From Pappus to Cayley, and Bach(arach)
Will Traves
Department of Mathematics
United States Naval Academy
USNA Basic Notions Seminar
23 OCT 2013
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Reference
THE AMERICAN MATHEMATICAL
MONTHLY
VOLUME 120, NO. 10
DECEMBER 2013
Quick, Does 23 = 67 Equal 33 = 97?
A Mathematician’s Secret from Euclid to Today
867
David Pengelley
Linear Algebra via Complex Analysis
877
Alexander P. Campbell and Daniel Daners
Rigorous Computer Analysis of the Chow–Robbins Game
893
Olle Häggström and Johan Wästlund
From Pascal’s Theorem to d-Constructible Curves
901
Will Traves
The Cuoco Configuration
916
Roger Howe
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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
P2 is a compactification of R2
P2 = R2 ∪ line at ∞
Parallel lines meet at infinity - one point at ∞ for each slope.
Line at infinity wraps twice around R2 .
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The linking numbers of triangles 124 and 356 is 0 because they are not linked;
apart from each other without being caught like a chain link.
Topology of the Projective Plane
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):
linked triangles in K6 )
If you add all of the linking numbers of all sets of triangles in this partic
find the sum to be 1 (246 and 135 are the only linked triangles). This is consist
Gordon and Sachs who claimed that as long as we are working in R3 then we w
linking number.
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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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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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Proof of Pascal’s Theorem
Theorem (Pascal)
If 6 points A, B, C, a, b, c on an irreducible conic are joined by lines
Ab, Bc, Ca, aB, bC, aC, then the red lines meet the blue lines in 3 new
collinear points.
Proof: Since conic goes
through 6 points of
Γ = R ∩ B, the line passes
through the remaining 3
points of Γ.
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Inscribing an Octagon in a Cubic Curve
Definition
An n-gon P with n edges is inscribed in a curve C if every edge of P
meets C only in 2-regular points, i.e. 2 edges of P pass through each
point.
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Cubics Admitting an n-gon
Question
Does every cubic admit an inscribed 8-gon? 10-gon? ...
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Cubics Admitting an n-gon
Question
Does every cubic admit an inscribed 8-gon? 10-gon? ...
Theorem (T-)
Almost every cubic admits an inscribed 2n-gon with n ≥ 4. In fact,
every elliptic curve admits such polygons.
Elliptic curves are smooth cubic curves
- carries a group law: A, B, C collinear ⇐⇒ A + B + C = 0
- important for both theoretical (FLT) and practical reasons (ECM, DH)
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Cubics Admitting an n-gon
Theorem (T-)
Every elliptic curve admits an inscribed 2n-gon with n ≥ 4.
Elliptic curve group law: A, B, C collinear ⇐⇒ A + B + C = 0.
Case n odd:
T
= −A0 − Bn−1 = (P1 + A1 ) + (Pn−1 + Bn−2 )
= (P1 − P2 + · · · − Pn−1 − An−1 ) + (Pn−1 − Pn−2 + · · · − P1 − B0 )
= −An−1 − B0 = Q.
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Case: n even
When n is even we get a pair of entwined n-gons:
This is reminiscent of a result due to Möbius (1848).
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A result of A. F. 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 points are collinear, then so is the last one.
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The 8 ⇒ 9 Theorem
Theorem (8 ⇒ 9)
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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An Important Vector Space
The set Pd of polynomials of degree ≤ d (together with 0) form a
vector space.
Each such polynomial determines a curve {p ∈ P2 : P(p) = 0}.
Given a point p0 ∈ P2 , the set I(p0 )Pd of polynomials P such that
P(p0 ) = 0 is a subspace of Pd .
In fact, each point imposes a single linear condition on Pd .
Define: I(p0 , . . . , pn )Pd = ∩0≤k≤n I(pk )Pd .
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Cayley-Bacharach Theorem
Theorem (Cayley-Bacharach)
Suppose that curves of degrees d1 and d2 meet in a set Γ of d1 d2
distinct points. Partition Γ = Γ0 ∪ Γ00 . The dimension of
I(Γ0 )Pd /I(Γ)Pd
equals the failure of the points in Γ00 to impose independent conditions
on curves of degree s = d1 + d2 − 3 − d.
The failure of Γ00 is the difference between |Γ00 | and the rank of the
linear conditions on Ps imposed by Γ00 .
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Cayley-Bacharach Implies the 8 ⇒ 9 Theorem
Suppose two cubics (d1 = d2 = 3) meet in 9 points. Let Γ0 consist of 8
of these points. Set d = 3.
dim
deg 3 through 8 points
= failure of 1 pt on degree 0 forms = 0
deg 3 through 9 pts
So any degree 3 poly through 8 points goes through the ninth.
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10 points on a cubic curve
I want to study a simple question about points on cubic curves using
the Cayley-Bacharach Theorem.
Question
When do 10 points lie on a cubic curve?
General equation for a cubic:
ax 3 + bx 2 y + cx 2 + dxy 2 + exy + fx + gy 3 + hy 2 + iy + j = 0.
Each point Pi (xi , yi ) imposes a linear constraint on the 10 coefficients:
axi3 + bxi2 yi + cxi2 + dxi yi2 + exi yi + fxi + gyi3 + hyi2 + iyi + j = 0.
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Matrix Formulation
We gather these equations in a matrix formulation:

x1 y1
x1 y13 y12 y1
x13 x12 y1 x12 x1 y12
 x 3 x 2y
2
2
x2
x2 y2
x2 y2
x2 y23 y22 y2
 2
2 2
 x 3 x 2y
x32 x3 y32
x3 y3
x3 y33 y32 y3
3
 3
 x 3 x32 y
2
2
x4
x4 y4
x4 y4
x4 y43 y42 y4
 4
4 4
 3
2
2
2
x5 y5 x5
x5 y5
x5 y5
x5 y53 y52 y5
 x5
 3
x62 y6 x62 x6 y62
x6 y6
x6 y63 y62 y6
 x6
 3
2
2
2
 x7
x7 y7
x7 y7
x7 y73 y72 y7
x7 y7 x7
 3
 x8
x82 y8 x82 x8 y82
x8 y8
x8 y83 y82 y8
 3
2
2
2
 x9
x9 y9 x9
x9 y9
x9 y9
x9 y93 y92 y9
3
2
2
2
3
2
x10 x10 y10 x10 x10 y10 x10 y10 x10 y10
y10
y10
   
1
0
a



1
b
   0

0
1
c





0
1
d




   
e  0
1 
= 



1  f 
0
 


1 g  0
   
h  0
1 



1 i  0
0
j
1
There is a non-zero solution when the determinant of the 10 × 10
matrix is zero.
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A Geometric Approach
2 blue conics and 2 red conics define quartics meeting in 16 points.
Cayley-Bacharach Theorem ⇒ original 10 points lie on a cubic iff the
six new points lie on a conic.
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A Geometric Approach: Cayley-Bacharach details
2 blue conics and 2 red conics define quartics meeting in 16 points.
Cayley-Bacharach Theorem ⇒ original 10 points lie on a cubic iff the
six new points lie on a conic:
dim
deg 3 through 10 points
= failure of 6 pts on degree 2 forms
deg 3 through 16 pts = {0}
LHS ≥ 1 precisely when 10 points lie on a cubic.
RHS ≥ 1 precisely when 6 points lie on a conic.
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Ruler and Compass Construction
This insight led David Wehlau and me to a ruler and compass
construction that checks whether 10 points lie on a cubic (we’re still
nailing down the details).
The construction ought to be much faster than computing the 10 × 10
determinant.
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Straightedge Construction
Meet and Join Algebra:
- algebra of points and lines studied by Grassmann and Cayley
- meet of two lines is their point of intersection
- join of two points is the line through the points - allows algebraic
formulation of straightedge-only constructions
Theorem (after Sturmfels and Whiteley)
There exists a straightedge construction to determine if 10 points lie on
a cubic (with about 100 million lines).
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