Number of Voronoi-relevant vectors in lattices with respect to arbitrary norms

Fakultät für Elektrotechnik, Informatik und Mathematik Arbeitsgruppe Codes und Kryptographie Number of Voronoi-relevant vectors in lattices with respect to arbitrary norms Master s Thesis in Partial Fulfillment of the Requirements for the Degree of Master of Science by Kathlén Kohn Fürstenallee 136 3310 Paderborn submitted to: Prof. Dr. Johannes Blömer and Prof. Dr. Friedhelm Meyer auf der Heide Paderborn, July 19, 015

Declaration Translation from German) I hereby declare that I prepared this thesis entirely on my own and have not used outside sources without declaration in the text. Any concepts or quotations applicable to these sources are clearly attributed to them. This thesis has not been submitted in the same or substantially similar version, not even in part, to any other authority for grading and has not been published elsewhere. Original Declaration Text in German: Erklärung Ich versichere, dass ich die Arbeit ohne fremde Hilfe und ohne Benutzung anderer als der angegebenen Quellen angefertigt habe und dass die Arbeit in gleicher oder ähnlicher Form noch keiner anderen Prüfungsbehörde vorgelegen hat und von dieser als Teil einer Prüfungsleistung angenommen worden ist. Alle Ausführungen, die wörtlich oder sinngemäß übernommen worden sind, sind als solche gekennzeichnet. City, Date Signature iii

Contents 1 Introduction 1 1.1 Motivation............................... 1 1. Summary of results.......................... 1 1.3 Problem definition.......................... 3 Two-dimensional lattices 7.1 Strictly convex norms......................... 7. Non-strictly convex norms...................... 14 3 Higher-dimensional lattices 7 3.1 Generalizations............................ 7 3. Consequences and comparisons................... 47 4 General shape of bisectors, Voronoi cells and their facets 51 4.1 Norms................................. 51 4. Bisectors................................ 58 4.3 Voronoi cells.............................. 67 4.4 Facets................................. 7 4.4.1 Two-dimensional lattices................... 74 4.4. Higher-dimensional lattices.................. 76 5 Conclusion 85 Bibliography 87 v

List of Figures.1 Lb 1, b ) with boundary of B 1,30)................. 16. H = 1 0, b 1 ), H = 1 0, b ) and H = 1 0, b b 1 )............. 17.3 V i) Lb 1, b ), 1 ) light gray) and V o) Lb 1, b ), 1 ) darker gray).................................. 18 3.1 B 3,10) intersecting different planes................. 8 3. Plane spanned by 0, b m,1 and b m, intersects ball at an edge such that line between 0 and b m,1 + mb m, lies on the edge cf. Figure 3.1h)................................ 9 3.3 Le 1, e, e 3 )............................... 30 3.4 Le 1, e, Me 3 ): Note that M is so large that this figure is not scaled properly................................. 31 3.5 R z Le 1, e, Me 3 )............................ 3 3.6 Λ m = R y R z Le 1, e, Me 3 )....................... 33 4.1 Convex bodies in two dimensions with different properties..... 58 4. Intersection of scaled and translated unit ball with H in a 1, a and a 3 for the case n = 3......................... ). 64 4.3 ϕp) for p as in Figure 4.c with B,1 0) and H a 1 ) + a 1 p a 1 p. 65 4.4 Illustration for the proof of Proposition 4.51: B and C yield not connected parts of the same facet................... 80 4.5 Illustration for the proof of Proposition 4.51: Situation which cannot occur................................ 81 vii

1 Introduction 1.1 Motivation An n-dimensional lattice is a discrete, additive subgroup of R n. Such a lattice can also be represented as the set of all integer linear combinations of some linearly independent vectors in R n, where the linearly independent vectors are called a basis of the lattice. There are many famous problems on lattices, including the shortest vector problem SVP) and the closest vector problem CVP). In the SVP, one is asked to find a shortest nonzero vector in a lattice generated by a given basis. In the CVP, one is given a basis as well as an arbitrary target vector x R n, and needs to find a vector in the lattice generated by the basis that is closest to the target x. Both problems have been shown to be NP-hard under randomized reductions in the case of SVP) [, 1]. Micciancio and Voulgaris gave an algorithm for both problems having On) time and space complexity with respect to the Euclidean norm [13]. The central part of their algorithm is solving a variant of the CVP where additionally to a lattice basis and a target vector one is given a description of the Voronoi cell of the lattice, i.e., the set of all points in R n that are at least as close to 0 as to any other lattice vector. It is clear that the Voronoi cell can be described as the intersection of all halfspaces H 0, v) for all lattice vectors v, where H 0, v) denotes the set of all points in R n that are at least as close to 0 as to v. Using the Euclidean norm, it is sufficient to consider all Voronoi-relevant vectors when taking this intersection. A lattice vector v 0 is called Voronoi-relevant if there is some x R n having the same distance to 0 as to v but a strictly larger distance to all other lattice vectors. The algorithm by Micciancio and Voulgaris uses the set of Voronoi-relevant vectors as a description of the Voronoi cell and thus relies on the fact that there are at most n 1) Voronoi-relevant vectors in a lattice when the Euclidean norm is used. This was shown in [1] with the crucial parallelogram identity that holds exactly for norms induced by a scalar product, e.g., the Euclidean norm. The open problems sections in [13] asks for an extension of the algorithm by Micciancio and Voulgaris to all p-norms, but for this one needs to find an adequate upper bound for the number of Voronoi-relevant vectors with respect to arbitrary p-norms. 1. Summary of results The main goal of this thesis is to analyze the number of Voronoi-relevant vectors in a lattice with respect to norms other than the Euclidean norm. 1

1 Introduction First, it will be shown that the upper bound n 1) shown in [1] still holds for strictly convex norms in the case n =, i.e., that a two-dimensional lattice has at most six Voronoi-relevant vectors with respect to an arbitrary strictly convex norm. A norm is called strictly convex if for all x, y R n with x y and x = y and all 0 < τ < 1 it holds that τx + 1 τ)y < x. For non-strictly convex norms an example of a two-dimensional lattice will be given where the Voronoi-relevant vectors are not sufficient to determine the Voronoi cell. Thus, the notion of generalized Voronoi-relevant vectors will be introduced. For two-dimensional lattices with non-strictly convex norms it will be shown that the number of generalized Voronoi-relevant vectors is not bounded by a constant, which yields that there is no generalization of the upper bound n 1) shown in [1] for non-strictly convex norms. Unfortunately, the same holds for strictly convex norms in dimensions higher than two. The main statement of this thesis is that an upper bound, solely depending on the lattice dimension, for the number of Voronoi-relevant vectors does not exist for general dimensions and general strictly convex norms. For this, a family of three-dimensional lattices will be constructed whose number of Voronoirelevant vectors is not bounded by a constant with respect to the 3-norm. Hence, the algorithm by Micciancio and Voulgaris in [13] cannot be easily extended to general p-norms, since they require that the upper bound n 1) shown in [1] only depends on the dimension n. Moreover, it will be shown for an arbitrary norm n that there are at most 1 + 4 µλ, ) λ 1 Λ, )) generalized) Voronoi-relevant vectors in an n-dimensional lattice Λ, where µλ, ) denotes the minimal d R >0 such that every x spanλ) has at most distance d to some lattice vector and λ 1 Λ, ) is the length of a shortest nonzero lattice vector. This bound, while depending exponentially on the dimension, is obviously also affected by other lattice properties. The number of generalized) Voronoi-relevant vectors of the lattice families constructed for the three-dimensional, strictly convex case as well as for the two-dimensional, non-strictly convex case will be compared with this bound. All these investigations raised the question how Voronoi cells and their facets look with respect to norms other than the Euclidean norm. Therefore, the last part of this thesis examines the set of vectors which is sufficient to determine the Voronoi cell of a given lattice as well as the n 1)-dimensional facets of the Voronoi cell and their connectedness. Two conjectures will be stated which have important implications: One conjecture yields that the Voronoi cell of a lattice with respect to a strictly convex norm is determined completely by the Voronoi-relevant vectors, and the other conjecture implies that under sufficiently nice norms there exists a bijection between the n 1)-dimensional facets of the Voronoi cell and the Voronoi-relevant vectors. The latter statement will be shown for two-dimensional lattices without using the conjectures. For two-dimensional lattices it also holds that these facets are connected, which is probably not true for higher dimensions, since there exist Voronoi cells of finite sets of points with

1.3 Problem definition unconnected facets when a p-norm for p N, p 3 is used. Additionally, one of the two conjectures leads to a fundamental result about bisectors: The bisector between two given points a, b R n with a b is defined as the set of all x R n having the same distance to a as to b. Under sufficiently nice norms and the assumption of the mentioned conjecture, it holds that the bisector between a and b intersected with the bisector between b and c is homeomorphic to R n as long as a, b, c are non-collinear. To the best of my knowledge, this result is only known for at most three dimensions [10] and it is known that each such bisector is homeomorphic to R n 1 [6]. 1.3 Problem definition Throughout this work the following notation will be used: N denotes the set of all strictly positive integers. For a ring R, elements in R n are column vectors. For x R n, the transposed row vector is denoted by x T. Analogously for x 1,..., x n R, the transposed x 1,..., x n ) T R n is a column vector. The dot product on R n is denoted by For p R with p 1,, : R n R n R, x, y) x T y. p : R n R 0, ) 1/p x 1,..., x n ) T xi p denotes the p-norm on R n. The -norm on R n is given by : R n R 0, x 1,..., x n ) T max { x i i {1,..., n}}. Let V R n be a subspace with norm : V R 0. For r R 0 and c V define B,r c) := {x V x c < r} and B,r c) := {x V x c r}. For a subset S R n and λ R as well as t R n, let λs := {λx x S} and S + t := {x + t x S}. 3

1 Introduction For a subset S R n, the linear span of S is denoted by { m } spans) := λ i v i m N, v i S, λ i R. For a subspace V R n let dimv ) denote its dimension. i=1 For a submanifold X R n of dimension m n with M X measurable, vol m M) := χ M x) dx denotes the m-dimensional volume of M, where { 1, if x M χ M : X {0, 1}, x 0, if x X \ M is the indicator function of M. An adjusted version of the signum function will be used which is given by X sgn : R {1, 1}, { 1, if x 0 x 1, if x < 0. Whenever m elements a 1,..., a m are listed, it is assumed that m N. The main object of study are lattices. Definition 1.1 Let b 1,..., b m R n be linearly independent. Then { m } Lb 1,..., b m ) := z i b i z 1,..., z m Z i=1 is a lattice with basis b 1,..., b m ). In addition, n is called the dimension and m the rank of the lattice. A classical result e.g., shown in [4] is that a subset Λ R n is a lattice if and only if it is a non-trivial discrete subgroup of R n, +). To state this formally, one needs to introduce the notion of a discrete subset of R n. Definition 1. A subset Λ R n is called discrete if there exists ε R >0 such that for every x, y Λ with x y it holds that x y ε. Proposition 1.3 Let Λ R n. There exist linearly independent b 1,..., b m R n with Λ = Lb 1,..., b m ) if and only if Λ, +) is a discrete subgroup of R n, +) with Λ {0}. 4

1.3 Problem definition Commonly investigated quantities of lattices are their successive minima as well as their covering radius. Here, mostly the first successive minimum is needed, which equals the length of a shortest nonzero lattice vector. The covering radius is the smallest d R 0 such that all vectors in the linear span of the given lattice are at most of distance d from a lattice vector. Definition 1.4 Let : R n R 0 be a norm and let Λ R n be a lattice of rank m. For i {1,..., m}, the i-th successive minimum of Λ with respect to is λ i Λ, ) := inf { r R 0 dim span Λ B,r 0) )) i }. The covering radius of Λ with respect to is µλ, ) := inf{d R 0 x spanλ) v Λ : x v d}. Apart from these classical notions, this thesis is mostly concerned with the Voronoi cell of a lattice. Such a Voronoi cell can be defined for every element from a given discrete set of points as the set of all vectors that are at least as close to this element as to any other point of the given set. If the given set of points forms a lattice, all Voronoi cells for the lattice points will be translates of the Voronoi cell of the lattice origin such that is enough to consider this specific Voronoi cell. Definition 1.5 Let : R n R 0 be a norm and let P R n be a discrete subset. The Voronoi cell of a P is defined as V,P a) := {x spanp) b P : x a x b }. If P is a lattice, VP, ) := V,P 0) denotes the Voronoi cell of the origin. Maybe the most important definition for this thesis is the notion of Voronoirelevant vectors since the main task of this work is to analyze the number of these vectors. Definition 1.6 Let : R n R 0 be a norm and let Λ R n be a lattice. A lattice vector v Λ\{0} is a Voronoi-relevant vector if there is some x spanλ) such that x = x v < x w holds for all w Λ \ {0, v}. The idea behind this notion is that one does not need to consider all lattice vectors in Definition 1.5 of the Voronoi cell when the Euclidean norm is used. Instead, the Voronoi cell of the origin of a lattice is already specified as the set of points in the linear span of the lattice which are at least as close to 0 as to all Voronoi-relevant vectors. This connection between Voronoi cell and Voronoirelevant vectors will be formally examined in Section 4.3. For both concepts it is natural to consider them in terms of bisectors and halfspaces. Definition 1.7 Let V R n be a subspace with norm : V R 0. For 5

1 Introduction a, b V, the bisector of a and b with respect to is defined as H = a, b) := {x V x a = x b }. The corresponding strict and non-strict halfspaces are denoted by H < a, b) := {x V x a < x b } and H a, b) := H< a, b) H= a, b). With this, the Voronoi cell of the origin of a lattice Λ can be written as ) VΛ, ) = spanλ) H 0, v), and a lattice vector v Λ \ {0} is Voronoi-relevant if and only if spanλ) H 0, = v) 0, w). v Λ w Λ\{0,v} During the investigations of all these geometrical objects, it will be differentiated if the underlying norm is strictly convex or not. As seen in the following chapters, strictly convex norms e.g., the Euclidean norm have a lot of nice properties. Definition 1.8 Let V R n be a subspace. A norm : V R 0 is called strictly convex if for all x, y V with x y and x = y and all τ 0, 1) it holds that τx + 1 τ)y < x. Note that for every norm it holds for all x, y V with x y and x = y and all τ [0, 1] that τx + 1 τ)y x. For strictly convex norms, this inequality is an equality if and only if τ {0, 1}. H < 6

Two-dimensional lattices Already in two dimensions, some different properties of strictly convex and nonstrictly convex norms will become clear. Whereas, every two-dimensional lattice has at most six Voronoi-relevant vectors with respect to every strictly convex norm, the Voronoi-relevant vectors are in general not sufficient in the case of a non-strictly convex norm to determine the Voronoi cell completely, such that a little broader notion needs to be introduced: Generalized Voronoi-relevant vectors. Moreover, it will be shown that every lattice of rank two has at most eight generalized Voronoi-relevant vectors with respect to every strictly convex norm, but the number of generalized Voronoi-relevant vectors is generally not upper bounded by a constant in lattices of rank two when a non-strictly convex norm is used..1 Strictly convex norms The ideas for this section were developed in collaboration with Prof. Dr. Johannes Blömer and David Teusner. It will be proven that every two-dimensional lattice has at most six Voronoi-relevant vectors with respect to arbitrary strictly convex norms. This is a direct consequence of Theorem.4 below and the following Proposition. Proposition.1 Let Λ R n be a lattice of rank one with basis b 1 ), and let : R n R 0 be an arbitrary norm. Then ±b 1 are Voronoi-relevant vectors. All other lattice vectors are not Voronoi-relevant. The proof of this proposition uses a lemma which holds for lattices of every rank. Lemma. Let : R n R 0 be a norm an let Λ R n be a lattice. For every v Λ and every k N, k it holds that kv is not Voronoi-relevant. Proof. Assume for contradiction that kv is Voronoi-relevant for some k. Then there exists some x spanλ) such that x kv = x < x w holds for all w Λ \ {0, kv}. This yields the contradiction x < x v = 1 k kx kv = 1 x kv + k 1)x k 1 k x kv + k 1) x ) = 1 x + k 1) x ) = x. k 7

Two-dimensional lattices Proof of Proposition.1. For every z 1 Z \ {0, 1} it holds that b 1 z 1b 1 = 1 z 1 b 1 3 b 1 > 1 b 1. Thus, b 1 is Voronoi-relevant, which directly implies that b 1 is Voronoi-relevant. By Lemma., kb 1 and kb 1 are not Voronoi-relevant for every k N, k. For Theorem.4, one needs the notion of a Gauss-reduced basis. Definition.3 Let : R n R 0 be a norm and let Λ R n be a lattice of rank two. A basis b 1, b ) of Λ is called Gauss-reduced with respect to if it holds that b 1 b b 1 b b 1 + b. In [7], Kaib and Schnorr show for an arbitrary norm that every lattice of rank two has a Gauss-reduced basis. In fact, they give and analyze an algorithm which computes a Gauss-reduced basis out of a given lattice basis. Hence, it can be assumed that a lattice of rank two is already given by a Gauss-reduced basis. Theorem.4 Let Λ R n be a lattice of rank two with Gauss-reduced basis b 1, b ), and let : R n R 0 be a strictly convex norm. Then ±b 1 and ±b are Voronoi-relevant vectors. In addition, ±b 1 b ) are Voronoi-relevant vectors if and only if b 1 b < b 1 +b. All other lattice vectors are not Voronoi-relevant. The proof of this theorem is based on multiple lemmata, which will be discussed below. Lemma.9 gives a superset of all Voronoi-relevant vectors under the same assumptions as in the above theorem. Furthermore, Lemmata.13,.14 and.15 consider all elements in this superset and determine if they are Voronoi-relevant or not. To prove these four lemmata, some further helpful statements will be shown. The first two of these statements are basic properties of Gauss-reduced bases and strictly convex norms. Remark.5 Let Λ R n be a lattice of rank two with Gauss-reduced basis b 1, b ), and let r R with r 0. Then rb 1, rb ) is a Gauss-reduced basis for r Λ. Lemma.6 Let : R n R 0 be a strictly convex norm. Moreover, let x, y R n with x y and x y, and let τ R with 0 < τ < 1. Then it holds that τx + 1 τ)y < y. Proof. For x = y, the desired inequality is directly given by the definition of strict convexity. If x < y, then it holds that τx + 1 τ)y τ x + 1 τ) y < y. 8

.1 Strictly convex norms The next lemma shows that a vector that lies above-right, above-left, below-right or below-left of the parallelogram Pb 1, b ) := {r 1 b 1 + r b r 1, r [ 1, 1]} is longer than the corresponding corner vector of this parallelogram if a strictly convex norm is used and b 1, b ) is a Gauss-reduced basis. Lemma.7 Let Λ R n be a lattice of rank two with Gauss-reduced basis b 1, b ), and let : R n R 0 be a strictly convex norm. Moreover, let r 1, r R. If r 1, r 1 and max{ r 1, r } > 1, then b 1 + sgnr 1 r )b < r 1 b 1 + r b. Proof. Let r 1, r R with r 1, r 1, and distinguish the following two cases. 1. r 1 r and r 1 > 1: Using b 1 +sgnr 1 r )b = 1 r 1 r 1 b 1 +r b )+sgnr 1 r ) b b 1 + sgnr 1 r )b leads to b 1 + sgnr 1 r )b 1 r 1 r 1b 1 + r b + 1 r 1 r 1b 1 + r b + 1 r r 1 1 r ) b r 1 1 1 r 1 ) b, r 1 and ) b 1 + sgnr 1 r )b. This shows b 1 + sgnr 1 r )b r 1 b 1 + r b. If b < b 1 + sgnr 1 r )b or r > 1, then this inequality is even strict, i.e., b 1 + sgnr 1 r )b < r 1 b 1 + r b. For r = 1 and b = b 1 + sgnr 1 r )b, Lemma.6 yields b 1 + sgnr 1 r )b = sgnr 1) 1 r 1 r 1b 1 + r b ) + sgnr 1 r ) < r 1 b 1 + r b.. r r 1 and r > 1: Using b 1 + sgnr 1 r )b = sgnr 1 r ) 1 r r 1 b 1 + r b ) + 1 1 r 1 ) b ) b 1, r 1 1 and 1 r 1 r b 1 b 1 + sgnr 1 r )b, analog arguments lead to the same conclusion. Using the above property for vectors outside of the parallelogram Pb 1, b ), it can be easily deduced in the next lemma that every vector having 0 as one of its closest lattice vectors must lie inside of Pb 1, b ). From this it follows that for every Voronoi-relevant vector v the parallelograms Pb 1, b ) and v + Pb 1, b ) intersect in their interior, which gives the superset of all Voronoi-relevant vectors as specified and formally proven in Lemma.9. 9

Two-dimensional lattices Lemma.8 Let Λ R n be a lattice of rank two with Gauss-reduced basis b 1, b ), and let : R n R 0 be a strictly convex norm. Moreover, let r 1, r R. If r 1 b 1 + r b r 1 b 1 + r b v holds for all v Λ, then r 1, r 1, 1). Proof. Let r 1, r R with r 1 / 1, 1) or r / 1, 1). The rest of this proof shows that there is some v Λ with r 1 b 1 + r b > r 1 b 1 + r b v. If r = 0, then r 1 b 1 sgnr 1 )b 1 = r 1 sgnr 1 ) b 1 < r 1 b 1 = r 1 b 1 follows due to r 1 / 1, 1). If r 1 = 0, then it holds analogously that r / 1, 1) and r b sgnr )b < r b. If r 1 = r 1, then r 1 b 1 + r b sgnr 1 )b 1 sgnr )b = r 1 sgnr 1 ) b 1 + sgnr 1 r )b < r 1 b 1 + sgnr 1 r )b = r 1 b 1 + r b holds. Hence, it can be assumed that r 1 0 r and that r 1 = r, and only two remaining cases need to be considered. 1. r 1 > r and r 1 1: By Remark.5, r b 1, r b ) is a Gauss-reduced basis for r Λ. Together with Lemma.7 applied to this basis, it follows that ) r b r r1 b 1 + sgn 1 r b < r 1 b 1 + r b. r If r 1 = 1, then r 1 b 1 + r b sgnr 1 )b 1 = r b < r 1 b 1 + r b holds. For r 1 > 1, Lemma.6 yields r 1 b 1 + r b sgnr 1 )b 1 = r 1 sgnr 1 ) r 1 b 1 + r b ) + 1 r ) 1 sgnr 1 ) r b < r 1 b 1 + r b. r 1 r 1. r > r 1 and r 1: Analog arguments lead to r 1 b 1 < r 1 b 1 + r b as well as r 1 b 1 + r b sgnr )b < r 1 b 1 + r b. Lemma.9 Let Λ R n be a lattice of rank two with Gauss-reduced basis b 1, b ), and let : R n R 0 be a strictly convex norm. Then all Voronoi-relevant vectors are contained in {±b 1, ±b, ±b 1 b ), ±b 1 + b )}. Proof. Let z 1, z Z and let z 1 b 1 + z b be a Voronoi-relevant vector. Then there are r 1, r R with r 1 b 1 +r b = r 1 b 1 +r b z 1 b 1 z b r 1 b 1 +r b v for all v Λ. In particular, Lemma.8 implies r 1, r 1, 1) as well as r 1 z 1, r z 1, 1). This leads to z 1, z, ) and thus to z 1, z { 1, 0, 1}. Hence, z 1 b + z b {±b 1, ±b, ±b 1 b ), ±b 1 + b )} follows. 10

.1 Strictly convex norms Now it is already shown that every lattice of rank two has at most eight Voronoirelevant vectors with respect to a strictly convex norm. Some of these eight candidates can be excluded, such that only four or six Voronoi-relevant vectors remain. For this, some further statements are needed, e.g., the next lemma shows that in every bisector between 0 and some x 0 the vector x is the unique shortest vector. Lemma.10 Let : R n R 0 be a strictly convex norm. Moreover, let x R n with x 0, and let y R n with y x and y = y x. Then x < y holds. Proof. Let x, y R n with y x 0 and y = y x, and assume for contradiction that x y. Consider w := 1 y + 1 x. Then it follows from Lemma.6 that w < x. With this, distinguish the following two cases. 1. w x w : The estimate x = x w + w w x + w w = y + x y + x x shows that w = x, which contradicts w < x.. w x > w : Now it follows that x w x + w < w x = y x) x y x + x x, which is a contradiction. Analogously to Lemma.7, it will be shown in the next lemma that a vector lying inside of the upper-right, upper-left, lower-right or lower-left quarter of the parallelogram Pb 1, b ) is shorter than the corresponding corner vector of Pb 1, b ). Lemma.11 Let Λ R n be a lattice of rank two with Gauss-reduced basis b 1, b ), and let : R n R 0 be a strictly convex norm. Moreover, let r 1, r R. If 0 < r 1, r 1 and min{ r 1, r } < 1, then r 1 b 1 + r b < b 1 + sgnr 1 r )b. Proof. Let r 1, r R with 0 < r 1, r 1. If r 1 < 1 = r, it holds by Lemma.6 that r 1 b 1 + r b = r 1 sgnr 1 )b 1 + r b ) + 1 r 1 )r b < sgnr 1 )b 1 + r b = b 1 + sgnr 1 r )b..1) 11

Two-dimensional lattices For r < 1 = r 1, the roles of r 1 and r can be exchanged in.1) to get r 1 b 1 + r b < b 1 + sgnr 1 r )b. If r 1, r < 1, Lemma.6 implies r 1 b 1 + r b = r r 1 b 1 + sgnr )b ) + 1 r )r 1 b 1 < max{ r 1 b 1 + sgnr )b, r 1 b 1 }. By.1), r 1 b 1 + sgnr )b < b 1 + sgnr 1 r )b. Since moreover it holds that r 1 b 1 < b 1 b 1 +sgnr 1 r )b, it follows that r 1 b 1 +r b < b 1 +sgnr 1 r )b. The next lemma gives an easy upper bound for the covering radius. With this bound at hand, the remaining three lemmata of this section will investigate the conditions under which the eight candidates for Voronoi-relevant vectors stated in Lemma.9 are indeed Voronoi-relevant. Lemma.1 Let Λ R n be a lattice of rank two with Gauss-reduced basis b 1, b ), and let : R n R 0 be a strictly convex norm. Then for every x spanλ) there exists v Λ such that x v b 1 +b. Proof. Let r 1, r R and x := r 1 b 1 + r b. Rounding the coefficients to the nearest integer yields z 1, z Z and s 1, s R with s 1, s 1 and x = z 1 + s 1 )b 1 + z + s )b. Defining v := z 1 b 1 + z b leads to x v = s 1 b 1 + s b. If s 1 = 0, then x v = s b 1 b 1 + b. If s = 0, x v 1 b 1 + b follows analogously. Hence, it can be assumed that s 1 0 s. If s 1 = s = 1, then x v = 1 sgns 1)b 1 + sgns )b 1 b 1 + b. If min{ s 1, s } < 1, Lemma.11 yields x v = s 1 b 1 + s b < b 1 + sgns 1 s )b b 1 + b. Lemma.13 Let Λ R n be a lattice of rank two with Gauss-reduced basis b 1, b ), and let : R n R 0 be a strictly convex norm. Then the vectors ±b 1 + b ) are not Voronoi-relevant. Proof. Assume for contradiction that b 1 + b is Voronoi-relevant. Then there is some x spanλ) with x = x b 1 b < x v for all v Λ \ {0, b 1 + b }. If x = b 1+b, then x b 1 = b 1 + b = 1 b 1 b 1 b 1 + b = x < x b 1, which is a contradiction. If x b 1+b, Lemma.10 implies b 1 +b < x. On the other hand, Lemma.1 shows that there is some v Λ with x v b 1 +b < x, which is a contradiction. 1

.1 Strictly convex norms Hence, b 1 + b cannot be Voronoi-relevant, which furthermore implies that b 1 b is not Voronoi-relevant. Lemma.14 Let Λ R n be a lattice of rank two with Gauss-reduced basis b 1, b ), and let : R n R 0 be a strictly convex norm. Then the vectors ±b 1 and ±b are Voronoi-relevant. Proof. To show that b 1 is Voronoi-relevant, it is sufficient to prove that b 1 < v b 1 holds for all v Λ \ {0, b1 }. Hence, consider z 1, z ) Z \ {0, 0), 1, 0)} and v := z 1 b 1 + z b. If z 0, then it follows from Lemma.7 that v b 1 = 1 z 1 1)b 1 + z b > 1 b 1 + sgnz 1 1)z )b b 1. If z = 0, then it is z 1 / {0, 1}, which implies that v b 1 = 1 z 1 1 b 1 3 b 1 > b 1. Therefore, b 1 and b 1 are Voronoi-relevant. By exchanging the roles of z 1 and z in the above inequalities, it follows analogously that b < v b holds for all v Λ \ {0, b }, yielding that b and b are Voronoi-relevant. Lemma.15 Let Λ R n be a lattice of rank two with Gauss-reduced basis b 1, b ), and let : R n R 0 be a strictly convex norm. Then the vectors ±b 1 b ) are Voronoi-relevant if and only if b 1 b < b 1 + b. Proof. First assume that b 1 b is Voronoi-relevant. Then there is some x spanλ) with x = x b 1 + b < x v for all v Λ \ {0, b 1 b }. If x b 1 b, then it holds by Lemmata.10 and.1 that b 1 b < x b 1 +b, which implies b 1 b < b 1 + b. If x = b 1 b, then it follows that b 1 b = x < x b 1 = b 1 + b. Now assume b 1 b < b 1 + b and show that b 1 b as well as b 1 + b are Voronoi-relevant. For this it is enough to prove that b 1 b < v b 1 + b holds for all v Λ \ {0, b 1 b }. For v {b 1, b } it holds that v b 1 + b = b 1 +b > b 1 b. Hence, consider z1, z ) Z \ {0, 0), 1, 1), 1, 0), 0, 1)} and v := z 1 b 1 + z b. Then it follows that z 1 / {0, 1} or z / { 1, 0}, leading to z 1 1 3 or z + 1 3. Thus, Lemma.7 implies v b 1 + b = 1 z 1 1)b 1 + z + 1)b > 1 b 1 + sgnz 1 1)z + 1))b b 1 b. 13

Two-dimensional lattices Proof of Theorem.4. This proof follows directly from Lemmata.9,.13,.14 and.15.. Non-strictly convex norms Consider the lattice Lb 1, b ) spanned by b 1 := 1, 1) T and b := 0, 3) T together with the 1-norm 1. In the following, the Voronoi cell VLb 1, b ), 1 ) of the origin will be constructed illustratively and afterwards a formal proof will be given. For the illustrative construction one can investigate the bisectors between 0 and all generalized Voronoi-relevant vectors v Lb 1, b ). Definition.16 Let : R n R 0 be a norm and let Λ R n be a lattice. A lattice vector v Λ \ {0} is a generalized Voronoi-relevant vector if there is some x spanλ) such that x = x v x w holds for all w Λ. The next lemma is helpful to determine which lattice vectors are generalized Voronoi-relevant vectors since it shows that these vectors cannot be too far away from the origin. Lemma.17 Let : R n R 0 be a norm and let Λ R n be a lattice. For every generalized Voronoi-relevant vector v Λ it holds that v µλ, ). Proof. Let v Λ \ {0} and x spanλ) with x = x v x w for all w Λ. Then it holds that x = x v µλ, ) and thus v = v x + x x v + x µλ, ). Next, a calculation of the covering radius of Lb 1, b ) with respect to 1 is needed in order to use the above lemma. Lemma.18 µlb 1, b ), 1 ) = 3. Proof. First prove µlb 1, b ), 1 ) 3. For this, let x = x 1, x ) T R, and show that x v 1 3 holds for some v Lb 1, b ). There are m Z and r [0, 1) with x 1 = m + r. Moreover, there is k Z such that m + 3k x < m + 3k + 1) holds. Now, show in each of the following four cases that x has distance at most to some lattice vector. 3 1. x m + 3k + 1 and r 1 : x mb 1 kb = x 1 m + x m 3k = r + x m 3k) 1 + 1 = 3. x m + 3k + 1 and r > 1 : x m + 1)b 1 kb = x 1 m 1 + x m 1 3k 14

. Non-strictly convex norms = 1 r) + m + 1 + 3k x ) < 1 + 1 = 3 3. x > m + 3k + 1 and x x 1 + 3k + 3: x m + 1)b 1 kb = x 1 m 1 + x m 1 3k = 1 + m x 1 ) + x m 1 3k) = x x 1 3k 3 4. x > m + 3k + 1 and x > x 1 + 3k + 3: x mb 1 k + 1)b = x 1 m + x m 3k + 1) = x 1 m) + m + 3k + 3 x ) = x 1 x + 3k + 3 < 3 To prove µlb 1, b ), 1 ) 3, consider x := 3 1, 1)T and compute all 4 v Lb 1, b ) with x v 1 3. For this, let z 1, z Z and v := z 1 b 1 + z b. Then it holds that x v 1 = 3 z 4 1 + 3 z 4 1 3z. Assuming x v 1 3 implies 3 z 4 1 3, leading to z 1 {0, 1, }. According to this, distinguish the following three cases. 1. z 1 = 0: From 3 x v 1 = 3 + 3 + 3z 4 4 it follows that 1+4z 1, which implies z = 0.. z 1 = 1: Here, 3 x v 1 = 1 + 7 + 3z 4 4 yields 7 + 1z 5 and thus z = 1. 3. z = : 3 In this case, x v 1 = 5 + 11 + 3z 4 4 must hold. 11 + 1z 1 and z = 1. This shows In all three cases it holds that x v = 3, which shows that every lattice vector has distance at least 3 to x. By combining both lemmata, it is enough to consider the bisectors between 0 and all v Lb 1, b ) \ {0} with v 1 3 in order to compute the desired Voronoi cell. An excerpt of the lattice Lb 1, b ) together with the boundary of B 1,30) is depicted in Figure.1, and the next lemma determines formally which lattice vectors have norm at most three. 15

Two-dimensional lattices Figure.1: Lb 1, b ) with boundary of B 1,30). Lemma.19 Let z 1, z Z. Then it holds that z 1 b 1 + z b 1 3 if and only if z 1, z ) { 3, 1),, 1), 1, 0), 1, 1), 0, 1), 0, 0), 0, 1), 1, 1), 1, 0),, 1), 3, 1)} =: I. Proof. If z 1, z ) I, easy calculation shows z 1 b 1 + z b 1 = z 1 + z 1 + 3z 3. Hence, it is left to consider z 1, z Z with z 1 b 1 + z b 1 3. From this it directly follows that z 1 3 holds. Due to symmetry, one can further assume that z 1 0. Now distinguish the remaining cases. 1. z 1 = 3: In this case, z 1 + 3z = 0 follows, which implies z = 1.. z 1 = : From z 1 + 3z 1 one deduces 3z { 3,, 1}, leading to z = 1. 3. z 1 = 1: Since z 1 + 3z must hold, 3z { 3,, 1, 0, 1} follows, yielding z { 1, 0}. 4. z 1 = 0: Now 3 z 3 needs to holds, which implies z { 1, 0, 1}. 16

. Non-strictly convex norms When computing the bisectors between 0 and the lattice vectors specified in the above lemma, it follows from the symmetry that only three different kinds of bisectors need to be examined. These are shown in Figure.. Since these bisectors are at the moment only used for illustration, a formal proof stating that the bisectors look exactly as depicted in the figure is omitted. Figure.: H = 1 0, b 1 ), H = 1 0, b ) and H = 1 0, b b 1 ). Intersecting these bisectors suggests that VLb 1, b ), 1 ) looks as given in Figure.3, where the Voronoi cell is partitioned into two parts VLb 1, b ), 1 ) = V i) Lb 1, b ), 1 ) V o) Lb 1, b ), 1 ). Definition.0 Let : R n R 0 be a norm and let Λ R n be a lattice. V i) Λ, ) := {x spanλ) v Λ \ {0} : x < x v } denotes the strict Voronoi cell of the origin of Λ with respect to. In addition, define V o) Λ, ) := VΛ, ) \ V i) Λ, ). Additionally to these figurative ideas, a formal proof to specify the Voronoi cell and its two parts will be given now. Proposition.1 It holds that V i) Lb 1, b ), 1 ) { ) x1 = R x x 1 < 1, x < 1, x 1 + x < 1, x 1 x < 3 } =: S i) and V o) Lb 1, b ), 1 ) = S o) 1 S o) S o) 3 S o) 4, where { ) } S o) x1 1 := ± R x x 1 + x = 1, x 1 0, 1), 17

Two-dimensional lattices Figure.3: V i) Lb 1, b ), 1 ) light gray) and V o) Lb 1, b ), 1 ) darker gray). { S o) := ± { S o) 3 := ± S o) 4 := { ± x1 x x1 x x1 x ) R x 1 x = 3 )} 1, x 1, 1, ) R x 1 1, x 0, x 1 x 3 }, ) R x 1 0, x 1, x 1 x 3 }. The proof of this proposition is split up in the three following lemmata. Lemma. VLb 1, b ), 1 ) S i) S o) 1 S o) S o) 3 S o) 4. Proof. Let x = x 1, x ) T R with x / S i) S o) 1 S o) S o) 3 S o) 4. It is left to show that x / VLb 1, b ), 1 ), i.e., that there is some v Lb 1, b ) such that x v 1 < x 1. Due to symmetry, one can assume that x 1 0. Now distinguish the three following cases. 1. x 0: Assume for contradiction that x 1 x 3. Then it follows from x / So) 3 that x 1 < 1. Analogously, x / S o) 4 implies x > 1. Hence, it holds that x 1 < 1, x < 1 and x 1 + x < 1. Since x / S i), x 1 x = x 1 x 3 must hold, which implies x 1 = 3 + x > 1. But now it follows that x So), which is a contradiction. Therefore it holds that x 1 = x 1 x > 3, and Lemma.18 shows that x v 1 3 < x 1 must hold for some v Lb 1, b ). 18

. Non-strictly convex norms. x > 0 and x 1 + x > 3 : In this case it directly follows from Lemma.18 that for some v Lb 1, b ) it is x 1 = x 1 + x > 3 x v 1. 3. x > 0 and x 1 + x 3 : First, assume for contradiction that x 1 = 0. Since x / S i), it must hold that x 1, but then x S o) 4 would follow, which is a contradiction. Thus, x 1 > 0 holds. Secondly, assume for contradiction that x 1 + x 1. From x 1 > 0 and x > 0 it follows that x 1 < 1 and x < 1. This further implies x 1 x < 1. Furthermore, x / S i) yields x 1 + x 1. Hence, x 1 + x = 1 and x S o) 1 follow, where the latter is a contradiction. Thus, x 1 = x 1 + x > 1 and x 1 > 0 hold. The rest of this proof shows that x b 1 1 = x 1 1 + x 1 < x 1. If x 1 < 1 and x < 1, then x b 1 1 = x 1 x < 1 < x 1 holds. For x 1 1, x 1 + x 3 implies x 1 and thus it follows due to x > 0 that x b 1 1 = x 1 x < x 1 + x = x 1. For x 1, one shows analogously x b 1 1 = x 1 + x < x 1 + x = x 1. Lemma.3 S i) V i) Lb 1, b ), 1 ). Proof. Let x = x 1, x ) T S i), and assume for contradiction that x / V i) Lb 1, b ), 1 ), i.e., that there is some v Lb 1, b ) \ {0} such that x v 1 x 1. In addition, let z 1, z Z with v = z 1 b 1 + z b. Due to symmetry, one can assume that x 1 0. Depending on x, two cases need to be distinguished. 1. x 0: In this case, it is x 1 z 1 + x z 1 3z = x v 1 x 1 = x 1 + x < 1 and thus x 1 z 1 < 1, which implies z 1 {0, 1}. Examine both of these cases independently as follows. a) z 1 = 0: x 3z < 1 leads to z = 0, which is a contradiction to v 0. b) z 1 = 1: Again, it follows from x 1 3z < 1 that z = 0, but this further implies that x v 1 = x 1 1 + x 1 = x 1 x > 1. This is a contradiction to x v 1 < 1. 19

Two-dimensional lattices. x < 0: It holds that x 1 z 1 + x z 1 3z = x v 1 x 1 = x 1 x < 3. Hence, it is x 1 z 1 < 3 and x z 1 3z < 3, and these two inequalities yield z 1, z 1 3z { 1, 0, 1, }. According to z 1, these four cases will be distinguished. a) z 1 = 1 : Then, it is 3z {, 1, 0, 1}, which leads to z = 0. Therefore, x > 1 yields the contradiction x v 1 = x 1 + 1 + x + 1 = + x 1 + x > x 1 x = x 1. b) z 1 = 0 : In this case, 3z { 1, 0, 1, } holds, which shows that z = 0, but this is a contradiction to v 0. c) z 1 = 1 : Now, 3z {0, 1,, 3} must hold, yielding z { 1, 0}. If z = 1, then x 1 x < 3 leads to x v 1 = x 1 1 + x + = 3 x 1 + x > x 1 x = x 1. If z = 0, then it follows from x 1 < 1 that x v 1 = x 1 1 + x 1 = x 1 x > x 1 x = x 1. Hence, both cases contradict x v 1 x 1. d) z 1 = : Here, 3z {1,, 3, 4} holds, which implies z = 1, but this gives the contradiction x v 1 = x 1 + x +1 = 3 x 1 +x > x 1 x = x 1. Lemma.4 S o) 1 S o) S o) 3 S o) 4 V o) Lb 1, b ), 1 ). In addition, for all z 1, z ) I \ {0, 0)} it holds that z 1 b 1 + z b is a generalized Voronoi-relevant vector with respect to 1. Proof. Let x = x 1, x ) T S o) 1 S o) S o) 3 S o) 4. It is left to show that x v 1 x 1 holds for all v Lb 1, b ), and that there is some w Lb 1, b ) with w 0 and x w 1 = x 1. For this, consider all v Lb 1, b ) with x v 1 x 1 and show that x v 1 = x 1 needs to hold. The proof distinguishes four cases according to the four sets S o) j for j = 1,, 3, 4, and finds in every case at least one lattice vector w 0 with x w 1 = x 1, which in particular shows that these lattice vectors are generalized Voronoi-relevant vectors. Hence, let z 1, z Z and v := z 1 b 1 + z b with x 1 z 1 + x z 1 3z = x v 1 x 1, and consider the four following cases. 1. x S o) 1 = { ± x1 x ) } R x 1 + x = 1, x 1 0, 1) : Due to symmetry, one can assume that x 1 0, 1). Then it follows that x 1 = x 1 + x = 1, implying x 1 z 1 x v 1 1. Thus, it is z 1 {0, 1}. Now distinguish these two cases as well. 0

. Non-strictly convex norms a) z 1 = 0: Since x 3z x v 1 1, it holds that z = 0 and v = 0. b) z 1 = 1:. x S o) = Now, x 1 3z 1 needs to hold, which implies again z = 0. This time it follows that v = b 1 and that x v 1 = x 1 1 + x 1 = x 1 x = 1 = x 1. Additionally, the vectors ±b 1 are generalized Voronoi-relevant. { ± x1 x ) R x 1 x = 3, x 1 1, 1)} : Due to symmetry, one can assume that x 1 1. Then it follows that x 1 = x 1 x = 3. This yields x 1 z 1 x v 1 3 and analogously x z 1 3z 3, which implies z 1, z 1 3z {0, 1, }. With this, distinguish the cases according to z 1. a) z 1 = 0: 3z {0, 1, } leads to z = 0 and v = 0. b) z 1 = 1: Here, it must hold that 3z {1,, 3} and thus z = 1. Then, it is v = b 1 b and x v 1 = x 1 1 + x + = 3 x 1 + x = 3 = x 1. c) z 1 = : 3. x S o) 3 = From 3z {, 3, 4} it follows again that z = 1. Hence, it is v = b 1 b and x v 1 = x 1 + x +1 = 3 x 1 +x = 3 = x 1. Additionally, this case and the upper case show that ±b 1 b ) and ±b 1 b ) are generalized Voronoi-relevant vectors. { ± x1 x ) R x 1 1, x 0, x 1 x 3 Due to symmetry, one can assume that x 1 1. Then it is x 1 = x 1 x 3. This yields x 1 z 1 x v 1 3 and analogously x z 1 3z 3, which implies z 1 {0, 1,, 3} and z 1 3z { 1, 0, 1, }. With this, distinguish the cases according to z 1. a) z 1 = 0: 3z { 1, 0, 1, } gives z = 0 and v = 0. b) z 1 = 1: 3z {0, 1,, 3} implies z { 1, 0}. If z = 1, then x 1 yields together with x 1 x = x 1 x v 1 = x 1 1 + x + = 1 + x 1 + x that x = 1, and from this it follows x 1 = 1 as well as x v 1 = 3 = x 1. If z = 0, then it is x v 1 = x 1 1 + x 1 = x 1 x = x 1 for every x S o) 3 with x 1 1. } : 1

Two-dimensional lattices c) z 1 = : Here, it must hold that 3z {1,, 3, 4}, which leads to z = 1. Furthermore, x 1 x 3 and x 1 x = x 1 x v 1 = x 1 + x + 1 = 3 x 1 + x imply x 1 x = 3, which gives x v 1 = 3 = x 1. d) z 1 = 3: 4. x S o) 4 = Now 3z {, 3, 4, 5} holds, which implies z = 1 and v = 3b 1 b. Moreover, it follows from x 1 3 and x 1 x = x 1 x v 1 = x 1 3 + x = 3 x 1 x that x 1 = 3 holds. Hence, x = 0 and x v 1 = 3 = x 1. In addition, this shows that ±3b 1 b ) are generalized Voronoi-relevant vectors. { ) x1 ± R x 1 0, x 1, x 1 x 3 x Due to symmetry, one can assume that x 1. Then it is x + b 1 S o) 3 with x 1 + 1 1. Furthermore, it is x + b 1 ) v + b 1 ) 1 = x v 1 x 1 = x 1 x = x + b 1 1. Hence, the above investigation for S o) 3 already shows that v + b 1 {0, b 1 b, b 1, b 1 b, 3b 1 b } which is equivalent to v { b 1, b, 0, b 1 b, b 1 b } and that x v 1 = x must hold in every case according to v. Additionally, in the case v = b 1 it follows that x v 1 = x 1 holds for every x S o) 4 with x 1, and the case v = b shows that ±b are generalized Voronoi-relevant vectors. } : These three lemmata give proof of Proposition.1 as well as a direct corollary specifying all generalized Voronoi-relevant vectors of Lb 1, b ) with respect to 1. Proof of Proposition.1. The proof follows directly from combining Lemmata.,.3 and.4 since V i) Lb 1, b ), 1 ) and V o) Lb 1, b ), 1 ) are disjoint by definition. Corollary.5 Let z 1, z Z. Then it holds that z 1 b 1 + z b is a generalized Voronoi-relevant vector with respect to 1 if and only if z 1, z ) I \ {0, 0)}. Proof. This statement is a direct consequence of Lemmata.17,.18,.19 and.4. Moreover, all Voronoi-relevant vectors can be computed as follows. Lemma.6 v Lb 1, b ) is Voronoi-relevant with respect to 1 if and only if v = ±b 1. Proof. Since every Voronoi-relevant vector is a generalized Voronoi-relevant vector, it holds for every Voronoi-relevant vector z 1 b 1 + z b Lb 1, b ) by Corollary.5 that z 1, z ) I \ {0, 0)}. Due to symmetry, only five different values for z 1, z ) need to be considered.

. Non-strictly convex norms 1. z 1, z ) = 0, 1): In this case, it will be shown that the vectors ±b are not Voronoi-relevant because for every x R with x 1 = x b 1 it is x + b 1 b 1 x 1. To see this, let x = x 1, x ) T R such that x 1 + x = x 1 = x b 1 = x 1 + x 3. Consequently, x = 3 holds cf. Figure.). Now distinguish the following three cases. a) x 1 0: Here, even the following equality holds. b) 1 < x 1 < 0: x + b 1 b 1 = x 1 + 3 = x 1 = x b 1.) It follows that x + b 1 b 1 = x 1 + 3 < x 1 + 3 = x 1. c) x 1 1: Then it holds that x + b 1 b 1 = x 1 1 < x 1 + 3 = x 1.. z 1, z ) = 3, 1): In this case, it will be shown that the vectors ±3b 1 b ) are not Voronoirelevant because for every x R with x 1 = x 3b 1 + b 1 it holds that x b 1 + b 1 x 1. This can be seen completely analogously to the case z 1, z ) = 0, 1) above by exchanging the roles of x 1 and x. In particular, it holds for x 0 that x b 1 + b 1 = x + 3 = x 1 = x 3b 1 + b 1..3) 3. z 1, z ) = 1, 1): In this case, it will be shown that the vectors ±b 1 b ) are not Voronoirelevant because for every x R with x 1 = x + b 1 b 1 there is some v {b 1, b, b 1 + b } such that x v 1 x 1 holds. To see this, let x = x 1, x ) T R such that x 1 + x = x 1 = x + b 1 b 1 = x 1 + 1 + x. Now distinguish the following three cases cf. Figure.). a) x 1 0: Then it must hold that x 1 = x, which leads to x = 3. Thus, in this case.) holds as well. b) 1 < x 1 < 0: 3

Two-dimensional lattices From x 1 = x + b 1 b 1 it follows that x x 1 1 = x and thus x x 1 = 3. Hence, x 1 = x 1 + x = 3 holds. If x 1 < 1, then it holds that x + b 1 b 1 = 3 + x 1 x = 3 = x 1. If x 1 1, then it is x b 1 1 = x 1 + x = 3 = x 1. c) x 1 1: Then it follows that x + 1 = x, which implies x = 1. Due to x 1 1, it moreover holds that x + b 1 b 1 = x 1 + + 1 x 1 + 1 = x 1. 4. z 1, z ) =, 1): In this case, it will be shown that the vectors ±b 1 b ) are not Voronoirelevant because for every x R with x 1 = x b 1 + b 1 there is some v {b 1, b 1 b, 3b 1 b } such that x v 1 x 1 holds. This can be seen completely analogously to the case z 1, z ) = 1, 1) above by exchanging the roles of x 1 and x, and using.3). 5. z 1, z ) = 1, 0): This final case will show that ±b 1 are Voronoi-relevant. For this, let x := 1b 1 to get x 1 = 1 = x b 1 1, and assume for contradiction that there is some v Lb 1, b ) \ {0, b 1 } with x v 1 1. Let z 1, z Z such that v = z 1 b 1 + z b. From x v 1 = 1 z 1 + 1 z 1 3z it follows that 1 z 1 1, which shows z1 {0, 1}. Thus it needs to hold that 1 3z 1 or 1 3z 1, but both cases imply z = 0 and v {0, b 1 }, which is a contradiction. This lemma as well as the above Voronoi cell lead to the observation that the Voronoi-relevant vectors are in general not sufficient to determine the Voronoi cell of the origin of a two-dimensional lattice completely when a non-strictly convex norm is used. This is expressed by the following corollary, and leads to the conclusion that all generalized Voronoi-relevant vectors need to be considered under the usage of a non-strictly convex norm. The relation between the Voronoi cell of the origin and the generalized) Voronoi-relevant vectors will be examined more formally in Section 4.3. Corollary.7 It holds that 7, 7 8 8 )T / VLb 1, b ), 1 ), although ) T 7 ) T 8, 7 7 v > 8 8, 7 8 1 holds for all v Lb 1, b ) that are Voronoi-relevant with respect to 1. 1 4

. Non-strictly convex norms Proof. 7, 7 8 8 )T / VLb 1, b ), 1 ) is a direct consequence from Proposition.1. Simple calculation shows 7, 7 8 8 )T b 1 1 = = 7, 7 8 8 )T + b 1 1 as well as 7, 7 8 8 )T 1 = 7, and the statement follows by Lemma.6. 4 Unfortunately, for generalized Voronoi-relevant vectors in two-dimensional lattices under non-strictly convex norms one cannot find an analogous result to Voronoi-relevant vectors in two-dimensional lattices under strictly convex norms. In fact, the number of generalized Voronoi-relevant vectors is not upper bounded by a constant in this setting. Theorem.8 For every m N there is a lattice Λ m R of rank two with at least ) m + 1 generalized Voronoi-relevant vectors with respect to 1. Proof. Let Λ m := Lb m,1, b m, ), where b m,1 = 1, 1) T and b m, = 0, m) T. Furthermore, consider x := 1b m,. The rest of this proof shows that for every v Λ m it holds that x v 1 m, and that there are exactly m + lattice vectors fulfilling the equality x v 1 = m. Hence, consider additionally v = z 1 b m,1 + z b m, with z 1, z Z as well as m x v 1 = z 1 + z1 + m z ) 1. This implies both z1 m and z 1 + m z ) 1 m z 1. With this, distinguish the following four cases. 1. z 1 0 and z 1: In this case, z 1 + m z 1 ) m z 1 holds, which implies m z 1 + mz mz m and thus z = 1 as well as z 1 = 0 follow. Therefore, x v 1 = m holds.. z 1 0 and z 0: Assume for contradiction that z 1 +m z ) 1 > 0. Then z1 > m 1 z ) m holds, which contradicts z 1 m. Hence, it holds that z 1 + m z ) 1 0, which leads to z1 + m 1 z ) m z 1. Thus, z 0 follows, yielding z = 0. For all z 1 [ ] 0, m it now holds that x v 1 = m. 3. z 1 < 0 and z 0: In this case, z 1 +m 1 z ) m +z 1 holds, which leads to the contradiction 0 < z 1 z 1 mz 0. Therefore, this case cannot occur. 4. z 1 < 0 and z 1: Assume for contradiction that z 1 +m z 1 ) < 0. Then z1 < m 1 z ) m holds, which contradicts z 1 m. Hence, it holds that z 1 + m z 1 ) 0, which yields z1 + m z 1 ) m + z 1. Thus, z 1 follows, leading to z = 1. For all z 1 [ m, 0) it now holds that x v 1 = m. 5

Two-dimensional lattices Summing up, these cases show that x v 1 = m holds if and only if z = 0 and z 1 [ ] 0, m or z = 1 and z 1 [ m, 0], and that x v 1 m holds for every v Λ m. Therefore, z 1 b m,1 + z b m, is a generalized Voronoi-relevant vector if z = 0, z 1 ] 0, m or z = 0, z 1 [ m, 0) or z = 1, z 1 [ m, 0] or z = 1, z 1 [ ] 0, m. In total, these are m ) + 1 generalized Voronoi-relevant vectors. In contrast to that, Lemma.9 and its proof also hold for generalized Voronoirelevant vectors instead of Voronoi-relevant vectors, which shows that every lattice of rank two has at most eight generalized Voronoi-relevant vectors with respect to an arbitrary strictly convex norm, but as shown in the above theorem the number of generalized Voronoi-relevant vectors for lattices of rank two is in general not bounded from above by a constant when working with non-strictly convex norms. 6

3 Higher-dimensional lattices Now one knows that the upper bound n 1) for the number of Voronoi-relevant vectors holds for the Euclidean norm in every dimension n, and that it further holds in the case n = for every strictly convex norm. Unfortunately, this is not true for arbitrary strictly convex norms in higher dimensions. Even worse, there is no upper bound at all which only depends on the lattice dimension. This is shown in the next section. An upper bound that also depends on other lattice properties and not only on the dimension is shown in Section 3.. This section also points out the consequences for the algorithm by Micciancio and Voulgaris [13] arising from the result that no upper bound for the number of Voronoi-relevant vectors can only depend on the lattice dimension when a general p-norm for p N, p 3 is considered. 3.1 Generalizations In the following, a family of three-dimensional lattices of rank three will be constructed such that their number of Voronoi-relevant vectors with respect to the 3-norm 3 is not bounded from above by a constant. The idea is to use a lattice of the form Le 1, e, Me 3 ), where e 1, e, e 3 ) denotes the standard basis of R 3 and M N is chosen sufficiently large, and to apply some rotations to this lattice. These rotations will depend on a parameter m N such that every lattice in the family is rotated differently. The basis vectors of the rotated lattices will be denoted by b m,1, b m, and b m,3 and will coincide with the rotated versions of e 1, e and Me 3, respectively. The intuition is to rotate Le 1, e, Me 3 ) such that the line between 0 and b m,1 + mb m, lies in an edge of a scaled and translated unit ball of the 3-norm when intersecting the plane spanned by 0, b m,1 and b m, with the ball. Figure 3.1 shows the unit ball of the 3-norm with and without intersections with different planes. Let the x-, y- and z-axis denote the axes of the standard threedimensional coordinate system which are spanned by e 1, e and e 3, respectively. As seen in Figures 3.1c and 3.1d, the intersection of the ball with a plane which is orthogonal to the z-axis e.g., the plane spanned by 0, e 1 and e ) yields a scaled unit ball of the 3-norm in two dimensions. But when such a plane is rotated around the y-axis by 45, as in Figures 3.1e to 3.1h, it intersects the three-dimensional unit ball of the 3-norm at one of its edges. These kinds of intersections are roughly speaking as less circular as possible, and the closer the plane is to the edge, the less circular the intersection is. Due to this, the plane spanned by 7

3 Higher-dimensional lattices 0, b m,1 and b m, should be of the form of the plane in Figures 3.1g and 3.1h. Moreover, the line between 0 and b m,1 + mb m, should lie directly on the edge of a scaled and translated unit ball such that all other lattice points in the plane spanned by 0, b m,1 and b m, lie outside of the ball. This is illustrated in Figure 3. for the case m = 3. If M is now chosen large enough, every lattice point of the form z 1 b m,1 + z b m, + z 3 b m,3 with z 1, z, z 3 Z, z 3 0 will be sufficiently far away from the plane spanned by 0, b m,1 and b m, such that it will also lie outside of the ball. Then 0 and b m,1 + mb m, are the only lattice points in the ball, and if they in fact lie on the boundary of the ball, it follows that b m,1 + mb m, is a Voronoi-relevant vector, where the center of the ball serves as x in Definition 1.6 of Voronoi-relevant vectors. a) B 3,10). b) B 3,10). c) B 3,10) and plane orthogonal to z-axis. d) Figure 3.1c from perspective orthogonal to plane. e) B 3,10) and plane intersecting the ball at an edge. f) Figure 3.1e from perspective orthogonal to plane. g) B 3,10) and plane intersecting the ball at an edge. h) Figure 3.1g from perspective orthogonal to plane. Figure 3.1: B 3,10) intersecting different planes. With these figurative ideas at hand, the rotations of Le 1, e, Me 3 ) will now be described formally. These modifications of the standard lattice are also illustrated in Figures 3.3 to 3.6 for the case m = 3. First, Le 1, e, Me 3 ) is rotated around the z-axis until e 1 + me lies on the y-axis, because all edges of the unit ball are 8

3.1 Generalizations b m,1 + mb m, b m, 0 b m,1 Figure 3.: Plane spanned by 0, b m,1 and b m, intersects ball at an edge such that line between 0 and b m,1 + mb m, lies on the edge cf. Figure 3.1h). parallel to the x-, y- or z-axis. This rotation is realized by the matrix R z := 1 m 0 +1 m m 0. +1 0 0 1 m m +1 1 m +1 Secondly, the resulting lattice R z Le 1, e, Me 3 ) is rotated around the y-axis by 45 such that after the rotation the plane formerly spanned by 0, e 1 and e intersects translated unit balls of the 3-norm at one of their edges. The second rotation is given by the matrix 1 1 0 R y := 0 1 0. 1 1 0 The resulting lattice Λ m := R y R z Le 1, e, Me 3 ) is spanned by b m,1 := R y R z e 1, b m, := R y R z e and b m,3 := MR y R z e 3. Using an appropriate scaling and translation of the unit ball, the situation in Figure 3. can be reached. As already mentioned, this can be used to show that b m,1 + mb m, is Voronoi-relevant if M is sufficiently large. Actually, Λ m has considerably more Voronoi-relevant vectors: In the following it is shown that choosing M := 5 m 5 implies that λ i Λ m, 3 ) = b m,i 3 holds for i {1,, 3} and that b m,1 + mb m, as well as b m,1 + kb m, for all k N, k [, m] are Voronoi-relevant with respect to 3. For this, two easy lemmata are shown first. Lemma 3.1 Let C, D R with D C. The function f : R R 0, x C + x 3 + D x 3 9

3 Higher-dimensional lattices e 3 e1 + me e 0 e 1 Figure 3.3: Le 1, e, e 3 ). has a global minimum at D C with function value 1 4 D + C 3. Proof. Since f x) = 3 sgnc + x)c + x) 3 sgnd x)d x), f x) = 0 holds if and only if sgnc + x)c + x) = sgnd x)d x). Now distinguish the following cases. 1. sgnc + x) = sgnd x): Then f x) = 0 holds if and only if C + x) = D x), which is equivalent to C + x = D x since X X is bijective on R 0 and R <0, respectively. Hence, f x) = 0 if and only if x = D C.. sgnc + x) = sgnd x): Then f x) = 0 holds if and only if C + x) = D x), which is equivalent to C+x = 0 = D x and thus to D = x = C. Since this contradicts the assumption D C, there is no x R with f x) = 0 and sgnc + x) = sgnd x). Hence, f x) = 0 if and only if x = D C. Since f x) = 6 C + x + 6 D x is everywhere strictly positive, D C is a global minimum of f with f ) D C = 1 D + 4 C 3. The next lemma computes the distance between some v R 3 and the plane spanned by 0, e 1 and e after this plane is translated along the z-axis and rotated around the y-axis by 45. 30

3.1 Generalizations Me 3 e 1 + me e 0 e 1 Figure 3.4: Le 1, e, Me 3 ): Note that M is so large that this figure is not scaled properly. Lemma 3. Let C R, v = α, β, γ) T R 3 and 1 0 E C := R y R 0 + R 1 + C 0 0 If v / E C, then v E C 3 3 = v R y α γ β C 3 3 0 0 1. = 1 4 C α γ 3. 31

3 Higher-dimensional lattices Me 3 R z e 1 + me ) R z e 0 R z e 1 Figure 3.5: R z Le 1, e, Me 3 ). Proof. First note that v E C is equivalent to α β γ = 1 C + A) B 1 C A) for some A, B R, which holds if and only if C α = C + γ. Hence under the assumptions v / E C, Lemma 3.1 can be applied as follows: v E C 3 C 3 = min A,B R + A 3 α + B β 3 + C A ) 3 γ = min B B R β 3 + 1 3 C min α + A 3 + C ) γ A 3 A R 3

3.1 Generalizations b m,3 b m,1 + mb m, b m, 0 b m,1 Figure 3.6: Λ m = R y R z Le 1, e, Me 3 ). = 1 1 3 4 C α γ 3 = 1 4 C α γ 3. Moreover, it follows from Lemma 3.1 that the minima in the above equality are reached for B = β and A = α γ. With this, it will be shown that the lengths of the basis vectors of Λ m coincide with the successive minima and that b m,1 + mb m, is Voronoi-relevant in Λ m. This yields for general strictly convex norms that the coefficients of Voronoi-relevant vectors, when they are represented in a successive minima basis, are not bounded from above by a bound that depends only on the lattice dimension. Note that this shows that Lemma.9 cannot be generalized to ranks higher than two. 33

3 Higher-dimensional lattices Proposition 3.3 For every m N, m and every i {1,, 3} it holds that λ i Λ m, 3 ) = b m,i 3. Proof. This statement follows from the following four intermediate steps: 1. b m,1 3 < b m, 3 : First note that b m,1 = R y R z b m, = R y R z 1 0 0 0 1 0 = = m m +1 1 m m +1 m +1 1 m +1 m m +1 1 m +1 and. Hence, b m,1 3 3 = m3 + m +1 3 and b m, 3 3 = m 3 +1 m +1 3. Since m, it holds that m 3 > 1 and thus m 3 +1 > m 3 +, which implies b m, 3 > b m,1 3.. b m, 3 < z 1 b m,1 + z b m, + z 3 b m,3 3 for all z 1, z, z 3 Z, z 3 0: For all z 1, z, z 3 Z with z 3 0 it holds that z 1 b m,1 + z b m, + z 3 b m,3 E Mz3 using the notation in Lemma 3.. Thus by the same lemma and the choice M = 5 m 5, z 1 b m,1 + z b m, + z 3 b m,3 3 3 0 E Mz3 3 3 = 1 4 Mz 3 3 50m 15. 3.1) The desired inequality follows since b m, 3 3 < m 3 + 1 < + 1)m 3. 3. b m, 3 < v 3 for all v Λ m \ {0, b m,1, b m,1, b m,, b m, }: By the above estimate, only v Λ m of the form v = z 1 b m,1 + z b m, for z 1, z Z are left to be considered. Assume that for such a v it holds that v 3 b m, 3. Since it is v 3 3 = 1 m +1 3 1 v = z 1 m z m +1 z m+z 1 m +1 z 1m z m +1, z 1 m z 3 + z m + z 1 3 ). Hence v 3 b m, 3 34

3.1 Generalizations implies that 1 z 1 m z 3 + z m + z 1 3 m 3 + 1. 3.) In particular it holds that z m + z 1 m due to z 1, z Z. If z, z m + z 1 m implies z 1 m. Thus it follows z 1 m z 3 m + ) 3 > m 6 > + 1)m 3 > m 3 + 1, which contradicts 3.). For z, it is z 1 m, which yields the same estimate with the same contradiction. If z = 1, m + z 1 m implies z 1 0. Thus it holds that z 3 1m 3 < z 1 m 1 3 m 3 + 1 < + 1)m 3, which leads to z 1 { 1, 0}. Since 3 1) < m, it follows 3 1)m + < m 3 + 3 + 1)m, which implies m 3 + 1 < m + 1) 3 + m 1) 3 = m 1 3 + m 1 3. This contradicts 3.) for the case z 1 = 1, yielding z 1 = 0. For z = 1, it is z 1 0, and z 3 1m 3 < z 1 m + 1 3 leads as above to z 1 {0, 1}, but the case z 1 = 1 contradicts 3.) using the same estimate as above. Hence it follows for z = 1 that z 1 = 0, i.e., v { b m,, b m, }. If z = 0, 3.) implies z 1 3 m 3 m 3 + 1 < + 1)m 3. Hence it holds that z 1 { 1, 0, 1}, i.e., v { b m,1, 0, b m,1 }. 4. b m,3 3 z 1 b m,1 + z b m, + z 3 b m,3 3 for all z 1, z, z 3 Z, z 3 0: Since b m,3 = MR y R z 0, 0, 1) T = 5m 5, 0, 5m 5 ) T, b m,3 3 3 = 50m 15. Together with 3.1), this shows the desired inequality. Proposition 3.4 For every m N, m it holds that b m,1 + mb m, is Voronoirelevant in Λ m with respect to 3. Proof. Define x := m 5 m +1 m 5. Then x 3 3 = m 15 + 1 8 m + 1 3 = b m,1 + mb m, x 3 3, and the following two estimates show that x 3 < x v 3 for all v Λ m \ {0, b m,1 + mb m, }, which completes the proof. 1. x 3 < z 1 b m,1 + z b m, + z 3 b m,3 x 3 for all z 1, z, z 3 Z, z 3 0: By Lemma 3. it follows for all z 1, z, z 3 Z with z 3 0 that z 1 b m,1 + z b m, + z 3 b m,3 x 3 3 x E Mz3 3 3 35

3 Higher-dimensional lattices = 1 4 Mz 3 m 5 3 = 1 4 m15 10z 3 3 1 4 m15 8 3 = 18m 15. The desired inequality follows since x 3 3 < m 15 + 1 8 m ) 3 < 3m 15.. x 3 < x v 3 for all v Λ m \ {0, b m,1 + mb m, }: By the first estimate, only v Λ m of the form v = z 1 b m,1 + z b m, for z 1, z Z have to be considered. Assume that for such a v it holds that x v 3 x 3. Define Z := z 1m z m +1. Then v = Z z m+z 1 m +1 Z and x v 3 3 m 5 Z 3 + m 5 + Z 3. If m 5 Z < 0, then Z > m 5 and x v 3 3 m 5 + Z ) 3 > 8m 15. If m 5 + Z < 0, then Z < m 5 and x v 3 3 m 5 Z ) 3 > 8m 15. Both cases lead to the contradiction 8m 15 < x v 3 3 x 3 3 < 3m 15. Hence it can be assumed that x v 3 3 m 5 Z ) 3 + m 5 + Z ) 3 = m 15 + 3Z m 5. Using x 3 3 = m 15 + 1 8 m + 1 3 implies the inequality 3Z m 5 8 1 m + 1 3. This leads to 576Z 4 m 10 m + 1) 3 and thus to Z 4 m +1) 3. By definition of Z, it holds that z 576m 10 1 m z 4 m +1) 5. 576m 10 Since m + 1) 5 = m 10 + 5m 8 + 10m 6 + 10m 4 + 5m + 1 < 3m 10 < 576m 10, it follows that z 1 m z < 1. Thus z 1 m z = 0 due to z 1, z Z. With this, v = 0, z 1 m + 1, 0) T and x v 3 3 = m 15 + m +1 3 z 8 1 1 3. From x v 3 x 3 it directly follows that z 1 1 1, which is equivalent to z 1 {0, 1}, i.e., v {0, b m,1 + mb m, }. The most important statement of this thesis will now be formulated: The above defined lattices Λ m do not only have b m,1 + mb m, as a Voronoi-relevant vector but also b m,1 + kb m, for all k N, k [, m]. Hence, it holds for the 3-norm that every Λ m has Ω m) Voronoi-relevant vectors, and that for every k N one can find a lattice that has at least k Voronoi-relevant vectors. This will be formalized in Corollary 3.6. 36

3.1 Generalizations Theorem 3.5 For every k N, k and all m N, m k b m,1 + kb m, is Voronoi-relevant in Λ m with respect to 3. it holds that Proof. For k, m N with m k define ) k x m,k := 1 b + 1 m 4 3 m,1 + kb m, ) + 0 Since 1 + 1 1 4 3 3, it holds that that k 4 + 1 3 x m,k 3 m k k 3 = 3 m + 1 + 4 + 1 3 ) 3 km + 1 + m + 1 k = 4 + 1 ) 3 k m 3 + 6 3 = b m,1 + kb m, x m,k 3 3. ) m = m k 3 + k m +1 + 1 4 3 km+1 m +1 k m 3 + k + 1 m +1 4 3 ) m ) m. ) k + 1 m 1 m k 4 3 3 3, which implies m +1 ) 3 m) + k m 3 m + 1 + k 4 + 1 3 ) ) 3 m 4 + 1 ) m k km + 1 m 3 3 + m + 1) m + 1 To complete this proof, the following three estimates show x m,k 3 < x m,k v 3 for all v Λ m \ {0, b m,1 + kb m, } and all m k. ) 3 1. x m,k 3 < z 1 b m,1 + z b m, + z 3 b m,3 x m,k 3 for all all m k and all z 1, z, z 3 Z, z 3 0: On the one hand, it is z 1 b m,1 +z b m, +z 3 b m,3 E Mz3 for all z 1, z, z 3 Z. On the other hand, x m,k / E Mz3, because otherwise Mz 3 = ) k m + 1 4 3 would hold, which is equivalent to 60m 4 z 3 = 3k +4 and hence implies z 3 > 0 leading to z 3 1 and the contradiction 60m 4 z 3 60k 4 > 4k 3k + 4 = 60m 4 z 3. By Lemma 3. it follows that z 1 b m,1 + z b m, + z 3 b m,3 x m,k 3 3 x m,k E Mz3 3 3 = 1 k 4 Mz3 4 + 1 ) 3 m 3 = 1 4 1000m15 z 3 1 k 5m 4 4 + 1 ) 3. 3 ) 1 k The prerequisite m k yields + 1 0, 1 5m 4 4 3 60]. This shows for 37

3 Higher-dimensional lattices z 3 Z\{0} that the inequality z 3 1 and that 5m 4 z 1 b m,1 + z b m, + z 3 b m,3 x m,k 3 3 50m 15 1 1 5m 4 ) k + 1 0 is equivalent to z 4 3 3 1, 50m 15 59 60 The desired inequality follows since m 4 + 1 3 m implies k 4 + 1 3 ) 3 > 00m 15. x m,k 3 3 m 9 + 6m 3 m m 3 + m + 1) + 1 m + 1 m 9 + 3 4 m3 + m 3 < 4m 15. ) 3 )) 3. x m,k 3 < z 1 b m,1 + z b m, x m,k 3 for all m k and all z 1, z ) Z \ {0, 1} {1,..., k 1}) {0, 0), 1, k)}): For v := z 1 b m,1 + z b m, with z 1, z Z it holds that If v x m,k 3 k 3 = 4 + 1 ) m z 3 1 1)m z k) 3 3 m + 1 k + 4 + 1 ) m + z 3 1 1)m z k) 3 3 m + 1 3 + z k)m + z 1 1) m + 1. ) m z 1 1)m z k) k 3 < 0 or + 1 m + z 1 1)m z k) m +1 4 3 3 m +1 ) k + 1 4 3 then v x m,k 3 3 > 8 k < 0, 4 + 1 3) 3 m 3. Assume in this case for contradiction that v x m,k 3 x m,k 3. This implies that 6 ) ) k 6 + 1 m k + ) km+1 3. m 4 3 3 m +1 m +1 Dividing by 6 k tiplying by 8 m + 1 3 k yields 8 + 1 4 3 < m k) m + 1 + km+1)3 m3k +4). k 4 + 1 3 3 + 1 4 3) m 3 < ) m and mul- ) m m + 1 3 Using m < m + 1 < m and km+1) < m3k +4) leads to 8 k4 16 m5 < m 3 +km+1) < m 3 +4k m. Hence 16m 3 k 4 m 3 < m + 8k < 4m + 8m follows, leading to 38

0 > 4m m 1 = 4 m 1+ 5 4 since 1 5 that 4 < 1+ 5 ) k 4 + 1 3 ) m 1 5 4 3.1 Generalizations ), but this is a contradiction < 1 and m. This shows v x 4 ) m,k 3 > x m,k 3 in case m z 1 1)m z k) k 3 < 0 or + 1 m + z 1 1)m z k) m +1 4 3 3 < 0. m +1 Hence it can be assumed in the following that v x m,k 3 3 = = k 4 + 1 ) ) m z 3 1 1)m z k) 3 3 m + 1 k + 4 + 1 ) ) m + z 3 1 1)m z k) 3 3 m + 1 3 + z k)m + z 1 1) m + 1 3 ) z k)m + z 1 1) k 3 + m 3 m + 1 ) m k + 6 4 + 1 3 4 + 1 3 z 1 1)m z k) 3 m + 1 ). 3.3) Thus, v x m,k 3 > x m,k 3 is equivalent to k 6 4 + 1 ) ) z 1 1)m z k) m 3 3 + m + 1 k > 6 4 + 1 ) m 3 and consequently to m k 3 m + 1) + km + 1 m + 1 z k)m + z 1 1) m + 1 fm, k, z 1, z ) k := 6 4 + 1 ) m m 3 + 1 z 1 1)m z k)) m k) ) + z k)m + z 1 1) 3 km + 1) 3 > 0. ) 3 3 With this, the following six cases can be distinguished: a) z 1 : 39

3 Higher-dimensional lattices If z 1 1)m z k) m k, then z k m + k and k fm, k, z 1, z ) 6 4 + 1 ) m m 3 + 1m k) + m + k)m + 3) 3 km + 1) 3 k 1 4 + 1 ) m 4 + km + 1) + m + 1)) 3 3 km + 1) 3 = 3k m 4 4m 4 + 6km + 1) m + 1) + 1km + 1)m + 1) + 8m + 1) 3 3k m 4 4m 4 + 6k m 4 + 1km 5 + 8m 6 > 0. 3.4) If z k)m + z 1 1) km + 1, then z k k < k holds, m implying z k 1 and z k k, which shows k fm, k, z 1, z ) 6 4 + 1 ) m m 3 + 1 3m k + ) m k) ) km + 1) 3 6 k 4 m m k) + m + 1)) m k) ) km + 1) 3 = 6k m m k)m + 1) + m + 1) ) km + 1) 3 1k m 4 + 1k m 3 + 3k m 7k 3 m 3 3km 1 5k m 4 + 1k m 3 + 8km > 0. 3.5) If z 1 1)m z k) > m k and z k)m + z 1 1) > km + 1, then it follows directly from the definition of f that fm, k, z 1, z ) > 0. Thus it holds for every z 1 that fm, k, z 1, z ) > 0 and this shows v x m,k 3 > x m,k 3. b) z 1 1: If z 1 1)m + z k) m k, then z k m k and the same estimate as in 3.4) holds. If z k)m z 1 1) km + 1, then z k) k m < k holds, implying z 1 and z k k, which leads to the same estimate as in 3.5). If z 1 1)m+z k) > m k and z k)m z 1 1) > km+1, 40

3.1 Generalizations then it follows directly from the definition of f that fm, k, z 1, z ) > 0. c) z 1 = 0 and z k: It holds that z k k and this implies k fm, k, z 1, z ) 6 4 + 1 ) m m 3 + 1 m + k) m k) ) + km 1) 3 km + 1) 3 3.6) 6k 3 m 3 6k m > 0. d) z 1 = 0 and z 1: In this case, it is z = a for some a N. This yields k fm, k, z 1, z ) = 6 4 + 1 ) m m 3 + 1 m + k + a) m k) ) + a + k)m + 1) 3 km + 1) 3 = m 3 6ak + 1a k + 8a 3 ) + m m + 1 6ak 8a) + m 1ak + 1a ) + m m + 16ak 3 + 8ak + 6a k + 8a ) + m 6a > m a m6k + 1ak + 8a ) ) m + 16k + 8). 3.7) Because of m k, it is obvious that m 4k6k + 8) + 144k ) = m 144k 3 + 19k + 144k ) 36k 4 + 96k + 64 = 6k + 8), and this is equivalent to m 6k + 8) + 1k) m + 1)6k + 8). Hence, m6k + 1ak + 8a ) m6k + 1k + 8) m + 16k + 8) and fm, k, z 1, z ) > 0 follow. e) z 1 = 1 and z 0: It holds that z k k, yielding the same estimate as in 3.6). f) z 1 = 1 and z k + 1: In this case, it is z = k + a for some a N. This yields the same estimate as in 3.7) and thus fm, k, z 1, z ) > 0. 3. x m,k 3 < z 1 b m,1 + z b m, x m,k 3 for all m k and all z 1, z ) {0, 1} {1,..., k 1}: For z 1, z ) {0, 1} {1,..., k 1} ) it holds that z 1 1)m z k) k m k) < m + k < 4 + 1 m < ) 3 k + 1 m m 4 3 4 3 + 1. Moreover, z 1 1)m z k) m k ) > ) 3 k m m + 1 4 + 1 3 41

3 Higher-dimensional lattices follows. These two inequalities show that z 1 b m,1 + z b m, x m,k 3 can be computed as in 3.3), and that z 1 b m,1 + z b m, x m,k 3 > x m,k 3 is equivalent to fm, k, z 1, z ) > 0. In addition, z k)m + z 1 1) < 0 is equivalent to z < 1 k z 1 ) 1 m, such that the following four cases need to be distinguished to show fm, k, z 1, z ) > 0: a) z 1 = 1 and 1 k 1 m) z < k: In this case, it follows with m < m + 1 < m + 1 that k fm, k, z 1, z ) = 6 4 + 1 ) m m 3 + 14z 4mz + 4km 4kz ) + 8m 3 z 3 k 3 m 3 + 1 1km 3 z + 6k m 3 z + 1m z + 6mz + 3k m 3km 1km z k 3 m 3 3k m 3km 1 6k + 8)m z + km) 6k + 8)mm + 1)m + k)z + 8m 3 z 3 + 1m 1km 3 )z + 6k m 3 1km + 6m)z k 3 m 3 6km = 8m 3 z 3 + 1km 3 + 6k m + 0m )z + 8m 3 6k 3 m 6k m 0km 8m )z + 6k 3 m 8km + 6m)z + 4k 3 m 3 + 8km 3 6km =: ϕ m,k) 1 z ) =: ϕ 1 z ). Hence, it is enough to show that ϕ m,k) 1 z ) > 0. For this consider x R with 0 < x k 1 and m k to get the estimate ϕ 1x) = 4m 3 x + 4km 3 + 1k m + 40m )x 8m 3 6k 3 m 6k m 0km 8m 6k 3 m 8km + 6m 4m 3 + 1k m + 40m )x 8m 3 6k 3 m 6k m 0km 8m 6k 3 m 8km + 6m 1k m + 40m )x 8m 3 6k 3 m 6k m 0km 8m 6k 3 m 8km + 6m 1k m x + 0km 8m 3 6k 3 m 6k m 8m 6k 3 m 8km + 6m 1k m x 4km 8m 3 6k m 8m 6k 3 m 8km + 6m 1k m x 4km 8m 3 6k m 8m 6k 3 m 10m < 0. 4

3.1 Generalizations This shows that ϕ m,k) 1 is strictly decreasing on 0, k 1] and thus ϕ m,k) 1 z ) ϕ m,k) 1 k 1) = 8k 1) 3 1kk 1) 8k 1) + 4k 3 + 8k)m 3 + 6k k 1) + 0k 1) 6k 3 k 1))m + 6k k 1) 0kk 1) 8k 1))m + 6k 3 k 1) 8kk 1) + 6k 1) 6k)m = 1km 3 + 1k 3 + 1k 8k + 8)m + 6k 4 + 6k 3 8k + 8k 6)m. Thus it is enough to show ψ 1 m) := ψ k) 1 m) := 1 Since it holds for y R with y k that ψ 1y) = 1ky 6k 3 + 6k 14k + 14 1 k 1) > 0. m ϕm,k) 6k 3 + 6k 14k + 14 6k + 10k + 14 > 0, ψ k) 1 is strictly increasing on [k, ) and thus ψ k) 1 m) ψ k) 1 k ) = 3k 4 11k 3 + 10k + 4k 3 = 3k k 5 ) k ) + 4k 3 4k 3 > 0. 3 b) z 1 = 1 and 1 z < 1 k 1 m) : Now, f can be estimated as k fm, k, z 1, z ) = 6 4 + 1 ) m m 3 + 14z 4mz + 4km 4kz ) 8m 3 z 3 + k 3 m 3 1 + 1km 3 z 6k m 3 z 1m z 6mz 3k m + 3km + 1km z k 3 m 3 3k m 3km 1 6k + 8)m z + km) 6k + 8)mm + 1)m + k)z 8m 3 z 3 + 1km 3 1m )z + 6k m 3 + 1km 6m)z 6k m = 8m 3 z 3 + 1km 3 + 6k m 4m )z + 1k m 3 8m 3 6k 3 m 6k m + 4km )z + 8m 6k 3 m 8km 6m)z + 6k 3 m 3 + 8km 3 6k m =: ϕ m,k) z ) =: ϕ z ). 43

3 Higher-dimensional lattices Therefore it is enough to show that ϕ m,k) z ) > 0. For this consider x R with 0 < x k to get the estimate ϕ x) = 4m 3 x + 4km 3 + 1k m 8m )x 1k m 3 8m 3 6k 3 m 6k m + 4km 8m 6k 3 m 8km 6m 4m 3 x 8m x 8m 3 6k m + 4km 8m 6k 3 m 8km 6m 4m 3 x 8m x 8m 3 8km 8m 6k 3 m 8km 6m < 0. This shows that ϕ m,k) is strictly decreasing on 0, ] k and thus ) ϕ m,k) z ) ϕ m,k) k = k 3 + 3k 3 6k 3 4k + 6k 3 + 8k)m 3 ) 3 + k4 k 3k 4 3k 3 4k + k 6k With + 3k 4 4k 3k)m = k 3 + 4k)m 3 + 3 ) k4 3k 3 5k 4k + 3k 4 4k 3k)m. ψ m) := ψ k) m) := 4k 3 + 8k)m + 3k 4 6k 3 10k 8k)m 6k 4 8k 6k it holds that ϕ m,k) k ) = m with y k it is ψk) m m m). Moreover, for every y R ψ y) = 8k 3 + 16k)y 3k 4 6k 3 10k 8k 8k 5 3k 4 + 10k 3 10k 8k 13k 4 + 10k 8k > 0, which implies that ψ k) is strictly increasing on [k, ). Thus, ψ k) m) ψ k) k ) = 4k 7 3k 6 + k 5 16k 4 8k 3 8k 6k 5k 6 1k 4 8k 3 8k 6k 8k 4 8k 3 8k 6k 44

3.1 Generalizations 8k 3 8k 6k 8k 6k 10k holds, yielding ϕ m,k) z ) ϕ m,k) c) z 1 = 0 and 1 z < 1 k + 1 m) : In this case, it is k fm, k, z 1, z ) = 6 4 + 1 3 k ) = m ψk) m) 5k 3 > 0. ) m m + 14z + 4m 4k)z 8m 3 z 3 + k 3 m 3 + 1 + 1km 3 z 6k m 3 z + 1m z 6mz + 3k m + 3km 1km z k 3 m 3 3k m 3km 1 6k + 8)m z + m)z 6k + 8)mm + 1)kz 8m 3 z 3 + 1m + 1km 3 )z + 6k m 3 1km 6m)z = 8m 3 z 3 + 1km 3 + 6k m + 0m )z + 8m 3 6k 3 m 0km 6k 3 m 8km 6m)z. Hence, it is enough to show that ϕ 3 z ) := ϕ m,k) 3 z ) : = 8m 3 z + 1km 3 + 6k m + 0m )z + 8m 3 > 0. 6k 3 m 0km 6k 3 m 8km 6m For this consider x R with 0 < x k+1 to get the estimate ϕ 3x) = 16m 3 x + 1km 3 + 6k m + 0m 4km 3 8m 3 + 6k m + 0m 6k m + 0m > 0. This shows that ϕ m,k) 3 is strictly increasing on ] 0, k+1 and thus ϕ m,k) 3 z ) ϕ m,k) 3 1) = 1km 3 + 6k 3 + 6k 0k + 0)m + 6k 3 8k 6)m ϕ m,k) 1 k 1) > 0. d) z 1 = 0 and 1 k + 1 m) z < k: Now, f can be estimated as k fm, k, z 1, z ) = 6 4 + 1 ) m m 3 + 14z + 4m 4k)z + 8m 3 z 3 k 3 m 3 1 1km 3 z + 6k m 3 z 45

3 Higher-dimensional lattices 1m z + 6mz 3k m 3km + 1km z k 3 m 3 3k m 3km 1 6k + 8)m z + m)z 6k + 8)mm + 1)kz + 8m 3 z 3 + 1m 1km 3 )z + 6k m 3 + 1km + 6m)z k 3 m 3 6k m 6km = 8m 3 z 3 + 1km 3 + 6k m 4m )z + 1k m 3 + 8m 3 6k 3 m + 4km 6k 3 m)z + 8km + 6m)z k 3 m 3 6k m 6km =: ϕ m,k) 4 z ) =: ϕ 4 z ). Therefore it is enough to show that ϕ m,k) 4 z ) > 0. For this consider x R with k x k 1 and m k 4 to get the estimate ϕ 4x) = 4m 3 x + 4km 3 + 1k m 8m )x + 1k m 3 + 8m 3 6k 3 m + 4km 6k 3 m 8km + 6m 1km 3 + 1k m 8m )x + 1k m 3 + 8m 3 6k 3 m + 4km 6k 3 m 8km + 6m 1km 3 + 40m )x + 1k m 3 + 8m 3 6k 3 m + 4km 6k 3 m 8km + 6m 40m x + 1km 3 + 8m 3 6k 3 m + 4km 6k 3 m 8km + 6m 40m x + 6k 3 m + 8m 3 + 4km 6k 3 m 8km + 6m 40m x + 18k 3 m + 8m 3 + 4km 8km + 6m 40m x + 18k 3 m + 8m 3 + 8km + 6m > 0. This shows that ϕ m,k) 4 is strictly increasing on [ k, k 1] and thus ϕ m,k) 4 z ) ϕ m,k) 4 ) k = k 3 3k 3 + 6k 3 + 4k k 3 )m 3 ) 3 + k4 k 3k 4 + k 6k + 3k 4 4k + 3k 6k)m = k 3 + 4k)m 3 + 3 ) k4 5k m + 3k 4 4k 3k)m m 46

ϕ m,k) ) k > 0. 3. Consequences and comparisons All these different cases together show that x m,k 3 < x m,k v 3 holds for all v Λ m \ {0, b m,1 + kb m, } and all m k. Combined with x m,k 3 = b m,1 + kb m, x m,k 3, this implies that b m,1 + kb m, is Voronoi-relevant in Λ m with respect to 3 if m k. Corollary 3.6 1. For every m N, m it holds that Λ m has at least m Voronoi-relevant vectors with respect to 3.. For every k N, k 3 it holds that Λ k vectors with respect to 3. has at least k Voronoi-relevant Proof. Since the second statement is a direct consequence of the first one, it is sufficient to show the first part of this corollary. For this, let m N with m. For all k N with k m it holds by Theorem 3.5 that b m,1 + kb m, is Voronoi-relevant in Λ m with respect to 3. Then b m,1 kb m, is also Voronoi-relevant for all k m. In addition, Proposition 3.4 gives that ±b m,1 + mb m, ) are Voronoi-relevant vectors, and the first statement follows. 3. Consequences and comparisons In the following, an upper bound for the number of generalized) Voronoi-relevant vectors in an arbitrary lattice with respect to an arbitrary norm is shown. This bound depends on the lattice dimension as well as the ratio of the covering radius to the length of a shortest non-zero lattice vector. Proposition 3.7 Let : R n R 0 be a norm and let Λ R n be a lattice. n Then Λ has at most 1 + 4 µλ, ) λ 1 Λ, )) generalized Voronoi-relevant vectors with respect to. Proof. From Lemma.17 it follows that RΛ, ) := {v Λ v generalized Voronoi-relevant w.r.t. } B,µΛ, ) 0). Since for every two lattice vectors v, w Λ with v w it holds that v w λ 1 Λ, ), it follows that B λ, 1 Λ, ) v) B λ, 1 Λ, ) w) =. Moreover, for every v RΛ, ) and every x B λ, 1 Λ, ) v) it is x x v + v < λ 1 Λ, ) + µλ, ). Thus B λ, 1 Λ, ) v RΛ, ) v) B λ,µλ, )+ 1 Λ, ) 0) 47

3 Higher-dimensional lattices holds, which implies ) RΛ, ) vol n B λ, 1 Λ, ) 0) = ) vol n B λ, 1 Λ, ) v) v RΛ, ) = vol n B λ, 1 Λ, ) v) ) 1 + 4 µλ, ) λ1 Λ, ) λ 1 Λ, ) v RΛ, ) vol n B λ,µλ, )+ 1 Λ, ) 0) Since µλ, )+ λ 1Λ, ) =, it is B λ,µλ, )+ 1 Λ, ) 0) = ) 1 + 4 µλ, ) λ 1 B λ Λ, ), 1 Λ, ) 0). For any positive constant r R >0 and any measurable M R n, integration by substitution for multiple variables yields x ) vol n rm) = χ rmx) dx = χ M dx = r n r R n R n R n = r n χ M y) dy = r n vol n M). Thus it follows that R n ) RΛ, ) vol n B λ, 1 Λ, ) 0) which implies RΛ, ) χ M x r ) n µλ, ) 1 + 4 vol n λ 1 Λ, ) 1 + 4 µλ, ) λ 1 Λ, )) n. ). ) 1 r n dx B λ, 1 Λ, ) 0) Corollary 3.8 Let : R n R 0 be a norm and let Λ R n be a lattice. Then n Λ has at most 6 µλ, ) λ 1 Λ, )) generalized Voronoi-relevant vectors with respect to. Proof. Assume for contradiction that µλ, ) < λ 1Λ, ). Then exists some x spanλ) such that x B λ, 1 Λ, ) 0) \ B,µΛ, ) 0). Due to the definition of the covering radius, there exists some v Λ \ {0} with x v < x < λ 1Λ, ). This gives the contradiction λ 1 Λ, ) v v x + x < λ 1 Λ, ). Hence, µλ, ) λ 1Λ, ) must hold, which implies 1 µλ, ). Plugging λ 1 Λ, ) this into Proposition 3.7 yields the desired upper bound. ), 48

3. Consequences and comparisons As seen in the last section, there is no upper bound for the number of Voronoirelevant vectors with respect to the 3-norm that only depends on the lattice dimension. Nevertheless, at least the upper bounds from Proposition 3.7 and Corollary 3.8 hold. Unfortunately, this does not help for the algorithm by Micciancio and Voulgaris [13] since they rely on the fact that the number of Voronoi-relevant vectors with respect to the Euclidean norm is in On), which is not true for the 3-norm. Hence, this algorithm cannot be easily extended to general p-norms, not even when only strictly convex p-norms i.e., for p 1, ) are considered. The rest of this section will compare the upper bound in Proposition 3.7 with the number of generalized) Voronoi-relevant vectors in the lattices that have been constructed to show that no upper bound for the number of generalized) Voronoi-relevant vectors can only depend on the lattice dimension. First, the two-dimensional lattices of Theorem.8 will be considered which do not have a constant number of generalized Voronoi-relevant vectors with respect to a nonstrictly convex norm. Secondly, the three-dimensional lattices of Corollary 3.6 will be investigated which do not have a constant number of Voronoi-relevant vectors with respect to a strictly convex norm. For m N, let Λ ) m := Lb ) m,1, b ) m,), where b ) m,1 := 1, 1) T and b ) m, := 0, m) T. Since b ) m,1 1 =, it follows that λ 1 Λ ) m, 1 ). The following calculation shows that {, if m λ 1 Λ ) m, 1 ) = 1, if m = 1. 3.8) Consider z 1, z ) Z \{0, 0)} with > z 1 b ) m,1+z b ) m, 1 = z 1 + z 1 +mz. Then it is clear that z 1 1. If z 1 = 1 would hold, z 1 + mz = 0 would follow, leading to 1 = m z and thus to m = 1 and z = z 1. For z 1 = 0 it follows from z 0 that 1 = m z, which implies m = 1 and z = 1. Both cases together show 3.8). The proof of Theorem.8 shows in particular that for every v Λ ) m it holds 1 b) m, v 1 m. This yields that µλ) m, 1 ) m. An easy upper bound for the covering radius follows from the fact that for every r R there exists some z Z with r z 1 : For every y spanλ) m ) there exists some v Λ ) m such that y v 1 1 b) m,1 1 + 1 b) m, = m + 1. Hence, µλ) m, 1 ) Θm) and the upper bound in Proposition 3.7 or Corollary 3.8, respectively, is in Θm ), which is asymptotically larger than the number of generalized Voronoi-relevant vectors in Theorem.8. An even bigger asymptotical gap is obtained from the lattices of Corollary 3.6. For m N with m, let Λ 3) m := Lb 3) m,1, b 3) m,, b 3) m,3) with m 1 b 3) m +1 1 m,1 := m +1, b 3) m +1 5 m 5 m m, :=, b 3) m,3 := 0. 5 m 5 m +1 m m +1 1 m +1 49

3 Higher-dimensional lattices It was shown in Proposition 3.3 that λ 1 Λ 3) m, 3 ) = b 3) m,1 3 = 3 m 3 +. From m +1 3 1)m 3 1)8 it follows that m + 1 3 m 3 m 3 +, which shows λ 1 Λ 3) m, 3 ) 1. A lower bound for the first successive minimum can be derived as follows: It holds that 8 )m 3 + 7 > 8 )m 3 8 )m 6 m 3 mm + 1), which implies 8m 3 + ) m + 1) 3 m + 1 3 and λ 1 Λ 3) m, 3 ) 1. Moreover, the proof of Proposition 3.4 shows that for every v Λ 3) m it holds that x v 3 x 3 = m 3 15 + 8 1 m + 1 3, where x := m 5, m +1, m 5 ) T. This shows that µλ 3) m, 3 ) x 3 3 m 5. As above for Λ ) m, one can deduce that for every y spanλ 3) m ) there exists some v Λ 3) m such that y v 3 1 b3) m,1 3 + 1 b3) m, 3 + 1 b3) m,3 3 3 b3) m,3 3 = 3 3 50m 5, where the last inequality follows from Proposition 3.3. Hence, µλ 3) m, 3 ) Θm 5 ) and the upper bound in Proposition 3.7 or Corollary 3.8, respectively, is in Θm 15 ), whereas the lower bound from Corollary 3.6 is only in Θ m). In both families of lattices discussed above, the number of generalized) Voronoirelevant vectors grows with the ratio of the covering radius to the first successive minimum, but there might be still room for a better upper bound than the one given in Proposition 3.7. 50

4 General shape of bisectors, Voronoi cells and their facets As seen in Section., it is not immediately clear which lattice vectors completely determine the Voronoi cell of a given lattice when arbitrary norms are considered. This general question will be discussed in Section 4.3. Moreover, such an n- dimensional Voronoi cell is bounded by its n 1)-dimensional facets, and for understanding the complexity of a Voronoi cell it is important to know the number of these facets. In Section 4.4, it will be examined how many n 1)-dimensional facets an n-dimensional Voronoi cell has, and if these facets are connected or not. To comprehend how Voronoi cells and their facets look like, it is inevitable to investigate bisectors and their intersections, since these facets are subsets of bisectors. This is done in Section 4.. In particular, a proof idea relying on some conjecture for the following fundamental and hopefully true) statement is given: The bisector of a and b intersected with the bisector of b and c is homeomorphic to R n as long as a, b, c R n are non-collinear and a sufficiently nice norm is used. This statement is already known for the case n 3 [10], and is further motivated by Horváth s result that bisectors are homeomorphic to R n 1 under a strictly convex norm [6]. The next section gives properties and definitions that are needed for the above mentioned sufficiently nice norms. 4.1 Norms In this section, it will be shown that every symmetric convex body in R n defines a norm, where its closed unit ball is the given body itself, and that the closed unit ball of a given norm in R n is a symmetric convex body such that its corresponding norm coincides with the given norm. Based on this, smooth norms will be defined and alternative definitions for strict convexity will be given. An intermediate result will be that all norms in R n are continuous with respect to the Euclidean norm. Definition 4.1 K R n is a convex body with center point c K if K is compact and convex, and c lies in the interior of K. K is symmetric with respect to c) if for every x K it holds that c x K. In the subsequent sections, convex bodies will be translated and scaled. Definition 4. Let K R n be a convex body with center point c K. K R n is called a uniformly scaled copy of K, if r R >0 with K = rk c) + c exists. 51

4 General shape of bisectors, Voronoi cells and their facets K R n is called a translated copy of K, if t R n with K = K + t exists. Every convex body defines a convex distance function, which is a metric if and only if the body is symmetric. This is shown in Lemmata 4.4 and 4.5. Furthermore as stated in Proposition 4.6 a symmetric convex body defines a norm such that its closed unit ball coincides with the convex body. Definition 4.3 Let K R n be a convex body with center point c K. For p, q R n with p q, let x p,q R n denote the unique intersection point of the boundary of K c + p and the ray from p through q, i.e., x p,q {sq + 1 s)p s R 0 }. The convex distance function based on K and c is defined as d K,c : R n R n R 0, p, q) { q p x p,q p, if p q 0, if p = q. Lemma 4.4 Let K R n be a convex body with center point c K. The convex distance function d K,c satisfies the following properties: 1. For every p, q R n it is d K,c p, q) 0, and d K,c p, q) = 0 if and only if p = q.. For every p R n it holds that p lies the interior of K if and only if d K,c c, p) < 1, p lies on the boundary of K if and only if d K,c c, p) = 1, p is not in K if and only if d K,c c, p) > 1. 3. For every p, q, r R n it is d K,c p r, q r) = d K,c p, q). 4. For every p R n and every s R >0 it holds that d K,c 0, sp) = sd K,c 0, p). 5. For every p, q, r R n it is d K,c p, q) d K,c p, r) + d K,c r, q). Proof. The first four assertions follow directly from Definition 4.3. For this, one only needs to consider x p,q in the different cases: First, it holds for p q that x p,q p such that d K,c p, q) > 0 follows. Moreover, x p r,q r = x p,q r and x 0,sp = x 0,p. For assertion five, note that for every b R n, b 0 it holds that and thus b d K,c 0,b) d K,c 0, a, b R n \ {0} with µ := ) b d K,c 0, b) = 1 d K,c 0, b) d K,c0, b) = 1, lies on the boundary of K c. Hence, the convexity of K yields for d K,c 0,a) d K,c 0,a)+d K,c 0,b) 0, 1) that µ a d K,c 0,a) + 1 µ) b d K,c 0,b) 5

4.1 Norms K c, which shows ) a b 1 d K,c 0, µ + 1 µ) d K,c 0, a) d K,c 0, b) a = d K,c 0, d K,c 0, a) + d K,c 0, b) + = 1 d K,c 0, a) + d K,c 0, b) d K,c0, a + b). b d K,c 0, a) + d K,c 0, b) Since assertion five is trivial as soon as some of the points p, q, r coincide, one can assume that p, q, r are pairwise distinct. Using the above inequality for the case a = r p and b = q r leads to d K,c p, q) = d K,c 0, r p) + q r)) d K,c 0, r p) + d K,c 0, q r) = d K,c p, r) + d K,c r, q). ) Lemma 4.5 Let K R n be a convex body with center point c K. Then it holds that K is symmetric if and only if for every p, q R n it is d K,c p, q) = d K,c q, p). Proof. First, assume that K is symmetric. For p, q R n with p q, this property implies that p x p,q K c + p. Thus, q + p x p,q K c + q lies on the ray from q through p, which leads to x q,p q q + p x p,q ) q = x p,q p and d K,c q, p) d K,c p, q). By exchanging the roles of p and q, the desired equality follows. Secondly, assume that d K,c p, q) = d K,c q, p) holds for every p, q R n, and let x K. Then 1 d K,c c, x) = d K,c x, c) = d K,c c, c x) implies c x K. Proposition 4.6 Let K R n c K. Then be a symmetric convex body with center point K,c : R n R 0, is a norm with unit ball B K,c,10) = K c. Proof. For s R <0 and x R n it is x d K,c 0, x) d K,c 0, sx) = d K,c sx, 0) = d K,c 0, sx) = sd K,c 0, x) by Lemmata 4.4 and 4.5. The same lemmata yield the remaining norm properties as well as B K,c,10) = K c. 53

4 General shape of bisectors, Voronoi cells and their facets Moreover, the unit ball of every norm in R n is a symmetric convex body and the norm defined by this body is the given norm. To prove this, it will be shown beforehand that every norm in R n is a continuous function with respect to the Euclidean norm. This result will be very useful throughout this chapter. Proposition 4.7 Every norm : R n R 0 is continuous with respect to the Euclidean norm. Proof. Let x = x 1,..., x n ) T R n and ε R >0. Denote by Rx) := { y 1,..., y n ) T R n i {1,..., n} : y i [x i 1, x i + 1] } the n-dimensional cube with side length two and x in its center. Then it holds that B,1x) Rx), because for every y 1,..., y n ) T R n \ Rx) there is some i {1,..., n} such that y i / [x i 1, x i +1], which implies y x y i x i > 1. In addition, it follows inductively that every y Rx) can be written as y = µ u x+u) with µ u [0, 1] for every u { 1, 1} n and µ u = 1. u { 1,1} n u { 1,1} n In fact, the definition of Rx) yields that for every y Rx) and every i {1,..., n} there is τ i [0, 1] such that y i = τ i x i 1) + 1 τ i )x i + 1), which directly gives the induction basis for n = 1. If µũ [0, 1] for ũ { 1, 1} n 1 with µũ = 1 and y 1,..., y n 1 ) T = µũ x1,..., x n 1 ) T + ũ ) ũ { 1,1} n 1 ũ { 1,1} n 1 are already found, µ u1,...,u n) T := { τn µ u1,...,u n 1 ) T, if u n = 1, 1 τ n ) µ u1,...,u n 1 ) T, if u n = 1 gives the desired equalities. Hence, for every y Rx) it holds that y µ u x + u max{ x + u u { 1, 1} n } =: M x. u { 1,1} n Define δ := 1 ε M x x +ε 0, ] 1, and let y B,δx). It is left to show that x y < ε. Thus assume for contradiction that x y ε. With this, two cases can be distinguished. 1. y x + ε: For z := 1y 1 1) x it holds that z x δ δ = 1 y x δ < 1, leading to z B,1x) Rx). The reverse triangle inequality and the definition of δ yield the contradiction M x z 1 ) 1 δ y δ 1 x x + ε ) 1 δ x + δ 1 ε > M x.. y x ε: 54

4.1 Norms Consider z := 1x 1 1) y. Then it follows z x δ δ = 1 1) x y δ 1 x y δ < 1 and z B,1x) Rx). Again, the reverse triangle inequality can be applied to get the contradiction M x z 1 ) ) 1 1 δ x δ 1 y x + δ 1 ε > M x. Corollary 4.8 Let : R n R 0 be a norm, and let a R n. Then F a : R n R 0, x x a is continuous with respect to the Euclidean norm. Proof. Let x R n and ε R >0. Then it follows from Lemma 4.7 that there is some δ R >0 such that x a y < ε holds for every y B,δx a). Since for every z B,δx) it holds that z a) x a) = z x < δ, it is F a x) F a z) = x a z a < ε. Definition 4.9 For n N {0}, the n-dimensional sphere is defined as S n := { x R n+1 x = 1 }. Proposition 4.10 Let : R n R 0 be a norm. Then B,1 0) is a symmetric convex body with center point 0, and for every p R n it is p B,1 0),0 = p. Proof. Since S n 1 is compact with respect to the Euclidean norm and for every x S n 1 it is x > 0, Proposition 4.7 implies the existence of m R >0 such that x m holds for every x S n 1. Hence, for each x B,1 0) with x 0 it is x 1 x = x x m x, which shows B,1 0) B,1/m0) and B,1 0) is bounded. For x = x 1,..., x n ) T R n with x > 1, let ε := x 1 > 0. With e 1,..., e n denoting the standard basis of R n, define M := max { e i i {1,..., n}} and δ := ε M > 0. Then it holds for each y = y n 1,..., y n ) T with y x < δ by the Cauchy Schwarz inequality y x n y i x i e i M i=1 n y i x i M n y x < M nδ = ε. i=1 This yields that B,δx) B,ε x) R n \ B,1 0), 55

4 General shape of bisectors, Voronoi cells and their facets and thus B,1 0) is closed and compact. For every x R n with x < 1, one can define ε := 1 x as well as M and δ as above to get analogously B,δx) B,ε x) B,1 0). This shows that B,1 0) is open, and in particular that 0 is in the interior of B,1 0). Moreover, B,1 0) is convex due to the triangle inequality which is satisfied by, and B,1 0) is symmetric because x = x holds for every x R n. For every{ x R n } with x = 1, and every ε R >0 it holds for µ 1 := max 1 ε x, 1 that µ 1 x B,εx) as well as µ 1 x < 1, and for µ := 1 + ε x that µ x B,εx) as well as µ x > 1. Therefore, the boundary of B,1 0) is {x R n x = 1}. For every p R n with p 0 it now follows that x 0,p = p, where the notation p from Definition 4.3 is used for B,1 0) with center point 0. This gives p B,1 0),0 = p x 0,p = p. Propositions 4.6 and 4.10 give two equivalent viewpoints on norms: Either one can directly consider the norm and derive properties of the unit ball from that, or one considers a symmetric convex body and deduces properties of the corresponding norm. In the introduction of this thesis, it is already defined what a strictly convex norm is. A convex body can also be strictly convex, and both definitions are compatible for symmetric convex bodies. Definition 4.11 Let K R n be a convex body with center point c K. K is called strictly convex if for every x, y K with x y and every τ 0, 1) it holds that τx + 1 τ)y lies in the interior of K. Proposition 4.1 Let : R n R 0 be a norm. Then it holds that is strictly convex if and only if B,1 0) is strictly convex. Proof. First, assume that is strictly convex, and let x, y B,1 0) with x y as well as τ 0, 1). By Lemma.6, it is τx + 1 τ)y < max{ x, y } 1, which gives τx + 1 τ)y B,1 0). Secondly, assume the strict convexity of B,1 0), and let x, y R n with x y and x = y =: m > 0 as well as τ 0, 1). Then it follows that x m, y m B,10) and the strict convexity of this unit ball implies that τ x m + 1 τ) y m B,10). Therefore, τx + 1 τ)y = m τ x m + 1 τ) y m < m. Another possibility to define strict convexity of convex bodies is given in the next statement. This variant is used in [10], and since this chapter refers several 56

4.1 Norms times to results in [10], it will be proven that this alternative definition is indeed equivalent to Definition 4.11. Proposition 4.13 Let K R n be a convex body with center point c K. Then it holds that K is strictly convex if and only if the boundary of K does not contain a line segment. Proof. The strict convexity of K directly implies that the boundary of K cannot contain a line segment. Thus, only the other direction needs to be shown. For this, assume that the boundary of K does not contain a line segment, and let x, y K with x y as well as τ 0, 1). By the convexity of K, it follows that a := τx+1 τ)y K. Assume for contradiction that a lies on the boundary of K. Then there exists some b 1 on the line segment between x and a which lies in the interior of K. In fact, this is trivially true for b 1 = x if x itself is in the interior of K. If x lies on the boundary of K, one finds b 1 as described above, since the boundary of K does not contain the line segment between x and a. Analogously, there exists b on the line segment between y and a that lies in the interior of K. Hence, d K,c c, b 1 ) < 1, d K,c c, b ) < 1 and for some λ 0, 1) it is a = λb 1 + 1 λ)b. This yields d K,c c, a) = d K,c 0, a c) = d K,c 0, λb 1 c) + 1 λ)b c)) d K,c 0, λb 1 c)) + d K,c 0, 1 λ)b c)) = λd K,c c, b 1 ) + 1 λ)d K,c c, b ) < 1 and a lies in the interior of K, which contradicts the assumption that a lies on the boundary of K. Sometimes strict convexity of a given norm is not enough, and one also wants the property that the unit ball has no sharp corners. Such a norm is called smooth. To define this notion formally, one needs to introduce supporting hyperplanes. Definition 4.14 Let S R n, and let s S lie on the boundary of S. A hyperplane H R n is a supporting hyperplane of S at s if s H and S is contained in one of the two closed halfspaces bounded by H. For every convex set it holds that each of its boundary points has a supporting hyperplane. This result is known as the supporting hyperplane theorem and can for example be found in [3]. Theorem 4.15 Supporting hyperplane theorem) Let S R n be convex, and let s S lie on the boundary of S. Then there exists a supporting hyperplane of S at s. In particular, every boundary point of a convex body has a supporting hyperplane, but these hyperplanes are not unique in general. The intuition is that non-unique supporting hyperplanes occur at sharp corners of the convex body. If a convex body has unique supporting hyperplanes everywhere on its boundary, it is called smooth. 57

4 General shape of bisectors, Voronoi cells and their facets Definition 4.16 Let K R n be a convex body with center point c K. K is called smooth if each point on its boundary has a unique supporting hyperplane. A norm : R n R 0 is called smooth if B,1 0) is smooth. A convex body can be strictly convex and smooth at the same time, or it can be neither strictly convex nor smooth, or it can have exactly one of the two properties. Illustrations for this are shown in Figure 4.1. a) Strictly convex and smooth. b) Strictly convex but not smooth. c) Smooth but not strictly convex. d) Neither smooth nor strictly convex. Figure 4.1: Convex bodies in two dimensions with different properties. 4. Bisectors In [6], Horváth shows that every bisector of two distinct points is homeomorphic to a hyperplane if a strictly convex norm is used. In the following, the intersection of two bisectors is examined, where one bisector is given by a 1 and a and the other bisector is given by a 1 and a 3. In other words, the set of all points having the same distance to a 1, a, a 3 R n is analyzed. For the case n = 3, it is shown in [10] cf. Lemma 3.1..6 and Corollary 3.1..7) that such a set is homeomorphic to a line under a strictly convex and smooth norm if a 1, a, a 3 are non-collinear. For strictly convex norms without smoothness it is also shown in [10] that this bisector intersection might be disconnected, but that each component is still homeomorphic to a line. As long as the underlying norm is strictly convex and smooth, I strongly conjecture for general dimension n that such a bisector intersection is homeomorphic to R n. One way to prove this relies on the following conjecture. Conjecture 4.17 Let : R n R 0 be a strictly convex and smooth norm, let a 1, a, a 3 R n be non-collinear, and let H be the plane spanned by a 1, a, a 3. Then it holds for every sequence p k ) k N H = a 1, a ) H = a 1, a 3 ) with lim p k a 1 = k 58

4. Bisectors that min{ p k h h H} lim k p k a 1 = 1. The precise statement regarding the appearance of the considered bisector intersection is formulated in the following theorem: Theorem 4.18 Let : R n R 0 be a strictly convex and smooth norm, and let V R n be a subspace of dimension m. If Conjecture 4.17 is true and a 1, a, a 3 V are non-collinear, then H = a 1, a ) H = a 1, a 3 ) V is homeomorphic to R m. If a 1, a, a 3 V are collinear and pairwise distinct, then H = a 1, a ) H = a 1, a 3 ) =. The proof of this theorem uses some topological concepts which will be introduced first. Definition 4.19 For a given topological space X, T ) and Y X, the subspace topology on Y is T Y := {O Y O T }. Lemma 4.0 Let X, T ) be a topological space, and let Z Y X. Then it is T Y ) Z = T Z. Proof. For O T Y ) Z there exists O Y T Y such that O = O Y Z. In addition, there exists O X T with O Y = O X Y. Thus, O = O X Y Z = O X Z T Z. Given O T Z, one finds O X T with O = O X Z = O X Y Z T Y ) Z. Definition 4.1 Let X, T ) be a topological space, and let Y X. X, T ) is compact if every open cover X = i I O i with O i T has a finite subcover X = O i1 O i... O in with i 1, i,..., i n I. Y is compact in X, T ) if every open cover Y O i with O i T i I has a finite subcover Y O i1 O i... O in with i 1, i,..., i n I. 59

4 General shape of bisectors, Voronoi cells and their facets Lemma 4. Let X, T ) be a topological space, and let Y X. compact in X, T ) if and only if Y, T Y ) is compact. Then Y is Proof. Assume that Y is compact in X, T ), and let Y = i I O i with O i T Y be an open cover. Then one finds for every i I an O X,i T such that O i = O X,i Y, and Y can be written as Y = ) O X,i Y ) = O X,i Y O X,i. i I i I i I The compactness of Y as a subset of X gives i 1,..., i n I such that Y O X,i1... O X,in leading to n ) n Y = O X,ij Y = OX,ij Y ) n = O ij. j=1 j=1 Now assume that Y, T Y ) is compact, and let Y i I O i with O i T be an open cover. This yields ) Y = O i Y = O i Y ) with O i Y T Y. i I i I The compactness of Y as a topological space gives i 1,..., i n I such that Y = n Oij Y ) n ) = O ij Y j=1 j=1 j=1 n O ij. j=1 Corollary 4.3 Let X, T ) be a topological space, and let Z Y X. Then Z is compact in Y, T Y ) if and only if Z is compact in X, T ). Proof. By Lemma 4., Z is compact in Y, T Y ) if and only if Z, T Y ) Z ) is compact, which is by Lemma 4.0 equivalent to Z, T Z ) being compact and thus again by Lemma 4. equivalent to Z being compact in X, T ). Definition 4.4 For two given topological spaces X, T ) and Y, S), a function f : X Y is proper if for every C Y compact in Y, S) the preimage f 1 C) is compact in X, T ). Additionally to these notions from topology, the proof needs orthogonal complements of real subspaces and projections. Furthermore, the proof specifies a homeomorphism under the assumption of Conjecture 4.17) from the bisector intersection to a projected open unit ball, such that it will be shown in advance that such an open unit ball is homeomorphic to its whole embedding space. 60

4. Bisectors Definition 4.5 For a given subspace V R n, the orthogonal complement of V is and the projection of R n on V is V := {w R n v V : v, w = 0}, P V : R n V, x w such that x w V. Lemma 4.6 Let V R n be a subspace with orthogonal complement W := V. Then P W is continuous with respect to the Euclidean norm. Proof. Let x 1 R n, ε R >0 and x B,εx 1 ). Then there are unique v 1, v V and w 1, w W such that x 1 = v 1 + w 1 and x = v + w. This yields P W x 1 ) P W x ) = w 1 w w 1 w, w 1 w + v 1 v, v 1 v = x 1 x, x 1 x = x 1 x < ε. Lemma 4.7 Let V R n be a subspace, and let : V R 0 be a norm. Then B,1 0) is homeomorphic to V. Proof. Define h : B,1 0) V, x x 1 x. First consider x 1, x B,1 0) with hx 1 ) = hx ). Then it holds x 1 = µx with µ := 1 x 1. In particular, x 1 x 1 = µ x which is equivalent to x 1 1 x ) = 1 x 1 ) x and further to x 1 = x. Thus, µ = 1 and x 1 = x. This shows that h is injective. For y V it holds that y 1+ y < 1, and due to 1 y 1+ y = 1 it is 1+ y ) y h = y. Hence, h is a bijection with inverse h 1 : V B 1+ y,1 0), x x. 1+ x In addition, Corollary 4.8 directly implies that h as well as h 1 are continuous. Lemma 4.8 Let : R n R be a norm, and let V R n be a subspace with orthogonal complement W := V. Then W : W R 0, w min{ w + v v V } is a norm with unit ball B W,10) = P W B,1 0) ). 61

4 General shape of bisectors, Voronoi cells and their facets Proof. First, one needs to consider why the above minimum is attained. For this, let w W. By Corollary 4.8, g w : V R 0, v w + v is continuous with respect to the Euclidean norm. This further implies that K w := {v V w + v w } is compact in R n, ) and thus by Corollary 4.3 also compact in V, ). Hence, g w attains its minimum on K w, i.e., there exists some ṽ K w such that g w ṽ) g w v) holds for all v K w. Furthermore, for every v V \ K w it is w + v > w = g w 0) g w ṽ). This shows that w + v w + ṽ holds for every v V. Secondly, it needs to be shown that W is indeed a norm. It is clear that 0 W = 0. For every w W with w W = 0 there exists some v V with w + v = 0, which implies w = v W V and thus w = 0. For every w W and every µ R \ {0} it holds µw W = min{ µw + v v V } = min{ µw + µv v V } = min{ µ w + v v V } = µ min{ w + v v V } = µ w W. For the triangle inequality let w 1, w W. For i {1, } there exists v i V such that w i W = w i + v i, which yields w 1 + w W w 1 + w + v 1 + v w 1 + v 1 + w + v = w 1 W + w W. Finally, B W,10) = P W B,1 0) ) can be shown as follows: On the one hand, for every w B W,10) there exists some v V with w + v < 1 such that w = P W w + v) P W B,1 0) ). On the other hand, for x B,1 0) with w := P W x) it holds that w W w + x w) = x < 1. Proof of Theorem 4.18. Let a 1, a, a 3 V be pairwise distinct. If a 1, a, a 3 are collinear, one can assume without loss of generality that a is in the middle, i.e., there exists τ 0, 1) such that a = τa 1 + 1 τ)a 3. For every p H = a 1, a 3 ) it holds that p a = τp a 1 )+1 τ)p a 3 ) < p a 1 and so p / H = a 1, a ). This shows H = a 1, a ) H = a 1, a 3 ) =. Hence, in the following a 1, a, a 3 are assumed to be non-collinear. For V one can choose an orthonormal basis v 1,..., v m ) with respect to the dot product on R n. Let e 1,..., e m ) denote the standard basis of R m. With this, one can define ψ : V R m, m m α i v i α i e i, i=1 as well as a norm ψ : R m R 0, x ψ 1 x), such that ψ is an isometric isomorphism from V, V ) to R m, ψ ) as well as an isometric isomorphism from V, ) to R m, ). Furthermore, ψ is a strictly convex and smooth i=1 6

4. Bisectors norm with unit ball B ψ,10) = ψ B,1 0) V ) satisfying ψ H = a 1, a ) H = a 1, a 3 ) V ) = H = ψ ψa 1 ), ψa )) H = ψ ψa 1 ), ψa 3 )). Thus, it is enough to show Theorem 4.18 for V = R n, since then it can be applied to ψ to get a homeomorphism from H = ψ ψa 1 ), ψa )) H = ψ ψa 1 ), ψa 3 )) to R m. Hence, assume in the following that V = R n. Now the desired homeomorphism will be first described informally and afterwards defined formally: Let H be the plane spanned by a 1, a, a 3, i.e., H := {a 1 + sa a 1 ) + ta 3 a 1 ) s, t R}. Then, H a 1 is a vector space with orthogonal complement W := H a 1 ). For p H = a 1, a ) H = a 1, a 3 ) it holds that a 1, a, a 3 are on the boundary of B,r p), where r := a 1 p > 0. This is illustrated for three dimensions in Figure 4.. The homeomorphism considers the intersection of this ball with H, and the relative position of this intersection in the unit ball. To be more specific, K p := B,r p) H and K p := 1 r K p p) = B,1 0) H a 1 ) + a ) 1 p r are strictly convex and smooth, and the position of K p in the unit ball is already determined by the projection of a 1 p on W. Therefore, the conjectured r homeomorphism as depicted in Figure 4.3 is ϕ : H a = 1, a ) H a = 1, a 3 ) W, ) a1 p p P W. a 1 p First, the injectivity of ϕ will be shown. Let p 1, p H = a 1, a ) H = a 1, a 3 ) with ϕp 1 ) = ϕp ) =: w. Then it holds that H a 1 ) + a 1 p 1 a 1 p 1 = H a 1) + w = H a 1 ) + a 1 p a 1 p, which shows K p1 = K p using the notation above. Lemma.1.1.1 and Theorem.1..3 in [10] show for strictly convex bodies in two-dimensions and for pairwise distinct points a, b, c that there is at most one uniformly scaled and translated copy of this body that has a, b, c on its boundary, and that exactly one such copy exists if a, b, c are non-collinear and the given body is smooth. From this it follows that a unique r R >0 and a unique p R n exist such that rk p1 +p H has a 1, a, a 3 on its boundary in H. Due to K p1 = a 1 p 1 K p1 + p 1 and K p = a 1 p K p1 + p, it must hold that a 1 p 1 = r = a 1 p and p 1 = p = p. Hence, ϕ is injective. Secondly, the image of ϕ will be calculated. It holds for every 63

4 General shape of bisectors, Voronoi cells and their facets a a 1 a 3 p a a 1 a 3 a) B,1 0). b) H. c) p H = a 1, a ) H = a 1, a 3 ) with B, a1 p p) and H. Figure 4.: Intersection of scaled and translated unit ball with H in a 1, a and a 3 for the case n = 3. ) p H = a 1, a ) H = a a 1, a 3 ) that ϕp) = P 1 p 1 W a 1 = P a1 +a p W a 1 p )) p and 1 a1 + a 1 p) = 1 a 1 p a 1 p a 1 p) + 1 a p) < 1, such that ϕp) P W B,1 0) ). For every x B,1 0) it holds that K := B,1 0) H a 1 ) + x) is strictly convex and smooth. Therefore, Lemma.1.1.1 and Theorem.1..3 in [10] give unique r R >0 and p R n such that rk + p H has a 1, a, a 3 on its boundary in H. Because of rk + p = B,r p) H a 1 ) + rx + p), it must hold that rk+p H H a 1 ) + rx + p), which implies H = H a 1 ) + rx + p and thus rx + p a 1 H a 1. Thus, x a 1 p r a p = a 3 p, which leads to p H = a 1, a ) H = a 1, a 3 ) and P W x) = ϕp). This shows cf. Figure 4.3) that ϕh = a 1, a ) H = a 1, a 3 )) = P W B,1 0) ) such that ϕ is a bijection H a 1 and P W x) = P W a 1 p r ) hold. Moreover, r = a 1 p = ϕ : H = a 1, a ) H = a 1, a 3 ) P W B,1 0) ). Combining Lemmata 4.7 and 4.8 yields that P W B,1 0) ) is homeomorphic to W and thus homeomorphic to R n. Therefore, it is only left to show that ϕ and ϕ 1 are continuous. The continuity of ϕ is a direct consequence of Corollary 4.8 and Lemma 4.6. Because of this, the rest of this proof will examine the continuity of ϕ 1. When assuming Conjecture 4.17, it can be shown that ϕ is proper: For this, 64