Geometry of Skeletal Structures and Symmetry Sets

Transcription

1 Geometry of Skeletal Structures and Symmetry Sets Azeb Zain Jafar Alghanemi Submitted in accordance with the requirements for the degree of Doctor of Philosophy The University of Leeds Department of Pure Mathematics June 2012 The candidate confirms that the work submitted is his own and that appropriate credit has been given where reference has been made to the work of others. This copy has been supplied on the understanding that it is copyright material and that no quotation from the thesis may be published without proper acknowledgement.

2 i Abstract In this thesis we study the geometry of symmetry sets and skeletal structures. The relationship between a symmetry point (skeletal point) and the associated midlocus point is studied and the impact of the singularity of the radius function on this relationship is investigated. Moreover, the concept of the centroid set associated to a smooth submanifold of R n+1 is introduced and studied. Also, the relationship between the shape operator of a skeletal structure at a smooth point and the shape operator of its boundary at the associated point is studied.

3 ii Introduction The idea of describing objects using the concept of medial axis was suggested by Harry Blum. Significant developments in describing many biological and physical objects using medial axis and symmetry set have been seen in the last century. In fact, the medial axis is a subset of a large set called the symmetry set. The concept of symmetry set and medial axis has been studied and developed by Peter Giblin, Bruce and others and a considerable mathematical investigation can be found in [4, 6, 11, 13]. In 2003, James Damon invented and developed the concept of skeletal structures of an object in an attempt to create its smooth boundary. A significant part of Damon s contribution is the radial shape operator which plays a central role in determining the differential geometry of the boundary of a skeletal structure [7, 8, 9]. In the real life, the symmetry set and the medial axis play a central role in many applications such as object recognition, object reconstruction and medical imaging and some of these applications can be found in [25]. In this thesis we focus only on the mathematical aspects of symmetry sets, medial axis, skeletal structures and their boundaries. The impact of the singularities of the radius function on the relationship between symmetry sets, medial axis, skeletal structures and their boundaries will be studied in this thesis. This thesis consists of six chapters and before describing those chapters we give the following definition. Definition: A map f : N n M m is singular at x 0 N n if the rank of the Jacobian matrix of f at x 0 is less than min (n, m). Now we give a brief description of each chapter of this thesis. In chapter one we give some basic definitions and theorems in the field of skeletal structures which will

4 iii be used in subsequent chapters. Chapter two deals with the symmetry set of a smooth hypersurface in R n+1. It consists of three parts. The first part deals with the creating of the symmetry set from its boundary. The second main part of chapter two deals with the reconstruction of the boundary using the information given by the symmetry set and the associated radius function. In this part we study the impact of the singularity of the radius function on the relationship between the symmetry point and its associated midlocus point. In fact, this study is a generalization of what Peter Giblin pointed out in the relationship between the normals of a plane curve at tangency points associated to a smooth point of its symmetry set [16]. The third main part of this chapter deals with creating the symmetry set from the associated midlocus and radius function and in this part we generalize what Peter Giblin and Paul Warder did in [16, 32]. Before giving the main result of this chapter we give the following definition. Definition B: Let x 0 be a non-singular point of the symmetry set of a region Ω in R n+1, with smooth boundary X. Let x 1 and x 2 be the tangency points of the boundary associated to x 0. Then the midlocus point is given by x m = 1 2 (x 1 + x 2 ). The main result of chapter two is the following theorem. Theorem A: Let S be the symmetry set of a region Ω in R n+1, with smooth boundary X. Let x 0 be a non-singular point of S. Then x 0 and the associated midlocus point x m coincide if and only if the radius function has a singularity at x 0. In chapter three we introduce the concept of the centroid set associated to a smooth submanifold M of R n+1 which is more general than the midlocus and depends on a multivalued radial vector field defined on M such that each value of the multivalued radial vector field forms a smooth radial vector field on M and each smooth radial vector

5 iv field has a smooth radius function. This chapter consists of four main parts. In the first part the centroid set associated to a smooth submanifold of R n+1 is defined and the impact of the singularities of the associated radius function is studied. The second part of this chapter deals with the impact of the singularities of the radius function on the relationship between a smooth skeletal point and its associated midlocus point. In the third part of chapter three we define the pre-medial axis in R n+1 and study the relationship between the parameters of the boundary of a medial axis at the tangency points associated to a smooth point of the medial axis. The fourth part of this chapter deals with the classification of the singularity of the midlocus of a skeletal structures in R 3. The main result of chapter three is the following. Theorem B: Let M be a smooth stratum of a skeletal structure (S, U) in R 3 containing a smooth point x 0 and r be the radius function with a singularity at x 0 and λ 1 and λ 2 be the eigenvalues of the Hessian of r, and w 1 and w 2 are the associated eigenvectors such that λ 1 λ 2, and r(x 0 ) = 1 λ 1, λ 1 0. Then the midlocus at x m associated to x 0 is A-equivalent to the crosscap if and only if λ 1 k x0 (w 1 ) 2 w 1 w2 r 2λ 2 τ g 3 w 1 r, where k x0 (w 1 ) is the normal curvature of M in the direction w 1, τ g is the geodesic torsion of M in the direction w 1, and wi r is the directional derivative of the radius function in the direction w i, i = 1, 2. The fourth chapter of this thesis deals with the relationship between the radial shape operator of a skeletal structure and the differential geometric shape operator of the associated boundary. In [8] James Damon expressed the matrix representing the differential geometric shape operator in terms of the matrix representing the radial shape operator. In this chapter we express the matrix representing the radial shape operator in terms of the matrix representing the differential geometric shape operator of the

6 v boundary. Also, the relationship between the principal radial curvatures, Gaussian radial curvature, mean radial curvature of a skeletal structure and the associated principal curvatures, Gaussian curvature, mean curvature of the boundary is pointed out through this chapter. The main result of this chapter is the following. Theorem C: Let (S, U) be a skeletal structure in R 3 such that for a choice of smooth value of U the associated compatibility 1-form η U vanishes identically on a neighbourhood of a smooth point x 0 of S, and 1 r is not an eigenvalue of the radial shape operator at x 0. Let x 0 = Ψ 1 (x 0 ) and V be the image of V under dψ 1 for a basis {v 1, v 2 }, then or equivalently S XV = S V = 1 r 2 K r 2rH r + 1 (S V rk r I) 1 r 2 K + 2rH + 1 (S XV + rki). Here S V is the matrix representing the radial shape operator, K r (resp. H r ) is the Gaussian (resp. mean) radial curvature of the skeletal structure, S XV is the matrix representing the differential geometric shape operator of the boundary and K (resp. H) is the Gaussian (resp. mean) curvature of the boundary. The fifth chapter of this thesis is devoted to study the relationship between the shape operator of skeletal structures and the associated shape operator of the boundary. In the first main part of this chapter we study the relationship between the curvature of a skeletal structure and the curvature of its associated boundary in the plane. Also, the relationship between the curvatures of the boundary at the tangency points associated to a smooth point of medial axis in the plane has been studied. In second main part of chapter five we express the matrix representing the shape operator of the medial axis in R n+1 at a smooth point in terms of the matrices representing the shape operators of the boundary at the associated tangency points. The main result of chapter five is given in the following.

7 vi Theorem D: Let (S, U) be a skeletal structure such that for a choice of smooth value of U the associated compatibility 1-form η U vanishes identically on a neighbourhood of a smooth point x 0 of S, and 1 is not an eigenvalue of the radial shape operator at x r 0. Let x 0 = Ψ 1 (x 0 ), and V be the image of V for a basis {v 1, v 2,..., v n }. Then the matrix S XV representing the differential geometric shape operator of the boundary is given by S T XV = 1 r {[I r(ht r + ρs T m 1 ρ dρdrt I 1 m + 1 ρ ST mdrdr T Im 1 1 )] I}, where S m is the matrix representing the differential geometric shape operator of S at x 0, I m is the first fundamental form of S at x 0 and H r is the matrix representing the Hessian radial operator at x 0. The last chapter of this thesis deals with the focal point of the boundary associated to a skeletal structure. In this chapter we define the radial focal point of a skeletal structure and we show that this point coincides with the focal point of the boundary. Moreover, the location of the focal point of the boundary associated to a Blum medial axis in R n+1 is investigated through this chapter.

8 vii This work is dedicated with great respect and deep affection to my parents, wife and lovely children.

9 viii Acknowledgements No scientific progress is possible without standing on the shoulders of giants, and it was my good fortune to have Dr. Kevin Houston as a supervisor. I am greatly indebted to him for his inspiring, enthusiastic supervision, willingness to help and devoting so much of his valuable time during the time of preparing this thesis. Also, it is my pleasure to thank Dr. Martin Speight for being a co-supervisor. Special thanks go to professor Peter Giblin for giving me some of his papers in symmetry sets and for invaluable mathematical discussion during singularity days and a conference in Bedlewo. I am also greatly indebted to my beloved parents and wife for their encouragement and support. Last but not the least, I gratefully acknowledge financial support from King Abdulaziz University.

10 ix Contents Abstract Introduction Dedication Acknowledgements Contents List of figures i ii vii viii ix xii 1 Background Whitney Stratification Skeletal Set and Skeletal Structure Radial Shape Operator Edge Radial Shape Operator Compatibility 1-Form and Compatibility Condition The Radial Map The Sufficient Conditions for Smooth Boundary Blum Medial Axis

11 CONTENTS x 2 Symmetry Set in R n Introduction Creating Symmetry Set from the Boundary Creating the Boundary from the Symmetry Set Singularity of the Radius Function at a Singular Point of the Symmetry Set Creating the Symmetry Set from the Midlocus Centroid Set, Skeletal Structure and the Singularity of the Radius Function Introduction Centroid Points The Skeletal Structure and the Singularity of the Radius Function Pre-medial Axis Singularity of the Midlocus in the Case of Skeletal Structure in R Relation Between Radial Geometry of Skeletal Structure and Differential Geometry of its Boundary Introduction Skeletal Structures in R n Skeletal Structures in R The Relationship Between the Differential Geometry of the Skeletal Structure and that of the Boundary Introduction

12 CONTENTS xi 5.2 The Differential Geometry of the Skeletal Structure and its Boundary in the Plane Shape Operator of the Blum Medial Axis and those of its Boundary at Tangency Points Blum Medial Axis and the Singularity of the Associated Midlocus Radial Focal Point of a Skeletal Set and Focal Point of the Boundary Introduction Location of the Focal Point of the Boundary Creating the Focal Points of the Boundary from Skeletal Structures Appendix 151 A B The Maple Code Used in Calculating the 2-jet of the Midlocus in Chapter The Maple Code Used in Calculating the 3-jet of the Midlocus in Chapter Bibliography 160

13 xii List of figures 1.1 The radial shape operator in 3D. The dashed line denotes projection onto T x0 S along U The edge shape operator in 3D. The dashed line denotes projection onto T x0 S N along U Skeletal set and associated boundary in example Local generic structure for Blum medial axis in R 3 and the associated radial vector fields The symmetry point Symmetry set and associated midlocus in example Symmetry set and associated midlocus in example The symmetry point and the associated tangency points in the case of curve Triangle formed by symmetry point and its associated tangency points Figure of example A schematic diagram of example The case when the radius function has a singularity

14 LIST OF FIGURES xiii 3.4 Crosscap or Whitney umbrella Figure of example Figure of example Figure of example Figure of example in the case, The possible positions of the radial vector field when the radius function has no singularity The Blum medial axis and associated boundary of example The Blum medial axis and associated boundary of example Skeletal set and associated boundary in example Skeletal set and associated boundary in example

15 1 Chapter 1 Background Skeletal structures of a smooth boundary has been studied by James Damon in several papers [7, 8, 9] In this chapter we give some basic definitions and theorems in the field of Whitney stratifications and skeletal structures which will be used in the subsequent chapters. Also, the radial and edge shape operators will be reviewed in this chapter. 1.1 Whitney Stratification Definition [18] Let S be a closed subset of a smooth manifold M and let S be decomposed into disjoint smooth submanifolds (possibly with boundary) S i called strata. Then the decomposition is called a Whitney Stratification if the following conditions are met 1. S i Sj ϕ if and only if S i S j for strata S i, S j with i j, this is called the frontier condition. 2. Whitney condition (a): if x i is a sequence of points in S a converging to y S b and T xi (S a ) converges to a plane τ (all this considered in the appropriate Grassmannian), then T y (S b ) τ.

16 Chapter 1. Background 2 3. Whitney condition (b): if x i and y i are two sequences in S a and S b respectively converge to y S b, l i denotes the secant line between x i and y i and l i converges to l then l τ. Remark Condition b implies condition a. triangulated (see [7]). Any Whitney stratified set can be Definition For a Whitney stratified set S we let 1. S reg denote the points in the top-dimensional strata and these points are the smooth points of S. 2. S sing denote the remaining strata. 3. S denote the subset of S sing consisting of points of S at which S is locally an n- manifold with boundary. We refer these points as edge points in order to distinguish between S and the boundary of the region of the skeletal structure. 4. S denote the closure of S. Definition let S be a Whitney stratified set and let x 0 S sing then we define the following 1. The complementary local components for x 0 are the connected components of B ε (x 0 ) \ S. 2. The neighbouring local components of x 0 are the connected components of B ε (x 0 ) S reg, where B ε (x 0 ) is a closed ball of radius ε about x 0 for sufficiently small ε > 0.

17 Chapter 1. Background Skeletal Set and Skeletal Structure Definition [7] An n-dimensional Whitney stratified set S R n+1 is a skeletal set if 1. For each local neighbouring component S α of x 0 S β there is a unique limiting tangent space T x0 S β from sequence of points in S α (by properties of Whitney stratified set T x0 S β T x0 S α ). 2. Locally in a neighbourhood of a singular point x 0, S may be expressed as a union of (smooth) n-manifolds with boundaries and corners S j, where two such intersect on boundary facets. 3. If x 0 S then those S j in 2 meeting S meet it in an (n 1)-dimensional facet. Facets means edges or faces in the triangulation of remark Definition [7] An edge coordinate parametrization at an edge point x 0 S consists of an open neighbourhood W of x 0 in S, an open neighbourhood W of 0 in R n + = {(x 1, x 2,..., x n ) R n : x n 0} and a differentiable homeomorphism Φ : W W such that: both Φ {(x 1, x 2,..., x n ) W : x n > 0} and Φ ( W R n 1 ) are diffeomorphisms on to their images. Definition Given an n-dimensional set S R n+1, a radial vector field U on S is nowhere zero multivalued vector satisfying the following conditions. 1. (Behaviour at smooth points) For each x 0 S reg, there are two values of U which are on opposite sides of T x0 S i.e., their dot products with a normal vector are non zero with opposite signs. Moreover, on a neighbourhood of a point of S reg, the values of U corresponding to one side form a smooth vector field.

18 Chapter 1. Background 4 2. (Behaviour at a non-edge singular point) Let x 0 be a non-edge singular point with S α a local component of x 0. Then both smooth values of U on S α extend smoothly to values U(x 0 ) on the stratum of x 0. If S α does not intersect S in a neighbourhood of x 0, then U(x 0 ) does not belong to T x0 S. Conversely to each value of U at a point x 0 S sing, there corresponds a local complementary component C i of S of S at x 0 such that the value U(x 0 ) locally points into C i in the following sense. The value U(x 0 ) extends smoothly to values U(x) on the local complementary components of S for x 0 in C i. For a neighbourhood W of x 0 and an ε > 0, x + tu(x) C i for 0 < t < ε and x (W S). 3. (Tangency behaviour at edge points) At edge points x 0 S there is a unique value for U tangent to the stratum of S reg containing x 0 in the closure which points away from S. Definition Given a skeletal set S and a smooth multivalued radial vector field U, the radial flow is defined by Ψ t (x) = x + tu(x), where x S and t [0, 1]. Definition A radial vector field U on a skeletal set S satisfies the local initial conditions if it satisfies the following. 1. (Local separation property) For a local complementary component C i of a non-edge point x 0 / S, let C i = S i denoting the local decomposition of C i into closed (in W ) n-manifolds with boundaries and corners. Then the set X = {x + tu(x) : x i S i, 0 t ε} is an embedded Whitney stratified set such that distinct int(s i ) and int(s j ) lie in separate connected components of the complement of C i \ X.

19 Chapter 1. Background 5 2. (Local edge property) For each edge closure point x 0 S there is a neighbourhood w of x 0 in S and ε > 0 so that for each smooth value of U, the radial flow Ψ(x, t) = x + tu is one-one on w [0, ε]. Definition For a radial vector field U, we put U = ru 1, for a positive multivalued function r, and a multivalued unit vector field U 1 on S. We will call r the radius function. Now suppose C i is a local complementary component of a singular point x 0. The local boundary of C i in a small open neighbourhood can be expressed as a union of n-manifolds with boundary and corner {S i, i = 1, 2,..., k}. The abstract boundary of C i consists of a copy of S i for each smooth value of U on S i pointing into C i [7]. Definition A skeletal structure (S, U) in R n+1 consists of an n-dimensional skeletal set and radial vector field U on S satisfying the local initial conditions, such that all abstract boundaries of local complementary components are homeomorphic to n-disks. Definition Given a skeletal structure (S, U), the associated boundary is defined by X = {x + U(x) : x S}, where the definition includes all values of U(x) for a given x. 1.3 Radial Shape Operator Definition Given a skeletal structure (S, U) in R n+1 we define for a regular point x 0 S and each smooth value of U defined in a neighbourhood of x 0 with associated unit vector field U 1, a radial shape operator S rad (v) = proj U ( U 1 v ), for v T x 0 S, (1.1) where U 1 v means vu 1 and proj U denotes projection onto T x0 S along U. Also, if {v 1, v 2,..., v n } is a basis for T x0 S then, U 1 v i = a i U 1 n s ji v j (1.2) j=1

20 Chapter 1. Background 6 which can be written in the vector form by U 1 V = A V.U 1 S T V V. (1.3) proj U U 1 v U 1 v U 1 T x0 S x 0 S rad (v) S Figure 1.1: The radial shape operator in 3D. The dashed line denotes projection onto T x0 S along U Definition For x 0 S reg and a given smooth value of U, we call the eigenvalues of the associated radial shape operator the principal radial curvatures at x 0 and denote them by κ ri. Example Let (S, U) be a skeletal structure in R 3 and let s 1 (x, y) = (x, y, 1) S reg such that {(x, y) R 2 x 2 + y 2 < 1 }. Now define the radial vector field on the image of 4 s 1 by ( U = (x 2 + y 2 + 1) 2x, 2y, ) 1 4(x 2 + y 2 ) = ru 1,

21 Chapter 1. Background 7 where r(x, y) = x 2 + y is the radius function and ( U 1 = 2x, 2y, ) 1 4(x 2 + y 2 ). Let v 1 = s 1 x = (1, 0, 0) and v 2 = s 1 y = (0, 1, 0), then the unit normal of s 1 is N = (0, 0, 1). It is clear that U 1 is a smooth unit vector field on the image of s 1 and Now we have U 1 x = ( 2, 0, 4x But from equation 1.4 we N = Therefore, Similarly U 1 = 2xv 1 2yv (x 2 + y 2 )N. (1.4) 1 1 4(x2 + y 2 ) U 1 + U 1 x = U 1 y = 1 4(x2 + y 2 ) ) = 2v 4x 1 1 4(x2 + y 2 ) N. 2x 1 4(x2 + y 2 ) v 1 + 4x 1 4(x 2 + y 2 ) U 2 8y 2 1 ( 1 4(x 2 + y 2 ) )v 1 4y 1 4(x 2 + y 2 ) U 1 8xy 1 4(x 2 + y 2 ) v 1 2y 1 4(x2 + y 2 ) v 2. 8xy 1 4(x 2 + y 2 ) v x 2 1 4(x 2 + y 2 ) v 2 Now we can apply definition to evaluate the radial shape operator, thus the matrix representing the radial shape operator is given by and S V = Now from definition we have 2 8y 2 1 4(x 2 +y 2 ) 8xy 1 4(x 2 +y 2 ) A V = 4x 1 4(x 2 +y 2 ) 4y 1 4(x 2 +y 2 ) 8xy 1 4(x 2 +y 2 ) 2 8x 2 1 4(x 2 +y 2 )., κ ri = 1 2 {tr(s V ) ± tr 2 (S V ) 4det(S V )}. After some calculations we get κ r1 = 2 and κ r2 = 2 1 4(x 2 +y 2 ).

22 Chapter 1. Background Edge Radial Shape Operator Definition Let (S, U) be a skeletal structure and let x 0 S and let N be the unit normal vector field to S in a neighbourhood of x 0. Then, the edge shape operator is defined by S E (v) = proj U ( U 1 ), (1.5) v for v T x0 S and proj U denotes projection onto T x0 S N. T x0 S N U 1 v N S U S E (v) T x0 S Figure 1.2: The edge shape operator in 3D. The dashed line denotes projection onto T x0 S N along U. Now given a basis {v 1, v 2,..., v n 1 } of T x0 S we choose a vector v n in the edge coordinate system at x 0 so that {v 1, v 2,..., v n } is a basis of T x0 S in the edge coordinate system and so that v n maps under the edge parametrization map to cu 1 (x 0 ) where c 0. Then we can compute a matrix representation for the edge shape operator. Let N be a unit normal vector field to S on a neighbourhood w of x 0 then we have U 1 v i n 1 = a i U 1 c i N b ji v j. (1.6) j=1

23 Chapter 1. Background 9 This equation can be written in vector form by U 1 V = A U U 1 C U N B UV Ṽ (1.7) or or U ( 1 V = A U U 1 U 1 V = A U U 1 S T EV ) B UV C U Ṽ Ṽ N N (1.8). (1.9) Therefore, S EV is a matrix representation of the edge shape operator. Here, A U and C U are n-dimensional column vectors, B UV is an n (n 1)-matrix, and Ṽ is the (n 1)-dimensional vector with entries v 1, v 2,..., v n 1. Remark The basis {v 1, v 2,..., v n } of T x0 S in the edge coordinate system is called a special basis of T x0 S. Definition The principal edge curvatures are the generalized eigenvalues of the pair (S EV, I n 1,1 ) where, I n 1,1 denotes the (n n)-diagonal matrix with 1 s in the first n 1 diagonal positions and 0 otherwise. 1.5 Compatibility 1-Form and Compatibility Condition Definition Given a skeletal structure (S, U) the compatibility 1-form η U is defined by η U (v) = v U 1 + dr(v), (1.10) v is a tangent vector to S. S satisfies the compatibility condition at x 0 S with smooth value U if η U 0 at x 0. The compatibility condition plays a central role in the investigation of the differential

24 Chapter 1. Background 10 geometry of the boundary. A radial vector field plays the role of a normal vector of the boundary when S satisfies the compatibility condition. Lemma [7] Let (S, U) be a skeletal structure. Suppose that S α is a local manifold component of x 0 on which is defined a smooth value of the radial vector field U. Suppose that either 1 r is an eigenvalue of the radial shape operator if S α is a non-edge component or 1 r is not a generalized eigenvalue of the pair (S EV, I n 1,1 ) if S α is an edge component. If the associated compatibility 1-form η U vanishes at x 0 then U(x 0 ) is orthogonal to the portion of the boundary X (given by Ψ 1 (S α )) at Ψ 1 (x 0 ). 1.6 The Radial Map Definition Given a skeletal structure (S, U) with boundary X then the radial map is given by: Ψ 1 (x) = x + r(x)u 1 (x), x S (1.11) Example With the skeletal structure (S, U) as in example 1.3.3, we will calculate the boundary of this skeletal structure using the radial map Ψ 1 ; in fact, we see that for any point x 0 s 1 the associated boundary point x 1 is given by ( x 1 = x 0 + ru 1 = (x, y, 1) + (x 2 + y 2 + 1) 2x, 2y, ) 1 4(x 2 + y 2 ), and after some calculations we obtain x 1 = ( 2x 3 2xy 2 x, 2y 3 2x 2 y y, 1 + (x 2 + y 2 + 1) ) 1 4(x 2 + y 2 ). Now we will check the compatibility condition and to do so we have to check the dot product x 1 x U 1 = x 1 y U 1 = 0. Now x 1 x = ( 6x 2 2y 2 1, 4xy, 12x3 12xy 2 2x 1 4(x2 + y 2 ) ),

25 Chapter 1. Background 11 and x 1 y = ( 4xy, 6y 2 2x 2 1, 12y3 12x 2 y 2y 1 4(x2 + y 2 ) Thus, this boundary is smooth and it is clear that x 1 x U 1 = x 1 y U 1 = 0. Therefore, using lemma the compatibility 1-form vanishes identically in the given domain. ) The boundary The skeletal set Figure 1.3: Skeletal set and associated boundary in example The Sufficient Conditions for Smooth Boundary James Damon Discussed in [7, 8, 9] the sufficient conditions for the skeletal structure (S, U) to have a smooth boundary. These conditions are 1. (Radial Curvature Condition) For all points of S off S r < min{ 1 κ ri } for all positive principal radial curvature κ ri.

26 Chapter 1. Background (Edge Condition) For all points of of S (closure of S) r < min{ 1 κ Ei } for all positive principal edge curvature κ Ei. 3. (Compatibility Condition) For all singular points of S (which includes edge points) η U = 0. Theorem [7] Let (S, U) be a skeletal structure which satisfies the above three conditions. Then 1. The associated boundary X is an immersed topological manifold which is smooth at all points except those point corresponding to points of S sing. 2. At points corresponding to points of S sing it is weakly C 1 (this implies that it is C 1 on the points which are in the images of strata of codimension 1). 3. At smooth points, the projection along lines of U will locally map X diffeomorphically onto the smooth part of S. 4. Also, if there is no nonlocal intersection, X will be an embedded manifold. 1.8 Blum Medial Axis Definition Given a region Ω R n+1 with smooth boundary X, then the Blum medial axis of Ω is the locus of centers of hyperspheres tangent to the boundary X at least two points or having a single degenerate tangency such that these hypersphers are contained in Ω. Definition [7] The pair (S, U) consisting of the Blum medial axis and associated multivalued radial vector field is a special case of a skeletal structure which satisfies the following

27 Chapter 1. Background 13 At each smooth point x 0, the two values U (1) and U (2) must satisfy U (1) = U (2) and U (1) U (2) is orthogonal to T x0 S. Proposition [7] If (S, U) is a medial axis and radial vector field of a region Ω R n+1 with generic smooth boundary X, then (S, U) satisfies both the radial curvature and edge conditions. edge Y branching fin 6 junction Figure 1.4: Local generic structure for Blum medial axis in R 3 and the associated radial vector fields.

28 14 Chapter 2 Symmetry Set in R n Introduction This chapter is focused on the symmetry set of a smooth hypersurface in R n+1. It is divided into three main parts. The first part deals with the creating of the symmetry set using the boundary. In this part, we define the symmetry set in R n+1 and we study the smoothness of the symmetry set using the information provided by the principal curvatures of the boundary (Theorem 2.2.3). Also, the necessary and sufficient condition for two points on the boundary to form a symmetry point is discussed (Theorem 2.2.4). In the second part we consider the inverse procedure to that in the first part. In fact, this part deals with the reconstruction of the boundary using the information given by the symmetry set and the radius function (Theorem 2.3.1). Furthermore, the impact of the singularity of the radius function on the relationship between the symmetry set and the associated midlocus is investigated (Theorem 2.3.4). The last part of this chapter is focused on the creating of the symmetry set using the information provided by the midlocus and the radius function.

29 Chapter 2. Symmetry Set in R n Creating Symmetry Set from the Boundary The symmetry set and its smooth boundary in R 2 and R 3 have been studied intensively by Giblin and others in several papers such as [4, 6, 11, 13, 14, 15]. In this section we define the symmetry set of a smooth boundary in R n+1 in the same way as Giblin and then we generalize some results to the higher dimensions. Definition Given a smooth hypersurface X in R n+1 the symmetry set S is the locus of centres of hyperspheres, bitangent to X. I.e., if x 1 = X(s) and x 2 = X(t) are two points of the tangency with a hypersphere then the corresponding point of the symmetry set S is given by x 0 = x 1 + rn 1 = x 2 + rn 2, where r is the radius function, N i, i = 1, 2 are the unit normals of X at x i, s = (s 1, s 2,..., s n ) and t = (t 1, t 2,..., t n ) pointing towards the centre of the hypersphere. X 1 x 1 rn 1 x 0 rn 2 x 2 X 2 Figure 2.1: The symmetry point.

30 Chapter 2. Symmetry Set in R n+1 16 Now let X 1 and X 2 be two pieces of smooth hypersurface X parametrized locally by s = (s 1, s 2,..., s n ) and t = (t 1, t 2,..., t n ) respectively. Define the function f : R 2n+1 R n+1 by f(s, t, r) = (X 1 (s) X 2 (t)) + r(n 1 (s) N 2 (t)). Then f = 0 when a hypersphere of radius r is tangent to X 1 and X 2. We expect f 1 (0) to be a smooth manifold with dimension n. Define the centre map C by: C : f 1 (0) R n+1 C(s, t, r) = X 2 (t) + rn 2 (t). Then clearly, C(f 1 (0)) is the symmetry set. Hence C projects f 1 (0) to R n+1, therefore the condition for f 1 (0) to project to a smooth hypersurface in R n+1 is C to be an immersion. Now since X 1 and X 2 are oriented, we can choose orthonormal bases for their tangent spaces using the principal directions. Proposition Assume as above, then 1. f 1 (0) is a smooth submanifold of R 2n+1 parametrized by s = (s 1, s 2,..., s n ) provided κ i 1, i = 1, 2,..., n. r 2. f 1 (0) is a smooth submanifold of R 2n+1 parametrized by t = (t 1, t 2,..., t n ) provided λ i 1 r, i = 1, 2,..., n, where κ i ( resp. λ i ) are the principal curvatures of X 2 (resp. X 1 ). Let p 0 and q 0 be tangency points (p 0 X 1 and q o X 2 ) let {v i }, i = 1, 2,..., n (resp. {u i }, i = 1, 2,..., n) be a basis for the tangent space of X 2 (resp. X 1 ) formed by the

31 Chapter 2. Symmetry Set in R n+1 17 principal directions. Now using the fact that differentiating the normal in the principal direction produces the principal curvature times the principal direction. Therefore, the Jacobian matrix of f has the following column vectors (1 rκ i )v i, (1 rλ i )u i, and (N 1 N 2 ). N 1 N 2 is parallel to p 0 q 0, thus N 1 N 2 0. Also, N 1 N 2, u 1, u 2,..., u n are linearly independent as well as N 1 N 2, v 1, v 2,..., v n. Therefore, using the implicit function theorem the result holds. Now we will give the necessary and sufficient condition for the tangency points to give a smooth point on the symmetry set. Theorem The symmetry set C(f 1 (0)) is a smooth hypersurface if κ i λ i 1, i = 1, 2,..., n. r 1 r and C(s, t, r) = X 2 + rn 2. The condition for f 1 (0) to project to a smooth hypersurface in R n+1 is that C is an immersion, or equivalently the kernel of Df intersects the kernel of DC in zero. Now let C(s, t, r) = X 2 + rn 2. Then the Jacobian matrix of C has the column vectors (1 rκ i )v i, 0 and N 2, and the Jacobian matrix of f has the column vectors (1 rκ i )v i, (1 rλ i )u i and N 1 N 2.

32 Chapter 2. Symmetry Set in R n+1 18 Now consider the following (1 rκ i)v i (1 rλ i )u i N 1 N 2 (1 rκ i )v i 0 N 2 ξ 1. ξ ṇ. = 0 ξ 2n+1 Therefore, the condition for C to be an immersion implies that ξ j = 0, j = 1, 2,..., n, n + 1,..., 2n, 2n + 1. Since N 1 N 2, we have ξ 2n+1 = 0. From proposition 2.2.2, we have f 1 (0) is smooth provided λ i 1 or κ r i 1. Now assume r that κ i 1 r, then n ξ i (1 rκ i )v i = 0. Now since v i are linearly independent then ξ i = 0, (i = 1, 2,..., n). Hence, if λ i 1 r then, ξ j = 0 where j = n + 1, n + 2,..., 2n, and by this the proof is completed. Now the natural question is: what is the necessary and sufficient condition for two points x 1 and x 2 on the boundary with normals N 1 and N 2 respectively to form a symmetry point? In [14] this matter has been investigated in the case of the plane curve and to generalize that result we will take the case when the normals of the boundary (which are oriented inside the object figure 2.1) intersect each other and this case is a generic. The answer of this question is given in the following theorem. Theorem Let x 1 and x 2 be two points on the boundary with unit normals N 1 and N 2 respectively. Suppose that N 1 N 2 and the normal lines intersect. Then a necessary

33 Chapter 2. Symmetry Set in R n+1 19 and sufficient condition for x 1 and x 2 to form a symmetry point is that (x 1 x 2 ) (N 1 + N 2 ) = 0. (2.1) First, assume that x 1 and x 2 form a symmetry point A, then A = x 1 + rn 1 = x 2 + rn 2. Therefore, So, x 1 x 2 = r(n 1 N 2 ). (x 1 x 2 ) (N 1 + N 2 ) = r(n 1 N 2 ) (N 1 + N 2 ) = 0. Second, assume that equation (2.1) holds since N 1 intersect, then there exist (a, b R) such that ±N 2 and the lines of normals x 1 + an 1 = x 2 + bn 2. Therefore Hence, x 1 x 2 + an 1 bn 2 = 0. (x 1 x 2 ).(N 1 + N 2 ) + (an 1 bn 2 ).(N 1 + N 2 ) = 0. Now using equation (2.1), we have (an 1 bn 2 ).(N 1 + N 2 ) = 0 So, Hence a b + an 1 N 2 bn 1 N 2 = 0. (a b)(1 + N 1 N 2 ) = 0. Since N 1 N 2, we have 1 + N 1 N 2 0 therefore, a b = 0 a = b hence x 1 and x 2 form a symmetry point and r = a.

34 Chapter 2. Symmetry Set in R n Creating the Boundary from the Symmetry Set In this section we turn to the reconstructing the boundary from its symmetry set, and we will investigate the relationship between the symmetry set and the associated midlocus. Furthermore, we will study the radius function and its singularity. Also, the relationship between the singularity of the radius function and that of the midlocus in the plane will be investigated. In the rest of this chapter S reg refers to the set of all smooth points of the the symmetry set that means the set of all points of type A 2 1 (an A 2 1 point of the symmetry set is the centre of a bitangent hypersphere which has ordinary contacts with the boundary). Theorem Let S be the symmetry set of a region Ω R n+1, with smooth boundary X and let S reg S. Then for any point x 0 S reg, the associated tangency points on the boundary X are given by such that (1 r 2 Riemannian gradient of r. X j = x 0 r r ± r 1 r 2 N, j = 1, 2 (2.2) 0), where N is the unit normal of S at x 0 and r is the Let S 1 be a smooth patch of the symmetry set containing x 0, and consider the function F = X S 1 2 r 2. (2.3) Now let {v 1, v 2,..., v n } be a basis of T x0 S 1, then F v j = 2v j (X S 1 ) 2rdr(v j ). Therefore, the envelope of hyperspheres centred on S 1 is given by {X R n+1 : F = F v j = 0, j = 1, 2,..., n}.

35 Chapter 2. Symmetry Set in R n+1 21 Therefore, v j (X S 1 ) = rdr(v j ). (2.4) Now since {v 1, v 2,..., v n, N}, where N is the unit normal of S 1 at x 0, is a basis for R n+1, every point Z R n+1 can be written as Therefore, Z = n λ i v i + λ n+1 N. X x 0 = n λ i v i + λ n+1 N, and from F = 0 we have λ n+1 = ± r 2 ( n λ i v i ) 2 N, thus n n X = x 0 + λ i v i ± r2 ( λ i v i ) 2 N. (2.5) But from equation (2.4), we have v j (X S 1 ) = rdr(v j ), where dr(v j ) is the directional derivative of r in the direction v j therefore dr(v j ) can be written as ( ) 1 n dr(v j ) = v j λ i v i. r Since S 1 is smooth, then as S 1 is a submanifold of R n+1 so it has a Riemannian structure. Therefore, dr can be written again as: Therefore, 1 r dr(v j ) = v j, 1 r n λ i v i. n λ i v i is the Riemannian gradient of the radius function r. Let r denotes n λ i v i = r. Therefore, the Riemannian gradient of r, then 1 r n λ i v i = r r. Hence by substitution in equation (2.5) the proof is completed.

36 Chapter 2. Symmetry Set in R n+1 22 Definition Given a smooth hypersurface X in R n+1 as in definition (2.2.1) the midlocus M of X is the closure of the set of midpoints of chord joining contact points of all hyperspheres bitangent to X. Thus if x 1 and x 2 are two points of tangency then the corresponding point of the midlocus is x m = 1(x x 2 ). Now from theorem we have the following. Corollary Assume as in theorem then, the midlocus point is given by x m = x 0 r r, where x 0 is a smooth point of the symmetry set and r is the Riemannian gradient of the radius function at x 0. The proof comes directly from theorem From corollary we find that if S is a smooth part of the symmetry set, then the associated midlocus M is given by M = S r r. Peter Giblin pointed out that in the case of 3D if the tangent planes of the boundary at the tangency points are parallel then the radius function has a singularity [11]. In fact, the radius function plays a central role in the relationship between the boundary, the symmetry and the midlocus. Also there is a very complicated relationship between the differential geometry of the boundary and that of the symmetry set involving the radius function and its derivatives. The following theorem gives the answer to the question: At what condition does the symmetry point coincide with the associated midlocus point?

37 Chapter 2. Symmetry Set in R n+1 23 Theorem Let S be the symmetry set of a region Ω in R n+1, with smooth boundary X. Let x 0 S reg, then x 0 and the associated midlocus point x m coincide if and only if the radius function has a singularity at x 0. Let x m = x 0, then r = 0 and since r is the Riemannian gradient, it can be written as r = g ij dr(v j )v i, where g ij is the inverse of the matrix representing the Riemannian metric, dr(v j ) is the partial derivative of the radius function r and v i is the basis of the tangent space of the symmetry set S at x 0. Therefore, r = 0 g ij dr(v j )v i = 0, hence dr(v j ) = 0. Conversely assume that the radius function r has a singularity, then r = 0, therefore, x 0 = x m which completes the proof. The above theorem tells us the impact of the singularity of the radius function on the relationship between a smooth symmetry point and its associated midlocus point. But what about the relation between a smooth symmetry point and its associated tangency points on the boundary. Does the singularity of the radius function affect it? The answer to this question is given in the following proposition. Proposition Let S be the symmetry set of a region Ω in R n+1 with smooth boundary X. Let x 0 S reg, then the tangency points associated to x 0 are given by: x j = x 0 ± rn, j = 1, 2 if and only if the radius function r has a singularity at x 0, where N is the unit normal of S at x 0. From theorem the tangency points associated to x 0 are given by x j = x 0 r r ± r 1 r 2 N, j = 1, 2.

38 Chapter 2. Symmetry Set in R n+1 24 Now assume that r has a singularity, then r = 0. Therefore, we get x j = x 0 ± rn, j = 1, 2. Conversely, assume that x j = x 0 ± rn, j = 1, 2. Then, r r = 0 and (1 r 2 ) = 1. Therefore, r = 0, hence the radius function r has a singularity, and by this the proof is completed. Lemma Let S be the symmetry set of a region Ω in R n+1 with smooth boundary X. Let x 0 be a smooth point in S. Then the Riemannian gradient of the radius function at x 0 is given by: r = 1 2 (N 1 + N 2 ), where N 1 and N 2 are the unit normals of the boundary at the tangency points. From definition 2.2.1, we have x 0 = x 1 + rn 1 = x 2 + rn 2 and from theorem we have x 1 = x 0 r r + r 1 r 2 N and, x 2 = x 0 r r r 1 r 2 N. Therefore, N 1 = r 1 r 2 N and N 2 = r + 1 r 2 N. Hence, r = 1 2 (N 1 + N 2 ).

39 Chapter 2. Symmetry Set in R n+1 25 Proposition Assume as in lemma Then r has a singularity if and only if N 1 points. = N 2. where N 1 and N 2 are the unit normals of the boundary at the tangency From lemma we have, r = 1 2 (N 1 + N 2 ). If the radius function has a singularity, then r = 0, and so N 1 = N 2. Now if N 1 = N 2 then r = 0 which implies that the radius function has a singularity. Example Let S(s, t) be the symmetry set of a smooth surface X in R 3, and r(s, t) be the radius function. If x 0 = S(s 0, t 0 ) be a smooth point, then the associated tangency points on the boundary X are given by r x j =x 0 ϵ 1 ϵ 2 {(r s ϵ 2 2 r t ϵ 1.ϵ 2 )ϵ 1 + (r t ϵ 1 2 r s ϵ 1.ϵ 2 )ϵ 2 } 1 ± r 1 ϵ 1 ϵ 2 {(r s ϵ r t ϵ 1.ϵ 2 )ϵ 1 + (r t ϵ 1 2 r s ϵ 1.ϵ 2 )ϵ 2 } 2 N, j = 1, 2, where ϵ 1 = S s (s 0,t 0 ) and, ϵ 2 = S t (s 0,t 0 ) and r s = r s, r t = r t. The midlocus point is given by: x m = x 0 r ϵ 1 ϵ 2 {(r s ϵ 2 2 r t ϵ 1.ϵ 2 )ϵ 1 + (r t ϵ 1 2 r s ϵ 1.ϵ 2 )ϵ 2 }. In the above example we give a general method to calculate the tangency points and the midlocus point associated to a smooth point on the symmetry set of a smooth surface X R 3. In the following we will present specific examples to illustrate the ideas of the theorems mentioned in this section.

40 Chapter 2. Symmetry Set in R n+1 26 Example Let S(s, t) = (s, t, st) and r(s, t) = s 2 + t 2 + 1, where 1 4 s 1 4 and 1 t 1. Then the midlocus is given by 4 4 M(s, t) = ( s + 2s 3 2st 2, t + 2t 3 2s 2 t, 3st ). It is clear that r s (0, 0) = r t (0, 0) = 0, which means that the radius function has a singularity at (0, 0). Also, it is obvious that S(0, 0) = M(0, 0) = (0, 0, 0). 0.2 The midlocus The symmetry set Figure 2.2: Symmetry set and associated midlocus in example Example Let S(s, t) = (s, t, s 2 + t 2 ) and r(s, t) = s 2 + t Then the midlocus is given by M(s, t) = ( ) 2s 3 + 2st 2 s 4(s 2 + t 2 ) + 1, 2t3 + 2s 2 t t 4(s 2 + t 2 ) + 1, 3(s 2 + t 2 ). 4(s 2 + t 2 ) + 1 It is clear that, r has a singularity at (0, 0) and S(0, 0) = M(0, 0) = (0, 0, 0).

41 Chapter 2. Symmetry Set in R n The symmetry set The midlocus Figure 2.3: Symmetry set and associated midlocus in example Example Let S be the symmetry set of a plane curve and let S be smooth at x 0. Now parameterize S by the arc-length then the Riemannian gradient of the radius function is given by r = r T = 1 2 (N 1 + N 2 ). Therefore, Also, Thus r 2 = 1 4 (N 1 + N 2 ) 2. r = 1 2 (T.N 1 + T.N 2 ). r = cos θ, where θ is the angle between N i and T, i = 1, 2. If N 1 N 2 then r 2 = 1 2. Let γ be a smooth plane curve and suppose that x 0 is the centre of a bitangent circle to γ at x 1 and x 2. Let γ 1 and γ 2 be small pieces of γ close to x 1 and x 2 respectively as shown

42 Chapter 2. Symmetry Set in R n+1 28 in the figure 2.4. The relation between the arc-lengths of the boundary was studied by Giblin. Our target is to study the relationship between the arc-lengths of the symmetry set and that of the boundary. γ 1 x 1 x 2 x 0 γ 2 Figure 2.4: The symmetry point and the associated tangency points in the case of curve. Lemma Let S be the symmetry set of a plane curve γ and x 0 S reg and let x 1 γ 1 and x 2 γ 2 be the associated tangency points on the boundary. Let s 1 and s 2 be the arc-lengths on γ 1 and γ 2 respectively. Then we have (1 rκ 1 ) ds 1 ds = (1 rκ 2) ds 2 ds where s is the arc-length of a smooth part of S close to x 0. From definition of symmetry set the part S 1 of S associated to γ 1 and γ 2 is given by Therefore we have S 1 (s) = γ 1 (s 1 ) + r(s 1 )N 1 (s 1 ) = γ 2 (s 2 ) + r(s 2 )N 2 (s 2 ). T = T 1 ds 1 ds + r N 1 rκ 1 T 1 ds 1 ds. (2.6)

43 Chapter 2. Symmetry Set in R n+1 29 Also, we have ds 2 T = T 2 ds + ds 2 r N 2 rκ 2 T 2 ds (2.7) where T, T 1 and T 2 are the unit tangents of the symmetry set and the boundary respectively. Now (2.6)-(2.7) gives the following equation (1 rκ 1 ) ds 1 ds T 1 (1 rκ 2 ) ds 2 ds T 2 + r (N 1 N 2 ) = 0. (2.8) Now the inner product on both sides of equation 2.8 with T 1 T 2 gives the following (1 rκ 1 )(1 T 1 T 2 ) ds 1 ds + (1 rκ 2)(1 T 1 T 2 ) ds 2 ds = 0. Since T 1 T 2 because of the orientation, then 1 T 1 T 2 0. Therefore, (1 rκ 1 ) ds 1 ds + (1 rκ 2) ds 2 ds = 0. Hence (1 rκ 1) ds 1 ds = (1 rκ 2) ds 2 ds, which completes the proof. Now we have the following corollary [13, 32]. Corollary Assume as in lemma Then ( ) ds 2 1 rκ1 =. ds 1 1 rκ 2 The proof of this corollary comes directly from the above lemma. In the next proposition we will give the relationship between the arc-length of the symmetry set and those on the boundary. First of all we need this lemma. Lemma [14] Let S be the symmetry set of a smooth plane curve γ. The tangents T 1 and T 2 and normals N 1 and N 2 of γ at the tangency points associated to a smooth point x 0 γ are given by T 1 = 1 r 2 T r N,

44 Chapter 2. Symmetry Set in R n+1 30 T 2 = 1 r 2 T r N, N 1 = r T 1 r 2 N, and N 2 = r T + 1 r 2 N where, T and N are the unit tangent and unit normal of the symmetry set at x 0. Proposition Let S be the symmetry set of a smooth plane curve γ and s be the arc-length on S reg. Then we have ds 1 1 r 2 ds = and 1 rκ 1 ds 2 ds = 1 r 2 1 rκ 2. Let S 1 be the smooth part of S associated to γ 1 and γ 2 parametrized by the arc-length s, thus S 1 = γ 1 + rn 1 which gives that But from lemma we have T = (1 rκ 1 ) ds 1 ds T 1 + r N 1. (2.9) T 1 = 1 r 2 T r N and N 1 = r T 1 r 2 N. Therefore, substituting in (2.9) we get the following equation ( T = (1 rκ 1 ) 1 r 2 ds ) ( 1 ds ) + r 2 T r (1 rκ 1 ) ds ) 1 ds + 1 r 2 N. Now equating the tangential part we obtain (1 rκ 1 ) 1 r 2 ds 1 ds + r 2 = 1. Therefore this equation gives ds 1 ds = 1 r 2 1 rκ 1. (2.10)

45 Chapter 2. Symmetry Set in R n+1 31 Also, from corollary we have (1 rκ 1 ) ds 1 ds = (1 rκ 2) ds 2 ds hence from this equation and equation 2.10 we find that ds 2 1 r 2 ds =, 1 rκ 2 and by this the proof is completed. In the following proposition we will turn to the relationship between the singularity of the radius function and that of the midlocus in the case of plane curve. Lemma [13, 32] Given a smooth curve γ with x 1 and x 2 being points of contact of bitangent circle, then the midlocus is smooth here provided T 1 T 2 or κ 1 + κ 2 2 r where, T i and κ i, i = 1, 2 are the tangents and curvatures of γ at x 1 and x 2. This lemma tells us conditions for the smoothness of the midlocus. At a smooth point of the symmetry set the condition T 1 T 2 means that the radius function has no singularity and this can be obtained from proposition Thus we can determine the conditions that allow the midlocus to have a singularity in terms of the singularity of the radius function. Proposition Let S be the symmetry set of a smooth plane curve γ. Let x 0 S reg. Then the midlocus is singular at the point associated to x 0 if and only if the radius function has a singularity at x 0 and r (x 0 ) = 1 r(x 0 ). Let S 1 be the smooth part of S close to x 0 parametrized by the arc-length s then the associated midlocus M is given by M = S 1 rr T.

46 Chapter 2. Symmetry Set in R n+1 32 Therefore M = (1 r 2 rr )T rr κn. Now assume that r = 0 and r (x 0 ) = 1 r(x 0 then M = ) 0 which means that the midlocus is singular. Conversely assume that the midlocus is singular, then the radius function has a singularity, so from the above equation we have 1 rr = 0 which implies that r = 1 r. The above proposition tells us the relationship between the singularity of the radius function and that of the midlocus. Also, it determines the type of the singularity of the radius function. Recall that a function f : R R is said to have an A 1 singularity at t 0 if f (t 0 ) = 0 and f (t 0 ) 0. Corollary Let S be the symmetry set of a smooth plane curve γ. Let x 0 S reg. If the midlocus is singular at the point associated to x 0, then the radius function has only an A 1 singularity at x 0. From proposition we have if the midlocus is singular then r = 1 r and r = 0 which means that the radius function has an A 1 singularity. Now we will end this section by calculating the area of the triangle formed by a smooth symmetry point and its associated tangency points on the boundary. Let x 0 S reg, then by theorem the tangency points are given by x 1 = x 0 r r + r 1 r 2 N, and, Therefore, x 2 = x 0 r r r 1 r 2 N. x 1 x 2 = 2r 1 r 2 N.

47 Chapter 2. Symmetry Set in R n+1 33 This implies x 1 x 2 = 2r 1 r 2. Also, the height of the triangle in figure 3.2 is given by h = S M = r r. Hence the area of this triangle is given by A = r 2 r 1 r 2. x1 r x0 r r xm r x2 Figure 2.5: Triangle formed by symmetry point and its associated tangency points. So, we can summarize this in the following proposition. Proposition Let S be the symmetry set of a region Ω R n+1 with smooth boundary X. Let x 0 be a smooth point of S, then the area of the triangle formed by x 0 and the associated tangency points is given by: A = r 2 r 1 r 2. Corollary Assume as in proposition singularity, then A = 0. If the radius function has a If the radius function has a singularity then we have r = 0. This implies that A = 0.

48 Chapter 2. Symmetry Set in R n Singularity of the Radius Function at a Singular Point of the Symmetry Set It was pointed out by Giblin that in the case of the plane curve if the symmetry set has a cusp then the radius function has a singularity. In this section we discuss this phonemena in the case of higher dimensions and we study the relationship between the singularity of the radius function and the coincidence of the symmetry point and its associated midlocus point. Before studying this phonemena we recall that a point x 0 of a symmetry set is of type A 1 A 1 = A 2 1 if the bitangent hypersphere has an ordinary contact with the boundary at the associated tangency points, i.e, the hypersphere is not the hypersphere of the curvature. A 1 A k 2 if the bitangent hypersphere has an ordinary contact with the boundary at one point (A 1 ) and it is the hypersphere of the curvature at the other tangency point of the boundary (A k 2 ). A 3 if the hypersphere has a single contact with the boundary. This point is a limiting case of the two points in the A 2 1 case above. Let S be the symmetry set of a smooth hypersurface X and x 1 and x 2 be the tangency points corresponding to x 0 S. Suppose that the contact at x 1 is of type A 1 and the contact at x 2 is of type A k 2 then, the symmetry set is singular at x 0. Now we will investigate the singularity of the radius function. Let X 1 X and X 2 X be two pieces of X around x 1 and x 2 respectively. Let s = (s 1, s 2,..., s n ) and t = (t 1, t 2,..., t n ) be the local parameters of X 1 and X 2 since the contact at x 1 is of type A 1 then using the implicit function theorem s is a smooth function of t. The part S 1 of the symmetry set associated

49 Chapter 2. Symmetry Set in R n+1 35 to X 1 and X 2 is given by From equation 2.11 we have S 1 = X 1 + rn 1 = X 2 + rn 2. (2.11) S 1 = X 2 + rn 2 differentiate this equation with respect to t i we get This equation can be written in vector form S 1 = v i + r N 2 + r N 2, v i = X 2. t i t i t i t i S 1 t = V rst x 2 + r t N 2 = (I rs T x 2 )V + r t N 2, (2.12) where S x2 is the matrix representation of the shape operator of the boundary at x 2, V = v 1 v 2. v n and r t = r t 1 r t 2. r t n. Now applying dot product with N 2 to each entry in equation 2.12, we obtain Also, from equation 2.11 we have Now differentiate X 1 with respect to t i we get S 1 t N 2 = r t. (2.13) S 1 = X 1 + rn 1. (2.14) X 1 = X 1 s 1 + X 1 s X 1 s ( 1 = t i s 1 t i s 2 t i s n t i s 1 t i s 2 t i s n t i ) v1 v2. v n,

50 Chapter 2. Symmetry Set in R n+1 36 where v i = X 1 s i. Therefore, X 1 t = s 1 t 1 s 2 t 1 s n t 1 s 1 s 2 s t 2 t 2 n t s 1 t n s 2 t n s n t n v1 v2. v n = ΓV. Similarly, N 1 t = ΓS T x 1 V, where S x1 is the matrix representation of the shape operator of the boundary at x 1. Now differentiating equation 2.14 with respect to t gives S 1 t = Γ(I rst x 1 )V + r t N 1. (2.15) Now applying dot product with N 1 to each entry in equation 2.15, we obtain S 1 t N 1 = r t. (2.16) Therefore, the radius function has a singularity if and only if N 1 and N 2 are perpendicular to the set of vectors Θ = S 1 t i, i = 1, 2,..., n. Therefore, we state the following. Theorem Let S be the symmetry set of a smooth hypersurface X R n+1. Let x 0 S be a singular point of type A 1 A k 2 of the symmetry set. Then the radius function has a singularity if and only if the unit normals of the tangency points corresponding to x 0 are perpendicular to the tangent space of the symmetry set at x 0. Example Let γ be a smooth closed curve and S be its symmetry set. Suppose that x 0 S be an A 1 A 2 point then the symmetry set is a cusp at x 0, therefore ds dt x 0 = (0, 0) which is perpendicular to any vector hence the unit normals are perpendicular to this vector. Thus the radius function has a singularity.

51 Chapter 2. Symmetry Set in R n+1 37 Remark The radius function has no singularity when the symmetry set has an edge point A 3 which is a limiting point of two points of type A 1. The normal of the boundary in this case is tangent the smooth stratum containing A 3. Now we will discuss what happens when the midlocus and the singular point of the symmetry set coincide. Proposition Let S be the symmetry set of a smooth hypersurface X R n+1. Let x 0 S be a singular point of type A 1 A k 2 of the symmetry set and x m be the associated midlocus. If x 0 and x m coincide then the radius function has a singularity. From the definition of the symmetry set we have x 0 = x 1 + rn 1 = x 2 + rn 2. Thus we get 2x 0 = x 1 + rn 1 + x 2 + rn 2 = 2x m + r(n 1 + N 2 ). Now if x 0 = x m, then we have N 1 + N 2 = 0 and from equations 2.13 and 2.16 we have 2 r = S 1 (N t t 1 + N 2 ) = 0. Thus r = 0. t 2.5 Creating the Symmetry Set from the Midlocus In this section we will discuss the possibility of creating the boundary using the information provided by the midlocus and the radius function. In fact, Peter Giblin and John Paul Warder [16, 32] created the symmetry set of a plane curve using the midlocus and radius function. Our task is to generalize this idea to the higher dimensions. Now if we are given the symmetry set S as a smooth hypersurface parametrized by (x 1, x 2,..., x n )

52 Chapter 2. Symmetry Set in R n+1 38 and let M be the associated smooth midlocus, then from corollary for each x S, the associated midlocus point x m is given by x m = x + n λ i v i, (2.17) where v i = S x i evaluated at x, also, we have Now from equation 2.17 we have Therefore, ( n ) v j λ i v i x x m = v i (x x m ) = v i = r j r, r j = r x j. n λ i v i. n λ i v i = r i r. (2.18) Now there are a lot of solutions of equation (2.18) i.e., there are many vectors v i R n+1 such that the equation holds, but there is only one solution on the form α(x x m ), where α R. This solution is of the form: r i r(x x m ) x x m 2 provided x x m. Therefore, we summarize this in the following theorem. Theorem Given the midlocus and the radius function (smooth function) describing the radius of the hypersphere generating each point of the midlocus M of a smooth hypersurface X in R n+1. Then the symmetry set associated to this midlocus is a solution of the PDEs provided S M, where r i = r x i. S = ri r(s M) x i S M 2

53 Chapter 2. Symmetry Set in R n+1 39 Now we discuss this theorem in R 2 and R 3. Now let M(t) = (m 1 (t), m 2 (t)) be the midlocus of a smooth curve in R 2 and r(t) be the smooth function describing the radius of each circle generating each point of M. The above theorem indicates that ( ds dt = ds1 dt, ds ) 2 = rr (S M) dt S M 2 ( rr ) (s 1 m 1 ) = (s 1 m 1 ) 2 + (s 2 m 2 ) 2, rr (s 2 m 2 ) (s 1 m 1 ) 2 + (s 2 m 2 ) 2. Now put X = s 1 m 1, Y = s 2 m 2 and rr = R, then we have X = RX X 2 + Y 2 m 1 RY and Y = X 2 + Y 2 m 2 which are the same ordinary differential equations obtained by Giblin and Warder [16, 32] and an interesting example can be found in [16]. Now we discuss this theorem in R 3 and we will have partial differential equations instead of ordinary differential equations. Let M(x, y) = (m 1, m 2, m 3 ) be the midlocus of a smooth surface in R 3 and r(x, y) be the radius function which is a smooth function describing the radius of each sphere generating each point of M. Our target is to find the associated symmetry set S(x, y) = (s 1, s 2, s 3 ), and from theorem we can create the partial differential equations which hold for the symmetry set, thus we have S x = rr x(s M) S M 2 and S y = rr y(s M) S M. 2 Therefore, s i x = rr x(s i m i ) and 3 (s j m j ) 2 j=1 s i y = rr y(s i m i ), i = 1, 2, 3. 3 (s j m j ) 2 j=1 Now put F i = s i m i, j = 1, 2, 3, then we have F i x = rr xf i 3 Fj 2 j=1 m i x and F i y = rr yf i 3 Fj 2 j=1 m i y. After solving these PDEs the required symmetry set is given by S = F + M = (F 1 + m 1, F 2 + m 2, F 3 + m 3 ).

54 40 Chapter 3 Centroid Set, Skeletal Structure and the Singularity of the Radius Function 3.1 Introduction The symmetry set of a hypersurface X R n+1 and its associated midlocus were studied in chapter 2 as well as the impact of the singularity of the radius function on the relationship between them. In this chapter we will study a more general concept than the midlocus. Precisely this chapter consists of four main parts. In the first part the centroid set associated to a smooth submanifold M of R n+1 will be defined. The centroid set is more general than the midlocus and it depends on a multivalued radial vector field U defined on M such that each value of U forms a smooth radial vector field on M and has associated radius function. The impact of the singularity of the radius function on the relationship between M and the associated centroid set will be studied in this part (Theorem and Theorem ). Moreover, the condition for the centroid set to have a singularity when the radius function has a singularity will be studied (Proposition ). The second main part of this chapter deals with the skeletal structure and the singularity of the radius function. In this part the relationship between the singularity of

55 Chapter 3. Centroid Set, Skeletal Structure and the Singularity of the Radius Function 41 the radius function and the orthogonality of the radial vector field on the tangent space of the skeletal structure will be studied (Proposition 3.3.2). Furthermore, the relationship between a smooth point of a skeletal structure and its associated midlocus point will be considered when the radius function has a singularity as well as the relationship between the tangent spaces of the boundary at the tangency points. The third part of this chapter deals with the pre-medial axis of a smooth hypersurface. In this part the pre-medial axis is defined and the relationship between the parameters of a skeletal structure in a neighbourhood of a smooth point and the parameters of the boundary in a neighbourhood of the associated point will be studied (Lemma 3.4.2). The main result of this part is (Proposition 3.4.3) which gives the relationship between the parameters of the boundary of a Blum medial axis at the tangency points associated to a smooth point. The fourth part of this chapter deals with the classification of the singularity of the midlocus associated to a skeletal structure in R 3. The impact of the eigenvalues of the Hessian of the radius function on the corank of the singularity of the midlocus will be studied (Theorem 3.5.6). The main result of this part is (Theorem ) which gives the necessary and sufficient condition for the midlocus to have a crosscap singularity. 3.2 Centroid Points Let M be a smooth submanifold of R n+1 such that on this submanifold we pick a multivalued vector field U = (u 1, u 2,..., u l ) such that each U i forms a smooth vector field on M. We put u i = r i U i where U i is a smooth unit vector field on M and r i is a smooth function on M i.e., r i : M R, and we assume that r i > 0. Now let T x0 M be the tangent space of M at x 0 and v T x0 M. For each smooth vector field U i we equip M with the 1-form η i (v) = dr i (v) + U i v,

56 Chapter 3. Centroid Set, Skeletal Structure and the Singularity of the Radius Function 42 where dr i (v) is the directional derivative of r i in the direction of v. Now since M is a smooth submanifold of R n+1 it has a Riemannian structure induced from R n+1 and the tangent space T x0 M of M at x 0 M is considered to be embedded in the tangent space T x0 R n+1 of R n+1 at x 0. Recall that the directional derivative of a smooth function on a Riemannian manifold in the direction of a tangent vector v j is given by dr i (v j ) = r i, v j, where r i is the Riemannian gradient of r i, and, is the Euclidean inner product. Therefore, η i can be written as η i (v j ) = r i, v j + U i, v j = r i + U i, v j. Definition Let M be a smooth k-dimensional submanifold of R n+1, then 1. The tangent space to M at x 0 M is the vector subspace T x0 M T x0 R n+1, which is defined by T x0 M := df p ({p} R k ) = df p (T p R k ) for a parametrization f : U M with f(p) = x 0, where U R k is an open set and df is the differential of f. The vector space T x0 M does not depend on the choice of f. 2. The normal space to M at x 0 M is the vector subspace N x0 M T x0 R n+1, which is the orthogonal complement of T x0 M: T x0 R n+1 = T x0 M N x0 M. Here denotes the orthogonal direct sum with respect to the Euclidean inner product. Lemma Let M be a smooth submanifold of R n+1 as above, then the 1-form η i vanishes at x 0 M if and only if r i + U i N x0 M, where N x0 M is the normal space of M at x 0.

57 Chapter 3. Centroid Set, Skeletal Structure and the Singularity of the Radius Function 43 Recall that a vector z T x0 R n+1 is a normal vector of a submanifold M at x 0 if and only if z, v j = 0 for all v j T x0 M. Thus, η i (v j ) = r i + U i, v j = 0 for all j if and only if ( r i + U i ) N x0 M, where N x0 M is the normal space of M at x 0. Remark Now let U i = U T i + U N i, where U T i is the tangential component of U i and U N i is the normal component. Then, the 1-form η i = 0 if and only if U T i = r i. Theorem Let (M, U) be a smooth submanifold of R n+1 and multivalued vector field as above such that η i = 0 at x 0 M. Then 1. r i has a singularity at x 0 if and only if U i (x 0 ) N x0 M. 2. If r i has a singularity at x 0 for all i, then l U i (x 0 ) N x0 M. 1. Since η i = 0, then r i + U i NM, by lemma r i has a singularity if and only if r i = 0 if and only if U i NM. 2. Follows trivially from 1. Corollary Let (M, U) be a smooth submanifold of R n+1 and multivalued vector field as above such that η i = 0 and r i = r for all i, then the following are equivalent 1. r has a singularity at x l U i (x 0 ) N x0 M.

58 Chapter 3. Centroid Set, Skeletal Structure and the Singularity of the Radius Function 44 (1 2) Since η i = 0, then ( r + u i ) NM. Thus (l r + l U i ) NM, hence r has a singularity if and only if r = 0 if and only if l U i NM. Definition Let M be a smooth submanifold of R n+1 such that for each x 0 M there exist a multivalued vector field U = (u 1, u 2,..., u l ) such that each u i = r i U i forms a smooth vector field on M, where U i is a smooth unit vector field on M and r i is a smooth real valued function on M. We define the centroid point associated to x 0 by x c = x l l r i (x 0 )U i (x 0 ). The centroid set of (M, U) is given by C(M, U) = {y R n+1 y = x + 1 l l r i (x)u i (x), for some x M}. Example Let S be the smooth part of the symmetry set of a smooth boundary X, and r be the radius function, then for each x 0 S, we can define the multivalued vector field U = (ru 1, ru 2 ) to be U 1 = r r + r 1 r 2 N on one side of S and U 2 = r r r 1 r 2 N on the other side as shown in the figure 3.1, where N is the unit normal of S at x 0 and r is the Riemannian gradient of the radius function r. It is obvious to observe that in this case l = 2, r 1 = r 2 and the centroid point is nothing but the midlocus point. An example of l = 3 is the Y-junction in a skeletal structure.

59 Chapter 3. Centroid Set, Skeletal Structure and the Singularity of the Radius Function 45 N U 1 = r + 1 r 2 N N Y junction U 2 U 1 x 0 x m = x c U 3 S U 2 = r 1 r 2 N Figure 3.1: Figure of example Now we will give the definition of the centroid set associated to a skeletal set. Definition Let (S, U) be a skeletal set and multivalued radial vector field with stratification of the form S = {M λ } λ Λ for some set Λ, and U = {U λ } λ Λ, then the centroid set associated to S is defined by C(S, U) = C(M λ, U λ ). λ Λ Example Let S = R R 2 such that S = {(x, 0) x R} with stratification S = {{0}, R \ {0}} and u R\{0} = ±(0, 1) and u {0} = {(0, 1), (0, 2)}. Then C({0}, u {0} ) = {(0, 1)} and C(R \ {0}, u R\{0} ) = R \ {0}. Thus C(S, U) = R \ {0} {(0, 1)}. Observe that C({0}, u {0} ) C(R \ {0}, u R\{0} ), where C denotes the closure of C. {0} C({0}, u {0} ) Figure 3.2: A schematic diagram of example

60 Chapter 3. Centroid Set, Skeletal Structure and the Singularity of the Radius Function 46 From example we can see that if the stratum X is in the closure of Y, then C(X, u X ) is not necessarily in the closure of C(Y, u Y ). Now we will study the impact of the singularity of the radius function on the relationship between a point x 0 M and its associated centroid point. In fact we will assume as in corollary 3.2.5, i.e., we will have the same radius function and in this case we have 1-form η on M. Theorem Let (M, U) be a smooth submanifold of R n+1 and multivalued vector field as in corollary Let x 0 M and x c be its associated centroid point, then 1. if x 0 = x c, then the radius function has a singularity at x x c x 0 N x0 M if and only if the radius function has a singularity at x Assume that x 0 = x c, then from the definition of the centroid set we have 1 l l r i U i = 0, but by our assumption we have r i = r and the 1-form η vanishes in a neighbourhood l of x 0. Now for any v j T x0 M, then r U l i v j = 0. Thus rdr(v j ) = 0 and hence the radius function has a singularity at x Since we have the same radius function, then the centroid point x c associated to x 0 is given by x c = x 0 r r + r l l Ui N. Thus x c x 0 N x0 M if and only if r = 0 if and only if r has a singularity at x 0. The above theorem is a generalization of proposition Now let M k (k indicates the dimension of M) be a smooth submanifold of R n+1. For any point x 0 M k we put {v 1, v 2,..., v k } as a basis for the tangent space of M k at x 0 and {w1 N, w2 N,..., wm} N is a

61 Chapter 3. Centroid Set, Skeletal Structure and the Singularity of the Radius Function 47 basis for the normal space of M k. In the following proposition the radius function is the same for each u i of multivalued radial vector field U = (u 1, u 2,..., u l ) and the sum of normal parts of u i is zero, i.e., the centroid point associated to x 0 is given by x c = x 0 r r. (3.1) Moreover, V is the matrix with i-th row entry v i, N is the matrix with i-th row wi N, H r is the Hessian matrix of r, β is the matrix of the normal coefficients of r, dr(v ) is a V column matrix with i-th entry r, and V c is the Jacobian matrix of the map v i x x r(x) r(x), x M k and all those terms are evaluated at x 0. Proposition Let (M k, U) be a smooth submanifold of R n+1 and multivalued vector field such that the radius function is the same for each u i and η = 0, and the sum of the normal parts of U = (u 1, u 2,..., u l ) is zero. Then the centroid is singular at x c associated to a point x 0 M k if and only if the rank of the matrix (V dr(v ) r rhr T V rβn) is less than k. In this case the centroid point associated to a given point x 0 M k is given by x c = x 0 r r. Thus if {v 1, v 2,..., v k } is a basis for the tangent space of M k at x 0, then v cj = v j r v j r r r v j, j = i, 2,..., k. Here v cj is the directional derivative of the map x x r(x) r(x), x M k in the direction of v j. This equation can be written in vector form as the following V c = V dr(v ) r rh T r V rβn. (3.2)

62 Chapter 3. Centroid Set, Skeletal Structure and the Singularity of the Radius Function 48 Thus the centroid set is singular if and only if the rank of the matrix V dr(v ) r rh T r V rβn is less than k. Corollary Let (M k, U) be a smooth submanifold of R n+1 as in proposition If the radius function has a singularity at x 0, then the centroid set is singular at x c if and only if 1 r is an eigenvalue of H r. If the radius function r has a singularity, then v j ( r) T x0 M k, thus β = 0. Therefore, V c = (I rh T r )V. Now since V is a (k (n + 1)) matrix with rank k and (I rh T r ) is a k k matrix, then the rank of V c is equal to the rank of (I rh T r ). Thus the centroid is singular if and only if the rank of (I rh T r ) is less than k if and only if 1 r is an eigenvalue of H r. Now using the new set-up of the centroid point the natural question is what is the impact of the singularity of the radius function on the relationship between a given point x 0 M k, and its associated centroid point? The answer of this question is given in the following theorem which is a generalization of theorem Theorem Let (M k, U) be a smooth submanifold of R n+1 and multivalued vector field as in proposition Let x 0 M k be any point and x c its associated centroid point, then the radius function r has a singularity at x 0 if and only if x 0, and x c coincide. The centroid point x c associated to a point x 0 M k is given by equation 3.1, thus using theorem the result holds.

63 Chapter 3. Centroid Set, Skeletal Structure and the Singularity of the Radius Function The Skeletal Structure and the Singularity of the Radius Function In the previous sections of this chapter the concept of the centroid set has been introduced and its singularity has been discussed in a general case. Also, the impact of the singularity of the radius function on the singularity of the centroid has been investigated and we found that the singularity of the radius function occurs when the radial vectors all lie in the normal space. In this section such a study will be carried out for the case of the skeletal structure. It is important to note that the centroid in the previous sections does not have a boundary but in the case of the midlocus associated to a skeletal structure with a smooth boundary it is subjected to the condition that allows the radial map to be a diffeomorphism. Lemma Let (S, U) be a skeletal structure of a region Ω in R n+1 with smooth boundary X and let x 0 S be a non-edge point. Let U be a smooth value (on a non-edge local manifold component S α ), for which 1 r is not an eigenvalue of S V radius function r has a singularity at x 0 if and only if Ψ 1 v i U 1 = v i U 1, i = 1, 2,..., n. at x 0. Then the James Damon pointed out in [7] that if 1 r is not an eigenvalue of S V at x 0, then Ψ 1 is a local diffeomorphism. Therefore, we choose a neighbourhood W of the local manifold component S α so that Ψ 1 is a diffeomorphism on W. Therefore, for v T x0 S α we have Ψ 1 v = v + dr(v)u 1 + r U 1 v. Therefore, dr(v) = Ψ 1 v U 1 v U 1.

64 Chapter 3. Centroid Set, Skeletal Structure and the Singularity of the Radius Function 50 Hence dr(v) = 0 if and only if Ψ 1 v U 1 = v U 1 which completes the proof. In the following proposition we will study the relationship between the singularity of the radius function and the orthogonality of the radial vector field on the skeletal structure. Also, the relationship between the tangent space of the skeletal structure and its associated tangent space of the boundary will be studied in the case when the radius function has a singularity. Proposition Suppose (S, U) is a skeletal structure and let x 0 be a non-edge point. Let U be a smooth value for which 1 is not an eigenvalue of the radial shape operator r and the compatibility 1-form η U vanishes at x 0. Then the following are equivalent. 1. The radius function has a singularity at x The radial vector field U is orthogonal to the tangent space T x0 S of S at x The space T x0 S is parallel to the associated tangent space of the boundary T x X. (1 2) Can be proved directly from theorem (1 3) Assume that 1 r is not an eigenvalue of S V at x 0, then Ψ 1 is a local diffeomorphism. Therefore, we choose a neighbourhood W of the local manifold component S α so that Ψ 1 is a diffeomorphism on W. Let B = {v 1, v 2,..., v n } be a basis for the tangent space T x0 S α. Then, {v 1, v 2,..., v n} such that v i = Ψ 1 v i = v i + dr(v i )U 1 + r U 1 v i (3.3) is a basis for the tangent space of the boundary. Now the dot product with U 1 for both sides of equation 3.3, gives v i U 1 = Ψ 1 v i U 1 = v i U 1 + dr(v i ) = η U (v i ).

65 Chapter 3. Centroid Set, Skeletal Structure and the Singularity of the Radius Function 51 From the definition, the compatibility 1-form vanishes if and only if η U (v) = v U + dr(v) = 0, which means that the radial vector field is perpendicular to the tangent space of the boundary. Now assume that the radius function has a singularity at x 0, then v i U 1 = 0 for i = 1, 2,..., n, thus the tangent spaces T x X and T x0 S are parallel. Conversely assume T x X and T x0 S are parallel, then U 1 is perpendicular to T x0 S, and from the compatibility condition the radius function has a singularity. Corollary Suppose Ω R n+1 is a region with smooth boundary X and Blum medial axis and radial vector field (S, U). Let x 1 X be a point for which the projection on the medial axis along the normal to X is a local diffeomorphism (with x 1 mapping to x 0 in S). Then the radius function r has a singularity if and only if U T x0 S. By the Blum condition we have U X and since the projection along normal is a local diffeomorphism, then its inverse, which is in this case Ψ 1, is a local diffeomorphism. Also, 1 is not an eigenvalue of S r V. Hence the result comes directly from proposition. Now we will generalize what Peter Giblin pointed out when the radius function has a singularity in the case of symmetry sets in R 3 [11] to skeletal structures in R n+1. In general the radius functions need not to be same at the both sides of a skeletal structure on a neighbourhood of a smooth point. But if we have the same radius function on both sides of the skeletal structure, does the singularity of the radius function affect the relationship between the skeletal point and its associated midlocus? Also, what is the relationship between the tangent spaces of the boundary at the tangency points in the case when the radius function has a singularity? The answer of these questions is given in the following.

66 Chapter 3. Centroid Set, Skeletal Structure and the Singularity of the Radius Function 52 Proposition Let (S, U) be a skeletal structure in R n+1 such 1 is not an eigenvalue r of the radial shape operators and the compatibility 1-form vanishes identically on a neighbourhood of a smooth point x 0 S and suppose that the radius function is the same for the two sides of S on a neighbourhood of x 0. Then the following are equivalent 1. The radius function has a singularity at x x 0 and the associated midlocus point x m coincide. 3. The tangent spaces of the boundary at the tangency points are parallel. (1 2) can be proved directly from theorem (1 3) Let Ψ 1 be the radial map on one side of S and Ψ 2 be the radial map on the other side of S. Since 1 r is not an eigenvalue of the radial shape operator on both sides then Ψ 1 and Ψ 2 are local diffeomorphisms at x 0. Therefore, we can choose a neighbourhood W of x 0 so that Ψ 1 and Ψ 2 are diffeomorphisms on W. Let B = {v 1, v 2,..., v n } be a basis for T x0 S then, B 1 = {v 1, v 2,..., v n} and B 2 = {v 1, v 2,..., v n} are bases for the tangent spaces of the boundary at the tangency points such that and v i = Ψ 1 v i = v i + dr(v i )U 1 + r U 1 v i, v i = Ψ 2 v i = v i + dr(v)u 2 + r U 2 v i, where U 1 is the smooth value of the unit radial vector field on one side of S and U 2 is the smooth value of the radial vector field on the other side. Now from proposition we have that T x0 S is parallel to the tangent spaces of the boundary at the tangency points, thus the tangent spaces of the boundary at the tangency points are parallel. From this proposition we can see the impact of the singularity of the radius function on

67 Chapter 3. Centroid Set, Skeletal Structure and the Singularity of the Radius Function 53 the relationship between the radial vector field and the normal of the skeletal structure at a smooth point (see figure 3.3). Corollary Suppose Ω R n+1 is a region with smooth boundary X and Blum medial axis and radial vector field (S, U). Let x 1 and x 2 be two points on X for which the projections onto the medial axis along normals are local diffeomorphisms (with x 1 and x 2 mapping to x 0 S). Then the following are equivalent 1. The radius function has a singularity. 2. x 0 and the associated midlocus x m coincide. 3. The tangent spaces of the boundary at x 1 and x 2 are parallel. By Blum condition we have U X and since the projections along the normals are local diffeomorphisms. Then, their inverses which are in this case Ψ 1 and Ψ 2 are local diffeomorphisms. Thus, 1 r is neither an eigenvalue of S V 1 nor S V2, where S V1 (resp. S V2 ) is the matrix representing the radial shape operator on one side (resp. the matrix representing the radial shape operator on the other side). proposition Therefore, we can apply N U 1 N = U 1 T x0 S x 0 The radius function has a singularity x 0 T x0 S N U 2 N = U 2 Figure 3.3: The case when the radius function has a singularity.

68 Chapter 3. Centroid Set, Skeletal Structure and the Singularity of the Radius Function 54 In proposition we discussed the effect of the singularity of the radius function on the relationship between a smooth skeletal point and the associated midlocus. But, we assume that the radius functions are same for the two sides of the skeletal set in a neighbourhood of a smooth point. The logical question is : given a smooth skeletal point, when does this point and its associated midlocus coincide? The answer of this question is given in the following proposition. Proposition Let (S, U) be a skeletal structure and x 0 S be a smooth point. Then, x 0 and the associated midlocus coincide if and only if r 1 (x 0 ) = r 2 (x 0 ) and U 1 (x 0 ) = U 2 (x 0 ). In this case the midlocus is nothing but the centroid point, thus x m = x (r 1(x 0 )U 1 (x 0 ) + r 2 (x 0 )U 2 (x 0 )). Now assume that x 0 and the associated midlocus coincide, then we have r 1 (x 0 )U 1 (x 0 ) + r 2 (x 0 )U 2 (x 0 ) = 0. Therefore, r 1 U 1 = r 2 U 2 which implies that r 1 U 1 = r 2 U 2. (3.4) Now since U 1 and U 2 are unit vectors and r 1 and r 2 are positive. Then equation 3.4 holds when r 1 = r 2 and U 1 = U 2. The converse is obvious. Lemma Let (S, U) be a skeletal structure such that the compatibility condition holds on a neighbourhood of a smooth point x 0 S. If the radius function has a singularity, then A V = 0.

69 Chapter 3. Centroid Set, Skeletal Structure and the Singularity of the Radius Function 55 Let {v 1, v 2,..., v n } be a basis for the tangent space of the skeletal set at x 0. James Damon shows in [7] that A V = SV T V U 1 and since the compatibility condition holds, then V U = dr(v ), thus A V = S T dr. If the radius function has a singularity we have A V = 0. Now we will define a function that plays a central role in the relationship between the matrices representing the radial shape operator and the geometric shape operator of the skeletal structure. Now let (S, U) be a skeletal structure in R n+1. Define the function ρ : S R by ρ = U 1 N, (3.5) where N is the unit normal of S at x 0 (x 0 is a non-edge point). Originally this function was introduced by James Damon [8] and he called it the normal component function for U 1. Let ω = {v 1, v 2,..., v n } be a basis for the tangent space of S at x 0 (for a non-edge singular point ω is a basis for the limiting tangent space). Differentiate equation 3.5 with respect to v i we obtain ρ = U 1 N + N U 1, i = 1, 2,..., n. v i v i v i This equation can be written in vector form by ρ V = U 1 V N + N V U 1 = (A V U 1 S T V V ) N S T mv U 1 = A V U 1 N S T mv U 1 = ρa V S T mv U 1

70 Chapter 3. Centroid Set, Skeletal Structure and the Singularity of the Radius Function 56 = (ρsv T Sm)V T U 1, where S V (resp. S m ) is the matrix representing the radial shape operator (resp. the differential geometric shape operator) on S and v 1 dρ(v 1 ) v V = 2 dρ(v and dρ(v ) = 2 )... dρ(v n ) v n Therefore, we can summarize this in the following proposition. Proposition Let (S, U) be a skeletal structure in R n+1. Let x 0 S (x 0 be a nonedge point ) and define ρ = U 1 N where N is the unit normal of S at x 0. Then ρ V = (ρst V S T m)v U 1, where S V and S m are the matrices representing the radial shape operator and the differential geometric shape operator of S at x 0 respectively. Corollary Let (S, U) be a skeletal structure such that for a choice of smooth value of the radial vector field U the compatibility 1-form η U vanishes in a neighbourhood of a non-edge point x 0. Define the function ρ as in proposition 3.3.8, then ρ V = dρ(v ) = (ρst V S T m)dr(v ). Since the compatibility condition holds, then η U = 0. Therefore, 0 = dr(v i ) + v i U 1, i = 1, 2,..., n. Thus dr(v ) = V U 1.

71 Chapter 3. Centroid Set, Skeletal Structure and the Singularity of the Radius Function 57 Hence dρ(v ) = (ρs T V S T m)dr(v ). Therefore, the proof is completed. Corollary Let (S, U) be a skeletal structure as in corollary If the radius function has a singularity then ρ has a singularity but the converse is not true. If the radius function has a singularity, then it is obvious that ρ has a singularity. The converse is not true (see example ). Corollary Assume as in corollary If the radius function r has no singularity and ρ has a singularity at x 0, then ρ(x 0 ) is a generalized eigenvalue of the pair (S m, S V ). Recall that a 0 is a generalized eigenvalue of the pair (A, B) if det(a ab) = 0. Now assume that the radius function has no singularity at x 0 and the function ρ has a singularity, then we have 0 = (ρsv T Sm)dr(V T ) or or 0 = (ρs T V S T m)dr(v ) 0 = (S T m ρs T V )dr(v ) and since dr(v ) 0, then the matrix (S m ρs V ) T is not invertible. i.e., det(s m ρs V ) = 0. Thus ρ is a generalized eigenvalue of the pair (S m, S V ).

72 Chapter 3. Centroid Set, Skeletal Structure and the Singularity of the Radius Function 58 Example Let (S, U) be a skeletal structure in R 3 and s 1 (x, y) = (x, y, 1 2 κ 1x κ 2y 2 + h.o.t.) S reg and let r(x, y) = r 0 + ax by2, (a, b R, s.t a 2 < 1) be the radius function we define the unit radial vector field by U 1 = r + 1 r 2 N, where r is the Riemannian gradient of r and N is the unit normal of s 1. In this case the compatibility condition holds. Now at the origin, direct calculations show that ρ = 1 a 2, dr = a, dρ = 0, Sm T = κ 1 0 and κ 2 S T V = κ 1 1 a b + κ 2 1 a 2. Now (ρs T V S T m)dr = κ 2 a 2 + b 1 a 2 a 0 = 0 0 = dρ. It is clear that if a 0, then the radius function has no singularity, but as shown from the calculations ρ has a singularity at the origin which means that the singularity of ρ does not imply the singularity of the radius function and this supports our result in corollary Moreover, at the origin ρ is a generalized eigenvalue of the pair (S m, S V ). Corollary Let (S, U) be a Blum medial axis and radial vector field of a region Ω R n+1 with smooth boundary X. Let x 0 S be a non-edge point then dρ(v ) = (ρs T V S T m)dr(v ).

73 Chapter 3. Centroid Set, Skeletal Structure and the Singularity of the Radius Function 59 Since S is a Blum medial axis, then by Blum condition the radial vector field is perpendicular to the boundary. Therefore, η U = 0. Thus, dr(v ) = V U 1 which implies that dρ(v ) = (ρsv T Sm)dr(V T ). Thus the proof is completed. 3.4 Pre-medial Axis The pre-symmetry sets of 2D and 3D shapes had been studied by Giblin and Diatta [10]. In this section we study the relationship between the parameters of the skeletal structure in a neighbourhood of a smooth point and the parameters of the boundary in a neighbourhood of the associated point. By this way we are able to transfer to the relationship between the parameters of the boundary. Definition Given a smooth hypersurface X R n+1, the pre-symmetry set is the closure of the set of pairs of distinct points (p, q) X X for which there exists a hypersphere tangent to X at p and at q. Lemma Let (S, U) be a skeletal structure of a region Ω R n+1 with smooth boundary X such that for a choice of smooth value of U the compatibility condition holds and 1 is not an eigenvalue of the radial shape operator S r rad. Let x 0 be a smooth point of S and ε 0 (x 0 ) S be a neighbourhood of x 0. Also, let ε 1 (x 1 ) X be a neighbourhood of x 1 = x 0 + ru 1. If ε 0 (x 0 ) parametrized by (s 1, s 2,..., s n ) and ε 1 (x 1 ) parametrized by (t 1, t 2,..., t n ), then the map: φ : (s 1, s 2,..., s n ) (t 1, t 2,..., t n )

74 Chapter 3. Centroid Set, Skeletal Structure and the Singularity of the Radius Function 60 is a local diffeomorphism. From the boundary point definition we have x 1 = x 0 +ru 1 and since 1 is not an eigenvalue r of the radial shape operator S rad, then the radial map is a diffeomorphism. As a notation let x 1 s i = ε 1 s 1 x1. Therefore, we have v 1 = x 1 s 1 = x 1 t 1 t 1 s 1 + x 1 t 2 t 2 s x 1 t n t n s 1 = v 1 + r s 1 U 1 + r U 1 s 1 or ( t 1 s 1 t 2 s 1... t n s 1 ) v 1 v 2... = v 1 + r s 1 U 1 + r U 1 s 1. v n Therefore, t 1 t 2 s 1 s 1 t 1 t 2 s 2 s t 1 t 2 s n s n t n s 1 t n s 2. t n s n v 1 v 2. v n = V + dr(v )U 1 + r U 1 V. Now since the radial map is a local diffeomorphism then the matrix t 1 t 2 t n s 1 s 1 s 1 v t 1 t 2 t 1 n v A = ΓV = s 2 s 2 s t 1 t 2 t n v n s n s n s n

75 Chapter 3. Centroid Set, Skeletal Structure and the Singularity of the Radius Function 61 has a maximal rank, i.e., rank(a) = n. Also, since the boundary X is smooth then the matrix V has rank n. Therefore, rank(a) = rank(γ) = n. Hence the map φ : (s 1, s 2,..., s n ) (t 1, t 2,..., t n ) is a local diffeomorphism. Proposition Let (S, U) be a Blum medial axis and radial vector field of a region Ω R n+1 with smooth boundary X. Let x 0 be a smooth point of S and ε 0 (x 0 ) S be a neighbourhood of x 0. Also, let ε 1 (x 1 ) X be a neighbourhood of x 1 = x 0 + ru 1. If ε 0 (x 0 ) parametrized by (s 1, s 2,..., s n ) and ε 1 (x 1 ) parametrized by (t 1, t 2,..., t n ), then the map: φ : (s 1, s 2,..., s n ) (t 1, t 2,..., t n ) is a local diffeomorphism. Since we are in the Blum case the compatibility condition holds and 1 is not an eigenvalue r of the radial shape operator. Therefore, we can apply lemma Lemma and proposition give us enough tools to study the relationship between the boundary parameters at the tangency points associated to a smooth point on the medial axis. Proposition Let (S, U) be a Blum medial axis and radial vector field of a region Ω R n+1 with smooth boundary X. Let x 1 and x 2 be the tangency points associated to a smooth point in x 0 S with neighbourhoods ε 1 (x 1 ) and ε 2 (x 2 ) respectively. If ε 1 (x 1 ) parametrized by (s 1, s 2,..., s n ) and ε 2 (x 2 ) parametrized by (t 1, t 2,..., t n ), then the map: φ : (s 1, s 2,..., s n ) (t 1, t 2,..., t n )

76 Chapter 3. Centroid Set, Skeletal Structure and the Singularity of the Radius Function 62 is a local diffeomorphism. Let ε 0 (x 0 ) be a neighbourhood of x 0 S parametrized by (z 1, z 2,..., z n ). Then by proposition φ 1 : (z 1, z 2,..., z n ) (s 1, s 2,..., s n ) and φ 2 : (z 1, z 2,..., z n ) (t 1, t 2,..., t n ) are local diffeomorphism. But φ = φ 2 φ 1 1. Therefore, the map φ : (s 1, s 2,..., s n ) (t 1, t 2,..., t n ) is a local diffeomorphism. 3.5 Singularity of the Midlocus in the Case of Skeletal Structure in R 3 In the rest of this chapter we focus on the singularity of the midlocus in R 3. Precisely we take a smooth point x 0 S reg and we take M(x, y) as a local parametrization of the smooth stratum containing x 0 around x 0. In fact, we will study corank one singularities and to do so we need a general form of the midlocus to deal with and lemma fulfils this need.

77 Chapter 3. Centroid Set, Skeletal Structure and the Singularity of the Radius Function 63 Note: In this section we are dealing with the case when the radius function is the same for both sides of M and the compatibility condition holds and in this case the midlocus point associated to x 0 is given by x m = x 0 r(x 0 ) r(x 0 ). Lemma Let M(x, y) = (x, y, f(x, y)) be a local parametrization of a smooth stratum of skeletal set around a smooth point x 0 S reg and r(x, y) be the radius function, then the midlocus is given by M(x, y) = (g, h, l), where g = x + xf x 2 + xfy 2 rr x rr x fy 2 + rr y f x f y, 1 + fx 2 + fy 2 and h = y + yf x 2 + yfy 2 rr y rr y fx 2 + rr x f x f y, 1 + fx 2 + fy 2 l = f + ff2 x + ffy 2 rr x f x rr y f y. 1 + fx 2 + fy 2 Let M(x, y) = (x, y, f(x, y)) and r be the radius function, then the midlocus point x m associated to x 0 M is given by x m = x 0 r r, r = dr T Im 1 V, where V = v 1 = v f x 0 1 f y, and I m = 1 + f x 2 f x f y. f x f y 1 + f 2 y Thus Im 1 = 1 + f 2 y 1 + fx 2 + fy 2 f x f y f x f y 1 + fx 2 + fy fx f 2 x + f 2 y 1 + f 2 x + f 2 y Therefore, ( rx + r x fy 2 r y f x f y r =, r y + r y fx 2 r x f x f y, r ) xf x + r y f y. 1 + fx 2 + fy fx 2 + fy fx 2 + fy 2

78 Chapter 3. Centroid Set, Skeletal Structure and the Singularity of the Radius Function 64 Thus, after some calculations the result follows. In the forthcoming results and examples we need some concepts from the theory of surface in R 3, these concepts are given in the following definition. Definition Let S be a regular surface parametrized by X(x, y), then 1. The shape operator (or Weingarten map ) at each point p S is defined by S p : T p S T p S, u u n, where n is the unit normal of the S. 2. The first and the second fundamental forms of S are the quadratic forms on the tangent plane defined by I(u, v) = u v and II(u, v) = u S p (v) respectively, they are represented by the matrices I = X x X x X x X y = E F and X x X y X y X y F G II = X xx n X xy n = L M. X yx n X yy n M N 3. The normal curvature k p in the tangent direction w = ax x + bx y is defined by k p (w) = II(w) I(w) = a2 L + 2abM + b 2 N a 2 E + 2abF + b 2 G. 4. The geodesic torsion in the direction of a unit vector w is defined by τ g (w) = II(w, w ), where w is the unit vector perpendicular to w. 5. If (cos θ, sin θ) is a direction on S with respect to a principal coordinate system ( A principal coordinate system is one where the x-axis and y-axis are always the principal directions ) then k p = κ 1 cos 2 θ + κ 2 sin 2 θ and τ g = (κ 2 κ 1 ) sin θ cos θ.

79 Chapter 3. Centroid Set, Skeletal Structure and the Singularity of the Radius Function 65 Definition Let f : R n R m be a smooth map, then the k-jet j k f at a point p is the Taylor expansion about p truncated at degree k. Definition Let f : R n R m be a smooth map, then the corank δ of f is defined by δ = min(n, m) rank(df), where df is the differential of f. Example Consider the f : R 2 R 3 such that (x, y) (x, xy, y 2 ), then the 1 0 differential of f is given by df = y x. 0 2y Theorem Let M be a smooth stratum of skeletal structure (S, U) in R 3 containing a smooth point x 0 S reg and r be a radius function with singularity at x 0. Let λ 1 and λ 2 be the eigenvalues of the Hessian of r at x 0 with r(x 0 ) = 1, λ 1 0 and let x m be the λ 1 associated midlocus point to x 0, then 1. The midlocus is parametrized by a corank two singularity at x m if and only if λ 1 = λ The midlocus is parametrized by a corank one singularity at x m if and only if λ 1 λ 2. Let M be a smooth stratum of skeletal structure (S, U) in R 3 containing a smooth point x 0 and r be the radius function with singularity at x 0 and r(x 0 ) = 1, λ 1 0 where λ 1 is λ 1 an eigenvalue of the Hessian of r at x 0. Now we parameterize M locally at x 0 such that (0, 0) x 0 = (0, 0, 0) and M is in Monge form i.e., M( x, ỹ) = ( x, ỹ, 1κ 2 1 x 2 + 1κ 2 2ỹ 2 + h.o.t) and the radius function is given by r( x, ỹ) = b b 20 x b 02 ỹ 2 + b 11 xỹ b 12 xỹ b 21 x 2 ỹ + h.o.t.

80 Chapter 3. Centroid Set, Skeletal Structure and the Singularity of the Radius Function 66 Now we rotate the new coordinates in the source by x cos t sin t = ỹ sin t cos t x y such that the radius function transforms to r(x, y) = b b 20x b 02y 2 + h.o.t. This rotation transforms M to M(x, y) = (x cos t + y sin t, x sin t + y cos t, 1 2 a 20x a 02y 2 + a 11 xy + h.o.t), where a 20 = κ 1 cos 2 t+κ 2 sin 2 t, a 0,2 = κ 1 sin 2 t+κ 2 cos 2 t, and a 11 = (κ 1 κ 2 ) sin t cos t. Now we rotate the coordinates in the target around z- axis by X cos( t) sin( t) 0 X Y = sin( t) cos( t) 0 Y, Z Z where X, Y, Z are the old coordinates and X, Y, Z are new coordinates. Thus, this transforms M to M(x, y) = (x, y, 1 2 a 20x a 02y 2 + a 11 xy + h.o.t). Now we use lemma to find the form of the midlocus, and by using Maple (see the linear parts of equations A.8, A.9 and A.10 in the appendix), we get j 1 M = ((1 b 00 b 20 )x, (1 b 00 b 02 )y, 0). Observe that, corollary tells us the midlocus is singular when 1 r is an eigenvalue of H r. This description of j 1 M allows a verification of this result in R 3. Now the Hessian matrix of the radius function is H r = b 20 0, without loss of 0 b 02 generality we put λ 1 = b 02, and λ 2 = b 20 and since b 00 = r(x 0 ) the 1-jet of the midlocus is now ( λ 1 λ 2 x, 0, 0). Therefore, the Jacobian matrix of the midlocus is λ 1 λ 1 λ dx m = λ 1. Thus, dx m has rank zero if and only if λ 1 = λ 2, and has 0 0 0

81 Chapter 3. Centroid Set, Skeletal Structure and the Singularity of the Radius Function 67 rank one if and only if λ 1 λ 2, hence the results have been proved. Now in the rest of this section we will study the singularity of the midlocus in R 3 in the case when it has corank one this means that the eigenvalues of the Hessian of the radius function are distinct. We will use the finite determinacy to study this singularity. First of all we state some needed definitions. Definition Two map-germs f i : (R n, 0) (R p, 0) (i = 1, 2) are A-equivalent if there exist germs of C -diffeomorphisms ϑ and φ such that φ f 1 = f 2 ϑ holds, where ϑ : (R n, 0) (R n, 0) and φ : (R p, 0) (R p, 0). Definition A map-germ f j k g(0) = j k f(0), then g is A-equivalent to f. : (R n, 0) (R p, 0) is k-determined if whenever Definition The crosscap or Whitney umbrella is a map-germ A-equivalent to (x, y) (x, xy, y 2 ) at the origin. Figure 3.4: Crosscap or Whitney umbrella. Since we are dealing with finite determinacy we focus on the 2-jet and 3-jet of the midlocus. First of all, we classify the second jet and for this we need the following theorem which was proved by Mond in [22]. Theorem The map-germ (x, y) (x, xy, y 2 ) is stable and 2-determined.

82 Chapter 3. Centroid Set, Skeletal Structure and the Singularity of the Radius Function 68 Lemma If the map-germ f : (R 2, 0) (R 3, 0) has a corank one singularity with j 2 f = (x, axy + by 2, cxy + dy 2 ), then f is A-equivalent to the crosscap if and only if ad cb 0. Assume that ad cb 0, then from the Mond classification in [22] (proposition 4.2) j 2 f transforms by an appropriate coordinate change to (x, xy, y 2 ) and since the crosscap is 2-determined thus f is A-equivalent to the crosscap. Conversely, assume that ad cb = 0 and f is A-equivalent to the crosscap. Therefore, j 2 f is A-equivalent to (x, xy, y 2 ). But since ad cb = 0, then from Mond classification j 2 f transforms to one element of the set {(x, 0, 0), (x, xy, 0), (x, y 2, 0)} and no one of these elements is A-equivalent to (x, xy, y 2 ) which is a contradiction. Thus, f is A-equivalent to the crosscap if and only if ad cb 0. Now we state a theorem which gives a necessary and sufficient conditions for a midlocus to be A-equivalent to the crosscap. Theorem Let M be a smooth stratum of a skeletal structure (S, U) in R 3 containing a smooth point x 0 and r be the radius function with a singularity at x 0 and λ 1 and λ 2 be the eigenvalues of the Hessian of r and w 1 and w 2 are the associated eigenvectors such that λ 1 λ 2, and r(x 0 ) = 1 λ 1, λ 1 0, then the midlocus at x m associated to x 0 is A-equivalent to the crosscap if and only if λ 1 k x0 (w 1 ) 2 w 1 w2 r 2λ 2 τ g 3 w 1 r, where k x0 (w 1 ) is the normal curvature of M in the direction w 1, τ g is the geodesic torsion of M in the direction w 1, and wi r is the directional derivative of the radius function in the direction w i, i = 1, 2.

83 Chapter 3. Centroid Set, Skeletal Structure and the Singularity of the Radius Function 69 We repeat the same procedure of the proof of theorem and thus we have r = b b 20x b 02y b 30x b 21x 2 y b 12xy b 03y 3 + b 40 x 4 + b 31 x 3 y + b 22 x 2 y 2 + b 13 xy 3 + b 04 y 4 + h.o.t and M(x, y) = (x, y, 1a 2 20x 2 + a 11 xy + 1a 2 02y 2 + h.o.t). As in the proof of theorem without loss of generality, we put λ 1 = b 02 and λ 2 = b 20 and by using lemma we find the form of the centroid set and using Maple in calculations ( see equations A.8, A.9 and A.10 in the appendix) we get j 2 M c = (p, q, s), where p = (1 b 00 b 20 )x b 00 b 30 x 2 b 00 b 21 xy 1 2 b 00b 12 y 2, s = q = 1 2 b 00b 21 x 2 b 00 b 12 xy b 00 b 03 y 2 and ( ) 1 2 a 20 b 00 b 20 a 20 x 2 b 00 b 20 a 11 xy 1 2 a 02y 2. Now consider the parameter change in the source x = u + b 12 uy + b 30 u 2 b 12 + b 02 b20 b 02 b 20 2(b 02 b 20 ) y2. This parameter change transforms j 2 M (see equations A.11, A.12 and A.13 in the appendix ) to (p, q, s), where and p = b 02 b 20 b 02 u, q = b 21 2b 02 u 2 b 12 b 02 uy b 03 b 02 y 2, s = a 20(b 02 b 20 ) 2b 02 u 2 a 11b 20 b 02 uy 1 2 a 02y 2. Now consider the coordinate change in the target b 02 X = X, Y = Y + b 21 b 02 ( ) b 02 b 20 2b 02 b 02 b 20 2 X 2, Z = Z a 20b 02 2(b 02 b 20 ) X2, where X, Y, Z are the old coordinates and X, Y, Z are new coordinates. This coordinate change transforms j 2 M into ( j 2 M = u, b 12 uy b 03 y 2, a 11b 20 uy 1 ) b 02 b 02 b 02 2 a 02y 2.

84 Chapter 3. Centroid Set, Skeletal Structure and the Singularity of the Radius Function 70 Now using lemma the midlocus is A-equivalent to the crosscap if and only if b 12 b 03 b 02 b 02 a 11b b 02 2 a 02 if and only if b 02 a 02 b 12 2b 20 a 11 b 03 0, if and only if b 02 a 02 b 12 2b 20 a 11 b 03 but b 02 = λ 1, b 20 = λ 2, a 02 = κ 2 cos 2 t+κ 1 sin 2 t = k x0 (w 1 ), a 11 = (κ 1 κ 2 ) sin t cos t = τ g, b 12 = r yyx (0, 0) = 2 w 1 w2 r, and b 03 = r yyy (0, 0) = 3 w 1 r, thus the result has been proved. Remark It is obvious from the above theorem that the conditions are generic and the geometry of the surface M plays a central role for the centroid to have a crosscap singularity. Corollary Assume as in theorem If x 0 is a planar point, i.e., κ 1 = κ 2 = 0, then the midlocus is not A-equivalent to the crosscap. If κ 1 = κ 2 = 0, then k x0 (w 1 ) = τ g = 0, thus the midlocus is not A-equivalent to the crosscap. Example Let M(x, y) = (x, y, y 2 ) and r(x, y) = 1 + x 3 + xy y2, then the midlocus is singular at the origin and direct calculation gives M(x, y) = (p, q, s), where p = x (xy y2 + 4x 3 y x2 y 2 + 3x 2 + 3x 3 ), q = y(7y2 4x 4x 4 2x 3 4x 2 y 2 4xy 2 ) 2 + 8y 2

85 Chapter 3. Centroid Set, Skeletal Structure and the Singularity of the Radius Function 71 and s = y2 (3y 2 4x 4x 4 2x 3 4x 2 y 2 4xy 2 1) 2 + 8y 2. Now we calculate the values of the terms are in the non-inequality of theorem It is clear that the Hessian matrix at the origin is given by H r = 0 0 and its 0 1 eigenvalues are λ 1 = 1 and λ 2 = 0 with associated eigenvectors w 1 = t and w 2 = t respectively. Now we do not need to calculate 3 w1r and τ g since λ 2 = 0. The tangent vector in the direction of w 1 is given by 0.M x + M y = M y, thus the normal curvature in the direction of w 1 is given by Now we calculate 2 w1 w2 r, k 0 (w 1 ) = II(w 1) I(w 1 ) = 2. w2 r = (r x, r y ) (1, 0) = r x = 3x 2 + y 2, w1 w2 r = (6x, 2y) (0, 1) = 2y and 2 w 1 w2 r = 2. Therefore, λ 1 k x0 (w 1 ) 2 w 1 w2 r 2λ 2 τ g 3 w 1 r. That is, the singularity of the midlocus is a crosscap.

86 Chapter 3. Centroid Set, Skeletal Structure and the Singularity of the Radius Function 72 Figure 3.5: Figure of example Example Let M(x, y) = (x, y, y 2 ) and consider the family of radius functions r(x, y) = y2 + µy 3 + xy x2. Now we discuss the conditions in theorem in this example and from the first look at this example someone could ask, " Do we need y 3 in the radius function to allow the midlocus to have a crosscap singularity?" The answer to this question will be given through the following geometric discussion. The Hessian matrix of the radius function at the origin is given by H r = 1 1 2, and the 1 2 eigenvalues are λ 1 = 5 and λ 2 2 = 0 and the associated eigenvectors are w 1 = t and w 2 = t 2 2 respectively. It is clear that r(0, 0) = 1 λ 1 and the tangent vector 1 of M in the direction of w 1 is given by w = M x + 2M y, hence k 0 (w 1 ) = II(w 1) I(w 1 ) = 8 5. Since λ 2 = 0, we do not need to calculate τ g and 3 w 1 r, so in this case the midlocus

87 Chapter 3. Centroid Set, Skeletal Structure and the Singularity of the Radius Function 73 is a crosscap if and only if 2 w 1 w2 r 0. Now direct calculations show that w2 r = (r x, r y ) ( 2, 1) = 2y x+x+2y+3µy 2 = 3µy 2, w1 w2 r = (0, 6µy) (1, 2) = 12µy, and finally 2 w 1 w2 r = (0, 12µ) (1, 2) = 24µ 0 if and only if µ 0. Thus, it is vitally important that the radius function should have a non zero coefficient for the y 3 term. Now take µ = 1, then the midlocus is A equivalent to the crosscap at the origin and the direct calculation gives M(x, y) = (g, h, l), where g = 4 5 x 1 8 x3 3 4 x2 y 2 5 y 3 2 xy2 y 3 y xy3, h = (4y + 40y3 24y 2 8x 100y 4 60xy 2 60y 5 80y 3 x 30x 2 y 15x 2 y 2 5x 3 ), 20(1 + 4y 2 ) and l = y(6y + 24y2 + 8x + 100y xy y xy x 2 y + 15x 2 y 2 + 5x 3 ). 10(1 + 4y 2 ) Remark Example shows that if r is R-equivalent to r, then we do not necessarily have the midlocus M associated to r is A-equivalent to M.

88 Chapter 3. Centroid Set, Skeletal Structure and the Singularity of the Radius Function 74 Figure 3.6: Figure of example Now we will study the singularity of the midlocus when it fails to have a crosscap. First of all, we state the following theorem. Theorem [22] A map germ (R 2, 0) (R 3, 0) whose 2-jet is equivalent to (x, y 2, 0) is equivalent to a germ of the form (x, y) (x, y 2, yp (x, y 2 )) for smooth P. Now we state the following theorem which indicates the form of the midlocus when it is not a crosscap.

89 Chapter 3. Centroid Set, Skeletal Structure and the Singularity of the Radius Function 75 Theorem Assume as in theorem and the midlocus is not A-equivalent to the crosscap. If k x0 (w 1 ) 0 or 3 w 1 r 0, then the midlocus parametrized locally by (u, v) (u, v 2, vp (u, v 2 )) for smooth P. From the proof of theorem we have ( j 2 M = u, b 12 uy b 03 y 2, a 11b 20 uy 1 ) b 02 b 02 b 02 2 a 02y 2. Since the midlocus is not A-equivalent to the crosscap, then b 12 b 03 b 02 b 02 a 11b 20 1 = 0. b 02 2 a 02 Now if k x0 (w 1 ) = a 02 0, then we can complete the square in the third component, and the corresponding change in the v variable then transforms j 2 M into (u, 0, v 2 ) and by the coordinate change X = X, Y = Z, Z = Y, we have j 2 M = (u, v 2, 0). Now if 3 w 1 r = b 0,3 0, then we can complete the square in the second component and the corresponding change in the v variable then transforms j 2 M into (u, v 2, 0). Thus in both case we apply theorem and the result holds. Example Let M(x, y) = (x, y, x 2 ) and r(x, y) = x2 + y 3, then the midlocus is singular at the origin and the direct calculation gives M(x, y) = (g, h, l), where g = x(7x2 2y 3 ) 2(1 + 4x 2 ), h = 1 2 y(3yx2 + 6y 4 + 6y 3) and l = x2 (3x 2 2y 3 1) 1 + 4x 2.

90 Chapter 3. Centroid Set, Skeletal Structure and the Singularity of the Radius Function 76 Figure 3.7: Figure of example Theorem gives the general form of the midlocus under the mentioned conditions. Now we will study the 3-jet of the midlocus when it fails to have a crosscap singularity and k x0 (w 1 ) 0. Particularly, we will discuss the conditions for the midlocus to have S 1 ± singularity, which is a map-germ A-equivalent to S 1 ± : (x, y) (x, y 2, x 2 y ± y 3 ) at the origin. Also, the conditions for the 3-jet of the midlocus to be A-equivalent to the cuspidal edge will be investigated. A cuspidal edge is a map-germ A-equivalent to CE : (x, y) (x, y 2, y 3 ) at the origin.

91 Chapter 3. Centroid Set, Skeletal Structure and the Singularity of the Radius Function 77 Lemma Let M be a smooth stratum of skeletal structure (S, U) in R 3 containing a smooth point x 0 and r be the radius function with a singularity at x 0. Let λ 1 and λ 2 (λ 1 λ 2 ) be the eigenvalues of the Hessian of r and w 1 and w 2 be the associated eigenvectors. Assume that r(x 0 ) = 1 λ 1, λ 1 0. If the midlocus fails to have a crosscap singularity and k x0 (w 1 ) 0, then the 3-jet of the midlocus is given by j 3 M = (u, v 2, w 1,2 u 2 v + w 3,0 v 3 ), where w 1,2 and w 3,0 are equations B.15 and B.14 respectively in appendix. We repeat the same procedure of theorems and and after using Maple in calculations we get j 3 M = (p, q, s) and we define a parameter change x = x(u, y) such that this parameter change transforms p into p = λ 1 λ 2 λ 1 u and again we define a parameter change y = y(u, v) such that this parameter change transforms s into and q into s = k 0,2 u 2 + k 0,3 u a 02v 2 q = w 0,2 u 2 + w 2,0 v 2 + w 1,1 uv + w 2,1 v 2 u + w 1,2 u 2 v + w 3,0 v 3 + w 0,3 u 3. Now we put j 3 M = (p, s, q), and we define the coordinate change in the target X = λ 1 λ 1 λ 2 X, Y = Y λ2 1k 0,2 (λ 1 λ 2 ) 2 X2 λ3 1k 0,3 (λ 1 λ 2 ) 3 X3, Z = Z λ2 1w 0,2 (λ 1 λ 2 ) 2 X2 λ3 1w 0,3 (λ 1 λ 2 ) 3 X3. This coordinate change transforms j 3 M into j 3 M = (u, 12 ) a 02v 2, w 1,1 uv + w 2,1 uv 2 + w 1,2 u 2 v + w 3,0 v 3 since the midlocus is not A-equivalent to the crosscap, then w 1,1 = 0. Now consider the coordinate change in the target X = X, Ỹ = 2 2w 2,1 Y, Z = Z + XY, a 0,2 a 02

92 Chapter 3. Centroid Set, Skeletal Structure and the Singularity of the Radius Function 78 where X, Y, Z are the old coordinates and X, Ỹ, Z are new coordinates. This coordinate change transforms j 3 M into j 3 M = (u, v 2, w 1,2 u 2 v + w 3,0 v 3 ) and thus the proof is completed. Proposition Assume as in lemma , then 1. If w 1,2 0 and w 3,0 0, then the 3-jet of the midlocus is equivalent to (u, v 2, u 2 v ± v 3 ) and consequently the midlocus is A-equivalent to an S 1 ± singularity. 2. If w 1,2 = 0 and w 3,0 0, then j 3 M = (u, v 2, v 3 ). 1. Assume that w 1,2 0 and w 3,0 0, then using the parameter change u = w 3,0 w 1,2 ũ, ( ) thus this transforms the 3-jet into j 3 M = w 3,0 w 1,2 ũ, v 2, w 3,0 (ũ 2 v ± v 3 ). Now consider the coordinate change in the target X = w 1,2 X, w 3,0 Y = Y, Z = 1 w 3,0 Z. Thus the 3-jet transformed by this to (ũ, v 2, ũ 2 v ± v 3 ), and since S ± 1 is 3-determined [17] the result has been proved. The second part is obvious. Now we will give an example when the midlocus is A-equivalent to S ± 1. Example Let M(x, y) = (x, y, ±y 2 ) and r(x, y) = 1 2 +y2 ±x 2 y 2 then the centroid set is A-equivalent to S ± 1 M(x, y) = (g ±, h ±, l ± ), where singularity at the origin and the direct calculation gives g ± = x(±2y 4 ± y 2 + 2x 2 y 4 1), h ± = y( 2y2 ± x 2 ± 4x 2 y 2 + 2x 4 y 2 ) 1 + 4y 2,

93 Chapter 3. Centroid Set, Skeletal Structure and the Singularity of the Radius Function 79 and l ± = y2 (2x 2 + 8x 2 y 2 ± 4x 4 y 2 ± 1) 1 + 4y 2. In the case of + the midlocus is equivalent to S 1, and in the case of the midlocus is equivalent to S + 1. Also, the direct calculations give that w ± 3,0 = 2 and w ± 1,2 = 1. Figure 3.8: Figure of example in the case, +.

94 80 Chapter 4 Relation Between Radial Geometry of Skeletal Structure and Differential Geometry of its Boundary 4.1 Introduction This chapter deals with the relationship between the radial geometry of the skeletal structure and the differential geometry of the associated boundary. In fact, James Damon studied this phenomenon in [8, 9] and he obtained a relationship between S V and S XV. Moreover, he expressed S XV in terms of S V and found out the link between the principal radial curvatures of the skeletal structure and the associated principal curvature of the boundary. In this chapter we express S V in terms of S XV (Proposition 4.2.4). Also, some algebraic properties between S V and S XV are investigated through out this chapter (Proposition 4.2.6). Moreover, the relationship between the Gauss radial curvature K r of a skeletal structure and its associated Gauss curvature K on the boundary has been studied as well as the relationship between the mean radial curvature H r and its associated mean curvature on the boundary (Proposition ). The final part of this chapter deals with

95 Chapter 4. Relation Between Radial Geometry of Skeletal Structure and Differential Geometry of its Boundary 81 the situation when K r and K coincide. We study the relationship between the radial skew curvature of a skeletal structure and the skew curvature of the boundary in situation when K r = K 0 in the case of skeletal structure in R Skeletal Structures in R n+1 Our aim in this section is to express the matrix of the radial shape operator in terms of the matrix of the differential geometric shape operator of the boundary. First of all, we give the following theorem which was proved by James Damon. Theorem ([8], Theorem 3.2 ) Let (S, U) be a skeletal structure such that for a choice of smooth value of U the associated compatibility 1-form η U vanishes identically on a neighbourhood of a non-edge point x 0 of S, and 1 r is not an eigenvalue of the radial shape operator at x 0. Let x 0 = Ψ 1 (x 0 ), and V be the image of V for a basis {v 1, v 2,..., v n }, then 1. The differential geometric shape operator S XV of the boundary X at x 0 has a matrix representation with respect to V given by S XV = (I rs V ) 1 S V. (4.1) 2. There is a bijection between the principal curvatures κ i of X at x 0 and the principal radial curvatures κ ri of S at x 0 (counted with multiplicities) given by κ i = κ ri 1 rκ ri or equivalently κ ri = κ i 1 + rκ i. (4.2) 3. The principal radial directions corresponding to κ ri are mapped by dψ 1 to the principal directions corresponding to κ i. One of the aims of this chapter is to express S V in terms of S XV the following lemmas. and to do so we need

96 Chapter 4. Relation Between Radial Geometry of Skeletal Structure and Differential Geometry of its Boundary 82 Lemma Let A be an (n n) matrix and suppose that 1, (α 0) is not eigenvalue α of A. If B = (I αa) 1 A, then B = 1 α [(I αa) 1 I], where I is the identity matrix. To prove this lemma it is enough to show that (I αa) 1 A 1 α [(I αa) 1 I] = 0. Now (I αa) 1 A 1 α (I αa) α I = (I αa) 1 (A 1 α I) + 1 α I = 1 α (I αa) 1 (I αa) + 1 α I = 1 α I + 1 α I = 0. Therefore, the proof is completed. Lemma Let A be an (n n) matrix and suppose that 1, (α 0) is not eigenvalue α of A. If B = (I αa) 1 A, then A = (I + αb) 1 B. From lemma we have B = 1 α [(I αa) 1 I]. Therefore, A = 1 α [I (I + αb) 1 ]. (4.3) Now our task is to show that (I + αb) 1 B A = 0. (4.4)

97 Chapter 4. Relation Between Radial Geometry of Skeletal Structure and Differential Geometry of its Boundary 83 Now (I + αb) 1 B A = (I + αb) 1 B 1 α [I (I + αb) 1 ] = 1 α (I + αb) 1 (I + αb) 1 α I = 1 α I 1 α I = 0. Hence equation 4.4 is satisfied. Now we are in the position to do express S V in terms of S XV. Using the above lemmas and theorem we have the following: Proposition Let (S, U) be a skeletal structure as in theorem 4.2.1, then the matrix S V representing the radial shape operator and the matrix S XV representing the differential geometric shape operator have the following relation S V = (I + rs XV ) 1 S XV or equivalently S V = 1 r [I (I + rs XV ) 1 ]. The proof of this theorem comes directly from theorem and the above lemmas. This proposition leads to the following corollary. Corollary Let (S, U) be a Blum medial axis and radial vector field of a region Ω R n+1 with smooth boundary X. Let x 0 = Ψ 1 (x 0 ) be the associated boundary point to a non-edge point x 0 S, and V be the image of V under dψ 1 for a basis {v 1, v 2,..., v n }. Then the matrix S V representing the radial shape operator at x 0 and the

98 Chapter 4. Relation Between Radial Geometry of Skeletal Structure and Differential Geometry of its Boundary 84 matrix S XV representing the differential geometric shape operator of the boundary at x 0 have the following relation S V = (I + rs XV ) 1 S XV. Since a Blum medial axis is a special case of the skeletal structure for which the compatibility 1-form vanishes and 1 r then, we can apply proposition to get the result. is not an eigenvalue of the radial shape operator The following proposition gives us two relations between S V and S V X conditions of theorem under the same Proposition Let (S, U) be a skeletal structure as in theorem 4.2.1, then 1. S XV S V = rs XV S V. 2. S XV S V = S V S XV, i.e., the operators commute. 1-From theorem we have S XV S V = (I rs V ) 1 S V S V = ((I rs V ) 1 I)S V = rs XV S V. 2- From (1) we have rs XV S V = S XV S V = S XV (I + rs XV ) 1 S XV (by proposition 4.2.5) = (I (I + rs XV ) 1 )S XV = rs V S XV (by proposition 4.2.5).

99 Chapter 4. Relation Between Radial Geometry of Skeletal Structure and Differential Geometry of its Boundary 85 Hence from above we have S V S XV = S XV S V. Remark If the S V is invertible then det(s XV ) = det((1 rs V ) 1 S V ) 0 thus S XV is invertible and vice versa. Thus if S V is invertible then, S 1 V S XV = S XV S 1 V and S 1 XV S V = S V S 1 XV. Now we will give an example to illustrate the results in proposition Example Let (S, U) be a skeletal structure in R 3 and let S 1 (x, y) = (x, y, 1 2 κ m1x κ 2 m2y 2 + h.o.t) S reg. and r = r ax by2 be the radius function on S 1 such that 1 r 0 / {κ m1 + a, κ m2 + b}. Now we define the unit radial vector field by U 1 = r + 1 r 2 N, where r is the Riemannian gradient of the radius function and N is the unit normal of S 1. Direct calculation shows that at the origin we have S V = κ m1 + a 0 0 κ m2 + b, and S XV = κ m1 + a 1 r 0 (κ m1 + a) 0 0 κ m2 + b 1 r 0 (κ m2 + b). It is clear that S V S XV = S XV S V = (κ m1 + a) 2 1 r 0 (κ m1 + a) 0 0 (κ m2 + b) 2 1 r 0 (κ m2 + b). Now S XV S V = r 0 (κ m1 + a) 2 1 r 0 (κ m1 + a) 0 0 r 0 (κ m2 + b) 2 1 r 0 (κ m2 + b) = r 0S V S XV.

100 Chapter 4. Relation Between Radial Geometry of Skeletal Structure and Differential Geometry of its Boundary 86 Definition Let S V skeletal structure and S XV be the matrix representation of the radial shape operator of a be the matrix representation of the differential geometric shape operator of the associated boundary, then the Gaussian radial curvature K r and the mean radial curvature H r are given by K r = det(s V ) and H r = 1 n tr(s V ), and the Gaussian curvature K and the mean curvature of the boundary are given by K = det(s XV ) and H = 1 n tr(s XV ). The i th mean radial curvature K ri is defined by ( ) 1 n K ri = κ rj1....κ rji i j 1 <j 2 <...<j i and the associated i th mean curvature of the boundary is defined by ( ) 1 n K i = κ j1....κ ji. i j 1 <j 2 <...<j i Now we will turn to the relation between the Gaussian radial curvature K r of the skeletal structure at a non-edge point and the Gaussian curvature K of the boundary at the associated point. Also, the relation between the mean radial curvature H r of the skeletal structure and the associated mean curvature H of the boundary will be investigated. In fact Anthony Pollitt studied in [24] the relationship between the principal radial curvatures and the associated principal curvatures on the boundary in the case of medial axis in R 3. In the following proposition we generalize the result obtained by Pollitt to the higher dimensions in the case of skeletal structure which is more general than medial axis and we give other results as well. Proposition Let (S, U) be a skeletal structure as in theorem 4.2.1, then 1. K = 1 rnh r + n 1 i=2 K r ( 1) i r i( n i ) Kri + ( 1) n r n K r.

101 Chapter 4. Relation Between Radial Geometry of Skeletal Structure and Differential Geometry of its Boundary H = 3. K r = 4. H r = H r rnh 2 r + r n i=2 n κ 2 ri + n j=3 ( 1) j 1 r j 1( n j) Kri. 1 rnh r + n 1 ( 1) i r i( ) n Kri + ( 1) n r i n K r K. 1 + rnh + n 1 r i( n i) Ki + r n K i=2 H + rnh r n 5. κ rl (1 + rnh + n 1 6. κ l (1 rnh r + n 1 7. If K r 0,then n κ 2 i + n i=2 j=3 r j 1( n j) Kj 1 + rnh + n 1 r i( n i) Ki + r n K i=2 i=2 H r H rnhr 2 r K r K = n. r i( ) n Ki + r n K) = κ i l + rnκ l H rκ 2 l + n ( n ) j r j 1 K j. j=3 ( 1) i r i( ) n i Kri +( 1) n r n K r ) = κ rl rnκ rl H r +rκ 2 rl + n ( 1) j 1 r j 1 K rj. j=3 n κ 2 ri n j=3 ( 1) j 1 r j 1( n j) Krj K r. 1. From theorem we have κ i = κ ri 1 rκ ri. Therefore, K = n κ ri 1 rκ ri = K r n. (4.5) (1 rκ ri ) Now from the theory of symmetric polynomials we can expand the denominator of the n above equation to get (1 rκ ri ) = 1 rh r + n 1 ( 1) i r i( n i) Kri + ( 1) n r n K r. Thus by substituting in equation 4.5 the result is proved. 2. We have nh = n κ i = n i=2 κ ri 1 rκ ri.

102 Chapter 4. Relation Between Radial Geometry of Skeletal Structure and Differential Geometry of its Boundary 88 Thus nh = κ r1 i=2 n (1 rκ ri ) + κ r2 (1 rκ r1 ) n n 1 (1 rκ ri ) κ rn (1 rκ ri ) i=3. 1 rh r + n 1 ) Kri + ( 1) n r n K r i=2 ( 1) i r i( n i Now we will simplify the numerator of this equation using the concept of symmetric polynomial. First of all we have κ r1 n i=2 (1 rκ ri ) = κ r1 (1 rκ r2 )(1 rκ r3 )...(1 rκ rn ) = κ r1 rnκ r1 H r + rκ 2 r1 + n ( ) n ( 1) j 1 r j 1 K rj. j j=3 Similarly, we have κ rl (1 rκ r1 )...(1 rκ rl 1 )(1 rκ rl+1 )...(1 rκ rn ) = κ rl rnκ rl H r + rκ 2 rl n ( ) n + ( 1) j 1 r j 1 K rj. j Therefore, after simplification the numerator becomes ( n H r rnhr 2 + r n n ( ) n κ 2 ri + ( 1) j 1 r )K j 1 rj. n j j=3 j=3 Thus H = H r rnh 2 r + r n i=2 n κ 2 ri + n j=3 ( 1) j 1 r j 1( n j) Kri. 1 rnh r + n 1 ( 1) i r i( ) n Kri + ( 1) n r i n K r

103 Chapter 4. Relation Between Radial Geometry of Skeletal Structure and Differential Geometry of its Boundary 89 Similarly we can prove 3 and We have n 1 ( ) n κ rl (1 + rnh + r i K i + r n K) = κ rl i i=2 Similarly we can prove If K r 0 then K = 0 and we have n = κ l 1 + rκ l H r H K r K = H H r rnhr 2 + r n r K r n rnhr 2 r κ 2 n ri n = Thus the proof is completed. (1 + rκ i ) n (1 + rκ i ) = κ l (1 + rκ 1 )...(1 + rκ l 1 )(1 + rκ l+1 )...(1 + rκ n ) n ( ) n = κ l + rnκ l H rκ 2 l + r j 1 K j. j j=3 n κ 2 ri + n K r j=3 j=3 ( 1) j 1 r j 1( n j) Krj K r. ( 1) j 1 r j 1( n j) Kri If the radius function is a constant on a smooth stratum S 1 of the skeletal structure containing a smooth point x 0, then this stratum and its associated boundary are parallel and the radial vector field becomes the normal of that stratum. Thus if we replace r in proposition by a constant, then K r and H r is the Gaussian curvature and mean curvature of S 1 at x 0 respectively. Thus proposition indicates that for each smooth point x 0 of the skeletal structure the smooth hypersurface containing x 0 and parallel to the boundary has Gaussian curvature K r and mean curvature H r. James Damon gave a relationship between the matrix representing the radial shape operator of the skeletal structure and the matrix representing the differential geometric

104 Chapter 4. Relation Between Radial Geometry of Skeletal Structure and Differential Geometry of its Boundary 90 shape operator of the skeletal structure (theorem 4.2.1). Our aim in this thesis is to find the relationship between the radial shape operator of the skeletal structure and the differential geometric shape operator of the associated boundary and we will look at this relation when the radius function has a singularity. Now let S rad denotes the radial shape operator and S Bond is the associated shape operator of the boundary, if the radius function has a singularity then this has strong consequence for the relationship between S rad and S Bond. In particular we have the following. Proposition Let (S, U) be a skeletal structure in R n+1 as in theorem If the radius function r has a singularity at x 0 then 1. S T XV V = S T V V, 2. S Bond (v ) = S rad (v), 3. V = (I + rs XV ) T V. 1. From equation 1.3 we have Also, the Jacobian matrix of the radial map is given by U 1 V = A V U 1 S T V V. (4.6) V = Ψ 1 V = (dr(v ) + ra V )U 1 + (1 rs V ) T V. (4.7) Now since the radius function r has a singularity then using lemma we have A V = 0. Thus, equation 4.7 becomes V = (I rs V ) T V. Also, since 1 r is not an eigenvalue of the radial shape operator we can solve for V to get V = (I rs T V ) 1 V. (4.8)

105 Chapter 4. Relation Between Radial Geometry of Skeletal Structure and Differential Geometry of its Boundary 91 From the proof of the proposition 2.1 in [8] we have Now substitute in equation (4.6) we have U 1 V = U 1 V. Now we will show that U 1 V = S T V (I rs T V ) 1 V = S T V V. To do so it is enough to show that Now S T V (I rs T V ) 1 = (I rs T V ) 1 S T V. S V (I rs V ) 1 1 r [(I rs V ) 1 I] = 0. [S V (I rs V ) 1 1 r [(I rs V ) 1 I] = S V (I rs V ) 1 1 r (I rs V ) r I Therefore, Hence S T XV V = S T V V. = 1 r I + 1 r I = 0. U 1 V = (I rs T V ) 1 S T V V. 2. The unit normal of ( the boundary ) is U 1 and the shape operator S Bond of the boundary U1 is given by proj U1, where proj v U1 is the projection along U 1 to the tangent ( ) U1 space of the boundary. Also, the radial shape operator S rad is given by proj U1 v where proj U1 is the projection ( along ) U 1 to the tangent space of the skeletal ( ) structure. U1 From 1 we have proj U1 = S T V V = S T XV V V = proj U1 U 1. Thus, V S Bond (v ) = S rad (v). 3. From equation (4.7) V = (I rs T V ) 1 V and from proposition it is easy to obtain that (I rs T V ) 1 = (I + rs XV ) T. Thus the proof is completed.

106 Chapter 4. Relation Between Radial Geometry of Skeletal Structure and Differential Geometry of its Boundary 92 Example Assume as in example 4.2.8, then at the origin the radius function has a singularity and V = v 1 = Thus v 2 S T V V = κ m1 + a 0 0 κ m2 + b and V = v 1 v = = 1 r 0(κ m1 + a) r 0 (κ m2 + b) 0 κ m1 + a κ m2 + b 0.. Also, S T XV V = = κ m1 + a 1 r 0 (κ m1 + a) 0 κ m1 + a κ m2 + b 0 0 κ m2 + b 1 r 0 (κ m2 + b). 1 r 0(κ m1 + a) r 0 (κ m2 + b) 0 Therefore, S T V V = S T XV V, S rad (v 1 ) = (κ m1 + a, 0, 0) = S Bond (v 1) and S rad (v 2 ) = (0, κ m2 + b, 0) = S Bond (v 2). Moreover, (I + r 0 S T )V = XV = 1 1 r 0 (κ m1 0 +a) r 0 (κ m2 +b) 1 r 0(κ m1 + a) r 0 (κ m2 + b) 0 = V.

107 Chapter 4. Relation Between Radial Geometry of Skeletal Structure and Differential Geometry of its Boundary 93 Now we turn to the relationship between the differential geometric shape operator of the skeletal structure at a smooth point and the differential geometric shape operator of the boundary at the associated point. First we have the following definition. Definition Let (S, U) be a skeletal structure and radial vector field such that the compatibility holds in a neighbourhood of a smooth x 0. We define the radial Hessian operator by H r : T x0 S T x0 S ( ) U1tan such that H r (v) = proj N, where proj v N denotes orthogonal projection onto T x0 S and U 1tan is the tangential component of the unit radial vector field U 1. Proposition ([8], proposition 4.1) Let (S, U) be a skeletal structure in R n+1 which satisfies the compatibility condition on an open set W S reg Let U be a smooth value on W. Then, on W there is the following relation S rad = ρs med + H r + Z, (4.9) where Z(v) = ρ 1 ( U 1 v N)U 1tan, N is the unit normal of the skeletal set and U 1tan is the tangential parts of the unit radial vector field U 1 and S med is the differential geometric shape operator of the skeletal structure. Damon discusses in [8] that the operator Z is difficult to work with and interpret. But again when the radius function has a singularity we get a situation with strong consequences for the various operators. Proposition Let (S, U) be a skeletal structure in R n+1 as in proposition If the radius function r has a singularity at x 0 then the differential geometric shape operator S med of S at x 0 and the differential geometric shape operator S Bond of the boundary at x 0 = Ψ 1 (x 0 ) are related by S Bond = S med + H r.

108 Chapter 4. Relation Between Radial Geometry of Skeletal Structure and Differential Geometry of its Boundary 94 If the radius function r has a singularity then the unit radial vector field is normal to the tangent space of the skeletal set. Thus Z = 0 and ρ = N U 1 = 1. From proposition we have S Bond = S rad. Therefore, equation (4.9) becomes S Bond = S med + H r, which completes the proof. This proposition gives the relationship between the geometric shape operator of the boundary and that of the skeletal structure and it is obvious from this proposition to obtain that the tangent space of the skeletal structure at x 0 is parallel to the tangent space of the boundary at the associated point. This means that the skeletal structure and the hypersurface containing x 0 and parallel to the boundary have the same tangent space at x 0. Example Assume as in example Then the radius function has a singularity at the origin and S med (v i ) = κ i v i, i = 1, 2. Also, H r (v 1 ) = av 1 and H r (v 2 ) = bv 2 and from example we have S Bond (v 1) = S rad (v 1 ) = (κ m1 + a, 0, 0) = κ m1 v 1 + av 1 = S med (v 1 ) + H r (v 1 ), and S Bond (v 2) = S rad (v 2 ) = (0, κ m2 + b, 0) = κ 2 v 2 + bv 2 = S med (v 2 ) + H r (v 2 ). 4.3 Skeletal Structures in R 3 In this section we will give a special form of the shape operator of the boundary in terms of the radial shape operator of the skeletal structure in R 3 and vice versa.

109 Chapter 4. Relation Between Radial Geometry of Skeletal Structure and Differential Geometry of its Boundary 95 Theorem Let (S, U) be a skeletal structure in R 3 smooth value of U the associated compatibility 1-form η U neighbourhood of a smooth point x 0 of S, and 1 r such that for a choice of vanishes identically on a is not an eigenvalue of the radial shape operator at x 0. Let x 0 = Ψ 1 (x 0 ) and V be the image of V under dψ 1 for a basis {v 1, v 2 } then or equivalently S XV = S V = 1 r 2 K r 2rH r + 1 (S V rk r I) (4.10) 1 r 2 K + 2rH + 1 (S XV + rki). (4.11) From lemma we have S XV = 1 r [(I rs V ) 1 I]. (4.12) Now let Therefore, (I rs V ) = S V = a c ra rc rb rd b d. = 1 ra rc rb 1 rd. Therefore, det(i rs V ) = (1 ra)(1 rd) r 2 cb = 1 r(a + b) + r 2 (ad cb) = 1 2rH r + r 2 K r. Now (I rs V ) 1 = 1 1 rd 1 2rH r + r 2 K r rc rb. 1 ra

110 Chapter 4. Relation Between Radial Geometry of Skeletal Structure and Differential Geometry of its Boundary 96 Thus (I rs V ) 1 I = hence this matrix becomes 1 1 2rH r + r 2 K r 1 rd rc rb 1 ra , 1 1 rd 1 + r(a + d) r2 K r rb 1 2rH r + r 2 K r rc 1 ra 1 + r(a + d) r 2 K r which gives the following (I rs V ) 1 r I = a rk r b 1 2rH r + r 2 K r c d rk r, Now by substituting in equation 4.12 we get = S XV = Similarly, we can prove equation r 1 2rH r + r 2 K r (S V rk r I) rH r + r 2 K r (S V rk r I). Now we have the following corollary from proposition Corollary Let (S, U) be a skeletal structure in R 3 as in theorem Then, the radial geometric factors ( the principal radial curvatures κ ri, the Gaussian radial curvature K r and mean radial curvature H r ) of S at x 0 and the differential geometric factors ( the principal curvatures κ i, the Gaussian curvature K and the mean curvature H) of the boundary at x 0 satisfy the following 1. K = 2. H = K r 1 2rH r + r 2 K r. H r rk r 1 2rH r + r 2 K r.

111 Chapter 4. Relation Between Radial Geometry of Skeletal Structure and Differential Geometry of its Boundary K r = 4. H r = K 1 + 2rH + r 2 K. H + rk 1 + 2rH + r 2 K. 5. κ rl (1 + 2rH + r 2 K) = κ l + rk, l = 1, κ l (1 2rH r + r 2 K r ) = κ rl rk r, l = 1, If K r 0 then, H r K r H K = r. The proof of this corollary comes directly from proposition just by taking n = 2. Example Assume as in example 4.2.8, then at the origin the principal radial curvatures, Gaussian radial curvature and mean radial curvature are κ r1 = κ m1 +a, κ r2 = κ m2 +b, K r = (κ m1 +a)(κ m2 +b) and H r = 1 2 (κ m1+κ m2 +a+b). Also the associated the principal curvatures, Gaussian curvature and mean curvature of the boundary are κ 1 = κ m1 + a 1 r 0 (κ m1 + a), κ 2 = κ m2 + b 1 r 0 (κ m2 + b), and K = H = (κ m1 + a)(κ m2 + b) 1 r 0 (κ m1 + κ m2 + a + b) + r 2 0(κ m1 + a)(κ m2 + b) 1 2 (κ m1 + κ m2 + a + b) r 0 (κ m1 + a)(κ m2 + b) 1 r 0 (κ m1 + a + κ m2 + b) + r 2 0(κ m1 + a)(κ m2 + b).

112 Chapter 4. Relation Between Radial Geometry of Skeletal Structure and Differential Geometry of its Boundary 98 Thus it is easy to check that this example satisfies relations 1, 2, 3, 4, 5 and 6 in corollary Now assume that K r 0, then we have H r H K r K = κ m1 + κ m2 + a + b 2(κ m1 + a)(κ m2 + b) κ m1 + κ m2 + a + b 2r 0 (κ m1 + a)(κ m2 + b) 2(κ m1 + a)(κ m2 + b) = 2r 0(κ m1 + a)(κ m2 + b) 2(κ m1 + a)(κ m2 + b) = r 0. Proposition Let (S, U) be a skeletal structure in R 3 as in corollary If K r 0 or equivalently K 0, then 1 K r S V 1 K S XV = ri. From equation 4.10 we have Therefore, S XV = 1 r 2 K r 2rH r + 1 (S V rk r I). 1 K r S V 1 K S XV = 1 K r S V 1 K ( 1 r 2 K r 2rH r + 1 (S V rk r I)). But from corollary we have Hence K = K r r 2 K 2rH r S V 1 K r K S XV = 1 S V r2 K r 2rH r + 1 ( K r K r 1 r 2 K r 2rH r + 1 S V rk r r 2 K r 2rH r + 1 I). Therefore, we have the following 1 K r S V 1 K S XV = 1 K r S V 1 K r S V + ri = ri.

113 Chapter 4. Relation Between Radial Geometry of Skeletal Structure and Differential Geometry of its Boundary 99 Therefore, the proof is completed. Now we will study the answer of the question: what happens if K r = K? Proposition Let (S, U) be a skeletal structure in R 3 as in corollary Then we have 1. If K r = K = 0, then or equivalently H = H r = H r 1 2rH r, H 1 + 2rH. 2. If K r = K 0, then H = H r = r 2 K 0. 1-From corollary we have H = H r rk r r 2 K r 2rH r + 1 and H r = Therefore, if K r = K = 0 we get the result. H + rk r 2 K + 2rH Assume that K r = K 0, then we have 1 + 2rH + r 2 K = 1 2rH r + r 2 K r which gives that r 2 K + 2rH = 0 and r 2 K r 2rH r = 0 thus the result holds. The second part of proposition indicates that if K r = K 0, then the boundary has non zero mean curvature. Example Assume as in example such that r 0 = 1, κ m1 = 2, κ m2 = 1 2 and a = b = 1, then at the origin we have S V = 3 0 and S XV =

114 Chapter 4. Relation Between Radial Geometry of Skeletal Structure and Differential Geometry of its Boundary 100 Thus K r = K = 9 2, H r = 9 4 and H = 9 4. In the next theorem we will discuss at what conditions does K r = K 0 in the case of the skeletal structure in R 3? Theorem Let (S, U) be a skeletal structure in R 3 smooth value of U the associated compatibility 1-form η U neighbourhood of a non-edge point x 0 of S, and 1 r such that for a choice of vanishes identically on a is not an eigenvalue of the radial shape operator at x 0. Let u r = 1 2 (κ r2 κ r1 ) = H 2 r K r is the radial skew curvature. If x 0 is not a radial umbilic point (i.e., κ r1 κ r2 ), then K r = K if and only if u = u r where u = 1 2 (κ 2 κ 1 ) = H 2 K. Let K r = K 0 then from proposition we have H = H r. Therefore, u r = Hr 2 K r = ( H) 2 K = u. Conversely, assume that u = u r then Therefore, κ 2 κ 1 = κ r2 κ r1 = κ rκ 2 κ rκ 1. κ 2 κ 1 = κ 2 κ 1 r 2 K + 2rH + 1. Since x 0 is not a radial umbilic point, then x 0 = Ψ 1 (x 0 ) is not an umbilic point on the boundary (i.e., κ 1 κ 2 ). Therefore, r 2 K + 2rH + 1 = 1 which gives that K r = K. Example Assume as in example 1.3.3, then the matrix representing the differential geometric shape operator is given by S XV = 1 r 2 K r 2rH r + 1 (S V rk r I),

115 Chapter 4. Relation Between Radial Geometry of Skeletal Structure and Differential Geometry of its Boundary 101 and after some calculations and simplifications we get 2 S XV = 1 + 2x2 + 6y 2 4xy. 4(x 2 + y 2 )[3(x 2 + y 2 ) + 2] + 1 4xy 1 + 6x 2 + 2y 2 Now the principal curvatures of the boundary are given by κ 1 = x 2 + 2y 2 and κ 2 = x 2 + 6y 2. Thus the Gaussian and mean curvatures of the boundary are given by K = 4 (1 + 6x 2 + 6y 2 )(1 + 2x 2 + 2y 2 ) and H = (2 + 8x 2 + 8y 2 ) (1 + 6x 2 + 6y 2 )(1 + 2x 2 + 2y 2 ). From direct calculation we have H r K r H K = x2 + y = r. From this example we can see that the radial curvature condition is not necessary for the smoothness of the boundary. For instance when the radius function has a singularity we have κ r1 = κ r2 = 2 but the radius function r = 1.

116 102 Chapter 5 The Relationship Between the Differential Geometry of the Skeletal Structure and that of the Boundary 5.1 Introduction In chapter 4 we studied the relationship between the radial shape operator of a skeletal structure and the differential geometric shape operator of its associated boundary. This chapter focuses on the relationship between the differential geometric shape operator of a skeletal structure and the differential geometric shape operator of its boundary. To find out this relationship we first study the relationship between the differential geometric shape operator of a skeletal structure and its radial shape operator. First we study this relationship in the case of a skeletal structure in R 2 (Theorem 5.2.4). After this we study the relationship between the curvatures of the boundary at the tangency points associated to a smooth point of a Blum medial axis in R 2 (Theorem ). Second we study the relationship between the radial shape operator of a skeletal structure in R n+1 and its differential geometric shape operator (Theorem ). This gives us enough tools

117 Chapter 5. The Relationship Between the Differential Geometry of the Skeletal Structure and that of the Boundary 103 to study the required relationship between the differential geometric shape operator of a skeletal structure and the differential geometric shape operator of its boundary which is given in theorem The Differential Geometry of the Skeletal Structure and its Boundary in the Plane In this section we will study the relationship between the curvature of the skeletal set and the curvature of its boundary. Let (S, U) be a skeletal structure in R 2 and let x 0 S be a smooth point such that the compatibility 1-form vanishes identically on a neighbourhood of x 0. Now let γ be the smooth stratum containing x 0 parametrized by the arc-length s. Define the following functions ρ 1 = U 1 N and ρ 2 = U 1 T, where N and T are the unit normal and the unit tangent such that U 1 and N oriented in the same direction. The smooth choice of the radial vector field is: U 1 = r T + 1 r 2 N and in this case ρ 1 = 1 r 2 and ρ 2 = r. Recall that at a smooth point x 0 U 1 T and if the radius function has no singularity at x 0, then the possible positions of U 1 are illustrated in figure 5.1. If the radius has a singularity at x 0, then U 1 = N. Remark Let ρ 1 and ρ 2 defined as above, then 1. ρ ρ 2 2 = 1; 2. if the radius function has no singularity at x 0 then ρ 1 has a singularity at x 0 if and only if ρ 2 has a singularity at x 0.

118 Chapter 5. The Relationship Between the Differential Geometry of the Skeletal Structure and that of the Boundary 104 N U 1 U 1 N T T x 0 γ x 0 γ Figure 5.1: The possible positions of the radial vector field when the radius function has no singularity. Lemma Let (S, U) be a skeletal structure in R 2 such that for a choice of smooth value of U, the associated compatibility 1-form vanishes identically on a neighbourhood of a smooth point x 0 S. If the function ρ 1 has a singularity at x 0 and the radius function has no singularity at x 0, then κ r = 1 κ m, ρ 1 where κ r (resp. κ m ) is the radial curvature of S at x 0 (resp. curvature of S at x 0 ). Let γ(s) be the smooth stratum containing x 0 parametrized by the arc-length s, then Therefore, U 1 s = au 1 κ r T. U 1 s U 1 = 0 = a κ r ρ 2 a = κ r ρ 2. Now the derivative of the function ρ 1 = U 1 N with respect to s is given by ρ 1 s = U 1 s N + N s U 1 = aρ 1 κ m ρ 2. Now since ρ 1 s = 0 this implies aρ 1 = κ m ρ 2 or a = ρ 2 κ m. But a = ρ 2 κ r therefore, ρ 1 ρ 2 κ r = ρ 2 κ m which gives κ r = 1 ρ ρ 1 κ m. 1

119 Chapter 5. The Relationship Between the Differential Geometry of the Skeletal Structure and that of the Boundary 105 The above lemma gives us a good tool to study the relationship between the curvature of the skeletal structure and the curvature of its boundary. This relation is given in the following proposition. Proposition Let (S, U) be a skeletal structure in R 2 such that for a choice of smooth value of U, the associated compatibility 1-form vanishes identically on a neighbourhood of a smooth point x 0 S and κ r 1 r. If the function ρ 1 has a singularity at x 0 and the radius function has no singularity, then κ = κ m ρ 1 rκ m, where κ m (resp. κ ) is the curvature of S at x 0 (resp. curvature of the associated boundary at x 0 = Ψ 1 (x 0 )), where Ψ 1 is the radial map. From lemma we have κ r = 1 ρ 1 κ m also from theorem we have κ = κ r 1 rκ r. Therefore, replacing κ r by 1 ρ 1 κ m gives the result. The previous proposition tells us the relationship between κ m and κ r under specific conditions depend on the singularity of ρ 1 when the radius function has no singularity. The next result gives this relation in general without controlling it by any conditions regarding to the singularity of ρ 1. Theorem Let (S, U) be a skeletal structure in R 2 such that for a choice of smooth value of U, the associated compatibility 1-form vanishes identically on a neighbourhood of a smooth point. Then κ r = ρ 1κ m dρ 2 ρ 2 1

120 Chapter 5. The Relationship Between the Differential Geometry of the Skeletal Structure and that of the Boundary 106 or equivalently κ r = 1 r 2 κ m + r 1 r 2, where κ m (resp. κ r ) is the curvature of S at x 0 (resp. radial curvature of S at x 0 ). Let γ(s) be the smooth stratum containing x 0 parametrized by the arc-length, then we have ρ 2 = T U 1. (5.1) Now differentiate equation (5.1) with respect to the arc-length we obtain dρ 2 = ρ 1 κ m + aρ 2 κ r. But from the proof of lemma we have a = ρ 2 κ r, thus dρ 2 = ρ 1 κ m + aρ 2 κ r = ρ 1 κ m + ρ 2 2κ r κ r. Therefore, or equivalently κ r = dρ 2 ρ 1 κ m ρ κ r = Therefore the proof is completed. = ρ 1κ m dρ 2 ρ r 2 κ m + r 1 r 2. Corollary Assume as in theorem If the radius function has a singularity, then κ r = κ m dρ 2 = κ m + r. From theorem we have κ r = ρ 1κ m dρ 2 ρ 2 1

121 Chapter 5. The Relationship Between the Differential Geometry of the Skeletal Structure and that of the Boundary 107 and since the radius function has a singularity then ρ 1 = 1 therefore, the above equation becomes κ r = κ m dρ 2 = κ m + r which completes the proof. Corollary Assume as in theorem If the radius function has a singularity, then ρ 2 has a singularity if and only if κ r = κ m. The proof comes directly from corollary Now we will turn to the relationship between the curvature of the skeletal structure and the curvature of its boundary. Theorem and theorem give enough information to discuss the requested relation in the following theorem. Theorem Let (S, U) be a skeletal structure in R 2 such that for a choice of smooth value of U, the associated compatibility 1-form vanishes identically on a neighbourhood of a smooth point x 0 S and κ r 1 r. Then the curvature κ of the boundary at x 0 = Ψ 1 (x 0 ) is given by or equivalently κ = κ = ρ 1 κ m dρ 2 ρ rdρ 2 rρ 1 κ m (5.2) 1 r 2 κ m + r. (5.3) 1 r 2 rr r 1 r 2 κ m From theorem we have κ = κ r 1 rκ r (5.4)

122 Chapter 5. The Relationship Between the Differential Geometry of the Skeletal Structure and that of the Boundary 108 and from theorem we have or equivalently κ r = κ r = ρ 1κ m dρ 2 ρ r 2 κ m + r 1 r 2. Now substituting in equation 5.4 the result holds. It can be seen from equation 5.3 that the type of the singularity of the radius function plays a central role in the relation between κ and κ m. Recall that a function f : R R is said to have an A k singularity at t 0 if f (t 0 ) = f (t 0 ) =... = f (k) (t 0 ) = 0, f (k+1) (t 0 ) 0. Corollary Assume as in theorem If the radius function has an A 1 singularity, then κ = 2. If the radius function has an A 2 singularity, then κ m + r 1 rr rκ m. (5.5) κ = The proof of this corollary comes directly from equation 5.3. κ m 1 rκ m. (5.6) In the rest of this section we will discuss the relationship between the curvature of the Blum medial axis and the curvatures κ 1 and κ 2 of the boundary at tangency points, on another words if we know the curvatures of the boundary, could we find the curvature of the Blum medial axis? Also, we will investigate the relationship between κ 1 and κ 2.

123 Chapter 5. The Relationship Between the Differential Geometry of the Skeletal Structure and that of the Boundary 109 Proposition Let (S, U) be a Blum medial axis and radial vector field of a region Ω R 2 with smooth boundary X. Let x 0 S be a smooth point. 1. The radial curvatures κ r1 and κ r2 are given by 1 r 2 κ m + r κ r1 = 1 r 2, (5.7) κ r2 = 1 r 2 κ m + r 1 r 2. (5.8) 2. The curvatures κ 1 and κ 2 of the boundary at the tangency points are given by 1 r 2 κ m + r κ 1 =, (5.9) 1 r 2 rr r 1 r 2 κ m 1 r 2 κ m + r κ 2 =. (5.10) 1 r 2 rr + r 1 r 2 κ m The proof of this proposition comes directly from theorems and Corollary Let (S, U) be a Blum medial axis and radial vector field of a region Ω R 2 with smooth boundary X. Let x 0 S be a smooth point, then This result comes by adding equations 5.7 and (κ r1 + κ r2 ) = r. (5.11) 2 1 r Proposition Let (S, U) be a Blum medial axis and radial vector field of a region Ω R 2 with smooth boundary X. Let x 0 S be a smooth point, then the curvature of S at x 0 is given by κ m = 1 ( κ1 κ ) 2 1 r 2 (5.12) rκ rκ 2 where κ 1 and κ 2 are the curvatures of the boundary at tangency points associated to x 0.

124 Chapter 5. The Relationship Between the Differential Geometry of the Skeletal Structure and that of the Boundary 110 From equations 5.7 and 5.8 we have κ r1 = 1 r 2 κ m + r 1 r 2 and Therefore, Thus κ r2 = 1 r 2 κ m + r 1 r 2. κ r1 κ r2 = 2 1 r 2 κ m 1 r 2. κ m = 1 2 (κ r1 κ r2 ) 1 r 2. But we have κ ri = κ i 1 + rκ i, i = 1, 2 and by substituting in the above equation the proof is completed. Corollary Assume as in proposition If the radius function has a singularity, then κ m = 1 2 ( κ1 κ ) rκ rκ 2 The proof is obvious. Now we are in the position to study the relationship between the curvatures of the boundary. This relation is given in the following theorem. Theorem Let ( S, U) be a Blum medial axis and radial vector field of a region Ω R 2 with smooth boundary X. Let x 0 S be a smooth point, then the curvatures of

125 Chapter 5. The Relationship Between the Differential Geometry of the Skeletal Structure and that of the Boundary 111 the boundary at the tangency points associated to x 0 are related by the following equation κ 1 = From corollary we have 2r κ 2 (1 r 2 2rr ) (1 r 2 2rr ) + 2rκ 2 (1 r 2 rr ). (5.13) κ r1 + κ r2 = 2r 1 r 2 or Thus Therefore, κ rκ 2 + κ rκ 2 = 2r 1 r 2. κ 1 + 2rκ 1 κ 2 + κ rκ 1 + rκ 2 + r 2 κ 1 κ 2 = 2r 1 r 2. κ 1 (1 + 2rκ 2 )(1 r 2 ) + κ 2 (1 r 2 ) = 2r + 2rr κ 1 (1 + rκ 2 ) + 2rr κ 2. Thus κ 1 (1 + 2rκ 2 r 2 2rr 2 κ 2 2rr 2r 2 r κ 2 ) = 2r + κ 2 (2rr 1 + r 2 ). Hence κ 1 [(1 r 2 2rr ) + 2rκ 2 (1 r 2 rr )] = 2r κ 2 (1 r 2 2rr ). Therefore, κ 1 = which completes the proof. 2r κ 2 (1 r 2 2rr ) (1 r 2 2rr ) + 2rκ 2 (1 r 2 rr ) Corollary Assume as in theorem If r = 0, then κ 1 = κ rκ 2.

126 Chapter 5. The Relationship Between the Differential Geometry of the Skeletal Structure and that of the Boundary 112 The proof comes directly from equation Shape Operator of the Blum Medial Axis and those of its Boundary at Tangency Points In this section we will turn to higher dimensions; in particular we will investigate the Hessian operator in terms of the radial shape operators of the Blum medial axis and then we are able to find the the expression of the Hessian operator in terms of the differential geometric shape operators of the boundary at the tangency points. Moreover, we are going to find out the relationship between the shape operator of the Blum medial axis and the the differential geometric shape operators of the boundary at the tangency points. Recall that for each smooth point x 0 S of skeletal structure we have two values of the radial vector field U which are on opposite sides of T x0 S. The values of U corresponding to one side form a smooth vector field. Also, for each side we have a radial shape operator. Theorem Let (S, U) be a Blum medial axis and radial vector field of a region Ω R n+1 with smooth boundary X. Let x 0 S be a smooth point and {v 1, v 2,..., v n } be a basis for the tangent space of S at x 0, then the radial Hessian operator is given by H r (V ) = 1 2 (S V 1 + S V2 ) T (I drdr T I 1 m )V, (5.14) where I m is the matrix representing the first fundamental form of S at x 0 and S Vi, i = 1, 2 are the matrices representing the radial shape operators, V is the matrix with i-th row entry v i and dr is a column matrix with i-row entry dr(v i ), where dr(v i ) is the directional derivative of the radius function in the direction of v i.

127 Chapter 5. The Relationship Between the Differential Geometry of the Skeletal Structure and that of the Boundary 113 The radial Hessian operator is a map such that H r : T x0 S T x0 S ( ) U1tan H r (v i ) = vi (U 1tan ) = proj N, v i where vi is the covariant derivative with respect to the basis of the tangent space of the Blum medial axis at a smooth point. But we have U 1tan = r = 1 2 (U 1 + U 2 ) and r is the Riemannian gradient of the radius function. Therefore, Thus H r (v i ) = vi ( r) = vi ( r) = proj N ( 1 2 Now using equation (1.2) we have ( U1 + U ) 2 = a1 i U 1 v i v i H r (v i ) = 1 ( 2 proj U1 N + U ) 2. v i v i n s1 ji v j + a2 i U 2 j=1 Now we write this equation in vector notation to get ) (U 1 + U 2 ). v i n s2 ji v j. j=1 U 1 V + U 2 V = A 1U 1 S T V 1V + A 2 U 2 S T V 2V, but from the proof of lemma we have A i = SV T idr, i = 1, 2. Therefore, U 1 V + U 2 V = (S V 1 + S V 2 ) T V S T V 1drU 1 S T V 2drU 2. But the possible choices for the radial vector fields are Thus U 1 = r + 1 r 2 N and U 2 = r 1 r 2 N. U 1 V + U 2 V = (S V 1 + S V 2 ) T V +(S V 1 + S V 2 ) T dr r 1 r 2 (S V 1 S V 2 ) T drn.

128 Chapter 5. The Relationship Between the Differential Geometry of the Skeletal Structure and that of the Boundary 114 Now the projection of this equation to the tangent space along the normal is given by ( U1 proj N V + U ) 2 = (S V 1 + S V 2 ) T (V dr r), (5.15) V but the Riemannian gradient of the radius function is given locally by r = dr T I 1 m V where I m is the first fundamental form of the Blum medial axis. Thus equation 5.15 becomes Therefore, ( U1 proj N V + U ) 2 = (S V 1 + S V 2 ) T (I drdr T Im 1 )V. V Hence by this the proof is completed. H r (V ) = 1 2 (S V 1 + S V2 ) T (I drdr T I 1 m )V. Now let H r be the matrix representing the radial Hessian operator H r. In the following corollary we express H r in terms of S V1 and S V2. Corollary Let (S, U) be a Blum medial axis and radial vector field of a region Ω R n+1 with smooth boundary X. Let x 0 S be a smooth point, then the matrix H r representing the radial Hessian operator is given by H r = 1 2 (S V 1 + S V2 ) T (I drdr T I 1 m ). (5.16) The proof of this corollary comes directly from equation Proposition Let (S, U) be a Blum medial axis and radial vector field of a region Ω R n+1 with smooth boundary X. Let x 0 S be a smooth point. If the radius function has a singularity at x 0, then the matrix H r representing the radial Hessian operator is given by H r = 1 2 (S V 1 + S V2 ) T. (5.17)

129 Chapter 5. The Relationship Between the Differential Geometry of the Skeletal Structure and that of the Boundary 115 Assume that the radius function has a singularity then equation 5.16 becomes H r = 1 2 (S V 1 + S V2 ) T. Hence the proof is completed. Now we define the mean Hessian curvature H by Using this we have the following. H = 1 n tr(h r). Corollary Assume as in proposition 5.3.3, then the mean Hessian curvature is given by H = 1 2 (H r1 + H r2 ), (5.18) where H r1 and H r2 are the mean radial curvatures of the Blum medial axis. If we take the trace for both sides of equation 5.17 we obtain tr(h r ) = 1 2 (tr(s V 1 ) + tr(s V2 )). Thus equation 5.18 is satisfied. Our task now is to find out the connection between the radial Hessian operator and the shape operators of the boundary at the tangency points. Theorem Let (S, U) be a Blum medial axis and radial vector field of a region Ω R n+1 with smooth boundary X. Let x 0 representing the radial Hessian operator is given by H r = 1 2 {(I + rs XV ) 1 S 1 XV 1 + (I + rs XV 2 S be a smooth point. Then the matrix H r ) 1 S XV } T (I drdr T Im 1 ), (5.19) 2

130 Chapter 5. The Relationship Between the Differential Geometry of the Skeletal Structure and that of the Boundary 116 where S XV and S 1 XV are the matrices representing the differential geometric shape 2 operators of the boundary at the tangency points associated to x 0. From corollary we have and from proposition we have H r = 1 2 (S V 1 + S V2 ) T (I drdr T I 1 m ) S V1 = (I + rs XV ) 1 S 1 XV 1 and S V2 = (I + rs XV ) 1 S 2 XV 2 and by substituting this in the above equation the proof is completed. Corollary Assume as in theorem If the radius function has a singularity at x 0, then H r = 1 2 {(I + rs XV ) 1 S 1 XV 1 + (I + rs XV 2 The proof of this result comes directly from equation ) 1 S XV } T. 2 Now we will give a special form for the matrix representing the radial Hessian operator in the case of a Blum medial axis in R 3. Theorem Let (S, U) be a Blum medial axis and radial vector field of a region Ω R 3 with smooth boundary X. Let x 0 H r representing the radial Hessian operator is given by H r = 1 2 { 1 r 2 K 1 + 2rH S XV 1 ( + S be a smooth point. Then the matrix 1 r 2 K 2 + 2rH S XV 2 rk 1 r 2 K 1 + 2rH rk 2 r 2 K 2 + 2rH ) I} T (I drdr T I 1 m ).

131 Chapter 5. The Relationship Between the Differential Geometry of the Skeletal Structure and that of the Boundary 117 From theorem we have and S V 1 = S V 2 = 1 r 2 K 1 + 2rH S rk 1 XV 1 r 2 K 1 + 2rH I, 1 r 2 K 2 + 2rH S rk 1 XV 2 r 2 K 1 + 2rH I. Now substitute by these in equation 5.16 the result holds. Corollary Assume as in theorem If the radius function has a singularity at x 0, then H r = 1 2 { 1 r 2 K 1 + 2rH S XV 1 ( + 1 r 2 K 2 + 2rH S XV 2 rk 1 r 2 K 1 + 2rH rk 2 r 2 K 2 + 2rH The proof of this corollary comes directly from theorem ) I} T. Example Let (S, U) be a Blum medial axis in R 3 and let S 1 (x, y) = (x, y, 1 2 κ m1x κ 2 m2y 2 + h.o.t) S reg and r = r ax by2 be the radius function on S 1 such that 1 r 0 / {κ m1 + a, κ m2 + b}. Now we define the unit radial vector fields by U 1 = r + 1 r 2 N and U 2 = r 1 r 2 N, where r is the Riemannian gradient of the radius function and N is the unit normal of S 1. The radius function has a singularity at the origin and direct calculations show that at the origin we have S V1 = κ m1 + a 0 0 κ m2 + b, S V2 = a κ m1 0 0 b κ m2,

132 Chapter 5. The Relationship Between the Differential Geometry of the Skeletal Structure and that of the Boundary 118 S XV2 = S XV1 = Now it is easy to check that a κ m1 1 r 0 (a κ m1 ) 0 κ m1 + a 1 r 0 (κ m1 + a) 0 0 b κ m2 1 r 0 (b κ m2 ) H r = 1 2 (S V 1 + S V2 ) T = 1 2 {(I + rs XV ) 1 S 1 XV 1 0 κ m2 + b 1 r 0 (κ m2 + b), and H r = + (I + rs XV 2 a 0 0 b. ) 1 S XV } T. 2 In the rest of this section we will focus on the relationship between the shape operator of the Blum medial axis and the shape operators of its boundary at tangency points corresponding to a smooth point on the medial axis. Now let (S, U) be a Blum medial axis and radial vector field of a region Ω R n+1 with smooth boundary X. Assume that x 1 and x 2 are the tangency points associated to a smooth point x 0 S then we have only two choices for the smooth value of the radial vector field U these choices are U 1 = r + 1 r 2 N and U 2 = r 1 r 2 N such that U 1 and N have the same direction and x 1 = x 0 + ru 1 and x 2 = x 0 + ru 2 and ρ = 1 r 2. Now it is clear that U 1 = U 2 + 2ρN. (5.20) Therefore, with this equation we have a good tool to investigate the relation mentioned above in particularly we have the following results. Theorem Let ( S, U ) be a Blum medial axis and radial vector field of a region Ω R n+1 with smooth boundary X. Let x 0 S be a smooth point and {v 1, v 2,..., v n } be a basis for the tangent space of S at x 0. Then the differential geometric shape operator S med of S at x 0 is given by S med (V ) = 1 2ρ (S V 1 S V 2 ) T (I drdr T I 1 m )V. (5.21)

133 Chapter 5. The Relationship Between the Differential Geometry of the Skeletal Structure and that of the Boundary 119 From equation 5.20 we have U 1 = U 2 + 2ρN. Let {v 1, v 2,..., v n } be a basis for the tangent space of S at x 0, now differentiate both sides of the above equation with respect to v i we get U 1 = U ρ N + 2ρ N, i = 1, 2,..., n. v i v i v i v i This equation can be written in vector forms as the following U 1 V = U 2 + 2dρ(V )N + 2ρ N V V or or A 1 U 1 S T V 1V = A 2 U 1 S T V 2V + 2dρN + 2ρ N V S T V 1(I drdr T I 1 m )V ρs T V 1drN = S T V 2(I drdr T I 1 m )V +ρs T V 2drN+2dρN+2ρ N V. Now apply the projection to the tangent space along normal ( proj N ) we obtain the following which completes the proof. S med (V ) = 1 2ρ (S V 1 S V 2 ) T (I drdr T I 1 m )V Corollary Let ( S, U ) be a Blum medial axis and radial vector field of a region Ω R n+1 with smooth boundary X. Let x 0 S be a smooth point. Then the matrix representing the shape operator of the Blum medial axis at x 0 is given by S T m = 1 2ρ (S V 1 S V 2 ) T (I drdr T I 1 m ). (5.22)

134 Chapter 5. The Relationship Between the Differential Geometry of the Skeletal Structure and that of the Boundary 120 The proof of this corollary comes directly from equation Corollary Assume as in corollary If the radius function has a singularity at x 0, then S T m = 1 2 (S V 1 S V 2 ) T. (5.23) If the radius function has a singularity, then ρ = 1. Therefore, equation 5.22 becomes S T m = 1 2 (S V 1 S V 2 ) T which completes the proof. Example Let (S, U) be a Blum medial axis in R 3 and let S 1 (x, y) = (x, y, y 2 x 2 ) S reg and r = 0.1+xy+y 2 be the radius function on S 1 such that x 2 +4xy+5y 2 < 1. At the origin the radius function has a singularity and we have S m = 2 0, S V1 = 2 1 and S V2 = It is clear that S T m = 1 2 (S V 1 S V 2 ) T.

135 Chapter 5. The Relationship Between the Differential Geometry of the Skeletal Structure and that of the Boundary 121 Figure 5.2: The Blum medial axis and associated boundary of example Now we will turn to one of the main aims of this chapter which is the relationship between the differential geometric shape operator of the Blum medial axis and the differential geometric shape operators of its boundary. In fact, corollary gives us a good tool as well as proposition to investigate this relationship which given in the following theorem. Theorem Let ( S, U ) be a Blum medial axis and radial vector field of a region Ω R n+1 with smooth boundary X. Let x 0 S be a smooth point. 1. The matrix representation of the shape operator of the Blum medial axis at x 0 is

136 Chapter 5. The Relationship Between the Differential Geometry of the Skeletal Structure and that of the Boundary 122 given by Sm T = 1 2ρ {(I + rs XV ) 1 S 1 XV 1 (I + rs XV 2 2. If the radius function has a singularity at x 0, then Sm T = 1 2 {(I + rs XV ) 1 S 1 XV 1 ) 1 S XV } T (I drdr T Im 1 ). (5.24) 2 (I + rs XV 2 ) 1 S XV } T, (5.25) 2 where S XV and S 1 XV are the matrices representing the differential geometric shape 2 operators of the boundary at the tangency points associated to x From proposition we have S V1 = (I + rs XV ) 1 S 1 XV 1 and S V2 = (I + rs XV 2 ) 1 S XV. 2 Now substitute by this in equation 5.22 the result holds immediately. 2. The proof is obvious. Example Let (S, U) be a Blum medial axis in R 3 and let S 1 (x, y) = (x, y, x 3 y 2 ) S reg and r = y 2 be the radius function on S 1 such that 4y 2 < 1. At the origin the radius function has a singularity and we have S m = 0 0, S XV = 0 0 and S V2 = Thus it is clear that Sm T = 1{(I + rs 2 XV ) 1 S 1 XV (I + rs XV 2 = (I + r 0 S XV 2 ) 1 S XV } T. 2 ) 1 S XV. 2

137 Chapter 5. The Relationship Between the Differential Geometry of the Skeletal Structure and that of the Boundary 123 Figure 5.3: The Blum medial axis and associated boundary of example Now we will give a special form for the matrix representing the differential geometric shape operator in the case of a Blum medial axis in R 3. Theorem Let (S, U) be a Blum medial axis and radial vector field of a region Ω R 3 with smooth boundary X. Let x 0 S be a smooth point. Then the matrix S m representing the differential geometric shape operator is given Sm T = 1 2ρ { 1 r 2 K 1 + 2rH S XV 1 ( + 1 r 2 K 2 + 2rH S XV 2 rk 1 r 2 K 1 + 2rH rk 2 r 2 K 2 + 2rH ) I} T (I drdr T I 1 m ).

138 Chapter 5. The Relationship Between the Differential Geometry of the Skeletal Structure and that of the Boundary 124 The proof of this theorem comes directly by applying theorem in corollary In the following we will give the exact relationship between the matrix representing the radial shape operator of a skeletal structure and the matrix representing the differential geometric shape operator of the skeletal set and then we will give the exact relationship between the matrix representing the differential geometric shape operator of the skeletal set and the matrix representing the differential geometric shape operator of the boundary at a point associated to a smooth point of the skeletal set. Let (S, U) be a skeletal structure in R n+1 such that the compatibility condition holds and let x 0 S reg be a smooth point and {v 1, v 2,..., v n } be a basis for the tangent space of S at x 0. Let r be the radius function, since x 0 is a smooth point and the compatibility condition holds then the unit radial vector field is given by [8]: U 1 = r + 1 r 2 N, where r is the Riemannian gradient of the radius function and N is the unit normal of the smooth stratum S containing x 0. We put ρ = 1 r 2, so we have the following U 1 = r + ρn. (5.26) From this equation we have the following equation N = 1 ρ U r. (5.27) ρ This equation is a useful tool to determine the coefficients of the unit radial vector field in U 1 v i. From equation 5.26, we have U 1 v i = r v i + ρ v i N + ρ N v i. (5.28) Now we put r v i = ( ) T ( ) N r r +, v i v i

139 Chapter 5. The Relationship Between the Differential Geometry of the Skeletal Structure and that of the Boundary 125 ( ) T ( ) N r r where (resp. ) is the tangential (resp. normal) part of r. Now v i v i v ( ) i N r since S is a hypersurface, then we put = β i N. Now using this together with v i equation 5.27, equation 5.28 becomes ( ) T U 1 r = + 1 ( ) ρ β i U ( ) ρ β i r + ρ N. (5.29) v i v i ρ v i ρ v i v i Now writing this equation in vector form we get From equation 1.3 we have U 1 V = 1 ρ (dρ β)u 1 (H T r + ρs T m 1 ρ (dρ β)drt I 1 m )V. (5.30) U 1 V = A V U 1 S T V. (5.31) Now since {U 1, v 1, v 2,..., v n } is a basis for R n+1, then from equations 5.30 and 5.31 we obtain A V = 1 ρ (dρ β) and ST V = H T r + ρs T m 1 ρ (dρ β)drt I 1 m, (5.32) where H r is the matrix representing the radial Hessian operator, S m is the matrix representing the differential geometric shape operator of the skeletal set, dρ is a column matrix with i-th entry ρ, dr is a column matrix with i-th entry r, β is a column matrix v i v i with i-th entry β i, I m is the first fundamental form of the skeletal set and V is the matrix with i-th row v i. Our task now is to find the exact expression of the matrix β. In [7] James Damon pointed out that A V = S T V V U 1 and since the compatibility condition holds, then A V = S T V dr. Thus from this and equation 5.32 we have SV T dr = 1 (dρ β). Thus ρ β = dρ + ρs T V dr. (5.33) Also, from corollary we have dρ = (ρs T V ST m)dr. Hence we obtain that β = S T mdr. Therefore, equation 5.32 can be rewritten as the following A V = 1 ρ (dρ ST mdr) and S T V = H T r + ρs T m 1 ρ (dρ ST mdr)dr T I 1 m. (5.34)

140 Chapter 5. The Relationship Between the Differential Geometry of the Skeletal Structure and that of the Boundary 126 Now we summarize the above discussion in the following theorem. Theorem Let (S, U) be a skeletal structure in R n+1 such the compatibility condition holds in a neighbourhood of smooth point x 0 S, then A V = 1 ρ (dρ ST mdr) (5.35) and S T V = H T r + ρs T m 1 ρ dρdrt I 1 m + 1 ρ ST mdrdr T I 1 m. (5.36) Example Let (S, U) be a skeletal structure in R n+1 suppose the image of s 1 (x 1, x 2,..., x n ) = (x 1, x 2,..., x n, 1) S reg such that {(x 1, x 2,..., x n ) R n n x 2 i < 1} 4 and let r(x 1, x 2,..., x n ) = 1 + n x 2 i be the radius function. Now the radial vector field is given by U 1 = r + ρn, where r is the Riemannian gradient, ρ = 1 4 n x 2 i and N is the unit normal of s 1. Now we will apply theorem to calculate S V and A V. Since s 1 is a hyperplane, then S m = 0. Also, r = (2x 1, 2x 2,..., 2x n, 0) and H r = 2I, where I in the (n n) identity matrix. From theorem we have For each j {1, 2,..., n} we have ρ x j = S T V = H T r 1 ρ dρdrt and A V = 1 ρ dρ. ρ x 1 ρ x 2. ρ x n = 4x j 1 4 n n x 2 i x 2 i. Thus x 1 x 2. x n.

141 Chapter 5. The Relationship Between the Differential Geometry of the Skeletal Structure and that of the Boundary 127 Therefore, A V = n x 2 i x 1 x 2. x n. Also, after simplifying the matrix H T r 1 ρ dρdrt we obtain that S V = n x 2 i 1 + 4x n x 2 i 4x 1 x 2 4x 1 x 3 4x 1 x n 4x 1 x x n x 2 i 4x 2 x 3 4x 2 x n 4x 1 x 3 4x 2 x x n x 2 i 4x 3 x n x 1 x n 1 4x 2 x n 1 4x 3 x n 1 4x n 1 x n 4x 1 x n 4x 2 x n 4x 3 x n 1 + 4x 2 n 4 n It is clear that at (x 1, x 2,..., x n ) = (0, 0,..., 0) we have S V = 2I. x 2 i. In theorem we expressed the matrix S V representing the radial shape operator of a skeletal structure in terms of H r, S m, dρ and dr. Our task now is to assume that if x 0 S is a smooth point then the smooth sheet of the skeletal set S say S 1 containing x 0 is in Monge form i.e., S 1 can be parametrized locally by (x 1, x 2,..., x n ) such that S 1 is n given by the graph S 1 (x 1, x 2,..., x n ) = (x 1, x 2,..., x n, 1 κ 2 mi x 2 i + h.o.t). Now let r be the radius function on S 1 i.e., it is a smooth function of (x 1, x 2,..., x n ). In the following we will calculate S V at the origin and to do so we just calculate H r, S m, dρ at the origin and substitute in theorem It is clear that at the origin we have S m = diag[κ mi ]. To calculate H r we will use the definition of Riemannian gradient which is given by r = g ij j rv i, where g ij is the inverse of the matrix representing the Riemannian metric, j is the partial derivative of the radius function and v i is a basis of the tangent space. Since S 1 is a smooth hypersurface in R n+1, we assume that the matrix I m representing the first fundamental form is the matrix g ij representing the Riemannian metric i.e., I m = g ij.

142 Chapter 5. The Relationship Between the Differential Geometry of the Skeletal Structure and that of the Boundary 128 Now we have and r x l = gij g ij g ij = g ij I m (5.37) j rv i + g ij ( j r)v i + g ij j r v i. (5.38) x l x l x l Now we calculate this at the origin and first of all we calculate the first fundamental form I m in general. It is clear that v 1 = S 1 x 1 = (1, 0, 0,..., 0, κ m1 x 1 + h.o.t), v 2 = S 1 x 1 = (0, 1, 0,..., 0, κ m2 x 2 + h.o.t),. Thus g ij = v n = S 1 x 1 = (0, 0, 0,..., 1, κ mn x n + h.o.t). 1 + κ 2 m1x h.o.t. κ m1 κ m2 x 1 x 2 + h.o.t. κ m1 κ mn x 1 x n + h.o.t. κ m1 κ m2 x 1 x 2 + h.o.t. 1 + κ 2 m2x h.o.t. κ m2 κ mn x 2 x n + h.o.t κ m1 κ mn x 1 x n + h.o.t. κ m2 κ mn x 2 x n + h.o.t. 1 + κ 2 mnx 2 n + h.o.t.. At the origin g ij = I and hence g ij = I. Also, g ij = 0 at the origin. In general (g ij g ij ) = ( g ij )g ij + g ij ( g ij ) = I = 0. Now since g ij = I at the origin thus g ij = 0 at the origin. Therefore, At the origin is easy to check that r x l = x l ( i r)v i + i r v i x l. (5.39) v i κ mi N if i = l = x l 0 if i l,

143 Chapter 5. The Relationship Between the Differential Geometry of the Skeletal Structure and that of the Boundary 129 where N is the unit normal of S 1. Thus r x1 x 1 r x1 x 2 r x1 x n r H r = x2 x 1 r x2 x 2 r x2 x n.....,. r xn x 1 r xn x 2 r xn x n where r xi x j = ( r x i x j dρ and to do so we put ). This matrix is given by [a kl ] = [r xk x l ]. Now we will calculate ρ 2 = 1 r r. Thus ρ ρ = r r. x l x l At the origin v i x l r = 0 for all l = 1, 2,..., n. Therefore, ρ ρ x l = x l ( i r)v i r. Thus ρdρ = H r v 1 r v 2 r. v n r. Also, at the origin we have r = (r x1, r x2,..., r xn, 0), v i r = r xi and ρ = 1 n rx 2 i. Hence dρ = 1H ρ rdr and after simplification we get dρ = 1 1 n rx 2 i n r xi r x1 x i n r xi r x2 x i.. n r xi r xnxi

144 Chapter 5. The Relationship Between the Differential Geometry of the Skeletal Structure and that of the Boundary 130 Now we will calculate A V which is given by A V = 1 ρ (dρ ST mdr) after direct calculation and simplification we get A V = 1 1 n rx 2 i Thus A V can be written as [b k1 ] = after some calculations we get 1 ρ dρdrt = 1 1 n rx 2 i This matrix can be given by Also, 1 ρ ST mdrdr T = 1 1 n r x1 r x1 r x1 n r xi r x1 x i + κ m1 r x1 1 n rx 2 i n r xi r x2 x i + κ m2 r x2 1 n rx 2 i. n r xi r xn x i + κ mn r xn 1 n 1 1 n rx 2 i [ n rx 2 i. r xi r xk x i + κ mk r xk 1 n rx 2 i ]. Now n n n r xi r x1 x i r x2 r xi r x1 x i r xn r xi r x1 x i n n n r xi r x2 x i r x2 r xi r x2 x i r xn r xi r x2 x i n n n r xi r xn x i r x2 r xi r xn x i r xn r xi r xn x i [c kl ] = rx 2 i r xl n r xi r xk x i 1 n. rx 2 i κ m1 rx 2 1 κ m1 r x1 r x2 κ m1 r x1 r xn κ m2 r x1 r x2 κ m2 r 2 x 2 κ m2 r x2 r xn κ mn r x1 r xn κ mn r x2 r xn κ mn rx 2 n This matrix can be written as [d kl ] = κ mk r xk r xl 1 n rx 2 i.

145 Chapter 5. The Relationship Between the Differential Geometry of the Skeletal Structure and that of the Boundary 131 Now let A = H r 1 ρ dρdrt + 1 ρ ST mdrdr T. Thus A is given by r xk x l (1 n rx 2 i ) + r xl [α kl ] = n r xi r xk x i + κ mk r xk r xl 1 n 1 n rx 2 i Now we summarize the above discussion in the following. rx 2 i. Proposition Let (S, U) be a skeletal structure in R n+1 such that the compatibility condition holds in a neighbourhood of a smooth point x 0 S. Assume that S 1 be the smooth sheet of S containing x 0 as the origin and S 1 is given in Monge form i.e., S 1 is n given by S 1 (x 1, x 2,..., x n ) = (x 1, x 2,..., x n, 1 κ 2 mi x 2 i + h.o.t), then at x 0 A V is given by the matrix and [b k1 ] = 1 1 n rx 2 i n n r xi r xk x i + κ mk r xk 1 where A is given by r xk x l (1 n rx 2 i ) + r xl [α kl ] = and Sm = diag [κ mk 1 n rx 2 i ]. S T V = A + S m, n r 2 x i r xi r xk x i + κ mk r xk r xl 1 n 1 n rx 2 i rx 2 i Corollary Let (S, U) be a skeletal structure in R 3 suppose the image of S 1 (x, y) = (x, y, 1 2 κ m1x κ m2y 2 + h.o.t) S reg and Let r be the radius function. Then, 1. S V and A V are given by 1 A V = 1 rx 2 ry 2 r xr xx + r y r xy + κ m1 r x 1 r 2 x r 2 y r x r xy + r y r yy + κ m2 r y 1 r 2 x r 2 y

146 Chapter 5. The Relationship Between the Differential Geometry of the Skeletal Structure and that of the Boundary 132 and SV T = a c b d where, a = r xx(1 ry) 2 + κ m1 (1 rx 2 ry) r x r y r xy + κ m1 rx 2 1 r 2 x ry 2, 1 rx 2 ry 2 b = r xy(1 rx) 2 + r x r y r xx + κ m1 r x r y 1 r 2 x ry 2, 1 rx 2 ry 2 and c = r xy(1 r 2 y) + r x r y r yy + κ m2 r x r y 1 r 2 x r 2 y 1 r 2 x r 2 y d = r yy(1 rx) 2 + κ m2 (1 rx 2 ry) r x r y r xy + κ m2 ry 2 1 r 2 x ry 2. 1 rx 2 ry 2 2. If the radius function has a singularity at the origin then A V = 0 and SV T = r xx + κ m1 r xy r xy r yy + κ m2 The proof of this corollary comes directly from proposition just by putting n = 2.. Example Let (S, U) be a skeletal structure in R 3 and let s 1 (x, y) = (x, y, x 2 ) S reg such that {(x, y) R < x < 0.45, 0.45 < y < 0.45}. We define the positive function r on s 1 by r(x, y) = 1 2 x + 1, and we define the unit vector field U 1 on s 1 by: U 1 = r + 1 r 2 N, where r is the Riemannian gradient of r and N is the unit normal of s 1. Thus after some calculations we obtain that ( 1 2x x U 1 = 2, 0, 2x + ) x 2. 2(1 + 4x 2 ) 2(1 + 4x 2 )

147 Chapter 5. The Relationship Between the Differential Geometry of the Skeletal Structure and that of the Boundary 133 It is clear that the compatibility condition holds and the associated boundary is given by ( X(x, y) = x (x + 2)(1 + 2x x 2 ), y, x 2 + (x + 2)( 2x + ) x 2 ). 4(1 + 4x 2 ) 4(1 + 4x 2 ) After some calculations we get and S V = A V = 8x(1 + 4x 2 ) 4(1 + 4x x 2 ) x 2 (1 + 4x 2 ) 2 (3 + 16x 2 ) 0 0 4x(1 + 4x 2 ) + 2(1 + 4x x 2 ) x 2 (1 + 4x 2 ) 2 (3 + 16x 2 ) 0 0. The radial principal curvatures are κ r1 = 8x(1 + 4x2 ) 4(1 + 4x x 2 ) x 2 (1 + 4x 2 ) 2 (3 + 16x 2 ) and κ r2 = 0. Thus the Gaussian radial curvature K r = The boundary The skeletal set Figure 5.4: Skeletal set and associated boundary in example Example Let (S, U) be a skeletal structure in R 3 and let s 1 (x, y) = (x, y, y 3 x 2 ) S reg and r = y 2 be the radius function such that 4y 2 < 1. At the origin the

148 Chapter 5. The Relationship Between the Differential Geometry of the Skeletal Structure and that of the Boundary 134 radius function has a singularity and A V = 0 and S V = Figure 5.5: Skeletal set and associated boundary in example Theorem gives the relationship between the matrix S V representing the radial shape operator and the matrix S m representing the differential geometric shape operator of the skeletal structure. In proposition we express the the matrix S V in terms of the matrix S XV representing the differential geometric shape operator of the boundary. Thus now we are able to find the exact relationship between S m and S XV, this relation is