Download Rationality of intersection points of a line and a quartic

Rationality of intersection points of a line and a quartic
Christophe Ritzenthaler
Joint work with Roger Oyono
Institut de Mathématiques de Luminy, CNRS
Istanbul, 27-30 June 2010
Christophe Ritzenthaler Joint work with Rationality
Roger Oyono
of intersection
(IML)
points of a line and
Istanbul,
a quartic
27-30 June 2010
1 / 12
Initial problem
How to add points on the Jacobian of non-hyperelliptic genus 3 curves C /k
?
Christophe Ritzenthaler Joint work with Rationality
Roger Oyono
of intersection
(IML)
points of a line and
Istanbul,
a quartic
27-30 June 2010
2 / 12
Initial problem
How to add points on the Jacobian of non-hyperelliptic genus 3 curves C /k
?
This problem was addressed and solved geometrically in Flon-Oyono-R. 08.
Christophe Ritzenthaler Joint work with Rationality
Roger Oyono
of intersection
(IML)
points of a line and
Istanbul,
a quartic
27-30 June 2010
2 / 12
Initial problem
How to add points on the Jacobian of non-hyperelliptic genus 3 curves C /k
?
This problem was addressed and solved geometrically in Flon-Oyono-R. 08.
Summary:
non-hyperelliptic genus 3 curves = smooth plane quartics.
Christophe Ritzenthaler Joint work with Rationality
Roger Oyono
of intersection
(IML)
points of a line and
Istanbul,
a quartic
27-30 June 2010
2 / 12
Initial problem
How to add points on the Jacobian of non-hyperelliptic genus 3 curves C /k
?
This problem was addressed and solved geometrically in Flon-Oyono-R. 08.
Summary:
non-hyperelliptic genus 3 curves = smooth plane quartics.
points on the Jacobian are represented by the sum D of 3 points on
the curve C .
Christophe Ritzenthaler Joint work with Rationality
Roger Oyono
of intersection
(IML)
points of a line and
Istanbul,
a quartic
27-30 June 2010
2 / 12
Initial problem
How to add points on the Jacobian of non-hyperelliptic genus 3 curves C /k
?
This problem was addressed and solved geometrically in Flon-Oyono-R. 08.
Summary:
non-hyperelliptic genus 3 curves = smooth plane quartics.
points on the Jacobian are represented by the sum D of 3 points on
the curve C .
Choice of a good divisor at infinity
condition (∗):
There is a rational line `∞ which crosses the quartic C in four
k-points P1∞ , P2∞ , P3∞ , P4∞ .
Christophe Ritzenthaler Joint work with Rationality
Roger Oyono
of intersection
(IML)
points of a line and
Istanbul,
a quartic
27-30 June 2010
2 / 12
A geometric addition algorithm
Aim: given D1 , D2 find D + such that D + + D ∞ ∼ D1 + D2 .
Christophe Ritzenthaler Joint work with Rationality
Roger Oyono
of intersection
(IML)
points of a line and
Istanbul,
a quartic
27-30 June 2010
3 / 12
A geometric addition algorithm
Aim: given D1 , D2 find D + such that D + + D ∞ ∼ D1 + D2 .
1
Take a cubic E which goes (with multiplicity) through the support of
D1 , D2 and P1∞ , P2∞ , P4∞ . This cubic also crosses C in the residual
effective divisor D3 .
Christophe Ritzenthaler Joint work with Rationality
Roger Oyono
of intersection
(IML)
points of a line and
Istanbul,
a quartic
27-30 June 2010
3 / 12
A geometric addition algorithm
Aim: given D1 , D2 find D + such that D + + D ∞ ∼ D1 + D2 .
1
2
Take a cubic E which goes (with multiplicity) through the support of
D1 , D2 and P1∞ , P2∞ , P4∞ . This cubic also crosses C in the residual
effective divisor D3 .
Take a conic Q which goes through the support of D3 and P1∞ , P2∞ .
This conic also crosses C in the residual effective divisor D + .
Christophe Ritzenthaler Joint work with Rationality
Roger Oyono
of intersection
(IML)
points of a line and
Istanbul,
a quartic
27-30 June 2010
3 / 12
A geometric addition algorithm
Aim: given D1 , D2 find D + such that D + + D ∞ ∼ D1 + D2 .
1
2
Take a cubic E which goes (with multiplicity) through the support of
D1 , D2 and P1∞ , P2∞ , P4∞ . This cubic also crosses C in the residual
effective divisor D3 .
Take a conic Q which goes through the support of D3 and P1∞ , P2∞ .
This conic also crosses C in the residual effective divisor D + .
Why ? Because (`∞ · C ) ∼ κ, (Q · C ) ∼ 2κ and (E · C ) ∼ 3κ where κ is
the canonical divisor on C .
Christophe Ritzenthaler Joint work with Rationality
Roger Oyono
of intersection
(IML)
points of a line and
Istanbul,
a quartic
27-30 June 2010
3 / 12
A chord construction
Christophe Ritzenthaler Joint work with Rationality
Roger Oyono
of intersection
(IML)
points of a line and
Istanbul,
a quartic
27-30 June 2010
4 / 12
Special forms of the curve
C admits an equation of the form
C : y 3 + h1 (x)y 2 + h2 (x)y = f4 (x),
deg(f4 ) ≤ 4,
Christophe Ritzenthaler Joint work with Rationality
Roger Oyono
of intersection
(IML)
points of a line and
Istanbul,
a quartic
27-30 June 2010
5 / 12
Special forms of the curve
C admits an equation of the form
C : y 3 + h1 (x)y 2 + h2 (x)y = f4 (x),
1
deg(f4 ) ≤ 4,
deg(h1 ) ≤ 2 and deg(h2 ) ≤ 3 if P1∞ = P2∞ (tangent case);
Christophe Ritzenthaler Joint work with Rationality
Roger Oyono
of intersection
(IML)
points of a line and
Istanbul,
a quartic
27-30 June 2010
5 / 12
Special forms of the curve
C admits an equation of the form
C : y 3 + h1 (x)y 2 + h2 (x)y = f4 (x),
deg(f4 ) ≤ 4,
1
deg(h1 ) ≤ 2 and deg(h2 ) ≤ 3 if P1∞ = P2∞ (tangent case);
2
deg(h1 ) ≤ 1 and deg(h2 ) ≤ 3 if P1∞ = P2∞ = P4∞ (flex case).
Christophe Ritzenthaler Joint work with Rationality
Roger Oyono
of intersection
(IML)
points of a line and
Istanbul,
a quartic
27-30 June 2010
5 / 12
Special forms of the curve
C admits an equation of the form
C : y 3 + h1 (x)y 2 + h2 (x)y = f4 (x),
deg(f4 ) ≤ 4,
1
deg(h1 ) ≤ 2 and deg(h2 ) ≤ 3 if P1∞ = P2∞ (tangent case);
2
deg(h1 ) ≤ 1 and deg(h2 ) ≤ 3 if P1∞ = P2∞ = P4∞ (flex case).
3
deg(h1 ) ≤ 1 and deg(h2 ) ≤ 2 if P1∞ = P2∞ = P3∞ = P4∞ (hyperflex
case). These curves are the C3,4 -curves.
Christophe Ritzenthaler Joint work with Rationality
Roger Oyono
of intersection
(IML)
points of a line and
Istanbul,
a quartic
27-30 June 2010
5 / 12
Special forms of the curve
C admits an equation of the form
C : y 3 + h1 (x)y 2 + h2 (x)y = f4 (x),
deg(f4 ) ≤ 4,
1
deg(h1 ) ≤ 2 and deg(h2 ) ≤ 3 if P1∞ = P2∞ (tangent case);
2
deg(h1 ) ≤ 1 and deg(h2 ) ≤ 3 if P1∞ = P2∞ = P4∞ (flex case).
3
4
deg(h1 ) ≤ 1 and deg(h2 ) ≤ 2 if P1∞ = P2∞ = P3∞ = P4∞ (hyperflex
case). These curves are the C3,4 -curves.
If char(k) 6= 3, C : y 3 = f4 (x) (Picard curves) iif P1∞ is a rational
Galois point.
Christophe Ritzenthaler Joint work with Rationality
Roger Oyono
of intersection
(IML)
points of a line and
Istanbul,
a quartic
27-30 June 2010
5 / 12
Special forms of the curve
C admits an equation of the form
C : y 3 + h1 (x)y 2 + h2 (x)y = f4 (x),
deg(f4 ) ≤ 4,
1
deg(h1 ) ≤ 2 and deg(h2 ) ≤ 3 if P1∞ = P2∞ (tangent case);
2
deg(h1 ) ≤ 1 and deg(h2 ) ≤ 3 if P1∞ = P2∞ = P4∞ (flex case).
3
4
deg(h1 ) ≤ 1 and deg(h2 ) ≤ 2 if P1∞ = P2∞ = P3∞ = P4∞ (hyperflex
case). These curves are the C3,4 -curves.
If char(k) 6= 3, C : y 3 = f4 (x) (Picard curves) iif P1∞ is a rational
Galois point.
: the more special, the faster.
Christophe Ritzenthaler Joint work with Rationality
Roger Oyono
of intersection
(IML)
points of a line and
Istanbul,
a quartic
27-30 June 2010
5 / 12
Special forms of the curve
C admits an equation of the form
C : y 3 + h1 (x)y 2 + h2 (x)y = f4 (x),
deg(f4 ) ≤ 4,
1
deg(h1 ) ≤ 2 and deg(h2 ) ≤ 3 if P1∞ = P2∞ (tangent case);
2
deg(h1 ) ≤ 1 and deg(h2 ) ≤ 3 if P1∞ = P2∞ = P4∞ (flex case).
3
4
deg(h1 ) ≤ 1 and deg(h2 ) ≤ 2 if P1∞ = P2∞ = P3∞ = P4∞ (hyperflex
case). These curves are the C3,4 -curves.
If char(k) 6= 3, C : y 3 = f4 (x) (Picard curves) iif P1∞ is a rational
Galois point.
: the more special, the faster.
Remark: Quartics with a hyperflex form a sub-variety of codimension 1. So
generically the two last cases do not occur.
Christophe Ritzenthaler Joint work with Rationality
Roger Oyono
of intersection
(IML)
points of a line and
Istanbul,
a quartic
27-30 June 2010
5 / 12
Study of the condition (∗) over Fq (Oyono-R.)
Theorem
Let C be a smooth plane quartic over the finite field Fq with q = p n
elements. If q ≥ 127, then there exists a line ` which intersects C at
rational points only.
Christophe Ritzenthaler Joint work with Rationality
Roger Oyono
of intersection
(IML)
points of a line and
Istanbul,
a quartic
27-30 June 2010
6 / 12
Study of the condition (∗) over Fq (Oyono-R.)
Theorem
Let C be a smooth plane quartic over the finite field Fq with q = p n
elements. If q ≥ 127, then there exists a line ` which intersects C at
rational points only.
Theorem
Let C be a smooth plane quartic over Fq . If q ≥ 662 + 1, then there exists
a tangent to C which intersects C at rational points only.
Christophe Ritzenthaler Joint work with Rationality
Roger Oyono
of intersection
(IML)
points of a line and
Istanbul,
a quartic
27-30 June 2010
6 / 12
Study of the condition (∗) over Fq (Oyono-R.)
Theorem
Let C be a smooth plane quartic over the finite field Fq with q = p n
elements. If q ≥ 127, then there exists a line ` which intersects C at
rational points only.
Theorem
Let C be a smooth plane quartic over Fq . If q ≥ 662 + 1, then there exists
a tangent to C which intersects C at rational points only.
Moreover, under a certain conjecture, the probability for a plane smooth
quartic over a finite field to have a rational flex is about 0.63.
Christophe Ritzenthaler Joint work with Rationality
Roger Oyono
of intersection
(IML)
points of a line and
Istanbul,
a quartic
27-30 June 2010
6 / 12
Proof of the generic case
Follow an idea of (Diem-Thomé 08).
Christophe Ritzenthaler Joint work with Rationality
Roger Oyono
of intersection
(IML)
points of a line and
Istanbul,
a quartic
27-30 June 2010
7 / 12
Proof of the generic case
Follow an idea of (Diem-Thomé 08).
1
Let P ∈ C (k). Consider the separable geometric cover
φ : C → |κ − P| = P1 of degree 3 induced by the linear system
|κ − P|. This parametrizes lines through the point P.
Christophe Ritzenthaler Joint work with Rationality
Roger Oyono
of intersection
(IML)
points of a line and
Istanbul,
a quartic
27-30 June 2010
7 / 12
Proof of the generic case
Follow an idea of (Diem-Thomé 08).
1
Let P ∈ C (k). Consider the separable geometric cover
φ : C → |κ − P| = P1 of degree 3 induced by the linear system
|κ − P|. This parametrizes lines through the point P.
2
Using effective Chebotarev’s density theorem for function fields, one
gets estimation on the number of completely split divisors in |κ − P|.
Christophe Ritzenthaler Joint work with Rationality
Roger Oyono
of intersection
(IML)
points of a line and
Istanbul,
a quartic
27-30 June 2010
7 / 12
Proof for the tangent case
Let T : C → Sym2 (C ), P 7→ TP (C ) · C − 2P be the tangential
correspondence. We associate to it its correspondence curve
XC = {(P, Q) ∈ C × C : Q ∈ T (P)}
which is defined over k.
Christophe Ritzenthaler Joint work with Rationality
Roger Oyono
of intersection
(IML)
points of a line and
Istanbul,
a quartic
27-30 June 2010
8 / 12
Proof for the tangent case
Let T : C → Sym2 (C ), P 7→ TP (C ) · C − 2P be the tangential
correspondence. We associate to it its correspondence curve
XC = {(P, Q) ∈ C × C : Q ∈ T (P)}
which is defined over k.
We want to prove that there is a rational point on XC when q is big
enough.
Christophe Ritzenthaler Joint work with Rationality
Roger Oyono
of intersection
(IML)
points of a line and
Istanbul,
a quartic
27-30 June 2010
8 / 12
Proof for the tangent case
Let T : C → Sym2 (C ), P 7→ TP (C ) · C − 2P be the tangential
correspondence. We associate to it its correspondence curve
XC = {(P, Q) ∈ C × C : Q ∈ T (P)}
which is defined over k.
We want to prove that there is a rational point on XC when q is big
enough.
Proposition (Aubry, Perret 95)
Let X /Fq be a geometrically irreducible curve of arithmetic genus πX .
Then
√
|#X (Fq ) − (q + 1)| ≤ 2πX q.
In particular if q ≥ (2πX )2 then X has a rational point.
Christophe Ritzenthaler Joint work with Rationality
Roger Oyono
of intersection
(IML)
points of a line and
Istanbul,
a quartic
27-30 June 2010
8 / 12
The question of the absolute irreducibility of XC
• The case of p 6= 2.
Let πi : XC → C be the projections on the first and second factors.
Christophe Ritzenthaler Joint work with Rationality
Roger Oyono
of intersection
(IML)
points of a line and
Istanbul,
a quartic
27-30 June 2010
9 / 12
The question of the absolute irreducibility of XC
• The case of p 6= 2.
Let πi : XC → C be the projections on the first and second factors. The
morphism π1 a degree 2-cover which ramification points are the bitangency
points.
Christophe Ritzenthaler Joint work with Rationality
Roger Oyono
of intersection
(IML)
points of a line and
Istanbul,
a quartic
27-30 June 2010
9 / 12
The question of the absolute irreducibility of XC
• The case of p 6= 2.
Let πi : XC → C be the projections on the first and second factors. The
morphism π1 a degree 2-cover which ramification points are the bitangency
points. Moreover if a bitangency point P is not a hyperflex, then the points
in the fiber of π1−1 (P) are smooth points on XC .
Christophe Ritzenthaler Joint work with Rationality
Roger Oyono
of intersection
(IML)
points of a line and
Istanbul,
a quartic
27-30 June 2010
9 / 12
The question of the absolute irreducibility of XC
• The case of p 6= 2.
Let πi : XC → C be the projections on the first and second factors. The
morphism π1 a degree 2-cover which ramification points are the bitangency
points. Moreover if a bitangency point P is not a hyperflex, then the points
in the fiber of π1−1 (P) are smooth points on XC .This implies that XC is
absolutely irreducible.
Christophe Ritzenthaler Joint work with Rationality
Roger Oyono
of intersection
(IML)
points of a line and
Istanbul,
a quartic
27-30 June 2010
9 / 12
The question of the absolute irreducibility of XC
• The case of p 6= 2.
Let πi : XC → C be the projections on the first and second factors. The
morphism π1 a degree 2-cover which ramification points are the bitangency
points. Moreover if a bitangency point P is not a hyperflex, then the points
in the fiber of π1−1 (P) are smooth points on XC .This implies that XC is
absolutely irreducible.
• The case p = 2 (not in the article, see Arxiv).
We prove that if XC is not absolutely irreducible, then XC is reducible and
each component is isomorphic to C .
Christophe Ritzenthaler Joint work with Rationality
Roger Oyono
of intersection
(IML)
points of a line and
Istanbul,
a quartic
27-30 June 2010
9 / 12
The question of the absolute irreducibility of XC
• The case of p 6= 2.
Let πi : XC → C be the projections on the first and second factors. The
morphism π1 a degree 2-cover which ramification points are the bitangency
points. Moreover if a bitangency point P is not a hyperflex, then the points
in the fiber of π1−1 (P) are smooth points on XC .This implies that XC is
absolutely irreducible.
• The case p = 2 (not in the article, see Arxiv).
We prove that if XC is not absolutely irreducible, then XC is reducible and
each component is isomorphic to C .Apply the previous bound to a
component.
Christophe Ritzenthaler Joint work with Rationality
Roger Oyono
of intersection
(IML)
points of a line and
Istanbul,
a quartic
27-30 June 2010
9 / 12
The question of the absolute irreducibility of XC
• The case of p 6= 2.
Let πi : XC → C be the projections on the first and second factors. The
morphism π1 a degree 2-cover which ramification points are the bitangency
points. Moreover if a bitangency point P is not a hyperflex, then the points
in the fiber of π1−1 (P) are smooth points on XC .This implies that XC is
absolutely irreducible.
• The case p = 2 (not in the article, see Arxiv).
We prove that if XC is not absolutely irreducible, then XC is reducible and
each component is isomorphic to C .Apply the previous bound to a
component.
Remark: the only case where XC is reducible is when C is geometrically
isomorphic to the Klein quartic x 3 y + y 3 z + z 3 x = 0 (in particular
Conjecture 1 in the paper is false).
Christophe Ritzenthaler Joint work with Rationality
Roger Oyono
of intersection
(IML)
points of a line and
Istanbul,
a quartic
27-30 June 2010
9 / 12
Computation of the arithmetic genus of XC
If p 6= 2 and there is no hyperflex on C , one can use Riemann-Hurwitz
formula with the degree 2-cover π1 : XC → C to get gXC .
Christophe Ritzenthaler Joint work with Rationality
Roger Oyono
of intersection
(IML)
points of a lineIstanbul,
and a quartic
27-30 June 2010
10 / 12
Computation of the arithmetic genus of XC
If p 6= 2 and there is no hyperflex on C , one can use Riemann-Hurwitz
formula with the degree 2-cover π1 : XC → C to get gXC . Then one uses a
flat family argument to get πXC in general.
Christophe Ritzenthaler Joint work with Rationality
Roger Oyono
of intersection
(IML)
points of a lineIstanbul,
and a quartic
27-30 June 2010
10 / 12
Computation of the arithmetic genus of XC
If p 6= 2 and there is no hyperflex on C , one can use Riemann-Hurwitz
formula with the degree 2-cover π1 : XC → C to get gXC . Then one uses a
flat family argument to get πXC in general.
When p = 2, wild ramification prevents to use Riemann-Hurwitz formula so
easily. We use instead intersection theory in C × C .
Christophe Ritzenthaler Joint work with Rationality
Roger Oyono
of intersection
(IML)
points of a lineIstanbul,
and a quartic
27-30 June 2010
10 / 12
Computation of the arithmetic genus of XC
If p 6= 2 and there is no hyperflex on C , one can use Riemann-Hurwitz
formula with the degree 2-cover π1 : XC → C to get gXC . Then one uses a
flat family argument to get πXC in general.
When p = 2, wild ramification prevents to use Riemann-Hurwitz formula so
easily. We use instead intersection theory in C × C .
In both cases, we find that πXC = 33.
Christophe Ritzenthaler Joint work with Rationality
Roger Oyono
of intersection
(IML)
points of a lineIstanbul,
and a quartic
27-30 June 2010
10 / 12
Computation of the arithmetic genus of XC
If p 6= 2 and there is no hyperflex on C , one can use Riemann-Hurwitz
formula with the degree 2-cover π1 : XC → C to get gXC . Then one uses a
flat family argument to get πXC in general.
When p = 2, wild ramification prevents to use Riemann-Hurwitz formula so
easily. We use instead intersection theory in C × C .
In both cases, we find that πXC = 33.
Remark : unfortunately, there is a mistake in a remark of our paper
(corrected in the Arxiv version). We claimed that πXC = 9 when p = 2.
This is due to a confusion between the degree of the dual curve and the
degree of the dual map π2 (which is inseparable in this case).
Christophe Ritzenthaler Joint work with Rationality
Roger Oyono
of intersection
(IML)
points of a lineIstanbul,
and a quartic
27-30 June 2010
10 / 12
The case of flexes
Let P14 be the linear system of all plane quartics over a field K and
I0 = {(P, `), P ∈ `} ⊂ P2 × (P2 )∗ .
Christophe Ritzenthaler Joint work with Rationality
Roger Oyono
of intersection
(IML)
points of a lineIstanbul,
and a quartic
27-30 June 2010
11 / 12
The case of flexes
Let P14 be the linear system of all plane quartics over a field K and
I0 = {(P, `), P ∈ `} ⊂ P2 × (P2 )∗ . Let I4 ⊂ P14 × I0 be the locus
I4 = {(C , (P, `)), C is smooth and P is a flex of C with tangent line `}.
Christophe Ritzenthaler Joint work with Rationality
Roger Oyono
of intersection
(IML)
points of a lineIstanbul,
and a quartic
27-30 June 2010
11 / 12
The case of flexes
Let P14 be the linear system of all plane quartics over a field K and
I0 = {(P, `), P ∈ `} ⊂ P2 × (P2 )∗ . Let I4 ⊂ P14 × I0 be the locus
I4 = {(C , (P, `)), C is smooth and P is a flex of C with tangent line `}.
Theorem (Harris 79)
The Galois group of the cover I4 → P14 over C is the full symmetric group
S24 .
Christophe Ritzenthaler Joint work with Rationality
Roger Oyono
of intersection
(IML)
points of a lineIstanbul,
and a quartic
27-30 June 2010
11 / 12
The case of flexes
Let P14 be the linear system of all plane quartics over a field K and
I0 = {(P, `), P ∈ `} ⊂ P2 × (P2 )∗ . Let I4 ⊂ P14 × I0 be the locus
I4 = {(C , (P, `)), C is smooth and P is a flex of C with tangent line `}.
Theorem (Harris 79)
The Galois group of the cover I4 → P14 over C is the full symmetric group
S24 .
If this result is still true over F̄p , using Chebotarev density theorem, one
gets that the probability to have a plane quartic with a rational flex tends to
1−
1
1
1
+ − ... −
≈ 1 − exp(−1) ≈ 0.63
2! 3!
24!
when q tends to infinity.
Christophe Ritzenthaler Joint work with Rationality
Roger Oyono
of intersection
(IML)
points of a lineIstanbul,
and a quartic
27-30 June 2010
11 / 12
Harris’ result over finite fields
General reduction arguments show that the Galois group remains S24 for
almost all p.
Christophe Ritzenthaler Joint work with Rationality
Roger Oyono
of intersection
(IML)
points of a lineIstanbul,
and a quartic
27-30 June 2010
12 / 12
Harris’ result over finite fields
General reduction arguments show that the Galois group remains S24 for
almost all p.However
Proposition
The Galois group over F̄3 is S8 .
Christophe Ritzenthaler Joint work with Rationality
Roger Oyono
of intersection
(IML)
points of a lineIstanbul,
and a quartic
27-30 June 2010
12 / 12
Harris’ result over finite fields
General reduction arguments show that the Galois group remains S24 for
almost all p.However
Proposition
The Galois group over F̄3 is S8 .
This is due to the fact that the flexes of C are on a conic, which is not the
case when p 6= 3.
Christophe Ritzenthaler Joint work with Rationality
Roger Oyono
of intersection
(IML)
points of a lineIstanbul,
and a quartic
27-30 June 2010
12 / 12
Harris’ result over finite fields
General reduction arguments show that the Galois group remains S24 for
almost all p.However
Proposition
The Galois group over F̄3 is S8 .
This is due to the fact that the flexes of C are on a conic, which is not the
case when p 6= 3.
We conjecture that p = 3 is the only exception.
Christophe Ritzenthaler Joint work with Rationality
Roger Oyono
of intersection
(IML)
points of a lineIstanbul,
and a quartic
27-30 June 2010
12 / 12