 12 is not true. For example, the element is an elliptic element of order 3, but does not contain a triangle. And also, by Theorem 5(3), does not contain a triangle either.
Remark 2 In , it is shown that the elliptic elements in correspond to circuits in the subgraph of the same order and vice versa. Here, in the case of , owing to Theorem 12 triangles in the subgraph correspond to elliptic elements in of order 3. But the converse is not true as shown in Remark 1.
Theorem 13 contains a 2-gon if and only if contains an elliptic element of order 2.
Proof If contains a 2-gon, then by Theorem 10, or 2. So, is an elliptic element of order 2 in both and .
Conversely, assume that contains an elliptic element of order 2. Then there is an element of of the form such that . From this we get or 2. Hence, the proof now follows from Theorem 10. □
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Dedicated to Professor Hari M Srivastava.
The authors declare that they have no competing interests.
All authors completed the paper together. All authors read and approved the final manuscript.
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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
- modular groups
- congruence subgroups
- suborbital graphs