Triple lines on curves

Transcription

1 Triple lines on curves Frank de Zeeuw EPFL (Switzerland) IPAM: Algebraic Techniques for Combinatorial Geometry - LA - May 13, 2014 Japanese-Swiss Workshop on Combinatorics - Tokyo - June 5, 2014

2 I will discuss the following articles: ES - György Elekes and Endre Szabó, On triple lines and cubic curves The Orchard Problem revisited, arxiv: RSZ - Orit Raz, Micha Sharir, and Frank de Zeeuw, Polynomials vanishing on Cartesian products: The Elekes-Szabó Theorem revisited, 2014, not yet public.

3 Main theorem of the talk Given n points in R 2, a triple line is a line with 3 of them. Theorem (Elekes-Szabó (2013) / Raz-Sharir-De Zeeuw (2014)) A set of n points on an algebraic curve C of degree d in R 2 determines O d (n 11/6 ) triple lines, unless C contains a cubic curve. Weaker version with O d (n 2 η ) was proved in Elekes-Szabó (2013) using a general theorem from Elekes-Szabó (2012), which bounds the intersection of an algebraic surface in R 3 with an n n n Cartesian product, with certain exceptions. In work with Orit Raz and Micha Sharir (not yet public), we improved the bound in this general theorem to O d (n 11/6 ). In this talk I will show how to deduce the theorem above from the general theorem (which I ll state later).

4 Background Orchard problem (Jackson (1812) / Sylvester (1867)): How many triple lines can n points in R 2 span? Double-counting gives the upper bound n 2 /6. Subgroups of elliptic curves give the lower bound n2 6 n This is exactly the best possible (at least for large n): Theorem (Green-Tao, 2013) Any n n 0 points span n2 6 n triple lines.

5 A cubic with many triple lines Green-Tao in fact show that if there are more than n2 6 cn triple lines, then most of the points must lie on a cubic curve. From Green-Tao (2013).

6 Conjecture Conjecture (Elekes) If n points in R 2 determine Ω(n 2 ) triple lines, then at least 10 of the points lie on a cubic curve. Note that any 9 points lie on some cubic, so 10 is the first nontrivial number. The original conjecture in fact went as high as n, but nothing has been proved in this direction. We will show the related statement that if the points with many triple lines lie on a low-degree algebraic curve, then that curve must be a cubic.

7 Cubics, collinearity, and groups Definition On a curve we say that collinearity is group-related around a collinear triple P 1, P 2, P 3, if there are three open neighborhoods U i P i on the curve, and analytic maps ϕ i : U i R, such that a triple Q 1 U 1, Q 2 U 2, Q 3 U 3 is collinear if and only if ϕ 1 (Q 1 ) + ϕ 2 (Q 2 ) + ϕ 3 (Q 3 ) = 0. U 1 P 1 Q 1 U 2 P 2 Q 2 U 3 P 3 Q 3

8 Cubic curves A cubic curve is one of the following: three lines, a conic and a line, a singular irreducible curve of degree 3, or a nonsingular irreducible curve of degree 3. Fact: On all cubics, collinearity is group-related around every collinear triple on the curve. We ll show this for one example.

9 Example: Parabola + line Take (in the projective plane) the parabola y 2 = xz and the line z = 0 at infinity. The line through [a : a 2 : 1] and [b : b 2 : 1] has slope a + b, and hits y z = 0 at [1 : a + b : 0]. [1 : c : 0] So choose [a : a 2 : 1] [b : b 2 : 1] z ϕ 1 ([a : a 2 : 1]) = a, ϕ 2 ([b : b 2 : 1]) = b ϕ 3 ([1 : c : 0]) = c. x Then [a : a 2 : 1], [b : b 2 : 1], [1 : c : 0] are collinear iff ϕ 1 (Q 1 ) + ϕ 2 (Q 2 ) + ϕ 3 (Q 3 ) = 0.

10 Elliptic curves On nonsingular irreducible cubic curves (a.k.a. elliptic curves), this actually turns the entire set of points into a group. Hence, finite subgroups give point sets with many triple lines. To the right is a subgroup isomorphic to Z/2Z Z/4Z. From Green-Tao (2013).

11 Proof sketch of main theorem Theorem A set of n points on an algebraic curve C of degree d in R 2 determines O d (n 11/6 ) triple lines, unless C contains a cubic curve. Proof sketch: Let C be defined by f (x, y) = 0. Three points (x i, y i ) C are collinear iff (x 1, y 1, x 2, y 2, x 3, y 3 ) R 6 satisfies 1 x 1 y 1 1 x 2 y 2 = 0, f (x 1, y 1 ) = 0, f (x 2, y 2 ) = 0, f (x 3, y 3 ) = 0. 1 x 3 y 3 Do a generic rotation in R 2, and projection from R 6 to (x 1, x 2, x 3 ). Then M triples map to M points of an n n n product in R 3, and they lie on some algebraic surface F (x 1, x 2, x 3 ) = 0.

12 So, starting with M collinear triples, we obtain M points of an n n n product on a surface F (x, y, z) = 0. Now we apply the main tool: Theorem (Elekes-Szabó (2012) / Raz-Sharir-De Zeeuw (2014)) Let A, B, C R of size n, let F R[x, y, z] with F x F y F z 0 and deg F = d, and define V = {(x, y, z) R 3 : F (x, y, z) = 0}. Then V (A B C) = O d (n 11/6 ), unless there are an open ball D R 3 and three analytic functions ϕ i : R R such that V D = {(x, y, z) D : ϕ 1 (x) + ϕ 2 (y) + ϕ 3 (z) = 0}. For us this implies that M = O d (n 11/6 ), unless collinearity on C is group-related around some collinear triple P 1, P 2, P 3.

13 Suppose collinearity on C is group-related around some collinear triple P 1, P 2, P 3, with corresponding U i and ϕ i. Pick a point Q on U 2 and start to build a cantilever ( ): U 1 P U 2 (P 2 ) 1 Q 2 4 U 3 P We can continue this as long as the U i allow.

14 In fact, given the first 9 points (not P 2 ), we can continue the cantilever without using the curves!

15 So given the curve C, the collinear triple P i, and a choice of Q (close enough to P 2 ), we get a 9-point configuration on C. Take a cubic curve D through these 9 points. Then, because on cubics collinearity is group-related around every collinear triple, any cantilever generated by these 9 points lies entirely on D. Now choose Q so close to P 2 that we can build a cantilever with d + 1 points on each U i. All of these are in C D. By Bézout s Theorem, every component of D is also a component of C, so D C.

16 Corollaries The proof in fact shows something more general: Theorem Three sets of n points on irreducible algebraic curves C 1, C 2, C 3 of degree d in R 2 span O d (n 11/6 ) triple lines with at least one point from each curve, unless C 1 C 2 C 3 is a cubic curve. There are some nice corollaries. Corollary 1: Quadruple lines A set of n points on an algebraic curve of degree d in R 2 spans O d (n 11/6 ) quadruple lines, unless it is a union of four lines. Erdős conjectured that if a set of n points in the plane spans cn 2 quadruple lines, then it must have five points on a line.

17 Corollary 2: Distinct directions on curves Ungar (1982): n non-collinear points in R 2 determine n 1 distinct directions. Conjecture (Elekes): If n non-collinear points determine O(n) directions, then at least 6 of the points are on a conic. Theorem (Elekes-Szabó (2013) / Raz-Sharir-De Zeeuw (2014)) A set of n points on an irreducible algebraic curve C of degree d in R 2 determines Ω d (n 4/3 ) distinct directions, unless C is a conic.

18 Corollary 2: Distinct directions on curves (For simplicity, I ll only prove a weaker bound here. For 4/3, I d have to state the unwieldy unbalanced version of the main theorem.) Weaker theorem A set of n points on an irreducible algebraic curve C in R 2 determines Ω(n 7/6 ) distinct directions, unless C is a conic. Proof sketch: Take C 1 = C 2 = C and C 3 the line at infinity, with n points on C 3 corresponding to the n most popular directions. If we have o(n 7/6 ) directions, we will get ω(n 11/6 ) triple lines on C 1 C 2 C 3, so C 1 C 2 C 3 is a cubic. Since C 3 is a line distinct from C 1 C 2 = C, C must be a conic.

19 References György Elekes and Endre Szabó, On triple lines and cubic curves The Orchard Problem revisited, arxiv: György Elekes and Endre Szabó, How to find groups? (and how to use them in Erdős geometry?), Combinatorica, Ben Green and Terence Tao, On Sets Defining Few Ordinary Lines, Discrete & Computational Geometry, (cantilever bridge in Holland)