Abstract
Higman has defined coset diagrams for . The coset diagrams are composed of fragments, and the fragments are further composed of two or more circuits. A condition for the existence of a certain fragment in a coset diagram is a polynomial f in , obtained by choosing a pair of words such that both and fix a vertex v in γ. Two pairs of words are equivalent if and only if they have the same polynomial. In this paper, we find distinct pairs of words that are equivalent. We also show there are certain fragments, which have the same orientations as those of their mirror images.
1. Introduction
The modular group [1–3] an interesting group of hyperbolic isometries and has a finite presentation where and are the linear fractional transformations defined by and , respectively. The modular group is of great interest in many fields of mathematics, for example, group theory, graph theory, and number theory. By adjoining a new element with and , we obtain a presentation of the extended modular group .
The Hecke group is the discrete group generated by , , where . It is well known that the modular group can be generalized to the Hecke groups. For , the resulting Hecke group is the modular group .
If is a power of a prime , then the projective line over the finite field , which is , is denoted by . The group has its customary meaning, a group of all linear fractional transformations such that and , while is its subgroup consisting of all those, where is a non-zero square in .
There is a well-known relationship between the action of on and continued fractions. There are many articles on the connection between the geodesics on the modular surface and continued fractions which are particularly important in the theory of approximation of real numbers by rational [4, 5]. Important results have been obtained using these ideas in a very well-written article [6]. A good account on relationship between continued fractions and indefinite binary quadratic forms is also given in [7]. In [8], Mushtaq has shown a relationship between reduced indefinite binary quadratic forms and orbits of the modular group. There is another interesting article [9] of Mushtaq and Hayat on the topic of Pell numbers, Pell-Lucas numbers, and modular group. In [10], Bong and Mushtaq determine the Fibonacci and Locus numbers through the action of modular group on real quadratic fields.
In , G. Higman introduced a new type of graph called a coset diagram for . The three cycles of are denoted by small triangles whose vertices are permuted counterclockwise by and any two vertices which are interchanged by are joined by an edge. The fixed points of and are denoted by heavy dots. Notice is equivalent to , which means that reverses the orientation of the triangles representing the three cycles of (as reflection does); because of this, there is no need to make the diagram more complicated by introducing -edges. For details about coset diagrams, one can refer to [11–14].
Two homomorphisms and from to are called conjugate if for some inner automorphism on . We call to be non-degenerate if neither of lies in the kernel of . In [15], it has been shown that the conjugacy classes of non-degenerate homomorphisms from to are in one to one correspondence with the elements of under the correspondence which maps each class to its parameter . As in [15], the coset diagram corresponding to the action of on via a homomorphism with parameter is denoted by . Each conjugacy class is represented by a unique coset diagram, unique in the sense that the diagram remains invariant except that labels of vertices change from to .
By a circuit in a coset diagram for an action of on , we will mean a closed path of triangles and edges. Let be a sequence of positive integers, the circuit which contains a vertex, fixed by for some ; we will mean the circuit in which triangles have one vertex inside the circuit and triangles have one vertex outside the circuit and so on. Since it is a cycle , so it does not make any difference if triangles have one vertex outside the circuit and triangles have one vertex inside the circuit and so on. The circuit of the type is called a periodic circuit and the length of its period is . A circuit that is not of this type is non-periodic. A circuit is called simple, if each vertex of the circuit is fixed by a unique word or its inverse .
2. Joining of Circuits
Consider two non-periodic and simple circuits and . Let be any vertex in fixed by a word and let be any vertex in fixed by a word . In order to connect these two circuits at and , we choose, without loss of generality , and apply on in such a way that ends at . Consequently, we get a fragment denoted by . In other words, by , we will mean a nonsimple fragment whose one vertex is fixed by a pair of words , , dented by .
Example 1. We connect the vertex , fixed by , in with the vertex , fixed by of , and compose a fragment as shown in Figures 1, 2, and 3.
One can see that the vertex in (Figure 3) is fixed by a pair of words
The action of on yields two components, namely, and . For sake of simplicity, let denote the complement .
The coset diagram is made of fragments. It is therefore necessary to ask when a fragment exists in . In [16], this question is answered in the following way.
Theorem 2. Given a fragment, there is a polynomial in such that(i)if the fragment occurs in , then ;(ii)if , then the fragment, or a homomorphic image of it occurs in or in .
In [16], the method of calculating a polynomial from a fragment is given. Here, we describe this method briefly. Since a fragment is composed of two non-periodic and connected circuits and with a common fixed vertex, say, , then there is a pair of words, for instance, such that and . Let and be the matrices corresponding to and of . Then, and can be expressed as where . By using (3.1) to (3.7) of [16], the matrices and can be expressed linearly as such that and , for , are polynomials in and , where is the trace of and is its determinant. Since is a common fixed vertex of and , therefore the matrices and have a common eigenvector. Then, by Lemma of [16], the algebra generated by and has dimension . The algebra contains , , , and and so these are linearly dependent. Using (3.1) to (3.7) of [16], the matrix is expressed as where , for , can be calculated in terms of the and , using to of [16]. The condition that , , , and are linearly dependent is expressed as If we calculate in terms of and and substitute in (8), we find that it is equivalent to
This gives a homogeneous equation in and . By substituting for , we get an equation in .
Two pairs of words and are equivalent if and only if they have the same polynomial. If two pairs of words and are equivalent, then we write .
In this paper, we find distinct pairs of words that are equivalent. We also show there are certain fragments, which have the same orientations as those of their mirror images.
3. Main Results
Theorem 3. In the above notation, the polynomial obtained from a fragment is unique.
Proof. Let and be any two vertices of a fragment , such that is fixed by a pair of words and is fixed by a pair of words . Suppose is a polynomial obtained by choosing and is a polynomial obtained by choosing . Now, if , then, by Theorem 2, the fragment or its homomorphic image occurs in or in . So, there exists a vertex in or in which is fixed by . Again, by Theorem 2, we have . Similarly, it can be proved that if , then . Hence, .
Remark 4. If a vertex is fixed by a word , then the vertex is fixed by the conjugate of .
We know that is a fragment created by joining a vertex of with the vertex of . Since the same polynomial is evolved for all the points of connection for the fragment , therefore it is important to know all the points of connection for the fragment . Following theorem is useful in finding all points of connection of and for the fragment .
Theorem 5. Let fragment be constructed by joining a vertex of with the vertex of . Then, is obtainable also, if the vertex of is joined with the vertex of .
Proof. Suppose vertex is fixed by a word and vertex is fixed by a word . Then, one vertex of the fragment is fixed by and . Now, we join the vertex of with the vertex of and compose a fragment . By Remark 4, the vertex of fragment is fixed by and , whereas the vertex of fragment is fixed by and . This shows that and are the same fragments.
Since and are the same fragments, and a unique polynomial is obtained from a fragment, hence .
Corollary 6. Let be the set of words such that for any , both vertices and lie on the circuits and . Let be the total number of points of connection of the circuits to compose . Then, .
Example 7. As in Example 1, the vertex in is connected with the vertex in , and a fragment is evolved. Then, one can see that is the set of words such that, for any , both vertices and lie on and . Since , there are nine points of connection of and composing . Hence, vertices of one circuit are merged in another circuit. In other words, triangles of one circuit are merged in another circuit.
The number of points of connection of the circuits for a fragment is always multiple of three and plays a significant role because they are directly related to the structure of the fragment. The following theorem illustrates this relation.
Theorem 8. Let be the total number of triangles in a fragment and the total number of points of connection of the circuits and . Then, .
Proof. Since circuits and have and number of triangles, respectively, and there are number of points of connection of and composing , therefore, by Corollary 6, there are number of words such that , lie on and , respectively. So, number of vertices of are merged in , and remaining number of vertices are included in to compose . So, the total number of vertices in is . This implies that .
Remark 9. Consider a fragment such that one vertex of is fixed by . Then, obviously, is fixed by . So, . Also, it is trivial that .
The following Theorem shows that and are not a unique pair of circuits for the fragment .
Theorem 10. The following pairs of words are equivalent:
Proof. Let , and be arbitrary vertices of the circuits , , and , respectively. Let the words , and fix the vertices , and , respectively. We join a vertex fixed by of the circuit with the vertex fixed by of the circuit to compose . Also, we join a vertex fixed by of the circuit with the vertex fixed by of the to compose a fragment . So, in , is a common fixed vertex of . The fragment is different from the fragment if and only if, in fragment , the word fixes the vertex in the following way: maps the vertex to some vertex , and then maps the vertex to the vertex ; that is, . Since is a composition of linear fractional transformations and , therefore . This implies that and are the same fragments, since a unique polynomial is obtained from a fragment. Hence, .
Similarly, it is easy to prove that
Since is an equivalence relation, hence
Example 11. Theorem 10 shows that the fragment , which has a vertex fixed by , and the fragment , which has a vertex fixed by , are the same. So, the fragment, which has a vertex fixed by , and the fragment, which has a vertex fixed by , are the same. Since the words , , and correspond to the circuits , , and , so the fragment (Figure 4) is obtained, not only by joining the vertex in (Figure 5) with the vertex in (Figure 6) but also by joining the vertex in (Figure 7) with the vertex in (Figure 8).
Theorem 12. In a coset diagram for the action of on , if a vertex is fixed by , then the vertex is fixed by .
Proof. Let be a word such that
Since , this implies that
Now, by using (14) and (15), we have . This shows that also fixes a point .
Theorem 13. If a vertex is fixed by a word in a circuit , then there exists a vertex in such that is fixed by .
Proof. Consider a simple circuit .
In Figure 9, one can see that the vertex is fixed by and is fixed by . So, . Also, the vertex is fixed by and is fixed by . So, . Similarly, , .
The vertices and are fixed by and , respectively, implying that . Also, the vertices and are fixed by and , respectively, implying that . Similarly, , .
Let the homomorphic image of the fragment (Figure 10) occur in the coset diagram . Since admits the axis of symmetry, the mirror image of the fragment under the permutation will also occur as shown in Figure 11.
Proposition 14. If the fragment has one vertex fixed by , then its mirror image has one vertex fixed by .
Proof. Let be the fragment with one vertex fixed by . Since is the mirror image of , therefore has one vertex . Then, by Theorem 12, the vertex is fixed by .
Theorem 15. The polynomials obtained from the fragment and its mirror image are the same.
Proof. Let be the polynomial obtained by choosing a pair of words from and let be the polynomial obtained by choosing a pair of words from . Let . Then, by Theorem 2, there is a vertex say in or in which is fixed by . Then, by Theorem 12, the vertex is fixed by . So, by Theorem 2, . Similarly, it can be proved that if , then . Hence, .
Since a unique polynomial is obtained from a fragment and its mirror image , hence .
Consider two circuits and . In general, fragment and its mirror image do not have the same orientation. There are certain fragments which have the same orientations as those of their mirror images. These kinds of fragments have vertical symmetry and may have fixed points of . The following Theorem shows how these fragments are composed.
Theorem 16. (i) The fragment composed by joining a vertex , fixed by
in , with the vertex , fixed by in , has the same orientation as that of its mirror image.
(ii) The fragment composed by joining a vertex , fixed by
in , with the vertex , fixed by in , has the same orientation as that of its mirror image.
(iii) The fragment composed by joining a vertex , fixed by
in , with the vertex , fixed by in , has the same orientation as that of its mirror image.
Proof. (i) Let be the fragment which has one vertex say , fixed by
Then, by Proposition 14, its mirror image fragment contains a vertex, say, , fixed by and . By Remark 4, the vertex of fragment is fixed by
and . This implies that . Hence fragment has the same orientation as that of its mirror image .
(ii) Let be the fragment which has one vertex, say, , fixed by
Then, its mirror image fragment has one vertex, say, , fixed by
The vertex of fragment is fixed by
and . This implies that . Hence, has the same orientation as that of its mirror image .
(iii) Let be the fragment which has one vertex, say, , fixed by
Then, its mirror image fragment has one vertex, say, , fixed by
The vertex of fragment is fixed by
and . This implies that . Hence, has the same orientation as that of its mirror image .
4. Motivation and Open Problems
If we join a pair of circuits at a certain point, we get a fragment and hence a polynomial. Since each such polynomial splits linearly in a suitable Galois [16] and corresponding to each zero, we get a triplet [15] which is a group. This shows that each pair of circuits can be related to a group. A pair of circuits has finitely many points of connections. Two distinct points of connection may or may not give the same fragment. So, it is important to ask the following question.
Problem 17. How many fragments (polynomials) are obtained if we join a pair of circuits and at all points of connection?
In order to give the answer of the above question, we first have to find those pair of words which are equivalent. In other words, we have to find those points of connection, at which we get the same polynomial. This issue is resolved in this paper. We will give the answer of the above question in an other paper. After that, one can establish a connection between a class of groups and a pair of circuits and , which is indeed a great development.
Conflict of Interests
The authors declare that there is no conflict of interests regarding the publication of this paper.
Acknowledgment
The authors would like to thank the referee for his very helpful comments, which have significantly improved the presentation of this paper.