Cohn triples of   matrices and their links with the theory of free groups of rank  were discovered in 1955 by Harvey Cohn. A lot of consequences were developed for the modular group SL (2, Z) and the free subgroup F2 In the present article, we deal with a new construction of such triples and resulting Diophantine equations.

1  Introduction

2  A new construction of the Cohn triples

3  Links with the Fibonacci numbers

4  Final result for the free group with two generators

5  Acknowledgments 

6  References

We deal in the present article with the Markoff spectrum M as defined by [1]. The minimum of an indefinite binary quadratic form  of an indefinite binary quadratic form

with real coefficients and positive discriminant   is

where the infimum is taken over all the pairs of integers not both zero. The set of values  not both zero. The set of values   is defined as being the Markoff spectrum M.

With the reduction of the form   to    we have the possibility of computing the values with doubly infinite sequences

We have 

We go from    and from   with    matrix with integer coefficients. Its determinant is often 1, but it will be possible to find another value -1 for the determinant of our matrices, and corresponding to a matrix of  More important, we will use the transpose of 

Also   (modular group) when  When multiplied, these matrices yield

Transposing the second equality and adding it to the first one, we find a relation similar to the Heisenberg relation:

In what follows, we generalize the computations made by Cohn [6], [7], [8], in à set of articles trying to approach Markoff’s forms through modular functions:

We see that the trace of the last commutator is not a multiple of 3. So we try to generalize the result quoted in the mémoire [9], saying that

Property 1. - For two matrices   the following are equivalent:

1/ The couple   generates the free group  

 2/ The triple  is a solution of the Markoff equation .

 3/ We have .

Moreover, if  is another generating system for   the free group generated by  there exists one and only one integer matrix and    up to a sign, verifying the conditions

if and only if we have  

Proof. See [17] (Chap. 6). Prop. 4.1 page 170 for Prop. 4.3 page 174 for Also [9] (Chap. 6). Prop. 6.0.1 page 57 for Prop. 6.0.2 page 57 for The equivalence is a consequence of the formula of Fricke ( [17] page 160). For the remaining part: ( [17] Chap. 6. prop. 5.1 page 175).  W

 2.1 Initial attempt to build a Cohn triple

We could choose, in order to have    

But this gives for C and A the same trace, which is limited enough, and a trace of B not a multiple of 3.Also:

We keep and put , two matrices being built with ,the matrices to be determined. Let us write, with and , and compute, where the interesting cases seem to be

We start with 

As    we have, because we suppose and

We can give new parameters defining , and new ones defining :

We write these two equations in dimension 4 and invert the matrices, after defining  and verifying that

Now we solve:

 This gives and for , so an expression of with two new parameters is as follows: and

For , it is also easy to write it with the same two parameters:

A simpler calculation is possible, here presented in order to confirm the previous one:

Hence, we obtain (2), (3), (5) more completely than (4), and with the expressions for and :

We find also some new expressions which are easy to establish: 

with                                                                                                   

then, defining K:

The asymmetric position of at the front of , not , provokes a question. Replacing by is the answer to the question. The condition for and to be in does not imply the same property for and . These matrices are in  

when their determinant is -1, and they are in . We have two cases, owing to the fact that can have two values, . In the two cases we exhibit a non trivial example.

We obtain a formula linking together the matrices and , situated in . With , this gives the triple introduced in ( [17] Chap. 6. page 162), associated to (5,2,1) W

We obtain a formula linking together the matrices , which are situated in .   

In both cases, we can evaluate :

Two cases have been defined owing to the fact that can have two values, . To determine , we got a Diophantine equation (4) which is easy to solve.

3.1   First case ( )

We find , and ,  We deal with the Diophantine equation

It has been already studied in [2], and we have:

Property 3.  The Diophantine equation has solutions if and only if .

Proof.  The references ([2], Theorem 6.3.1. p. 150) [21], [16]) give all that is needed about the solutions.  W

Any solution of this equation corresponds by a 1 to 1 to  correspondence to , a solution of the equation . We have only to look at our equation , to get all the solutions of the other. Moreover, the matrices and are in . It is interesting to realize that with ,

Applying Fricke’s formula ([17] p. 160, Prop. 2) and simplifying

This equation is solvable in integers with a method obtained from the classical Markoff theory. The solutions are written with the Fibonacci sequence (OEIS A000045). We find in [2], [16], all the solutions: (1,1),(-1,-1) and for all : 

For n=1, we get 

For all the couples of solutions, if is one of them is another:

We find a figure with four sequences connected at the end of each other, at the singular solution (1,1). But with bisequences , defined as indexed by Z, the Fibonacci bisequence gives

Hence we can write, on the upper infinite dihedral group , a bisequence to name the nodes of , but this constrains us to use the second group for the other bisequence . We will give another notation in the sequel, where will be replaced by if and only n is even, and so on for the three couples obtained by permutation of  and multiplication of the two terms of the couple by -1. With this method we find a bisequence of pairs of positive Fibonacci numbers which are the positive solutions of the equation This corresponds to the infinite cyclic group , in the upper position in Figure 1.

We see with the negative Fibonacci numbers a structure of a group isomorphic to . We can also say that the matrix operates on the set of solutions.

Remark 1. We have given in [20] the relation

Together with the same relation for

The commutative group , operates on solutions of the equation It is not the corresponding infinite dihedral group , not studied here.  W

But a formula such as that one which will be true for is more complicated, and does not seem to be given in any of the numerous articles written about Fricke’s formula. Working on this, we found:

This equation has been studied in [2] (pp. 130–150), and we have:

Property 4. The Diophantine equation has solutions if and only if .     

Proof.  The reference [2] gives all that is needed about the solutions. 

We deal with matrices situated in but not in . We give the continued fraction of , the root of  :

We produce then the classical table of the values of the associated form. For we have , and we find with the following table two classes of solutions, couples of Fibonacci numbers up to signs, of the equation of the first case:

We see that the matrix plays an important role for the transportation of the period of : 

This gives all the solutions of the equation , with a sign corresponding to the cycle and the infinite cycle given by :

In the present case, . We would like to be able to apply some relation similar to Fricke’s formula, for example, the last expression of [17] (p. 28).

But a formula such as that one which will be true for is more complicated, and does not seem to be given in any of the numerous articles written about Fricke’s formula. Working on this, we found:

Property 5. For any matrices , we have the generalized Fricke’s formula:

Proof. Let

Then

With

and so we find 

Now we combine:

and we get

Here, the commutator to deal with is . And we are in the parabolic case if and only if  .

Example 3.  With and for example

We are in the case of the positive Fricke’s relation, linking together the matrices , situated in . With , the triple introduced in ([17] Chap. 6. page 162), is associated to (5,2,1). W

Example 4.  With and for example  

We obtain a formula linking together the matrices , which are situated in .   

Remark 2. The cases with which we deal in Property 2 and Property 5 are different. In the first case,  are linked with strong constraints by the common coefficients , and their positions inside these matrices. On the contrary, Property 5 is true for any matrices .

Example 5.  With and for example 

we are not in the parabolic case. Moreover, we verify Property 5:

We face the fact that the group generated by  is free. By Property 1 and ,  generate the free group in . This group countains and .

Property 6.  The subgroup of is free and isomorphic to , but not equal to .

Proof. The group generated by  is a subgroup of , hence by the theorem of Nielsen–Schreier ([14] p. 92), it is a free subgroup of . But confirms with Property 1 that is not a system of generators of . 

 A confirmation that is a free group is not given by the properties of the commutator of

and not through Property 1, because

The rank, which is the number of generators of a free group, is 2 for . The index of the subgroup of , denoted by  may be used:

 1/ Suppose k is infinite. We are in a situation where is a free group and , not a group with one element, has infinite index in . Then is of infinite rank ([4] p. 355). But this is false, because this group has two generators  hence a rank less than 2. This case is impossible.

Note that in , the derived group, 

has infinite rank ([4] Théorème (9.39) p. 355 or [15] prop. 3.1. p. 13):

2/ Suppose k is finite. We have ([15] Proposition 3.9. p. 16):

Because our free groups and have two generators, the former relations give : 

The conclusion is that , not . It would be more comforting if  could be written with words in . The conclusion would be an equality. But this does not happen: only the isomorphism is sure. W

Property 1 is verified with , not , and we have (6). If we could write as a word of , we could write the same way, and conversely

We would like to conclude that and so . But this is not true, and we have only

Remark