University:
Amherst College
Course:
MATH 378 | Differential Geometry of Curves and Surfaces
Academic year:

2009
Views:

376

Pages:

67
Author:

Rafael Werner

Topics in Curve and Surface Implicitization Outline Curves: Moving Lines & µ-Bases Moving Curve Ideal & the Rees Algebra Adjoint Curves Outline Curves: Moving Lines & µ-Bases Moving Curve Ideal & the Rees Algebra Adjoint Curves Surfaces: Parametrized Surfaces Moving Planes & Syzygies Affine, Projective & Bihomogeneous The Resultant of a µ-Basis Curve Implicitization Turn a parametrization into an equation. Affine: Turn a(t) x= , c(t) into F (x, y) = 0. b(t) y= c(t) Curve Implicitization Turn a parametrization into an equation. Affine: Turn a(t) x= , c(t) b(t) y= c(t) into F (x, y) = 0. Projective: Turn x = a(s, t), y = b(s, t), z = c(s, t) into F (x, y, z) = 0. Moving Lines A moving line is an equation A(s, t)x + B(s, t)y + C(s, t)z = 0. Moving Lines A moving line is an equation A(s, t)x + B(s, t)y + C(s, t)z = 0. A moving line follows a parametrization x = a(s, t), y = b(s, t), z = c(s, t) if (a(s, t), b(s, t), c(s, t)) lies on the line A(s, t)x + B(s, t)y + C(s, t)z = 0 for all s, t. Moving Lines A moving line is an equation A(s, t)x + B(s, t)y + C(s, t)z = 0. A moving line follows a parametrization x = a(s, t), y = b(s, t), z = c(s, t) if (a(s, t), b(s, t), c(s, t)) lies on the line A(s, t)x + B(s, t)y + C(s, t)z = 0 for all s, t. If two moving lines follow a parametrization, their intersection is the parametrization. Moving Line Picture Here are two moving lines for an ellipse: Moving Line Picture Here are two moving lines for an ellipse: Moving Line Picture Here are two moving lines for an ellipse: Together they define the ellipse: Moving Line Picture Here are two moving lines for an ellipse: Together they define the ellipse: Moving Line Picture Here are two moving lines for an ellipse: Together they define the ellipse: This construction is due Steiner in 1832. µ-Bases Theorem: a, b, c ∈ k[s, t] homogeneous, deg = n, gcd(a, b, c) = 1. There exist moving lines p, q with: 1. Every moving line can be uniquely written up + vq, where u and v are homogeneous of degree m − deg(p) and m − deg(q). 2. deg(p) + deg(q) = n. µ-Bases Theorem: a, b, c ∈ k[s, t] homogeneous, deg = n, gcd(a, b, c) = 1. There exist moving lines p, q with: 1. Every moving line can be uniquely written up + vq, where u and v are homogeneous of degree m − deg(p) and m − deg(q). 2. deg(p) + deg(q) = n. Definition: p, q are a µ-basis when µ = deg(p) ≤ deg(q) = n − µ. Proof Let R = k[s, t] and I = ha, b, ci ⊂ R. The Hilbert Syzygy Theorem and a Hilbert polynomial computation give a free resolution 3 (a,b,c) 0 → R(−n−µ)⊕R(−2n+µ) → R(−n) −→ I → 0. The kernel of (a, b, c) is Syz(a, b, c). It consists of all triples (A, B, C) ∈ R3 such that Aa + Bb + Cc = 0. These give the moving lines Ax + By + Cz that follow the curve. QED Some History 1832 – Steiner describes conic sections using moving lines. Some History 1832 – Steiner describes conic sections using moving lines. 1887 – Meyer describes the syzygies of a, b, c ∈ k[s, t] and makes a conjecture for the syzygies of a1 , . . . , am . Some History 1832 – Steiner describes conic sections using moving lines. 1887 – Meyer describes the syzygies of a, b, c ∈ k[s, t] and makes a conjecture for the syzygies of a1 , . . . , am . 1890 – Hilbert proves the Syzygy Theorem and proves Meyer’s conjecture. Some History 1832 – Steiner describes conic sections using moving lines. 1887 – Meyer describes the syzygies of a, b, c ∈ k[s, t] and makes a conjecture for the syzygies of a1 , . . . , am . 1890 – Hilbert proves the Syzygy Theorem and proves Meyer’s conjecture. 1995 – Sederberg and Chen interpret moving lines in terms of syzygies. Moving Curves Moving lines are not the full story. Let R = k[s, t]. A polynomial P F = i+j+l=m Aijl (s, t) xi y j z l ∈ R[x, y, z] is called a moving curve of degree m. Moving Curves Moving lines are not the full story. Let R = k[s, t]. A polynomial P F = i+j+l=m Aijl (s, t) xi y j z l ∈ R[x, y, z] is called a moving curve of degree m. A moving curve follows our parametrization if P i j l A a i+j+l=m ijl b c ≡ 0 in R Moving Curves Moving lines are not the full story. Let R = k[s, t]. A polynomial P F = i+j+l=m Aijl (s, t) xi y j z l ∈ R[x, y, z] is called a moving curve of degree m. A moving curve follows our parametrization if P i j l A a i+j+l=m ijl b c ≡ 0 in R The moving curve ideal M C ⊂ R[x, y, z] is generated by these moving curves. The Rees Algebra I = ha, b, ci ⊂ R = k[s, t] has Rees algebra L∞ i i R[I] = i=0 I e ⊂ R[e]. Rees algebras are important in commutative algebra. The Rees Algebra I = ha, b, ci ⊂ R = k[s, t] has Rees algebra L∞ i i R[I] = i=0 I e ⊂ R[e]. Rees algebras are important in commutative algebra. The map (x, y, z) 7→ (ae, be, ce) gives a surjection R[x, y, z] −→ R[I]. The kernel is M C. Thus the moving curve ideal gives the defining relations of the Rees algebra! Example 1 a = 6s2 t2 − 4t4 , b = 4s3 t − 4st3 , c = s4 The moving curve ideal has five generators: • Two moving lines of degree 2 in s, t: p = st x + ( 12 s2 − t2 )y − 2st z q = s2 x − st y − 2t2 z • Two moving conics of degree 1 in s, t: s xy − t y 2 − 2t xz − s yz + 4t z 2 s x2 − t xy + 21 s y 2 − 2s xz + t yz • The implicit equation: y 4 + 4x3 z + 2xy 2 z − 16x2 z 2 − 6y 2 z 2 + 16xz 3 Example 1 Picture Here is the curve of Example 1: 2 1 0 0 -1 -2 0.5 1 1.5 2 Compute Example 1 To generate M C, begin with the moving lines: p = st x + ( 12 s2 − t2 )y − 2st z q = s2 x − st y − 2t2 z • s, t2 give p = (tx + 12 sy − 2tz)s + (−y)t2 q = (sx − ty)s + (−2z)t2 The Sylvester form is 1 tx + 2 sy − 2tz −y det sx − ty −2z This is the first moving conic generator of M C! Compute Example 1 • s2 , t give the second moving conic generator! • s, t and the moving conic generators give (xy − yz)s + (4z 2 − y 2 − 2xz)t (x2 + 12 y 2 − 2xz)s + (yz − xy)t The Sylvester form is 2 2 xy − yz 4z − y − 2xz det 2 1 2 x + 2 y − 2xz yz − xy This is the implicit equation! • The implicit equation is also Res(p, q). Example 2 a = 3s3 t − 3s2 t2 , b = 3s2 t2 − 3st3 , c = (s2 + t2 )2 The moving curve ideal has five generators: • Two moving lines of degree 1,3 in s, t: p = tx − sy q = s3 x + (2s2 t + t3 )y + (3st2 − 3s2 t)z • One moving conic of degree 2 in s, t: s2 x2 + (2s2 + t2 )y 2 + (3st − 3s2 )yz • One moving cubic of degree 1 in s, t: s x3 + 2s xy 2 + t y 3 − 3s xyz + 3s y 2 z • The implicit equation: x4 + 2x2 y 2 + y 4 − 3x2 yz + 3xy 2 z Example 2 Picture Here is the curve of Example 2: 2 1.5 1 0.5 0 -2 -1.5 -1 -0.5 0 0.5 -0.5 -1 1 Compute Example 2 To generate M C, begin with the moving lines: p = tx − sy q = s3 x + (2s2 t + t3 )y + (3st2 − 3s2 t)z • The moving lines give p = (−y)s + (x)t q = (s2 x)s + ((2s2 + t2 )y + (3st − 3s2 )z)t The Sylvester form is −y x det 2 s x (2s2 + t2 )y + (3st − 3s2 )z This is −1 times the moving conic generator! Compute Example 2 • p = t x − s y and the moving conic give −y x det sx2 + 2sy 2 − 3syz ty 2 + 3syz This is −1 times the moving cubic generator! • p = t x − s y and the moving cubic give −y x det 3 x + 2xy 2 − 3xyz + 3y 2 z y 3 This is −1 times is the implicit equation! • The implicit equation is also Res(p, q). PASI 2009 – p.18/41 Theorems There are theorems that explain these examples, plus results on parametrized curves in Pn . Some of the people involved: CC-, Hoffman, and Wang Busé Hoon, Simis, and Vasconcelos Kustin, Polini, and Ulrich Cortadellas Benítez and D’Andrea Goldman, Jia, and Wang I will explain some of this tomorrow. Rational Plane Curves Theorem: If C ⊂ P2 is defined by an irreducible equation of degree n, then C is rational ⇐⇒ P (n − 1)(n − 2) = p νp (νp − 1), where the sum is over all singular points p of C and νp is the multiplicity of C at p. Classical Proof: We will use adjoint curves. A curve D of degree m is adjoint to C if: • At all singular points p of C with multiplicity νp , the curve D has multiplicity at least νp − 1. The Classical Proof Lemma: For m ∈ {n − 1, n − 2}, ∃ a 1-dim linear system of plane curves whose general member D is adjoint to C and meets C in mn − (n − 1)(n − 2) − 1 fixed smooth points of C. Consequences: P • By Bezout, mn = D · C = p (νp − 1)νp + mn − (n − 1)(n − 2) − 1 + one more point. The Classical Proof Lemma: For m ∈ {n − 1, n − 2}, ∃ a 1-dim linear system of plane curves whose general member D is adjoint to C and meets C in mn − (n − 1)(n − 2) − 1 fixed smooth points of C. Consequences: P • By Bezout, mn = D · C = p (νp − 1)νp + mn − (n − 1)(n − 2) − 1 + one more point. • This moving point gives a parametrization! The Classical Proof Lemma: For m ∈ {n − 1, n − 2}, ∃ a 1-dim linear system of plane curves whose general member D is adjoint to C and meets C in mn − (n − 1)(n − 2) − 1 fixed smooth points of C. Consequences: P • By Bezout, mn = D · C = p (νp − 1)νp + mn − (n − 1)(n − 2) − 1 + one more point. • This moving point gives a parametrization! • Linear system ⇒ the curve is rational! QED The Classical Proof Lemma: For m ∈ {n − 1, n − 2}, ∃ a 1-dim linear system of plane curves whose general member D is adjoint to C and meets C in mn − (n − 1)(n − 2) − 1 fixed smooth points of C. Consequences: P • By Bezout, mn = D · C = p (νp − 1)νp + mn − (n − 1)(n − 2) − 1 + one more point. • This moving point gives a parametrization! • Linear system ⇒ the curve is rational! QED Also: we get the parametrization by a resultant. Example Consider the affine curve defined by F (x, y) = y 4 + 4x3 + 2xy 2 − 16x2 − 6y 2 + 16x = 0. Note (4 − 1)(4 − 2) = 3 [sing pts] · 2 [mult] · (2 − 1). To parametrize, use the linear system of conics Gs,t(x, y) = sx2 − txy + 12 sy 2 − 2sx + ty = 0. These adjoint curves all go through the origin. Observation: G is one of our moving conics! Picture 2 1 0 0 0.5 1 1.5 2 -1 -2 These conics go through the singular points and the origin, plus one more point that moves. The Parametrization Compute the resultants: Res(F, G, y) = x(x − 1)4 (x − 2)2 (s4 x − 6t2 s2 + 4t4 ) Res(F, G, x) = y 3 (y 2 − 2)2 (s3 y − 4ts2 + 4t3 ) The constant factors show that G = 0 goes through the origin and the singular points of F = 0. The other factors give 6s2 t2 − 4t4 4s3 t − 4st3 x= , y= . 4 4 s s This is an affine version of our original parametrization of F = 0! Theorem Theorem: Given a proper parametrization of degree n, there are elements of the moving curve ideal M C of degree one in s, t and degree n − 1 or n − 2 in x, y, z can be chosen to be adjoint linear systems on the rational curve defined by the parametrization. • Proof by Busé; µ = 1 by C-, Hoffman, Wang. • Theorem based on an observation of Sendra. • Moving lines: small deg x, y, z, large deg s, t. Adjoint curves: large deg x, y, z, small deg s, t. Surfaces Example: The Steiner surface is given by a(s, t) 2st x = = 2 d(s, t) s + t2 + 1 b(s, t) 2t y = = 2 d(s, t) s + t2 + 1 c(s, t) 2s z = = 2 d(s, t) s + t2 + 1 Surfaces Example: The Steiner surface is given by 0 0.10.2 0.3 1 0.75 0.5 0.25 1 0 0.75 0.5 a(s, t) 2st x = = 2 d(s, t) s + t2 + 1 b(s, t) 2t y = = 2 d(s, t) s + t2 + 1 c(s, t) 2s z = = 2 d(s, t) s + t2 + 1 Boeing 777 Boeing 777 The Boeing 777 was designed using 50 million surface patches. Guggenheim Bilbao Gehry Sketch Bilbao Close-Up Commutative Algebra Affine Case: A moving plane Ax + By + Cz + D = 0, A, B, C, D ∈ k[s, t] follows a, b, c, d iff Aa + Bb + Cc + Dd = 0. Thus moving planes live in the syzygy module Syz(a, b, c, d) ⊂ k[s, t]4 . Commutative Algebra Affine Case: A moving plane Ax + By + Cz + D = 0, A, B, C, D ∈ k[s, t] follows a, b, c, d iff Aa + Bb + Cc + Dd = 0. Thus moving planes live in the syzygy module Syz(a, b, c, d) ⊂ k[s, t]4 . Theorem: The syzygy module Syz(a, b, c, d) is a free k[s, t]-module of rank 3. Commutative Algebra Affine Case: A moving plane Ax + By + Cz + D = 0, A, B, C, D ∈ k[s, t] follows a, b, c, d iff Aa + Bb + Cc + Dd = 0. Thus moving planes live in the syzygy module Syz(a, b, c, d) ⊂ k[s, t]4 . Theorem: The syzygy module Syz(a, b, c, d) is a free k[s, t]-module of rank 3. Proof: Auslander-Buchsbaum & Quillen-Suslin! Commutative Algebra Projective Case: More complicated! • I = ha, b, c, di ⊂ R = k[s, t, u] homogeneous • φ : P2 99K P3 rational map • Basepoints V(a, b, c, d) ⊂ P2 • S = image ⊂ P3 parametrized surface P 2 • deg φ · deg S = n − p ep • ep = Hilbert-Samuel Multiplicity Projective Case The following are equivalent: • Syz(a, b, c, d) is free • pd(R/I) = 2 • R/I is Cohen-Macaulay • I is saturated. Example: Cubic surface in P3 has a, b, c, d deg 3: • Syz(a, b, c, d): 3 moving planes deg 1 in s, t, u. • Basepoints: Six. Also: No basepoints ⇒ Syz(a, b, c, d) not free. The Bihomogeneous Case Geometric Modeling often uses rectangular surfaces patches, built from polynomials in s, t whose Newton polygon is a rectangle. This leads naturally to a parametrization φ : P1 × P1 → P3 (assuming no basepoints), where φ is given by bihomogeneous polynomials of bidegree (n, m). Bigraded commutative algebra is very different! I will give an example on Thursday. The Affine Case A basis of Syz(a, b, c, d) over k[s, t] is a µ-basis. The Affine Case A basis of Syz(a, b, c, d) over k[s, t] is a µ-basis. Write the µ-basis as p=A x+B y+C z+D = 0 q = A′ x + B ′ y + C ′ z + D ′ = 0 ′′ ′′ ′′ ′′ r =A x+B y+C z+D = 0 The Affine Case A basis of Syz(a, b, c, d) over k[s, t] is a µ-basis. Write the µ-basis as p=A x+B y+C z+D = 0 q = A′ x + B ′ y + C ′ z + D ′ = 0 ′′ ′′ ′′ ′′ r =A x+B y+C z+D = 0 By Cramer, a, b, c, d are the 3×3 minors of     p A B C D    ′  ′ ′ ′ q  = A B C D  r A′′ B ′′ C ′′ D′′ Resultant of a µ-Basis For surfaces, the resultant of an affine µ-basis almost gives the implicit equation. Resultant of a µ-Basis For surfaces, the resultant of an affine µ-basis almost gives the implicit equation. Analysis: At a point (x, y, z) where Res(p, q, r) = 0, the equations p=A x+B y+C z+D = 0 q = A′ x + B ′ y + C ′ z + D ′ = 0 r = A′′ x + B ′′ y + C ′′ z + D′′ = 0 have a solution s, t (possibly at ∞). No Basepoints     p A B C D    ′  ′ ′ ′ q  = A B C D  r A′′ B ′′ C ′′ D′′ has rank 3 since a, b, c, d are the 3 × 3 minors. So no basepoints ⇒ the moving planes always have a unique point of intersection! p=0 q=0 r=0 Basepoints At a basepoint, the parameter values “blow up” to an exceptional curve on the surface. These curves come in three flavors: A line. A plane curve. A space curve. Basepoints At a basepoint, the parameter values “blow up” to an exceptional curve on the surface. These curves come in three flavors: A line. A plane curve. A space curve. These cases correspond to the rank of     p A B C D    ′  ′ ′ ′ q  = A B C D  . r A′′ B ′′ C ′′ D′′ Rank 2 Basepoints Here, the moving planes intersect in a line: q=0 p=0 r=0 Furthermore: ↑ exceptional curve The resultant Res(p, q, r) vanishes exactly on the surface, at least for s, t finite. A basepoint has rank two ⇐⇒ it is LCI! Rank 1 Basepoints Here, the moving planes coincide: ↑ p=q=r=0 exceptional curve Furthermore: The resultant Res(p, q, r) has an extraneous factor = the equation of the plane to the power ep − dp , dp = dimk Op /ha, b, c, di. A basepoint has rank one ⇐⇒ ha, b, c, di is almost LCI. Rank 0 Basepoints Here, the moving “planes” are the ambient space, since we have a space curve. Thus: The resultant Res(p, q, r) vanishes identically. A basepoint has rank zero ⇐⇒ locally ha, b, c, di requires four generators. Hence Res(p, q, r) requires a truly bad basepoint before it vanishes!

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