# Generalizing Pascal’s Mystic Hexagon Theorem Will Traves Mathematics and Statistics Colloquium Dalhousie University ```Generalizing Pascal’s Mystic Hexagon Theorem
Will Traves
Department of Mathematics
Mathematics and Statistics Colloquium
Dalhousie University
Halifax
18 APR 2013
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Ulterior Motives
Really in Halifax to honor my father as he nears the end of his term as
President at Dalhousie.
Tom Traves
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Start at the Beginning
Line Arrangement due to Pappus of Alexandria (Synagogue; c. 340)
Richter-Gebert: 9 proofs in Perspectives on Projective Geometry, 2011
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Pascal’s Mystic Hexagon Theorem
Pascal: placed the 6 intersection points on a conic (1639)
Converse: Braikenridge-Maclaurin Theorem
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Why Mystic?
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The Projective Plane
Figure 16
The linking numbers of triangles 124 and 356 is 0 because they are not linked;
2 obtained by attaching a line at infinity
P2 is a compactification
of Rwithout
apart from each other
being caught like a chain link.
Thicken line at infinity:
P2 = disk ∪ Möbius band
The linking numbers of triangles 246 and 135 is 1 because they are linked toge
P2 can’t be embedded in R3
(Conway, Gordon, Sachs (1983):
If you add all of the linking numbers of all sets of triangles in this partic
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Bézout’s Theorem
Compactness of P2 allows us to count solutions:
Theorem (Bézout)
Any two curves, without common components, defined by the
vanishing of polynomials of degrees d1 and d2 meet in d1 d2 points in
P2 , suitably interpreted.
Lines meeting an
y=3
y=2
y=1
y=0
4x2+9y2=36
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Line m
points
Doubl
tangen
4x2 +
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The 8 ⇒ 9 Theorem
Theorem (Cayley-Bacharach-Chasles)
If two cubics C1 and C2 meet in 9 distinct points, then any other cubic
C through 8 of these points goes through the ninth too.
Elementary proof on Terry Tao’s blog (July 15, 2011)
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Proof of Pascal’s Theorem
C1 : deg 3 curve consisting of red lines
C2 : deg 3 curve consisting of blue lines
C: deg 3 curve consisting of conic and
line through 1 and 2
Since C goes through 8 points of C1 ∩ C2 , C goes through the ninth
point too (by CBC Theorem).
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From Pascal to Möbius
Theorem (Möbius)
Inscribe a polygon with 4k + 2 sides into a conic and consider the
2k + 1 points where opposite edges meet. If 2k of these points are
collinear, then so is the last one. Inscribe two 2k -gons in a conic and
associate edges cyclicly. The associated edges meet in 2k points and
if 2k − 1 of these point are collinear, then so is the last one.
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Folklore
Theorem
Suppose that k red lines meet k blue lines in a set Γ of k 2 distinct
points. If S = 0 is an irreducible curve of degree d through kd points of
Γ then the remaining points lie on a unique curve C of degree
t = k − d.
Proof (Existence):
R: deg k poly defining red lines.
B: deg k poly defining blue lines.
Pick P ∈ S \ Γ. Choose
Fa,b = aR + bB
to vanish at P. Then S = 0 is a
component of Fa,b = 0 (by Bézout
and S irred) and Fa,b /S defines C.
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d-Constructible Curves
Definition
A curve C of degree t is d-constructible if there exist k = d + t red
lines and k blue lines so that:
(1) red and blue lines meet in a set Γ of k 2 distinct points
(2) dk points in Γ lie on a degree-d curve S
(3) the remaining tk points of Γ lie on C.
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Density: definition
Definition
A curve C of degree t is d-constructible if there exist k = d + t red
lines and k blue lines so that:
(1) red and blue lines meet in a set Γ of k 2 distinct points
(2) dk points in Γ lie on a degree-d curve S
(3) the remaining tk points of Γ lie on C.
Definition
The d-construction is dense in degree t if almost every degree-t
curve is d-constructible (i.e. there is a nonempty Zariski-open set
U ⊂ P(C[x, y, z]t ) so that every degree-t curve in U is d-constructible).
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First Example
Example
For any d, all lines are d-constructible.
Arrange d + 1 points on the line and red and blue lines through these
points (with distinct intersection points).
Folklore Theorem ⇒ ∃ complementary curve of degree d.
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Second Example
Example
For any d, all irreducible conics are d-constructible.
Arrange 2(d + 2) points, joined cyclicly by alternating red and blue
lines, ensuring distinct intersection points.
Folklore Theorem ⇒ ∃ complementary curve of degree d.
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Counting Dimensions
Theorem (T-)
If d ≥ 3, then the d-construction is not dense in degrees d + 4 or
higher. The 2-construction is not dense in degrees five or higher. The
1-construction is not dense in degrees six or higher.
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Elliptic Curves
Elliptic curves are genus 1 smooth curves (isomorphic to a real torus).
The points on the curve form a group.
Any elliptic curve can be
embedded in P2 as a smooth cubic
E with 0E a point of inflection at
infinity.
Group law:
A + B + C = 0 ⇐⇒
A, B, C are collinear pts of E.
Can use CBC Theorem to check that this group law is associative.
The smooth cubics form a dense set of all cubics in P2 .
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Elliptic Curves are d-constructible
Theorem (T-)
All elliptic curves are d-constructible so the d-construction is dense in
degree 3.
Need to arrange 3(d + 3) points on E with a blue and red line through
each of them; construction has d + 4 parameters.
Case d even:
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Elliptic Curves are d-constructible: Case d odd
Theorem (T-)
All elliptic curves are d-constructible so the d-construction is dense in
degree 3.
Case d odd (c.f. Möbius):
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Density
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1-Constructible Curves
Theorem (Roth and T-)
The 1-construction is dense in degrees 4 and 5.
deg 4 pf: Find f = (f1 , . . . , f15 ) : C17 → {1-constr deg-4 curves}.
Thm of generic fiber: dim(imf ) + dim(gen fiber) = dim(domain).
Suppose P ∈ fiber f −1 (c); for unit tangent vector v to fiber,
∂fi Dv f (P) =
v = Jac(f )(P)v = 0.
xj x=P
Rank Jac(f )(P0 ) = 15 ⇒ dim(fiber) = 2 ⇒ dim(gen fiber) ≤ 2.
Thus dim(im f) ≥ 15 = dim(deg 4 curves).
Chevalley’s Thm ⇒ im(f) ⊃ dense open set of 1-constr deg-4 curves.
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Density
Conjecture
The d-construction is dense in degree 4 for all d.
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Strongly Inscribed Polygons
Call a polygon P strongly inscribed in a curve C if every edge,
extended to a line, meets C only in points p with exactly two edge lines
passing through p.
The d-construction is dense in degree t if and only if a dense set of
degree-t curves admits a strongly inscribed polygon with 2(d + t)
edges.
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Strongly Inscribed Polygon Results
Theorem (T-)
Every elliptic curve admits a strongly inscribed polygon with 2k ≥ 8
edges.
Easy to see that if degree C is odd then only polygons with an even
number of edges can be strongly inscribed.
Question
Which degree 4 curves admit a strongly
inscribed polygon with an odd number
of sides?
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