Rationality questions concerning Poncelet’sclosure theorem

Jaap Top

Bernoulli Institute & DIAMANT

June 19th, 2018

(math colloquium, Groningen)

1

Jean-Victor Poncelet

(1788–1867)

2

1812-1814 Poncelet was prisoner of war in Russia

During this time he wrote texts (7 notebooks) on projective

geometry, published in 1862

But already in 1822 he published a book loosely based on his

notebooks

3

4

5

pictures from the 1822 book:

6

7

Definition: consider a polygon with sides

`1, `2, . . . , `n

and vertices

P1 = `n ∩ `1, P2 = `1 ∩ `2, . . . , Pn = `n−1 ∩ `n.

It is called “Poncelet figure ” if conics C,D exist such that all

Pj ∈ C and all `j tangent to D.

It is called “trivial Poncelet figure ” if moreover either n = 2k−2

is even and Pk ∈ C ∩D, or n = 2k− 1 is odd and `k is tangent to

both D and C.

8

Theorem. Given smooth conics C 6= D. If a nontrivial Ponceletfigure using C,D exists and consisting of n vertices, then forevery P ∈ C there is one having P as vertex. If it is nontrivialthen it also has exactly n vertices.

9

Poncelet’s “proof” led to disputes, but Jacobi (1828, using el-

liptic functions) published in Crelle a new and complete proof:

10

Carl Gustav Jacob Jacobi (1804–1851)

11

• The result has consequences for discrete integrable systems(“QRT-maps”, book by Duistermaat), and to “billiard balltrajectories on an elliptical table”:

(picture by Alfonso Sorrentino)

12

• In 1976 Phillip A. Griffiths published a purely algebraic geo-

metric proof of Poncelet’s result (“Variations on a theorem

of Abel”)

13

• aim of today’s talk: show how the algebraic geometric ap-proach motivates number theoretic questions.

• For this, we first sketch the proof by Griffiths;

• next, we show how over Q it naturally leads to a reformulationof a famous result on torsion points of elliptic curves;

• then we “turn the proof around” and construct examplesstarting from an elliptic curve with a given torsion point;

• finally we show that the proposed reformulation of the torsionpoints result is in a sense more general. . .

14

joint work with Johan Los and Tiemar Mepschen (bachelor’s

projects), Majken Roelfszema (master’s project), and Nurdagul

Anbar

15

Sketch of proof. Given conics C,D ⊂ P2 .

Lines ` ⊂ P2 tangent to D are the points of the “dual conic” D∨.

In this way

X := {(P, `) : P ∈ C, ` tangent to D, P ∈ `}

defines an algebraic curve embedded in C ×D∨.

Throughout, we restrict to the general case, namely where C

and D intersect without multiplicities (hence in 4 points).

The curve X comes with additional structure:

16

The morphism X → C given by (P, `) 7→ P has degree 2: for

almost all P , there are precisely 2 lines `, `′ containing P and

tangent to D.

The morphism X → D∨ given by (P, `) 7→ ` also has degree 2:

almost all tangent lines ` to D at Q intersect C in precisely two

points P, P ′.

The two involutions of X given by, respectively, τ : (P, `)↔ (P, `′)and σ: (P, `) ↔ (P ′, `) each have 4 fixpoints (since we assume

#C ∩D = 4).

Hence X has genus 1, comes with two involutions, and the quo-

tient by any of them has genus 0.

17

Fixing any point on X, it becomes an elliptic curve E.

Having as quotient a curve of genus 0, the involutions must be

given as τ :x 7→ x1 − x resp. σ:x 7→ x2 − x for certain x1, x2 ∈ E.

The composition τσ therefore equals x 7→ (x1−x2)+x, translation

over x1 − x2 ∈ E.

If a Poncelet figure with edges `j and vertices Pj would exist for

C,D, then τσ acts as

(P1, `1) 7→ (P2, `2) 7→ . . . 7→ (Pn, `n) 7→ (P1, `1),

so (στ)n = id. This implies the closure theorem!

18

Now number theory, which means we take C,D (and hence X

and the involutions σ, τ) defined over Q.

Natural questions:

1. What are the possibilities for ord(στ) i this case?

2. Starting from an elliptic curve E/Q with a rational point of

order n, how to obtain a Poncelet figure with n vertices and

C,D over Q from it?

3. Construct, given a possible n, an example X/Q with X(Q) = ∅and ord(στ) = n.

19

1. What are, over Q, the possibilities for ord(στ)?

Thm. Given X/Q of genus 1, and involutions σ, τ ∈ AutQ(X)

such that n := ord(στ) <∞. Then n ∈ {1,2,3,4,5,6,7,8,9,10,12}.

The proof uses the “jacobian” of X/Q plus a famous 1977 result

by Barry Mazur.

20

2. Obtain Poncelet figures over Q from an E/Q?

Construction: Start with E/Q with rational points t and p, and

t of exact order n.

Aim: rewrite as a curve X as earlier, coming from conics C,D

over Q, such that the sets {ap+mt : 1 ≤ m ≤ n} correspond to

Poncelet figures {(Pm, `m) ∈ X(Q) : 1 ≤ m ≤ n}.

21

Example: elliptic curve E with point t = (0,0) of order 5

22

Start: put E : y2 + uxy + vy = x3 + vx2 and t = (0,0) ∈ E.

E is an elliptic curve; in particular, an abelian group.

Involutions on E: σ: p 7→ −p and τ : p 7→ t − p; composition τσ is

translation over t.

Quotient by σ is E → P1 given by (x, y) 7→ c := x.

Quotient by τ is E → P1 given by (x, y) 7→ b := (y + v)/x.

Then b, c satisfy b2c+ ubc− vb− c2 − vc− uv = 0.

23

Now turn this into a curve X ⊂ C ×D∨:

First use b to parametrize a conic C, say

b 7→ Pb :=

(b2 − 1

b2 + 1,

2b

b2 + 1

).

Next we construct rational functions α(c), β(c) such that the

lines `c : y = α(c)x+ β(c) are the points of a dual conic D∨:

The desired condition Pb ∈ `c should come from the equation

b2c+ ubc− vb− c2 − vc− uv = 0 we found.

24

This works if one puts

α(c) =c2 + vc+ c+ uv

v − uc, β(c) =

c2 + vc− c+ uv

uc− v.

The corresponding lines `c are tangent to the conic D (enveloping

curve) given by

(v2 + 2v + 1− 4uv)x2 + (2uv + 2u+ 4v)xy + u2y2

+(8uv − 2v2 + 2)x+ (2u− 2uv + 4v)y= 4uv − v2 + 2v − 1.

This finishes the construction!

25

Example. (n = 9)

E : y2 − 47xy − 624y = x3 − 624x2.

In this case t = (0,0) ∈ E has order 9 and

p := (−6,90) ∈ E has infinite order.

Hence this yields Poncelet figures with 9 vertices, all Q-rational.

Explicitly, with C : x2 + y2 = 1 and D given by

270817x2 + 56066xy + 2209y2 = 544126x+ 61246y − 273313,

one obtains from (p + mt)m≥1 on E the Poncelet figure with

points . . .

26

(3960/3961,89/3961)↓

(3843/3845,124/3845)↓

(2952/2977,−385/2977)↓

(17472/17497,935/17497)↓

(24/25,7/25)↓

(1155/1157,68/1157)↓

(12/13,5/13)↓

(39456/39505,1967/39505)↓

(43488/43537,−2065/43537).27

Many similar examples, including ones with points in a fieldQ(√m) and up to 18 vertices, are in the bachelor’s theses of

Johan and of Tiemar, and in our paper:

28

3. Examples over Q with X(Q) = ∅ and στ of given order n.

(This is part of the master’s project of Majken Roelfszema.)

First attempt: find Poncelet figures with n vertices, starting from

two circles C : x2 + y2 = 1 and D : (x − δ)2 + y2 = ρ defined

over Q.

So need δ, ρ ∈ Q such that this results in Poncelet figures with

precisely n vertices, and such that moreover X(Q) = ∅.

Some 18th(!!!) century literature helps here . . . . . .

29

(Classical) Thm. C : x2 + y2 = 1 and D : (x − δ)2 + y2 = ρ

yield a Poncelet figure with n ∈ {3,4,6,8} vertices if and only if

n=3:

ρ =(δ2 − 1)2

4

n=4:

ρ =(δ2 − 1)2

2(δ2 + 1)

n=6:

ρ =(δ2 − 1)2

4δ2

n=8:

ρ =(δ2 − 1)2

4δ

The n = 3 case was shown in 1746 by William Chapple, and versions of the

other cases appeared in two papers (1797 and 1802) by Nicolaus Fuss.

30

Using the (standard) rational parametrization of the circle C, the

curve X turns out to be birational to the one given by

y2 = (x2 + 1) · ((δ2 − ρ+ 1)x2 + 4δx− δ2 + ρ− 1) · ρ.

The involutions σ, τ act on this model as well; Chapple and Fuss

in fact provide necessary and sufficient conditions on δ, ρ such

that any divisor P − σ(τ(P )) on X is torsion of order n. This

observation leads to a simple modern proof.

Given δ, ρ ∈ Q, one readily decides using the above model whether

X(Qp) = ∅ for some prime p (or even X(R) = ∅).

31

Example n = 3

C : x2 + y2 = 1

D : (x− 2)2 + y2 = 94

X can be written as

y2 = (x2 +1) ·(

9916x

2 − 18x+ 9916

)Then X(Q2) = ∅, hence also

X(Q) is empty.

32

Example n=6

C : x2 + y2 = 1

D :(x− 1

2

)2+ y2 = 9

16

Here, same X as above.

So X(Q) = ∅ and AutQ(X) con-

tains a dihedral group of order 6

as well as one or order 12 . . .

33

Example n=5

C : x2 + y2 = 1

D : (x− 3)2 + (y − 1)2 = 8116

X:

y2 = (x2 +1) ·(63x2 − 192x+ 127

),

X(Q2) = ∅, so again no rational

points over Q.

34

Example n=10

C : x2 + y2 = 1

D :(x− 3

10

)2+(y − 1

10

)2= 81

160

Here same X as n = 5 example,

So AutQ(X) contains a dihedral

group of order 10 as well as one

or order 20 . . .

35

The given examples motivate and illustrate a small result:

Proposition: Let C,D be conics over Q such that the corre-

sponding X/Q and σ, τ ∈ AutQ(X) satisfy ord(στ) = n is odd.

If C and D have an axis of symmetry in common, and this axis is

defined over Q, then there are also involutions σ′, τ ′ ∈ AutQ(X)

such that σ′τ ′ has order 2n.

(An algebraic proof in the special case of two circles is due to

B. Mirman (2012). We present a geometric argument here :)

36

Idea of proof:

Let µ denote reflection in the

common symmetry axis. Then

µ ∈ AutQ(X) has fixed points,

and commutes with σ and τ .

Hence σµ “is” translation over

a point of order 2.

So because n is odd, the com-

position (σµ)(στ) has order 2n.

This finishes the proof.

37

Corollary. A Poncelet figure of odd order n ≥ 7 coming from

conics C,D over Q which have an axis of symmetry defined over

Q in common, does not exist.

Proof. Indeed, using the proposition, the Jacobian of the associ-

ated genus 1 curve X would be an elliptic curve E/Q, containing

a rational torsion point of order 2n ≥ 14. This is impossible by

Mazur’s 1977 theorem.

38

So to obtain an “order 7 or order 9 example over Q” we need to

avoid the symmetry that the earlier examples have.

Example n=7:

C : y = x2

D : x2 − 23xy + 1

13y2 = −52

81

This results in

X : y2 = 1053x4 − 9126x3 + 13689x2 + 8788.

In this case X(Q13) = ∅, hence in particular there are no Q-

rational points.

39

To conclude, by searching in some families of pairs C,D, and

also by ‘twisting’ appropriate examples, we constructed X/Q of

genus 1 with X(Q) = ∅, admitting involutions σ, τ ∈ AutQ(X)

such that ord(στ) = n, for each n ∈ {1,2,3,4,5,6,7,8,10,12}.

Remaining challenge:

what about the only case left, namely n = 9??, i.e.,

Find X/Q of genus one with X(Q) = ∅ and AutQ(X) containing

a dihedral group of order 18 . . .

40