Sciresol Sciresol https://indjst.org/author-guidelines Indian Journal of Science and Technology 10.17485/IJST/v13i25.66 Primitive Representations and the Modular Group Nadeem Bari Muhammad drnadeembari@gmail.com 1 Malik Muhammad Aslam 1 Department of Mathematics, University of the Punjab Lahore, Pakistan, 54590 13 25 2020 Abstract

Objectives: Primitive representations are useful to explore the modular group action on real quadratic field. Methods/Statistical Analysis: By using primitive representations structure of G-orbit are obtained. Finding: Conditions on  n and a,  b,  c are determined when αG= (α-)G,  αG= (-α-)G,  αG= (-α)G,  αG= (α-)G= (-α-)G= (-α)G and αG (α-)G (-α-)G (-α)G, where α= a+nc with b=a2-nc is real quadratic irrational number. We also find some elements of modular group PSL(2, ) that moves α to α- , α to -α- and α to -α. Applications: By using these conditions, we can construct the structure of the G-orbit. These results are verified by suitable examples.

Keywords Primitive Representations coset diagram modular group quadratic field None
Introduction

Binary quadratic form is one of the subjects treated in elementary number theory. Another subject treated in elementary number theory is the possibility of representing a positive integer as a sum of two squares and difference of two squares. The representations n= x2+y2 and n= x2-y2 which are of our interest are special cases of general binary quadratic form fx, y=ax2+bxy+cy2 and the representation n= x2+y2 is primitive representation if (x, y)=1.

Let n= k2 m, where k   N and m is a square free positive integer. Take Q*n={a+nc:a, b=a2-nc,cZ, c0 and a,b,c=1} and Qred* n=αQ*n: α>1 and -1<α-<0.  Then (Q( m)Q)=  UkNQ*k2m contain Q*n and Qred*n as G-subset and subsets respectively.

If α = a+nc  Q * n, if α and α¯ have different signs, then αis said to be an ambiguous number. A quadratic irrational number α is said to be reduced if α > 1 and -1 < α¯  < 0 . The modular group PSL(2, ) is the group of all linear fractional transformations zsz+tuz+v with sv-tu=1 , where s, v, t, u are integers.

This group can be presented as G=x,y:x2=y3=1 , where x:z-1z,y:zz-1z Modular group can be written in the matrix form as it is the set of 2 × 2 matrices with integral entries and determinant 1. It is generated by two matrices X= 0-110, Y = 1-110 of orders 2 and 3 respectively.

Now the product of two transformations is the same as the product of corresponding matrices. For the sake of simplicity, we use matrices instead of transformations.﻿

A coset diagram is a graph consisting of vertices and edges. It depicts a permutation representation of the modular group G, the 3-cycles of y are denoted by three vertices of a triangle permuted anticlockwise by y and the two vertices which are interchanged by x are joined by an edge.

In 1, 2, types of length 4, 6 satisfying exactly one of the conditions namely αG= (α-)G, αG= (-α-)G,  αG= (-α)G, αG= (α-)G= (-α-)G= (-α)G have been determined.

In 3, 4 formula for total numbers of ambiguous numbers in Q*(n) is determined. In 5 it is explored that if p1(mod 4) then p+pG include circuit of length 2 and in which αG= (α-)G= (-α-)G= (-α)G. In 6 it is describe that if p3(mod 4) then p+pG contains circuit of length 2 and in which αG= (-α-)G.

Materials and Methods

Lemma 2.1 7 Let α= a+nc be an ambiguous number. Then xα,  yα, y2α are always ambiguous numbers.

Lemma 2.2 8 If a natural number n can be written as sum of two squares of two rational numbers, then n can be written as sum of two squares of two integers.

Lemma 2.3 9 Any two elements of the same order are conjugate in a group G.

Lemma 2.4 6 gα-= gα¯ for all gG and α Q*n.

Results and Discussion

For α=a+ncQ*(n) , the elements α, α¯, -α and - α¯ play an important role in the study of modular group action on Q(m)Q=UkNQ*(k2m) .

In this section we determine the elements of G and conditions on a, b, c when αG= (α-)G,  αG= (-α-)G.

In the following theorem, we describe the elements of G that moves real quadratic irrational numbers to their conjugates.

Theorem 3.1: If α = a+nc Q*n is such that αG= (α-)G, then the element g of G such that g(α)= α- is of the form g= (g1)-1 xg1 for some g1  G.

Proof: Let α = a+nc Q*n be such that αG= (α-)G, then there exists an element g= stuv in G, which satisfy sα+tuα+v=α-.

That is sα+t=uα+vα-.

This implies that sα+t=uαα-+vα-.

This can be written as sa+nc+t=ua2-nc2+s-a+n-c.

This gives as+ct=bu+av,  s= -v.

So, we have g = st2as+ctb-s.

Then

g2 = st2as+ctb-s st2as+ctb-s=s2+2ast+ct2b00s2+2ast+ct2b = 1001 = I

Since g is an element of order 2, but any two elements of same order are conjugate by lemma 2.3. So, g is of the form g = (g1)-1 xg1.Example 3.1: If  α=-3+29-10 , then α- = 3+2910. The elements which moves α to α- are y2xy and xy4xy2x4 seeFigure 1.

Orbit

But both elements can be written as y2xy=y-1xy and xy4xy2x4=  y2x-4xy2x4. Both elements of G are in Cx=  {g-1xg : g  G}.

Corollary 3.2: If α= a+nc Q*n, then xα= α- if and only if b= -c.

Proof: As xa+nc= -a+n-c implies that -a+nb = -a+n-c. So, b=-c.

Conversely, if b=-c, then xa+nc = -a+nb = -a+n-c.

Corollary 3.3: If α= a+nc Q*n,  then xα= α- if and only if n has a primitive representation.

Proof: It has been proved in 5, that x(α)=α- if and only if n=a2+c2. It remains only to show that this representation is primitive.

As α= a+nc Q*n, then a, b, c=1. Now by Lemma 3.2, xα= α- if and only if b= -c. Thus a, b, c=a,-c,c=a,c=1. As required.

Remark 3.4 Corollary 3.3 holds only when n has primitive representation.

Example 3.5: Consider n=22+62, then this representation is not primitive. By using corollary 3.3, we have α = 2+406 corresponding this representation. Then xα= x2+406= -2+40-6= α- . But α = 2+406 = 1+103  Q * 40

Corollary 3.6: If α= a+nc Q*n and xα= α-. Then x-α= -α-.

Proof: If xα= α- then by lemma 3.2 b= -c.

Now x-α=xa+n-c=-a+na2-n-c=-a+n-b = -a+nc =- α¯.

Corollary 3.7: If α= a+nc Q*n and xα= α-. Then c+nax=-c+n-a .

Proof: It has been proved in 5, that x(α)=α- if and only if n=a2+c2.

Also n=c2+a2 if and only if xc+na=-c+n-a.

Corollary 3.8: If α= nc Q*n , then xα α- .

Proof: We prove this result by contradiction.

On contrary, we suppose that α= α-.

Then, xnc= n-c. This implies that nc= n-c.

That is, n-nc= n-c.

Thus n= c2, a contradiction. So, α α-.

Example 3.9: If α= 2, then α-= 2-1 and x2 2-1 .

Corollary 3.10: If α= a+ncQ*n  and xα=α-, then   γαGsuch that xγ= γ-.

Proof: If xα= α-, then by theorem 3.1, the elements of G which moves α to α- are x and g-1xg see example 3.1. One element is in anticlockwise direction, other element is in clockwise direction and g depends on the type of circuit of αG. Now g-1xg α= α- this implies that xg α=gα-. By substituting g α= γ  and using Lemma 2.4, we have xγ= γ-.

In the following theorem we determine condition on , b, c when αG= (-α-)G and this result is verified by a suitable example.

Theorem 3.2: If α= a+nc Q*n is such that either -2ab or -2ac is integer, then αG= (-α-)G.  Proof: Case I. If  -2ac, we show αG= (-α-)G.

Consider (yx)-2acα= α-2ac because (yx)lα = α+l.

This implies that, (yx)-2acα = a+nc-2a c  .

That is, (yx)-2acα= -a+nc= -α-. So, αG= (-α-)G.

Case II. If -2ab, we show αG= -α-G.

Consider (y2x)-2abα= α-2a(α)b+1  because (y2x)lα= αlα+1.

That is

(y2x)-2abα = a+nc-2aba+nc+1

After simplification, we have

(y2x)-2abα = ba+n-2a2-2an+bc

After rationalization, we have

(y2x)-2abα = b-2a3+abc+2an+bcn-2a2+bc2-4a2n

This can be written as

(y2x)-2abα = b-2aa2-n+abc+bcn4a4+b2c2-4a2bc-4a2n

After simplification, we have

y2x-2ab(α)=b(-abc+bcn)b2c2=-a+nc=-α¯. So, αG=(-α¯)G

Following corollary is an immediate consequence of the above result.

Corollary 3.11: If α= a+nc Q*n is such that b or  c divides-2a, then αG= (-α-)G. Proof: As in such cases -2ab or -2ac becomes integer.Example 3.12: In the orbit (2+61)G as shown inFigure 2 we have α= 2+61 with a=2, c=1,  b=-2.

Orbit

Now

-2ac=-2(2)1=-4. So, 2+61G=-2+61G

Similarly, for α= 1+6-5 with a=1, c=-5,  b=1.

As

-2ab=-2(1)1=-2, so 1+6-5G=-1+6-5G

In 1, 2 types of lengths 4, 6 have been determined in which all the four orbits αG, -αG, (α-)G and (-α-)G are distinct.

The following corollary follows from theorem 3.2 and corollary 3.2.

Corollary 3.13: If α= a+nc Q*n is such that -2ac is integer and b= -c, then αG= (α-)G=  (-α)G= (-α-)G.

Example 3.14: In the orbit  2+51G as shown inFigure 3, we have α= 2+51 with  a=2, c=1,  b=-1.

Now -2ac= -221=-4 and   b=-c=-1. So,

2+51G= -2+51G=2+5-1G= -2+5-1G

Orbit

Corollary 3.15: If α= a+nc Q*n is such that -2ac  and b= -c, then the element of G which moves α to -α is of the form xyx-2ac.

Proof: In theorem 3.2, it is derived that if -2ac , then yx-2acα= -a+nc.  This implies that xyx-2acα=x-a+nc= a+nb=a+n-c . As required.

Corollary 3.16: If α= nc Q*n then the element g which moves α to -α is of the form g=(g1)-1xg1 for some g1G.

Proof: If α= nc Q*n then in this case α-= -α, so by theorem 3.1 the element g which moves α to -α is of the form g=  (g1)-1xg1 for some g1G.

Corollary 3.17: If α= ncQ*n is such that  ncG= n-cG, then αG= (α-)G=  (-α)G= (-α-)G.

Proof: Here α= nc then -2ac=0, so by theorem 3.2, we have αG= (-α-)G. Also ncG= n-cG, then αG= (α-)G=  (-α)G= (-α-)G.

Converse of above result is not hold because

(1+52)G= (-1+52)G=(1+5-2)G= (-1+5-2)G

But the orbit does not contain these ambiguous numbers 51 , 5-1 , 55 and 5-5 .

Corollary 3.18: If the orbit  αG is such that αG (α-)G  (-α)G (-α-)G, then all ambiguous numbers which lies on G-circuit neither satisfy -2ac nor  b= -c .

Proof: By taking contrapositive to corollary 3.13, we get this result.

It has been proved in 5, that (α)x=α- if and only if n=a2+c2. In the following theorem, we generalize this result. In particular, we describe the condition on n when αG= (α-)G.

Theorem 3.3: If α= a+nc Q*n is such that   αG= (α-)G, then n can be written as the sum of two squares and this representation is primitive.

Proof: Let a+nc Q*n be such that   αG= (α-)G, then there exists an element g= stuv in G, which satisfy sα+tuα+v= α-.

That is sα+t=uα+vα-.

This implies that   sα+t=uαα-+vα-.

This can be written as

sa+nc+t=ua2-nc2+v-a+n-c

This gives as+ct=bu+av,  s= -v.

Combining both equations, we have as+ct=ub-as.

After simplification, we obtain  -t= 2as-ubc.

But sv-tu=1.

By substitution, we have   -s2+ 2as-ubuc =1.

This can be written as -cs2+2asu-bu2=c.

After substituting, the value of b, we have

-cs2+2asu-a2-ncu2=c.

After simplification, we obtain

-cs2+2asu-a2u2c+nu2c=c.

This can be written as

n=cu2+-a+csu2

In this expression u0, because if u=0 then s=-v and sv-tu=1 implies that s2=-1 which is not possible.

By Lemma 2.2 if a natural number n can be written as sum of two squares of two rational numbers, then n can be written as sum of squares of two integers. It is enough to prove this representation is primitive.

Let d=(cu,-a+csu). Then d|cu and d|(-a+csu).

This shows that ud|c and ud|(-au+cs). That is ud|cs and ud|(-au+cs).

This implies that ud|(-au+cs-cs). So, d|a.

Also, d|c and d2|n From equation 1. Thus, d2|(a2-bc), as d2|a2.

This implies that d2|bc, but d|c. So, d|b.

Thus d|a, b, c, but ( a, b, c)= 1. So, d=1.

Example 3.19 :

In the orbit  -1+13-6G, the element of G which moves -1+13-6 to 1+136 is y2 xy as shown inFigure 4 .

Orbit

Now corresponding element in matrix form is given by:

y2xy=0-11-10-1101-110=-11-21

Here s=-1,  t=1,  u= -2,  v=1 and a=-1,  c=-6,  b=2.

Now n=cu2+-a+csu2.

After substituting the values of  s,  t,  u,  v,  a,  b,  c, we get

n=-6-22+-(-1)+-6(-1)-22=32+22.

As required.

In the following theorem, we generalize the results of 5. In particular, we describe the condition on n when αG= (-α)G.

Theorem 3.4: If α= a+nc Q*n is such that αG= (-α)G, then n can be written as the sum of two squares and this representation is primitive.

Proof: Let α= a+nc Q*n be such that  αG= (-α)G, then there exists an element g= stuv in G, which satisfy sα+tuα+v= -α.

That is sα+t=-uα+vα.

This implies that  sα+t=-uα2-vα.

This can be written as

sa+nc+t=-ua+nc2-va+nc.

Which gives asc+t=-ua2+nc2-vac and cs= -2au-vc.

Combining both equations, we have

asc+t= -ua2+nc2-a-sc-2auc2

After simplification, we obtain   asc+t= acs-un+a2uc2.

This implies that -t=-ubc. But sv-tu=1.

By substituting the value of v and  t, we have s-2auc-s-u2bc=1.

After substituting the value of  b, we obtain -s2-2ausc-u2a2-nc2=1.

After some simplification, we have u2n= c2s2+2acus+u2a2+c2.

This can be written as

n=cu2+cs+auu2

In this expression u0, because if u=0 then s=-v and sv-tu=1 implies that s2=-1 which is not possible.

By Lemma 2.2 if a natural number n can be written as sum of two squares of two rational numbers then n can be written as sum of two squares of two integers. It is enough to prove this representation is primitive.

Let  d=(cu,a+csu). Then d|cu and d|(a+csu).

This shows that ud|c and  ud|(au+cs). This can be written ud|cs and ud|(au+cs).

This implies that  ud|(au+cs-cs). So, d|a.

Also, d|c and d2|n From equation 2. Thus d2|(a2-bc), as d2|a2.

This implies that d2|bc, but d|c. So, d|b.

Thus d|a, b, c, but ( a, b, c)= 1. So, d=1.

Example 3.20:

In the orbit 3+172G, the element of G which moves 3+172 to 3+17-2 is xy2x3yxy2 x as shown inFigure 5.

Orbit

Now corresponding element in matrix form is given by:

xy2x3yxy2x=0-110103111011011=-7-421

Here s=-7,  t=-4,  u= 2,  v=1 and a=3,  c=2,  b=-4.

Now n=cu2+a+csu2.

After substituting the values of s,  t,  u,  v,  a,  b,  c, we get

n=222+(3)+2(-7)22=12+42.

As required.

Theorem 3.5 If α= a+nc Q*n is such that   αG= (-α-)G, then n can be written as the difference of two squares of two rational numbers.

Proof: Let a+nc Q*n be such that   αG= (-α-)G, then there exists an element g= stuv in G , which satisfy sα+tuα+v= -α-.

That is sα+t=-uα+vα-.

This implies that sα+t=-uαα--vα-.

This can be written as

sa+nc+t=-ua2-nc2-v-a+n-c.

This gives as+ct=-bu-av,  s= v.

Combining both equations, we have 2as+ct+ub=0.

After simplification, we obtain -t= 2as+ubc.

But sv-tu=1. By substitution, we have s2+ 2as+ubuc =1.

This can be written as cs2+2asu+bu2=c.

After substituting, the value of b, we have

cs2+2asu+a2-ncu2=c.

After simplification, we obtain

cs2+2asu+a2u2c-nu2c=c.

This can be written as

n=a+csu2-cu2

If u=0, then t0. Otherwise g= 1001.

In the similar way, by eliminating s and u we can obtain n=a+bvt2-bt2. As required.

Example 3.21: In the orbit 2+81G, the element of G which moves 2+81 to -2+81 is y2 x as shown inFigure 6.

Orbit

Now corresponding element in matrix form is given by   y2x= 1011.

Here s=1,  t=0,  u= 1,  v=1 and a=2,  c=1, b=-4.

Now n=a+csu2-cu2.

After substituting the values of s,  t,  u,  v,  a,  b,  c in equation 3, we get

n=(2)+1(1)12-112=32-12. As required.

The element of G which moves 1+81 to -1+81 is yxy2 xyx as shown in Figure 6.

Now corresponding element in matrix form is given by

yxy2xyx=110110111101=2312

Here s=2,  t=3,  u= 1,  v=2 and a=1,  c=1,  b=-7.

Now n=a+csu2-cu2.

After substituting the values of s,  t,  u,  v,  a,  b,  c in equation 3, we get

n=(1)+1(2)12-112=32-12

As required.

Conclusion

The idea of study the elements that moves α to α- , α to -α- and  α to -α given in this paper is new and original. We have determined the conditions on n and a, b,  c when αG= (-α-)G, αG= (-α)G, αG= (α-)G, αG= (-α-)G= (-α)G= (α-)Gand αG (-α-)G(-α)G(α-)G, where αQ*n under the action of modular group G. These results are verified by some suitable examples.

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