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Rationality questions concerning Poncelet’sclosure theorem
Jaap Top
Bernoulli Institute & DIAMANT
June 19th, 2018
(math colloquium, Groningen)
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Jean-Victor Poncelet
(1788–1867)
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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
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pictures from the 1822 book:
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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.
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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.
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Poncelet’s “proof” led to disputes, but Jacobi (1828, using el-
liptic functions) published in Crelle a new and complete proof:
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Carl Gustav Jacob Jacobi (1804–1851)
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• 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)
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• In 1976 Phillip A. Griffiths published a purely algebraic geo-
metric proof of Poncelet’s result (“Variations on a theorem
of Abel”)
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• 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. . .
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joint work with Johan Los and Tiemar Mepschen (bachelor’s
projects), Majken Roelfszema (master’s project), and Nurdagul
Anbar
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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:
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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.
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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!
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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.
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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.
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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}.
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Example: elliptic curve E with point t = (0,0) of order 5
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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.
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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.
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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!
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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 . . .
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(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
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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:
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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 . . . . . .
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(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.
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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) = ∅).
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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.
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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 . . .
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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.
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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 . . .
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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 :)
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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.
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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.
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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.
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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 . . .
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