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Colloquium, School of Mathematcs University of Minnesota, October 5, 2006 Computing the Genus of a Curve Numerically Andrew Sommese University of Notre Dame In collaboration with Daniel Bates, IMA Christopher Peterson, Colorado State University Charles Wampler, General Motors R&D Center
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2 Reference on Numerical Algebraic Geometry up to 2005: A.J. Sommese and C.W. Wampler, Numerical solution of systems of polynomials arising in engineering and science, (2005), World Scientific Press. Website with more information and some articles: www.nd.edu/~sommese
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3 Overview What is an algebraic curve? What is the genus of a curve? An example from applications Four-bar-planar linkage coupler curves Background on Numerical Algebraic Geometry How to represent Positive Dimensional Solution Sets Numerical approach by using Hurwitz’s formula
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4 What is an algebraic curve? Curves are ancient, beautiful, and useful. They are pervasive in mathematics: in number theory, algebraic, analytic, and differential geometry, complex analysis, numerical analysis, topology,... The arise everywhere in applications.
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5 x 2 + y 2 = 1
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6 y 2 - x(x-2)(x-4)
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7 Real points of a High Order Plane Curve
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8 What is the genus of a curve? Let’s start with x 2 + y 2 = 1:
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9 genus 0 curve
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10 y 2 - x(x-2)(x-4)
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11 genus 1 curve
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12 x 2 + y 2 = 1 and x 4 + y 4 = 1
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13 genus 3 curve
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14 Interlude between slides Including blackboard discussion of the examples; the Euler characteristic; and Hurwitz’s formula.
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15 Some examples from applications One dimensional component of a polynomial system on :
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16 Planar four-bar coupler curves A four-bar planar linkage is a planar quadrilateral with a rotational joint at each vertex. They are useful for converting one type of motion to another. They occur everywhere.
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17 A simple four-bar
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18 More Abstractly
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19 D′D′ P δ λ µ u b v C D x y P0P0 C′C′
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20 P δ µ b-δ v C y P0P0 θ ye iθ b v = y – b ve iμ = ye iθ - (b - δ) = ye iθ + δ - b C′C′
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21 We use complex numbers (as is standard in this area) Summing over vectors we have two equations plus their two conjugates
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22 This gives four equations: in the variables
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23 Multiplying each side by its complex conjugate and letting we get three equations in the four variables
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24 We can solve for in the first two equations using Cramer’s rule, and substitute into the last Equation. This gives us an equation of degree (3,3) in We replace by
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25 Background on Numerical Algebraic Geometry Find all isolated solutions in of a system on n polynomials :
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26 Solving a system Homotopy continuation is our main tool: Start with known solutions of a known start system and then track those solutions as we deform the start system into the system that we wish to solve.
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27 Path Tracking This method takes a system g(x) = 0, whose solutions we know, and makes use of a homotopy, e.g., Hopefully, H(x,t) defines “nice paths” x(t) as t runs from 1 to 0. They start at known solutions of g(x) = 0 and end at the solutions of f(x) at t = 0.
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28 The paths satisfy the Davidenko equation To compute the paths: use ODE methods to predict and Newton’s method to correct.
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29 Solutions of f(x)=0 Known solutions of g(x)=0 t=0 t=1 H(x,t) = (1-t) f(x) + t g(x) x 3 (t) x 1 (t) x 2 (t) x 4 (t)
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30 Newton correction prediction { t x j (t) x* 0 1
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31 Continuation’s Core Computation Given a system f(x) = 0 of n polynomials in n unknowns, continuation computes a finite set S of solutions such that: any isolated root of f(x) = 0 is contained in S; any isolated root “occurs” a number of times equal to its multiplicity as a solution of f(x) = 0; S is often larger than the set of isolated solutions.
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32 Positive Dimensional Solution Sets We now turn to finding the positive dimensional solution sets of a system
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33 How to represent positive dimensional components? S. + Wampler in ’95: Use the intersection of a component with generic linear space of complementary dimension.
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34 Use a generic flag of affine linear spaces to get witness point supersets This approach has 19 th century roots in algebraic geometry, e.g., adjunction theory and the use of hyperplane sections to study varieties.
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35 Numerical approach by using Hurwitz’s formula Given an irreducible curve X that is a component of V(f) of a polynomial system, X is represented by a generic hyperplane L = 0 plus W, the set of degree X points where L meets X. L = 0 is the fiber of a general projection p from C N to C.
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36 g = -deg X + 1 + Steps: Find a finite set B of points of C containing all singular points of X plus all branch points of X. Do monodromy of the fiber of p around sufficiently small loops around the images R of B under p. Compute the number of sets r which the monodromy breaks the fiber over r in R into. The sum of the (deg X - r ) as r runs over R equals for the induced map from the desingularization of X down to C.
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37 Fermat curves: x d + y d = 1 we use a general linear projection. So we have a degree d projection; d(d-1) branchpoints in different fibers all contributing 1 to : the genus g is d(d-3)/2 + 1. So g = 6 for d = 5; g = 10 for d = 6; and g = 15 for d = 7.
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38 d = 5: 20 branch points
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39 d = 6: 30 branch points
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40 d = 7: 42 branch points
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41 16y 6 + 4x 4 - 24y 4 - 4x 2 + 9y 2 = 0
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42 genus 0: 10 contributions of one each to zero contributions 7 double points 1 sextuple point g = -6 + 10/2 + 1 = 0
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43 A general planar four-bar coupler curve C C is a (3,3) curve in P 1 x P 1. Arithmetic genus at most 4 with at least 3 singularities: so geometric genus is at most 1. We treat it as a degree 6 curve in P 2. We compute 30 potential branch points counting multiplicities; 2 potential branch points with multiplicity 6 each make a local contribution 0 to ; 3 potential branch points with multiplicity 2 each make a contribution of 0 to ; 12 potential branch points contribute one each to
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44 The geometric genus is 1
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45 Summary The numerical approach gives a direct method of computing the geometric genus of one-dimensional irreducible components of the solutions set of a system of polynomials.
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