From Pappus to Cayley, and Bach(arach) Will Traves Department of Mathematics United States Naval Academy USNA Basic Notions Seminar 23 OCT 2013 Traves (USNA) Pappus-Cayley-Bacharach Basic Notions, 23 OCT 2013 1 / 25 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 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 Traves (USNA) Pappus-Cayley-Bacharach Basic Notions, 23 OCT 2013 3 / 25 Pascal’s Mystic Hexagon Theorem Pascal: placed the 6 intersection points on a conic (1639) Converse: Braikenridge-Maclaurin Theorem Traves (USNA) Pappus-Cayley-Bacharach Basic Notions, 23 OCT 2013 4 / 25 Why Mystic? Traves (USNA) Pappus-Cayley-Bacharach Basic Notions, 23 OCT 2013 5 / 25 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 . Traves (USNA) Pappus-Cayley-Bacharach Basic Notions, 23 OCT 2013 6 / 25 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. Traves (USNA) Pappus-Cayley-Bacharach Basic Notions, 23 OCT 2013 7 / 25 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 Traves (USNA) Pappus-Cayley-Bacharach Line m points Doubl tangen 4x2 + Basic Notions, 23 OCT 2013 8 / 25 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. Traves (USNA) Pappus-Cayley-Bacharach Basic Notions, 23 OCT 2013 9 / 25 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. Traves (USNA) Pappus-Cayley-Bacharach Basic Notions, 23 OCT 2013 10 / 25 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. Traves (USNA) Pappus-Cayley-Bacharach Basic Notions, 23 OCT 2013 11 / 25 Cubics Admitting an n-gon Question Does every cubic admit an inscribed 8-gon? 10-gon? ... Traves (USNA) Pappus-Cayley-Bacharach Basic Notions, 23 OCT 2013 12 / 25 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) Traves (USNA) Pappus-Cayley-Bacharach Basic Notions, 23 OCT 2013 12 / 25 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. Traves (USNA) Pappus-Cayley-Bacharach Basic Notions, 23 OCT 2013 13 / 25 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). Traves (USNA) Pappus-Cayley-Bacharach Basic Notions, 23 OCT 2013 14 / 25 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. Traves (USNA) Pappus-Cayley-Bacharach Basic Notions, 23 OCT 2013 15 / 25 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) Traves (USNA) Pappus-Cayley-Bacharach Basic Notions, 23 OCT 2013 16 / 25 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 . Traves (USNA) Pappus-Cayley-Bacharach Basic Notions, 23 OCT 2013 17 / 25 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 . Traves (USNA) Pappus-Cayley-Bacharach Basic Notions, 23 OCT 2013 18 / 25 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. Traves (USNA) Pappus-Cayley-Bacharach Basic Notions, 23 OCT 2013 19 / 25 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. Traves (USNA) Pappus-Cayley-Bacharach Basic Notions, 23 OCT 2013 20 / 25 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. Traves (USNA) Pappus-Cayley-Bacharach Basic Notions, 23 OCT 2013 21 / 25 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. Traves (USNA) Pappus-Cayley-Bacharach Basic Notions, 23 OCT 2013 22 / 25 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. 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 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. 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 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). Traves (USNA) Pappus-Cayley-Bacharach Basic Notions, 23 OCT 2013 25 / 25