Anais do CNMAC v.2 ISSN 1984-820X

Curvature (k2, k3) formulas for implicit in n-dimensions

Osmar Alessio´ Depto de Matematica,´ FEIS, UNESP, 15054-000, Ilha Solteira, SP E-mail: [email protected]

Abstract: We derive (k2, k3) for transversal intersections of intersection curves of (n − 1) implicit surfaces in Rn. Keywords: (n − 1 )-surface intersection, Transversal intersection, Geometric properties, Implicit curves.

1 Introduction

The purpose of this work is provide cuvatures (k2, k3) formula for implicit curves in (n)-dimension, that is, to curves generated by the intersection of (n − 1) implicit equation: F1 (x1, . . . , xn) = 0 ∩ ... ∩ F(n−1) (x1, . . . , xn) = 0. Initially, the motivation of this paper was an open problem, proposed by R. Goldman. In his paper, he proposed the following open problem: Problem 1: Derive closed formulas for higher curvatures for implicit curves in (n + 1)-dimensions. While differential geometry of a parametric in R3 can be found in textbooks such as in (Struik, 1950; Wilmore, 1959; Stoker, 1969; Spivak, 1975; do Carmo, 1976), differential geometry of a parametric curve in Rn can be found in textbook such as in (Klingenberg, 1978), there is little literature on differential geometry of intersection curves in R3 and, rarely, in R4 and Rn. Willmore (1959) describes how to obtain the k and the torsion τ of the intersection curve of two implicit surfaces in R3. Hartmann (1996) provides formulas for computing the curvature k of the intersection curves for all three types of intersection problems in R3, using the theorem. Ye and Maekawa (1999) provides k, τ, using the vector α00 as linear combination of the vectors of the surfaces and α000 as linear combination of the vector and normal vectors of the surfaces. Goldman (2005) provides formulas for computing the curvature k 3 and torsion τ of intersection curve of two implicit surfaces in R and curvature one (k1) of intersection curve in (n)-dimensions. Alessio´ in (2006) provides formulas for computing the curvature k and torsion τ of the intersection curves of two implicit surface in R3, using the implicit function theorem. Alessio´ in (2009) provides formulas for computing the curvature k1, k2, k3 of the intersection curves of two implicit surface in R4, using the implicit function theorem and generalizing the method of X. Ye and T. Maekawa for 4-dimension. In this work, we derive curvatures k2 and k3 for implicit curvesµ in¶(n)- ··· ···· dimensions, operating T ∗ with ∇ two ∇ (∇T ∗) and three times ∇ (∇ (∇T ∗)), for obtain β, β and

using outer product for obtain (k2, k3).

2 Review the of Differential Geometry and Outer Product

Where ∧ is outer product of the three and four vectors in (n) −dimensional, × is cross product of the (n − 1) vectors in (n) −dimensional and ∗ product between matrices. The extension of the cross product (×) to n-dimensions that generates a vector perpendicular to a collection of n − 1 vectors is given by a determinant. Let e = (e1, e2, ··· , en) be the canonical basis for n R . Then     e e1 e2 ··· en  ∂F1 ∂F1 ∂F1   ∇F1   ∂x ∂x · ·· ∂x     1 2 n  ∇F1 × ... × ∇Fn−1 = Det  .  = Det  . . .   .   . . · ·· .  ∂Fn−1 ∂Fn−1 ∂Fn−1 ∇Fn−1 · ·· ∂x1 ∂x2 ∂xn

— 1132 — The outer product (∧) of the r vectors in n−dimensional P i P i Let (e1, ..., en) be an orthonormal basis. If u1 = α1ei, ..., ur = αrei, the outer product of i i P J r-vectors is u1 ∧ ... ∧ ur = det(α )eJ where ej = ej1 ∧ ... ∧ ejr ,J = {j1, ..., jr}, 1 ≤ j1 ≤ · · · ≤ ¯ J ¯ ¯ αj1 αj1 . . . αj1 ¯ ¯ 1 2 r ¯ ¯ j2 j2 j2 ¯ ¯ α1 α2 . . . αr ¯ j ≤ (n). Then αJ =¯ . . . ¯ is the matrix r × r. r ¯ . . .. . ¯ ¯ . . . . ¯ ¯ jr jr jr ¯ α1 α2 . . . αr

2.0.1 The Local Theory of Curves Parametrized Arbitrary. Definition 1 Let β :(a, b) → Rn be a regular curve, and let α :(c, d) → Rn be a unit-speed parametrization of β. Write β(t) = α(s(t)) ( where s(t) is just the arc length function).© Denote byª ki, i = 1, 2, 3, ..., n the curvatures. Also, let {b1, ..., bn} be the Frenet frame field of α and b1, ..., bn be the Frenet frame field of β. Then we define ki (t) = ki (s (t)) , i = 1, ..., n and bi (t) = bi (s (t)) , i = 1, ..., n. ° ° ° ° Definition 2 Let β :(a, b) → Rn be a regular curve whit speed v (t) = °β˙(t)° =s ˙ (t) and let α :(c, d) → Rn be a unit-speed parametrization of β. Write β (t) = α (s (t)) where s(t) is just the arc length function. Then, the Frenet formulas of β are: d ¡ ¢ b (t) = −v (t) κ (t) b (t) + v (t) κ (t) b (t) , dt i (i−1) i−1 i i+1 where κ0 = κn = 0, b0 = bn+1 = 0.

The following properties characterize some special curves: i+1 κi (t) = 0, i ∈ {1, ··· , n − 1} if and only if β is lying in a i−dimensional subspace of R ...... Lemma 1 The vectors β,˙ β¨, β, and β of a regular curve β are given by ˙ β(t) = vb1 (t) (1) dv β¨(t) = b (t) + v2k (t) b (t) (2) dt 1 1 2 µ 2 ¶ µ ¶ ... d v 2 dv dk (t) β(t) = − v3k (t) b (t) + 3v k (t) + v2 1 b (t) + v3k (t) k (t) b (t) (3) dt2 1 1 1 dt 1 dt 2 1 2 3 .... 4 β (t) = a14 (t) b1 (t) + a24 (t) b2 (t) + a34 (t) b3 (t) + v k1 (t) k2 (t) k3 (t) b4 (t) , (4) where v denotes the speed of β.

2.1 Implicit Surface Definition 3 (Implicit Surface) Let f : D ⊂ Rn −→ R be a differentiable mapping of an open set D. Given c ∈ R, we remember that the c of the f is the set defined as −1 −1 f (c) = {(x1, ..., xn) ∈ D; f (x1, ..., xn) = c}, i.e., f (c) is the set of the solutions in D of the equations f (x1, ..., xn) = c.

Proposition 1 [Regular Surfaces] If f : U ⊂ Rn → R is a differentiable function and c ∈ f (U) −1 n is a regular³ value of´ f, then f (c) is a regular surface in R . The implicit surface f is regular if ∇f = ∂f , ..., ∂f 6= 0. ∂x1 ∂xn

−1 −1 Definition 4 Transversal Intersection Curve Three regular surfaces S1 = F1 (c) ,S2 = F2 (c) , ..., −1 F1 S(n−1) = Fn−1 (c) intersect each other transversally if, whenever p ∈ S1 ∩S2 ∩...∩Sn−1, then N (p), N F2 (p), ..., N Fn−1 (p) are not parallel, i.e., linearly independent.

— 1133 — 3 Curves in Surface

n Consider an implicit surface represented by F1,F2, ..., Fn−1 : R −→ R. A curve x1 = x1 (t), x2 = x2(t), ..., xn = xn(t), in the n-dimensional space, defines a implicit curve β(t) = {(x1 (t) , ..., xn(t)) | F1 (β(t)) = 0 ∩ ... ∩ Fn−1 (β(t)) = 0} on (n)-dimesional space. Then, we have ˙ β (t) = ∇F1 × ... × ∇Fn−1, (5) ¨ β (t) = (∇F1 × ... × ∇Fn−1) ∗ ∇ (∇F1 × ... × ∇Fn−1) , (6) ... β (t) = T ∗ ∗ ∇ (T ∗) ∗ ∇ (T ∗) + T ∗ ∗ ∇ (∇ (T ∗)) ∗ (T ∗)T , (7) ...... β (t) = β (t) ∗ ∇ (T ∗) + 3β¨ (t) ∗ ∇ (∇ (T ∗)) ∗ (T ∗)T + (T ∗) ∗ [T ∗ ∗ ∇ (∇ (∇ (T ∗)))] ∗ (T ∗)T (8) £ ¤ ....T £ ...... ¤ where β˙T (x) = x˙ x˙ ··· x˙ , ··· , β (x) = x x ··· x , £ 1 2 ¤ n 1 2 n ∗ ∗ ∗ ∗ ∇f = fx1 fx2 ··· fxn ,T = ∇F1 × ... × ∇Fn−1 = (T11,T12, ··· ,T1n).

∗ ∗ ∗ T ∗ Here ∇ (T ) means apply ∇ to each element T1k of the vector T to generate a column, ∇ (T1k) =  ∗   ∗ ∗  (T ) (T ) ··· (T ) ··· (T1n) 1k x1 11 x1 1k x1 x1 ∗ ∗ ∗ ∗  (T )   (T11) ··· (T ) ··· (T1n)   1k x2   x2 1k x2 x2   . , then ∇ (T ∗) =  . . . . .  ,  .   . . . . .  (T ∗ ) (T ∗ ) ··· (T ∗ ) ··· (T ∗ ) 1k xn 11 xn 1k xn 1n xn and ∇ (∇ (T ∗)) means apply ∇ to each column vector of the matrix ∇ (T ∗) to generate a matrix,  ³ ´  ∗  ∗  ∇ (T )  ∗ ∗ ∗  (T ) 1k x1 (T ) (T ) ··· (T ) 1k x1  ³ ´  1k x1x1 1k x1x2 1k x1xn  ∗   ∗   ∗ ∗ ∗  ¡ ¢ (T1k)  ∇ (T1k)  (T1k)x x (T1k)x x ··· (T1k)x x T ∗  x2   x2   2 1 2 2 2 n  H1k = ∇ ∇ (T1k) = ∇  .  = =  . . .  ,  .   .   . . .. .  .  .  . . . . ∗  ³ ´  ∗ ∗ ∗ (T ) ∗ (T ) (T ) ··· (T ) 1k xn ∇ (T ) 1k xnx1 1k xnx2 1k xnxn £ ¤1k xn ∗ then ∇ (∇ (T )) = H11 ··· H1k ··· H1n , is 1 × n matrices, and ∇ (∇ (∇ (T ∗))) means apply ∇ to each element H (matrix) of the matrix ∇ (∇ (T ∗)) , to generate   1k (H1k) x1  (H1k)  T  x2  a column J1k = ∇ (H1k) =  .  , of n × 1 matrices, then  .  (H1k) xn   (H11) ··· (H1k) ··· (H1n) x1 x1 x1  (H11) ··· (H1k) ··· (H1n)  ∗ £ ¤  x2 x2 x2  ∇ (∇ (∇ (T ))) = J11 ··· J1k ··· J1n =  . . . . .  , is n × n ma-  . . . . .  (H11) ··· (H ) ··· (H1n) xn 1k xn xn trices. Proof. To be simpler the calculations, we go to make for the case of the curve β : I → R3. The Eq. ( 6) is a consequence

³ ´ ¨ d ˙ β (t) = dt β (t) ¨ d β (t) = dt (∇F1 × ∇F2) ¨ d d β (t) = (∇F1) × ∇F2 + ∇F1 × (∇F2) ³¡ dt¢ ´ ³dt¡ ¢ ´ ¨ 0 T 0 T β (t) = α (s) HF1 × ∇F2 + ∇F1 × α (s) HF2     1 2 1 2 1 2 1 2 ∗ T Fx x Fx − Fx x Fx · · · Fx x Fx − Fx x Fx T11  2 1 3 3 1 2 1 1 2 2 1 1  ¨  ∗   1 2 1 2 1 2 1 2  β (t) = T Fx x Fx − Fx x Fx · · · Fx x Fx − Fx x Fx + 12∗  2 2 3 3 2 2 1 2 2 2 2 1  T13 F 1 F 2 − F 1 F 2 · · · F 1 F 2 − F 1 F 2  x2x3 x3 x3x3 x2 x1x3 x2 x2x3 x1   1 2 1 2 1 2 1 2 ∗ T Fx Fx x − Fx Fx x · · · Fx Fx x − Fx Fx x T11  2 3 1 3 2 1 1 2 1 2 1 1   ∗   1 2 1 2 1 2 1 2  T Fx Fx x − Fx Fx x · · · Fx Fx x − Fx Fx x 12∗  2 3 2 3 2 2 1 2 2 2 1 2  T13 F 1 F 2 − F 1 F 2 · · · F 1 F 2 − F 1 F 2  x2 x3x3 x3 x2x3 x1 x2x3 x2 x1x3    1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 ∗ T Fx x Fx + Fx Fx x − Fx x Fx − Fx Fx x · · · Fx x Fx + Fx Fx x − Fx x Fx − Fx Fx x T11  2 1 3 2 3 1 3 1 2 3 2 1 1 1 2 1 2 1 2 1 1 2 1 1  ¨  ∗   1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2  β (t) = T Fx x Fx + Fx Fx x − Fx x Fx − Fx Fx x · · · Fx x Fx + Fx Fx x − Fx x Fx − Fx Fx x 12∗  2 2 3 2 3 2 3 2 2 3 2 2 1 2 2 1 2 2 2 2 1 2 1 2  T13 F 1 F 2 + F 1 F 2 − F 1 F 2 − F 1 F 2 · · · F 1 F 2 + F 1 F 2 − F 1 F 2 − F 1 F 2 x2x3 x3  x³2 x3x3 x3x3 x´2 x3 ³x2x3 x1x´3 x2 ³ x1 x2x3 x2x3´ x1  x2 x1x3 F 1 F 2 − F 1 F 2 − F 1 F 2 − F 1 F 2 F 1 F 2 − F 1 F 2    x2 x3 x3 x2 x1 x3 x3 x1 x1 x2 x2 x1  T ∗ T  ³ ´x1 ³ ´x1 ³ ´x1  11∗  1 2 1 2 1 2 1 2 1 2 1 2  β¨ (t) =  T   Fx Fx − Fx Fx − Fx Fx − Fx Fx Fx Fx − Fx Fx  12∗  2 3 3 2 x 1 3 3 1 x 1 2 2 1 x  T  ³ ´ 2 ³ ´ 2 ³ ´ 2  13 F 1 F 2 − F 1 F 2 − F 1 F 2 − F 1 F 2 F 1 F 2 − F 1 F 2 x2 x3 x3 x2 x1 x3 x3 x1 x1 x2 x2 x1 x3  ¡ ¢ ¡ ¢ x3¡ ¢  x3 T ∗ T ∗ T ∗ 11 x1 12 x1 13 x1 £ ¤  ¡ ∗ ¢ ¡ ∗ ¢ ¡ ∗ ¢  β¨ (t) = T ∗ T ∗ T ∗  T T T  11 12 13  ¡ 11¢x2 ¡ 12¢x2 ¡ 13¢x2  T ∗ T ∗ T ∗ 11 x3 12 x3 13 x3 ¨ β (t) = (∇F1 × ∇F2) ∗ ∇¡(∇F¢ 1 × ∇F2) β¨ (t) = T ∗ ∗ ∇ T ∗

— 1134 — d ∗ ∗ ∗ ∗ The dt (T ) = T ∗ ∇ (T ) is valid for T = ∇F1 × ... × ∇Fn−1. d ∗ ∗ ∗ T d ∗ ∗ ∗ We have dt [∇ (T )] = ∇ (∇ (T )) ∗ (T ) and dt [∇ (∇ (T ))] = T ∗ ∇ (∇ (∇ (T ))) . The Eq. ( 7) is a consequence of ... h h ii β (t) = d d β˙ (t) ... dt dt d ∗ ∗ ... β (t) = dt [(T ) ∗ ∇ (T )] β (t) = d [(T ∗)] ∗ ∇ (T ∗) + (T ∗) ∗ d [∇ (T ∗)] ... dt dt β (t) = (T ∗) ∗ ∇ (T ∗) ∗ ∇ (T ∗) + (T ∗) ∗ ∇ (∇ (T ∗)) ∗ (T ∗)T The Eq. ( 8) is a consequence of

.... £... ¤ β (t) = d β (t) .... h dt i d ∗ ∗ ∗ ∗ ∗ ∗ T β (t) = dt (T ) ∗ ∇ (T ) ∗ ∇ (T ) + (T ) ∗ ∇ (∇ (T )) ∗ (T ) .... h i d ∗ ∗ ∗ d ∗ ∗ ∗ T β (t) = dt [(T ) ∗ ∇ (T ) ∗ ∇ (T )] + dt (T ) ∗ ∇ (∇ (T )) ∗ (T ) ...... β (t) = β (t) ∗ ∇ (T ∗) + 3β¨ (t) ∗ ∇ (∇ (T ∗)) ∗ (T ∗)T + (T ∗) ∗ [T ∗ ∗ ∇ (∇ (∇ (T ∗)))] ∗ (T ∗)T

3.1 Curvature Formulas Outer Product

Theorem 1 (Goldman) First Curvature of n-1 implicit hypersurfaces

kT ∗ ∧ [T ∗ ∗ ∇ (T ∗)]k k1 = (9) kT ∗k3

Proof. The outer product of β˙(t) ∧ β¨(t)is

3 ¡ ¢ β˙(t) ∧ β¨(t) = v k1 (t) b1 (t) ∧ b2 (t) ° ° ° ° ° ° 3 ¡ ¢ °β˙(t) ∧ β¨(t)° = °v k1 (t) b1 (t) ∧ b2 (t) ° ° ° 3 = v k1 (t) °b1 (t) ∧ b2 (t)° , kT ∗ ∧ [T ∗ ∗ ∇ (T ∗)]k then k1 = . kT ∗k3

Theorem 2 Second Curvature of n-1 implicit hypersurfaces ° h i° ° ° °T ∗ ∧ [T ∗ ∗ ∇ (T ∗)] ∧ T ∗ ∗ ∇ (T ∗) ∗ ∇ (T ∗) + T ∗ ∗ ∇ (∇ (T ∗)) ∗ (T ∗)T ° k2 = (10) ∗ 6 2 kT k k1 ... Proof. The outer product of β˙(t) ∧ β¨(t)∧β(t) is ... ¡ ¢ ˙ ¨ 6 2 β(t) ∧ β(t) ∧ β(t) = v k1 (t) k2 (t) b1 (t) ∧ b2 (t) ∧ b3 (t) ° ... ° ° ¡ ¢° ° ˙ ¨ ° ° 6 2 ° °β(t) ∧ β(t) ∧ β(t)° = v k1 (t) k2 (t) b1 (t) ∧ b2 (t) ∧ b3 (t) ° ° 6 2 ° ° = v k1 (t) k2 (t) b1 (t) ∧ b2 (t) ∧ b3 (t) , ° h i° ° ° °T ∗ ∧ [T ∗ ∗ ∇ (T ∗)] ∧ T ∗ ∗ ∇ (T ∗) ∗ ∇ (T ∗) + T ∗ ∗ ∇ (∇ (T ∗)) ∗ (T ∗)T ° then k2 = . ∗ 6 2 kT k k1

— 1135 — Theorem 3 Third Curvature of n-1 implicit hypersurfaces ° ° ° h i ···· ° ° ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ T ° °T ∧ [T ∗ ∇ (T )] ∧ T ∗ ∇ (T ) ∗ ∇ (T ) + T ∗ ∇ (∇ (T )) ∗ (T ) ∧β (t)° k3 = . (11) ∗ 10 3 2 kT k k1 (t) k2 (t) ...... Proof. The outer product of β˙(t) ∧ β¨(t)∧β(t)∧ β (t)is

...... ¡ ¢ ˙ ¨ 10 3 2 β(t) ∧ β(t) ∧ β(t) ∧ β (t) = v k1 (t) k2 (t) k3 (t) b1 (t) ∧ b2 (t) ∧ b3 (t) ∧ b4 (t) ° ...... ° ° ¡ ¢° ° ˙ ¨ ° ° 10 3 2 ° °β(t) ∧ β(t) ∧ β(t) ∧ β (t)° = v k1 (t) k2 (t) k3 (t) b1 (t) ∧ b2 (t) ∧ b3 (t) ∧ b4 (t) ° ° 10 3 2 ° ° = v k1 (t) k2 (t) k3 (t) b1 (t) ∧ b2 (t) ∧ b3 (t) ∧ b4 (t)

° h i ° ° T .... ° °T ∗ ∧ [T ∗ ∗ ∇ (T ∗)] ∧ T ∗ ∗ ∇ (T ∗) ∗ ∇ (T ∗) + T ∗ ∗ ∇ (∇ (T ∗)) ∗ (T ∗) ∧ β (t)° Then k3 = . ∗ 10 3 2 kT k k1 (t) k2 (t)

3 4 Examples (The intersection F1 ∩ F2 ∩ F3 ∩ F4 is Helix in R )

Example 1 The implicit surface F1 ,F2,F3 and F4 are given by , F1(x, y, z, w, u) = u, F2(x, y, z, w, u) = w, 2 2 2 F3(x, y, z, w, u) = x + y − w − u − a ³z ´ F (x, y, z, w, u) = y − x tan . 4 b √ √ The point of the intersection curve is p = (a 2 , a 2 , b π , 0, 0) ∈ SF1 ∩ SF2 ∩ SF3 ∩ SF4 . We have 0 2 2 4 Ã √ ! ¡ √ √ ¢ a 2 ∇F = (0, 0, 0, 0, 1) , ∇F = (0, 0, 0, 1, 0) , ∇F = a 2, a 2, 0, −1, −1 and ∇F = −1, 1, − , 0, 0 . 1 2 3 4 b ... Compute β˙ (t) , β¨ (t) , β (t) Compute β˙ (t)

· β = ∇F1 × ∇F2 × ∇F3 × ∇F4 where × is cross product. µ ¶ · 2a2 2a2 √ β = , − , −2 2a, 0, 0 , b b Compute β¨ (t) ·· β = T ∗ ∗ ∇T ∗ where T ∗ = (T 11,T 12,T 13,T 14,T 15)    √ √  T 11 T 12 T 13 T 14 T 15 2 2a − 4 2a −2 0 0 x x x x x  √b b   T 11 T 12 T 13 T 14 T 15  2 2a  y y y y y   b 0 −2 0 0  ∗  2 2 √  ∇ (T ) =  T 11z T 12z T 13z T 14z T 15z  =  4a 4a 2 2a    2 − 2 − 0 0    b b b  T 11w T 12w T 13w T 14w T 15w  0 0 0 0 0  T 11u T 12u T 13u T 14u T 15u 0 0 0 0 0  √ √  2 2a − 4 2a −2 0 0 √b b µ ¶  2 2a  ·· 2 2  b 0 −2 0 0  ∗ ∗ 2a 2a √  2 2 √  β= T ∗ ∇T = − −2 2a 0 0  4a − 4a − 2 2a 0 0  b b  b2 b2 b   0 0 0 0 0  0 0 0 0 0 µ √ ¶ ·· 8 2a3 8a2 β= − , 0, , 0, 0 b2 b ... Compute β (t)

— 1136 — ... β (t) = T ∗ ∗ ∇ (T ∗) ∗ ∇ (T ∗) + T ∗ ∗ ∇ (∇ (T ∗)) ∗ (T ∗)T ∗ = ∇ (∇T)    T 11xx T 11xy T 11xz T 11xw T 11xu T 15xx T 15xy T 15xz T 15xw T 15xu      T 11yx T 11yy T 11yz T 11yw T 11yu   T 15yx T 15yy T 15yz T 15yw T 15yu      = T 11zx T 11zy T 11zz T 11zw T 11zu  , ··· ,  T 15zx T 15zy T 15zz T 15zw T 15zu   T 11wx T 11wy T 11wz T 11ww T 11wu   T 15wx T 15wy T 15wz T 15ww T 15wu  T 11ux T 11uy T 11uz T 11uw T 11uu T 15ux T 15uy T 15uz T 15uw T 15uu  √   4 4 2a  √    0 0 0 8 8 2a 0 0 0 0 0 b √b2 − b 0 − 2 0 0    b  4   4 4 2a  0 0 0 0 0  0 0 − 0 0    b 0 2 0 0   √   √b    √ √ b   2   4 4 2a   = 4 2a 4 2a 16a2  ,  − 8 2a 0 − 16a 0 0  ,  0 − − 0 0  , [0] , [0] , where  0 0   b2 b3   b b    b2 b2 b3    0 0 0 0 0  0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0  0 0 0 0 0 0 0 0 0 0  0 0 0 0 0  ³ √ ´   ... 4 3   96a 64 2a [0] =  0 0 0 0 0 , then β (t) = b3 , 0, − b2 , 0, 0 .  0 0 0 0 0  0 0 0 0 0 .... Compute β (t) ···· ... ∗ ¨ ∗ ∗ T ∗ ∗ ∗ ∗ T β (t) = β (t) ∗ ∇ (T ) + 3β (t) ∗ ∇ (∇ (T )) ∗ (T ) + (T ) ∗ [T ∗ ∇ (∇ (∇ (T )))] ∗ (T ) (H ) 1k x1  (H1k)  £ ¤  x2  ∗ T ∇ (∇ (∇ (T )))= J J J J J ,J = ∇ (H ) =  (H1k) , 11 12 13 14 15 1k 1k  x3   (H )  1k x4 (H1k)  ³ ´  x5 ∇ (T ∗ ) 1k x1    ³ ´  (T ∗ ) (T ∗ ) (T ∗ ) (T ∗ ) (T ∗ )  ∗  1k x1x1 1k x1x2 1k x2x3 1k x1x4 1k x1x5  ∇ (T )  ∗ ∗ ∗ ∗ ∗ 1k x2  (T ) (T ) (T ) (T ) (T )   ³ ´   1k x2x1 1k x2x2 1k x2x3 1k x2x4 1k x2x5   ∗  ∗ ∗ ∗ ∗ ∗ H1k = ∇J1k =  ∇ (T )  =  (T ) (T ) (T ) (T ) (T )  , 1k x3  1k x3x1 1k x3x2 1k x3x3 1k x3x4 1k x3x5   ³ ´   ∗ ∗ ∗ ∗ ∗   ∗  (T ) (T ) (T ) (T ) (T )  ∇ (T )  1k x4x1 1k x4x2 1k x4x3 1k x4x4 1k x4x5 1k x4 ∗ ∗ ∗ ∗ ∗  ³ ´  (T1k)x x (T1k)x x (T1k)x x (T1k)x x (T1k)x x ∗ 5 1 5 2 5 3 5 4 5 5 ∇ (T1k)  x5  (H ) (H ) (H ) (H ) (H ) 11 x1 12 x1 13 x1 14 x1 15 x1  (H11) (H12) (H13) (H14) (H15)   x2 x2 x2 x2 x2  ∗ ∇ (∇ (∇ (T )))=  (H11) (H12) (H13) (H14) (H15)  ,  x3 x3 x3 x3 x3   (H ) (H ) (H ) (H ) (H )  11 x4 12 x4 13 x4 14 x4 15 x4 (H ) (H ) (H ) (H ) (H ) 11 x5 12 x5 13 x5 14 x5 15 x5 ∇ (∇ (∇ (T ∗)))=       0 0 0 0 0 0 0 − 16 0 0   b2 0 0 0 0 0   0 0 8 0 0       2   0 0 0 0 0   0 0 0 0 0     b√   √      8 16 2a   16 32 2a   0 0 0 0 0  [0] [0]  0 2 3 0 0 − 2 0 − 3 0 0    b b   b b   0 0 0 0 0     0 0 0 0 0   0 0 0 0 0    0 0 0 0 0   0 0 0 0 0 0 0 0 0 0       8       0 0 2 0 0 0 0 0 0 0 0 0 0 0 0    b     0 0 0 0 0   0 0 0 0 0   0 0 0 0 0     √         8 32 2a   0 0 0 0 0   0 0 − 8 0 0  [0] [0]    2 0 − 3 0 0  b2    b b   0 0 0 0 0       0 0 0 0 0  0 0 0 0 0   0 0 0 0 0 0 0 0 0 0   0 0 0√ 0 0      √      8 16 2a 16 32 2a   0 2 3 0 0 − 0 − 0 0 0 0 0 0 0    b b√  b2 b3 8    8 16 2a     0 0 − 0 0   0 0 0  0 0 0 0 0   b2    b2 b3   √   √     √ √ 2   32 2a 80a2   8 16 2a  [0] [0]    16 2a 16 2a 80a  − 0 − 0 0 0 − 2 − 3 0 0    3 3 4 0 0   b3 b4   b b     b b b     0 0 0 0 0    0 0 0 0 0 0 0 0 0 0   0 0 0 0 0   0 0 0 0 0 0 0 0 0 0   [0] [0] [0] [0] [0]  [0] [0] [0] [0] [0]   0 0 0 0 0  0 0 0 0 0    where [0] =  0 0 0 0 0   0 0 0 0 0  0 0 0 0 0 ···· √ √ 384 2a5 128 2a3 128(12a6+11a4b2) β (t) = ( b4 , b2 , b5 , 0, 0).

Compute Curvatures k1 (t) , k2 (t) and k3 (t) Compute for using outer product

— 1137 — Compute k1 (t) v u ¯ ¯2 r ¯ j1 j1 ¯ q P u P ¯ α α ¯ ku ∧ u k = det(αJ )2 = t ¯ 1 2 ¯ = kuk2 kvk2 − (u · v)2, 1 2 ¯ j2 j2 ¯ J j1

∗ ∗ ∗ kT ∧ [T ∗ ∇ (T )]k a k1 = = 2 2 kT ∗k3 a +b

Compute k2 (t) v u ¯ ¯2 ¯ j1 j1 j1 ¯ r u ¯ α α α ¯ P u P ¯ 1 2 3 ¯ ku ∧ u ∧ u k = det(αJ )2 = u ¯ αj2 αj2 αj2 ¯ , 1 2 3 t ¯ 1 2 3 ¯ J j1

Compute k3 (t) There ∧ is outer product of the four vectors in (5) −dimensional is the same × is cross product of the four vectors in (5) −dimensional. ° ° ° ° ° ... ···· ° ° ... ···· ° ° ˙ ¨ ° ° ˙ ¨ ° °β(t) ∧ β(t) ∧ β(t) ∧β (t)° °β(t) × β(t) × β(t) ×β (t)° k3 = = = 0 ∗ 10 3 2 ∗ 10 3 2 kT k k1 (t) k2 (t) kT k k1 (t) k2 (t)

References

[1] O. Alessio.´ Differential geometry of intersection curves in R4 of three implicit surfaces , Computer Aided Geometric Design, Volume 26, Issue 4, Pages 455-471 (May 2009) .

[2] O. Alessio.´ Geometria Diferencial de Curvas de Intersecc¸ao˜ de Duas Superf´ıcies Impl´ıcitas, TEMA Vol.7, No.2, 169-178, 2006.

[3] R. Goldman. Curvatura formulas for implicit curves and surfaces, Computer Aided Geometric De- sign, Volume 22, Issue 7, Pages 632-658 (October 2005) .

[4] E. Hartmann. G2 Interpolation and blending on surfaces, The Visual Computer Volume 12, Number 4, (April, 1996 ).

[5] T. J. Willmore. An Introduction to Differential Geometry, Clarendon Press,Oxford, 1959.

[6] X. Ye and T. Maekawa. Differential geometry of intersection curves of two surfaces, Computer Aided Geometric Design,Volume 16, Issue 8, Pages 767-788 (September 1999) .

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