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Connectedness of a suborbital graph for congruence subgroups
Journal of Inequalities and Applications volume 2013, Article number: 117 (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.
1 Introduction
Let \stackrel{\u02c6}{\mathbb{Q}}=\mathbb{Q}\cup \{\mathrm{\infty}\} be the extended rationals and \mathrm{\Gamma}=PSL(2,\mathbb{Z}) be the modular group acting on \stackrel{\u02c6}{\mathbb{Q}} as with the upper halfplane \mathcal{H}=\{z\in \mathbb{C}:Imz>0\}:
where a, b, c, and d are rational integers and adbc=1.
Jones, Singerman, and Wicks [1] used the notion of the imprimitive action [2–4] for a Γinvariant equivalence relation induced on \stackrel{\u02c6}{\mathbb{Q}} by the congruence subgroup {\mathrm{\Gamma}}_{0}(n)=\{g\in \mathrm{\Gamma}:c\equiv 0\phantom{\rule{0.25em}{0ex}}(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 Γinvariant equivalence relation by using the congruence subgroup {\mathrm{\Gamma}}_{1}(n) instead of {\mathrm{\Gamma}}_{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 {\mathrm{\Gamma}}_{1}(n). As Γ only has finite order elements of orders 2 and 3, the same is true for {\mathrm{\Gamma}}_{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 {\mathrm{\Gamma}}_{1}(n)=\{g\in \mathrm{\Gamma}:a\equiv d\equiv 1\phantom{\rule{0.25em}{0ex}}(modn),c\equiv 0\phantom{\rule{0.25em}{0ex}}(modn)\}, which is one of the congruence subgroups of Γ. Then {\mathrm{\Gamma}}_{\mathrm{\infty}}<{\mathrm{\Gamma}}_{1}(n)\u2a7d\mathrm{\Gamma} for each n, where {\mathrm{\Gamma}}_{\mathrm{\infty}} is the stabilizer of ∞ generated by the element \left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right), and second inclusion is strict if n>1.
Since, by [1], Γ acts transitively on \stackrel{\u02c6}{\mathbb{Q}}, any reduced fraction \frac{r}{s} in \stackrel{\u02c6}{\mathbb{Q}} equals g(\mathrm{\infty}) for some g\in \mathrm{\Gamma}. Hence, we get the following imprimitive Γinvariant equivalence relation on \stackrel{\u02c6}{\mathbb{Q}} by {\mathrm{\Gamma}}_{1}(n):
where g=\left(\begin{array}{cc}r& \ast \\ s& \ast \end{array}\right) and h is similar.
Here, as in [1], the imprimitivity means that the above relation is different from the identity relation (a\sim b if and only if a=b) and the universal relation (a\sim b for all a,b\in \stackrel{\u02c6}{\mathbb{Q}}).
From the above, we can easily verify that
The equivalence classes are called blocks and the block containing \frac{x}{y} is denoted by [\frac{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 {\mathrm{\Gamma}}_{1}(n) instead of {\mathrm{\Gamma}}_{0}(n). The main reason of changing the equivalence relation lies in the fact that in the case of {\mathrm{\Gamma}}_{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 \stackrel{\u02c6}{\mathbb{Q}}\times \stackrel{\u02c6}{\mathbb{Q}} through g:(\alpha ,\beta )\to (g(\alpha ),g(\beta )). The orbits are called suborbitals. From the suborbital O(\alpha ,\beta ) containing (\alpha ,\beta ) we can form the suborbital graph G(\alpha ,\beta ) whose vertices are the elements of \stackrel{\u02c6}{\mathbb{Q}} and edges are the pairs (\gamma ,\delta )\in O(\alpha ,\beta ), which we will denote by \gamma \to \delta and represent them as hyperbolic geodesics in ℋ.
Since Γ acts transitively on \stackrel{\u02c6}{\mathbb{Q}}, every suborbital O(\alpha ,\beta ) contains a pair (\mathrm{\infty},\frac{u}{n}) for \frac{u}{n}\in \stackrel{\u02c6}{\mathbb{Q}}, n\ge 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 [\mathrm{\infty}]=[\frac{1}{0}], which is the set \{\frac{x}{y}\in \stackrel{\u02c6}{\mathbb{Q}}\mid x\equiv 1\phantom{\rule{0.25em}{0ex}}(modn)\text{and}y\equiv 0\phantom{\rule{0.25em}{0ex}}(modn)\}. The following two results were proved in [1].
Theorem 1 There is an edge \frac{r}{s}\to \frac{x}{y} in {G}_{u,n} if and only if either

1.
x\equiv ur\phantom{\rule{0.25em}{0ex}}(modn)
, y\equiv us\phantom{\rule{0.25em}{0ex}}(modn) and rysx=n or

2.
x\equiv ur\phantom{\rule{0.25em}{0ex}}(modn)
, y\equiv us\phantom{\rule{0.25em}{0ex}}(modn) and rysx=n.
Lemma 1 {G}_{u,n}={G}_{v,m} if and only if n=m and u\equiv v\phantom{\rule{0.25em}{0ex}}(modn).
The suborbital graph F:={G}_{1,1} is the familiar Farey graph with \frac{a}{b}\to \frac{c}{d} if and only if adbc=\pm 1.
As it is illustrated in Figure 1, the pattern is periodic of period 1. That is, if x\to y is an edge, then x+1\to y+1 is an edge as well.
Lemma 2 No edges of F cross in ℋ.
Theorem 1 clearly gives the following.
Theorem 2 Let \frac{r}{s} and \frac{x}{y} be in [\mathrm{\infty}]. Then there is an edge \frac{r}{s}\to \frac{x}{y} in {H}_{u,n} if and only if

1.
x\equiv ur\phantom{\rule{0.25em}{0ex}}(modn)
, rysx=n, or

2.
x\equiv ur\phantom{\rule{0.25em}{0ex}}(modn)
, rysx=n.
Theorem 3 Let \frac{r}{s} and \frac{x}{y} be in [\mathrm{\infty}]. Then there is an edge \frac{r}{s}\to \frac{x}{y} in {H}_{u,n} if and only if

1.
u=0
and rysx=1 or

2.
u=1
and rysx=n or

3.
u=n1
and rysx=n.
Proof Let \frac{r}{s}\to \frac{x}{y} be an edge in {H}_{u,n}. Since \frac{r}{s} and \frac{x}{y} are in [\mathrm{\infty}], x,r\equiv 1\phantom{\rule{0.25em}{0ex}}(modn). Therefore, according to Theorem 2, we have 1\equiv u\phantom{\rule{0.25em}{0ex}}(modn), rysx=n or 1\equiv u\phantom{\rule{0.25em}{0ex}}(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 x\equiv r(n1)\equiv 0\phantom{\rule{0.25em}{0ex}}(modn). This, by Theorem 2, completes the proof. □
Theorem 4 {\mathrm{\Gamma}}_{1}(n) permutes the vertices and the edges of {H}_{u,n} transitively.
Proof

1.
Let v and w be vertices in {H}_{u,n}. Then w=g(v) for some g\in \mathrm{\Gamma}. Since v\sim \mathrm{\infty}, g(v)\sim g(\mathrm{\infty}), that is, w\sim g(\mathrm{\infty}). Therefore, g(\mathrm{\infty}) lies in the block [\mathrm{\infty}] and so g is in {\mathrm{\Gamma}}_{1}(n).

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 \varphi :{H}_{u,n}\to {H}_{v,m}.
Theorem 5

1.
If m\mid n, then \varphi (v)=\frac{nv}{m} is a homomorphism from {H}_{u,n} to {H}_{u,m}.

2.
Let m\mid n and m\ne n; then the homomorphism in (1) is not an isomorphism.

3.
Let \varphi :{H}_{1,n}\to {H}_{n1,n}, given by \varphi (a)=a for all vertices and \varphi (a\to b)=b\to a, be an isomorphism.
Proof

1.
Let \frac{r}{sn}\to \frac{x}{yn} be in {H}_{u,n}. To see \frac{r}{sm}\to \frac{x}{ym} is in {H}_{u,m} is an easy consequence of Theorem 2.

2.
Conversely, suppose that h:{H}_{u,n}\to {H}_{u,m}, h(v)=\frac{nv}{m} is an isomorphism. Then there exists a vertex v in {H}_{u,n} such that h(v)=\frac{m+1}{m}. Therefore, v=\frac{m+1}{n}. But, since m\mid n and m\ne n, m+1\not\equiv 1\phantom{\rule{0.25em}{0ex}}(modn). That is, \frac{m+1}{n} is not a vertex in {H}_{u,n}. This gives the proof.

3.
Since the subgraphs {H}_{1,n} and {H}_{n1,n} have same set of vertices, ϕ is well defined. Now suppose \frac{r}{sn}\to \frac{x}{yn} is an edge in {H}_{1,n}. Then, by Theorem 3(2), rysx=n. So, sxry=n. That is, using Theorem 3(3), \frac{x}{yn}\to \frac{r}{sn} is an edge in {H}_{n1,n}. □
Corollary 1 If m\mid n, then {H}_{u,n}\to {H}_{u,m}, v\to \frac{nv}{m} is an isomorphism if and only if m=n.
Corollary 2 \varphi :{H}_{u,n}\to F, given by v\to nv, 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 ℋ.
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 ℋ. 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\in \mathbb{N}, m\ge 2, let {v}_{1},{v}_{2},\dots ,{v}_{m} be vertices of {H}_{u,n}. The configuration {v}_{1}\to {v}_{2}\to \cdots \to {v}_{m}\to {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 2gon.
A graph is called a forest if it contains no circuits other than 2gons.
As in examples \mathrm{\infty}\to \frac{1}{2}\to \mathrm{\infty} is a 2gon in {H}_{1,2} and \mathrm{\infty}\to 1\to 2\to \mathrm{\infty} is a triangle in {H}_{1,1} and furthermore we will see below that \mathrm{\infty}\to {v}_{1}\to {v}_{2}\to \cdots never becomes a circuit in {H}_{1,n} for n\ge 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 \mathrm{\infty}\to \frac{1}{3} is an edge in {H}_{1,3} and {H}_{1,3} is periodic with period 1, we just need consider the strip \frac{1}{3}\le Rez\le \frac{4}{3}. It is clear that ∞ is adjacent to \frac{1}{3} and \frac{4}{3} in {H}_{1,3}, but to no intermediate vertices. We will show that no vertices of {H}_{1,3} in the interval [\frac{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 \frac{19}{27}, in [\frac{2}{3},1].
As in Figure 2, suppose that the edge \frac{a}{3b}\to \frac{c}{3d} in {H}_{1,3} crosses Rez=\frac{2}{3}. Then Corollary 2 implies that \frac{a}{b}\to \frac{c}{d} is an edge in F and furthermore \frac{a}{b}<2<\frac{c}{d}. This proves that the edges \frac{a}{b}\to \frac{c}{d} and \mathrm{\infty}\to 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.
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\ge 3.
Proof Since {H}_{1,n} is periodic with period 1, we can, again, work in the strip \frac{1}{n}\le Rez\le \frac{n+1}{n}. Note that ∞ is adjacent to \frac{1}{n} and \frac{n+1}{n} in {H}_{1,n}, but to no intermediate vertices. We will show that no vertices in {H}_{1,n}, between \frac{2}{n} and 1, are adjacent to vertices outside this interval. We note that there are vertices of {H}_{1,n} in (\frac{2}{n},1) for n\in \mathbb{N}. Indeed as in Figure 3, if n is odd, take the vertex \frac{6n+1}{9n} in (\frac{2}{n},1) and if n is even, take the vertex \frac{n+1}{2n} in (\frac{2}{n},1).
Suppose that an edge crosses Rez=\frac{2}{n}, whence that it joins v=\frac{a}{nb} to w=\frac{c}{nd}. 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 \frac{2}{n} and 1 are not adjacent to ∞, it follows that {H}_{1,n} is not connected.
Consequently, since there is no circuit like \mathrm{\infty}\to \frac{1}{n}\leftarrow {v}_{1}\leftarrow \cdots \leftarrow \frac{n+1}{n}\leftarrow \mathrm{\infty} in {H}_{1,n}, {H}_{1,n} is not connected for n\ge 3. □
Theorem 8 {H}_{u,n} is connected if and only if n\le 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=n1. Using Theorem 5(3), we may only work in {H}_{1,n}. By Theorem 4, we may suppose that D has the form \mathrm{\infty}\to {v}_{1}\to {v}_{2}\to \mathrm{\infty} or \mathrm{\infty}\to {v}_{1}\leftarrow {v}_{2}\to \mathrm{\infty}. Let us do calculations only for the first triangle. We easily see that {v}_{1}=\frac{x}{n} and {v}_{2}=\frac{y}{n} for some x,y\in \mathbb{Z}. If \frac{x}{n}<\frac{y}{n}, then xy=1. Since \frac{x}{n} and \frac{y}{n}\in [\mathrm{\infty}], xy\equiv 0\phantom{\rule{0.25em}{0ex}}(modn). So, n=1. If \frac{x}{n}>\frac{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 \frac{1}{0}\to \frac{1}{1}\to \frac{2}{1}\to \frac{1}{0}. □
Theorem 10 {H}_{u,n} contains a 2gon if and only if n=1 or 2.
Proof Suppose \frac{x}{kn}\to \frac{y}{ln}\to \frac{x}{kn} is a 2gon 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 \frac{1}{0}\to \frac{1}{n}\to \frac{1}{0} is a 2gon. □
We now give one of our main theorems.
Theorem 11 If n\ge 2, then {H}_{1,n} and {H}_{n1,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 2gon, in {H}_{1,2}. By Theorem 4 and Theorem 3, we may assume that D has the form \mathrm{\infty}\to {v}_{1}\to \cdots \to {v}_{k}\to \mathrm{\infty}, where the vertices {v}_{1},{v}_{2},\dots ,{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 [\frac{1}{2},\frac{3}{2}]. By Theorem 3, {v}_{1}=\frac{1}{2} or \frac{3}{2}. If {v}_{1}=\frac{1}{2}, then {v}_{k}=\frac{2a+1}{2}\in [\frac{1}{2},\frac{3}{2}] and {v}_{1}\ne {v}_{k} give that {v}_{k}=\frac{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}=\frac{3}{2}. That is, {H}_{1,2} is a forest.
Now let n\ge 3. If {H}_{1,n} is not a forest, then, as we will see now by Theorem 3, D must be of the form \mathrm{\infty}\to {v}_{1}\leftarrow \cdots \leftarrow {v}_{k}\leftarrow \mathrm{\infty} or \mathrm{\infty}\to {v}_{1}\to \cdots \to {v}_{k}\leftarrow \mathrm{\infty}. As above, we choose the vertices in the finite interval [\frac{1}{n},\frac{n+1}{n}]. By Theorem 3, {v}_{1}=\frac{1}{n} or \frac{n+1}{n}. If {v}_{1}=\frac{1}{n}, then, as above, {v}_{k} must be \frac{n+1}{n}. In this case D has the form \mathrm{\infty}\to \frac{1}{n}\leftarrow \cdots \leftarrow \frac{n+1}{n}\leftarrow \mathrm{\infty}. 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}=\frac{n+1}{n}, then there does not exist a circuit D like \mathrm{\infty}\to \frac{n+1}{n}\to \cdots \to \frac{1}{n}\leftarrow \mathrm{\infty}. Therefore {H}_{1,n} is a forest. Using Theorem 5(3), we see that the subgraph {H}_{n1,n} is a forest as well. Therefore the proof is completed. □
Theorem 12 If {H}_{u,n} contains a triangle, then {\mathrm{\Gamma}}_{1}(n) contains an elliptic element of order 3.
Proof If {H}_{u,n} contains a triangle, then by Theorem 9, n=1. So, {\mathrm{\Gamma}}_{1}(1)=\mathrm{\Gamma} and \left(\begin{array}{cc}1& 1\\ 3& 2\end{array}\right)\in {\mathrm{\Gamma}}_{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 \left(\begin{array}{cc}1& 1\\ 3& 2\end{array}\right)\in {\mathrm{\Gamma}}_{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 {\mathrm{\Gamma}}_{0}(n) correspond to circuits in the subgraph {F}_{u,n} of the same order and vice versa. Here, in the case of {\mathrm{\Gamma}}_{1}(n), owing to Theorem 12 triangles in the subgraph {H}_{u,n} correspond to elliptic elements in {\mathrm{\Gamma}}_{1}(n) of order 3. But the converse is not true as shown in Remark 1.
Theorem 13 {H}_{u,n} contains a 2gon if and only if {\mathrm{\Gamma}}_{1}(n) contains an elliptic element of order 2.
Proof If {H}_{u,n} contains a 2gon, then by Theorem 10, n=1 or 2. So, \left(\begin{array}{cc}1& 1\\ 2& 1\end{array}\right) is an elliptic element of order 2 in both {\mathrm{\Gamma}}_{1}(1) and {\mathrm{\Gamma}}_{1}(2).
Conversely, assume that {\mathrm{\Gamma}}_{1}(n) contains an elliptic element of order 2. Then there is an element of {\mathrm{\Gamma}}_{1}(n) of the form \left(\begin{array}{cc}1+an& b\\ cn& 1+dn\end{array}\right) 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/10.1186/1029242X2013117
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DOI: https://doi.org/10.1186/1029242X2013117
Keywords
 modular groups
 congruence subgroups
 suborbital graphs