Fronts of Whitney umbrella a differential geometric approach via blowing up

Transcription

1 Journal of Singularities Volume 4 (202), received 3 October 20 in revised form 9 January 202 DOI: /jsing.202.4c Fronts of Whitney umbrella a differential geometric approach via blowing up Toshizumi FUKUI and Masaru HASEGAWA Abstract We investigate the differential geometric ingredients for Whitney umbrella, which is known as the only stable singularity of surface to 3-dimensional Euclidean space. We obtain several criteria of the singularity types of fronts of Whitney umbrella in terms of differential geometric language we discuss. Contents Introduction 35 2 Differential geometry for Whitney umbrella The unit normal vectors Normal form of Whitney umbrella Principal curvatures and principal directions Ridge points and sub-parabolic points over Whitney umbrella Constant principal curvature lines Singularities of the distance squared unfolding Focal conics Versality of distance squared unfolding Singularities of caustics and fronts of Whitney umbrella 6 Introduction H. Whitney [28] has found Whitney umbrella (also known as the cross-cap) as singularities which are not avoidable by small perturbation. This is very important singularity type, since it is the only singularity of a map of surface to 3-dimensional Euclidean space which is stable under small deformations. This singularity is fundamental in the context of differential topology but it does not seems that Whitney umbrella is a subject of differential geometry, Mathematics Subject Classification (200). Primary 53A05; Secondary 58K20. Keywords and phrases. Whitney umbrella, Cross-cap, Singularities, Distance squared function, Blowing up, Front. 35

2 at least before C. Gutierrez and J. Sotomayor s paper [0]. Motivated by Darboux s classification ([5]), they aimed to determine the configuration of lines of curvature near Whitney umbrella, and complete it in [9]. J. W. Bruce and J. M. West [3] investigated functions on Whitney umbrella using singularity theory. In [8], we show that a unified treatment for differential geometric properties for regular and singular maps g : (R 2, 0) (R 3, 0) and show that Whitney umbrella is not a bad singularity ([8, proposition 4.2]) from the view point of investigating singularities of distance squared functions. In other words, Whitney s umbrella is as good as (or as bad as) Darbouxian umbilics (i.e., both are characterised by the condition that rank R(g, 0) = 4 in the notation of [8]). Several authors continue to investigate the configuration of the solution curves of particular binary differential equations (i.e. lines of curvature, asymptotic and characteristic curves) in [20, 24]. In [9], the classification of parabolic lines of Whitney umbrella is used to investigate the projections of smooth surface in R 4 to 3-spaces. When we consider parallel surfaces of a regular surface, we are not able to avoid singularities. These singularities are often called front and this subject is investigated by M. Kokubu, W. Rossman, K. Saji, M. Umehara and K. Yamada [6]. They mean by a front a map g : (R 2, 0) R 3 such that there exists a well-defined normal n : (R 2, 0) S 2 R 3 so that (g, n) : (R 2, 0) (R 3, 0) S 2 is an immersion. Cuspidal edges and swallowtails are typical singularity types of fronts. In [6], criteria for the these singularities are given. Furthermore, criteria for the cuspidal lips and the cuspidal beaks are given in [5], criterion for the cuspidal butterfly is given in [4], and criteria for the D 4 -singularities are given in [23]. Whitney umbrella is not a front in their sense, since any unit normals defined at regular points near the singular point cannot extend continuously to the singular point. Physically, the wave propagation is described by Huygens s principle: every point to which a luminous disturbance reached becomes a source of a spherical wave, and the sum of these secondary waves determines the form of the wave front at any subsequent time. We remark that this does not require the notion of unit normal vectors. Mathematically, a wave front is the envelope of the spherical waves, and this requires us to investigate the singularities of the members of the family of functions: Φ : (R 2, 0) R 3 R, (u, v) (x, y, z) 2 ( (x, y, z) g(u, v) 2 t 0 2 ) (.) where t 0 is a constant. The family Φ is an unfolding of the distance squared function ϕ(u, v) = Φ(u, v, x 0, y 0, z 0 ) where (x 0, y 0, z 0 ) is a point in R 3, and the discriminant set D(Φ) of Φ is a (wave) front of g at distance t 0 where D(Φ) = {(x, y, z) R 3 ; Φ = Φ u = Φ v = 0 for some (u, v) (R 2, 0)}. For regular surfaces, we investigate the distance squared unfolding Φ and show several criteria for singularity types of parallel surfaces in terms of differential geometric language (principal curvatures, ridge points, sub-parabolic points, etc.) in [7]. In this paper, we investigate singularities of the distance squared unfolding for Whitney s umbrella, and show similar criteria for versality (Theorems 3.7). To investigate the singularities of the distance squared unfolding for Whitney s umbrella, we need several differential geometric languages of Whitney umbrella. 36

3 We also investigate the focal sets (caustics) of Whitney umbrella, since the bifurcation set B(Φ) of Φ represents the focal set where B(Φ) = {(x, y, z) R 3 ; Φ u = Φ v = Φ uu Φ vv Φ uv 2 = 0 for some (u, v) (R 2, 0)}. In Section 2, we introduce some differential geometric ingredients (principal curvatures, ridge, sub-parabolic points, etc.) for Whitney umbrella. For regular surfaces in Euclidean 3-space, several authors investigate ridge points and sub-parabolic points; see for example [2], [4], [8], [2], and [22]. The ridge points were first studied in details by I. Porteous [2] in terms of singularities of distance squared functions. The ridge line is the locus of points where one principal curvature has an extremal value along lines of the same principal curvature. The sub-parabolic points were studied in details by J. W. Bruce and T. C. Wilkinson [4] in terms of singularities of folding maps. The sub-parabolic line is the locus of points where one principal curvature has an extremal value along lines of the other principal curvature. Recently, in the case of the hyperbolic space, the analogous notion to the ridge point of hypersurfaces is introduced in [3], and the analogous notion to the sub-parabolic point of smooth surfaces is introduced in [2]. We develop the differential geometric ingredients over Whitney umbrella, which seem to be missing pieces of knowledge of the people who work on singularity theory and differential geometry. Since Whitney umbrella is a singularity of rank one, the tangent planes of nearby point degenerate to the tangent line at the singularity, and the normal lines are developed to the normal plane at the singular point. This means that we have a chance to have a bounded normal curvature in one direction at singular point. In Section 2., we first show that for Whitney umbrella there is a well-defined unit normal via the double oriented blowing-up ([, example (a) in p. 22]): π : R S R 2, (r, θ) (r cos θ, r sin θ). (.2) Then we are able to talk about principal curvatures and principal directions via π, and we discuss their asymptotic behaviours in Section 2.3. Let M denote the quotient space of R S with identification (r, θ) ( r, θ + π). Then we obtain a natural map π : M R 2, [(r, θ)] (r cos θ, r sin θ), (.3) which we usually call a blow up. We remark that M is topologically a Möbius strip. It is a natural problem to ask configurations of the parabolic line, ridge lines, sub-parabolic lines etc. on M near the exceptional set X = π (0, 0). We show () The parabolic line intersects with X in at most two points (Proposition 2.3), (2) Along an arc which reaches the singularity of Whitney umbrella, one principal curvature κ is bounded if the arc is not tangent to the double point locus, and the other principal curvature κ 2 tends to infinity (Lemma 2.2), (3) The ridge line with respect to κ intersects with X in at most four points (Lemma 2.6, see Lemma 2.7 also) in generic context, (4) The ridge line with respect to κ 2 intersects with X at two points (Proposition 2.0), 37

4 (5) The sub-parabolic line with respect to κ intersects with X in at most three points (Lemma 2.), and (6) A constant principal curvature (CPC) line intersects with X in at most four points (Proposition 2.6). In Section 3, we investigate singularities of the distance squared unfolding Φ defined by (.). We define the focal conic in the normal plane as a counterpart of focal points, and discuss versality of the unfolding Φ of ϕ, which is one of the fundamental notion in singularity theory. As a consequence, we are able to determine singularity types of caustics and fronts of Whitney umbrella in Section 4. We summarise our results as follows: () If (x 0, y 0, z 0 ) is on the focal conic, then ϕ is at least A 2 -singularity. (2) If (x 0, y 0, z 0 ) g(0, 0) does not correspond to the ridge over Whitney umbrella, then ϕ has an A 2 -singularity and Φ is an R + -versal unfolding (and a K-versal unfolding). The caustic is nonsingular at (x 0, y 0, z 0 ), and the front has the cuspidal edge at (x 0, y 0, z 0 ). (3) If (x 0, y 0, z 0 ) g(0, 0) corresponds to the first-order ridge over Whitney umbrella, then ϕ has an A 3 -singularity and Φ is R + -versal. Thus the caustic has the cuspidal edge at (x 0, y 0, z 0 ). Moreover if (x 0, y 0, z 0 ) does not correspond to the sub-parabolic point over Whitney umbrella, then Φ is a K-versal unfolding. We thus conclude that the front has the swallowtail at (x 0, y 0, z 0 ) (4) If (x 0, y 0, z 0 ) g(0, 0) corresponds to the first-order ridge and the sub-parabolic over Whitney umbrella, and the CPC line has definite (resp. indefinite) Morse singularity on X, then the front is the cuspidal lips (resp. cuspidal beaks) at (x 0, y 0, z 0 ). (5) If (x 0, y 0, z 0 ) g(0, 0) corresponds to the second-order ridge and does not correspond to the sub-parabolic point over Whitney umbrella, then the front is the cuspidal butterfly at (x 0, y 0, y 0 ). See Theorem 3.7 and Theorem 4.3 for a precise statement. We remark that there are no D 4 -singularities (or worse) for distance squared function ϕ at Whitney umbrella. 2 Differential geometry for Whitney umbrella We consider a smooth map g : U R 3 given by g(u, v) = (g (u, v), g 2 (u, v), g 3 (u, v)) which defines a surface in R 3, where U R 2 is an open subset. The map g possibly has singularities. The map g : (R 2, 0) (R 3, 0) has a Whiney umbrella singularity at (0, 0) if it is A-equivalent to the map germ: (R 2, 0) (R 3, 0), (u, v) (u, uv, v 2 ). We remark that some authors distinguish between Whitney umbrellas and cross-caps as follows: the Whitney umbrella is the zero-set of the function x 2 z y 2 = 0; the cross-cap is the image of the map that is A-equivalent to (u, v) (u, uv, v 2 ) (see, for example, [3] and [24]). But the authors prefer to use the word Whitney umbrella for map germs with respect for H. Whitney s work. 38

5 Away from singularities a unit normal vector is defined by n = (g u g v )/ g u g v, and the first and second fundamental forms for g are given by respectively, where I = E du 2 + 2F du dv + G dv 2, II = L du 2 + 2M du dv + N dv 2, E = g u, g u, F = g u, g v, G = g v, g v, L = g uu, n, M = g uv, n, N = g vv, n. The principal curvatures κ and κ 2 are the roots of the equation L κ i E M κ i F M κ i F N κ i G = 0. If a non-zero vector v i = (ξ i, η i ) (i =, 2) is the principal vector with principal curvature κ i, then ( ) ( ) ( L κi E M κ i F ξi 0 = (2.) M κ i F N κ i G η i 0) We can choose (ξ i, η i ) so that the tangent vector ξ i g u + η i g v is of unit length. We investigate the asymptotic behaviour of these ingredients near Whitney umbrella. 2. The unit normal vectors Now we suppose that g has a rank one singularity at (0, 0). Take the image of dg 0 to be the x-axis. Then we may write g as ( ) g(u, v) = u, 2 (a 02u 2 + 2a uv + a 02 v 2 ) + O(u, v) 3, 2 (b 20u 2 + 2b uv + b 02 v 2 ) + O(u, v) 3. We consider the unit normal vector ñ = n π in the coordinates (r, θ), where π is as in (.2). By a straightforward calculation we show that the unit normal vector ñ is expressed as follows: ñ(r, θ) = (0 + O(r), b cos θ b 02 sin θ + O(r), a cos θ + a 02 sin θ + O(r)). (a 2 + b 2 ) cos 2 θ + 2(a a 02 + b b 02 ) cos θ sin θ + (a b 2 02 ) sin 2 θ If the singular point of g is a Whitney umbrella, then a a 02 b b We thus conclude the unit normal vector ñ is well-defined on {r = 0}, since (a a 02 + b b 02 ) 2 (a 2 + b 2 )(a b 02 2 ) = (a b 02 a 02 b ) 2. 39

6 2.2 Normal form of Whitney umbrella For a regular surface, we can take the z-axis as the normal line, and, after suitable rotation if necessary, we can express the surface in the Monge normal form: (u, v) ( u, v, 2 (k u 2 + k 2 v 2 ) + O(u, v) 3) For Whitney umbrella we can perform similar computations and obtain the following normal form theorem of Whitney umbrella. Proposition 2.. Let g : (R 2, 0) (R 3, 0) be a smooth map with a Whitney umbrella at (0, 0). Then there are a rotation T : R 3 R 3 and a diffeomorphism φ : (R 2, 0) (R 2, 0) so that where T g φ(u, v) = B(v) = k i=3 ( u, uv + B(v) + O(u, v) k+, b i i! vi, and A j (u, v) = j i=0 k ) A j (u, v) + O(u, v) k+ j=2 a i,j i i!(j i)! ui v j i with a The result was first proved in [27], but we repeat the proof for completeness. (k 3), Proof. Take the image of dg 0 to be the x-axis. Then we may write g as g(u, v) = ( u, i+j=2 b ij i!j! ui v j + O(u, v) 3, i+j=2 a ij i!j! ui v j + O(u, v) 3) with Take θ so that (cos θ, sin θ) = (a 02, b 02)/ a b 2 02, and set a rotation of R 3 T = cos θ sin θ 0 sin θ cos θ b b 02 a a and a change of coordinates ψ(u, v) = u, b b 02 a a 02 ( 2 b 20 b 02 a 20 a 02 ) u + a b 022 v. Then we have T g ψ(u, v) = u, uv + i+j=3 β ij i!j! ui v j + O(u, v) 4, i+j=2 α ij i!j! ui v j + O(u, v) 3 40

7 for some constants α ij and β ij. Setting B k = i+j=k b iju i v j /(i!j!) (k 3) and replacing v by v + i+j=k c iju i v j /(i!j!), we have uv + B k + O(u, v) k+ = uv + i+j=k ic i,j + b ij u i v j + O(u, v) k+. i!j! For a suitable choice of c ij (i + j = k ), we can reduce this to b (0,k) v k /k! + O(u, v) k+. Hence, we obtain the result. 2.3 Principal curvatures and principal directions Coefficients of the first and second fundamental forms. Throughout the rest of the paper, we suppose that g is given in the normal form of Whitney umbrella: where B(v) = g(u, v) = 4 i=3 ( u, uv + B(v) + O(u, v) 5, b i i! vi, and A j (u, v) = j i=0 4 A j (u, v) + O(u, v) ), 5 (2.2) j=2 a i,j i i!(j i)! ui v j i with a Then we have g u = g v = (, v + O(u, v) 4, 4 (A j ) u + O(u, v) 4), j=2 ( 0, u + B v + O(u, v) 4, 4 (A j ) v + O(u, v) 4) j=2 and thus have E = + v 2 + (A 2u ) 2 + 2A 3u A 2u + O(u, v) 4, F = uv + A 2u A 2v + A 3u A 2v + A 3v A 2u + 2 b 3v 3 + O(u, v) 4, (2.3) G = u 2 + (A 2v ) 2 + 2A 3v A 2v + b 3 uv 2 + O(u, v) 4. Since 3 ( ) g u g v = b 3 v (Aj ) v u (A j ) u 2 v2 A 2u + O(u, v) 4, j=2 4 (A j ) v + O(u, v) 4, u + 2 b 3v b 4v 3 + O(u, v) 4, j=2 4

8 we have where It follows that g u g v 2 = λ 2 + λ 3 + λ 4 + O(u, v) 5, λ 2 = u 2 + A 2 2v, λ 3 = 2A 3v A 2v + b 3 uv 2, λ 4 = 2A 4v A 2v + A 2 3v + (ua 2u va 2v ) b 4uv b 3 2 v 4. (2.4) ñ(0, θ) = A (0, a cos θ a 02 sin θ, cos θ), (2.5) where A = A(θ) = cos 2 θ + (a cos θ + a 02 sin θ) 2. Since a 02 0, ñ(0, θ) defines an isomorphism from real projective line P (R) P (R) P (R), θ ñ(0, θ). We set Since l = g uu, g u g v, m = g uv, g u g v, n = g vv, g u g v. g uu = 0, 0, 4 (A j ) uu + O(u, v) 3, j=2 g uv = 0, + O(u, v) 3, 4 (A j ) uv + O(u, v) 3, j=2 g vv = 0, B vv + O(u, v) 3, 4 (A j ) vv + O(u, v) 3, j=2 l, m, and n are expressed as follows: l = l + l 2 + l 3 + O(u, v) 4, m = m + m 2 + m 3 + O(u, v) 4, n = n + n 2 + n 3 + O(u, v) 4, where l = a 20 u, l 2 = ua 3uu + 2 a 20b 3 v 2, l 3 = ua 4uu + 2 b 3A 3uu v a 20b 4 v 3 m = a 02 v, m 2 = ua 3uv A 3v + 2 a b 3 v 2, m 3 = ua 4uv A 4uv + 2 b 3A 3uv + 6 a b 4 v 3 (2.6) n = a 02 u, n 2 = ua 3vv + 2 b 3(a 02 v 2 2vA 2v ), n 3 = ua 4vv + 2 b 3v(vA 3vv 2A 3v ) + 6 b 4v 2 (a 02 v 3A 2v ). 42

9 By using the Taylor series for L(ru, rv), M(ru, rv), and N(ru, rv) in r we obtain L = a 20u + 2l 2λ 2 l λ 3 λ2 2λ 3/2 2 M = a 02v + 2m 2λ 2 m λ 3 λ2 2λ 3/2 2 N = a 02u + 2n 2λ 2 n λ 3 λ2 2λ 3/2 2 Principal curvatures. r + 8l 3λ 2 2 4l 2 λ 2 λ 3 4l λ 2 λ 4 + l λ 3 r 2 + O(r 3 ), 8λ 5/2 2 r + 8m 3λ 2 2 4m 2 λ 2 λ 3 4m λ 2 λ 4 + m λ 3 r 2 + O(r 3 ), 8λ 5/2 2 r + 8n 3λ 2 2 4n 2 λ 2 λ 3 4n λ 2 λ 4 + n λ 3 r 2 + O(r 3 ). 8λ 5/2 2 Lemma 2.2. The principal curvatures κ i = κ i π are expressed as follows: (2.7) κ (r, θ) = k 0 (θ) + k (θ)r + k 2 (θ)r 2 + O(r 3 ), κ 2 (r, θ) = r 2 [ k20 (θ) + k 2 (θ)r + O(r 2 ) ], (2.8) where k 0 = A 2 sec θ A, (2.9) k = ( ) 6a02 à 3 à 2v tan θ + 2Ã3u(a à 2v + cos θ) cos θ A 3 Ã3vÃ2v(a 20 cos 2 θ + a 02 sin 2 + b 3A 2Ã2 2v tan 2 θ θ) sec θ 2A 3, (2.0) [ k 2 = 2a 02 A 5 24a 02 A 4 à 4 sec θ 2a 02 A 2 à 3 à 3v à 2v sec θ 2a 02 A 2 à 4v (Ã2uÃ2v + A 2 tan θ + cos θ sin θ) ( ) 2 à 2 2v(a 2 a 20 a 02 ) + 4a à 2v cos θ Ã2 3v +(a 20 a ) cos 2 θ a 02 sin 2 cos θ θ ] 8a 02 A 4 A 2Ã2 2 sec 3 θ 8a 02 A 4 A 2 sin θ tan θ a 02 A 2 A 23 sec θ, (2.) k 20 = a 02 cos θ A 3, (2.2) k 2 = ] [( 3a A 5 02 à 3v à 2v + A 2 à 3vv ) cos θ + [ ] (2.3) 2A 5 b 3 ( 2A 2 à 2v + a 02 A 2 sin θ 3a 02 cos 2 θ sin θ) sin θ. Here A 2 = a 20 cos 2 θ a 02 sin 2 θ, à 2v = A 2v π r=, à 3v = A 3v π r=, and so on. Proof. The principal curvatures κ i are the roots of the equation (EG F 2 )k 2 (EN 2F M + GL)k + (LN M 2 ) = 0. From (2.3), (2.4), (2.6), and (2.7), it follows that (EG F 2 ) π = a 2 r 2 + a 3 r 3 + a 4 r 4 + O(r 5 ), 43

10 where (EN 2F M + GL) π = b 0 + b r + b 2 r 2 + O(r 3 ), (LN M 2 ) π = c 0 + c r + c 2 r 2 + O(r 3 ), a 2 = A 2, b 0 = a 02 cos θ A, c 0 = a 02A 2 A 2, and the coefficients a 3, a 4, b, b 2, c, and c 2 are the trigonometric polynomials in the coefficients appearing in the terms of degree four or less in the normal form of Whitney umbrella. Therefore, we obtain κ = c 0 + b c 0 b 0 c b 2 r + b 2 c 0 b 0 b 2 c 0 + a 2 c 2 0 b 0 b c + b 2 0 c 2 0 b 0 κ 2 = r 2 ( b 0 a 2 + a 3b 0 a 2 b a 2 3 r + O(r 2 ) We thus obtain (2.9) (2.3) by a straightforward calculation. ). b 0 3 r 2 + O(r 3 ), Gaussian curvature. Since the Gaussian curvature K is the product of the principal curvatures, it does not depend on the choice of the unit normal vector. From (2.9) and (2.2), the Gaussian curvature is expressed as follows: ( ) K(r, θ) = K π(r, θ) = a 02 (a 20 cos 2 θ a 02 sin 2 θ) r 2 A(θ) 4 + O(r), where π is as in (.3). By this expression, we say that a point (0, θ 0 ) on the Möbius strip M is elliptic, hyperbolic, or parabolic point over Whitney umbrella if r 2 K(0, θ0 ) is positive, negative, or zero, respectively. We often omit the phrase over Whitney umbrella if no confusion is possible from the context. We immediately have the following proposition. Proposition 2.3. a 20 a 02 < 0. () There is no parabolic point over Whitney umbrella if and only if (2) There is one parabolic point over Whitney umbrella if and only if a 20 = 0. (3) There are two parabolic points over Whitney umbrella if and only if a 20 a 02 > 0. Furthermore, in terms of the parabolic line in a domain, Whitney umbrella is classified into three types. We say that Whitney umbrella is hyperbolic, elliptic, or parabolic if the parabolic line has an A + -singularity (an isolated point), A -singularity (a pair of smooth curves intersecting transversally), or A 2 -singularity (a cusp), respectively. In the case of the hyperbolic Whitney umbrella all non-singular points near the Whitney umbrella singularity are hyperbolic. In the case of the elliptic Whitney umbrella the parabolic line divides the surface into hyperbolic and elliptic regions (see [27] for details). From (2.6), the parabolic line in the Möbius strip M is expressed by the equation a 02 A 2+ [ (a 20 Ã 3vv +a 02 Ã 3uu ) cos 2 θ+2a 02 (Ã3uv cos θ Ã3v) sin θ a b 3 A 2 sin θ ] r+o(r 2 ) = 0. 44

11 If a 20 a 02 < 0, then the parabolic line dose not meet with the exceptional set X = π (0, 0) on M, in which case the surface is the hyperbolic Whitney umbrella. If a 20 a 02 > 0, then the parabolic line meets with X at two parabolic point over Whitney umbrella, in which case the surface is the elliptic Whitney umbrella. If a 20 = 0, then the parabolic line meets with X at one parabolic point (r, θ) = (0, 0), which is the sub-parabolic point over Whitney umbrella (See Section 2.4). Calculating the tangent vector of the parabolic line at (0, 0) on M, we show that the parabolic line meets tangentially with X at (0, 0) if and only if a In this case, we have the parabolic Whitney umbrella (This classification according to the coefficients of the normal form of Whitney umbrella is also obtained in [20]). Moreover, the parabolic point (0, 0) is the singular point of the parabolic line if and only if a 30 = 0, equivalently this point is the ridge point over Whitney umbrella (See Section 2.4). In this case, this point is of Morse type if and only if 3a a 40 a Principal directions. Lemma 2.4. The unit principal vectors ṽ i in the coordinates (r, θ) are expressed as follows: ( ) ṽ = (sec θ + O(r)) r + 2Ã3v + b 3 Ã 2v sin θ tan θ + O(r) 2a 02 ṽ 2 = [( ) ( ) ] sin θ cos θ r 2 A r + O(r2 ) r + A + O(r). θ Proof. From the equation (2.), one of the vectors along the principal vectors v i in the coordinates (u, v) are given by ξ i u + η i v = (N κ ig) u + ( M + κ if ) v. (2.4) Since u = cos θ r sin θ and r θ v = sin θ r + cos θ r θ, the vector (2.4) can be lifted by π and we obtain ξ i r + η [ ] i θ = (N π κ i G π) cos θ + ( κ i F π M π) sin θ r + [ ( κ i F π M π) cos θ + ( κ i G π N π) sin θ] r θ. From (2.3), (2.4), (2.6), and (2.7), we have θ, where F π = F 2 r 2 + F 3 r 3 + O(r 4 ), G π = G 2 r 2 + G 3 r 3 O(r 4 ), M π = M 0 + M r + M 2 r 2 + O(r 3 ), N π = N 0 + N r + N 2 r 2 + O(r 3 ), F 2 = cos θ sin θ + Ã2uÃ2v, F 3 = Ã3uÃ2v + Ã3vÃ2u + 2 b 3 sin 3 θ, G 2 = cos 2 θ + Ã2 2v, G 3 = 2Ã3vÃ2v + b 3 cos θ sin 2 θ, M 0 = a 02 sin θ A, N 0 = a 02 cos θ A, 45

12 and coefficients L, M, N, L 2, M 2, and N 2 are the trigonometric polynomials in the coefficients appearing in the terms of degree four or less in the normal form of Whitney umbrella. It follows that ξ i and η i are expressed as follows: ξ = N 0 cos θ M 0 sin θ + (N cos θ M sin θ)r + O(r 2 ), η = M cos θ N sin θ + [(F 2 k 0 M 2 ) cos θ + (G 2 k 0 N 2 ) sin θ] r + O(r 2 ), ξ 2 = ( G 2 k 20 + N 0 ) cos θ + ( F2 k 20 M 0 ) sin θ + [( G 3 k 20 G 2 k 2 + N ) cos θ + (F 3 k 20 + F 2 k 2 M ) sin θ] r + O(r 2 ), η 2 = r [k 20(F 2 cos θ + G 2 sin θ) + ((F 3 k 20 + F 2 k 2 M ) cos θ + (G 3 k 20 + G 2 k 2 N ) sin θ)r + O(r 2 ) ]. After a long calculation, it follows that ξ i and η i are expressed as follows: ξ = ξ 0 + ξ r + O(r 2 ), η = η 0 + η r + O(r 2 ), ξ 2 = ξ 20 + ξ 2 r + O(r 2 ), η 2 = r [ η20 + η 2 r + O(r 2 ) ], where ξ 0 = a 02 A, η 0 = ( 2 à 3v cos θ + b 3 à 2v sin 2 θ ), 2A ξ = [ 2A 3 2 ( à 3v (a 02 à 2v cos θ a à 2v sin θ cos θ sin θ) a 02 à 3uv à 2v + Ã3vv(a à 2v + cos θ) ) ] b 3 à 2v (a à 2v + A 2 cos 2 θ + cos θ) sin θ, η = 2A 3 [ 24A 2 à 4v cos θ + 2Ã2 3vÃ2v cos θ + 24Ã2Ã2vA 2 cos θ + 2a 02 à 2 2u(3a à 2v cos θ + a 02 2 sin 2 θ) cos θ sin θ 24a a 02 à 3 2v sin 3 θ + 2a 02 3 à 2 2u cos θ sin 3 θ + 2a 20 a 3 à 2u cos 5 θ + 60a 2 a 02 2 à 2v cos θ sin 4 θ + 24A 2 A 2 cos θ + 2a 20 a 4 cos 5 θ a 5 02 sin 5 θ tan θ + 4b 4 A 2 à 2v sin 3 θ ] 3b 2 3 à 2v cos θ sin 4 θ 6b 3 à 3v (Ã2 2v cos 2 θ) sin 2 θ, ξ 20 = 2a 02 A 3 (Ã2Ã2v + cos 2 θ sin θ) sin θ, η 20 = ξ 20 cot θ, η 2 = ξ 2 cot θ, ξ 2 = A 5 [3a a 02 A 2 à 3 + a 02 2 A 2 à 3u sin θ + Ã3v(2a 02 à 2 à 2 2v a A 2 A 2 8a a 02 cos θ sin θ) cos θ a 02 à 3uv (2Ã2Ã2 2v + a à 2 2v cos θ sin θ + 3a 02 cos 2 θ sin 2 θ) ] + Ã3vv(2Ã2Ã2v + a à 2u à 2 2v + a à 2 2v sin θ + 2A 2 sin θ 3a 2 02 sin 3 θ) cos θ sin θ b [ 3 2A 5 2Ã2Ã2v(a à 2 2v + Ã2v cos θ 2a cos 2 θ) + 3Ã2uÃ2 2v cos 2 θ 46

13 ] + 5a à 2 2v cos 2 θ sin θ + 3Ã2v cos 3 θ sin θ + a 02 cos 3 θ sin 2 θ sin 2 θ. The unit principal vectors ṽ i are given by ṽ i = Ẽ ξ2 i + 2 F ξ i η i + G η i 2 ( ξ i r + η i θ ), (2.5) where Ẽ, F, and G are the coefficients of the first fundamental form of g in the coordinates (r, θ). We calculate that ( ) Ẽ = cos 2 θ + 4 cos 2 θ sin 2 θ + Ã2 2 r 2 + O(r 3 ), F = r cos θ sin θ [ + 2 (cos 2 θ sin 2 θ) cos θ sin θ + Ã2(Ã2v cos θ Ã2u sin θ )] r 3 + O(r 4 ), [ G = r 2 sin 2 θ + (cos 2 θ sin 2 θ) 2 + ( à 2v cos θ Ã2u sin θ ) ] 2 r 4 + O(r 5 ). Hence, we obtain Ẽ ξ F ξ η + G η 2 = a 02 2 cos 2 θ A 2 + O(r), Ẽ ξ F ξ 2 η 2 + G η 2 2 = 4a 02 2 (Ã2Ã2v + cos 2 θ sin θ) 2 A 4 r 2 + O(r 3 ). This completes proof together with (2.5). Remark 2.5. The (unit) principal vector ṽ is extendible on {(r, θ) ; r 0 or cos θ 0} and thus the principal field defined by v is extendible on the Möbius strip M except on the set {(r, θ) ; r = 0, cos θ = 0}. The principal curvature vector r 2 ṽ 2 is extendible over R S even though ṽ 2 is not. So the principal field defined by v i is extendible over M. 2.4 Ridge points and sub-parabolic points over Whitney umbrella Ridge points. follows: By the computation in the previous subsection, we can express ṽ i κ i as ṽ κ (r, θ) =R 0 (θ) + R (θ)r +, ṽ 2 κ 2 (r, θ) = r 4 (R 20(θ) + R 2 (θ)r +, where ṽ i κ i denotes the directional derivative of the principal curvature κ i in the principal vector ṽ i. We say that a point (r 0, θ 0 ) is a ridge point relative to the principal vector ṽ (resp. ṽ 2 ) if ṽ κ (r 0, θ 0 ) = 0 (resp. r 4 ṽ 2 κ 2 (r 0, θ 0 ) = 0). If the ridge point (r 0, θ 0 ) is over Whitney umbrella (that is, r 0 = 0) this is equivalent that R 0 (θ 0 ) = 0 (resp. R 20 (θ 0 ) = 0). It is possible that R i0 (θ) has multiple roots. We say that (0, θ 0 ) is a multiple ridge point relative to ṽ i if θ 0 is a multiple root of R i0 (θ). We say that a point (0, θ 0 ) is a n-th order ridge point relative to ṽ (resp. ṽ 2 ) over Whitney umbrella if R im0 (θ 0 ) = 0 ( m n) and R i,n+,0 (θ 0 ) 0, where ṽ (m) κ (r, θ) =R m0 (θ) + R m (θ)r +, ṽ (m) 2 κ 2 (r, θ) = r 2+2m (R 2m0(θ) + R 2m (θ)r +. 47

14 Here, ṽ (n) i κ i denotes the n-th time directional derivative of κ i in the direction ṽ i. The ridge line relative to ṽ i near the exceptional set X = π (0, 0) is expressed by the equation: R i0 (θ) + R i (θ)r + = 0. In terms of the normal form of Whitney umbrella, we have κ (0, θ) tends to infinity as θ approaches ±π/2, by (2.9), and, after some calculations, we obtain where ṽ κ (r, θ) = Γ 3(θ) sec 3 θ A(θ) ṽ 2 κ (r, θ) = ( r 2 a 02Γ ) 3(θ) sec θ A(θ) 4 + O(r) + O(r), (2.6) ṽ 2 κ 2 (r, θ) = ( r 4 3a 02 2 (a cos θ + a 02 sin θ) cos θ A(θ) ) 6 ṽ κ 2 (r, θ) = r 3 ( 2a 02 A(θ) 3 + O(r), (2.7) ) + O(r), (2.8), (2.9) Γ 3 (θ) = 6Ã3 cos θ b 3 à 2v sin 3 θ, Γ 3(θ) = 2Ã2Ã2v + 2 cos 2 θ sin θ. Lemma 2.6. A point (0, θ 0 ) with cos θ 0 0 is a ridge point relative to ṽ if and only if Γ 3 (θ 0 ) = 0. Moreover, the point (0, θ 0 ) is a first order ridge point relative to ṽ if and only if Γ 4 (θ 0 ) 0, where Γ 4 (θ) = 24a 02 à 4 cos 2 θ 2Ã2 3v cos 2 θ 2a 02 A 2Ã2 2 2a 02 A 2 cos 2 θ sin 2 θ a 02 b 4 à 2v cos θ sin 4 θ + 2b 3 2 à 3v à 2v cos θ sin 2 θ 3b 3 à 2 2v sin 4 θ. Proof. The expansion (2.6) implies the first assertion. Since cos θ 0 0, the condition Γ 3 (θ 0 ) = 0 is equivalent to a 30 = [ cos θ 0 (3a 2 cos 2 θ 0 sin θ 0 + 3a 2 cos θ 0 sin 2 θ 0 + a 03 sin 3 θ 0 ) b 3 sin 3 θ 0 (a cos θ 0 + a 02 sin θ 0 ) ] sec 4 θ 0. Using the above relation, we can reduce ṽ 2 κ (0, θ 0 ) to and the proof is complete. ṽ 2 κ (0, θ 0 ) = Γ 4(θ 0 ) sec 5 θ 0 a 02 A(θ 0 ) 3, Lemma 2.7. If b 3 0, then there are at most four ridge points relative to ṽ over Whitney umbrella. 48

15 Proof. The ridge points relative to ṽ over Whitney umbrella are given by Γ 3 (θ) = 0; equivalently, a a 2 tan θ + 3a 2 tan 2 θ + (a 30 a b 3 ) tan 3 θ a 02 b 3 tan 4 θ = 0. This implies assertions. Remark 2.8. When b 3 = a 03 = a 2 = a 2 = a 30 = 0, the multiplicity of ridge points is not defined but the order is defined. In fact, we have Γ 4 (θ) 0. Proposition 2.9. Suppose that a point (0, θ 0 ) is a ridge point relative to ṽ over Whitney umbrella, and that the ridge line relative to ṽ is non-singular at (0, θ 0 ). Then the ridge line is tangent to X at (0, θ 0 ) if and only if (0, θ 0 ) is the multiple ridge point relative to ṽ. Proof. It follows form (2.6) that if and only if Γ 3(θ 0 ) = 0, the ridge line relative to ṽ is tangent to X at (0, θ 0 ). Hence, we have proved the proposition. From (2.8), we have the following proposition. Proposition 2.0. There are two simple ridge points relative to ṽ 2 over Whitney umbrella. That is, the ridge line relative to ṽ 2 is not tangent to X. Sub-parabolic points. ṽ i κ j (i j) as follows: By the computation in the previous subsection, we can express ṽ κ 2 (r, θ) = r 3 (P 0(θ) + P (θ)r + ), ṽ 2 κ (r, θ) = r 2 (P 20(θ) + P 2 (θ)r + ). We say that a point (r 0, θ 0 ) is a sub-parabolic point relative to the principal vector ṽ (resp. ṽ 2 ) if r 3 ṽ κ 2 (r 0, θ 0 ) = 0 (resp. r 2 ṽ 2 κ (r 0, θ 0 ) = 0). When the sub-parabolic point is over Whitney umbrella (that is, r 0 = 0), we obtain P i0 (θ 0 ) = 0. A point (0, θ 0 ) is said to be a multiple sub-parabolic point relative to v i over Whitney umbrella if θ 0 is a multiple root of P i0 (θ) = 0. The sub-parabolic line relative to v i near X is expressed by the equation: From (2.7) we have the following lemma. P i0 (θ) + P i (θ)r + = 0. Lemma 2.. A point (0, θ 0 ) is a sub-parabolic point relative to ṽ 2 if and only if Γ 3(θ 0 ) = 0 holds. From (2.9), we have the following proposition. Proposition 2.2. There is no sub-parabolic point relative to ṽ over Whitney umbrella. Lemma 2.3. There is at least one and at most three sub-parabolic points relative to ṽ 2 over Whitney umbrella. 49

16 Proof. The sub-parabolic points relative to ṽ 2 over Whitney umbrella are given by Γ 3(θ) = 0; equivalently, a 20 a + (2 + a 20 a a 2 ) tan θ + 3a a 02 tan 2 θ + a 02 2 tan 3 θ = 0. Since a 02 0, the equation has at least one and at most three roots and we have completed the proof of the Lemma. It follows from (2.7) that we obtain the following proposition. Proposition 2.4. Suppose that a point (0, θ 0 ) is a sub-parabolic point relative to ṽ 2 over Whitney umbrella, and that the sub-parabolic line relative to ṽ 2 is non-singular at (0, θ 0 ). Then the sub-parabolic line is tangent to X at (0, θ 0 ) if and only if (0, θ 0 ) is the multiple sub-parabolic point relative to ṽ 2. Example 2.5. We set (a 20, a, a 02 ) = ( 3, 0, ) in the normal form of Whitney umbrella, then we have Γ 3(θ) = cos 2 θ sin θ + sin 3 θ. The roots of Γ 3(θ) = 0 are θ = ±π/4 and θ = 0. Hence, we have the distinct three simple sub-parabolic points (0, ±π/4) and (0, 0) relative to ṽ 2 over Whitney umbrella. () We take (a 30, a 2, a 2, a 03, b 3 ) = (, 0, 0/9, 0, ). Then we have Γ 3 (θ) = cos 4 θ cos2 θ sin 2 θ sin 4 θ. The roots of Γ 3 (θ) = 0 are θ = ±π/3 and θ = ±π/6. Hence, we have four distinct simple ridge points (0, ±π/3) and (0, ±π/6) relative to ṽ over Whitney umbrella. (2) We take (a 30, a 2, a 2, a 03, b 3 ) = ( 3, 0, 4/3, 0, ). Then we have Γ 3 (θ) = 3 cos 4 θ + 4 cos 2 θ sin 2 θ sin 4 θ. The roots of Γ 3 (θ) = 0 are θ = ±π/3 and θ = ±π/4. Hence, we have four distinct simple ridge points (0, ±π/3) and (0, ±π/4) relative to ṽ over Whitney umbrella. Remark that the points (0, ±π/4) are ridge relative to ṽ and sub-parabolic relative to ṽ Constant principal curvature lines We set Σ k (resp. Σ2 k ) (k 0) as the image of {(r, θ) R S ; κ (r, θ) = ±k} (resp. {(r, θ) R S ; κ 2 (r, θ) = ±k}) by the double covering R S M. We call Σ k = Σ k Σ2 k by the constant principal curvature (CPC) line with a constant value of k. Remark that X Σ 2 k =. Remark also that Σ 0 is nothing but the parabolic line, which we already described in Proposition 2.3. Remember that M is non-orientable and the induced image of ñ, by π : R S R 2, covers all possible unit normals. We remark that Σ k (k > 0) is the singular set of the front of g at distance /k. 50

17 Proposition 2.6. A CPC line Σ k (k 0) intersects with X in at most four points. Proof. The number of the intersection points of Σ k and X equals the number of roots of the equation κ (0, θ) = k. From (2.9), we obtain the equation (a 20 cos 2 θ a 02 sin 2 θ) sec θ = k. cos2 θ + (a cos θ + a 02 sin θ) 2 Squaring both sides and setting the equation to 0, we get [ a20 2 k 2 (a 2 +) ] cos 4 θ 2a 02 a k 2 cos 3 θ sin θ a 02 (2a 20 a 02 k) cos 2 θ sin 2 θ+a 02 2 sin 4 θ = 0. Dividing both sides by cos 4 θ, we obtain a 20 2 k 2 (a 2 + ) 2a 02 a k 2 tan θ a 02 (2a 20 a 02 k 2 ) tan 2 θ + a 02 2 tan 4 θ = 0. Since a 02 0, this equation is quartic in tan θ and we have thus completed the proof. Setting C(r, θ) = κ (r, θ) 2 k 2 (k 0) and by (2.8), we have C(r, θ) = k 2 + k 0 (θ) 2 + 2k 0 (θ)k (θ)r + ( 2k 0 (θ)k 2 (θ) + k (θ) 2) r 2 +. (2.20) From (2.6) and (2.8), the principal vectors ṽ i can be written in ṽ (r, θ) = (x 0 (θ) + x (θ)r + ) r + (y 0(θ) + y (θ)r + ) θ, ṽ 2 (r, θ) = [ (x2 r 2 (θ)r + x 22 (θ)r 2 + ) r + (y 20(θ) + y 2 (θ)r + ) ]. θ Note that x 0 (θ) 0 and y 20 (θ) 0. Therefore, the directional derivatives of κ (r, θ) by ṽ i are expressed as follows: ṽ κ (r, θ) = x 0 (θ)k (θ) + y 0 (θ)k 0(θ) + (2x 0 (θ)k 2 (θ) + x (θ)k (θ) + y 0 (θ)k (θ) + y (θ)k 0(θ))r +, ṽ 2 κ (r, θ) = r 2 [ y20 (θ)k 0(θ) + (x 2 (θ)k (θ) + y 20 (θ)k (θ) + y 2 (θ)k 0(θ))r + ]. (2.2) The following lemma provides the criterion for the singularity of the CPC line intersecting with X in terms of the configurations of the ridge line and the sub-parabolic line. Lemma 2.7. Suppose that a point (0, θ 0 ) is not parabolic and that the CPC line Σ k meets X at (0, θ 0 ). Then the CPC line Σ k is singular at (0, θ 0 ) if and only if (0, θ 0 ) is the ridge point relative to ṽ and the sub-parabolic point relative to ṽ 2. In this case, the singularity is of Morse type if and only if the ridge line relative to ṽ and the sub-parabolic line relative to ṽ 2 intersect transversely at (0, θ 0 ). Proof. Let us use expansions (2.20) and (2.2). Now we have k 0 (θ 0 ) 0. The CPC line Σ k is singular at (0, θ 0 ) if and only if C r (0, θ 0 ) = C θ (0, θ 0 ) = 0. By computation, we have C r (0, θ 0 ) = 2k 0 (θ 0 )k (θ 0 ) and C θ (0, θ 0 ) = 2k 0 (θ 0 )k 0(θ 0 ). 5

18 It follows that Σ k has singularity at (0, θ 0 ) if and only if k 0(θ 0 ) = k (θ 0 ) = 0. From (2.2), we deduce that (0, θ 0 ) is the ridge point relative to ṽ and the sub-parabolic point relative to ṽ 2 if and only if k 0(θ 0 ) = k (θ 0 ) = 0. This completes the proof of the first assertion. We show the second assertion. Assume that (0, θ 0 ) is a singularity of Σ k. Then (0, θ 0 ) is a Morse singularity if and only if C rr (0, θ 0 ) C rθ (0, θ 0 ) C rθ (0, θ 0 ) C θθ (0, θ 0 ) 0. This is equivalent to 2k 0(θ 0 )k 2 (θ) k (θ 0 ) 2 0. It follows from (2.2) that the ridge line relative to ṽ and the sub-parabolic line relative to ṽ 2 intersect transversely at (0, θ 0 ) if and only if equivalently, 2x 0 (θ 0 )k 2 (θ 0 ) + y 0 (θ 0 )k (θ 0 ) x 0 (θ 0 )k (θ 0 ) + y 0 (θ 0 )k 0(θ) y 20 (θ 0 )k (θ 0 ) y 20 (θ 0 )k 0(θ 0 ) x 0 (θ 0 )y 20 (θ 0 )(2k 0(θ 0 )k 2 (θ) k (θ 0 ) 2 ) 0. We thus have completed of the proof of the second assertion. 0; Lemma 2.8. Suppose that a point (0, θ 0 ) is not parabolic and that the CPC line Σ k meets X at (0, θ 0 ). Then the CPC line Σ k is tangent to X at (0, θ 0 ) if and only if (0, θ 0 ) is a sub-parabolic point relative to ṽ 2 which is not a ridge point relative to ṽ. Proof. Let us consider expansions (2.20) and (2.2). From (2.20), the equation of the tangent line of Σ k at (0, θ 0 ) is then k (θ 0 )r + k 0(θ 0 )(θ θ 0 ) = 0. It follows that the CPC line Σ k is tangent to X if and only if k 0(θ 0 ) = 0 and k (θ 0 ) 0. From (2.2), we show that the point (0, θ 0 ) is a sub-parabolic point relative to ṽ 2 which is not a ridge point relative to ṽ if and only if k 0(θ 0 ) = 0 and k (θ 0 ) 0. Lemma 2.9. Assume that the CPC line Σ k and the ridge line relative to ṽ meet at a point (0, θ 0 ) which is not parabolic over Whitney umbrella. Then () These two curves intersect transversely at the point (0, θ 0 ) if and only if the point (0, θ 0 ) is a first order ridge point relative to ṽ. (2) These two curves are tangent at the point (0, θ 0 ) if and only if the point (0, θ 0 ) is a second or higher order ridge point relative to ṽ. Proof. Let us consider expansions (2.20) and (2.2). Remark that k 0 (θ 0 ) 0. The ridge line relative to ṽ passes through (0, θ 0 ), that is, (0, θ 0 ) is a ridge point relative to ṽ over Whitney umbrella if and only if x 0 (θ 0 )k (θ 0 ) + y 0 (θ 0 )k 0(θ 0 ) = 0. 52

19 Since x 0 0, this is equivalent to k (θ 0 ) = y 0(θ 0 )k 0(θ 0 ). (2.22) x 0 (θ 0 ) The CPC line Σ k and the ridge line relative to ṽ intersect transversely at (0, θ 0 ) if and only if the determinant k (θ 0 ) 2x 0 (θ 0 )k 2 (θ 0 ) + x (θ 0 )k (θ 0 ) + y 0 (θ 0 )k (θ 0 ) + y (θ 0 )k 0(θ 0 ) k 0(θ 0 ) x 0 (θ 0 )k (θ 0 ) + x 0(θ 0 )k (θ 0 ) + y 0 (θ 0 )k 0(θ 0 ) + y 0(θ 0 )k 0(θ 0 ) (2.23) is not zero. Otherwise, these two curves are tangent at (0, θ 0 ) if and only if the determinant (2.23) is zero. By using (2.22), the determinant (2.23) is expanded as x 0 (θ 0 ) 2 k 0(θ 0 ) [ 2x 0 (θ 0 ) 3 k 2 (θ 0 ) + 2x 0 (θ 0 ) 2 y 0 (θ 0 )k (θ 0 ) + x 0 (θ 0 ) 2 y (θ 0 )k 0(θ 0 ) + x 0 (θ 0 )y 0 (θ 0 ) 2 k 0(θ 0 ) x 0 (θ 0 )x (θ 0 )y 0 (θ 0 )k 0(θ 0 ) and we obtain ṽ 2 κ (0, θ 0 ) = + x 0 (θ 0 )y 0 (θ 0 ) 2 k 0(θ 0 ) y 0 (θ 0 ) 2 x 0(θ 0 )k 0(θ 0 ) ], [ 2x 0 (θ 0 ) 3 k 2 (θ 0 ) + 2x 0 (θ 0 ) 2 y 0 (θ 0 )k x 0 (θ 0 ) (θ 0 ) + x 0 (θ 0 ) 2 y (θ 0 )k 0(θ 0 ) + x 0 (θ 0 )y 0 (θ 0 ) 2 k 0(θ 0 ) x 0 (θ 0 )x (θ 0 )y 0 (θ 0 )k 0(θ 0 ) ] + x 0 (θ 0 )y 0 (θ 0 ) 2 k 0(θ 0 ) y 0 (θ 0 ) 2 x 0(θ 0 )k 0(θ 0 ). Hence, we conclude that the determinant (2.23) is zero (resp. non-zero) if and only if (0, θ 0 ) is a first order (resp. second or higher order) ridge point relative to ṽ, and we have completed the proof. 3 Singularities of the distance squared unfolding We assume that g : (R 2, 0) (R 3, 0) is given by (2.2). In this section, we investigate the singularities of the members of the family of the distance squared function: Φ : (R 2, 0) R 3 R, (u, v) (x, y, z) 2 ( (x, y, z) g(u, v) 2 t 0 2 ) (3.) where t 0 is a constant. We set ϕ(u, v) = Φ(u, v, x 0, y 0, z 0 ) where (x 0, y 0, z 0 ) is a point in R 3, and take t 0 so that ϕ(0, 0) = 0, that is, t 0 2 = x y z 0 2. Now we recall the definition of the normal plane. When the map g : (R 2, 0) (R 3, 0) has Whitney umbrella at (0, 0), the image of dg (0,0) is a line in R 3. We call the plane perpendicular to this line the normal plane at Whitney umbrella. Proposition 3.. The following conditions are equivalent: () The function ϕ has at least an A -singularity at (0, 0); 53

20 (2) The point (x 0, y 0, z 0 ) is on the normal plane at Whitney umbrella; (3) There exist ρ 0 R and θ 0 with θ 0 [ π/2, π/2] such that (x 0, y 0, z 0 ) = ρ 0 ñ(0, θ 0 ). Proof. The unfolding Φ can be expressed as follows: We have Φ(u, v, x, y, z) = c 00 (x, y, z) + xu + 4 i+j=2 i!j! c ij(x, y, z)u i v j + O(u, v) 5. (3.2) c 00 = 2 (t 0 2 x 2 y 2 z 2 ), c 20 = a 20 z, c = y + a z, c 02 = a 02 z, c 30 = a 30 z, c 2 = a 2 z, c 2 = a 2 z, c 03 = a 03 z + b 3 y, c 40 = 3a a 40 z, c 3 = 3a 20 a + a 3 z, c 22 = 2 2a 2 a 20 a 02 + a 22 z, c 3 = a a 02 + a 3 z, c 04 = 3a b 4 y + a 04 z. Then ϕ can be written in the form ϕ(u, v) = x 0 u + 4 i+j=2 i!j! c0 iju i v j + O(u, v) 5, (3.3) where c 0 ij = c ij(x 0, y 0, z 0 ). It follows that ϕ has at least an A -singularity at (0, 0) if and only if x 0 = 0. Directly from the definition of the normal form we obtain that the image of dg (0,0) is the x-axis. Hence, the normal plane is the yz-plane. Thus () and (2) are equivalent. Next, suppose (2) holds. Since a 02 0, ñ(0, θ) = (0, a cos θ a 02 sin θ, cos θ) cos2 θ + (a cos θ + a 02 sin θ) 2 expands in all direction in the yz-plane. Hence, there exist ρ 0 R and θ 0 with θ 0 [ π/2, π/2] such that (x 0, y 0, z 0 ) = ρ 0 ñ(0, θ 0 ). Finally, suppose (3) holds. Then we have x 0 = 0. Thus (3) implies (). 3. Focal conics Proposition 3.2. The points (x 0, y 0, z 0 ) at which ϕ has at least A 2 -singularity (valid for the rest of the paper an A k -singularity ) at (0, 0) form a conic in the normal plane. Proof. Assume that ϕ has at least an A -singularity at (0, 0), that is, x 0 = 0. Then the determinant of the Hessian of ϕ at (0, 0) is given by ϕ uu (0, 0) ϕ uv (0, 0) ϕ uv (0, 0) ϕ vv (0, 0) = c0 20 c 0 c 0 c 0 02 Therefore, the locus of the equation = y 0 2 2a y 0 z 0 + (a 20 a 02 a 2 )z 0 2 a 02 z 0. y 2 2a yz + (a 20 a 02 a 2 )z 2 a 02 z = 0 is the set of the points at which ϕ is an A k -singularity at (0, 0). proof. Thus we complete the 54

21 Lemma 3.3. The function ϕ has an A k -singularity at (0, 0) if and only if (x 0, y 0, z 0 ) = (0, 0, 0) or (x 0, y 0, z 0 ) = ( κ (0, θ 0 )ñ(0, θ 0) with κ (0, θ 0 ) 0, where θ 0 π 2, π ). 2 Proof. Suppose that ϕ is at least an A -singularity at (0, 0). By Proposition 3., we have (x 0, y 0, z 0 ) = ρ 0 ñ(0, θ 0 ) where ρ 0 R and θ 0 [ π/2, π/2]. Substituting this into the equation y 2 0 2a y 0 z 0 + (a 20 a 02 a 2 )z 2 0 a 02 z 0 = 0, we obtain cos2 θ 0 + (a cos θ 0 + a 02 sin θ 0 ) ρ 0 = 0, or ρ 0 = 2 (a 20 cos 2 θ 0 a 02 sin 2. θ 0 ) sec θ 0 When ρ 0 = 0, the point (x 0, y 0, z 0 ) coincides with (0, 0, 0). In the later case, ρ 0 coincides with the principal radius / κ (0, θ 0 ). For this reason, we call the set of the points (x, y, z) at which ϕ has an A k -singularity at (0, 0) the focal conic of Whitney umbrella. Focal conics are classified into three types as shown in Figure. The following proposition provides a classification of focal conics. Proposition 3.4. () The focal conic is an ellipse if and only if a 20 a 02 < 0. (2) The focal conic is a hyperbola if and only if a 20 a 02 > 0, in which case its asymptotes are parallel to y + (a ± a 20 a 02 )z = 0. (3) The focal conic is a parabola if and only if a 20 = 0, in which case its axis of symmetry is parallel to y + a z = 0. Proof. Remark that the focal conic is the zero locus of F C(y, z) = y 2 2a yz + (a 20 a 02 a 2 )z 2 a 02 z. Firstly, we assume that a Replacing y and z by y a /(2a 20 ) and z + /(2a 20 ), respectively. Then the equation F C(y, z) = 0 has the form a 02 4a 20 (y + a z) 2 + a 20 a 02 z 2 = 0. This form implies the assertion () and (2). Next, we assume that a 20 = 0. Then the equation F C(y, z) = 0 reduces to This implies the assertion (3). (y + a z) 2 a 02 z = 0. The following proposition provides properties of the focal conic. It is easy to verify this proposition and we omit its proof. Proposition 3.5. Let g be given in the normal form of Whitney umbrella and let C k denote the circle centred at the origin with radius /k in the normal plane. 55

22 Figure. : The three types of focal conics: focal ellipse left, focal parabola center, focal hyperbola right. () The y-axis is tangent to the focal conic at the origin. (2) The circle C k is a subset of the front at distance /k. (3) For k > 0 we have #Σ k X = #C k {(y, z) ; F C(y, z) = 0} where X is the exceptional set X = π (0, 0) on the Möbius strip M. Since #C k {(y, z) ; F C(y, z) = 0} is at most four, #Σ k X is also at most four (cf. Proposition 2.6). Moreover, we have C k {(y, z) ; F C(y, z) = 0} = Rñ(0, θ) {(y, z) ; F C(y, z) = 0} \ {(0, 0)} = θ: κ (0,θ)=k θ: κ (0,θ)=k ñ(0, θ). k If ñ(0, θ) is parallel to the axis of symmetry of the focal parabola or the asymptotes of the focal hyperbola, then Rñ(0, θ) {(y, z) ; F C(y, z) = 0} \ {(0, 0)} = and κ (0, θ) = 0, that is, (0, θ) is a parabolic point over Whitney umbrella. (4) The circle C k is tangent to the focal conic at ñ(0, θ)/k if and only if (0, θ) a is subparabolic point relative to ṽ 2 over Whitney umbrella. For any focal conic, there exists at least one and at most three values of k such that C k is tangent to the focal conic. This implies that the number of the sub-parabolic points relative to ṽ 2 over Whitney umbrella is at least one and at most three (cf. Lemma 2.3). (5) The origin of the normal plane is the vertex of the focal conic if and only if a = 0. In this case, we have κ (0, θ) = κ (0, θ), and (r, θ) = (0, 0) is a sub-parabolic point relative to ṽ 2. (6) When the focal conic is a parabola, (r, θ) = (0, 0) is parabolic (in fact, κ (0, 0) = 0) and sub-parabolic relative to ṽ 2 over Whitney umbrella (i.e., Γ 3(0) = 0). We obtain the following corollary by Propositions 2.3 and

23 Corollary 3.6. () There is no parabolic point over Whitney umbrella, that is, the parabolic line dose not meet with X, if and only if the focal conic is an ellipse, in which case we have the hyperbolic Whitney umbrella. (2) There is one parabolic point over Whitney umbrella, that is, the parabolic line meets with X at one point, if and only if the focal conic is a parabola. In this case the intersection point of the parabolic line and X is a sub-parabolic point relative to ṽ 2. Furthermore, the parabolic line is tangent to X at the point if and only if the point is not a ridge point relative to ṽ. In this case, we have the parabolic Whitney umbrella. (3) There are two parabolic points, that is, the parabolic line meets with X at two points, if and only if the focal conic is a hyperbola, in which case we have the elliptic Whitney umbrella. 3.2 Versality of distance squared unfolding We do not repeat here the definition of versal unfolding, which is fundamental in singularity theory. Please refer to [] for elegant explanation, [7] for elementary introduction, and [25] for carefully prepared survey. The notation in [25] becomes the standard in singularity theory. Theorem 3.7. Suppose that g : (R 2, 0) (R 3, 0) is given in the normal form of Whitney umbrella. Assume that Φ : (R 2, 0) R 3 R is the distance squared function defined by (3.) and that ϕ(u, v) = Φ(u, v, x 0, y 0, z 0 ) where (x 0, y 0, z 0 ) is a point in R 3. () Suppose that (x 0, y 0, z 0 ) = ñ(0, θ 0 )/ κ (0, θ 0 ) (0, 0, 0) with κ (0, θ 0 ) 0 and t 0 2 = x y z 0 2, where θ 0 ( π/2, π/2). (a) The function ϕ(u, v) has an A 2 -singularity at (0, 0) if and only if (0, θ 0 ) is not a ridge point relative to ṽ over Whitney umbrella. In this case, Φ is R + and K-versal unfolding of ϕ. (b) The function ϕ has an A 3 -singularity at (0, 0) if and only if (0, θ 0 ) is a first order ridge point relative to ṽ over Whitney umbrella. In this case, Φ is an R + -versal unfolding of ϕ. Moreover, Φ is a K-versal unfolding of ϕ if and only if (0, θ 0 ) is not a sub-parabolic point relative to ṽ 2 over Whitney umbrella. (2) Suppose that (x 0, y 0, z 0 ) = (0, 0, 0) and t 0 = 0. Then ϕ has an A 3 -singularity at (0, 0). In this case, Φ is neither an R + -versal nor a K-versal unfolding of ϕ. Proof. Let us use expansions of Φ and ϕ as in (3.2) and (3.3), respectively. We remark that the coefficient a 02 appearing in the normal form of Whitney umbrella is not zero. () We first prove the condition for the point (0, 0) to be an A 2 or A 3 -singularity of ϕ. Lemma 3.3 now shows that ϕ has an A k -singularity at (0, 0). By (2.5) and (2.9), we have ( (x 0, y 0, z 0 ) = 0, (a cos θ 0 + a 02 sin θ 0 ) cos θ 0 cos 2 ) θ 0 a 20 cos 2 θ a 02 sin 2, θ 0 a 20 cos 2 θ 0 a 02 sin 2 (0, 0, 0). θ 0 Simple calculations show that ( ) ( c 0 20 c 0 a 02 c 0 c 0 = 02 a 20 cos 2 θ 0 a 02 sin 2 θ 0 57 sin 2 θ 0 cos θ 0 sin θ 0 cos θ 0 sin θ 0 cos 2 θ 0 ).

24 Taking s and (ξ, η) so that ( ) ( ) η 2 ξη c 0 s ξη ξ 2 = 20 c 0 c 0 c 0, 02 we obtain s = a 02 a 20 cos 2 θ 0 a 02 sin 2 θ 0 and (ξ, η) = (cos θ 0, sin θ 0 ). Setting c = 2s(c 0 2ξ 2 + 2c 0 2ξη + c 0 03η 2 )/ξ 4 and replacing v by v + (η/ξ)u cu 2, we have where Φ = c 00 (x, y, z) + xu + 4 i+j=2 i!j!ĉij(x, y, z)u i v j + O(u, v) 5, ĉ 20 = ξ 2 (c 20ξ 2 + 2c ξη + c 02 η 2 ), ĉ = ξ (c ξ + c 02 η), ĉ 02 = c 02, ĉ 30 = ( ) sξ 5 sξ 2 (c 30 ξ 3 + 3c 2 ξ 2 η + 3c 2 ξη 2 + c 03 η 3 ) 3(c ξ + c 02 η)(c 0 2ξ 2 + 2c 0 2ξη + c 0 02η 2 ) ĉ 2 = [ sξ 2 sξ 4 (c 2 ξ 2 + 2c 2 ξη + c 02 η 2 ) c 02 (c 0 2ξ 2 + 2c 0 2ξη + c 0 02η 2 ) ], ĉ 2 = ξ (c 2ξ + c 03 η), ĉ 03 = c 03, ĉ 40 = ξ 4 (c0 40ξ 4 + 4c 0 3ξ 3 η + 6c 0 22ξ 2 η 2 + 4c 0 3ξ 3 η + c 0 04η 4 ) 3 sξ 6 (c0 2ξ 2 + 2c 0 2ξη + c 0 03η 2 ) [ 2(c 2 ξ 2 + 2c 2 ξη + c 02 η 2 ) (c 0 2ξ 2 + 2c 0 2ξη + c 0 03η 2 ) ]. Therefore, ϕ is expressed as ϕ = 2ĉ0 02v (ĉ0 30u 3 + 3ĉ 0 2uv 2 + ĉ 0 03v 3 ) + i+j=4 i!j!ĉ0 iju i v j + O(u, v) 5, where ĉ 0 ij = ĉ ij(x 0, y 0, z 0 ). By this form, ϕ does not have D 4 or worse singularities. This form also shows that ϕ has an A 2 or A 3 singularity for (x 0, y 0, z 0 ) at (0, 0) if and only if ĉ , or ĉ 0 30 = 0 and ĉ , respectively. After some computations, we extract that ĉ 0 30 = Γ 3 (θ 0 ) sec 2 θ 0 a 20 cos 2 θ 0 a 02 sin 2 and ĉ 0 Γ 4 (θ 0 ) sec 4 θ 0 40 = θ 0 a 02 (a 20 cos 2 θ 0 a 02 sin 2 θ 0 ). Therefore, from Lemma 2.6 it follows that ϕ has an A 2 or A 3 -singularity at (0, 0) if and only if (0, θ 0 ) is not a ridge point relative to ṽ or a first order ridge point relative to ṽ, respectively. We now turn to the versality of Φ. Firstly, we prove Case (a). Suppose that (0, 0) is an A 2 -singularity. We remark that A 2 -singularity is 3-determined. To show the R + -versality and the K-versality of Φ, it is enough to verify that, respectively, E 2 = ϕ u, ϕ v E2 + Φ x R 2 p 0, Φ y R 2 p 0, Φ z R 2 p 0 R + R + u, v 4 E 2, and (3.4) 58

25 E 2 = ϕ u, ϕ v, ϕ E2 + Φ x R2 p 0, Φ y R2 p 0, Φ z R2 p 0 R + u, v 4 E 2, (3.5) where p 0 = (x 0, y 0, z 0 ). The coefficients of u i v j of functions appearing in (3.4) and (3.5) are given by the following table: u v u 2 uv v 2 u 3 u 2 v uv 2 v 3 Φ x Φ y y α 20 0 α 30 α 2 α 2 α 03 Φ z z β 20 β β 02 β 30 β 2 β 2 β 03 ϕ u ĉ ĉ0 2 2ĉ0 03 ϕ v 0 0 ĉ ĉ 0 2 ϕ ĉ0 02 6ĉ0 40 2ĉ0 3 2ĉ0 22 6ĉ0 3 2ĉ0 22 2ĉ0 3 6 c ĉ0 2 6ĉ0 3 6ĉ0 04 6ĉ0 03 uϕ u ĉ ĉ0 2 0 vϕ u ĉ uϕ v ĉ ĉ 0 2 vϕ v ĉ ĉ 0 2 2ĉ0 2 2ĉ u 2 ϕ v ĉ uvϕ v ĉ v 2 ϕ v ĉ ĉ0 03 Here, α ij = ĉ ij y (x 0, y 0, z 0 ), β ij = ĉ ij z (x 0, y 0, z 0 ), and boxed elements are non-zero. Hence, Gauss s elimination method using boxed elements as pivots leads to that the matrix presented by this table is full rank. Thus the equality (3.5) holds. The case of (3.4) is similar. Next, we consider Case (b). Suppose that (0, 0) is an A 3 -singularity. We remark that A 3 -singularity is 4-determined. To show the R + -versality and the K-versality of Φ, it is enough to verify that E 2 = ϕ u, ϕ v E2 + Φ x R 2 p 0, Φ y R 2 p 0, Φ z R 2 p 0 R + R + u, v 5, and (3.6) E 2 = ϕ u, ϕ v, ϕ E2 + Φ x R2 p 0, Φ y R2 p 0, Φ z R2 p 0 R + u, v 5, (3.7) respectively. The coefficients of u i v j of functions appearing in (3.6) and (3.7) are given by 59