A New Method for Finding the Shape Operator of a Hypersurface in Euclidean 4-Space

Home , Darboux frame

Filomat 32:17 (2018), 5827–5836 Published by Faculty of Sciences and Mathematics, https://doi.org/10.2298/FIL1817827U University of Nis,ˇ Serbia Available at: http://www.pmf.ni.ac.rs/filomat

Bahar Uyar D ¨uld¨ula

aYıldız Technical University, Education Faculty, Department of Mathematics and Science Education, Istanbul,˙ Turkey

Abstract. In this paper, by taking a Frenet curve lying on a parametric or implicit hypersurface and using the extended Darboux frame ﬁeld of this curve, we give a new method for calculating the shape operator’s matrix of the hypersurface depending on the extended Darboux frame curvatures. This new method enables us to obtain the Gaussian and mean curvatures of the hypersurface depending on the geodesic torsions of the curve and the normal curvatures of the hypersurface.

1. Introduction

In differential geometry, the shape of a geometric object (curve, surface, etc.) has always been a point of interest. In Euclidean space, along a space curve we have only one direction to move. This direction is determined by the tangent vector of the curve. The curvature of the curve measures the rate of change of its tangent vector, and together with the torsion of the curve they give us some information to its shape. However, along a surface we have different directions to move. These directions constitute a plane at a point, called the tangent plane of the surface at that point. When we move along a given direction, the tangent plane and thereby the normal vector of this plane, that is, the unit normal vector of the surface will change its direction except for some special cases. It is well-known that the rate of change of the unit normal vector of a surface has been measured by the shape operator. That’s why computing the shape operator of a surface is important in surface theory. The calculation of the shape operator of a surface in Euclidean 3-space is well-known [3, 4]. The shape operator’s matrix of a surface can be expressed by depending on the first and second fundamental form coefficients of the surface by using the Weingarten equations [3].

In this paper, we give a method which enables us to compute the shape operator of a hypersurface in Euclidean 4-space. We consider a Frenet curve lying on a hypersurface, and use the extended Darboux frame of the Frenet curve in Euclidean 4-space while constructing this new method. Our method includes two cases in which one can obtain directly the shape operator’s matrix of a hypersurface along a Frenet curve. The elements of obtained new matrix are composed of the geodesic torsions of ﬁrst and second order of the curve and the normal curvatures of the hypersurface.

2010 Mathematics Subject Classification. Primary 53A07; Secondary 53A04, 53A55 Keywords. Shape operator; hypersurface; Darboux frame Received: 15 February 2018; Accepted: 20 November 2018 Communicated by Ljubica Velimirovic´ Email address: [email protected] (Bahar Uyar Duld¨ ul)¨ B. Uyar Düldül / Filomat 32:17 (2018), 5827–5836 5828

2. Preliminaries Definition 2.1. Let be an orientable hypersurface with the unit normal vector field N. The shape operator M N 4 : χ( ) χ( ) of is defined by (X) = X for X χ( ), where is the Riemannian connection of E , [5].S M → M M S −D ∈ M D

4 Definition 2.2. Let be an orientable hypersurface in E and P be a point of . Then K(P) = det P and 1 M M S H(P) = trace( P) are called the Gaussian and the mean curvatures of at P, respectively, [5]. 3 S M 4 Definition 2.3. Let be an orientable hypersurface in E , P be a point of , and XP TP( ). The normal M M ∈ M (XP), XP curvature at P of in the direction of XP is defined by κn(XP) = hS i,[3]. M XP, XP h i 4 4 P Definition 2.4. Let e1, e2, e3, e4 be the standard basis of R . The ternary product of the vectors u = uiei, { } i=1 P4 P4 v = viei, and w = wiei is defined by the vector [7] i=1 i=1

e1 e2 e3 e4

u1 u2 u3 u4 u v w = , ⊗ ⊗ v1 v2 v3 v4

w1 w2 w3 w4 and satisﬁes the following properties:

u v w

x y (u v w) = u, y v, y w, y , ⊗ ⊗ ⊗ ⊗ h i h i h i u, x v, x w, x h i h i h i x, u x, v x, w D E h i h i h i x y z, u v w = y, u y, v y, w . ⊗ ⊗ ⊗ ⊗ h i h i h i z, u z, v z, w h i h i h i 2.1. Curves on a parametric hypersurface in E4 Let be a regular hypersurface given by its parametric equation R = R(u , u , u ) and β : I R M 1 2 3 ⊂ → M be a Frenet curve with arc-length parametrization. Since is regular, R1 R2 R3 , 0 at every point of ∂R M ⊗ ⊗ , where Ri = . Then, the unit normal vector ﬁeld of is given by M ∂ui M R R R N = 1 ⊗ 2 ⊗ 3 . (1) R R R || 1 ⊗ 2 ⊗ 3|| Moreover, since the curve β(s) lies on , we may write β(s) = R u (s), u (s), u (s) . Then we have M 1 2 3 X3 β0(s) = Riui0, (2) i=1

X3 X3 β00(s) = Riui00 + Rijui0u0j, (3) i=1 i,j=1 X3 X3 X3 β000(s) = Riui000 + 3 Rijui00u0j + Rijkui0u0juk0 , i=1 i,j=1 i,j,k=1

∂2R ∂3R where Rij = , and Rijk = , 1 i, j, k 3. ∂uj∂ui ∂uk∂uj∂ui ≤ ≤ B. Uyar D¨uld¨ul / Filomat 32:17 (2018), 5827–5836 5829

2.2. Curves on an implicit hypersurface in E4 Let be a regular hypersurface given by f (x, y, z, w) = 0, and β(s) = (x(s), y(s), z(s), w(s)) be a Frenet M f curve on . Since is regular, at every point of we have N(s) = ∇ (s) for the unit normal vector ﬁeld M M M f of along β. Besides, we have [1] ||∇ || M

f , β0 = 0, h∇ i t f , β00 = β0H f β0 , (4) h∇ i − t d(H f ) t f , β000 = 3β0H f β00 β0 β0 , h∇ i − − ds where β0 = [x0 y0 z0 w0], β00 = [x00 y00 z00 w00], β000 = [x000 y000 z000 w000], f = [ fx fy fz fw], and ∇    fxx fxy fxz fxw    d H f h i  fyx fyy fyz fyw  ∂H f t ∂H f t H f =   , = β0 β0 ,  fzx fzy fzz fzw  ds ∂x ∂w   ··· fwx fwy fwz fww      fxxx fxyx fxzx fxwx   fxxw fxyw fxzw fxww  ∂H   ∂H   f  fyxx fyyx fyzx fywx  f  fyxw fyyw fyzw fyww  =   , , =   . ∂x  fzxx fzyx fzzx fzwx  ··· ∂w  fzxw fzyw fzzw fzww      fwxx fwyx fwzx fwwx fwxw fwyw fwzw fwww

2.3. The extended Darboux frame(ED-frame) field in E4 Let E4 be an orientable hypersurface and β : I R be a Frenet curve in Euclidean 4-space E4. LetMT ⊂and N denote the unit tangent vector field of⊂ the→ curve M β and the unit normal vector field of restricted to the curve β, respectively. Then the extended Darboux frame (ED-frame) field along β is denotedM by T, E, D, N [2], where { }

β00 β00, N N E = − h i , if N, T, β00 is linearly independent (Case 1), (5) β β , N N { } || 00 − h 00 i ||

β000 β000, N N β000, T T E = − h i − h i , if N, T, β00 is linearly dependent (Case 2) β β , N N β , T T { } || 000 − h 000 i − h 000 i || and

D = N T E, in both cases. ⊗ ⊗ The diﬀerential equations for ED-frame ﬁeld have the form

   1    T0  0 κ1 0 κn  T    1 2 1     E   κ1 0 κ1 τ1   E   0  =     (Case 1),  D   − 2 2   D   0   0 κ1 0 τ1       − 1 2    N0 κn τ τ 0 N − − 1 − 1       T0  0 0 0 κn  T    2 1     E   0 0 κ1 τ1   E   0  =     (Case 2),  D   2   D   0   0 κ1 0 0       − 1    N0 κn τ 0 0 N − − 1 i i where κ1 and τ1 are the geodesic curvature and the geodesic torsion of order i,(i = 1, 2), respectively [2]. B. Uyar D¨uld¨ul / Filomat 32:17 (2018), 5827–5836 5830

3. Shape Operator’s Matrix of a Hypersurface

Let us consider an orientable hypersurface in 4-dimensional Euclidean space and take a Frenet curve β(s) lying on this hypersurface. Let us denote theM extended Darboux frame field along β with T, E, D, N . It is obvious from the definition that T, E, D constitutes a basis for χ( ) along β. Our aim is now{ to obtain} the matrix of the shape operator along{ the} curve and to give the GaussianM and the mean curvatures of the hypersurface depending on the geodesic torsions of first and second order of the curve and the normal curvatures in the directions of E, D and E + D of the hypersurface for each case.

3.1. Case 1 Proposition 3.1. Let be an orientable hypersurface, β be a Frenet curve lying on , and T, E, D, N denotes the ED-frame ﬁeld of β. Then,M the shape operator’s matrix of along β with respect to theM basis {T, E, D is} obtained by M { }  κ (T) τ1 τ2   n 1 1   1  =  τ1 κn(E) Ψ  , (6) S  2  τ1 Ψ κn(D) where 1 Ψ = κn(E + D) κn(E) + κn(D) . (7) − 2 Proof. Let   (T) = a T + a E + a D  S 11 12 13  (E) = a T + a E + a D . (8)  21 22 23  S(D) = a T + a E + a D S 31 32 33 Then, we obtain

a = (T), T = κn(T), a = (E), E = κn(E), a = (D), D = κn(D), 11 hS i 22 hS i 33 hS i 1 2 a = a = (T), E = E0, N = τ , a = a = (T), D = D0, N = τ . 12 21 hS i h i 1 13 31 hS i h i 1 On the other hand, we may write

(E + D), E + D = (E), E + (D), D + 2 (E), D hS i hS i hS i hS i i.e.

2κn(E + D) = κn(E) + κn(D) + 2 (E), D . hS i Thus, we have

1 a = a = (E), D = κn(E + D) κn(E) + κn(D) =: Ψ. 23 32 hS i − 2 Substituting the obtained results into (8), we find the shape operator’s matrix as desired. S 1 2 The computations of κn(T), τ1 and τ1 are known for not only parametric hypersurfaces [6] but also for implicit hypersurfaces [2] in Euclidean 4-space. To make the matrix of the shape operator in the Proposition 3.1 understandable, we now show how the normal curvatures in the directions of E, D and E + D can be obtained for both parametric and implicit hypersurfaces. B. Uyar Düldül / Filomat 32:17 (2018), 5827–5836 5831

Lemma 3.2. Let be an orientable hypersurface given by R(u1, u2, u3) and β(s) = R(u1(s), u2(s), u3(s)) be a Frenet curve lying on M. Denoting the ED-frame field of β by T, E, D, N , the normal curvature of in the direction of T is (also in CaseM 2) given by [6] { } M X3 κn(T) = hijui0u0j, i,j=1 where hij = Rij, N , (1 i, j 3), denotes the second fundamental form coefficients of . h i ≤ ≤ M Lemma 3.3. Let be an orientable hypersurface given by R(u1, u2, u3), and β(s) = R(u1(s), u2(s), u3(s)) be a Frenet curve with its EDM-frame field T, E, D, N . Then, the normal curvature of in the direction of E is found as { } M X3 κn(E) = hijaiaj, (9) i,j=1 where

1 n 2 o ai = (1jj1kk 1 ) Ri, E + (1ik1jk 1ij1kk) Rj, E + (1ij1jk 1ik1jj) Rk, E , (10) ∆2 − jk h i − h i − h i i,j,k=1,2,3 (cyclic), ∆2 = R R R 2 = 1 1 1 + 21 1 1 1 12 1 12 1 12 , (11) k 1 ⊗ 2 ⊗ 3k 11 22 33 12 13 23 − 11 23 − 22 13 − 33 12 3 3 1 n X X o Rm, E = 1imu00 + Rij, Rm u0u0 , m = 1, 2, 3, h i µ i h i i j i=1 i,j=1

3 3 3 3 1 n X X X X 2o 2 µ = 1iju00u00 + 2 Rij, Rk u0u0u00 + Rij, Rk u0u0u0 u0 hiju0u0 , i j h i i j k h ì i j k ` − i j i,j=1 i,j,k=1 i,j,k,`=1 i,j=1 and 1ij = Ri, Rj , (1 i, j 3), denotes the first fundamental form coefficients of . h i ≤ ≤ M Proof. We may write X3 E = Riai. (12) i=1

Taking the scalar product both hand sides of (12) with Rj, j = 1, 2, 3, we obtain the linear equations P3 Rj, E = 1ijai. These equations constitute a linear system which has nonzero coeﬃcients determinant h i i=1 2 2 ∆ = R R R . By solving this system, we ﬁnd the scalars ai as k 1 ⊗ 2 ⊗ 3k 1 n 2 o ai = (1jj1kk 1 ) Ri, E + (1ik1jk 1ij1kk) Rj, E + (1ij1jk 1ik1jj) Rk, E , ∆2 − jk h i − h i − h i i,j,k=1,2,3 (cyclic). If we use (5) and (3), we get

3 3 1 n X X o Rm, E = 1imu00 + Rij, Rm u0u0 , 1 m 3, h i µ i h i i j ≤ ≤ i=1 i,j=1 where 2 1 µ = β00 β00, N N = β00, β00 β00, N 2 nk 3 − h i k {h3 i − h i } 3 3 2o 1 P 1 P P P 2 = ijui00u00j + 2 Rij, Rk ui0u0juk00 + Rij, Rk` ui0u0juk0 u`0 hijui0u0j . i,j=1 i,j,k=1h i i,j,k,`=1h i − i,j=1

Since κn(E) = (E), E , using Lemma 3.2, the normal curvature of in the direction of E is obtained as desired. hS i M B. Uyar D¨uld¨ul / Filomat 32:17 (2018), 5827–5836 5832

Lemma 3.4. Let be an orientable hypersurface given by R(u1, u2, u3) and β(s) = R(u1(s), u2(s), u3(s)) be a Frenet curve lying on M. Denoting the ED-frame ﬁeld of β by T, E, D, N , the normal curvature of in the direction of D is obtained by M { } M

X3 κn(D) = hijAiAj, (13) i,j=1 where    3 3  1  X X  Ai = Rj, E 1kmu0 Rk, E 1jmu0 , i, j, k = 1, 2, 3(cyclic). (14) ∆ h i m − h i m  m=1 m=1 

P3 Proof. Let D = RiAi. Since D = N T E, using (1) we get i=1 ⊗ ⊗

3 1 X n o D = Rj, E Rk, T Rj, T Rk, E Ri, i, j, k = 1, 2, 3(cyclic). ∆ h ih i − h ih i i=1

Then, using (2), we ﬁnd Ai as given in (14). So, Lemma 3.2 yields the desired normal curvature.

Lemma 3.5. Let be an orientable hypersurface given by R(u1, u2, u3), and β(s) = R(u1(s), u2(s), u3(s)) be a Frenet curve with the EDM-frame ﬁeld T, E, D, N lying on . Then, the normal curvature of in the direction of E + D is given by { } M M

3 1 X κ (E + D) = h B B , (15) n 2 ij i j i,j=1 where Bi = ai + Ai.

P3 P3 Proof. We may write E + D = Ri(ai + Ai) = RiBi. Then it follows from i=1 i=1 1 κn(E + D) = (E + D), E + D . 2hS i

Lemma 3.6. Let : f (x, y, z, w) = 0 be an orientable hypersurface and β(s) be a Frenet curve lying on . Denoting the ED-frame ﬁeldM of β by T, E, D, N , the normal curvature of in the direction of T is given by [2] M { } M

1 t κ (T) = − β0H (β0) , n f f ||∇ || where H f denotes the Hessian of f .

Lemma 3.7. Let : f (x, y, z, w) = 0 be an orientable hypersurface, β(s) be a Frenet curve lying on and T, E, D, N denotesM the ED-frame ﬁeld of β. Then the normal curvature of in the direction of E is obtainedM by { } M

1 n t t t 2 to κn(E) = − β00H f (β00) + λβ00H f ( f ) + λ f H f (β00) + λ f H f ( f ) , δ2 f ∇ ∇ ∇ ∇ ||∇ || where

1 t 2 t 2 2 λ = β0H f (β0) , δ = β00(β00) λ f . f 2 − ||∇ || ||∇ || B. Uyar D¨uld¨ul / Filomat 32:17 (2018), 5827–5836 5833

f ∇ Proof. If we substitute N = f into (5), and use (4), we ﬁnd ||∇ || 1 E = (β00 + λ f ), (16) δ ∇ where

1 1 t λ = β00, f = β0H f (β0) , − f 2 h ∇ i f 2 ||∇ || ||∇ ||

2 2 t 2 2 δ = β00, β00 β00, N = β00(β00) λ f . h i − h i − ||∇ || Then, the desired result is obtained from Lemma 3.6.

Lemma 3.8. Let : f (x, y, z, w) = 0 be an orientable hypersurface and T, E, D, N denotes the ED-frame ﬁeld of M { } the Frenet curve β(s) = (β1(s), β2(s), β3(s), β4(s)) lying on . Then, the normal curvature of in the direction of D is found as M M

1 t κn(D) = − ΦH f Φ , δ2 f 3 ||∇ || where Φ = [φ φ φ φ ], f = ( f , f , f , f ), and 1 2 3 4 ∇ 1 2 3 4

j k ` φi = ( 1) fj β0 β00 β00β0 + ( 1) fk β0β00 β00β0 + ( 1) f β0β00 β00β0 , − k ` − k ` − j ` − j ` − ` j k − j k i, j, k, ` = 1, 2, 3, 4(cyclic).

f 1 Proof. Substituting N = ∇ and E = (β00 + λ f ) into D = N T E gives f δ ∇ ⊗ ⊗ ||∇ ||

e1 e2 e3 e4

1 1 f1 f2 f3 f4 1 D = f T β00 = = Φ, (17) δ f ∇ ⊗ ⊗ δ f β10 β20 β30 β40 δ f ||∇ || ||∇ || ||∇ || β100 β200 β300 β400 where Φ = [φ1 φ2 φ3 φ4]. Then, the normal curvature of in the direction of D follows from Lemma 3.6. M

Lemma 3.9. Let : f (x, y, z, w) = 0 be an orientable hypersurface, β(s) be a Frenet curve lying on and T, E, D, N denotesM the ED-frame ﬁeld of β. Then the normal curvature of in the direction of E + D is obtainedM by { } M

1 t κn(E + D) = − ΩH f Ω , 2δ2 f ||∇ || where

1 Ω = β00 + λ f + Φ. ∇ f ||∇ || Proof. The result can be seen by using (16) and (17). B. Uyar D¨uld¨ul / Filomat 32:17 (2018), 5827–5836 5834

3.2. Case 2 The proofs of the following results can be done similar to the proofs given in Case 1.

Proposition 3.10. Let be an orientable hypersurface, β be a Frenet curve lying on , and T, E, D, N denotes the ED-frame ﬁeld of β.M Then, the shape operator’s matrix of along β with respect to theM basis {T, E, D is} found as M { }   κ (T) τ1 0  n 1   1  =  τ1 κn(E) Ψ  , (18) S   0 Ψ κn(D)

1 where Ψ = κn(E + D) κn(E) + κn(D) . − 2

Lemma 3.11. Let be an orientable hypersurface given by R(u1, u2, u3), and β(s) = R(u1(s), u2(s), u3(s)) be a Frenet curve with its ED-frameM ﬁeld T, E, D, N . Then, the normal curvature of in the direction of E is given by { } M X3 κn(E) = hijaiaj, i,j=1

2 where ai and ∆ are as given in (10) and (11),

n 3 3 3 E 1 P 1 P P Rn, = ν inui000 + 3 Rij, Rn ui00u0j + Rijk, Rn ui0u0juk0 h i i=1 i,j=1h i i,j,k=1h i 3 3 3 3 o (19) P 1 P 1 P P inui0 ijui000u0j + 3 Rij, Rk ui00u0juk0 + Rijk, R` ui0u0juk0 u`0 , − i=1 i,j=1 i,j,k=1h i i,j,k,`=1h i n = 1, 2, 3, and

2 2 1 ν = β000 β000, N N β000, T T = β000, β000 β000, N β000, T 2 n|| 3 − h i − h 3 i || {h i − h 3 i − h i } P 1 P P = ijui000u000j + 6 Rij, Rk ui000u00j uk0 + 9 Rij, Rk` ui00u00j uk0 u`0 i,j=1 i,j,k=1h i i,j,k,`=1h i

X3 X3 +2 Rijk, R u000u0u0 u0 + 6 Rijk, R m u00u0u0 u0 u0 h `i i j k ` h ` i i j k ` m i,j,k,`=1 i,j,k,`,m=1

3 3 3 X X X 2 + Rijk, R mn u0u0u0 u0 u0 u0 3 hiju00u0 + Rijk, N u0u0u0 h ` i i j k ` m n − i j h i i j k i,j,k,`,m,n=1 i,j=1 i,j,k=1

3 3 3 1 X X X 2o 2 1iju000u0 + 3 Rij, Rk u00u0u0 + Rijk, R u0u0u0 u0 . − i j h i i j k h `i i j k ` i,j=1 i,j,k=1 i,j,k,`=1

Lemma 3.12. Let be an orientable hypersurface given by R(u1, u2, u3) and β(s) = R(u1(s), u2(s), u3(s)) be a Frenet curve lying on .MT, E, D, N denotes ED-frame ﬁeld of β, then the normal curvature of in the direction of D is obtained by M { } M

X3 κn(D) = hijAiAj, i,j=1 where Ai is as given in (14), and Rj, E is as given in (19). h i B. Uyar D¨uld¨ul / Filomat 32:17 (2018), 5827–5836 5835

Lemma 3.13. Let be an orientable hypersurface given by R(u1, u2, u3), and β(s) = R(u1(s), u2(s), u3(s)) be a Frenet curve lying on .M Let T, E, D, N denotes the ED-frame ﬁeld of β. Then, the normal curvature of in the direction of E + D is foundM as { } M

3 1 X κ (E + D) = h B B , n 2 ij i j i,j=1 where Bi = ai + Ai. Lemma 3.14. Let : f (x, y, z, w) = 0 be an orientable hypersurface, β(s) be a Frenet curve lying on and T, E, D, N denotesM the ED-frame ﬁeld of β. Then the normal curvature of in the direction of E is given byM { } M 1 n t t t t κn(E) = − β000H f (β000) + ζ β000H f ( f ) + ζ β000H f (β0) + ζ f H f (β000) 2 f 1 ∇ 2 1∇ ||∇ || 2 t t t t 2 to +ζ f H f ( f ) + ζ ζ f H f (β0) + ζ β0H f (β000) + ζ ζ β0H f ( f ) + ζ β0H f (β0) , 1∇ ∇ 1 2∇ 2 1 2 ∇ 2 where

1 t d(H f ) t ζ = 3β0H (β00) + β0 (β0) , 1 f 2 f ds ||∇ || t ζ = β0(β000) , 2 − 2 t 2 2 2 = β000(β000) ζ f ζ . − 1||∇ || − 2 Lemma 3.15. Let : f (x, y, z, w) = 0 be an orientable hypersurface and T, E, D, N denotes the ED-frame ﬁeld of M { } the Frenet curve β(s) = (β1(s), β2(s), β3(s), β4(s)) lying on . Then the normal curvature of in the direction of D is found as M M

1 t κn(D) = − ΓH f Γ , 2 f 3 ||∇ || j k ` where Γ = [γ γ γ γ ] and γi = ( 1) fj β0 β000 β000β0 + ( 1) fk β0β000 β000β0 + ( 1) f β0β000 β000β0 , 1 2 3 4 − k ` − k ` − j ` − j ` − ` j k − j k i, j, k, ` = 1, 2, 3, 4(cyclic). Lemma 3.16. Let : f (x, y, z, w) = 0 be an orientable hypersurface, β(s) be a Frenet curve lying on and T, E, D, N denotesM the ED-frame ﬁeld of β. Then the normal curvature of in the direction of E + D is obtainedM by { } M 1 t κn(E + D) = − ΛH f Λ , 22 f ||∇ || 1 where Λ = β000 + ζ1 f + ζ2β0 + f Γ. ∇ ||∇ ||

If we use the matrices for the shape operator in each case, we may give the following corollaries: Corollary 3.17. Let be an orientable hypersurface, β be a Frenet curve lying on , and T, E, D, N denotes the ED-frame ﬁeld of β. Then,M the Gaussian curvature and the mean curvature of the hypersurfaceM { along β }can be given by

2 2 2 1 2 1 2 Case 1: K s = κn(T)κn(E)κn(D) Ψ κn(T) (τ ) κn(E) (τ ) κn(D) + 2τ τ Ψ, β( ) − − 1 − 1 1 1 2 1 2 Case 2: K s = κn(T)κn(E)κn(D) Ψ κn(T) (τ ) κn(D), β( ) − − 1 and in both cases 1 H = κ (T) + κ (E) + κ (D) . β(s) 3 n n n B. Uyar D¨uld¨ul / Filomat 32:17 (2018), 5827–5836 5836

Corollary 3.18. Let be an orientable hypersurface, β be a line of curvature on . Then T corresponds to the M M principal direction, i.e. (T) = k T. If the principal curvature k , 0 and E, D correspond to principal directions, S 1 1 then we have Ψ = 0 in each case. In this case, κn(E + D) corresponds to the arithmetic mean of the normal curvatures κn(E) and κn(D).

4. Examples In this part, we calculate the Gaussian curvatures and the mean curvatures of the hypersurfaces depending on the ED-frame curvatures by ﬁnding the matrices of the shape operators of hypersurfaces given by parametric or implicit equations.

Example 4.1. Let us consider the hypercylinder given by its parametric equation R(u1, u2, u3) = (cos u1 cos u2, Ms s s sin u1 cos u2, sin u2, u3) and the curve β(s) = R , , cos lying on . Since Case 1 is valid along β, at √2 √2 √2 M 1 2 the point β(0) = (1, 0, 0, 1) we obtain κn(T) = 1, τ = τ = 0 [6]. We also ﬁnd ∆ = 1, R , E = R , E = 0, − 1 1 h 1 i h 2 i R3, E = 1. Substituting these values into (10) gives a1 = a2 = 0, a3 = 1. So, from (9) we obtain κn(E) = 0. h i − 1 − If we use (14), we get A1 = A2 = , A3 = 0. Then, from (13) we have κn(D) = 1. Besides, we ﬁnd − √2 − 1 1 B1 = B2 = , B3 = 1 and from (15), we get κn(E + D) = . So, from (7) we obtain Ψ = 0. − √2 − − 2 Finally, the shape operator’s matrix of at β(0) = (1, 0, 0, 1) is obtained from (6) as M  1 0 0     −0 0 0  β(0) =   S  0 0 1  − which yields the Gaussian curvature and mean curvature of as K = 0 and H = 2 , respectively. M (β(0)) (β(0)) − 3 √3 s 1 s s √3 Example 4.2. Let us consider the unit speed curve α(s) = cos , cos , sin , s lying on the 2 2 2 2 2 2 hypercylinder given by its implicit equation x2 + y2 + z2 = 1. Since Case 2 is valid along α, at the point M √3 1 α(0) = , , 0, 0 we have 2 2

1 √3 1 3 7 τ = , κn(T) = , κn(E) = , κn(D) = 1, κn(E + D) = , Ψ = 0. 1 − 4 −4 −4 − −8 The shape operator’s matrix is then follows from (18) as

 1 √3   4 4 0   −√ −  α(0) =  3 3 0  . S  4 4   −0− 0 1  − We again have 2 K = 0, H = . (α(0)) (α(0)) −3

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