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3264 Conics in a Second 3264 Conics in a Second Paul Breiding, Bernd Sturmfels, and Sascha Timme This article and its accompanying web interface present infinity, provided 퐴 and 푈 are irreducible and not multi- Steiner’s conic problem and a discussion on how enumerative ples of each other. This is the content of B´ezout’s theorem. and numerical algebraic geometry complement each other. To take into account the points of intersection at infinity, The intended audience is students at an advanced under- algebraic geometers like to replace the affine plane ℂ2 with 2 grad level. Our readers can see current computational the complex projective plane ℙℂ. In the following, when tools in action on a geometry problem that has inspired we write “count,” we always mean counting solutions in scholars for two centuries. The take-home message is that projective space. Nevertheless, for our exposition we work numerical methods in algebraic geometry are fast and reli- with ℂ2. able. A solution (푥, 푦) of the system 퐴 = 푈 = 0 has multiplic- We begin by recalling the statement of Steiner’s conic ity ≥ 2 if it is a zero of the Jacobian determinant 2 problem. A conic in the plane ℝ is the set of solutions to 휕퐴 휕푈 휕퐴 휕푈 2 ⋅ − ⋅ = 2(푎1푢2 − 푎2푢1)푥 a quadratic equation 퐴(푥, 푦) = 0, where 휕푥 휕푦 휕푦 휕푥 (3) 2 2 +4(푎1푢3 − 푎3푢1)푥푦 + ⋯ + (푎4푢5 − 푎5푢4). 퐴(푥, 푦) = 푎1푥 + 푎2푥푦 + 푎3푦 + 푎4푥 + 푎5푦 + 푎6. (1) Geometrically, the conic 푈 is tangent to the conic 퐴 if (1), If there is a second conic (2), and (3) are zero for some (푥, 푦) ∈ ℂ2. For instance, Fig- 2 2 푈(푥, 푦) = 푢1푥 + 푢2푥푦 + 푢3푦 + 푢4푥 + 푢5푦 + 푢6, (2) ure 1 shows a red ellipse and five other blue conics, which then the two conics intersect in four points in ℂ2, count- are tangent to the red ellipse. Steiner’s conic problem asks ing multiplicities and counting intersections at points at the following question: How many conics in the plane are tangent to five Paul Breiding is a postdoctoral researcher at Technische Universität Berlin. His given conics in general position? email address is [email protected]. The number is five, because each tangency condition re- Bernd Sturmfels is a director of the Max Planck Institute for Mathematics in moves one of the five degrees of freedom in a conic. the Sciences, Leipzig, and a professor of mathematics, statistics, and computer science at University of California, Berkeley. His email address is bernd@mis The present article concerns the following two subject .mpg.de. areas and how they approach Steiner’s problem: Sascha Timme is a PhD candidate at Technische Universität Berlin. His email Enumerative algebraic geometry: address is [email protected]. How many conics are tangent to five conics? Communicated by Notices Associate Editor Daniel Krashen. Numerical algebraic geometry: For permission to reprint this article, please contact: How do we find all conics tangent to five conics? [email protected]. The first question is the original conic problem, first asked DOI: https://doi.org/10.1090/noti2010 in 1848 by Steiner, who suggested the answer 7776. That 30 NOTICES OF THE AMERICAN MATHEMATICAL SOCIETY VOLUME 67, NUMBER 1 푎1, … , 푎6, 푢1, … , 푢6 of 퐴 and 푈: 4 2 2 4 4 2 2 3 풯(퐴, 푈) = 256푎1푎3푢3푢6 − 128푎1푎3푢3푢5푢6 4 2 4 2 4 2 2 4 (5) + 16푎1푎3푢5 푢6 + ⋯ + 푎5 푎6푢1푢2. The polynomial 풯 is a sum of 3210 terms. It is of degree six in the variables 푎1, … , 푎6 and of degree six in 푢1, … , 푢6. Known classically as the tact invariant, it vanishes precisely when the two conics are tangent. If the coefficients are general, we can assume that each conic 푈 that is tangent to 퐴, 퐵, 퐶, 퐷, 퐸 has nonzero con- stant term 푢6. We can then set 푢6 = 1. Steiner’s problem for the conics 퐴, 퐵, 퐶, 퐷, 퐸 now translates into a system of five polynomial equations in five unknowns, 푢1, 푢2, 푢3, 푢4, 푢5. Each of the five tangency constraints is an equation of degree six: 풯(퐴, 푈) = 풯(퐵, 푈) = ⋯ = 풯(퐸, 푈) = 0. (6) Steiner used B´ezout’s theorem to argue that these equa- tions have 65 = 7776 solutions. However, this number overcounts, because there is a Veronese surface of extraneous Figure 1. The red ellipse is tangent to four blue ellipses and solutions 푈, namely, the squares of linear forms. These de- one blue hyperbola. generate conics have the form 푈(푥, 푦) = (푥, 푦, 1) ⋅ ℓ푇 ℓ ⋅ (푥, 푦, 1)푇 , number turned out to be incorrect. In the year 1864 3264 3 Chasles gave the correct answer of . This was further where ℓ = (ℓ1, ℓ2, ℓ3) is a row vector in ℂ . Since 푈(푥, 푦) = 2ᵆ ᵆ ᵆ developed by Schubert, whose 1879 book led to Hilbert’s 1 2 4 푇 (푥, 푦, 1) ( ᵆ2 2ᵆ3 ᵆ5 ) (푥, 푦, 1) , the condition for 푈 to be a 15th problem and thus to the twentieth-century develop- ᵆ4 ᵆ5 2ᵆ6 ment of enumerative algebraic geometry. The number square is equivalent to 3264 appears prominently in the title of the textbook by 2푢 푢 푢 Eisenbud and Harris [EH16]. A delightful introduction 1 2 4 rank ( 푢 2푢 푢 ) ≤ 1. (7) to Steiner’s problem was presented by Bashelor, Ksir, and 2 3 5 푢 푢 2푢 Traves in [BKT08]. 4 5 6 Numerical algebraic geometry is a younger subject. It This discussion leads us to the following algebraic refor- started about forty years ago, going back at least to [GZ79]. mulation of Steiner’s conic problem: The textbook by Sommese and Wampler [SW05] is a stan- Find all solutions 푈 of the equations (6) dard reference. It focuses on numerical solutions to poly- (8) nomial equations. The field is now often seen as a branch such that the matrix in (7) has rank ≥ 2. of applied mathematics. But, as we demonstrate in this Ronga, Tognoli, and Vust [RTV97] proved the existence article, its methodology can be used in pure mathemat- of five real conics whose 3264 conics all have real coeffi- ics too. cients. In their argument they do not give an explicit in- An instance of our problem is given by a list of 30 = 5×6 stance but rather show that in the neighborhood of some coefficients in ℝ or ℂ: particular conic arrangement there must be an instance that has all of the 3264 conics real. Hence, this raises the 2 2 퐴(푥, 푦) = 푎1푥 + 푎2푥푦 + 푎3푦 + 푎4푥 + 푎5푦 + 푎6, following problem: 2 2 퐵(푥, 푦) = 푏1푥 + 푏2푥푦 + 푏3푦 + 푏4푥 + 푏5푦 + 푏6, Find an explicit instance 퐴, 퐵, 퐶, 퐷, 퐸 such 퐶(푥, 푦) = 푐 푥2 + 푐 푥푦 + 푐 푦2 + 푐 푥 + 푐 푦 + 푐 , (9) 1 2 3 4 5 6 (4) that the 3264 solutions 푈 to (8) are all real. 2 2 퐷(푥, 푦) = 푑1푥 + 푑2푥푦 + 푑3푦 + 푑4푥 + 푑5푦 + 푑6, Using numerical algebraic geometry we discovered the 퐸(푥, 푦) = 푒 푥2 + 푒 푥푦 + 푒 푦2 + 푒 푥 + 푒 푦 + 푒 . 1 2 3 4 5 6 solution in Figure 2. We claim that all the 3264 conics that are tangent to those five conics are real. By eliminating the two unknowns 푥 and 푦 from the three equations (1), (2), and (3), we can write the tangency Proposition 1. There are 3264 real conics tangent to those condition directly in terms of the 12 = 6 + 6 coefficients given by the 5 × 6 matrix in Figure 2. JANUARY 2020 NOTICES OF THE AMERICAN MATHEMATICAL SOCIETY 31 10124547 8554609 5860508 −251402893 −25443962 1 ⎡ 662488724 755781377 2798943247 1016797750 277938473 ⎤ 푎 푎 푎 푎 푎 푎 ⎢ 520811 2183697 9030222 −12680955 −24872323 ⎥ ⎡ 1 2 3 4 5 6⎤ ⎢ 1⎥ 푏1 푏2 푏3 푏4 푏5 푏6 1788018449 542440933 652429049 370629407 105706890 ⎢ ⎥ ⎢ 6537193 −7424602 6264373 13097677 −29825861 ⎥ ⎢푐1 푐2 푐3 푐4 푐5 푐6 ⎥ = ⎢ 1⎥ . ⎢ ⎥ ⎢ 241535591 363844915 1630169777 39806827 240478169 ⎥ 푑 푑 푑 푑 푑 푑 13173269 4510030 2224435 33318719 92891037 ⎢ 1 2 3 4 5 6⎥ ⎢ 1⎥ ⎣푒1 푒2 푒3 푒4 푒5 푒6 ⎦ ⎢ 2284890206 483147459 588965799 219393000 755709662 ⎥ ⎢ 8275097 −19174153 5184916 −23713234 28246737 ⎥ 1 ⎣ 452566634 408565940 172253855 87670601 81404569 ⎦ Figure 2. The five conics from Proposition 1. open source Julia package described in [BT18]. Those playing with the web interface need not worry about the inner workings. But, if you are curious, please read our section titled “How Does This Work?” Shortly after the user submits their instance, by entering real coefficients, the web interface reports whether thein- stance was generic enough to yield 3264 distinct complex solutions. These solutions are computed numerically. The Figure 3. The five blue conics in the central picture are those browser displays the number of real solutions, along with in Proposition 1. Shown in red is one of the 3264 real conics a picture of the instance and a rotating sample of real so- that are tangent to the blue conics. Each blue conic looks like lutions. As promised in our title, the computation of all a pair of lines, but it is a thin hyperbola whose branches are close to each other. The two pictures on the sides show solutions takes only around one second. closeups around two of the five points of tangency. The red Remark 2. We always assume that the five given conics are conic is tangent to one of the two branches of the blue 3264 hyperbola. real and generic. This ensures that there are complex solutions, and these conics are tangent to the given conics We provide an animation showing all the 3264 real con- at 5 × 3264 distinct points.
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