Transversal Intersection of Hypersurfaces in R^ 5

arXiv:1601.04252v1 [math.DG] 17 Jan 2016 Keywords: dependent. linearly norma are the point intersection nontransversal in t as at while surfaces pendent, the of normals the intersection transversal In [email protected] ercadipii ufcs o htrao ifrn met different inte reason surface g that implicit-implicit in For or used parametric-parametric most surfaces. surfaces implicit of types and representation two metric the The in animations. useful is and It jects system. 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Patrikalakis, 1988; al., et aj c R c,apoce fspro order superior of approaches ncy, 7) oCro(96,differen- (1976), Carmo do 975), oamdHsnShahid Hasan Mohammad , oshv engvnfreither for given been have hods 3 cpoete fteintersection the of properties ic ln Frn 02 oce and Hoschek 2002; (Farin, lling feeta emti properties geometric ifferential R a efudi etok such textbooks in found be can ftedsg fcmlxob- complex of design the of oercdsgigaepara- are designing eometric rcs eddi modeling in needed process 5 h nescincre in curves intersection the intersection. l 204 India 121004, 1005 India 025, -110 Mhma Jamali), (Mohammad nso nintersection an of ints R 3 h numerical The . inter- l d, ∗ R R 5 3 . Lasser 1993), but there is only a scarce of literature on the differential geometry of intersection curves. Willmore (1959) and Aléssio (2006) presented algorithms to obtain the unit tangent, unit principle normal, unit binormal, curvature and torsion of the transversal intersection curve of two implicit surfaces. Hartmann (1996) provided formulas for computing the curvature of the intersection curves for all types of intersection problems in R3. Ye and Maekawa (1999) presented algorithms for computing the differential geometric properties of both transversal and tangential intersection curves of two surfaces. Aléssio (2009) formulated the algorithms for obtaining the geometric properties of intersection curves of three implicit hypersurfaces in R4. Based on the work of Aléssio (2009), Mustufa Düldül (2010) worked with three parametric hypersurfaces in R4 to derive the algorithms for differential geometric properties of transversal intersection. Abdel-All et al. (2012) provided algorithms for geometric properties of implicit- implict-parametric and implicit-parametric-parametric hypersurfaces in R4. Naeim-Badr et al. (2014) obtained algorithms for differential geometric properties of non-transversal intersection curves of three parametric hypersurfaces in R4. Recently Naeim Badr, Abdel-All et al. (2015) derived the algorithms for non-transversal intersection curves of implicit-parametric- parametric and implicit-implicit-parametric hypersurfacres in R4. To obtain the first geodesic κSi τSi curvature ( 1g) and the first geodesic torsion ( 1g) for the transversal intersection curve of 4 parametric hypersurfaces in R5, we need to derive the Darboux frame U Mi , ,U Mi . The 1 ··· 5 Darboux frame is obtained by using the Gram-Schmidt orthogonalizationn process. o In this paper we extended the methods of Mustufa Düldül[8], to obtain the Frenet frame

t,n,b1,b2,b3 and curvatures κ1,κ2,κ3,κ4 of transversal intersection curve of four para- { } { } metric hypersurfaces in R5. In section 2 we introduce some notations and reviews of the differential geometry of curves and surfaces in R5. In section 3 we ﬁnd the formulas for computing the properties of transversal intersection of four parametric hypersurfaces in R5. In section 4 we derive the formulas for obtaining the geodesic curvature and geodesic torsion of the intersecting curve with respect to four hypersurfaces. Finally, to be more constructive we present an example in section 5. Moreover in addition to the use of classical results of differential geometry we will also make use of Matlab/Mathematica.

2. Preliminaries

Definition 2.1. Let e1,e2,e3,e4,e5 be the standard basis of five dimensional Euclidean space 5 ∑5 ∑5 ∑5 E . The vector product of the vectors x = i=1 xiei, y = i=1 yiei, z = i=1 ziei and w = ∑5 i=1 wiei is defined by

e1 e2 e3 e4 e5 x1 x2 x3 x4 x5

x y z w = y y y y y (1) ⊗ ⊗ ⊗ 1 2 3 4 5

z1 z2 z3 z4 z5

w1 w2 w3 w4 w5

The vector product x y z w yields a vector that is orthogonal to x, y, z, w. ⊗ ⊗ ⊗ 2 5 let R E be a regular hypersurface given by Φ = Φ(u1,u2,u3,u4) and α : I R Φ ba ⊂ ⊂ → an arbitrary curve with arc length parametrisation. If t,n,b1,b2,b3 is the Frenet Frame along α

t′ = κ1n

n′ = κ1t + κ2b1 − b′ = κ2n + κ3b2 (2) 1 − b′ = κ3b1 + κ4b3 2 − b′ = κ4b2 3 −

Where t, n, b1, b2 and b3 denote the tangent, the principal normal, the ﬁrst binormal, the second binormal and third binormal vector ﬁelds. The normal vector n is the normalised acceleration

vector α ′′ . The unit vector b1 is determined such that n′ can be decomposed into two components, a tangent one in the direction of t and a normal one in the direction of b1. The unit vector ′ b2 is determined such that b1 can be decomposed into two components, a normal one and b2. The unit vector b3 is the unique unit vector field perpendicular to four dimensional subspace t,n,b1,b2 . The functions κ1, κ2, κ3 and κ4 are the first, second, third and fourth curvatures { } of α(s). The first, second, third and fourth curvatures measures how rapidly the curve pulls away in a neighbourhood of s, from the tangent line, from planar curve, from three dimensional curve and from the four dimensional curve at s, respectively. Now, using the Frenet Frame we have the derivatives of α as

2 α ′ = t, α ′′ = t′ = κ1n, α ′′′ = κ t + κ ′ n + κ1κ2b1 (3) − 1 1 (4) 3 2 α = 3κ1κ ′ t +( κ + κ ′′ κ1κ )n +(2κ ′ κ2 + κ1κ ′ )b1 + κ1κ2κ3b2 (4) − 1 − 1 1 − 2 1 2 (5) 2 4 2 2 2 2 α = ( 3(κ ′ ) 4κ1k′′ + κ + κ κ )t +( 6κ κ ′ + κ ′′′ κ ′ κ − 1 − 1 1 1 2 − 1 1 1 − 1 2 2 3 3 2 3κ1κ2κ ′ 2κ ′ κ )n +(κ κ2 κ1κ + 3κ ′′ κ2 + 3κ ′ κ ′ + κ1κ ′′ κ1κ2κ )b1 − 2 − 1 22 1 − 2 1 1 2 2 − 3 κ ′ κ κ κ κ ′ κ κ κ ′ κ κ κ κ ′ κ κ κ κ (3 1 2 3 + 2 1 2 3 + 2 1 2 3 + 1 2 3)b2 + 1 2 3 4b3 (5)

∂Φ Also since Φ is regular, the partial derivatives Φ1, Φ2, Φ3, Φ4, where (Φi = ) are linearly ∂ui independent at every point of Φ , i.e., Φ1 Φ2 Φ3 Φ4 = 0. Thus, the unit normal vector of ⊗ ⊗ ⊗ 6 Φ is given by Φ1 Φ2 Φ3 Φ4 N = ⊗ ⊗ ⊗ . Φ1 Φ2 Φ3 Φ4 k ⊗ ⊗ ⊗ k Furthermore, the ﬁrst, second, and the third binormal vectors of the curve are given by

α ′ α ′′ α ′′′ α(4) b α ′ α ′′ α ′′′ b b α ′ α ′′ b = , b = 3 , b = 2 3 (6) 3 ⊗ ⊗ ⊗ (4) 2 ⊗ ⊗ ⊗ 1 ⊗ ⊗ ⊗ α ′ α ′′ α ′′′ α b3 α ′ α ′′ α ′′′ b2 b3 α ′ α ′′ k ⊗ ⊗ ⊗ k k ⊗ ⊗ ⊗ k k ⊗ ⊗ ⊗ k and the curvatures are obtained with

(4) (5) α ′′′ ,b1 α ,b2 α ,b3 κ1 = α ′′ , κ2 = h i, κ3 = h i, κ4 = h i (7) k k κ1 κ1κ2 κ1κ2κ3

3 On the other hand, since the curve α(s) lies on Φ, we may write

α(s) = Φ(u1(s),u2(s),u3(s),u4(s)). We then have 4 α ′ Φ ′ (s) = ∑ iui (8) i=1 4 4 α ′′ Φ ′′ Φ ′ ′ (s) = ∑ iui + ∑ ijuiu j (9) i=1 i, j=1 4 4 4 α ′′′ Φ ′′′ Φ ′′ ′ Φ ′ ′ ′ (s) = ∑ iui + 3 ∑ ijui u j + ∑ i jkuiu juk (10) i=1 i, j=1 i, j,k=1

4 4 4 4 4 α(4) Φ ( ) Φ ′′′ ′ Φ ′′ ′′ Φ ′′ ′ ′ (s) = ∑ iui + 4 ∑ ijui u j + 3 ∑ ijui u j + 6 ∑ i jkui u juk + i=1 i, j=1 i, j=1 i, j,k=1 4 Φ ′ ′ ′ ′ ∑ ijkluiu jukul (11) i, j,k,l=1

4 4 4 4 5 4 α(5) Φ ( ) Φ ( ) ′ Φ ′′′ ′′ Φ ′′′ ′ ′ = ∑ iui + 5 ∑ ijui u j + 10 ∑ ijui u j + 10 ∑ i jkui u juk i=1 i, j=1 i, j=1 i, j,k=1 4 4 Φ ′′ ′′ ′ Φ ′′ ′ ′ ′ +15 ∑ i jkui u juk + 10 ∑ ijkjui u jukul i, j,k=1 i, j,k,l=1 4 Φ ′ ′ ′ ′ ′ + ∑ ijklruiu jukulur (12) i, j,k,l,r=1

3. The curvature of the transversal intersection of hypersurfaces

Let R1, R2, R3 and R4 be four regular transversally intersecting hypersurfaces given by Φi Φi i i i i parametric equations = (u1,u2,u3,u4), (i = 1,2,3,4). Then the unit normal vector of these hypersurfaces are

Φi Φi Φi Φi N = 1 ⊗ 2 ⊗ 3 ⊗ 4 , i = 1,2,3,4 i Φi Φi Φi Φi k 1 ⊗ 2 ⊗ 3 ⊗ 4k Since the intersection is transversal, the normal vectors of these hypersurfaces at the inter-

section points are linearly independent, i.e., N1 N2 N3 N4 = 0. It is assumed that the ⊗ ⊗ ⊗ 6 intersection is a smooth curve say α, in E5. Let the intersection curve α be parameterised by

arc lenght function s. Then, at the intersection point α(s0) = P, the unit tangent vector t of the intersection curve α can be found by the vector product of the normal vectors at P.

N1 N2 N3 N4 t = ⊗ ⊗ ⊗ (13) N1 N2 N3 N4 k ⊗ ⊗ ⊗ k 4 3.1. First curvature of the transversal intersection curve Now, we ﬁnd the ﬁrst curvature of the intersection curve at P. Since t′ is orthogonal to t, we may write 4 α ′′ = t′ = ∑ aiNi, ai R,i = 1,2,3,4 (14) i=1 ∈

Thus, we need to calculate the scalars ai to ﬁnd α ′′ . If we take the dot product of both hand sides of (14) with Ni, we have

1 N1,N1 a1 + N2,N1 a2 + N3,N1 a3 + N4,N1 a4 = κ , h i h i h i h i n 2 N1,N2 a1 + N2,N2 a2 + N3,N2 a3 + N4,N2 a4 = κ , h i h i h i h i n (15) 3 N1,N3 a1 + N2,N3 a2 + N3,N3 a3 + N4,N3 a4 = κ , h i h i h i h i n 4 N1,N4 a1 + N2,N4 a2 + N3,N4 a3 + N4,N4 a4 = κ h i h i h i h i n i Where, κ = t′,Ni , (i = 1,2,3,4) and , is the scalar product. Hence we must compute n h i h i κ1 κ2 κ3 κ4 n , n , n , n at P to ﬁnd the scalars ai On using (9), we obtain

κi Πi i ′ 2 Πi i ′ 2 Πi i ′ 2 Πi i ′ 2 Πi i ′ i ′ Πi i ′ i ′ n = 11(u1 ) + 22(u2 ) + 33(u3 ) + 44(u4 ) + 2( 12u1 u2 + 13u1 u3 (16) Πi i ′ i ′ Πi i ′ i ′ Πi i ′ i ′ Πi i ′ i ′ + 14u1 u4 + 23u2 u3 + 24u2 u4 + 34u3 u4 ) Πi Where lm, l,m = 1,2,3,4 are the second fundamental form coefﬁcients of the hypersurfaces Φi. Since the unit tangent is known from (13) and Φi Φi Φi Φi = 0, the scalar k 1 ⊗ 2 ⊗ 3 ⊗ 4k 6 Φi Φi Φi Φi multiplication of both hand sides of (8) with 1, 2, 3 and 4, respectively yields a linear system of four equations as

ϒi ui ′ + ϒi ui ′ + ϒi ui ′ + ϒi ui ′ = t,Φi 11 1 12 2 13 3 14 4 h 1i ϒi i ′ ϒi i ′ ϒi i ′ ϒi i ′ Φi 21u1 + 22u2 + 23u3 + 24u4 = t, 2 h i (17) ϒi ui ′ + ϒi ui ′ + ϒi ui ′ + ϒi ui ′ = t,Φi 31 1 32 2 33 3 34 4 h 3i ϒi ui ′ + ϒi ui ′ + ϒi ui ′ + ϒi ui ′ = t,Φi 41 1 42 2 43 3 44 4 h 4i

′ ′ ′ ′ ϒi with respect to u1, u2, u3 and u4 where lm, l,m = 1,2,3,4 are the first fundamental form coefficients of the hypersurface Φi. Substituting the solutions of these systems into (16) gives κi us n, i = 1,2,3,4. Then using matlab/mathematica the system of linear equations in (15) can be solved for a1, a2, a3 and a4.Thus the first curvature of the intersection curve at P is obtained from (14) and the first equation of (7). Remark 3.1. If the normal vectors are mutually orthogonal to each other at the intersection κ 2 2 2 2 point, then the first curvature is given by 1 = a1 + a2 + a3 + a4. q 3.2. Second curvature To compute the second curvature we have to find the third derivative of the intersection curve α. 5 Since Ni,i = 1,2,3,4 are orthogonal to t, we may write c1N1 + c2N2 + c3N3 + c4N4 instead of κ ′ κ κ α ′′′ terms 1n + 1 2b1 in , i.e.,

2 α ′′′ = κ t + c1N1 + c2N2 + c3N3 + c4N4. (18) − 1

Taking the dot product of both hand sides of above equation with Ni, we obtain

N1,N1 c1 + N2,N1 c2 + N3,N1 c3 + N4,N1 c4 = µ1 h i h i h i h i N1,N2 c1 + N2,N2 c2 + N3,N2 c3 + N4,N2 c4 = µ2 h i h i h i h i (19) N1,N3 c1 + N2,N3 c2 + N3,N3 c3 + N4,N3 c4 = µ3 h i h i h i h i N1,N4 c1 + N2,N4 c2 + N3,N4 c3 + N4,N4 c4 = µ4 h i h i h i h i

Where µi = α ′′′ ,Ni ,i = 1,2,3,4 and ci R. Now, let us ﬁnd the unknown scalars µi. Using h i ∈ (10), we have

4 4 µ Φr ′′ ′ Φr ′ ′ ′ r = 3 ∑ ij,Nr ui u j + ∑ i jk,Nr uiu juk (20) i, j=1h i i, j,k=1h i

′ ′′ ′′ Since the components ui are known from the system (17), we have to ﬁnd ui . To obtain ui we use (9) and write 4 ∆ Φr ′ ′ r = ∑ ijuiu j, r = 1,2,3,4 (21) i, j=1 Then we have

i i ′′ i i ′′ i i ′′ i i ′′ i ϒ u + ϒ u + ϒ u + ϒ u = t′ ∆i,Φ 11 1 12 2 13 3 14 4 h − 1i ϒi i ′′ ϒi i ′′ ϒi i ′′ ϒi i ′′ ′ ∆ Φi 21u1 + 22u2 + 23u3 + 24u4 = t i, 2 h − i (22) i i ′′ i i ′′ i i ′′ i i ′′ i ϒ u + ϒ u + ϒ u + ϒ u = t′ ∆i,Φ 31 1 32 2 33 3 34 4 h − 3i i i ′′ i i ′′ i i ′′ i i ′′ i ϒ u + ϒ u + ϒ u + ϒ u = t′ ∆i,Φ 41 1 42 2 43 3 44 4 h − 4i

Which gives us required derivatives. Thus from (20), we find the values of µi, which finally helps us to find the value of ci in system (19). Thus the second curvature can be found from the second equation of (7), untill we find b1. On using (7) we obtain κ ′ = α ′′′ ,n . 1 h i 3.3. Third curvature To find the third curvature, we need to find the fourth derivative of the intersection curve α at P. 3 Since Ni is orthogonal to t, we may write d1N1 + d2N2 + d3N3 + d4N4 instead of( κ + κ ′′ − 1 1 − κ κ2 κ ′ κ κ κ ′ κ κ κ α(4) 1 2 )n +(2 1 2 + 1 2)b1 + 1 2 3b2 in , i.e.,

(4) α = 3κ1κ ′ t + d1N1 + d2N2 + d3N3 + d4N4 (23) − 1 6 Taking the dot product of (23) with N1, N2, N3 and N4, we get

N1,N1 d1 + N2,N1 d2 + N3,N1 d3 + N4,N1 d4 = ξ1 h i h i h i h i N1,N2 d1 + N2,N2 d2 + N3,N2 d3 + N4,N2 d4 = ξ2 h i h i h i h i (24) N1,N3 d1 + N2,N3 d2 + N3,N3 d3 + N4,N3 d4 = ξ3 h i h i h i h i N1,N4 d1 + N2,N4 d2 + N3,N4 d3 + N4,N4 d4 = ξ4 h i h i h i h i (4) Where ξi = α ,Ni ,i = 1,2,3,4 and di R h i ∈ (4) Now to ﬁnd di we have to ﬁnd ξi. For that taking the dot product of α with Ni, we obtain

4 4 4 ξ Φr ′′′ ′ Φr ′′ ′′ Φr ′′ ′ ′ r = 4 ∑ ij,Nr ui u j + 3 ∑ ij,Nr ui u j + 6 ∑ i jk,Nr ui u juk + i, j=1h i i, j=1h i i, j,k=1h i 4 Φr ′ ′ ′ ′ ∑ ijkl,Nr uiu jukul (25) i, j,k,l=1h i

′ ′′ ξ ′′′ Since ui, ui are already known, so to find r we have to find ui . These derivatives can be found Φi Φi Φi Φi by taking the product of both hand side of (10) with 1, 2, 3 and 4, respectively.Hence, we can compute the Frenet vectors at P of the intersection curve by finding b3, b2 and b1 - the third, second and first binormal vectors from the equations in (6) as now α ′ , α ′′ , α ′′′ and α(4) are at our disposal. Thus on using the binormal vector b1 and (18), the second curvature of the intersection curve at P is obtained from the second equation of (7). Since κ1, κ2 and b2 are already known, the third curvature can be now found from the third equation of (7).

3.4. Fourth curvature

To obtain the forth curvature κ4, we need to ﬁnd the ﬁfth derivative of the intersection curve of α at P . Similar to third and fourth derivative of the curve α, we may write

(5) 2 (4) 2 2 α = 3(κ ′ ) 4κ1κ ′′ + κ + κ κ t + m1N1 + m2N2 + m3N4 + n4N4 (26) {− 1 − 2 1 1 2 } Where, the system of equations for unknowns is

N1,N1 m1 + N2,N1 m2 + N3,N1 m3 + N4,N1 m4 = η1 h i h i h i h i N1,N2 m1 + N2,N2 m2 + N3,N2 m3 + N4,N2 m4 = η2 h i h i h i h i (27) N1,N3 m1 + N2,N3 m2 + N3,N3 m3 + N4,N3 m4 = η3 h i h i h i h i N1,N4 m1 + N2,N4 m2 + N3,N4 m3 + N4,N4 m4 = η4 h i h i h i h i (5) and ηi = α ,Ni . Projecting (12) onto the unit vector Ni, respectively, we obtain ηi depending 4 h i 4 4 ( ) ′′′ ′′ ′ ( ) ( ) on ui besides ui , u2 and u1. Except ui all are known. So toﬁnd ui taking the scalar product Φi Φi Φi Φi of (11) with 1, 2, 3, 4, respectively. Consequently, the fourth curvature of the intersection curve can be found from the last equation of (7).

7 4. Darboux Frame, First Geodesic Curvature and First, Second and Third Geodesic Tor- sion.

i i i i i In this section, we derive the Darboux frame U1,U2,U3,U4,U5 , the ﬁrst geodesic curva- κi τi , , ture ( 1g) and the ﬁrst, second and third geodesic torsion ( jg), j = 1 2 3 for the transversal intersection curve of 4 parametric hypersurfaces in R5.

5 Deﬁnition 4.1. Let Mi be a regular hypersurface in R and α be a curve on Mi. Then for each i,1 i 4, the function ≤ ≤ κig : I R deﬁned for s I by κ −→ ∈ ig(s) = Ui′(s),Ui+1(s) h th i th is called the i geodesic curvature function of the curve α and κig(s) is called the i geodesic curvature of the curve α at α(s).

4.0.1. Darboux Frame:

We are able to obtain a natural frame for the intersection curve-hypersurface pair (α(s),Mi), i i i i i i.e., the frame U1,U2,U3,U4,U5 by using the Gram-Schmidt orthogonalization process. By (4) assumption the sets α (s), α (s), α (s) and α (s), and N1(p),N2(p),N3(p),N4(p) are ′ ′′ ′′′ { } linearly independent. i α i Fixing U1 = ′(p) and U5 = Ni, we have The natural frame (Darboux frame) U i,U i,U i,U i,U i is obtained, with 1 i 4. 1 2 3 4 5 ≤ ≤ i α U1 = ′ (p) i U5 = Ni i i α (p),Ni Ni α (p),U U +α (p) i −h ′′ i − ′′ 1 1 ′′ U2 = α hα i i i α , ′′(p),Ni Ni ′′(p),U1 U1+ ′′(p) −h i − i i i i α (pk),Ni Ni α (p),Uh U αi (p),U Uk+α (p) i −h ′′′ i − ′′′ 2 2− ′′′ 1 1 ′′′ U3 = α h α ii i h α ii i α , ′′′(p),Ni Ni ′′′(p),U2 U2 ′′′(p),U1 U1+ ′′′(p (4) −h (4)i − i i (4) − i i (4) i i (4) α (p)k,Ni Ni α (p),Uh U α i (p),hU U αi (p),U Uk +α (p) i − − 3 3− 2 2− 1 1 U4 = hα(4) i hα(4) i i i hα(4) i i i hα(4) i i i α(4) . (p),Ni Ni (p),U3 U3 (p),U2 U2 (p),U1 U1+ (p) k−h i −h κi i −h i −h i k The j-th geodesic curvature jg associated with i-th hypersurface is

κi = U i ′ ,U i = U i ′ ,U i , j 1,2,3 . (28) jg j j+1 − j+1 j ∈{ } τi The j-th geodesic torsion jg associated with i-th hypersurface is

τi = U i ′ ,U i = U i ′ ,U i , j 1,2,3 . (29) jg − 5 j+1 j+1 5 ∈{ } D E 4.0.2. First geodesic and j-th torsion geodesic Formulas Theorem 4.1. First Geodesic Curvature and the j-th geodesic torsion of the intersection curve of 4 parametric hypersurfaces is

4 κi i 1g = ∑ a j Nj,U2 (30) j=1 8 (Ni)u u +(Ni)u u +(Ni)u u +(Ni)u u τi = 1 1′ 2 2′ 3 3′ 4 4′ ,U i , j = 1,2,3. (31) jg − Φi Φi Φi Φi j+1 * 1 × 2 × 3 × 4 +

Proof. For First Geodesic Curvature

κi i i α i 1g = U1 ′ ,U2 = ′′(s),U2 . (32) D E Now using (14), Eq. (30) follows easily. Φi Φi Φi Φi For the j-th geodesic torsion, we need derivative of U i = N = 1 × 2 × 3 × 4 . If 5 i Φi Φi Φi Φi 1 2 3 4 deﬁned × × ×

Ni Φi Φi Φi Φi Ni = , where Ni = 1 2 3 4. Hence we derive Ni × × × dNi d Ni Ni Ni d i dNi ds ds U (s) = = − ds 5 ds 2 Ni dN τi = i ,U i , j = 1,2 ,3. jg − ds j+1 d Ni dNi Ni i ds Ni ds i τ = ,U , j = 1,2,3. jg 2 2 j+1 −* N i − Ni +

dNi N τi ds i i i jg = 2 ,Uj+1 , j = 1,2,3., because Ni,Uj+1 = 0, j = 1,2,3. −* N i + D E dNi τi ds i jg = ,Uj+1 , j = 1,2,3. −* Ni + dNi =(Ni)u u′ +(Ni)u u′ +(Ni)u u′ +(Ni)u u4′ ds 1 1 2 2 3 3 4 i i i i i i i i (Ni)u = Φ Φ Φ Φ + Φ Φ Φ Φ + i 1i × 2 × 3 × 4 1 × 2i × 3 × 4 . Φi Φi Φi Φi + Φi Φi Φi Φi 1 × 2 × 3i × 4 1 × 2 × 3 × 4i 5. Example

Let M1, M2, M3 and M4 be the hypersurfaces given by, respectively

1 1 X (u1,u2,u3,u4)=(√2sinu3 cosu2 cosu1,√2sinu3 sinu2 sinu1,√2sinu3 cosu2,√2u4 cosu3, sinu3) √2

2 X (u1,u2,u3,u4)=(u3 cosu1 cosu2,u3 sinu1 cosu2,u3 sinu2,u3,cosu2 cosu4) 3 X (u1,u2,u3,u4)=(u3 cosu1 cosu2,u3sinu1 cosu2,u4 sinu1,u3,sinu1 sinu2) 1 1 1 u X 4(u ,u ,u ,u )=( + cosu , sinu ,u ,u , 4 ) 1 2 3 4 2 2 1 2 1 2 3 2

9 let us ﬁnd the Frenet vectors and the curvatures of the intersection curve at the intersection point

π π π π π π π π π √2 1 1 √2 1 p = X 1 , , ,1 = X 2 , ,1, = X 3 , ,1,1 = X 4 , ,1,1 = , , ,1, 4 4 4 4 4 4 4 4 2 2 ! 2 2 2 2! The unit normals of these hypersurfaces are

1 1 2 1 1 1 1 N1 = , ,0,0, ,N2 = , , , ,0 √6 √6 − 3 −2√2 −2√2 −2 √2 r ! 2 1 2 ,N3 = ,0,0, , ,N4 =(0,1,0,0,0) −3 3 −3 The non-vanishing ﬁrst fundamental coefﬁcients are

1 9 1 1 g1 = g1 = 1 g1 = g1 = 1 g1 = g1 = g1 = 1 11 2 22 33 4 44 12 2 23 −2 34 − 1 5 1 1 5 3 g2 = , g2 = g2 = 2, g2 = , g2 = , g3 = , g3 = , 11 2 22 4 33 44 4 24 4 11 4 22 4 3 1 1 1 1 1 g3 = , g3 = , g3 = , g3 = , g3 = , g4 = ,g4 = 1, 33 2 44 2 12 4 14 2 23 −2 11 4 22 1 g4 = 1, g4 = 33 44 4 The unit tangent at the intersection point is found by

N1 N2 N3 N4 2 2 6 1 t = ⊗ ⊗ ⊗ = ,0, 5 , , N1 N2 N3 N4 −√91 − 91 −√91 −√91 k ⊗ ⊗ ⊗ k r ! The non-vanishing second fundamental coefﬁcients are

1 1 1 1 1 2 2 h1 = , h1 = , h1 = , h2 = , h2 = , h3 = , h3 = 11 −√6 22 −√6 12 −√6 11 2√2 22 √2 11 3 22 3 2 1 1 h3 = , h3 = , h4 = 12 −3 13 3 11 −2 From the linear system of equations in (17), we obtain

(u1)′ = 0.628971, (u1)′ = 0.838628, (u1)′ = 0.209657, (u1)′ = 0.838628 1 − 2 3 − 4 − (u2)′ = 0.209657, (u2)′ = 0.419314, (u2)′ = 0.628971, (u2)′ = 0.628971 1 2 − 3 − 4 (u3)′ = 0.209657, (u3)′ = 0.419314, (u3)′ = 0.628971, (u3)′ = 1.25794 1 2 − 3 − 4 − (u4)′ = 0.419314, (u4)′ = 0.741249, (u4)′ = 0.628971, (u4)′ = 0.209651 1 2 − 3 − 4 − Thus, we obtain κ1 = 0.879304, κ2 = 0.139867, κ3 = 0.117216, κ4 = 0.0879121. Hence, n − n n n − we have

α ′′ = 1.321618N1 0.469981N2 + 0.698466N3 0.285472N4 − − 10 − Or, α ′′ =( 0.839029, 0.658857,0.234991, 0.0995048,0.613453) − − − Thus, κ1 = α ′′ = 1.25679 k k Also the unit normal vector is n=( 0.524421, 0.411807,0.146877, 0.621938,0.383428) − − − From (21), we get

1 =( 1.05495, 1.14286, 0.279735, 0.395605, 0.021978), △ − − − − − 2 =( 0.32967, 0.417583,0.248653,0, 0.549451) △ − − − 3 =( 0.32967, 0.417583, 0.404061,0, 0.197802), 4 =(0, 0.0879121,0,0,0) △ − − − − △ − From the linear system of equations in (22), we obtain

1 ′′ 1 ′′ 1 ′′ 1 ′′ (u1) = 0.106614, (u2) = 0.161474, (u3) = 0.89026, (u4) = 1.05091 (u2)′′ = 0.268086, (u2)′′ = 0.366087, (u2)′′ = 0.309769, (u2)′′ = 2.69189 1 2 3 − 4 − (u3)′′ = 0.454481, (u3)′′ = 0.795236, (u3)′′ = 0.141793, (u3)′′ = 0.450137 1 2 3 − 4 (u4)′′ = 1.67806, (u4)′′ = 0.2356, (u4)′′ = 0.234991, (u4)′′ = 1.22691 1 2 3 − 4 Which yields

µ1 = 0.4544, µ2 = 0.5105, µ3 = 1.4338, µ4 = 1.0554 − − − Then, we have

2 α ′′′ = (1.25679) t + 1.84188N1 + 0.451035N2 2.14772N3 1.64788N4 − − − α ′′′ =(2.35545, 1.0554,0.945301,0.596496,0.0935034) − κ ′ = α ′′′ ,n = 0.996915 1 h i − Using α ′′′ , we obtain

(u1)′′′ = 5.90731, (u1)′′′ = 2.14186, (u1)′′′ = 1.51393, (u1)′′′ = 2.11042 1 − 2 3 4 (u2)′′′ = 3.76546, (u2)′′′ = 2.58181, (u2)′′′ = 0.76716, (u2)′′′ = 2.76882 1 − 2 3 4 − (u3)′′′ = 0.88196, (u3)′′′ = 3.60164, (u3)′′′ = 1.14366, (u3)′′′ = 2.21882 1 − 2 − 3 − 4 (u4)′′′ = 3.33833, (u4)′′′ = 0.545801, (u4)′′′ = 0.596496, (u4)′′′ = 0.187007 1 − 2 3 4 Hence, we get

ξ1 = 21.7826, ξ2 = 3.3273, ξ3 = 8.8342, ξ4 = 1.3710 − − − From (23), we have

(4) 2 2 6 1 α = 3(1.25679)( 0.996915) ,0, 5 , , 41.7712N1 − − −√91 − r91 −√91 −√91! − 29.6654N2 + 34.1873N3 + 5.19372N4 − 11 Or, α(4) =( 30.1443, 1.371,12.0465, 11.945,10.9205) − − − Thus,

α ′ α ′′ α ′′′ α(4) b3 = ⊗ ⊗ ⊗ =(0.235522,0.439366, 0.161734, 0.0297594,0.851143), α ′ α ′′ α ′′′ α(4) − − k ⊗ ⊗ ⊗ k

b3 α ′ α ′′ α ′′′ b2 = ⊗ ⊗ ⊗ =(0.0469859, 0.0544597,0.64055, 0.763297, 0.0435799) b3 α ′ α ′′ α ′′′ − − − k ⊗ ⊗ ⊗ k Also,

b2 b3 α ′ α ′′ b1 = ⊗ ⊗ ⊗ =(0.648319, 0.724303, 0.161293, 0.0530185,0.16199) b2 b3 α ′ α ′′ − − − k ⊗ ⊗ ⊗ k Now curvature,

(4) α ′′′ ,b1 α ,b2 κ2 = h i = 1.68888, κ3 = h i = 7.07463 κ1 κ1κ2

κ ′′ Also from (4) we have, 1 = 35.3284 Using α(4), we obtain

(u1)(4) = 44.1618, (u1)(4) = 21.5916, (u1)(4) = 4.64151, (u1)(4) = 16.5865 1 2 − 3 − 4 − (u2)(4) = 22.5702, (u2)(4) = 33.779, (u2)(4) = 7.12615, (u2)(4) = 55.62 1 2 3 − 4 (u3)(4) = 26.8139, (u3)(4) = 22.2084, (u3)(4) = 4.81273, (u3)(4) = 9.77667 1 2 3 − 4 − (u4)(4) = 0.6116, (u4)(4) = 16.2234, (u4)(4) = 11.945, (u4)(4) = 21.841 1 − 2 3 − 4 Hence, we have

η1 = 194.9662, η2 = 27.5036, η3 = 56.7000, η4 = 29.6466 − − Thus we have

α(5) = 3(1.25679)2 4(1.25679)(35.3284)+(1.25679)4 +(1.25679)2(1.68888)2 t − − + 313.53N1 + 145.818N2 210.771N3 46.7969N4 − − Or, α(5) =(253.719,29.6467,57.0616,143.136, 97.1016) − Thus, the fourth curvature is given by

(5) α ,b3 κ4 = h i = 1.55521 κ1κ2κ3 − κ1 τ1 κ1 Now, to ﬁnd 1g and jg, j = 1,2,3 for hypersurface M1, we have from (30) and (31), 1g = 0.584888,τ1 = 0.774977,τ1 = 0.0496875,τ1 = 0.276372. Similarly we can ﬁnd κi 1g − 2g − 3g − 1g τi and jg for M2,M3,M4. 12 References

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