# Connectedness of a suborbital graph for congruence subgroups

- Yavuz Kesicioğlu
^{1}Email author, - Mehmet Akbaş
^{2}and - Murat Beşenk
^{2}

**2013**:117

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

© Kesicioğlu et al.; licensee Springer 2013

**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 groups congruence subgroups suborbital graphs## 1 Introduction

where *a*, *b*, *c*, and *d* are rational integers and $ad-bc=1$.

Jones, Singerman, and Wicks [1] used the notion of the imprimitive action [2–4] for a Γ-invariant 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 Γ-invariant 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 $n>1$.

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 $a,b\in \stackrel{\u02c6}{\mathbb{Q}}$).

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., $y\equiv us\phantom{\rule{0.25em}{0ex}}(modn)$$x\equiv ur\phantom{\rule{0.25em}{0ex}}(modn)$
*and*$ry-sx=n$*or* - 2., $y\equiv -us\phantom{\rule{0.25em}{0ex}}(modn)$$x\equiv -ur\phantom{\rule{0.25em}{0ex}}(modn)$
*and*$ry-sx=-n$.

**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 $ad-bc=\pm 1$.

**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., $ry-sx=n$,$x\equiv ur\phantom{\rule{0.25em}{0ex}}(modn)$
*or* - 2., $ry-sx=-n$.$x\equiv -ur\phantom{\rule{0.25em}{0ex}}(modn)$

**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*$ry-sx=1$*or* - 2.$u=1$
*and*$ry-sx=n$*or* - 3.$u=n-1$
*and*$ry-sx=-n$.

*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 $1\equiv u\phantom{\rule{0.25em}{0ex}}(modn)$, $ry-sx=n$ or $1\equiv -u\phantom{\rule{0.25em}{0ex}}(modn)$, $ry-sx=-n$. 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 $ry-sx=-n$. Then $x\equiv -r(n-1)\equiv 0\phantom{\rule{0.25em}{0ex}}(modn)$. 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*$\varphi :{H}_{1,n}\to {H}_{n-1,n}$,*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 ${H}_{n-1,n}$ 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), $ry-sx=n$. So, $sx-ry=-n$. That is, using Theorem 3(3), $\frac{x}{yn}\to \frac{r}{sn}$ is an edge in ${H}_{n-1,n}$. □

**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 2-gon.

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

As in examples $\mathrm{\infty}\to \frac{1}{2}\to \mathrm{\infty}$ is a 2-gon 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 $[\frac{2}{3},1]$.

*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 $u=n-1$. Using Theorem 5(3), we may only work in ${H}_{1,n}$. By Theorem 4, we may suppose that *D* has the form $\mathrm{\infty}\to {v}_{1}\to {v}_{2}\to \mathrm{\infty}$ 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 $x-y=-1$. Since $\frac{x}{n}$ and $\frac{y}{n}\in [\mathrm{\infty}]$, $x-y\equiv 0\phantom{\rule{0.25em}{0ex}}(modn)$. So, $n=1$. If $\frac{x}{n}>\frac{y}{n}$, 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 $\frac{1}{0}\to \frac{1}{1}\to \frac{2}{1}\to \frac{1}{0}$. □

**Theorem 10** ${H}_{u,n}$ *contains a* 2-*gon if and only if* $n=1$ *or* 2.

*Proof* Suppose $\frac{x}{kn}\to \frac{y}{ln}\to \frac{x}{kn}$ 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 $\frac{1}{0}\to \frac{1}{n}\to \frac{1}{0}$ is a 2-gon. □

We now give one of our main theorems.

**Theorem 11** *If* $n\ge 2$, *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 $\mathrm{\infty}\to {v}_{1}\to \cdots \to {v}_{k}\to \mathrm{\infty}$, 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 ${H}_{n-1,n}$ 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* 2-*gon if and only if* ${\mathrm{\Gamma}}_{1}(n)$ *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, $\left(\begin{array}{cc}1& -1\\ 2& -1\end{array}\right)$ 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. □

## Declarations

### Acknowledgements

Dedicated to Professor Hari M Srivastava.

## Authors’ Affiliations

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