From Pappus to Cayley, and Bach(arach) - USNA · From Pappus to Cayley, and Bach(arach) Will Traves...
Transcript of From Pappus to Cayley, and Bach(arach) - USNA · From Pappus to Cayley, and Bach(arach) Will Traves...
From Pappus to Cayley, and Bach(arach)
Will Traves
Department of MathematicsUnited States Naval Academy
USNA Basic Notions Seminar23 OCT 2013
Traves (USNA) Pappus-Cayley-Bacharach Basic Notions, 23 OCT 2013 1 / 25
Reference
THE AMERICAN MATHEMATICAL
MONTHLYVOLUME 120, NO. 10 DECEMBER 2013
867Quick, Does 23 = 67 Equal 33 = 97?
A Mathematician’s Secret from Euclid to Today
David Pengelley
877Linear Algebra via Complex Analysis
Alexander P. Campbell and Daniel Daners
893Rigorous Computer Analysis of the Chow–Robbins Game
Olle Haggstrom and Johan Wastlund
901From Pascal’s Theorem to d-Constructible Curves
Will Traves
916The Cuoco Configuration
Roger Howe
NOTES
924A Generalization of the Leibniz Rule
Ulrich Abel
929Hamiltonian Cycles on Archimedean Solids Are Twisting Free
Richard Ehrenborg
933The Parbelos, a Parabolic Analog of the Arbelos
Jonathan Sondow
940Inequalities for Gamma Function Ratios
G. J. O. Jameson
945PROBLEMS AND SOLUTIONS
REVIEWS
953Real Analysis Through Modern InfinitesimalsBy Nader Vakil
James M. Henle with the assistance of Michael G. Henle
958EDITOR’S ENDNOTES
000INDEX TO VOLUME 120
An Official Publication of the Mathematical Association of America
Traves (USNA) Pappus-Cayley-Bacharach Basic Notions, 23 OCT 2013 2 / 25
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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Topology of the Projective Plane
Thicken line at infinity:P2 = disk ∪ Mobius band
P2 can’t be embedded in R3
(Conway, Gordon, Sachs (1983):linked triangles in K6)
! 20 !
Projective space is also the union of a disc in R2 and a Möbius strip, and is equivalent to the sphere S2 with a blow up at one point.
The Relationship between RP2 and R3.
Earlier we claimed that RP2 is not a subset of R3, that it does not “fit” into R3. For this proof we call upon Conway, Gordon and Sachs’ 1983 result (ams.org) that 6 points all cannot be linked to one another such that the total linking number of all triangles formed is even, in R3. We will give an example of 6 points linked in RP2 , K6, such that the linking number of the set is even. First, to demonstrate Conway, Gordon and Sachs’ proof, an example of K6, 6 points linked in R3:
Figure 16
The linking numbers of triangles 124 and 356 is 0 because they are not linked; they could be pulled apart from each other without being caught like a chain link.
The linking numbers of triangles 246 and 135 is 1 because they are linked together.
If you add all of the linking numbers of all sets of triangles in this particular linking, you will
find the sum to be 1 (246 and 135 are the only linked triangles). This is consistent with Conway, Gordon and Sachs who claimed that as long as we are working in R3 then we will have an odd total linking number.
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Bezout’s Theorem
Compactness of P2 allows us to count solutions:
Theorem (Bezout)Any two curves, without common components, defined by thevanishing of polynomials of degrees d1 and d2 meet in d1d2 points inP2, suitably interpreted.Lines meeting an Ellipse
y=0
4x2+9y2=36
y=1 y=2 y=3 Line meets curve in two
points (possibly imaginary).
Double point: tangency when y=2 4x2 + 9(4) = 36 so 4x2 = 0
It seems that such a curve of degree 1 always meets such a curve of degree 2 in (2)(1) points, if we count them properly.
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Folklore
TheoremSuppose that k red lines meet k blue lines in a set Γ of k2 distinctpoints. If S = 0 is an irreducible curve of degree d through kd points ofΓ then the remaining points lie on a unique curve C of degreet = 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 + bBto vanish at P. Then S = 0 is acomponent of Fa,b = 0 (by Bezoutand 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 linesAb,Bc,Ca,aB,bC,aC, then the red lines meet the blue lines in 3 newcollinear points.
Proof: Since conic goesthrough 6 points ofΓ = R ∩ B, the line passesthrough the remaining 3points of Γ.
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Inscribing an Octagon in a Cubic Curve
DefinitionAn n-gon P with n edges is inscribed in a curve C if every edge of Pmeets C only in 2-regular points, i.e. 2 edges of P pass through eachpoint.
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Cubics Admitting an n-gon
QuestionDoes 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
QuestionDoes 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 Mobius (1848).
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A result of A. F. Mobius
Theorem (Mobius)Inscribe a polygon with 4k + 2 sides into a conic and consider the2k + 1 points where opposite edges meet. If 2k of these points arecollinear, then so is the last one. Inscribe two 2k-gons in a conic andassociate edges cyclicly. The associated edges meet in 2k points andif 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 cubicC 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 avector 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 thatP(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≤nI(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 d1d2distinct points. Partition Γ = Γ′ ∪ Γ′′. The dimension of
I(Γ′)Pd/I(Γ)Pd
equals the failure of the points in Γ′′ to impose independent conditionson curves of degree s = d1 + d2 − 3− d.
The failure of Γ′′ is the difference between |Γ′′| and the rank of thelinear conditions on Ps imposed by Γ′′.
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Cayley-Bacharach Implies the 8⇒ 9 Theorem
Suppose two cubics (d1 = d2 = 3) meet in 9 points. Let Γ′ consist of 8of these points. Set d = 3.
dimdeg 3 through 8 points
deg 3 through 9 pts= failure of 1 pt on degree 0 forms = 0
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 usingthe Cayley-Bacharach Theorem.
QuestionWhen do 10 points lie on a cubic curve?
General equation for a cubic:
ax3 + bx2y + cx2 + dxy2 + exy + fx + gy3 + hy2 + iy + j = 0.
Each point Pi(xi , yi) imposes a linear constraint on the 10 coefficients:
ax3i + bx2
i yi + cx2i + dxiy2
i + exiyi + fxi + gy3i + hy2
i + iyi + j = 0.
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Matrix Formulation
We gather these equations in a matrix formulation:
x31 x2
1 y1 x21 x1y2
1 x1y1 x1 y31 y2
1 y1 1x3
2 x22 y2 x2
2 x2y22 x2y2 x2 y3
2 y22 y2 1
x33 x2
3 y3 x23 x3y2
3 x3y3 x3 y33 y2
3 y3 1x3
4 x24 y4 x2
4 x4y24 x4y4 x4 y3
4 y24 y4 1
x35 x2
5 y5 x25 x5y2
5 x5y5 x5 y35 y2
5 y5 1x3
6 x26 y6 x2
6 x6y26 x6y6 x6 y3
6 y26 y6 1
x37 x2
7 y7 x27 x7y2
7 x7y7 x7 y37 y2
7 y7 1x3
8 x28 y8 x2
8 x8y28 x8y8 x8 y3
8 y28 y8 1
x39 x2
9 y9 x29 x9y2
9 x9y9 x9 y39 y2
9 y9 1x3
10 x210y10 x2
10 x10y210 x10y10 x10 y3
10 y210 y10 1
abcdefghij
=
0000000000
There is a non-zero solution when the determinant of the 10× 10matrix 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 thesix 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 thesix new points lie on a conic:
dimdeg 3 through 10 points
deg 3 through 16 pts = {0}= failure of 6 pts on degree 2 forms
LHS ≥ 1 precisely when 10 points lie on a cubic.RHS ≥ 1 precisely when 6 points lie on a conic.
Traves (USNA) Pappus-Cayley-Bacharach Basic Notions, 23 OCT 2013 23 / 25
Ruler and Compass Construction
This insight led David Wehlau and me to a ruler and compassconstruction that checks whether 10 points lie on a cubic (we’re stillnailing down the details).
The construction ought to be much faster than computing the 10× 10determinant.
Traves (USNA) Pappus-Cayley-Bacharach Basic Notions, 23 OCT 2013 24 / 25
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 algebraicformulation of straightedge-only constructions
Theorem (after Sturmfels and Whiteley)There exists a straightedge construction to determine if 10 points lie ona cubic (with about 100 million lines).
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