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Research Article | DOI: https://doi.org/10.31579/2690-0440/012
Rue du Bon Pasteur, 57070 Metz, France.
*Corresponding Author: Serge Perrine, Rue du Bon Pasteur, 57070 Metz, France.
Citation: Serge Perrine. (2023). Construction of Cohn Triples and Applications. J. Mathematical Methods in Engineering, 4(1): Doi:10.31579/2690-0440.2022/012
Copyright: © 2023 Serge Perrine, this is an open-access article distributed under the terms of the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original author and source are credited.
Received: 01 July 2022 | Accepted: 17 September 2022 | Published: 01 April 2023
Keywords: cohn triples; fricke relations; fibonacci numbers
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
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