The flat trefoil and other oddities

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1 The flat trefoil and other oddities Joel Langer Case Western Reserve University ICERM June, 2015

2 Plane curves with compact polyhedral geometry Thm Assume: C CP 2 is an irreducible, real algebraic curve of degree d 4. Then the geometry (C, dx 2 + dy 2 ) is compact if and only if C is a Bernoulli lemniscate.

3 Plane curves with compact polyhedral geometry Thm Assume: C CP 2 is an irreducible, real algebraic curve of degree d 4. Then the geometry (C, dx 2 + dy 2 ) is compact if and only if C is a Bernoulli lemniscate. Cor Let x(s), y(s) parameterize an arc of C (as above) by unit speed. If x(s), y(s) extend meromorphically to all s C, then C is a line, a circle, or.

4 Plane curves with compact polyhedral geometry Thm Assume: C CP 2 is an irreducible, real algebraic curve of degree d 4. Then the geometry (C, dx 2 + dy 2 ) is compact if and only if C is a Bernoulli lemniscate. Cor Let x(s), y(s) parameterize an arc of C (as above) by unit speed. If x(s), y(s) extend meromorphically to all s C, then C is a line, a circle, or. Thm (, Singer, 2015) Let C be as above but with d < 8. Then (C, dx 2 + dy 2 ) is compact and flat if and only if C is the sextic trefoil.

5 Sextics studied by Euler, Serret, Liouville and others. Rational sextics with meromorphic arc length parameterizations: Left: 4(x 2 + y 2 ) (x 2 + y 2 ) (x 4 y 4 ) = 27 Middle: (x 2 + y 2 )(6 3 3x + x 2 + y 2 ) 2 = 4 Right: 4(x 2 + y 2 ) 3 + ( x)(x 2 + y 2 ) 2 12(x 2 + y 2 ) = 1

6 The Euler-Serret sextic (x 2 + y 2 )(6 3 3x + x 2 + y 2 ) 2 = 4 x(t) = 3 6t + 7 3t 2 16t t 4 6t 5 + 3t 6 (1 + t 2 ) 2 (1 3t + t 2 ) y(t) = 1 2 3t + 3t 2 3t t 5 t 6 3s (1 + t 2 ) 2 (1 ; t = tan 3t + t 2 ) 6

7 Polyhedral geometry of a quadratic differential Q = q(u)du 2 = u8 +14u 4 +1 u 2 (1 u 4 ) 2 du 2

8 Polyhedral geometry of Q = q(u)du 2 g = Q = λ 2 dudū has curvature K = log λ λ 2.

9 Polyhedral geometry of Q = q(u)du 2 g = Q = λ 2 dudū has curvature K = log λ For Q = (u u 0 ) n du 2 : u u 0 <ɛ λ 2. KdA = nπ.

10 Polyhedral geometry of Q = q(u)du 2 g = Q = λ 2 dudū has curvature K = log λ For Q = (u u 0 ) n du 2 : u u 0 <ɛ λ 2. KdA = nπ. Horizontal geodesics Q > 0: q(α(t))α (t) 2 > 0. n = 1 n = 1 n = 2

11 A surface of genus 28: 4g 4 = Z (Q) P(Q) Q = q(u)du 2 has 216 simple zeros and 54 double poles.

12 The Euclidean quadratic differential Q = ds 2 Isotropic coordinates: u = x + iy, v = x iy.

13 The Euclidean quadratic differential Q = ds 2 Isotropic coordinates: u = x + iy, v = x iy. Plane curve: 0 = f (x, y) = p(u, v).

14 The Euclidean quadratic differential Q = ds 2 Isotropic coordinates: u = x + iy, v = x iy. Plane curve: 0 = f (x, y) = p(u, v). Q = dx 2 + dy 2 = dudv = pu p v du 2 = pv p u dv 2.

15 The Euclidean quadratic differential Q = ds 2 Isotropic coordinates: u = x + iy, v = x iy. Plane curve: 0 = f (x, y) = p(u, v). Q = dx 2 + dy 2 = dudv = pu p v du 2 = pv p u dv 2. Natural equations: du ds = i pv p u, dv ds = i pu p v

16 The Euclidean quadratic differential Q = ds 2 Isotropic coordinates: u = x + iy, v = x iy. Plane curve: 0 = f (x, y) = p(u, v). Q = dx 2 + dy 2 = dudv = pu p v du 2 = pv p u dv 2. Natural equations: du ds = i pv p u, dv ds = i pu p v Rational natural equations: du ds = τ, dv ds = 1 τ, dτ ds = kn = p2 2 p 11 2p 1 p 2 p 12 + p 2 1 p 22 2p 2 1 p 2

17 Unit speed ellipse

18 Total ellipse

19 Total ellipse on one sheet Decomposition of the ellipse into five Euclidean subdomains.

20 Circular points c ±, isotropic projections and foci Isotropic lines: u = u 0, v = v 0

21 Circular points c ±, isotropic projections and foci Isotropic lines: u = u 0, v = v 0 Zeros of Q = dudv are isotropic tangent points.

22 Circular points c ±, isotropic projections and foci Isotropic lines: u = u 0, v = v 0 Zeros of Q = dudv are isotropic tangent points. Poles of Q are ideal points of C.

23 Circular points c ±, isotropic projections and foci Isotropic lines: u = u 0, v = v 0 Zeros of Q = dudv are isotropic tangent points. Poles of Q are ideal points of C. Poles of order n = 1, 2, 3 are circular ideal points of C.

24 Half of the unit speed Neumann quartic A totally circular rational quartic: Z P = 4 8 = 4g 4

25 Biflecnodal circular point Two inflectional tangents (u u 1 )(u u 2 ) = 0.

26 Complex Points on the Lemniscate Parallel curves projected to the real plane.

27 Squaring the Circle via Lemniscatic Sine

28 Unit speed lemniscate

29 Triunduloidal circular point Three unduloidal tangents T (u) = (u u 1 )(u u 2 )(u u 3 ) = 0.

30 Foci and double points for a pencil of sextics Bôcher-Grace locates nodes of p(u, v) = T (u) T (v) λ.

31 A trio of related curves parameterized by Dixon functions Fermat cubic x 3 + y 3 = 1, trihyperbola u 3 + v 3 = 1, and trefoil u 3 + v 3 = u 3 v 3 (u = x + iy, v = x iy).

32 Equal areas in equal times Centro-affine arc length along the Fermat cubic: x = sm t, y = cm t.

33 Group structure under Cremona transformation q s p q p r p q r s p q s Addition on the cubic H and sextic T = σ(h).

34 Conformal mapping by Dixon sine function Conformal mapping from disk to triangle.

35 Period parallelogram for Dixon functions sm z, cm z Lattice of inflections zeros and poles of sm z or cm z.

36 Symmetries of trefoil parameterization Tiling the plane by trihexagons; subdivision of a tile into 108 subtiles ( triangles).

37 The trefoil, a torus, covers the Riemann sphere 3-1 Unit speed trefoil: u = sm is, v = sm( is)

38 Uniform Subdivision of the Circle Theorem (Gauss, 1801; Wantzel, 1837) The regular n-gon may be constructed by straightedge and compass iff n = 2 j p 1 p 2... p k (p i distinct Fermat primes ).

39 Uniform Subdivision of the Lemniscate Theorem (Abel, 1827) The lemniscate may be evenly subdivided by straightedge and compass for precisely the same integers n = 2 j p 1 p 2... p k.

40 Uniform Subdivision of the Clover by Origami 1 Theorem (Cox, Shurman, 2005) The clover r 3/2 = cos( 3 2 θ) The clover can be divided into n equal lengths by origami if and only if n = 2 a 3 b p 1 p n where a, b 0 and p 1,..., p n are distinct Pierpont primes such that p i = 5, p i = 17, or p i 1 (mod 3).

41 Uniform Subdivision of the Trefoil by Origami 1 Theorem (, Singer, 2012) The trefoil r 3 = cos(3θ) The trefoil can be n-subdivided by origami iff n = 2 i 3 j p 1 p k (distinct Pierpont primes p m = 5, 17, or p m 1 mod 3).

Solutions to Practice Final

Solutions to Practice Final s to Practice Final 1. (a) What is φ(0 100 ) where φ is Euler s φ-function? (b) Find an integer x such that 140x 1 (mod 01). Hint: gcd(140, 01) = 7. (a) φ(0 100 ) = φ(4 100 5 100 ) = φ( 00 5 100 ) = (