Modular J-invariant function - Tata Institute of …mathstud/seminar/events/2016-01-25...2016/01/25...

Modular J-invariant function

N. Saradha

T.I.F.R.

Jan 25, 2016Student’s Seminar

TIFR, Mumbai

Elliptic function

Let L = {mω1 + nω2 : m,n ∈ Z} be a lattice spanned by the twolinearly independent complex numbers ω1 and ω2. An ellipticfunction with respect to L is a meromorphic function f on Cwhich satisfies

f (z + ω) = f (z) for all ω ∈ L and z ∈ C.

D = {sω1 + tω2 : 0 ≤ s, t < 1}

is called the fundamental parallelogram and any translate ofthis is referred to as a fundamental domain. L is called theperiod lattice.

Elliptic function

Let L = {mω1 + nω2 : m,n ∈ Z} be a lattice spanned by the twolinearly independent complex numbers ω1 and ω2.

An ellipticfunction with respect to L is a meromorphic function f on Cwhich satisfies


D = {sω1 + tω2 : 0 ≤ s, t < 1}


Elliptic function



D = {sω1 + tω2 : 0 ≤ s, t < 1}


Elliptic function



D = {sω1 + tω2 : 0 ≤ s, t < 1}

is called the fundamental parallelogram and any translate ofthis is referred to as a fundamental domain.

L is called theperiod lattice.

Elliptic function



D = {sω1 + tω2 : 0 ≤ s, t < 1}


The Weierstrass elliptic function ℘(z)

Here is an explicit example of an elliptic function.The Weierstrass ℘− function associated with L is defined bythe series

℘(z) =1z2 +

∑ω∈L′

{1

(z − ω)2 −1ω2

},

where L′ denotes the set of non-zero periods. It convergesabsolutely and uniformly on every compact subset of C/L. It isa meromorphic function having a double pole at every point of Land no other poles.


Here is an explicit example of an elliptic function.

The Weierstrass ℘− function associated with L is defined bythe series

℘(z) =1z2 +

∑ω∈L′

{1

(z − ω)2 −1ω2

},




℘(z) =1z2 +

∑ω∈L′

{1

(z − ω)2 −1ω2

},

where L′ denotes the set of non-zero periods.

It convergesabsolutely and uniformly on every compact subset of C/L. It isa meromorphic function having a double pole at every point of Land no other poles.



℘(z) =1z2 +

∑ω∈L′

{1

(z − ω)2 −1ω2

},

where L′ denotes the set of non-zero periods. It convergesabsolutely and uniformly on every compact subset of C/L.

It isa meromorphic function having a double pole at every point of Land no other poles.



℘(z) =1z2 +

∑ω∈L′

{1

(z − ω)2 −1ω2

},


Eisenstein series

Let us denote by

G2k = G2k (L) =∑ω∈L′

ω−2k

the associated Eisenstein series of weight 2k.Then ℘(z) can be expressed as

℘(z) =1z2 +

∞∑k=1

(2k + 1)G2k+2z2k .

Eisenstein series

Let us denote by

G2k = G2k (L) =∑ω∈L′

ω−2k

the associated Eisenstein series of weight 2k.

Then ℘(z) can be expressed as

℘(z) =1z2 +

∞∑k=1

(2k + 1)G2k+2z2k .

Eisenstein series

Let us denote by

G2k = G2k (L) =∑ω∈L′

ω−2k

the associated Eisenstein series of weight 2k.Then ℘(z) can be expressed as

℘(z) =1z2 +

∞∑k=1

(2k + 1)G2k+2z2k .

The differential equation for ℘(z)

It can be shown that for all z ∈ C, z 6∈ L we have

℘′(z)2 = 4℘(z)3 − 60G4℘(z)− 140G6.

Thus (℘(z), ℘′(z)) lies on the curve defined by the equation

E : y2 = 4x3 − g2x − g3

where g2 = 60G4,g3 = 140G6. This cubic polynomial has thediscriminant

g32 − 27g2

3 .

It turns out that this discriminant is non-zero. Such curves arecalled elliptic curves (E).



℘′(z)2 = 4℘(z)3 − 60G4℘(z)− 140G6.


E : y2 = 4x3 − g2x − g3

where g2 = 60G4,g3 = 140G6.

This cubic polynomial has thediscriminant

g32 − 27g2

3 .




℘′(z)2 = 4℘(z)3 − 60G4℘(z)− 140G6.


E : y2 = 4x3 − g2x − g3


g32 − 27g2

3 .




℘′(z)2 = 4℘(z)3 − 60G4℘(z)− 140G6.


E : y2 = 4x3 − g2x − g3


g32 − 27g2

3 .

It turns out that this discriminant is non-zero.

Such curves arecalled elliptic curves (E).



℘′(z)2 = 4℘(z)3 − 60G4℘(z)− 140G6.


E : y2 = 4x3 − g2x − g3


g32 − 27g2

3 .


We can parametrize

Now we start with such an equation. Given (x , y) ∈ C2, lying onthe curve, can we find z such that x = ℘(z), y = ℘′(z)? If so, wecan always parametrize such curves with the Weierstarsselliptic functions. Suppose ℘(z) = x has no solution. Then1/(℘(z)− x) is an elliptic function which is holomorphic on L.By periodicity, it is entire and bounded. Hence must be aconstant- a contradiction. Hence y = ±℘′(z). One can checkthat ℘ is an even function and hence ℘′ is an odd function. Sowe can adjust the sign so that (x , y) = (℘(z), ℘′(z)).

We can parametrize

Now we start with such an equation.

Given (x , y) ∈ C2, lying onthe curve, can we find z such that x = ℘(z), y = ℘′(z)? If so, wecan always parametrize such curves with the Weierstarsselliptic functions. Suppose ℘(z) = x has no solution. Then1/(℘(z)− x) is an elliptic function which is holomorphic on L.By periodicity, it is entire and bounded. Hence must be aconstant- a contradiction. Hence y = ±℘′(z). One can checkthat ℘ is an even function and hence ℘′ is an odd function. Sowe can adjust the sign so that (x , y) = (℘(z), ℘′(z)).

We can parametrize

Now we start with such an equation. Given (x , y) ∈ C2, lying onthe curve, can we find z such that x = ℘(z), y = ℘′(z)?

If so, wecan always parametrize such curves with the Weierstarsselliptic functions. Suppose ℘(z) = x has no solution. Then1/(℘(z)− x) is an elliptic function which is holomorphic on L.By periodicity, it is entire and bounded. Hence must be aconstant- a contradiction. Hence y = ±℘′(z). One can checkthat ℘ is an even function and hence ℘′ is an odd function. Sowe can adjust the sign so that (x , y) = (℘(z), ℘′(z)).

We can parametrize

Now we start with such an equation. Given (x , y) ∈ C2, lying onthe curve, can we find z such that x = ℘(z), y = ℘′(z)? If so, wecan always parametrize such curves with the Weierstarsselliptic functions.

Suppose ℘(z) = x has no solution. Then1/(℘(z)− x) is an elliptic function which is holomorphic on L.By periodicity, it is entire and bounded. Hence must be aconstant- a contradiction. Hence y = ±℘′(z). One can checkthat ℘ is an even function and hence ℘′ is an odd function. Sowe can adjust the sign so that (x , y) = (℘(z), ℘′(z)).

We can parametrize

Now we start with such an equation. Given (x , y) ∈ C2, lying onthe curve, can we find z such that x = ℘(z), y = ℘′(z)? If so, wecan always parametrize such curves with the Weierstarsselliptic functions. Suppose ℘(z) = x has no solution.

Then1/(℘(z)− x) is an elliptic function which is holomorphic on L.By periodicity, it is entire and bounded. Hence must be aconstant- a contradiction. Hence y = ±℘′(z). One can checkthat ℘ is an even function and hence ℘′ is an odd function. Sowe can adjust the sign so that (x , y) = (℘(z), ℘′(z)).

We can parametrize

Now we start with such an equation. Given (x , y) ∈ C2, lying onthe curve, can we find z such that x = ℘(z), y = ℘′(z)? If so, wecan always parametrize such curves with the Weierstarsselliptic functions. Suppose ℘(z) = x has no solution. Then1/(℘(z)− x) is an elliptic function which is holomorphic on L.

By periodicity, it is entire and bounded. Hence must be aconstant- a contradiction. Hence y = ±℘′(z). One can checkthat ℘ is an even function and hence ℘′ is an odd function. Sowe can adjust the sign so that (x , y) = (℘(z), ℘′(z)).

We can parametrize

Now we start with such an equation. Given (x , y) ∈ C2, lying onthe curve, can we find z such that x = ℘(z), y = ℘′(z)? If so, wecan always parametrize such curves with the Weierstarsselliptic functions. Suppose ℘(z) = x has no solution. Then1/(℘(z)− x) is an elliptic function which is holomorphic on L.By periodicity, it is entire and bounded. Hence must be aconstant- a contradiction.

Hence y = ±℘′(z). One can checkthat ℘ is an even function and hence ℘′ is an odd function. Sowe can adjust the sign so that (x , y) = (℘(z), ℘′(z)).

We can parametrize

Now we start with such an equation. Given (x , y) ∈ C2, lying onthe curve, can we find z such that x = ℘(z), y = ℘′(z)? If so, wecan always parametrize such curves with the Weierstarsselliptic functions. Suppose ℘(z) = x has no solution. Then1/(℘(z)− x) is an elliptic function which is holomorphic on L.By periodicity, it is entire and bounded. Hence must be aconstant- a contradiction. Hence y = ±℘′(z).

One can checkthat ℘ is an even function and hence ℘′ is an odd function. Sowe can adjust the sign so that (x , y) = (℘(z), ℘′(z)).

We can parametrize

Now we start with such an equation. Given (x , y) ∈ C2, lying onthe curve, can we find z such that x = ℘(z), y = ℘′(z)? If so, wecan always parametrize such curves with the Weierstarsselliptic functions. Suppose ℘(z) = x has no solution. Then1/(℘(z)− x) is an elliptic function which is holomorphic on L.By periodicity, it is entire and bounded. Hence must be aconstant- a contradiction. Hence y = ±℘′(z). One can checkthat ℘ is an even function and hence ℘′ is an odd function.

Sowe can adjust the sign so that (x , y) = (℘(z), ℘′(z)).

We can parametrize

Now we start with such an equation. Given (x , y) ∈ C2, lying onthe curve, can we find z such that x = ℘(z), y = ℘′(z)? If so, wecan always parametrize such curves with the Weierstarsselliptic functions. Suppose ℘(z) = x has no solution. Then1/(℘(z)− x) is an elliptic function which is holomorphic on L.By periodicity, it is entire and bounded. Hence must be aconstant- a contradiction. Hence y = ±℘′(z). One can checkthat ℘ is an even function and hence ℘′ is an odd function. Sowe can adjust the sign so that (x , y) = (℘(z), ℘′(z)).

Converse is also true

Suppose we are given an elliptic curve of the formy2 = 4x3 − Ax − B (thus the discriminant is non-zero) , thereexist a unique lattice L such that g2(L) = A and g3(L) = B andhence the complex points of the curve are parametrized by theWeierstrass ℘ function of L.

A Theorem of Schneider

Schneider(1937): Let L be a lattice and suppose that g2,g3 arealgebraic. Then for any algebraic α 6∈ L, ℘(α) is transcendental.As a consequence, we have that any non-zero period of anelliptic curve defined over Q is transcendental.Reason: Since ℘′ is an odd function,℘′(ω1/2) = ℘′(ω2/2) = ℘′((ω1 + ω2)/2) = 0. Hence℘(ω1/2), ℘(ω2/2), ℘((ω1 + ω2)/2) are algebraic (known as twodivision points) which shows that non-zero periods aretranscendental.One can also show by standard complex analytic argument,that the above two division points are distinct. And this impliesthat the discriminant

∆ := g32 − 27g2

3 6= 0.


Schneider(1937): Let L be a lattice and suppose that g2,g3 arealgebraic. Then for any algebraic α 6∈ L, ℘(α) is transcendental.

As a consequence, we have that any non-zero period of anelliptic curve defined over Q is transcendental.Reason: Since ℘′ is an odd function,℘′(ω1/2) = ℘′(ω2/2) = ℘′((ω1 + ω2)/2) = 0. Hence℘(ω1/2), ℘(ω2/2), ℘((ω1 + ω2)/2) are algebraic (known as twodivision points) which shows that non-zero periods aretranscendental.One can also show by standard complex analytic argument,that the above two division points are distinct. And this impliesthat the discriminant

∆ := g32 − 27g2

3 6= 0.


Schneider(1937): Let L be a lattice and suppose that g2,g3 arealgebraic. Then for any algebraic α 6∈ L, ℘(α) is transcendental.As a consequence, we have that any non-zero period of anelliptic curve defined over Q is transcendental.

Reason: Since ℘′ is an odd function,℘′(ω1/2) = ℘′(ω2/2) = ℘′((ω1 + ω2)/2) = 0. Hence℘(ω1/2), ℘(ω2/2), ℘((ω1 + ω2)/2) are algebraic (known as twodivision points) which shows that non-zero periods aretranscendental.One can also show by standard complex analytic argument,that the above two division points are distinct. And this impliesthat the discriminant

∆ := g32 − 27g2

3 6= 0.


Schneider(1937): Let L be a lattice and suppose that g2,g3 arealgebraic. Then for any algebraic α 6∈ L, ℘(α) is transcendental.As a consequence, we have that any non-zero period of anelliptic curve defined over Q is transcendental.Reason: Since ℘′ is an odd function,℘′(ω1/2) = ℘′(ω2/2) = ℘′((ω1 + ω2)/2) = 0.

Hence℘(ω1/2), ℘(ω2/2), ℘((ω1 + ω2)/2) are algebraic (known as twodivision points) which shows that non-zero periods aretranscendental.One can also show by standard complex analytic argument,that the above two division points are distinct. And this impliesthat the discriminant

∆ := g32 − 27g2

3 6= 0.


Schneider(1937): Let L be a lattice and suppose that g2,g3 arealgebraic. Then for any algebraic α 6∈ L, ℘(α) is transcendental.As a consequence, we have that any non-zero period of anelliptic curve defined over Q is transcendental.Reason: Since ℘′ is an odd function,℘′(ω1/2) = ℘′(ω2/2) = ℘′((ω1 + ω2)/2) = 0. Hence℘(ω1/2), ℘(ω2/2), ℘((ω1 + ω2)/2) are algebraic (known as twodivision points) which shows that non-zero periods aretranscendental.

One can also show by standard complex analytic argument,that the above two division points are distinct. And this impliesthat the discriminant

∆ := g32 − 27g2

3 6= 0.


Schneider(1937): Let L be a lattice and suppose that g2,g3 arealgebraic. Then for any algebraic α 6∈ L, ℘(α) is transcendental.As a consequence, we have that any non-zero period of anelliptic curve defined over Q is transcendental.Reason: Since ℘′ is an odd function,℘′(ω1/2) = ℘′(ω2/2) = ℘′((ω1 + ω2)/2) = 0. Hence℘(ω1/2), ℘(ω2/2), ℘((ω1 + ω2)/2) are algebraic (known as twodivision points) which shows that non-zero periods aretranscendental.One can also show by standard complex analytic argument,that the above two division points are distinct.

And this impliesthat the discriminant

∆ := g32 − 27g2

3 6= 0.


Schneider(1937): Let L be a lattice and suppose that g2,g3 arealgebraic. Then for any algebraic α 6∈ L, ℘(α) is transcendental.As a consequence, we have that any non-zero period of anelliptic curve defined over Q is transcendental.Reason: Since ℘′ is an odd function,℘′(ω1/2) = ℘′(ω2/2) = ℘′((ω1 + ω2)/2) = 0. Hence℘(ω1/2), ℘(ω2/2), ℘((ω1 + ω2)/2) are algebraic (known as twodivision points) which shows that non-zero periods aretranscendental.One can also show by standard complex analytic argument,that the above two division points are distinct. And this impliesthat the discriminant

∆ := g32 − 27g2

3 6= 0.

modular property of g2,g3 and ∆.

Let H denote the upper half plane and take z ∈ H. Let Lz be thelattice spanned by 1 and z. Then the corresponding g2,g3 and∆ are functions of z.

Suppose(

a bc d

)∈ SL2(Z). Then

(a bc d

)· z =

(az + bcz + d

).

And

g2

(az + bcz + d

)= (cz + d)4g2(z), g3

(az + bcz + d

)= (cz + d)6g3(z),

∆

(az + bcz + d

)= (cz + d)12∆(z).


Let H denote the upper half plane and take z ∈ H.

Let Lz be thelattice spanned by 1 and z. Then the corresponding g2,g3 and∆ are functions of z.

Suppose(

a bc d

)∈ SL2(Z). Then

(a bc d

)· z =

(az + bcz + d

).

And

g2

(az + bcz + d

)= (cz + d)4g2(z), g3

(az + bcz + d

)= (cz + d)6g3(z),

∆

(az + bcz + d

)= (cz + d)12∆(z).


Let H denote the upper half plane and take z ∈ H. Let Lz be thelattice spanned by 1 and z.

Then the corresponding g2,g3 and∆ are functions of z.

Suppose(

a bc d

)∈ SL2(Z). Then

(a bc d

)· z =

(az + bcz + d

).

And

g2

(az + bcz + d

)= (cz + d)4g2(z), g3

(az + bcz + d

)= (cz + d)6g3(z),

∆

(az + bcz + d

)= (cz + d)12∆(z).



Suppose(

a bc d

)∈ SL2(Z). Then

(a bc d

)· z =

(az + bcz + d

).

And

g2

(az + bcz + d

)= (cz + d)4g2(z), g3

(az + bcz + d

)= (cz + d)6g3(z),

∆

(az + bcz + d

)= (cz + d)12∆(z).



Suppose(

a bc d

)∈ SL2(Z).

Then

(a bc d

)· z =

(az + bcz + d

).

And

g2

(az + bcz + d

)= (cz + d)4g2(z), g3

(az + bcz + d

)= (cz + d)6g3(z),

∆

(az + bcz + d

)= (cz + d)12∆(z).



Suppose(

a bc d

)∈ SL2(Z). Then

(a bc d

)· z =

(az + bcz + d

).

And

g2

(az + bcz + d

)= (cz + d)4g2(z), g3

(az + bcz + d

)= (cz + d)6g3(z),

∆

(az + bcz + d

)= (cz + d)12∆(z).

The modular j− invariant function

j = j(z) =1728g3

2

g32 − 27g2

3

is called the modular invariant of the elliptic curve E .Why modular invariant:

j(

az + bcz + d

)= j(z)

so invariant under the action of SL2(Z).


j = j(z) =1728g3

2

g32 − 27g2

3

is called the modular invariant of the elliptic curve E .

Why modular invariant:

j(

az + bcz + d

)= j(z)



j = j(z) =1728g3

2

g32 − 27g2

3

is called the modular invariant of the elliptic curve E .Why modular invariant:

j(

az + bcz + d

)= j(z)


Known facts

(i) Two elliptic curves over C are isomorphic if and only if theyhave the same j .(ii) The j function takes every complex value exactly once onthe quotient space H/SL2(Z).

(iii) j(i) = 1728; j((1 +√−3)/2) = 0.

Known facts

(i) Two elliptic curves over C are isomorphic if and only if theyhave the same j .

(ii) The j function takes every complex value exactly once onthe quotient space H/SL2(Z).

(iii) j(i) = 1728; j((1 +√−3)/2) = 0.

Known facts

(i) Two elliptic curves over C are isomorphic if and only if theyhave the same j .(ii) The j function takes every complex value exactly once onthe quotient space H/SL2(Z).

(iii) j(i) = 1728; j((1 +√−3)/2) = 0.

Fourier expansion of j

We shall denote from now on ω2/ω1 = τ and assume thatIm(τ) > 0. Put q = e2πiτ , J(e2πiτ ) = j(τ). Then

J(q) =1q

(1 + 240

∞∑m=1

m3 qm

1− qm

)3 ∞∏n=1

(1− qn)24

=1q

+ 744 + 196884q + 21493760q2 + .....


We shall denote from now on ω2/ω1 = τ and assume thatIm(τ) > 0.

Put q = e2πiτ , J(e2πiτ ) = j(τ). Then

J(q) =1q

(1 + 240

∞∑m=1

m3 qm

1− qm

)3 ∞∏n=1

(1− qn)24

=1q

+ 744 + 196884q + 21493760q2 + .....


We shall denote from now on ω2/ω1 = τ and assume thatIm(τ) > 0. Put q = e2πiτ ,

J(e2πiτ ) = j(τ). Then

J(q) =1q

(1 + 240

∞∑m=1

m3 qm

1− qm

)3 ∞∏n=1

(1− qn)24

=1q

+ 744 + 196884q + 21493760q2 + .....


We shall denote from now on ω2/ω1 = τ and assume thatIm(τ) > 0. Put q = e2πiτ , J(e2πiτ ) = j(τ).

Then

J(q) =1q

(1 + 240

∞∑m=1

m3 qm

1− qm

)3 ∞∏n=1

(1− qn)24

=1q

+ 744 + 196884q + 21493760q2 + .....


We shall denote from now on ω2/ω1 = τ and assume thatIm(τ) > 0. Put q = e2πiτ , J(e2πiτ ) = j(τ). Then

J(q) =1q

(1 + 240

∞∑m=1

m3 qm

1− qm

)3 ∞∏n=1

(1− qn)24

=1q

+ 744 + 196884q + 21493760q2 + .....

Complex Multiplication

All analytic endomorphisms of C/L denoted as End(E) (whereE is the associated elliptic curve) are of the form αz + L forsome α satisfying αL ⊆ L. In particular, for Lτ , we have

α = aτ + b;ατ = cτ + d .

Thus τ is an algebraic integer of degree at most two over Q.End(E) is either Z or an order in an imaginary quadratic field k.The latter case arises if and only if τ is a quadratic number.Then the field of endomorphisms is k = Q(τ) and we say E hascomplex multiplication.


All analytic endomorphisms of C/L denoted as End(E) (whereE is the associated elliptic curve) are of the form αz + L forsome α satisfying αL ⊆ L.

In particular, for Lτ , we have

α = aτ + b;ατ = cτ + d .




α = aτ + b;ατ = cτ + d .




α = aτ + b;ατ = cτ + d .

Thus τ is an algebraic integer of degree at most two over Q.

End(E) is either Z or an order in an imaginary quadratic field k.The latter case arises if and only if τ is a quadratic number.Then the field of endomorphisms is k = Q(τ) and we say E hascomplex multiplication.



α = aτ + b;ατ = cτ + d .

Thus τ is an algebraic integer of degree at most two over Q.End(E) is either Z or an order in an imaginary quadratic field k.

The latter case arises if and only if τ is a quadratic number.Then the field of endomorphisms is k = Q(τ) and we say E hascomplex multiplication.



α = aτ + b;ατ = cτ + d .


An interesting fact in the CM case

In CM casej(τ) IS AN ALGEBRAIC INTEGERITS DEGREE = h− THE CLASS NUMBER OF THEQUADRATIC FIELD k = Q(τ).

Suppose τ ∈ H with j(τ) algebraic. Then

τ algebraic if and only if τ is imaginary quadratic.

Hilbert’s 7th problem

Gelfond and Schneider (1934) independently showed:For α and β algebraic numbers with α 6= 0 and β 6∈ Q and forany choice of logα 6= 0, the number αβ is transcendental.

Note: (i) eπ is transcendental. So are 2√

2,eπ√

163.

(ii) The functions ez and eβz are algebraically independent. Butcannot take algebraic values at logα.Schneider (1937) using elliptic analogue of the above resultshowed: Suppose τ ∈ H is algebraic and not an imaginaryquadratic, then the value j(τ) is transcendental. Schneiderhimself asked if this can be proved using modular functions.


Gelfond and Schneider (1934) independently showed:

For α and β algebraic numbers with α 6= 0 and β 6∈ Q and forany choice of logα 6= 0, the number αβ is transcendental.


2,eπ√

163.





2,eπ√

163.




Note: (i) eπ is transcendental.

So are 2√

2,eπ√

163.





2,eπ√

163.





2,eπ√

163.

(ii) The functions ez and eβz are algebraically independent. Butcannot take algebraic values at logα.

Schneider (1937) using elliptic analogue of the above resultshowed: Suppose τ ∈ H is algebraic and not an imaginaryquadratic, then the value j(τ) is transcendental. Schneiderhimself asked if this can be proved using modular functions.




2,eπ√

163.

(ii) The functions ez and eβz are algebraically independent. Butcannot take algebraic values at logα.Schneider (1937) using elliptic analogue of the above resultshowed:

Suppose τ ∈ H is algebraic and not an imaginaryquadratic, then the value j(τ) is transcendental. Schneiderhimself asked if this can be proved using modular functions.




2,eπ√

163.

(ii) The functions ez and eβz are algebraically independent. Butcannot take algebraic values at logα.Schneider (1937) using elliptic analogue of the above resultshowed: Suppose τ ∈ H is algebraic and not an imaginaryquadratic, then the value j(τ) is transcendental.

Schneiderhimself asked if this can be proved using modular functions.




2,eπ√

163.


Diophantine Approximation via the j function

eπ√

163 = 262537412640768743.9999999999992......Suppose τ = 1+i

√163

2 . Then q = −e−π√

163.

j(τ) = −(640320)3

and ∣∣∣∣j(τ)− 1q− 744

∣∣∣∣ < 10−12

i.e| − (640320)3 − 744 + eπ

√163| < 10−12

i.e j(τ) is a good approximation to eπ√

163.

So why is this happening? What is special about eπ√

163?


eπ√

163

= 262537412640768743.9999999999992......Suppose τ = 1+i

√163

2 . Then q = −e−π√

163.

j(τ) = −(640320)3

and ∣∣∣∣j(τ)− 1q− 744

∣∣∣∣ < 10−12

i.e| − (640320)3 − 744 + eπ

√163| < 10−12


163.


163?


eπ√

163 = 262537412640768743.9999999999992......

Suppose τ = 1+i√

1632 . Then q = −e−π

√163.

j(τ) = −(640320)3

and ∣∣∣∣j(τ)− 1q− 744

∣∣∣∣ < 10−12

i.e| − (640320)3 − 744 + eπ

√163| < 10−12


163.


163?


eπ√

163 = 262537412640768743.9999999999992......Suppose τ = 1+i

√163

2 .

Then q = −e−π√

163.

j(τ) = −(640320)3

and ∣∣∣∣j(τ)− 1q− 744

∣∣∣∣ < 10−12

i.e| − (640320)3 − 744 + eπ

√163| < 10−12


163.


163?


eπ√

163 = 262537412640768743.9999999999992......Suppose τ = 1+i

√163

2 . Then q = −e−π√

163.

j(τ) = −(640320)3

and ∣∣∣∣j(τ)− 1q− 744

∣∣∣∣ < 10−12

i.e| − (640320)3 − 744 + eπ

√163| < 10−12


163.


163?


eπ√

163 = 262537412640768743.9999999999992......Suppose τ = 1+i

√163

2 . Then q = −e−π√

163.

j(τ) = −(640320)3

and ∣∣∣∣j(τ)− 1q− 744

∣∣∣∣ < 10−12

i.e| − (640320)3 − 744 + eπ

√163| < 10−12


163.

So why is this happening?

What is special about eπ√

163?


eπ√

163 = 262537412640768743.9999999999992......Suppose τ = 1+i

√163

2 . Then q = −e−π√

163.

j(τ) = −(640320)3

and ∣∣∣∣j(τ)− 1q− 744

∣∣∣∣ < 10−12

i.e| − (640320)3 − 744 + eπ

√163| < 10−12


163.


163?

163

We have alreday seen thatj(τ) IS AN ALGEBRAIC INTEGER AND ITS DEGREEEQUALS h− THE CLASS NUMBER OF THE QUADRATICFIELD k = Q(τ). Let us take

τ =

{i√

d if d = 2 or 3 (mod 4)

(1 + i√

d)/2 if d ≡ 1(mod 4).

Then

q =

{e−2π

√d if d ≡ 2 or 3(mod 4)

−e−π√

d if d ≡ 1(mod 4).

If h = h(d) = 1 then the transcendental number 1/q which iseither e2π

√d or −eπ

√d is close to the rational integer

J(q)− 744. The error is ∼ 196884q which is small if d is large.And this is what is happening when d = 163 since h(163) = 1.This good approximation can be used to give effectiveimprovement over Liouville’s theorem for some numbers.

163

We have alreday seen that

j(τ) IS AN ALGEBRAIC INTEGER AND ITS DEGREEEQUALS h− THE CLASS NUMBER OF THE QUADRATICFIELD k = Q(τ). Let us take

τ =

{i√

d if d = 2 or 3 (mod 4)

(1 + i√

d)/2 if d ≡ 1(mod 4).

Then

q =

{e−2π

√d if d ≡ 2 or 3(mod 4)

−e−π√

d if d ≡ 1(mod 4).


√d or −eπ



163

We have alreday seen thatj(τ) IS AN ALGEBRAIC INTEGER AND ITS DEGREEEQUALS h− THE CLASS NUMBER OF THE QUADRATICFIELD k = Q(τ).

Let us take

τ =

{i√

d if d = 2 or 3 (mod 4)

(1 + i√

d)/2 if d ≡ 1(mod 4).

Then

q =

{e−2π

√d if d ≡ 2 or 3(mod 4)

−e−π√

d if d ≡ 1(mod 4).


√d or −eπ



163


τ =

{i√

d if d = 2 or 3 (mod 4)

(1 + i√

d)/2 if d ≡ 1(mod 4).

Then

q =

{e−2π

√d if d ≡ 2 or 3(mod 4)

−e−π√

d if d ≡ 1(mod 4).


√d or −eπ



163


τ =

{i√

d if d = 2 or 3 (mod 4)

(1 + i√

d)/2 if d ≡ 1(mod 4).

Then

q =

{e−2π

√d if d ≡ 2 or 3(mod 4)

−e−π√

d if d ≡ 1(mod 4).


√d or −eπ


J(q)− 744. The error is ∼ 196884q which is small if d is large.

And this is what is happening when d = 163 since h(163) = 1.This good approximation can be used to give effectiveimprovement over Liouville’s theorem for some numbers.

163


τ =

{i√

d if d = 2 or 3 (mod 4)

(1 + i√

d)/2 if d ≡ 1(mod 4).

Then

q =

{e−2π

√d if d ≡ 2 or 3(mod 4)

−e−π√

d if d ≡ 1(mod 4).


√d or −eπ


J(q)− 744. The error is ∼ 196884q which is small if d is large.And this is what is happening when d = 163 since h(163) = 1.

This good approximation can be used to give effectiveimprovement over Liouville’s theorem for some numbers.

163


τ =

{i√

d if d = 2 or 3 (mod 4)

(1 + i√

d)/2 if d ≡ 1(mod 4).

Then

q =

{e−2π

√d if d ≡ 2 or 3(mod 4)

−e−π√

d if d ≡ 1(mod 4).


√d or −eπ



Diophantine equations

We get a large integral solution (x , y) to the equation

163y2 = x3 + 1728

viz.,

x = −(j(τ))1/3 = 640320; y = 23337.11.19.127 = 40133016.

Dividing by 26 we get a large solution to

163y2 = x3 + 27.

Dividing by 33, we get

3 · 163y2 = x3 + 1.

Dividing by 63, we get

6 · 163y2 = x3 + 8.

Class number 1

There are 9 imaginary quadratic fields having class number1.They are

d = −3,−4,−7,−8,−11,−19,−43,−67,−163.

The Schläfli modular function is given by

f (z) = q−1/48∞∏

n=1

(1 + qn−1/2).

Then f (√

d) is an algebraic integer having degree 3h(d). It isfound that f (

√−163) is the real root of the equation

f 3 − 6f 2 + 4f − 2 = 0.

What is the connection to the j function?Let F (z) = f 8(z). Then F = F (

√d) satisfies

F 3 + j

(−3 +

√d

2

)1/3

F 2 − 256 = 0.

Class number 1

There are 9 imaginary quadratic fields having class number1.

They are

d = −3,−4,−7,−8,−11,−19,−43,−67,−163.


f (z) = q−1/48∞∏

n=1

(1 + qn−1/2).

Then f (√



f 3 − 6f 2 + 4f − 2 = 0.


√d) satisfies

F 3 + j

(−3 +

√d

2

)1/3

F 2 − 256 = 0.

Class number 1


d = −3,−4,−7,−8,−11,−19,−43,−67,−163.


f (z) = q−1/48∞∏

n=1

(1 + qn−1/2).

Then f (√



f 3 − 6f 2 + 4f − 2 = 0.


√d) satisfies

F 3 + j

(−3 +

√d

2

)1/3

F 2 − 256 = 0.

Class number 1


d = −3,−4,−7,−8,−11,−19,−43,−67,−163.


f (z) = q−1/48∞∏

n=1

(1 + qn−1/2).

Then f (√

d) is an algebraic integer having degree 3h(d).

It isfound that f (


f 3 − 6f 2 + 4f − 2 = 0.


√d) satisfies

F 3 + j

(−3 +

√d

2

)1/3

F 2 − 256 = 0.

Class number 1


d = −3,−4,−7,−8,−11,−19,−43,−67,−163.


f (z) = q−1/48∞∏

n=1

(1 + qn−1/2).

Then f (√



f 3 − 6f 2 + 4f − 2 = 0.


√d) satisfies

F 3 + j

(−3 +

√d

2

)1/3

F 2 − 256 = 0.

Class number 1


d = −3,−4,−7,−8,−11,−19,−43,−67,−163.


f (z) = q−1/48∞∏

n=1

(1 + qn−1/2).

Then f (√



f 3 − 6f 2 + 4f − 2 = 0.

What is the connection to the j function?

Let F (z) = f 8(z). Then F = F (√

d) satisfies

F 3 + j

(−3 +

√d

2

)1/3

F 2 − 256 = 0.

Class number 1


d = −3,−4,−7,−8,−11,−19,−43,−67,−163.


f (z) = q−1/48∞∏

n=1

(1 + qn−1/2).

Then f (√



f 3 − 6f 2 + 4f − 2 = 0.


√d) satisfies

F 3 + j

(−3 +

√d

2

)1/3

F 2 − 256 = 0.

More equations

Similarly it was found that

f (√−3) satisfies x3 − 2 = 0

f (√−11) satisfies x3 − 2x2 + 2x − 2 = 0

f (√−19) satisfies x3 − 2x − 2 = 0

f (√−43) satisfies x3 − 2x2 − 2 = 0

f (√−67) satisfies x3 − 2x2 − 2x − 2 = 0

f (√−163) satisfies x3 − 6x2 + 4x − 2 = 0.

And some of the partial quotients in their continued fractionexpansions are very large giving rise to very good rationalapproximations to these numbers. This was first noticed byBrillhart (1964) and the connection to Class number 1 wasanalyzed by Stark (1970). Using the good approximationproperty Chudnovsky (1983) could give effective improvementover Liouville’s theorem for these numbers.

More equations


f (√−3) satisfies x3 − 2 = 0

f (√−11) satisfies x3 − 2x2 + 2x − 2 = 0

f (√−19) satisfies x3 − 2x − 2 = 0

f (√−43) satisfies x3 − 2x2 − 2 = 0

f (√−67) satisfies x3 − 2x2 − 2x − 2 = 0

f (√−163) satisfies x3 − 6x2 + 4x − 2 = 0.

And some of the partial quotients in their continued fractionexpansions are very large giving rise to very good rationalapproximations to these numbers.

This was first noticed byBrillhart (1964) and the connection to Class number 1 wasanalyzed by Stark (1970). Using the good approximationproperty Chudnovsky (1983) could give effective improvementover Liouville’s theorem for these numbers.

More equations


f (√−3) satisfies x3 − 2 = 0

f (√−11) satisfies x3 − 2x2 + 2x − 2 = 0

f (√−19) satisfies x3 − 2x − 2 = 0

f (√−43) satisfies x3 − 2x2 − 2 = 0

f (√−67) satisfies x3 − 2x2 − 2x − 2 = 0

f (√−163) satisfies x3 − 6x2 + 4x − 2 = 0.

And some of the partial quotients in their continued fractionexpansions are very large giving rise to very good rationalapproximations to these numbers. This was first noticed byBrillhart (1964) and the connection to Class number 1 wasanalyzed by Stark (1970).

Using the good approximationproperty Chudnovsky (1983) could give effective improvementover Liouville’s theorem for these numbers.

More equations


f (√−3) satisfies x3 − 2 = 0

f (√−11) satisfies x3 − 2x2 + 2x − 2 = 0

f (√−19) satisfies x3 − 2x − 2 = 0

f (√−43) satisfies x3 − 2x2 − 2 = 0

f (√−67) satisfies x3 − 2x2 − 2x − 2 = 0

f (√−163) satisfies x3 − 6x2 + 4x − 2 = 0.

And some of the partial quotients in their continued fractionexpansions are very large giving rise to very good rationalapproximations to these numbers. This was first noticed byBrillhart (1964) and the connection to Class number 1 wasanalyzed by Stark (1970). Using the good approximationproperty Chudnovsky (1983) could give effective improvementover Liouville’s theorem for these numbers.

Ramanujan Functions

In his famous paper on " On certain Arithmetical Functions"Ramanujan (1916) introduced the following functions whichwere later recognized as special cases of normalizedEisenstein series.

P(q) = 1− 24∞∑

n=1

nqn

1− qn ,Q(q) = 1 + 240∞∑

n=1

n3qn

1− qn ,

R(q) = 1− 504∞∑

n=1

n5qn

1− qn .

Thus P(q) = E2(τ),Q(q) = E4(τ),R(q) = E6(τ) where

E2k (τ) =1

2ζ(2k)

∑(m,n)6=(0,0)

1(mτ + n)2k for k ≥ 2.

J in terms of Ramanujan Functions

Taking τ = ω2/ω1, the invariants g2,g3 satisfy

g2 =43

(π

ω1

)4

E4(τ),g3 =8

27

(π

ω1

)6

E6(τ).

The discriminant ∆ and J invariant function are related to theRamanujan functions as

∆ =Q3 − R2

123 = q∏

(1− qn)24 and J = Q3/∆.

Conjecture of Mahler and Manin

Manin’s Conjecture(1971): Let a be an algebraic numberdistinct from 0 and 1. Then for any ξ ∈ H at least one of thenumbers

aξ, j(ξ)

is transcendental.This is still open.Mahler proposed this in 1969 when a = e2πi . Mahler’sconjecture was resolved by K. Barré- Sirieix, G. Diaz, F.Gramain and G. Philibert in 1996. Thus we haveFor any τ ∈ C with Im τ > 0, at least one of the numberse2πiτ , j(τ) is transcendental.In other words,For any algebraic q with 0 < |q| < 1, the value J(q) istranscendental.



aξ, j(ξ)

is transcendental.This is still open.

Mahler proposed this in 1969 when a = e2πi . Mahler’sconjecture was resolved by K. Barré- Sirieix, G. Diaz, F.Gramain and G. Philibert in 1996. Thus we haveFor any τ ∈ C with Im τ > 0, at least one of the numberse2πiτ , j(τ) is transcendental.In other words,For any algebraic q with 0 < |q| < 1, the value J(q) istranscendental.



aξ, j(ξ)

is transcendental.This is still open.Mahler proposed this in 1969 when a = e2πi .

Mahler’sconjecture was resolved by K. Barré- Sirieix, G. Diaz, F.Gramain and G. Philibert in 1996. Thus we haveFor any τ ∈ C with Im τ > 0, at least one of the numberse2πiτ , j(τ) is transcendental.In other words,For any algebraic q with 0 < |q| < 1, the value J(q) istranscendental.



aξ, j(ξ)

is transcendental.This is still open.Mahler proposed this in 1969 when a = e2πi . Mahler’sconjecture was resolved by K. Barré- Sirieix, G. Diaz, F.Gramain and G. Philibert in 1996. Thus we haveFor any τ ∈ C with Im τ > 0, at least one of the numberse2πiτ , j(τ) is transcendental.

In other words,For any algebraic q with 0 < |q| < 1, the value J(q) istranscendental.



aξ, j(ξ)

is transcendental.This is still open.Mahler proposed this in 1969 when a = e2πi . Mahler’sconjecture was resolved by K. Barré- Sirieix, G. Diaz, F.Gramain and G. Philibert in 1996. Thus we haveFor any τ ∈ C with Im τ > 0, at least one of the numberse2πiτ , j(τ) is transcendental.In other words,For any algebraic q with 0 < |q| < 1, the value J(q) istranscendental.

Results of Nesterenko

Their method was used by Nesterenko to prove algebraicindependence results for the values of the Ramanujanfunctions. For instance, it can be shown thatIf q ∈ C with 0 < |q| < 1, then at least three of the numbersq,P(q),Q(q),R(q) are algebraically independent over Q.From this result one can derive thatIf τ ∈ H not equivalent to i or e2πi/3 then at least three numbersamong q, J(q), δJ(q), δ2J(q) are algebraically independent overQ. Here δ = z d

dz .

Further he also showed thatfor any natural number d, the numbers π and eπ

√d are

algebraically independent.


Their method was used by Nesterenko to prove algebraicindependence results for the values of the Ramanujanfunctions.

For instance, it can be shown thatIf q ∈ C with 0 < |q| < 1, then at least three of the numbersq,P(q),Q(q),R(q) are algebraically independent over Q.From this result one can derive thatIf τ ∈ H not equivalent to i or e2πi/3 then at least three numbersamong q, J(q), δJ(q), δ2J(q) are algebraically independent overQ. Here δ = z d

dz .


√d are



Their method was used by Nesterenko to prove algebraicindependence results for the values of the Ramanujanfunctions. For instance, it can be shown that

If q ∈ C with 0 < |q| < 1, then at least three of the numbersq,P(q),Q(q),R(q) are algebraically independent over Q.From this result one can derive thatIf τ ∈ H not equivalent to i or e2πi/3 then at least three numbersamong q, J(q), δJ(q), δ2J(q) are algebraically independent overQ. Here δ = z d

dz .


√d are



Their method was used by Nesterenko to prove algebraicindependence results for the values of the Ramanujanfunctions. For instance, it can be shown thatIf q ∈ C with 0 < |q| < 1, then at least three of the numbersq,P(q),Q(q),R(q) are algebraically independent over Q.

From this result one can derive thatIf τ ∈ H not equivalent to i or e2πi/3 then at least three numbersamong q, J(q), δJ(q), δ2J(q) are algebraically independent overQ. Here δ = z d

dz .


√d are



Their method was used by Nesterenko to prove algebraicindependence results for the values of the Ramanujanfunctions. For instance, it can be shown thatIf q ∈ C with 0 < |q| < 1, then at least three of the numbersq,P(q),Q(q),R(q) are algebraically independent over Q.From this result one can derive that

If τ ∈ H not equivalent to i or e2πi/3 then at least three numbersamong q, J(q), δJ(q), δ2J(q) are algebraically independent overQ. Here δ = z d

dz .


√d are




dz .


√d are




dz .

Further he also showed that

for any natural number d, the numbers π and eπ√

d arealgebraically independent.



dz .


√d are


Coefficients of the j function

Let us write

J(q) =∞∑

n=−1

c(n)qn with c(−1) = 1, c(0) = 744 and so on.

Surprisingly, little is known about the behaviour of thecoefficients of J(q) modulo a prime. Lehner (1949): made anextensive study on the congruence properties of thecoefficients. In particular he showed that

c(n) ≡ 0(mod p) for p =,2,3,5,7,11 and n ≡ 0(mod p).

Thus for all even integers n, we have c(n) even. In fact, it is wellknown (we will see in a while) that c(n) is even whenevern 6≡ 7(mod 8). What can we say when n ≡ 7(mod 8)? Thereare some contributions by Ono, Ramsey and Alfes.


Let us write

J(q) =∞∑

n=−1


Surprisingly, little is known about the behaviour of thecoefficients of J(q) modulo a prime.

Lehner (1949): made anextensive study on the congruence properties of thecoefficients. In particular he showed that




Let us write

J(q) =∞∑

n=−1


Surprisingly, little is known about the behaviour of thecoefficients of J(q) modulo a prime. Lehner (1949): made anextensive study on the congruence properties of thecoefficients.

In particular he showed that




Let us write

J(q) =∞∑

n=−1






Let us write

J(q) =∞∑

n=−1




Thus for all even integers n, we have c(n) even.

In fact, it is wellknown (we will see in a while) that c(n) is even whenevern 6≡ 7(mod 8). What can we say when n ≡ 7(mod 8)? Thereare some contributions by Ono, Ramsey and Alfes.


Let us write

J(q) =∞∑

n=−1




Thus for all even integers n, we have c(n) even. In fact, it is wellknown (we will see in a while) that c(n) is even whenevern 6≡ 7(mod 8).

What can we say when n ≡ 7(mod 8)? Thereare some contributions by Ono, Ramsey and Alfes.


Let us write

J(q) =∞∑

n=−1




Thus for all even integers n, we have c(n) even. In fact, it is wellknown (we will see in a while) that c(n) is even whenevern 6≡ 7(mod 8). What can we say when n ≡ 7(mod 8)?

Thereare some contributions by Ono, Ramsey and Alfes.


Let us write

J(q) =∞∑

n=−1





A recent result

The following result is due to Ram Murty and Thangadurai(2015):There are infinitely many integers n ≡ 7(mod 8) for which c(n)is odd; there are infinitely many integers n ≡ 7(mod 8) forwhich c(n) is even. Further, the number of such n ≤ x is� log log x in each case. It is expected that for half ofn ≡ 7(mod 8) we should have c(n) odd.

A recent result

The following result is due to Ram Murty and Thangadurai(2015):There are infinitely many integers n ≡ 7(mod 8) for which c(n)is odd;

there are infinitely many integers n ≡ 7(mod 8) forwhich c(n) is even. Further, the number of such n ≤ x is� log log x in each case. It is expected that for half ofn ≡ 7(mod 8) we should have c(n) odd.

A recent result

The following result is due to Ram Murty and Thangadurai(2015):There are infinitely many integers n ≡ 7(mod 8) for which c(n)is odd; there are infinitely many integers n ≡ 7(mod 8) forwhich c(n) is even.

Further, the number of such n ≤ x is� log log x in each case. It is expected that for half ofn ≡ 7(mod 8) we should have c(n) odd.

A recent result

The following result is due to Ram Murty and Thangadurai(2015):There are infinitely many integers n ≡ 7(mod 8) for which c(n)is odd; there are infinitely many integers n ≡ 7(mod 8) forwhich c(n) is even. Further, the number of such n ≤ x is� log log x in each case.

It is expected that for half ofn ≡ 7(mod 8) we should have c(n) odd.

A recent result

The following result is due to Ram Murty and Thangadurai(2015):There are infinitely many integers n ≡ 7(mod 8) for which c(n)is odd; there are infinitely many integers n ≡ 7(mod 8) forwhich c(n) is even. Further, the number of such n ≤ x is� log log x in each case. It is expected that for half ofn ≡ 7(mod 8) we should have c(n) odd.

Proof of the result of Ram and Thanga

From the expression for the j− function,

J(q)∆(q) = Q3(q) ≡ 1(mod 2).

Now

∆(q) = q∞∏

n=1

(1− qn)24 ≡ q∞∏

n=1

(1 + q8n + q16n + q24n)(mod 2).

Hencec(n) ≡ 0(mod 2) if n + 1 6≡ 0(mod 8)

proving that c(n) is even whenever n 6≡ 7(mod 8).

Use of Jacobi Identity

We again consider ∆(q) ≡ q∏∞

n=1(1− q8n)3(mod 2).The wellknown Jacobi identity:

∞∏n=1

(1− qn)3 =∞∑

k=0

(−1)k (2k + 1)qk(k+1)/2.

Hence

∆(q) ≡ q∞∑

n=0

q8n(n+1)/2 ≡∞∑

n=0

q(2n+1)2(mod 2).

Thus we get∞∑

n=0

∑k≥0

c(n − (2k + 1)2)qn ≡ 1(mod 2).

Comparing the coefficients of qn we get∑m≥0

c(n − (2m + 1)2) ≡ 0(mod 2) for n ≥ 1.



n=1(1− q8n)3(mod 2).

The wellknown Jacobi identity:

∞∏n=1

(1− qn)3 =∞∑

k=0

(−1)k (2k + 1)qk(k+1)/2.

Hence

∆(q) ≡ q∞∑

n=0

q8n(n+1)/2 ≡∞∑

n=0

q(2n+1)2(mod 2).

Thus we get∞∑

n=0

∑k≥0

c(n − (2k + 1)2)qn ≡ 1(mod 2).


c(n − (2m + 1)2) ≡ 0(mod 2) for n ≥ 1.



n=1(1− q8n)3(mod 2).The wellknown Jacobi identity:

∞∏n=1

(1− qn)3 =∞∑

k=0

(−1)k (2k + 1)qk(k+1)/2.

Hence

∆(q) ≡ q∞∑

n=0

q8n(n+1)/2 ≡∞∑

n=0

q(2n+1)2(mod 2).

Thus we get∞∑

n=0

∑k≥0

c(n − (2k + 1)2)qn ≡ 1(mod 2).


c(n − (2m + 1)2) ≡ 0(mod 2) for n ≥ 1.

Proof contd..

Take n = 4a(a + 1)− 1 with a ≥ 1. Then the previouscongruence becomes∑m≥0

c(n−(2m+1)2) ≡∑m≥0

c(4a(a+1)−4m(m+1)−1) ≡ 0(mod 2).

Here when m = a we get the term c(−1) = 1. When m < a,observe that

4a(a + 1)− 4m(m + 1) > a.

Thus if for all the integers m in the interval (a,4a(a + 1)− 1],c(m) is even, the above sum gives a contradiction. Hence forsome m in this interval, c(m) is odd. As already seen,m ≡ 7(mod 8).

Proof contd..

Take n = 4a(a + 1)− 1 with a ≥ 1.

Then the previouscongruence becomes∑m≥0

c(n−(2m+1)2) ≡∑m≥0

c(4a(a+1)−4m(m+1)−1) ≡ 0(mod 2).


4a(a + 1)− 4m(m + 1) > a.


Proof contd..


c(n−(2m+1)2) ≡∑m≥0

c(4a(a+1)−4m(m+1)−1) ≡ 0(mod 2).


4a(a + 1)− 4m(m + 1) > a.


Proof contd..


c(n−(2m+1)2) ≡∑m≥0

c(4a(a+1)−4m(m+1)−1) ≡ 0(mod 2).

Here when m = a we get the term c(−1) = 1.

When m < a,observe that

4a(a + 1)− 4m(m + 1) > a.


Proof contd..


c(n−(2m+1)2) ≡∑m≥0

c(4a(a+1)−4m(m+1)−1) ≡ 0(mod 2).


4a(a + 1)− 4m(m + 1) > a.


Proof contd..


c(n−(2m+1)2) ≡∑m≥0

c(4a(a+1)−4m(m+1)−1) ≡ 0(mod 2).


4a(a + 1)− 4m(m + 1) > a.

Thus if for all the integers m in the interval (a,4a(a + 1)− 1],c(m) is even, the above sum gives a contradiction.

Hence forsome m in this interval, c(m) is odd. As already seen,m ≡ 7(mod 8).

Proof contd..


c(n−(2m+1)2) ≡∑m≥0

c(4a(a+1)−4m(m+1)−1) ≡ 0(mod 2).


4a(a + 1)− 4m(m + 1) > a.


Counting n ≤ x with c(n) odd

Put a0 = 1,a1 = 7 and

ak = 4ak−1(ak−1 + 1)− 1 for k ≥ 2.

Let ` be the largest integer such that a` ≤ x . Then each one ofthe intervals [a0,a1], (a1,a2], · · · , (a`, x ] has an integer m suchthat c(m) is odd. Thus there are at least ` integers ≤ x forwhich c(m) is odd. Thus we need to get a lower bound for `.Using the recurrence relation satisfied by ak ’s, we get

ak < 8a2k−1 for k ≥ 0.

By iterating, we get

ak ≤ 8ka2k−1

1 ≤ 82kfor k ≥ 0.

Since a`+1 ≥ x we get `� log log x . The proof that there are� log log x number of n ≤ x with c(n) even is similar.


Put a0 = 1,a1 = 7 and

ak = 4ak−1(ak−1 + 1)− 1 for k ≥ 2.

Let ` be the largest integer such that a` ≤ x .

Then each one ofthe intervals [a0,a1], (a1,a2], · · · , (a`, x ] has an integer m suchthat c(m) is odd. Thus there are at least ` integers ≤ x forwhich c(m) is odd. Thus we need to get a lower bound for `.Using the recurrence relation satisfied by ak ’s, we get

ak < 8a2k−1 for k ≥ 0.


ak ≤ 8ka2k−1

1 ≤ 82kfor k ≥ 0.



Put a0 = 1,a1 = 7 and

ak = 4ak−1(ak−1 + 1)− 1 for k ≥ 2.

Let ` be the largest integer such that a` ≤ x . Then each one ofthe intervals [a0,a1], (a1,a2], · · · , (a`, x ] has an integer m suchthat c(m) is odd.

Thus there are at least ` integers ≤ x forwhich c(m) is odd. Thus we need to get a lower bound for `.Using the recurrence relation satisfied by ak ’s, we get

ak < 8a2k−1 for k ≥ 0.


ak ≤ 8ka2k−1

1 ≤ 82kfor k ≥ 0.



Put a0 = 1,a1 = 7 and

ak = 4ak−1(ak−1 + 1)− 1 for k ≥ 2.

Let ` be the largest integer such that a` ≤ x . Then each one ofthe intervals [a0,a1], (a1,a2], · · · , (a`, x ] has an integer m suchthat c(m) is odd. Thus there are at least ` integers ≤ x forwhich c(m) is odd.

Thus we need to get a lower bound for `.Using the recurrence relation satisfied by ak ’s, we get

ak < 8a2k−1 for k ≥ 0.


ak ≤ 8ka2k−1

1 ≤ 82kfor k ≥ 0.



Put a0 = 1,a1 = 7 and

ak = 4ak−1(ak−1 + 1)− 1 for k ≥ 2.

Let ` be the largest integer such that a` ≤ x . Then each one ofthe intervals [a0,a1], (a1,a2], · · · , (a`, x ] has an integer m suchthat c(m) is odd. Thus there are at least ` integers ≤ x forwhich c(m) is odd. Thus we need to get a lower bound for `.

Using the recurrence relation satisfied by ak ’s, we get

ak < 8a2k−1 for k ≥ 0.


ak ≤ 8ka2k−1

1 ≤ 82kfor k ≥ 0.



Put a0 = 1,a1 = 7 and

ak = 4ak−1(ak−1 + 1)− 1 for k ≥ 2.


ak < 8a2k−1 for k ≥ 0.


ak ≤ 8ka2k−1

1 ≤ 82kfor k ≥ 0.



Put a0 = 1,a1 = 7 and

ak = 4ak−1(ak−1 + 1)− 1 for k ≥ 2.


ak < 8a2k−1 for k ≥ 0.


ak ≤ 8ka2k−1

1 ≤ 82kfor k ≥ 0.

Since a`+1 ≥ x we get `� log log x .

The proof that there are� log log x number of n ≤ x with c(n) even is similar.


Put a0 = 1,a1 = 7 and

ak = 4ak−1(ak−1 + 1)− 1 for k ≥ 2.


ak < 8a2k−1 for k ≥ 0.


ak ≤ 8ka2k−1

1 ≤ 82kfor k ≥ 0.


Modular J-invariant function - Tata Institute of …mathstud/seminar/events/2016-01-25...2016/01/25...

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