ELLIPSES ANDELLIPTIC CURVES

M. Ram MurtyQueen’s University

Planetary orbits are elliptical

What is an ellipse?

x2

a2 + y2

b2 = 1

An ellipse has two foci

From: gomath.com/geometry/ellipse.php

Metric mishap causes loss of Mars orbiter (Sept. 30, 1999)

How to calculate arc length

Let f : [0;1] ! R2 be given by t 7! (x(t);y(t)).

By Pythagoras, ¢ s =p

(¢ x)2 + (¢ y)2

ds =

r¡

dxdt

¢2 +³

dydt

´2dt

Arc length =R1

0

r¡

dxdt

¢2 +³

dydt

´2dt

Circumference of an ellipseWecan parametrizethepointsof an ellipsein the¯rst quadrant byf : [0;¼=2]! (asint;bcost):

Circumference =4

R¼=20

pa2 cos2 t + b2 sin2 tdt:

Observe that if a = b, we get 2¼a.

Wecan simplify this as follows.Let cos2 µ= 1¡ sin2 µ.The integral becomes4a

R¼=20

p1¡ ¸ sin2 µdµ

Where¸ = 1¡ b2=a2.

Wecan expand thesquare root usingthebinomial theorem:4a

R¼=20

P 1n=0

³1=2n

´(¡ 1)n¸n sin2n µdµ:

We can use the fact that2

R¼=20

sin2n µdµ= ¼(1=2)(1=2)+1)¢¢¢((1=2)+(n¡ 1))n!

The final answerThe circumference is given by2¼aF (1=2;¡ 1=2;1;¸)whereF (a;b;c;z) =

P 1n=0

(a)n (b)nn!(c)n

zn

is the hypergeometric series and(a)n = a(a+ 1)(a+ 2) ¢¢¢(a+ n ¡ 1):

We now use Landens transformation:F (a;b;2b; 4x

(1+x)2 ) = (1+ x)2aF (a;a ¡ b+ 1=2;b+ 1=2;x2).

The answer can now be written as ¼(a+ b)F (¡ 1=2;¡ 1=2;1;x2).

We put a = ¡ 1=2;b= 1=2 and x = (a¡ b)=(a+ b).

Approximations

• In 1609, Kepler used the approximation (a+b). The above formula shows the perimeter is always greater than this amount.

• In 1773, Euler gave the approximation 2√(a²+b²)/2.

• In 1914, Ramanujan gave the approximation (3(a+b) -√(a+3b)(3a+b)).

What kind of number is this?

For example, if a and bare rational,is the circumeference irrational?

In case a = bwe have a circleand the circumference 2¼ais irrational.

In fact, ¼is transcendental.

What does this mean?

In 1882, Ferdinand von Lindemannproved that ¼is transcendental.

This means that ¼does not satisfyan equation of the typexn +an¡ 1xn¡ 1 +¢¢¢+a1x +a0 = 0with ai rational numbers.

1852-1939

This means that you can’t square the circle!

• There is an ancient problem of constructing a square with straightedge and compass whose area equals .

• Theorem: if you can construct a line segment of length α then α is an algebraic number.

• Since is not algebraic, neither is √.

Some other interesting numbers

• The number e is transcendental.

• This was first proved by Charles Hermite (1822-1901) in 1873.

Is ¼+ e transcendental?

Answer: unknown.Is ¼e transcendental?Answer: unknown.

Not both can be algebraic!

• Here’s a proof.

• If both are algebraic, then(x ¡ e)(x¡ ¼) = x2 ¡ (e+ ¼)x + ¼eis a quadratic polynomial with algebraic coe±cients.This implies both e and ¼are algebraic,a contradiction to the theorems ofHermite and Lindemann.

Conjecture:¼and e are algebraically independent.

But what about the case of the ellipse?

Lets look at the integral again.

Yes.This is a theorem ofTheodor Schneider (1911-1988)

Is the integralR¼=20

pa2 cos2 t + b2 sin2 tdt

transcendentalif a and bare rational?

Putting u = sint the integral becomes

Set k2 = 1¡ b2=a2:

The integral becomes

aR1

0

q1¡ k2u2

1¡ u2 du:

Put t = 1¡ k2u2:

R10

qa2¡ (a2¡ b2)u2

1¡ u2 du:

12

R11¡ k2

tdtpt(t¡ 1)(t¡ (1¡ k2))

Elliptic Integrals

Integrals of the formRdxp

x3+a2x2+a3x+a4are called elliptic integrals of the ¯rst kind.

Integrals of the formRxdxp

x3+a2x2+a3x+a4are called elliptic integrals of the second kind.

Our integral is of thesecond kind.

Elliptic Curves

The equation

is an example of an elliptic curve.

y2 = x(x ¡ 1)(x ¡ (1¡ k2))

0 1¡ k2 1

One can write theequation of such a curve asy2 = 4x3 ¡ ax ¡ b.

Elliptic Curves over C

Let L bea latticeof rank 2 over R.

This means that L = Z! 1 ©Z! 2

We attach the Weierstrass }-function to L:

}(z) = 1z2 +

P06=! 2L

³1

(z¡ ! )2 ¡ 1! 2

´

! 1

! 2

The} function is doubly periodic:This means}(z) = }(z + ! 1) = }(z + ! 2).

e2¼i = 1Theexponential function is periodic sinceez = ez+2¼i :The periods of the exponential functionconsist of multiples of 2¼i.

That is, the period \ lattice; ;

is of the form Z(2¼i).

The Weierstrass function

This means that (}(z);};(z))is a point on the curvey2 = 4x3¡ g2x¡ g3:

The }-function satis̄ esthe following di®erential equation:(};(z))2 = 4}(z)3¡ g2}(z)¡ g3;whereg2 = 60

P06=! 2L ! ¡ 4

andg3 = 140

P06=! 2L ! ¡ 6:

The Uniformization Theorem

Conversely, every complex point on the curvey2 = 4x3 ¡ g2x ¡ g3is of the form (}(z);};(z))for some z 2 C.

Given any g2;g3 2 C,there is a } function such that(};(z))2 = 4}(z)3¡ g2}(z)¡ g3:

Even and Odd Functions

Recall that a function f is even iff (z) = f (¡ z).For example, z2 andcosz are even functions.

A function is called odd iff (z) = ¡ f (¡ z).For example, z andsinz are odd functions.

}(z) is even since }(z) = 1z2 +

P06=! 2L

³1

(z¡ ! )2 ¡ 1! 2

´

};(z) is odd since };(z) = ¡ 2z3 ¡ 2

P06=! 2L

³1

(z¡ ! )3

´.

Note that };(¡ ! =2) = ¡ };(! =2)

This means };(! =2) = 0:

In particular};(! 1=2) = 0};(! 2=2) = 0 and};((! 1 + ! 2)=2) = 0.

This means the numbers}(! 1=2);}(! 2=2);}((! 1 + ! 2)=2)are roots of the cubic4x3¡ g2x¡ g3 = 0:

One can show these roots are distinct.In particular, if g2;g3 are algebraic,the roots are algebraic.

Schneider’s Theorem

If g2;g3 arealgebraicand } is theassociatedWeierstrass }-function, thenfor ®algebraic,}(®) is transcendental.

This is theelliptic analog of theHermite-Lindemann theoremthat says if ®is a non-zero algebraic number,then e® is transcendental.

Note that weget ¼transcendental by setting ®= 2¼i

Since}(! 1=2),}(! 2=2) arealgebraicIt follows that theperiods! 1;! 2 mustbe transcendental wheng2;g3 arealgebraic.

Why should this interest us?

Let y = sinx.Then(dy

dx )2 + y2 = 1.

Thus dyp1¡ y2

= dx.

Integrating both sides, we getRsin b0

dyp1¡ y2

= b.

Let y = }(x).Then³

dydx

´2= 4y3 ¡ g2y ¡ g3.

Thus dyp4y3 ¡ g2y¡ g3

= dx:

Integrating both sides, we getR}((! 1+! 2)=2)}(! 1=2)

dyp4y3¡ g2y¡ g3

= ! 22

Let’s look at our formula for the circumference of an ellipse again.

R11¡ k2

tdtpt(t¡ 1)(t¡ (1¡ k2))

wherek2 = 1¡ b2

a2 :The cubic in the integrand is not in Weierstrass form.

It can be put in this form.But let us look at the case k = 1=

p2.

Putting t = s + 1=2, the integral becomesR1=2

02s+1p4s3¡ s

ds:

The integralR1=20

dsp4s3¡ s

is a period of theelliptic curvey2 = 4x3¡ x:

But what aboutR1=20

sdsp4s3¡ s

?

Let us look at the Weierstrass ³-function:

³(z) = 1z +

P06=! 2L

³1

z¡ ! + 1! + z

! 2

´.

Observe that ³ ; (z) = ¡ }(z).

is not periodic!

• It is “quasi-periodic’’.

• What does this mean?

³(z + ! ) = ³(z) + ´(! ):

Put z = ¡ ! =2 to get³(! =2) = ³(¡ ! =2) + ´(! ).Since ³ is an odd function, we get´(! ) = 2³(! =2).

This is called a quasi-period.

What are these quasi-periods?Observe that³(! 1=2) = ³(¡ ! 1=2+! 1) = ³(¡ ! 1=2) +´(! 1)But ³ is odd, so ³(¡ ! 1=2) = ¡ ³(! 1=2) so that2³(! 1=2) = ´(! 1):

Similarly, 2³(! 2=2) = ´(! 2)

Since ´ is a linear function on the period lattice,we get2³((! 1 + ! 2)=2) = ´(! 1) + ´(! 2).

Thusd³ = ¡ }(z)dz.Recall thatd} =

p4}(z)3 ¡ g2}(z) ¡ g3dz:

Hence,d³ = ¡ }(z)d}p

4}(z)3¡ g2}(z)¡ g3

Hence, our original integral is a sum of aperiod and a quasi-period.

Wecan integrateboth sidesfrom z = ! 1=2to z = (! 1 +! 2)=2 to get´2 = 2

Re2

e1

xdxpx3¡ g2x¡ g3

wheree1 = }(! 1) ande2 = }((! 1 +! 2)=2):

Are there triply periodic functions?

• In 1835, Jacobi proved that such functions of a single variable do not exist.

• Abel and Jacobi constructed a function of two variables with four periods giving the first example of an abelian variety of dimension 2.

1804-1851

1802 - 1829

What exactly is a period?

• These are the values of absolutely convergent integrals of algebraic functions with algebraic coefficients defined by domains in Rn given by polynomial inequalities with algebraic coefficients.

• For example is a period.

¼=RR

x2+y2 · 1dxdy

Some unanswered questions

• Is e a period?

• Probably not.

• Is 1/ a period?

• Probably not.

• The set of periods P is countable but no one has yet given an explicit example of a number not in P.

NµZµQµAµP