On the linking of number lattices

TL;DR

This paper explores the structure and properties of number lattices using topological network ideas, introducing linking operations, duality theorems, and algorithms for constructing and analyzing lattices in a mathematical framework.

Contribution

It introduces a linking operation for generalized number lattices, proves duality and inversion theorems, and develops algorithms for constructing and analyzing lattices with applications in network theory.

Findings

Linked number lattices can be combined and inverted using the proposed operations.

Duality theorems enable construction of self-dual lattices from existing ones.

Algorithms are provided for basis reduction in lattice constructions involving unimodular matrices.

Abstract

In this paper we study ideas which have proved useful in topological network theory in the context of lattices of numbers. A number lattice $L_S$ is a collection of row vectors, over $\mathbb{Q}$ on a finite column set $S,$ generated by integral linear combination of a finite set of row vectors. A generalized number lattice $K_S$ is the sum of a number lattice $L_S$ and a vector space $V_S$ which has only the zero vector in common with it. The dual $K^d_S$ of a generalized number lattice is the collection of all vectors whose dot product with vectors in $K_S$ are integral and is another generalized number lattice. We consider a linking operation ('matched composition`) between generalized number lattices $K_{SP},K_{P}$ (regarded as collections of row vectors on column sets $S\cup P, P,$ respectively with $S,P,$ disjoint) defined by $K_{SP}\leftrightarrow K_{P}\equiv \{f_S:((f_S,g_P)\in K_{SP}, g_P \in K_{P}\}.$ We show that this operation together with contraction and restriction, and the results, the implicit inversion theorem (which gives simple conditions for the equality $K_{SP}\leftrightarrow (K_{SP}\leftrightarrow K_S)= K_S,$ to hold) and implicit duality theorem ($(K_{SP}\leftrightarrow K_{P})^d= K_{SP}^d\leftrightarrow K_{P}^d$)), are both relevant and useful in suggesting problems concerning number lattices and their solutions. Using the implicit duality theorem, we give simple methods of constructing new self dual lattices from old. We also give new and efficient algorithms for the following. Given $V_{SP},K_P,$ such that $V_{SP}\leftrightarrow (V_{SP}\leftrightarrow K_P)= K_P,$ where $V_{SP}$ is a vector space with a totally unimodular basis matrix, to construct reduced bases for the number lattice part of $V_{SP}\leftrightarrow K_P, K_P^d, (V_{SP}\leftrightarrow K_P)^d,$ from a reduced basis for the number lattice part of $K_P.$

Figures

Equations

$${\mathcal{K}}_{SP}+{\mathcal{K}}_{PQ}$$

$$\begin{array}{rl}{\mathcal{K}}_{SP}\cap {\mathcal{K}}_{PQ}\equiv \{f_{SPQ}& :f_{SPQ}=(f_S,h_P,g_Q),\\ & \text{ where }(f_S,h_P)\in {\mathcal{K}}_{SP},(h_P,g_Q)\in {\mathcal{K}}_{PQ}.\}.\end{array}$$

$$\displaystyle\mathcal{K}_{SP}\cap\mathcal{K}_{PQ}\equiv(\mathcal{K}_{SP}\oplus\mbox{$\cal F$}_{Q})\cap(\mbox{$\cal F$}_{S}\oplus\mathcal{K}_{PQ}).$$

$$\left(\begin{array}{c}{C}_{SP}\end{array}\right)\equiv \left(\begin{array}{ccc}{C}_{1S}& ⋮& 0_{1P}\\ C_{2S}& ⋮& C_{2P}\end{array}\right),$$

$$\left(\begin{array}{c}{0}_{1P}\\ C_{2P}\end{array}\right),$$

$$\left(\begin{array}{c}{C}_{SP}\end{array}\right)\equiv \left(\begin{array}{ccc}{C}_{1S}& ⋮& 0_{1P}\\ C_{2S}& ⋮& C_{2P}\end{array}\right),$$

$${\mathcal{K}}_S^d\equiv \{g_S:⟨f_S,g_S⟩\text{mbox}aninteger,f_S\in {\mathcal{K}}_S\}.$$

$${\left(\begin{array}{c}{B}_2\\ C_2\\ D_2\end{array}\right)}^T={\left(\begin{array}{c}{B}_1\\ C_1\\ D_1\end{array}\right)}^{-1}.$$

$$({\mathcal{K}}_S+{\hat{\mathcal{K}}}_S{)}^d={\mathcal{K}}_S^d\cap {\hat{\mathcal{K}}}_S^d.$$

$$({\mathcal{K}}_S\cap {\hat{\mathcal{K}}}_S{)}^d={\mathcal{K}}_S^d+{\hat{\mathcal{K}}}_S^d.$$

$$({\mathcal{K}}_S\oplus {\mathcal{K}}_P{)}^d={\mathcal{K}}_S^d\oplus {\mathcal{K}}_P^d.$$

$$({\mathcal{K}}_S+{\mathcal{K}}_P{)}^d$$

$$=({\mathcal{K}}_S\oplus 0_{P\setminus S}{)}^d\cap ({\mathcal{K}}_P\oplus 0_{S\setminus P}{)}^d$$

$$=({\mathcal{K}}_S^d\oplus (0_{(P\setminus S)}{)}^d)\cap ({\mathcal{K}}_P^d\oplus (0_{(S\setminus P)}{)}^d)$$

$$\displaystyle=({\cal K}_{S}^{d}\oplus\mbox{$\cal F$}_{(P\setminus S)})\cap({\cal K}_{P}^{d}\oplus\mbox{$\cal F$}_{(S\setminus P)})$$

$$={\mathcal{K}}_S^d\cap {\mathcal{K}}_P^d,$$

$$({\mathcal{K}}_{SP}\circ S{)}^d$$

$$({\mathcal{K}}_{SP}\times S{)}^d$$

$$C_{SP}\equiv \left[\begin{array}{ccc}{C}_{1S}& ⋮& C_{1P}\\ C_{2S}& ⋮& C_{2P}\end{array}\right],$$

$${\hat{C}}_{SP}\equiv R{C}_{SP}=\left[\begin{array}{ccc}{\hat{C}}_{1S}& ⋮& {\hat{C}}_{1P}\\ 0_{2S}& ⋮& 0_{2P}\end{array}\right],$$

$$\left(\begin{array}{c}{C}_{SP}\end{array}\right)\equiv \left(\begin{array}{ccc}{C}_{1S}& ⋮& 0_{1P}\\ C_{2S}& ⋮& C_{2P}\end{array}\right).$$

$$C_{S_1{S}_2}\equiv \left(\begin{array}{ccc}{C}_{1S_1}& ⋮& 0_{1S_2}\\ C_{2S_1}& ⋮& C_{2S_2}\end{array}\right),$$

$${\hat{C}}_{S_1{S}_2}\equiv \left(\begin{array}{ccc}{C}_{1S_1}& ⋮& 0_{1S_2}\\ 0_{2S_1}& ⋮& C_{2S_2}\end{array}\right),$$

$${\mathcal{K}}_{SP}\leftrightarrow {\mathcal{K}}_{PQ}$$

$${\mathcal{K}}_{SP}\rightleftharpoons {\mathcal{K}}_{PQ}$$

$${\cal K}_{S}\circ(S-T)={\cal K}_{S}\leftrightarrow\mbox{$\cal F$}_{T},{\cal K}_{S}\times(S-T)={\cal K}_{S}\leftrightarrow{\mathbf{0}}_{T},T\subseteq S.$$

$$({\mathcal{K}}_{SPZ}\leftrightarrow {\mathcal{K}}_S)\leftrightarrow {\mathcal{K}}_P=({\mathcal{K}}_{SPZ}\leftrightarrow {\mathcal{K}}_P)\leftrightarrow {\mathcal{K}}_S={\mathcal{K}}_{SPZ}\leftrightarrow ({\mathcal{K}}_S\oplus {\mathcal{K}}_P).$$

$${\mathcal{K}}_{SPZ}\times PZ\circ Z={\mathcal{K}}_{SPZ}\circ SZ\times Z.$$

$$({\mathcal{K}}_{SP}\leftrightarrow {\mathcal{K}}_{PQ})=({\mathcal{K}}_{SP}+{\mathcal{K}}_{(-P)Q})\times (S\cup Q)$$

$$({\mathcal{K}}_{SP}\cap {\mathcal{K}}_{PQ})\circ (S\cup Q).$$

$$({\mathcal{K}}_{SP}\rightleftharpoons {\mathcal{K}}_{PQ})=({\mathcal{K}}_{SP}+{\mathcal{K}}_{PQ})\times (S\cup Q)$$

$$({\mathcal{K}}_{SP}\cap {\mathcal{K}}_{(-P)Q})\circ (S\cup Q).$$

Full text

We consider a linking operation (’matched composition‘) between generalized number lattices ${\cal K}_{SP},{\cal K}_{PQ}$ (regarded as collections of row vectors on column sets $S\cup P,P\cup Q,$ respectively with $S,P,Q$ disjoint) defined by ${\cal K}_{SP}\leftrightarrow{\cal K}_{PQ}\equiv\{(f_{S},h_{Q}):((f_{S},g_{P})\in{\cal K}_{SP},(g_{P},h_{Q})\in{\cal K}_{PQ}\},$ and another (’skewed composition‘) defined by ${\cal K}_{SP}\rightleftharpoons{\cal K}_{PQ}\equiv\{(f_{S},h_{Q}):((f_{S},g_{P})\in{\cal K}_{SP},(-g_{P},h_{Q})\in{\cal K}_{PQ}\}.$

We show that these basic operations together with contraction and restriction, and the results, the implicit inversion theorem (which gives simple conditions for the equality ${\cal K}_{SP}\leftrightarrow({\cal K}_{SP}\leftrightarrow{\cal K}_{S})={\cal K}_{S},$ to hold) and implicit duality theorem ($({\cal K}_{SP}\leftrightarrow{\cal K}_{PQ})^{d}=({\cal K}_{SP}^{d}\rightleftharpoons{\cal K}_{PQ}^{d}$)), are both relevant and useful in suggesting problems concerning number lattices and their solutions. While sets of generalized number lattices are closed under matched and skewed composition and the dualization operation, not all sets of number lattices are. However, it is easy to extract, from a generalized number lattice, its number lattice part.

Using the implicit duality theorem, we give simple methods of constructing new self dual lattices from old.

We also give new and efficient algorithms for the following problems:

To construct a block diagonal basis for a number lattice, if it exists.

• 2.

Given \mbox{\cal V}_{SP},{\cal K}_{P}, such that \mbox{\cal V}_{SP}\leftrightarrow(\mbox{\cal V}_{SP}\leftrightarrow{\cal K}_{P})={\cal K}_{P}, where \mbox{\cal V}_{SP} is a vector space with a totally unimodular basis matrix, to construct

1. reduced bases for the number lattice part of \mbox{\cal V}_{SP}\leftrightarrow{\cal K}_{P},{\cal K}_{P}^{d},(\mbox{\cal V}_{SP}\leftrightarrow{\cal K}_{P})^{d}, from a reduced basis for the number lattice part of ${\cal K}_{P};$

2. to construct approximate shortest vectors of $L_{P}$ from approximate shortest vectors of the number lattice part of \mbox{\cal V}_{SP}\leftrightarrow L_{P}, when L_{P}\subseteq{\cal V}_{SP}\leftrightarrow(\mbox{\cal V}_{SP}\leftrightarrow L_{P}).

††journal: Linear Algebra and its Applications

1 Introduction

Number lattices are established areas of research in mathematics and computer science ([4], [12]). The notion of ‘short vectors’ has been of importance for number lattices both from a theoretical as well as a computational point of view ([11],[2]). They are often studied in terms of dual pairs and there are fundamental results relating the lengths of short vectors in such pairs ([3],[10],[23],[13],[21]). However, ‘linked’ number lattices, related to each other through linear relations, do not appear to have been paid attention to, in the literature.

In this paper, we introduce techniques for studying number lattices related to each other through linear relations and through dualization. These techniques have been used successfully over many years to study electrical networks ([18]). In this paper we use them, for instance, to efficiently construct reduced bases for a number lattice using another such, for a related number lattice. We also bring out some analogies that exist between electrical networks and linked number lattices, for instance, that dual number lattices are similar to dual electrical networks and therefore that reciprocal networks are analogous to self-dual number lattices.

The conventional notion of a dual number lattice is natural for the case of full dimensional lattices - the dual lattice is simply the collection of all vectors whose dot product with the vectors of the original lattice is integral. For the case where it is not full dimensional, we need to introduce an additional condition that we work only within the span of the original number lattice. If we drop this additional condition, the dual of a number lattice would, in general, be the sum of a number lattice and a vector space (Theorem 1).

The linking operation that we use is the ‘matched composition’ between ${\cal K}_{SP},{\cal K}_{PQ}$ (regarded as collections of row vectors on column sets $S\cup P,P\cup Q,$ respectively with $S,P,Q$ disjoint) defined by

${\cal K}_{SP}\leftrightarrow{\cal K}_{PQ}\equiv\{(f_{S},h_{Q}):((f_{S},g_{P})\in{\cal K}_{SP},(g_{P},h_{Q})\in{\cal K}_{PQ}\}.$ This operation is usually performed with ${\cal K}_{SP}$ as a vector space, which does the linking, and with ${\cal K}_{PQ}\equiv{\cal K}_{P},$ which gets linked to ${\cal K}_{SP}\leftrightarrow{\cal K}_{P}.$ We use the ‘$\leftrightarrow$’ operation because it occurs naturally in many physical systems, such as electrical networks, systems defined through linear block diagrams, etc. It also lends itself to being treated implicitly, without eliminating variables. But, in general, even if ${\cal K}_{P}$ is a number lattice, when ${\cal K}_{SP}$ is a vector space, ${\cal K}_{SP}\leftrightarrow{\cal K}_{P}$ would be the sum of a vector space and a number lattice.

We therefore choose to work with a ‘generalized number lattice’ ${\cal K}_{P},$ defined to be the sum of a number lattice $L_{P}$ and a vector space ${\cal V}_{P}.$ There is no loss in generality if we assume ${\cal V}_{P}$ and $L_{P}$ to be orthogonal. In this case, both ${\cal V}_{P}$ and $L_{P}$ are unique for ${\cal K}_{P}$ and are also easy to extract from it. Further, if ${\cal K}_{S}={\cal V}_{SP}\leftrightarrow{\cal K}_{P},$ and ${\cal K}_{S}={\cal V}_{S}+L_{S},$ with ${\cal V}_{S},L_{S},$ orthogonal, under some weak conditions on ${\cal V}_{SP},$ one can algorithmically relate vectors in $L_{S}$ uniquely to those in $L_{P}$ (Theorems 9, 25).

If we take the vectors in the generalized number lattice ${\cal K}_{P}$ to be row vectors, ${\cal V}_{SP}\leftrightarrow{\cal K}_{P}$ can be regarded as a generalization of post multiplication by a matrix. There are essential differences: every vector of ${\cal K}_{P}$ would not get acted upon and a vector, in general, would get ‘transformed’ to a non trivial affine subspace. Linking has the technical advantage, over post multiplication by a matrix, that it can be handled implicitly. It is often associated with graphs, which are algorithmically easy to process. We choose to work, more generally, in terms of a ‘regular’ vector space \mbox{\cal V}_{SP}, which is defined to be spanned by the rows of a totally unimodular matrix (such as the incidence matrix of a graph), with ${\cal K}_{P}$ being a generalized number lattice. The greater generality, thus available, is theoretically more convenient while algorithmically the situation is almost as good as working with graphs.

The theme of this paper is that the linking operation using regular vector spaces, has some of the properties of post multiplication by a $0,\pm 1$ matrix, which are desirable from the point of view of number lattices, such as short vectors getting linked. Finding the shortest (least norm) vector in a number lattice is known to be a hard problem ([1],[9],[12]). Therefore, if we have somehow found a short vector in the number lattice part of ${\cal K}_{P},$ it is interesting to note that a related vector in the number lattice part of ${\cal V}_{SP}\leftrightarrow{\cal K}_{P}$ is also short, the length being within a factor $|S|\times|P|,$ when ${\cal V}_{SP}$ is regular.

The outline of the paper follows.

Section 2 is on preliminary definitions and results.

Section 3 is on basic operations used in the paper. These are, sum and intersection of generalized number lattices, generalized to include their being defined on different sets, restriction and contraction and the dualization of generalized number lattices.

Section 4 deals with algorithms for constructing a basis for a number lattice, given a generating set for it. We mainly use the algorithms available for building the Hermite normal form (HNF) of an integral matrix. An instance of the usefulness of HNF - to detect if the given number lattice is the direct sum of smaller number lattices - is also described.

Section 5 is on the fundamental results available for linking or dualizing generalized number lattices. These are:

the implicit inversion theorem which gives necessary and sufficient conditions for the statement

‘${\cal K}_{SP}\leftrightarrow({\cal K}_{SP}\leftrightarrow{\cal K}_{S})={\cal K}_{S}$’ to hold and

the implicit duality theorem which states that ‘$({\cal K}_{SP}\leftrightarrow{\cal K}_{P})^{d}={\cal K}_{SP}^{d}\leftrightarrow{\cal K}_{P}^{d}$’.

The implicit inversion theorem is used to show that, when ${\cal V}_{SP}\leftrightarrow{\cal K}_{P}={\cal K}_{S}$ and certain simple conditions on ${\cal V}_{SP},{\cal K}_{P}$ are satisfied, the number lattice parts of ${\cal K}_{P}$ and ${\cal K}_{S}$ have invertible maps between them.

The implicit duality theorem is used to build new self dual number lattices from old.

Section 6 is a discussion of how linkages permit wide ranging generalizations of the notion of maps.

Section 7 contains results on lengths of corresponding vectors in the number lattice parts of generalized number lattices ${\cal K}_{P},{\cal K}_{S},$ linked through a regular vector space ${\cal V}_{SP},$ which satisfies some simple conditions in relation to them. Among other results, it is proved that a shortest vector in one of the number lattices can be transformed into a vector, in the other, whose length is no more than the shortest vector of that lattice, by a factor of $|S|\times|P|.$

Section 8 discusses how LLL-reduced basis for a dual number lattice can be built from an LLL-reduced basis for the primal, efficiently.

Section 9 summarizes the results on approximate shortest vectors in number lattices related to each other through linking by regular vector spaces and through dualization.

Section 10 relates shortest vectors of special kinds to the vectors closest to related number lattices.

Section 11 is on conclusions.

2 Preliminaries

The preliminary results and the notation used are from [18]. A vector $\boldsymbol{f}$ on a finite set $X$ over $\mathbb{F}$ is a function $f:X\rightarrow\mathbb{F}$ where $\mathbb{F}$ is a field. In this paper, we work only with the rational field $\mathbb{Q}.$ The length of a vector $x$ is the Euclidean norm $||x||$ of $x.$

The size of a set $X$ is denoted by $\boldsymbol{|X|}.$ When $X$, $Y$ are disjoint, $\boldsymbol{X\uplus Y}$ denotes the disjoint union of $X$ and $Y.$ A vector $f_{X\uplus Y}$ on $X\uplus Y$ would be written as $\boldsymbol{f_{XY}}.$ The sets on which vectors are defined would always be finite. When a vector $x$ figures in an equation, we use the convention that $x$ denotes a column vector and $x^{T}$ denotes a row vector such as in ‘$Ax=b,x^{T}A=b^{T}$’. Let $f_{Y}$ be a vector on $Y$ and let $X\subseteq Y$. The restriction $f_{Y}|_{X}$ of $f_{Y}$ to $X$ is defined as follows: $f_{Y}|_{X}\equiv g_{X},\textrm{ where }g_{X}(e)=f_{Y}(e),e\in X.$

When $f$ is on $X$ over $\mathbb{F}$, $\lambda\in\mathbb{F},$ then the scalar multiplication $\boldsymbol{\lambda f}$ of $f$ is on $X$ and is defined by $(\lambda f)(e)\equiv\lambda[f(e)]$, $e\in X$. When $f$ is on $X$ and $g$ on $Y$ and both are over $\mathbb{F}$, we define $\boldsymbol{f+g}$ on $X\cup Y$ by

$(f+g)(e)\equiv f(e)+g(e),e\in X\cap Y,\ (f+g)(e)\equiv f(e),e\in X\setminus Y,\ (f+g)(e)\equiv g(e),e\in Y\setminus X.$

When $X,Y,$ are disjoint, $f_{X}+g_{Y}$ is written as $\boldsymbol{(f_{X},g_{Y})}.$ When $f,g$ are on $X$ over $\mathbb{F},$ the dot product $\langle f,g\rangle$ of $f$ and $g$ is defined by $\langle f,g\rangle\equiv\sum_{e\in X}f(e)g(e).$

We say $f$, $g$ are orthogonal (orthogonal) iff $\langle f,g\rangle$ is zero.

An arbitrary collection of vectors on $X$ with $0_{X}$ as a member would be denoted by $\boldsymbol{\mathcal{K}_{X}}$.

A collection ${\cal K}_{X}$ is a vector space on $X$ iff it is closed under addition and scalar multiplication. For any collection ${\cal K}_{X},$ $span({\cal K}_{X})$ is the collection of all linear combinations of vectors in it.

For a vector space \mbox{\cal V}_{X}, since we take $X$ to be finite, any maximal independent subset of \mbox{\cal V}_{X} has size less than or equal to $|X|$ and this size can be shown to be unique. A maximal independent subset of a vector space \mbox{\cal V}_{X} is called its basis and its size is called the dimension of \mbox{\cal V}_{X} and denoted by {\boldsymbol{dim}(\mbox{\cal V}_{X})} or by {\boldsymbol{r}(\mbox{\cal V}_{X})}. For any collection of vectors ${\cal K}_{X},$ $\boldsymbol{r}({\cal K}_{X})$ is defined to be $dim(span({\cal K}_{X})).$ The collection of all linear combinations of the rows of a matrix $A$ is a vector space that is denoted by $row(A).$

For any collection of vectors $\mathcal{K}_{X},$ the collection $\boldsymbol{\mathcal{K}_{X}^{\perp}}$ is defined by ${\mathcal{K}_{X}^{\perp}}\equiv\{g_{X}:\langle f_{X},g_{X}\rangle=0\},$ It is clear that $\mathcal{K}_{X}^{\perp}$ is a vector space for any $\mathcal{K}_{X}.$ When $\mathcal{K}_{X}$ is a vector space, and the underlying set $X$ is finite, it can be shown that $({\mathcal{K}_{X}^{\perp}})^{\perp}=\mathcal{K}_{X}$ and $\mathcal{K}_{X},{\mathcal{K}_{X}^{\perp}}$ are said to be complementary orthogonal. The symbol $0_{X}$ refers to the zero vector on $X$ and $\boldsymbol{0_{X}}$ refers to the zero vector space on $X.$ The symbol \boldsymbol{\mbox{\cal F}_{X}} refers to the collection of all vectors on $X$ over the field in question. The notation \boldsymbol{\mbox{\cal V}_{X}} would always denote a vector space on $X.$

A collection $\mathcal{K}_{X}$ on $X$ is a number lattice iff there exists a set of vectors $b_{X}^{1},\cdots,b_{X}^{n}$ in ${\mathbb{Q}}^{m},$ such that $\mathcal{K}_{X}\equiv\{\lambda_{1}b_{X}^{1}+\cdots+\lambda_{n}b_{X}^{n},\lambda_{i}\in\mathbb{Z}\}.$ Such a set is said to be a generating set for the number lattice. If the generating set of vectors is independent, it is called a basis of the number lattice. Starting from a generating set for a number lattice, it is possible to build a basis for it through efficient algorithms.

A collection $\mathcal{K}_{X}$ on $X$ is a generalized number lattice iff there exists a set of vectors

$b_{X}^{1},\cdots,b_{X}^{n},c_{X}^{1},\cdots,c_{X}^{k}$ in ${\mathbb{Q}}^{m},$ such that $\mathcal{K}_{X}\equiv\{\lambda_{1}b_{X}^{1}+\cdots+\lambda_{n}b_{X}^{n}+\mu_{1}c_{X}^{1}+\cdots\mu_{k}c_{X}^{k},\lambda_{i}\in\mathbb{Z},\mu_{j}\in{\mathbb{Q}}\}.$ The set of vectors $\{b_{X}^{1},\cdots,b_{X}^{n},c_{X}^{1},\cdots,c_{X}^{k}\}$ is called a set of generating vectors for ${\cal K}_{X}.$ We say ${\cal K}_{X}$ is full dimensional iff $dim(span({\cal K}_{X}))=|X|.$ Let $c_{X}^{1},\cdots,c_{X}^{k},$ in the above definition, span the vector space \mbox{\cal V}_{X}. We call this the vector space part of ${\cal K}_{X}$. Clearly it is unique for ${\cal K}_{X},$ being the collection of all vectors $c_{X}\in{\cal K}_{X},$ such that $\alpha c_{X}\in{\cal K}_{X},\alpha\in\mathbb{Q}.$

Each $b_{X}^{i}$ in the above generating set can be resolved into components $b_{X}^{i1},b_{X}^{i2}$ in \mbox{\cal V}^{\perp}_{X},\mbox{\cal V}_{X} respectively. Therefore $\mathcal{K}_{X}=\{\lambda_{1}b_{X}^{11}+\cdots+\lambda_{n}b_{X}^{n1}+\mu_{1}c_{X}^{1}+\cdots\mu_{k}c_{X}^{k},\lambda_{i}\in\mathbb{Z},\mu_{j}\in{\mathbb{Q}}\}.$ Let \mbox{L}_{X} be the number lattice generated by $b_{X}^{11},\cdots,b_{X}^{n1}.$ It is clear that \mbox{L}_{X}\subseteq\mbox{\cal V}^{\perp}_{X}. The generating set for ${\cal K}_{X},$ can be taken without loss of generality to be the set $\{b_{X}^{1},\cdots,b_{X}^{n},c_{X}^{1},\cdots,c_{X}^{k}\},$ where the $b_{X}^{i},i=1,\cdots n,$ form a basis for \mbox{L}_{X} and the $c_{X}^{i},i=1,\cdots,k,$ form a basis for \mbox{\cal V}_{X}. Thus, we can take {\cal K}_{X}\equiv L_{X}+\mbox{\cal V}_{X}, where L_{X},\mbox{\cal V}_{X}, are orthogonal. In this case, both $L_{X}$ and \mbox{\cal V}_{X} would be unique for the given ${\cal K}_{X}.$

Thus generalized number lattices are generalizations of finite dimensional vector spaces as well as of number lattices. They arise naturally from number lattices through simple operations like dualization.

It is clear that a generalized number lattice that does not contain a nontrivial vector space (i.e., has the vector space part as the zero vector space) is a number lattice.

We use the symbol $\boldsymbol{L_{X}}$ for the number lattice on $X$ as opposed to $\mathcal{K}_{X}$ for arbitrary collections of vectors on $X$ with a zero vector as a member.

A matrix of full row rank, whose rows generate a vector space \mbox{\cal V}_{X}, is called a representative matrix for \mbox{\cal V}_{X}. A representative matrix which can be put in the form $(I\ |\ K)$ after column permutation is called a standard representative matrix.

If the rows of a matrix generate a number lattice $L_{X},$ by integral linear combination, then the matrix is called a generating matrix for $L_{X}.$ If further the rows are linearly independent, the generating matrix is called a basis matrix for $L_{X}.$

When $X$, $Y$ are disjoint we usually write $\mathcal{K}_{XY}$ in place of $\mathcal{K}_{X\uplus Y}$.

The collection $\{(f_{X},\lambda f_{Y}):(f_{X},f_{Y})\in\mathcal{K}_{XY}\}$ is denoted by $\mathcal{K}_{X(\lambda Y)}.$ When $\lambda=-1$ we would write more simply $\mathcal{K}_{X(-Y)}.$ Observe that $(\mathcal{K}_{X(-Y)})_{X(-Y)}=\mathcal{K}_{XY}.$

We say sets $X$, $X^{\prime}$ are copies of each other iff they are disjoint and there is a bijection, usually clear from the context, mapping $e\in X$ to $e^{\prime}\in X^{\prime}$. When $X,X^{\prime}$ are copies of each other, the vectors $f_{X}$ and $f_{X^{\prime}}$ are said to be copies of each other with $f_{X^{\prime}}(e^{\prime})\equiv f_{X}(e),e\in X.$ The copy ${\cal K}_{X^{\prime}}$ of ${\cal K}_{X}$ is defined by ${\cal K}_{X^{\prime}}\equiv\{f_{X^{\prime}}:f_{X}\in{\cal K}_{X}\}.$

2.1 Complexity convention

For any $g,h:\Re^{m}\rightarrow\Re,$ $g=\tilde{O}(h)$ denotes $g=O(h\log^{c}(h)),$ for some constant $c>0.$

We take the multiplication of two $n\times n$ matrices over a ring $R$ to be $O(n^{\theta})$ ring operations over $R.$ It is known that $\theta\leq 2.373.$

3 Basic operations

The basic operations we use in this paper are as follows:

3.1 Sum and Intersection

Let $\mathcal{K}_{SP}$, $\mathcal{K}_{PQ}$ be collections of vectors on sets $S\cup P,$ $P\cup Q,$ respectively, where $S,P,Q,$ are pairwise disjoint. The sum $\boldsymbol{\mathcal{K}_{SP}+\mathcal{K}_{PQ}}$ of $\mathcal{K}_{SP}$, $\mathcal{K}_{PQ}$ is defined over $S\cup P\cup Q,$ as follows:

[TABLE]

When $S$, $Q,$ are pairwise disjoint, $\mathcal{K}_{S}+\mathcal{K}_{Q}$ is usually written in this paper as $\boldsymbol{\mathcal{K}_{S}\oplus\mathcal{K}_{Q}}$ and is called the direct sum. Thus,

[TABLE]

The intersection $\boldsymbol{\mathcal{K}_{SP}\cap\mathcal{K}_{PQ}}$ of $\mathcal{K}_{SP}$, $\mathcal{K}_{PQ}$ is defined over $S\cup P\cup Q,$ where $S,P,Q,$ are pairwise disjoint, as follows:

[TABLE]

Thus,

[TABLE]

It is immediate from the definition of the sum operation that sum of generalized number lattices is a generalized number lattice. In Subsection 3.3, we show, by using the notion of dualization, that intersection of generalized number lattices is also a generalized number lattice.

3.2 Restriction and contraction

The restriction of $\boldsymbol{\mathcal{K}_{SP}}$ to $S$ is defined by $\boldsymbol{\mathcal{K}_{SP}\circ S}\equiv\{f_{S}:(f_{S},f_{P})\in\mathcal{K}_{SP}\}.$ The contraction of $\boldsymbol{\mathcal{K}_{SP}}$ to $S$ is defined by $\boldsymbol{\mathcal{K}_{SP}\times S}\equiv\{f_{S}:(f_{S},0_{P})\in\mathcal{K}_{SP}\}.$

Here again $\boldsymbol{\mathcal{K}_{SPZ}\circ SP}$, $\boldsymbol{\mathcal{K}_{SPZ}\times SP}$, respectively when $S,P,Z,$ are pairwise disjoint, denote

$\boldsymbol{\mathcal{K}_{SPZ}\circ(S\uplus P)}$, $\boldsymbol{\mathcal{K}_{SPZ}\times(S\uplus P)}.$

3.2.1 Visibility of restriction and contraction in bases of \mbox{\cal V}_{SP}

Let $(B_{S}\vdots B_{P})$ be a representative matrix for vector space \mbox{\cal V}_{SP}. By invertible row operations on $(B_{S}\vdots B_{P}),$ we can obtain a matrix of the form

[TABLE]

where the rows of $(C_{SP})$ form a basis for \mbox{\cal V}_{SP} (therefore the rows of $(C_{1S})$ are linearly independent), and the rows of $(C_{2P})$ are linearly independent. Whenever $(f_{S},f_{P})$ is a vector in \mbox{\cal V}_{SP}, it is clear that $f_{P}$ is linearly dependent on the rows of $(C_{2P}).$ Since these rows are independent, if $(f_{S},0_{P})$ is a vector in \mbox{\cal V}_{SP}, $f_{S}$ must be linearly dependent on the rows of $(C_{1S}).$

We conclude that $(C_{1S})$ is a representative matrix for \mbox{\cal V}_{SP}\times S and that $(C_{2P})$ is a representative matrix for \mbox{\cal V}_{SP}\circ P and say that these latter are visible in the representative matrix $(C_{SP})$ of \mbox{\cal V}_{SP}.

3.2.2 Visibility of restriction and contraction in bases of

$L_{SP}$

The visibility of restriction and contraction in basis matrices of number lattices is similar to the case of vector spaces except that instead of invertible linear operations on the rows, we have to perform ‘integral invertible row operations’. These are integral operations whose inverses are also integral and correspond to premultiplication by integral matrices whose inverses are also integral. Such matrices are said to be unimodular.

Given a generating matrix $(C_{P}),$ for a number lattice $L_{P},$ it is well known that, by integral invertible row operations, we can get a matrix of the form

[TABLE]

in which the rows of $(C_{2P})$ are linearly independent (see Section 4).

Starting from a basis matrix $(C_{S}\vdots C_{P})$ for the number lattice $L_{SP},$ by integral invertible row operations on $(C_{S}\vdots C_{P}),$ we can obtain a basis matrix for $L_{SP},$ of the form

[TABLE]

where the rows of $(C_{2P})$ are linearly independent. Whenever $(f_{S},f_{P})$ is a vector in $L_{SP},$ it is clear that $f_{P}$ is integrally linearly dependent on the rows of $(C_{2P}).$ If $(f_{S},0_{P})$ is a vector in \mbox{\cal V}_{SP}. it can be expressed as an integral linear combination of the rows of $(C_{SP}).$ This integral linear combination cannot involve the rows of $(C_{2S}\vdots C_{2P}),$ since the rows of $(C_{2P})$ are linearly independent. Therefore, $f_{S}$ is integrally linearly dependent on the rows of $(C_{1S}).$ Further, we know that the rows of $(C_{1S})$ are linearly independent.

We conclude that $(C_{1S})$ is a basis matrix for $L_{SP}\times S$ and $(C_{2P})$ is a basis matrix for $L_{SP}\circ P$ and say that these are visible in the basis matrix $(C_{SP})$ of $L_{SP}.$

3.3 Dualization

For any generalized number lattice $\mathcal{K}_{S},$ $\boldsymbol{\mathcal{K}_{S}^{d}}$ is defined by

[TABLE]

We have the following useful characterization of ${\mathcal{K}_{S}^{d}}$ ([18]).

Theorem 1**.**

Let {\cal K}_{S}\equiv\mbox{L}^{(1)}_{S}+\mbox{\cal V}^{(1)}_{S}, where \mbox{L}^{(1)}_{S} is a number lattice and \mbox{\cal V}^{(1)}_{S}, a vector space orthogonal to it. Let $B_{1}$ be a basis matrix for the number lattice \mbox{L}^{(1)}_{S}, let $C_{1},D_{1}$ be representative matrices respctively for the vector spaces \mbox{\cal V}^{(1)}_{S}, (\mbox{L}^{(1)}_{S}+\mbox{\cal V}^{(1)}_{S})^{\perp}.

Let

[TABLE]

Then

Rows of $B_{1},B_{2}$ span the same vector space and $D_{2}$ is the representative matrix for \mbox{\cal V}^{(2)}_{S}\equiv(\mbox{L}^{(1)}_{S}+\mbox{\cal V}^{(1)}_{S})^{\perp}. 2. 2.

${\mathcal{K}_{S}^{d}}$* is the generalized number lattice which is equal to \mbox{L}^{(2)}_{S}+\mbox{\cal V}^{(2)}_{S}, where $B_{2}$ is a basis matrix for the number lattice \mbox{L}^{(2)}_{S}.* 3. 3.

$({\mathcal{K}_{S}^{d}})^{d}={\mathcal{K}}_{S}.$**

Proof.

Part 1 is straightforward.

2. It is clear that {\mathcal{K}_{S}^{d}}\supseteq\mbox{L}^{(2)}_{S}+\mbox{\cal V}^{(2)}_{S}, Let $y_{S}\in{\mathcal{K}_{S}^{d}}.$ Since \langle x_{S},y_{S}\rangle,x_{S}\in\mbox{\cal V}^{(1)}_{S}, is an integer we must have y_{S}\in(\mbox{\cal V}^{(1)}_{S})^{\perp}=span(\mbox{L}^{(2)}_{S}+\mbox{\cal V}^{(2)}_{S}). We can therefore write y_{S}=y^{1}_{S}+y^{2}_{S},y^{1}_{S}\in span(\mbox{L}^{(2)}_{S}),y^{2}_{S}\in span(\mbox{\cal V}^{(2)}_{S})=(\mbox{L}^{(1)}_{S}+\mbox{\cal V}^{(1)}_{S})^{\perp}. By part 1, rows of $B_{2}$ generate span(\mbox{L}^{(1)}_{S}). Let $(y^{1}_{S})^{T}=\lambda^{T}B_{2}.$ We will show that $\lambda^{T}$ is integral. If x^{1}_{S}\in\mbox{L}^{(1)}_{S},\mbox{\ i.e,\ }x^{1}_{S}=(B_{1})^{T}\mu,\mu\ \mbox{integral}, then since $\langle y^{1}_{S}+y^{2}_{S},x^{1}_{S}\rangle=\langle y^{1}_{S},x^{1}_{S}\rangle$ must be an integer, we must have $\langle y^{1}_{S},x^{1}_{S}\rangle=\lambda^{T}B_{2}(B_{1})^{T}\mu=\lambda^{T}(I)\mu=\lambda^{T}\mu,$ integral for arbitrary $\mu.$ This is possible only if $\lambda^{T}$ is integral. Thus y_{S}\in L^{(2)}_{S}+\mbox{\cal V}^{(2)}_{S}. We conclude that {\mathcal{K}_{S}^{d}}=\mbox{L}^{(2)}_{S}+\mbox{\cal V}^{(2)}_{S}.

3. This is immediate from part 2 above. ∎

The following corollary is immediate

Corollary 2**.**

When ${\cal K}_{S}$ is a full dimensional number lattice (i.e., dimension = $|S|$), $\mathcal{K}_{S}^{d}$ is also a full dimensional number lattice. 2. 2.

When ${\cal K}_{S}$ is a vector space, $\mathcal{K}_{S}^{d}$ is the space ${\cal K}_{S}^{\perp}$ complementary orthogonal to it.

The following result is easy to see.

If ${\cal K}_{S},\widehat{{\cal K}}_{S}$ are generalized number lattices,

[TABLE]

Using $({\cal K}^{d})^{d}={\cal K},$ we get

[TABLE]

We will show that this result is true even if the generalized number lattices are defined on different sets with appropriate modification of the dualization operation.

When $S$, $P$ are disjoint, and ${\cal K}_{S},{\cal K}_{P}$ are generalized number lattices, it is easily verified that

[TABLE]

(Here we abuse notation for better readability, The ‘$d$’ on the left hand side is with respect to $S\uplus P$ while the ones on the right hand side are with respect to $S$ and $P,$ respectively.)

When $S$, $P$ are not disjoint, ${\cal K}_{S}+{\cal K}_{P}\equiv({\cal K}_{S}\oplus{\mathbf{0}}_{P\setminus S})+({\cal K}_{P}\oplus{\mathbf{0}}_{S\setminus P})$. If ${\cal K}_{S},{\cal K}_{P}$ are generalized number lattices on $S,P,$ respectively, we have

[TABLE]

by the definition of intersection of generalized number lattices on two distinct sets.

Using $({\cal K}^{d})^{d}={\cal K}$ for generalized number lattices, we have that ${\cal K}_{S}\cap{\cal K}_{P}=({\cal K}_{S}^{d}+{\cal K}_{P}^{d})^{d},$ is a generalized number lattice when ${\cal K}_{S},{\cal K}_{P}$ are generalized number lattices.

The following results for generalized number lattices can also be easily verified:

[TABLE]

The above pair of results will be referred to as the dot-cross duality.

When ${\cal K}_{SP}$ is a generalized number lattice, it is immediate from the definition that so is ${\cal K}_{SP}\circ S.$ It follows that ${\cal K}_{SP}\times S=({\cal K}^{d}_{SP}\circ S)^{d}$ is also a generalized number lattice.

4 Constructing a basis from a generating set

A convenient basis for a number lattice, for many purposes, is the one in Hermite Normal Form (HNF). This is unique for a given number lattice. Given an integral generating matrix of a number lattice, one can construct an HNF basis for the latter, by building the HNF of the generating matrix. When the matrix is rational, but not integral, one multiplies all the entries by an integer $k,$ so that they become integers, builds the HNF of the resulting integral matrix, and divides all the entries by $k.$ We will call the resulting matrix, which is unique for the given rational generating matrix, its HNF.

4.1 Hermite Normal Form

The Hermite Normal form (HNF) of an integral matrix is the number lattice analogue of the row reduced echelon form for matrices over ${\mathbb{Q}}.$ Our definition is row based. The column based HNF can be defined similarly. We remind the reader that a square matrix is said to be unimodular iff it has integral entries and has determinant equal to one. It is clear that the inverse of a unimodular matrix is unimodular and that product of unimodular matrices is also unimodular. We say that two matrices are integrally row equivalent iff each can be obtained from the other by integral row operations. When the matrices have the same number of rows, we say that each can be obtained from the other by integral invertible operations iff each can be obtained from the other by premultiplication by a unimodular matrix. HNF is defined as follows:

Definition 3**.**

*An integral matrix of full column rank is said to be in the Hermite Normal Form iff it has the form \left[\begin{array}[]{lll}\mbox{{\bf B}}\\ {\bf 0}\end{array}\right] where ${\bf B}$ satisfies the following:

i. it is an upper triangular, integral, nonnegative matrix;

ii. its diagonal entries are positive and have the unique highest magnitude in their columns.*

When a matrix $K$ has dependent columns, one first picks the sequence of columns $c_{i_{1}},\cdots,c_{i_{k}},$ such that, scanning from the left, $c_{i_{1}}$ is the first nonzero column, $c_{i_{j}}$ is independent of all columns occurring before it and all columns occurring after $c_{i_{k}}$ are dependent on $c_{i_{1}},\cdots,c_{i_{k}}.$ We will call such a column basis lexicographically earliest.

Let the submatrix composed of these columns be $M$ and let $T$ be a unimodular matrix such that $TM$ is in HNF form. Then $TK$ is said to be the HNF for $K.$ Column $c_{j}$ will have no nonzero entries in rows after $r$ if $j<i_{r+1}.$

The HNF of a matrix can be seen to be unique for a given matrix, by observing that once a matrix is in the HNF form, it cannot be put in another such form by integral invertible row operations. If two matrices with independent rows are integrally row equivalent, it is clear that they have the same HNF matrix. Therefore all basis matrices of a number lattice have the same HNF.

A naive algorithm for constructing the HNF of a matrix is as follows. (It is naive because it does not guarantee that numbers encountered during intermediate states of the algorithm do not grow exponentially large in terms of the size of the matrix ([8]).)

Let $c_{i_{1}},\cdots,c_{i_{k}},$ be the lexicographically earliest column basis for the matrix. Let us suppose columns $c_{i_{1}},\cdots,c_{i_{r}},r<k,$ of this matrix satisfy properties $(i)$ and $(ii)$ above of HNF matrices. Note that the submatrix of this matrix composed of columns $t<i_{r+1}$ has rows $j>r$ as zero rows.

Let $c^{\prime}_{i_{r+1}},$ denote the column vector composed of the entries $(j,i_{r+1}),j\geq r+1,$ of the column $c_{i_{r+1}}.$ Perform integral invertible operations on the rows $j\geq r+1$ of the matrix to bring the gcd, say $d,$ of the entries of $c^{\prime}_{i_{r+1}}$ to the $(r+1,i_{r+1})$ position, all other entries being zero. Subtract integral multiples of the present $(r+1)^{th}$ row of the matrix from earlier rows so that entries $(j,i_{r+1}),j<r+1,$ are all less than $d.$ (Note that these operations do not disturb the submatrix composed of columns $t<i_{r+1}.$) This completes the processing of column $c_{i_{r+1}},$ of the matrix. It can be seen that $c_{i_{1}},\cdots,c_{i_{r+1}}$ of the resulting matrix satisfy properties $(i)$ and $(ii)$ above of HNF matrices.

The present fastest algorithm for computing the HNF $H$ of an $m\times n,$ rank $n$ integral matrix $A,$ appears to be the one in [25]. This algorithm uses the ideas in [7] and has complexity $\tilde{O}(mn^{\theta}log(max(|A_{ij}|))$ time.

The algorithm also produces, in addition to the HNF $H,$ a unimodular matrix $R,$ such that $H=RA.$ The matrix $R$ has entries of bit size $\tilde{O}(nlog(max(|A_{ij}|))).$ We will call this the SL-algorithm for HNF.

Next, let us consider the case of a number lattice that is not full dimensional. Let $(C_{1S}|C_{1P})$ be an integral matrix with linearly independent rows and maximal independent columns corresponding to set $S.$ If now we have to find a basis for the number lattice $L_{SP}$ generated by rows of the $(m\times n)$ matrix

[TABLE]

where the second set of rows are linearly dependent upon the first, we can use the SL-algorithm on the set of columns $S$ and obtain the $(m\times m)$ unimodular matrix $R$ in $\tilde{O}(m|S|^{\theta}log(\kappa))$ time, where $\kappa=max(|C_{SP}(i,j)|).$ The bit size of entries in $R$ will be $\tilde{O}(|S|\times log(\kappa)).$ We can premultiply the matrix $C_{SP}$ by $R$ to obtain the matrix

[TABLE]

where the first set of rows of the matrix constitute a basis matrix for the number lattice $L_{SP}.$ Multiplication of the set of columns of $C_{SP}$ by $R$ can be carried out in $\tilde{O}(m|S|^{\theta}log(\kappa))+\tilde{O}(m^{2}|P|\times|S|log(\kappa))$ time.

In Subsection 3.2.2 we have discussed the usefulness of putting, by integral invertible row operations, a basis matrix $(C_{S}\vdots C_{P})$ for the number lattice $L_{SP}$ into the form

[TABLE]

We have shown there that, if the rows of $(C_{2P})$ are linearly independent, then rows of $(C_{1S}),(C_{2P})$ respectively form bases for the number lattices $L_{SP}\times S$ and $L_{SP}\circ P.$ By using the SL-algorithm, the basis of the form in Equation 3 can be computed in $\tilde{O}(m|P|^{\theta}log(\kappa))$

• $\tilde{O}(m^{2}n|P|log(\kappa))$ time, where $(C_{S}\vdots C_{P})$ is an $m\times n$ matrix and $log(\kappa)$ is the maximum bit size of entries in $C_{SP}.$ The second term in the complexity calculation is that of the multiplication $RC_{SP}.$ Here, $R$ has bit size of entries $\tilde{O}(|P|log(\kappa)).$

A contains a discussion of a few common variations on the problem of finding a basis of a number lattice.

4.1.1 Connectedness of number lattices through HNF

A number lattice $L_{S}$ can sometimes be regarded as the direct sum $\bigoplus_{i}L_{S_{i}},i=1,2,\cdots,k.$ of number lattices defined over the blocks of a partition $\{S_{1},\cdots,S_{k}\}$ of $S.$ This would clearly make computations with the lattice much easier. In this section, we discuss how to recognize this situation through the use of the Hermite normal form of a basis matrix for $L_{S}.$

From the definition of contraction and restriction it is clear that if $L_{S}=\bigoplus_{i}L_{S_{i}},i=1,2,\cdots,k,$ we must have $L_{S}\times S_{i}=L_{S}\circ S_{i},i=1,\cdots k.$

We now prove the converse. Suppose $L_{S}=L_{S_{1}}\oplus L_{S_{2}}.$ Let us suppose, wlog, that the elements of $S$ are ordered such that those in $S_{1}$ occur before those in $S_{2}.$ It is clear by the discussion in Subsection 3.2.2, that we can build a basis matrix for $L_{S}=L_{S_{1}S_{2}},$ which has the form

[TABLE]

with rows of $(C_{2S_{2}})$ linearly independent. We know that $(C_{1S_{1}})$ is a basis matrix for $L_{S}\times S_{1}=L_{S}\circ S_{1}.$ Therefore by integral invertible row operations, we can reduce $C_{S_{1}S_{2}}$ to the form

[TABLE]

Thus $L_{S}=L_{S}\times S_{1}\oplus L_{S}\times S_{2}=L_{S}\circ S_{1}\oplus L_{S}\circ S_{2}.$ The result follows by induction on $k.$

In general, we will be working with a column permuted version of the matrix in Equation 4. Therefore it is clear that $L_{S}=L_{S_{1}}\oplus\cdots\oplus L_{S_{m}}$ iff it has a basis matrix where no row contains nonzero entries from columns corresponding to more than one of the $S_{i}.$ We will say that such a basis matrix has sets of columns $S_{i},i=1,\cdots m,$ decoupled.

We will now show that it is adequate for us to check if the HNF of any basis matrix of $L_{S}$ has this property.

Let $c_{i_{1}},\cdots,c_{i_{k}},$ be the lexicographically earliest column basis for a basis matrix of $L_{S},$ which has sets of columns $S_{i},\ i=1,\cdots,m,$ decoupled.

Suppose, in the conversion to HNF, as we scan the columns of the lexicographically earliest column basis from the left, we have processed columns up to and including $c_{i_{j}}.$ Let the resulting matrix have sets of columns $S_{i},i=1,\cdots m,$ decoupled. We will show that this situation continues when we process $c_{i_{j+1}}$ also.

Let $r_{z}$ denote the $z^{th}$ row of the matrix at the current stage. Let $c_{i_{j+1}}$ correspond to $e_{i_{j+1}}\in S_{h}$ and let $e_{p}\in S_{q},\ h\neq q.$

If column $c_{i_{j+1}},$ has a zero entry in $r_{t},$ or in $r_{s},$ we would not be adding a multiple of one of them to the other. Therefore we need only consider the case where the column has both the entries nonzero.

But in this case, the column $c_{p}$ corresponding to $e_{p}$ must have zero entry in both rows. So when we add an integral mutiple of say $r_{t}$ to $r_{s}$ the column $c_{p}$ will continue to have zero entries in both rows. So sets of columns $S_{i},i=1,\cdots m,$ remain decoupled after this operation too.

Since processing column $c_{i_{j+1}}$ is made up of only such elementary operations we conclude sets of columns $S_{i},i=1,\cdots m,$ remain decoupled even after this processing is complete. It is therefore clear that during the conversion of a basis matrix of $L_{S}$ which has sets of columns $S_{i},i=1,\cdots m,$ decoupled, into the HNF, the matrices encountered will all have $S_{i},i=1,\cdots m,$ decoupled and this would be true for the HNF too.

But the HNF of all basis matrices of a number lattice has to be the same. Therefore if some basis matrix of $L_{S}$ has sets of columns $S_{i},i=1,\cdots m,$ decoupled, so will be the HNF basis matrix of $L_{S}.$

We say $S_{1}\subseteq S$ is a separator for $L_{S},$ iff $L_{S}=L_{S_{1}}\oplus L_{S-S_{1}}.$ A separator of $L_{S}$ is elementary iff it does not contain another separator of $L_{S}$ as a proper subset. The following simple algorithm detects the elementary separators of $L_{S}$ from its HNF.

Build the graph ${\cal G}_{L_{S}}$ from the HNF of $L_{S}$ as follows. Let $S\equiv\{e_{1},\cdots,e_{n}\}.$ Take $S$ to be the set of nodes of ${\cal G}_{L_{S}}.$ Join $e_{i},e_{j}$ by an edge if there is a row in the HNF where both columns have nonzero entries. Let $S_{i},\ i=1,\cdots,m$ be the node sets of the connected components of ${\cal G}_{L_{S}}.$ Then $S_{i},\ i=1,\cdots,m$ are the elementary separators of $L_{S}.$

5 Linking generalized number lattices

In this section, we introduce two basic operations, viz., matched and skewed composition, for ‘linking’ generalized number lattices. We prove two basic results ‘implicit inversion (IIT)’ and ‘implicit duality (IDT)’ (Theorems 8 and 11) involving these operations. From IIT, we are able to show that number lattice parts of linked generalized number lattices have similar properties. From IDT, among other things, we develop a technique for building new self dual lattices from old.

5.1 Matched and Skewed Composition

Let ${\cal K}_{SP},{\cal K}_{PQ},$ be collections of vectors respectively on $S\uplus P,P\uplus Q,$ with $S,P,Q,$ being pairwise disjoint. Further, let ${\mathbf{0}}_{SP}\in{\cal K}_{SP},{\mathbf{0}}_{PQ}\in{\cal K}_{PQ}.$

The matched composition $\boldsymbol{\mathcal{K}_{SP}\leftrightarrow\mathcal{K}_{PQ}}$ is on $S\uplus Q$ and is defined as follows:

[TABLE]

Matched composition is referred to as matched sum in [18].

The skewed composition $\boldsymbol{\mathcal{K}_{SP}\leftrightarrow\mathcal{K}_{PQ}}$ is on $S\uplus Q$ and is defined as follows:

[TABLE]

When $S$, $Y$ are disjoint, both the matched and skewed composition of ${\cal K}_{S},{\cal K}_{Y},$ correspond to the direct sum ${\cal K}_{S}\oplus{\cal K}_{Y}$.

It is clear from the definition of matched composition and that of restriction and contraction, that

[TABLE]

When $\mathcal{K}_{SP}$, $\mathcal{K}_{P}$ are generalized number lattices, observe that $(\mathcal{K}_{SP}\leftrightarrow\mathcal{K}_{P})=(\mathcal{K}_{SP}\rightleftharpoons\mathcal{K}_{P}).$

When $S,P,Z,$ are pairwise disjoint, we have

[TABLE]

When $\mathcal{K}_{S}\equiv{\mathbf{0}}_{S},\mathcal{K}_{P}\equiv\mathcal{K}_{SPZ}\circ P,$ the above reduces to

[TABLE]

Such an object is called a minor of $\mathcal{K}_{SPZ}.$

In the special case where $Y\subseteq S$, the matched composition ${\cal K}_{S}\leftrightarrow{\cal K}_{Y},$ is called the generalized minor of $\mathcal{K}_{S}$ with respect to $\mathcal{K}_{Y}$.

We have already seen (Subsection 3.3) that sum, intersection, restriction and contraction of generalized number lattices are also generalized number lattices. We therefore have the following result.

Theorem 4**.**

Let ${\cal K}_{SP},{\cal K}_{PQ}$ be generalized number lattices with $S,P,Q$ disjoint. From the definition of matched composition, we then have

[TABLE]

and also equal to

[TABLE]

Similarly from the definition of skewed composition,

[TABLE]

and also equal to

[TABLE] 2. 2.

If ${\cal K}_{SP},{\cal K}_{PQ}$ are generalized number lattices with $S,P,Q$ disjoint, then ${\cal K}_{SP}\leftrightarrow{\cal K}_{PQ},{\cal K}_{SP}\rightleftharpoons{\cal K}_{PQ},$ are also generalized number lattices. 3. 3.

Let \mbox{\cal E}({\cal K}_{S},\cdots,{\cal K}_{Q}) be an expression involving generalized number lattices,

and the operations $+,\cap,\leftrightarrow,\rightleftharpoons.$ Then the expression yields a generalized number lattice.

5.2 Vector spaces and number lattices

We have seen in Theorem 4, that matched and skewed composition of generalized number lattices yield other generalized number lattices. When ${\cal K}_{S}={\cal K}_{SP}\leftrightarrow{\cal K}_{P},$ we say ${\cal K}_{S},{\cal K}_{P}$ are linked through ${\cal K}_{SP}.$ Below, we prove two basic results related to linking, viz., ‘implicit inversion theorem’ and ‘implicit duality theorem’. These enable us to relate properties of generalized number lattices that are linked, under certain relatively weak conditions, in the case of the former and, very generally, in the case of the latter. The most natural situation is when linking is done through vector spaces although the theorems are valid more generally.

Vector spaces are useful for linking number lattices as the following result indicates.

Lemma 5**.**

Let rows of $(B_{S}\vdots B_{P}),(C_{P}\vdots C_{Q})$ be bases for vector space \mbox{\cal V}_{SP} and number lattice $L_{PQ}$ respectively.

*Let $B_{S}=B_{P}N,$ for some matrix $N,$ equivalently, let ${\cal V}_{SP}\times S={\mathbf{0}}_{S}.$ Then ${\cal V}_{SP}\leftrightarrow{L}_{PQ}$ is a number lattice on $S\uplus Q.$ * 2. 2.

Let $B_{S}=B_{P}N$ and let every row of $C_{P}$ be a linear combination of rows of $B_{P},$ equivalently, let ${L}_{PQ}\circ P\subseteq{\cal V}_{SP}\circ P.$ Then $(C_{P}N\vdots C_{Q})$ is a basis matrix for the number lattice ${\cal V}_{SP}\leftrightarrow{L}_{PQ}.$

Proof.

1. Let $(f_{S},g_{Q})\in{\cal V}_{SP}\leftrightarrow{L}_{PQ}.$ Then there exist $(f_{S},h_{P})\in{\cal V}_{SP}$ and $(h_{P},g_{Q})\in{L}_{PQ}.$ Therefore $(f_{S},h_{P})=\sigma^{T}(B_{S}\vdots B_{P})=\sigma^{T}(B_{P}N\vdots B_{P})$ for some rational vector $\sigma$ and $(h_{P},g_{Q})=\lambda^{T}(C_{P}\vdots C_{Q})$ for some integral vector $\lambda.$ We thus have $\sigma^{T}B_{P}=\lambda^{T}C_{P}$ so that $\sigma^{T}B_{P}N=\lambda^{T}C_{P}N.$ Therefore $(f_{S},g_{Q})=\lambda^{T}(C_{P}N\vdots C_{Q}).$ Thus the rows of $(C_{P}N\vdots C_{Q})$ generate all vectors in ${\cal V}_{SP}\leftrightarrow{L}_{PQ}$ through integral linear combination. Therefore ${\cal V}_{SP}\leftrightarrow{L}_{PQ}$ does not contain a nontrivial vector subspace. On the other hand ${\cal V}_{SP}\leftrightarrow{L}_{PQ}$ is a generalized number lattice. Since it does not contain a nontrivial vector subspace we conclude that it is a number lattice.

2. We saw in the proof of part 1, that the number lattice generated by the rows of $(C_{P}N\vdots C_{Q})$ contains the number lattice ${\cal V}_{SP}\leftrightarrow{L}_{PQ}.$ Now let $(f_{S},g_{Q})=\lambda^{T}(C_{P}N\vdots C_{Q}),$ where $\lambda^{T}$ is an integral vector and let $h_{P}\equiv\lambda^{T}C_{P}$ so that $(h_{P},g_{Q})=\lambda^{T}(C_{P}\vdots C_{Q}).$ By the hypothesis, there exists a rational vector $\sigma$ such that $\sigma^{T}B_{P}=\lambda^{T}C_{P}.$

Now $\sigma^{T}(B_{S}\vdots B_{P})=\sigma^{T}(B_{P}N\vdots B_{P})=\lambda^{T}(C_{P}N\vdots C_{P})=(f_{S},h_{P}).$ Hence, $(f_{S},h_{P})\in{\cal V}_{SP},(h_{P},g_{Q})\in{L}_{PQ}$ so that $(f_{S},g_{Q})\in{\cal V}_{SP}\leftrightarrow{L}_{PQ}.$ ∎

Remark 1**.**

*Lemma 5, is intended to motivate the notion of linking as well as the definition of generalized number lattice. Its hypothesis is strong and the conclusion, not surprising. However, if we allow the result of the linking to be a generalized number lattice, by using the implicit inversion theorem (IIT) (Theorem 8), we can obtain Theorem 9, where with weaker hypothesis, we obtain a result of greater applicability. *

Example 6**.**

Let $Z\equiv S\uplus P$ and let \mbox{\cal V}_{Z} be a vector space on $Z.$ Let $Z\equiv\{e_{1},\cdots e_{m}\}$ and let $Z^{\prime}\equiv\{e^{\prime}_{1},\cdots e^{\prime}_{m}\}$ be a disjoint copy of $Z.$ Let \mbox{\cal V}_{Z} be a vector space on $Z$ and \mbox{\cal V}_{Z^{\prime}} be a copy on $Z^{\prime}.$ The simplest such vector space is the coboundary space spanned by the incidence matrix of a graph ${{\cal G}},$ with directed edges $Z.$ Let \mbox{\cal V}_{ZZ^{\prime}}\equiv\mbox{\cal V}_{Z}\oplus\mbox{\cal V}^{\perp}_{Z^{\prime}}.

Let $(Q_{S}|Q_{P})$ be a representative matrix for \mbox{\cal V}_{Z}, and let $(B_{S}|B_{P})$ be a representative matrix for \mbox{\cal V}^{\perp}_{Z}. Let a maximal linearly independent set of columns of $(Q_{S})$ be also a maximal linearly independent set of columns of $(Q_{S}|Q_{P}).$ Further let the columns of $(Q_{P})$ be linearly independent. It can be shown that a maximal linearly independent set of columns of $(B_{S})$ will also be a maximal linearly independent set of columns of $(B_{S}|B_{P})$ and the columns of $(B_{P})$ will be linearly independent. If \mbox{\cal V}_{Z} is the coboundary space of graph ${\cal G},$ the above conditions mean that the edges in $P$ can be included in a tree as well as in the complement of a tree (i.e., $P$ contains no loops or cutsets of ${\cal G}$).

A consequence is that there will exist matrices $M,N$ such that $(Q_{P})=(Q_{S})M$ and $(B_{P})=(B_{S})N.$ Therefore \mbox{\cal V}_{Z}\times P={\mathbf{0}}_{P} and \mbox{\cal V}^{\perp}_{Z}\times P={\mathbf{0}}_{P}, i.e., \mbox{\cal V}^{\perp}_{Z^{\prime}}\times P^{\prime}={\mathbf{0}}_{P^{\prime}}, so that \mbox{\cal V}_{ZZ^{\prime}}\times PP^{\prime}={\mathbf{0}}_{PP^{\prime}}.

If \mbox{L}_{SS^{\prime}} is a number lattice on $S\uplus S^{\prime},$ by Theorem 5 , \mbox{\cal V}_{ZZ^{\prime}}\leftrightarrow\mbox{L}_{SS^{\prime}} is a number lattice on $P\uplus P^{\prime}.$

5.2.1 Implicit Inversion

Consider the equation

[TABLE]

In this subsection we examine when, given ${\cal K}_{SP},{\cal K}_{SQ},$ the equation has some solution ${\cal K}_{PQ},$ and when the solution is unique.

The following lemma is a generalization of a result (Problem 7.5) in [18]. The proof is relegated to the appendix.

Lemma 7**.**

1. Let \mbox{{\cal K}{SP}},\mbox{{\cal K}{SQ}} be collections of vectors on $S\uplus P,S\uplus Q$ respectively. Then there exists a collection of vectors ${\cal K}_{PQ}$ on $P\uplus Q$ s.t. {\bf 0}_{PQ}\in\mbox{{\cal K}{PQ}} and \mbox{{\cal K}{SP}}\leftrightarrow\mbox{{\cal K}{PQ}}=\mbox{{\cal K}{SQ}}, only if

[TABLE]

*2. Let ${\cal K}_{SP}$ be a collection of vectors closed under subtraction on $S\uplus P$ and let ${\cal K}_{SQ}$ be a collection of vectors on $S\uplus Q,$ closed under addition. Further let \mbox{{\cal K}{SP}}\circ S\supseteq\mbox{{\cal K}{SQ}}\circ S and let \mbox{{\cal K}{SP}}\times S\subseteq\mbox{{\cal K}{SQ}}\times S. Then, {\mathbf{0}}_{SQ}\in\mbox{{\cal K}{SQ}}, and the collection of vectors \mbox{{\cal K}{SP}}\leftrightarrow\mbox{{\cal K}{SQ}}, is closed under addition with ${\bf 0}_{PQ}$ as a member. Further we have that \mbox{{\cal K}{SP}}\leftrightarrow(\mbox{{\cal K}{SP}}\leftrightarrow\mbox{{\cal K}{SQ}})=\mbox{{\cal K}_{SQ}}.

3. Let ${\cal K}_{SP}$ be a collection of vectors closed under subtraction, and let ${\cal K}_{SQ}$ satisfy the conditions, closure under addition, \mbox{{\cal K}{SP}}\circ S\supseteq\mbox{{\cal K}{SQ}}\circ S and \mbox{{\cal K}{SP}}\times S\subseteq\mbox{{\cal K}{SQ}}\times S. Then the equation*

[TABLE]

where ${\cal K}_{PQ}$ has to satisfy closure under addition, as well as the conditions

[TABLE]

*has \hat{{\cal K}}_{PQ}\equiv\mbox{{\cal K}{SP}}\leftrightarrow\mbox{{\cal K}{SQ}}, as the unique solution. *

Remark 2**.**

We note that the collections of vectors in Lemma 7 can be over rings rather than over fields - in particular over the ring of integers.

• 2.

The hypotheses of Lemma 7 are clearly true when \mbox{{\cal K}{SP}},\mbox{{\cal K}{PQ}},\mbox{{\cal K}_{SQ}} are generalized number lattices. So the lemma holds for them.

• 3.

*Only ${\cal K}_{SP}$ has to be closed over subtraction. The other two collections \mbox{{\cal K}{PQ}},\mbox{{\cal K}{SQ}} have to be only closed over addition with a zero vector as a member. In particular ${\cal K}_{SP}$ could be a vector space over rationals while \mbox{{\cal K}{PQ}},\mbox{{\cal K}{SQ}} could be cones over rationals. *

We specialize Lemma 7 to the case of generalized number lattices in the following theorem. We will call this the implicit inversion theorem(IIT) for generalized number lattices.

Theorem 8**.**

Consider the equation

[TABLE]

*where ${\cal K}_{SP},{\cal K}_{PQ},{\cal K}_{SQ}$ are generalized number lattices.Then *

given ${\cal K}_{SP},{\cal K}_{SQ},\exists{\cal K}_{PQ},$ satisfying the equation iff ${\cal K}_{SP}\circ S\supseteq{\cal K}_{SQ}\circ S$ and ${\cal K}_{SP}\times S\subseteq{\cal K}_{SQ}\times S.$ 2. 2.

given ${\cal K}_{SP},{\cal K}_{PQ},{\cal K}_{SQ}$ satisfying the equation, we have ${\cal K}_{SP}\leftrightarrow{\cal K}_{SQ}={\cal K}_{PQ}$ iff ${\cal K}_{SP}\circ P\supseteq{\cal K}_{PQ}\circ P$ and ${\cal K}_{SP}\times P\subseteq{\cal K}_{PQ}\times P.$ 3. 3.

given ${\cal K}_{SP},{\cal K}_{SQ},$ assuming that the equation ${\cal K}_{SP}\leftrightarrow{\cal K}_{PQ}={\cal K}_{SQ}$ is satisfied by some $\hat{{\cal K}}_{PQ}$ it is unique under the condition ${\cal K}_{SP}\circ P\supseteq{\cal K}_{PQ}\circ P$ and ${\cal K}_{SP}\times P\subseteq{\cal K}_{PQ}\times P.$

The following theorem, which relates number lattice parts of linked generalized lattices, is a consequence.

Theorem 9**.**

Let ${\cal V}_{SP}$ be a vector space, ${\cal K}_{P}$ be a generalized number lattice with ${\cal K}_{P}={\cal V}_{P}+L_{P},{\cal V}_{P}\cap L_{P}={\mathbf{0}}_{P}.$ Let ${\cal K}_{S}\equiv{\cal V}_{SP}\leftrightarrow{\cal K}_{P},$ and let ${\cal V}_{S}\equiv{\cal V}_{SP}\leftrightarrow{\cal V}_{P}.$ Let ${\cal V}_{SP}\times P\subseteq{\cal V}_{P}$ and let ${\cal V}_{SP}\circ P\supseteq{\cal K}_{P}.$ Then

${\cal V}_{SP}\leftrightarrow{\cal K}_{S}={\cal K}_{P};$* ${\cal V}_{SP}\leftrightarrow{\cal V}_{S}={\cal V}_{P}$ ;* 2. 2.

Let $B_{P}$ be a basis matrix for $L_{P}$ with rows $x_{1P},\cdots,x_{nP}.$ Then there exist vectors $\hat{x}_{iS},i=1,\cdots,n,$ such that $(\hat{x}_{iS},x_{iP})\in{\cal V}_{SP},i=1,\cdots,n.$

Let $B_{S}$ be the matrix with $\hat{x}_{iS},i=1,\cdots,n,$ as the $i^{th}$ row. Then $B_{S}$ has independent rows and is a basis matrix for a number lattice $L_{S}\subseteq{\cal V}_{SP}\leftrightarrow L_{P},$ such that $L_{P}\subseteq{\cal V}_{SP}\leftrightarrow L_{S},{\cal V}_{S}\cap L_{S}={\mathbf{0}}_{S},$ and such that ${\cal K}_{S}={\cal V}_{S}+L_{S}.$ 3. 3.

Let $L^{\prime}_{S}\subseteq{\cal V}_{SP}\leftrightarrow L_{P}$ be such that $L^{\prime}_{S}\cap{\cal V}_{S}={\mathbf{0}}_{S}$ and such that $L_{P}\subseteq{\cal V}_{SP}\leftrightarrow L^{\prime}_{S}.$

(a) If $x_{P}\in L_{P},$ and $x_{P}\neq 0_{P},$ then there is a unique $\hat{x}_{S}\in L^{\prime}_{S},$ such that $(\hat{x}_{S},x_{P})\in{\cal V}_{SP}.$ Further, $\hat{x}_{S}\neq 0_{S}.$

(b) Let $B_{P}$ be a basis matrix for $L_{P}$ with rows $x_{1P},\cdots,x_{nP}.$ Let $\hat{x}_{iS}\in L^{\prime}_{S},i=1,\cdots,n,$ be such that $(\hat{x}_{iS},x_{iP})\in{\cal V}_{SP},i=1,\cdots,n.$ Let $B_{S}$ be the matrix with $\hat{x}_{iS},i=1,\cdots,n,$ as the $i^{th}$ row. Then $B_{S}$ is a basis matrix for $L^{\prime}_{S}.$

(c) ${\cal K}_{S}=L^{\prime}_{S}+{\cal V}_{S}.$

Proof.

1. We have ${\cal V}_{SP}\times P\subseteq{\cal V}_{P}\subseteq{\cal K}_{P}$ and ${\cal V}_{SP}\circ P\supseteq{\cal K}_{P}\supseteq{\cal V}_{P}.$ Therefore, by Theorem 8,

[TABLE]

We have, $L_{P}\subseteq{\cal K}_{P}\subseteq{\cal V}_{SP}\circ P.$ Therefore, if $x_{ip}\in L_{P},i=1,\cdots,n,$ there exist $\hat{x}_{iS},i=1,\cdots,n,$ such that $(\hat{x}_{iS},x_{ip})\in{\cal V}_{SP}.$

Let $L_{S}$ be the number lattice generated by $\hat{x}_{iS},i=1,\cdots,n.$ By the definition of the $\leftrightarrow$ operation, $L_{S}\subseteq{\cal V}_{SP}\leftrightarrow L_{P}$ and since we also have $\hat{x}_{iP},i=1,\cdots,n,$ as basis vectors for $L_{P},$ we must have $L_{P}\subseteq{\cal V}_{SP}\leftrightarrow L_{S}.$ Suppose a nontrivial linear combination of the vectors $\hat{x}_{iS}$ lies in ${\cal V}_{S}.$ Then, since ${\cal V}_{SP}\leftrightarrow{\cal V}_{S}={\cal V}_{P},$ the same linear combination of $\hat{x}_{iP},i=1,\cdots,n,$ would lie in ${\cal V}_{P}.$ But ${\cal V}_{P}\cap L_{P}={\mathbf{0}}_{P}$ so that the nontrivial linear combination yields a zero vector, a contradiction. Therefore, $L_{S}\cap{\cal V}_{S}={\mathbf{0}}_{S}$ and $\hat{x}_{iS},i=1,\cdots,n,$ are basis vectors for $L_{S}.$

It is clear that ${\cal K}_{S}\supseteq L_{S}+{\cal V}_{S}.$ To see the reverse containment, let $\tilde{x}_{S}\in{\cal K}_{S}.$

There exists $\tilde{x}_{P}\in{\cal V}_{SP}\leftrightarrow{\cal K}_{S}={\cal K}_{P},$ such that $(\tilde{x}_{S},\tilde{x}_{P})\in{\cal V}_{SP}.$ Since $\tilde{x}_{P}\in{\cal K}_{P}={\cal V}_{P}+L_{P},$ there exist $x^{\prime}_{P},x"_{P},$ such that $\tilde{x}_{P}=x^{\prime}_{P}+x"_{P},x^{\prime}_{P}\in{\cal V}_{P},x"_{P}\in L_{P}.$ Let $x^{\prime}_{S},x"_{S},$ be such that $(x^{\prime}_{S},x^{\prime}_{P}),(x"_{S},x"_{P})\in{\cal V}_{SP}.$ We know that there exists $\hat{x}_{S}\in L_{S},$ such that $(\hat{x}_{S},x"_{P})\in{\cal V}_{SP}.$

Therefore, we must have $(\hat{x}_{S}-x"_{S})\in{\cal V}_{SP}\times S\subseteq{\cal V}_{SP}\leftrightarrow{\cal V}_{P}={\cal V}_{S},$ so that, $x"_{S}\in L_{S}+{\cal V}_{S}.$

Further, $x^{\prime}_{S}\in{\cal V}_{SP}\leftrightarrow{\cal V}_{P}={\cal V}_{S}.$ Therefore $x^{\prime}_{S}+x"_{S}\in L_{S}+{\cal V}_{S}.$

We have that $(x^{\prime}_{S}+x"_{S},\tilde{x}_{P}),(\tilde{x}_{S},\tilde{x}_{P})\in{\cal V}_{SP},$ so that $\tilde{x}_{S}\in(x^{\prime}_{S}+x"_{S})+{\cal V}_{SP}\times S,$

i.e., $\tilde{x}_{S}\in L_{S}+{\cal V}_{S}+{\cal V}_{SP}\times S=L_{S}+{\cal V}_{S}.$

(a) Let $x_{P}\in L_{P},$ with $x_{P}\neq 0_{P}.$ Since ${\cal V}_{SP}\leftrightarrow L^{\prime}_{S}\supseteq L_{P},$ there exists $x_{S}\in L^{\prime}_{S}$ such that $(x_{S},x_{P})\in{\cal V}_{SP}.$ The uniqueness of $x_{S}$ follows from the fact that $L^{\prime}_{S}\cap({\cal V}_{SP}\times S)\subseteq L^{\prime}_{S}\cap{\cal V}_{S}={\mathbf{0}}_{S}.$ Next, if $x_{S}=0_{S},$ since ${\cal V}_{SP}\leftrightarrow{\cal V}_{S}={\cal V}_{P},$ we must have $x_{P}\in{\cal V}_{P}.$ But $L_{P}\cap{\cal V}_{P}=0_{P}$ so this means $x_{P}=0_{P},$ a contradiction. We conclude that $x_{S}\neq 0_{S}.$

(b) Next, let $\hat{x}_{iS},i=1,\cdots,n,$ be as in the hypothesis. Since $L^{\prime}_{S}\subseteq{\cal V}_{SP}\leftrightarrow L_{P},$ and since $x_{1P},\cdots,x_{nP},$ generate $L_{P}.$ the vectors $\hat{x}_{iS}$ generate $L^{\prime}_{S}.$ Since ${\cal V}_{SP}\leftrightarrow{\cal V}_{S}={\cal V}_{P},$ if a nontrivial linear combination of $\hat{x}_{iS},i=1,\cdots,n,$ lies in ${\cal V}_{S},$ the same combination of $x_{1P},\cdots,x_{nP}$ will lie in ${\cal V}_{P}.$ But $L_{P}\cap{\cal V}_{P}={\mathbf{0}}_{P}$ and $x_{1P},\cdots,x_{nP}$ are given to be independent. Hence $\hat{x}_{iS},i=1,\cdots,n,$ are independent and $B_{S}$ is a basis matrix for $L^{\prime}_{S}.$

(c) The proof is as in that of part 2 of the theorem. ∎

Definition 10**.**

Let ${\cal K}_{S},{\cal K}_{P},{\cal V}_{SP}$ be such that ${\cal K}_{S}={\cal V}_{SP}\leftrightarrow{\cal V}_{P},{\cal K}_{P}={\cal V}_{SP}\leftrightarrow{\cal V}_{S}.$ Let $L_{S}\subseteq{\cal K}_{S},L_{P}\subseteq{\cal K}_{P}$ be such that for each $x_{P}\in L_{P}$ there exists a unique $x_{S}\in L_{S}$ such that $(x_{S},x_{P})\in{\cal V}_{SP}$ and for each $x_{S}\in L_{S}$ there exists a unique $x_{P}\in L_{P}$ such that $(x_{S},x_{P})\in{\cal V}_{SP}.$ We then say that ${\cal K}_{S},{\cal K}_{P}$ are invertibly linked through ${\cal V}_{SP}$ and so are $L_{S},L_{P}.$

5.2.2 Implicit Duality

Theorem 11, is from [18]. In the present paper, it is used to derive results on self duality and short vectors of duals. The proof in this subsection is new. It generalizes naturally to the matroid case [24].

From the definition of matched composition, when $\mathcal{K}_{SP},\mathcal{K}_{PQ}$ are generalized number lattices and $(S,P,Q)$ are disjoint,

[TABLE]

and also equal to

[TABLE]

Similarly from the definition of skewed composition,

[TABLE]

and also equal to

[TABLE]

Hence we have

[TABLE]

In particular,

[TABLE]

since $\mathcal{K}_{P}=\mathcal{K}_{(-P)}$. To summarize, noting further that if ${\cal V}_{SP}$ is a vector space ${\cal V}_{SP}^{d}={\cal V}_{SP}^{\perp},$

Theorem 11**.**

Let $\mathcal{K}_{SP},\mathcal{K}_{PQ}$ be generalized number lattices. Then $(\mathcal{K}_{SP}\leftrightarrow\mathcal{K}_{PQ})^{d}\ \hskip 2.84544pt=\hskip 2.84544pt\ (\mathcal{K}_{SP}^{d}\rightleftharpoons\mathcal{K}_{PQ}^{d}).$

In particular,

$(\mathcal{K}_{SP}\leftrightarrow\mathcal{K}_{P})^{d}\ \hskip 2.84544pt=\hskip 2.84544pt\ (\mathcal{K}_{SP}^{d}\leftrightarrow\mathcal{K}_{P}^{d}).$**

• 2.

$(\mathcal{V}_{SP}\leftrightarrow\mathcal{K}_{P})^{d}\ \hskip 2.84544pt=\hskip 2.84544pt\ (\mathcal{V}_{SP}^{\perp}\leftrightarrow\mathcal{K}_{P}^{d}).$**

The above result will be referred to as the implicit duality theorem (IDT) for generalized number lattices.

Observe that the dot-cross duality is also a consequence of the implicit duality theorem since

[TABLE]

By Corollary 2, we know that the dual of a full dimensional number lattice is also full dimensional. When $\mathcal{K}_{SP},\mathcal{K}_{PQ}$ are full dimensional number lattices, it is easily verified that $\mathcal{K}_{SP}\circ S,\mathcal{K}_{SP}\times S$ $(=(\mathcal{K}^{d}_{SP}\circ S)^{d})$ are full dimensional number lattices on $S$ and $\mathcal{K}_{SP}+\mathcal{K}_{PQ},$ $\mathcal{K}^{d}_{SP}+\mathcal{K}^{d}_{PQ},$ $\mathcal{K}_{SP}\cap\mathcal{K}_{PQ}$ $(=(\mathcal{K}^{d}_{SP}+\mathcal{K}^{d}_{PQ})^{d})$ are full dimensional number lattices on $S\uplus P\uplus Q.$

From the definition of matched and skewed composition, we therefore have

Theorem 12**.**

*Let ${L}_{SP},{L}_{PQ}$ be full dimensional number lattices. Then ${L}_{SP}\leftrightarrow{L}_{PQ}$ is a full dimwnsional number lattice and therefore, $({L}_{SP}\leftrightarrow{L}_{PQ})^{d}$ is a full dimensional number lattice and is equal to

${L}_{SP}^{d}\rightleftharpoons{L}_{PQ}^{d}.$ In particular, $({L}_{SP}\leftrightarrow{L}_{P})^{d}\ \hskip 2.84544pt=\hskip 2.84544pt\ {L}_{SP}^{d}\leftrightarrow{L}_{P}^{d}.$*

The situation where \mbox{\cal V}_{SP}\leftrightarrow\mbox{L}_{PQ} is a full dimensional number lattice is also of some interest and is dealt with in the following theorem.

Theorem 13**.**

Let \mbox{\cal V}_{SP} be a vector space such that \mbox{\cal V}_{SP}\circ S=\mbox{\cal F}_{S}, \mbox{\cal V}_{SP}\times S={\mathbf{0}}_{S}, and let \mbox{L}_{PQ} be a number lattice such that \mbox{L}_{PQ}\times Q is full dimensional and span(\mbox{L}_{PQ}\circ P)=\mbox{\cal V}_{SP}\circ P. Then \mbox{\cal V}_{SP}\leftrightarrow\mbox{L}_{PQ} is a full dimensional number lattice and therefore so is ({\mbox{\cal V}}_{SP}\leftrightarrow{L}_{PQ})^{d}\ \hskip 2.84544pt=\hskip 2.84544pt\ {\mbox{\cal V}}_{SP}^{d}\leftrightarrow{L}_{PQ}^{d}={\mbox{\cal V}}_{SP}^{\perp}\leftrightarrow{L}_{PQ}^{d}.

Proof.

Let rows of $(B_{S}\vdots B_{P}),(C_{P}\vdots C_{Q})$ be bases for vector space \mbox{\cal V}_{SP} and number lattice $L_{PQ}$ respectively. Since ${\cal V}_{SP}\times S={\mathbf{0}}_{S},$ there exists a matrix $N,$ such that $B_{S}=B_{P}N.$

By Theorem 5, since span(\mbox{L}_{PQ}\circ P)=\mbox{\cal V}_{SP}\circ P, we must have that ${\cal V}_{SP}\leftrightarrow{L}_{PQ}$ is a number lattice for which $(C_{P}N\vdots C_{Q})$ is a basis matrix. We will show that the columns of $(C_{P}N\vdots C_{Q})$ are linearly independent. Let $(C"_{Q})$ be a basis matrix for \mbox{L}_{PQ}\times Q. Therefore there exists a generating matrix (see Subsubsection 3.2.2)

[TABLE]

for the number lattice ${\cal V}_{SP}\leftrightarrow{L}_{PQ}.$ Now \mbox{L}_{PQ}\times Q is full dimensional. So columns of $(C"_{Q})$ are linearly independent. The matrix $(C_{SQ})$ is row equivalent to $(C_{P}N\vdots C_{Q}).$ Since span(\mbox{L}_{PQ}\circ P)=\mbox{\cal V}_{SP}\circ P, $(C_{P}),(B_{P})$ are row equivalent and therefore so also $(C_{P}N),(B_{P}N).$ Now \mbox{\cal V}_{SP}\circ S=\mbox{\cal F}_{S}. Therefore, columns of $(B_{S})=(B_{P}N)$ are linearly independent and therefore also those of $(C_{P}N).$ It follows that columns of $(C^{\prime}_{P}N)$ are linearly independent and therefore also columns of $(C_{SQ}).$ Since $(C_{SQ}),$ $(C_{P}N\vdots C_{Q})$ are row equivalent, the columns of $(C_{P}N\vdots C_{Q})$ are also linearly independent so that ${\cal V}_{SP}\leftrightarrow{L}_{PQ}$ is a full dimensional number lattice. ∎

Implicit duality theorem and its applications are dealt with in detail in [14], [16], [15], [18]. More recently, there has been interest in the applications of this result in [6], in the context of ‘Pontryagin duality’. The above proof is similar to the one in [15]. Versions for other contexts are available in [18], for operators in [19], [20], for matroids in [24].

5.2.3 Self duality

Let $Q\equiv\{e_{1},\cdots e_{m}\}.$ Let $p(\cdot)$ be a permutation of the elements of $Q$ such that $p^{2}(\cdot)=\ identity.$

Define, for a vector $f_{e_{1},\cdots,e_{m}}$ on $Q,$ the vector $f_{p(e_{1}),\cdots,p(e_{m})}$ as follows:

$f_{p(e_{1}),\cdots,p(e_{m})}(p(e_{i}))\equiv f_{e_{1},\cdots,e_{m}}(e_{i}),\ i=1,\cdots,m.$

Note that $\langle f_{p(e_{1}),\cdots,p(e_{m})},g_{e_{1},\cdots,e_{m}}\rangle=\Sigma_{i}f_{p(e_{1}),\cdots,p(e_{m})}(p(e_{i}))g_{e_{1},\cdots,e_{m}}(e_{i}).$

Let ${\cal K}_{pQ}\equiv\{f_{p(e_{1}),\cdots,p(e_{m})},f_{e_{1},\cdots,e_{m}}\in{\cal K}_{Q}\}.$

We say the generalized number lattice ${\cal K}_{Q}$ is self dual relative to the permutation $p(\cdot)$ iff ${\cal K}^{d}_{Q}={\cal K}_{pQ}.$

We now have, as a corollary to Theorem 11, a way of constructing new self dual (generalized) number lattices from old.

Corollary 14**.**

Let $p(\cdot)$ be a permutation of $S\uplus P$ such that $p^{2}(\cdot)=\ identity,$ $p(S)=S$ and $p(P)=P.$

Let ${{\cal K}}_{SP},{{\cal K}}_{P}$ be self dual generalized number lattices relative to $p(\cdot),p|_{P}(\cdot),$ respectively. Then so is ${{\cal K}}_{SP}\leftrightarrow{{\cal K}}_{P}$ relative to $p|_{S}(\cdot).$ 2. 2.

Let ${L}_{SP},{L}_{P}$ be self dual number lattices relative to $p(\cdot),p|_{P}(\cdot),$ respectively. Then so is ${L}_{SP}\leftrightarrow{L}_{P}$ relative to $p|_{S}(\cdot).$ 3. 3.

Let {\mbox{\cal V}}_{SP} be a vector space and let ${L}_{P}$ be a number lattice such that {\mbox{\cal V}}_{SP}\leftrightarrow{L}_{P} is a number lattice. Further, let {\mbox{\cal V}}_{SP} be self dual relative to the the permutation $p(\cdot)$ and let ${L}_{P}$ be self dual relative to $p|_{P}(\cdot).$ Then {\mbox{\cal V}}_{SP}\leftrightarrow{L}_{P} is a self dual number lattice relative to $p|_{S}(\cdot).$

Proof.

Let ${\cal K}_{S}\equiv{{\cal K}}_{SP}\leftrightarrow{{\cal K}}_{P}.$ We then have

[TABLE] 2. 2.

From part 1 above, ${L}_{SP}\leftrightarrow{L}_{P}$ is a self dual generalized number lattice relative to $p|_{S}(\cdot).$

We have that ${L}_{SP},{L}_{P}$ are full dimensional number lattices and so also (by Theorem 11) is ${L}_{SP}\leftrightarrow{L}_{P}.$ Therefore $({L}_{SP}\leftrightarrow{L}_{P})^{d}$ cannot contain a vector space. It is therefore a self dual number lattice. 3. 3.

From part 1 above, it is clear that {\mbox{\cal V}}_{SP}\leftrightarrow{L}_{P} is a self dual generalized number lattice relative to $p|_{S}(\cdot).$ We know that {\cal K}_{S}\equiv{\mbox{\cal V}}_{SP}\leftrightarrow{L}_{P} is a number lattice, and therefore ${\cal K}^{d}_{S}={\cal K}_{pS}$ cannot contain a nontrivial vector space. Therefore {\mbox{\cal V}}_{SP}\leftrightarrow{L}_{P} is a self dual number lattice.

∎

Corollary 14 suggests some simple ways of constructing new self dual lattices from old.

Suppose $S_{1}\uplus P_{1},\cdots,S_{k}\uplus P_{k}$ are identical copies of the set $S_{1}\uplus P_{1}.$ Let $S\equiv S_{1}\uplus\cdots\uplus S_{k}$ and let $P\equiv P_{1}\uplus\cdots\uplus P_{k}.$ We will suppose that the sets $P$ and $S_{1}$ are ‘small’ and that we have self dual number lattices $L_{S_{1}P_{1}}$ and $L_{P}.$ Let $L_{S_{i}P_{i}},i=2,\cdots,k$ be copies of the number lattice $L_{S_{1}P_{1}}.$ It is clear that $L_{S_{1}P_{1}}\oplus\cdots\oplus L_{S_{k}P_{k}}$ is self dual, so that $L_{S_{1}P_{1}}\oplus\cdots\oplus L_{S_{k}P_{k}}\leftrightarrow L_{P}$ is self dual. If now we define a permutation $p(\cdot)$ on $S\uplus P,$ such that $p^{2}(\cdot)=identity,$ $p(S_{i})=S_{i},p(P_{i})=P_{i},i=1,\cdots,k,$ and $L_{S_{i}P_{i}},i=1,\cdots,k$ and $L_{P}$ are self dual relative to the permutation $p(\cdot),$ so would $L_{S_{1}P_{1}}\oplus\cdots\oplus L_{S_{k}P_{k}}\leftrightarrow L_{P}$ be.

2. We describe a way of building self dual number lattices analogous to the manner in which reciprocal electrical networks are built by connecting smaller reciprocal networks (see for instance [17] ).

Let \mbox{\cal V}_{Z},\mbox{\cal V}_{Z^{\prime}} be as in Example 6.

Let $p(\cdot)$ be the permutation on $Z\uplus Z^{\prime}$ defined by $p(e_{i})\equiv e_{i}^{\prime},p(e_{i}^{\prime})\equiv e_{i},i=1,\cdots,m.$ Let \mbox{\cal V}_{ZZ^{\prime}}\equiv\mbox{\cal V}_{Z}\oplus\mbox{\cal V}^{\perp}_{Z^{\prime}}.

Clearly \mbox{\cal V}^{d}_{ZZ^{\prime}}=\mbox{\cal V}^{\perp}_{ZZ^{\prime}}=(\mbox{\cal V}_{Z}\oplus\mbox{\cal V}^{\perp}_{Z^{\prime}})^{\perp}=\mbox{\cal V}^{\perp}_{Z}\oplus\mbox{\cal V}_{Z^{\prime}}=\mbox{\cal V}_{p(ZZ^{\prime})}. Therefore, \mbox{\cal V}_{ZZ^{\prime}} is self dual relative to the permutation $p(\cdot).$

As in Example 6, let $S\equiv\{e_{1},\cdots,e_{n}\},n\leq m,$ $P\equiv\{e_{n+1},\cdots,e_{m}\},$ $S^{\prime}\equiv\{e^{\prime}_{1},\cdots,e^{\prime}_{n}\},$ $P^{\prime}\equiv\{e^{\prime}_{n+1},\cdots,e^{\prime}_{m}\}.$ Let $Z\equiv S\uplus P,Z^{\prime}\equiv S^{\prime}\uplus P^{\prime}.$ Let \mbox{\cal V}_{Z} be the coboundary space of a graph ${\cal G},$ and let the edges in $P$ be such that they can be included in a tree as well as in the complement of a tree (i.e., $P$ contains no loops or cutsets of ${\cal G}$).

If \mbox{L}_{SS^{\prime}} is a number lattice on $S\uplus S^{\prime},$ then by Theorem 5, as we saw in Example 6, \mbox{\cal V}_{ZZ^{\prime}}\leftrightarrow\mbox{L}_{SS^{\prime}} is also a number lattice.

Suppose $S_{1},\cdots,S_{k}$ are identical copies of the set $S_{1}.$ Let $S\equiv S_{1}\uplus\cdots\uplus S_{k}.$ We will suppose that the set $S_{1}$ is ‘small’ and that we have a self dual number lattice $L_{S_{1}S_{1}^{\prime}}.$ Let $L_{S_{i}S_{i}^{\prime}},i=2,\cdots,k$ be copies of the number lattice $L_{S_{1}S_{1}^{\prime}}.$ It is clear that $L_{SS^{\prime}}\equiv L_{S_{1}S_{1}^{\prime}}\oplus\cdots\oplus L_{S_{k}S_{k}^{\prime}}$ is self dual, so that \mbox{\cal V}_{ZZ^{\prime}}\leftrightarrow L_{SS^{\prime}} is self dual relative to the permutation $p(\cdot).$

6 Number lattices with a specified partition of the underlying set

In this section we compare composition of ‘maps’ with matched and skewed composition of the corresponding objects, viz., ‘linkages’. A linkage is a generalized number latice $\mathcal{K}_{S}$ with a partition $\{S_{1},\cdots,S_{k}\}$ of $S,$ specified. It will be denoted by $\mathcal{K}_{S_{1}\cdots S_{k}}.$ When the linkage is a vector space it will be referred to as a vs-linkage $\mathcal{V}_{S_{1}\cdots S_{k}}.$

Observe that a map $x^{T}K=y^{T}$ can be regarded as a vs-linkage $\mathcal{V}_{SP},$ with a typical vector $(x^{T},y^{T})$, whose basis is the set of rows of $\begin{bmatrix}I&K\end{bmatrix},$ the first set of columns of the matrix being indexed by $S$ and the second set by $P.$

The matched composition operation ‘$\leftrightarrow$’ can be regarded as a generalization of the composition operation for ordinary maps.

Thus if $x^{T}K=y^{T},y^{T}P=z^{T},$ it can be seen that $\mathcal{V}_{SP}\leftrightarrow\mathcal{V}_{PQ}$ would have as the basis, the rows of $\begin{bmatrix}I&KP\end{bmatrix},$ the first set of columns of the matrix being indexed by $S$ and the second set by $Q.$

Remark 3**.**

When we compose matrices, the order is important. In the case of linkages, since vectors are indexed by subsets, order does not matter. Thus ${\cal K}_{SP}={\cal K}_{PS}$ and ${\cal K}_{SP}\leftrightarrow{\cal K}_{PQ}\hskip 2.84544pt=\hskip 2.84544pt{\cal K}_{PQ}\leftrightarrow{\cal K}_{PS}.$ 2. 2.

$\mathcal{V}_{SP}$* mimics the map $K$ by linking row vectors $x^{T},y^{T}$ which correspond to each other when $x^{T}$ is operated upon by $K,$ into the long vector $(x^{T},y^{T}).$ The collection of all such vectors forms a vector space which contains all the information that the map $K$ and $K^{-1}$ (if it exists) contain and represents both of them implicitly.*

But there are two essential differences: a map takes vectors in ${\cal F}_{X}$ to vectors in ${\cal F}_{Y},$ but takes the vector $0_{S}$ only to the vector $0_{P}.$ In the case of vs-linkages only a subspace of ${\cal F}_{X}$ may be involved as $\mathcal{V}_{SP}\circ S$ and the vector $0_{S}$ may be linked to a nontrivial subspace of ${\cal F}_{Y}.$

6.1 Expressions of linkages

The ‘$\leftrightarrow$’ operation is not inherently associative. For instance, consider the expression

[TABLE]

If we treat this expression to be ${\cal K}_{BC}\leftrightarrow({\mathbf{0}}_{AB}\leftrightarrow{\mathbf{0}}_{BQ}),$ it will reduce to ${\cal K}_{BC}\leftrightarrow{\mathbf{0}}_{AQ}={\cal K}_{BC}\oplus{\mathbf{0}}_{AQ}.$

If, however, we treat it to be $({\cal K}_{BC}\leftrightarrow{\mathbf{0}}_{AB})\leftrightarrow{\mathbf{0}}_{BQ},$ it will reduce to ${\cal K}_{BC}\times C\oplus{\mathbf{0}}_{ABQ}.$

The two reduced expressions are clearly different if ${\cal K}_{BC}\circ C\neq{\cal K}_{BC}\times C.$

The problem here is that the index set $B$ occurs more than twice.

Next consider the expression

[TABLE]

Here, no index set occurs more than twice, but if we regard the expression as $({\cal K}_{AB}\leftrightarrow{\cal K}_{BC}\leftrightarrow{\cal K}_{CA})\leftrightarrow{\cal K}_{PQ},$ we get a subexpression ${\cal K}_{AB}\leftrightarrow{\cal K}_{BC}\leftrightarrow{\cal K}_{CA},$ where the index set becomes null and the ‘$\leftrightarrow$’ operation is not defined for such a situation. We will call such expressions ‘null’ expressions.

Let us only look at expressions where no index set occurs twice and which do not contain null subexpressions.

In the case of such expressions, the brackets can be got rid of without ambiguity. Consider for instance the expression

[TABLE]

where all the $A_{i},B_{j}$ are mutually disjoint. It is clear that this expression has a unique meaning (evaluation), namely, the collection of all $(f_{A_{2}},g_{B_{1}})$ such that there exist $h_{A_{1}},k_{B_{2}}$ with

[TABLE]

Essentially the terms of the kind $l_{D_{i}}$ survive if they occur only once and they get coupled through the terms which occur twice. These latter dont survive in the final expression. Such an interpretation is valid even if we are dealing with signed index sets of the kind ${-A_{i}}.$ (We remind the reader that ${\cal K}_{(-A)B}$ is the space of all vectors $(-f_{A},g_{B})$ where $(f_{A},g_{B})$ belongs to ${\cal K}_{AB}.$ ) If either ${-A_{i}},$ or ${A_{i}}$ occurs, it counts as one occurrence of the index set ${A_{i}}.$ We state this result as a theorem but omit the routine proof.

Let us define a regular expression of linkages to be one of the form $\leftrightarrow_{A_{i},B_{j},C_{k},\cdots}({\cal K}_{(\pm A_{i})(\pm B_{j})(\pm C_{k})\cdots}),$ where the index sets $\pm A_{i},\pm B_{j},\pm C_{k},\cdots$ are all mutually disjoint, where no index set occurs more than twice, and that contains no null subexpressions.

We thus have,

Theorem 15**.**

A regular expression of linkages has a unique evaluation as a linkage ${\cal K}_{\cdots,{\pm D_{i}},\cdots}$ where $D_{i}$ are the index sets that occur only once in the expression.

Remark 4**.**

*The condition ‘no null subexpressions’ can be relaxed, if we permit a special ‘object’ ${\cal K}_{\emptyset}$ such that

1. ${\cal K}_{\emptyset}={\cal K}_{S}\leftrightarrow{\cal K}^{\prime}_{S},$ for any set $S$ and any ${\cal K}_{S},{\cal K}^{\prime}_{S},$ and

2. ${\cal K}_{\emptyset}\leftrightarrow{\cal K}_{P}={\cal K}_{P},$ for any ${\cal K}_{P}.$*

A regular expression of linkages is best represented by means of a diagram, where individual linkages are represented by nodes, with edges corresponding to index sets common to two linkages.

This diagram would look different from the usual diagram for maps. In the case of maps, the usual diagram would simply be a directed line going through a sequence of nodes, say $T_{i},i=1,\cdots,k,$ corresponding to the composed map $T_{1}\times\cdots\times T_{k}.$ The diagram of the dual would correspond to the composed map $T^{T}_{k}\times\cdots\times T^{T}_{1}$ and the line would be directed in the opposite direction.

In the case of linkages there is no need of a directed line for describing simple matched composition, the index set specification being adequate. We need only distinguish ‘$\leftrightarrow$’ from ‘$\rightleftharpoons$’. But ${\cal K}_{SP}\rightleftharpoons{\cal K}_{PQ}$ can be treated as ${\cal K}_{SP}\leftrightarrow{\cal K}_{(-P)Q}.$ The dual of ${\cal K}_{SP}\leftrightarrow{\cal K}_{PQ}$ would be ${\cal K}_{SP}^{d}\leftrightarrow{\cal K}^{d}_{(-P)Q}.$

If the maps were regarded as linkages, there would be undirected bold edges corresponding to ‘$\leftrightarrow$’ and undirected but, say, dotted edges corresponding to ‘$\rightleftharpoons$’. If a particular edge is bold in the diagram of the primal expression, it would be dotted in the diagram of the dual expression.

Linkages permit much greater flexibility while losing none of the essential properties of dualization. Linkages can simultaneously relate vectors over several sets: consider, for instance, a regular expression of the kind ${\cal K}_{SPQ}\leftrightarrow{\cal K}_{STW}\rightleftharpoons{\cal K}_{P}.$ There is no natural analogue of this expression in the case of maps. But the dual is still simple : $({\cal K}_{SPQ}\leftrightarrow{\cal K}_{STW}\rightleftharpoons{\cal K}_{P})^{d}=({\cal K}_{SPQ}\leftrightarrow{\cal K}_{STW}\leftrightarrow{\cal K}_{(-P)})^{d}={\cal K}^{d}_{(-S)PQ}\leftrightarrow{\cal K}^{d}_{STW}\leftrightarrow{\cal K}^{d}_{P}.$

Directed lines are needed corresponding to situations where the implicit inversion theorem (Theorem 8) is applicable. C contains a description of such diagrams and their essential invariance under dualization.

7 Number lattices invertibly linked through regular vector spaces

7.1 Regular vector spaces

We saw in Subsection 5.2.1, that when generalized number lattices are invertibly linked as in Theorem 9, a vector in the number lattice part of each is uniquely linked to a vector in that of the other. We show in this section that when the vector space linking the two generalized number lattices has special properties (such as having totally unimodular representative matrices), the length of a vector in one number lattice bounds the length of the corresponding vector in the other. In particular ‘short’ vectors of one are linked to ‘short’ vectors of the other. The incidence matrix of a graph is totally unimodular. The ideas of this section are immediately applicable when the linking vector spaces are spanned by or are orthogonal to the rows of the incidence matrix of a graph.

We need some preliminary definitions and lemmas.

The column matroid of a matrix on column set $S$ is the family of independent subsets of $S.$ A set of columns is linearly independent in a matrix iff it is linearly independent in any other matrix that is row equivalent to it. Therefore, any two representative matrices of a vector space \mbox{\cal V}_{S} have the same column matroid. We call this the matroid of \mbox{\cal V}_{S} and denote it by \mbox{\cal M}(\mbox{\cal V}_{S}). A maximal independent subset of columns of a representative matrix of \mbox{\cal V}_{S} is called a base of the matroid \mbox{\cal M}(\mbox{\cal V}_{S}).

A matrix is said to be totally unimodular iff all its subdeterminants are $0,\pm{1}.$ A vector space which has a totally unimodular representative matrix is called a regular vector space. We remind the reader that a standard representative matrix is a representative matrix which can be put in the form $(I\ |\ K)$ after column permutation. It is clear that corresponding to every base $B$ of \mbox{\cal M}(\mbox{\cal V}_{S}), there is a standard representative matrix of \mbox{\cal V}_{S} with the unit submatrix corresponding to $B.$ We will call this the standard representative matrix of \mbox{\cal V}_{S} with respect to $B.$

Definition 16**.**

Let $C_{S}$ be a representative matrix of ${\cal V}_{S}.$ We say a base $B$ of \mbox{\cal M}({\cal V}_{S}) is picked according to a priority sequence $(S_{1},\cdots,S_{k}),S_{i}\subseteq S,S_{i}\cap S_{j}=\emptyset,i\neq j,$ iff $B$ is a maximal independent set of columns $C_{S}$ with elements from $S_{1},\cdots,S_{i},i<k,$ picked before elements from $S_{i+1}.$ The standard representative matrix built with respect to a base of \mbox{\cal M}({\cal V}_{S}) picked according to a priority sequence $(S_{1},\cdots,S_{k})$ would simply be referred to as being built according to the priority sequence $(S_{1},\cdots,S_{k}).$

The following lemma is well known and its routine proof is omitted ([27]).

Lemma 17**.**

Every standard representative matrix of a regular vector space is totally unimodular. If $(I\ |\ K),$ is a standard representative matrix of ${\cal V}_{S},$ then $(-K^{T}\ |\ I)$ is a standard representative matrix of ${\cal V}_{S}^{\perp}.$ Therefore, if ${\cal V}_{S}$ is regular, so is ${\cal V}_{S}^{\perp}.$

The following is an easy consequence

Lemma 18**.**

*Let a standard representative matrix of a regular vector space ${\cal V}_{S}$ with respect to a base $B_{1}$ of \mbox{\cal M}(\mbox{\cal V}_{S}) be given. Let $|B_{1}|=dim({\cal V}_{S})=m.$ Then we can compute the standard representative matrix of ${\cal V}_{S}$ with respect to any other base $B_{2}$ of \mbox{\cal M}(\mbox{\cal V}_{S}), which could be specified in terms of a priority sequence of elements from $S,$ in $O(m^{2}|S|)$ time with all the numbers occurring during intermediate computations being equal to $0,\pm 1.$ *

The proof is given in D.

The following simple result is useful for relating length of vectors in generalized number lattices linked through regular vector spaces.

Lemma 19**.**

Let $Q\equiv(Q_{T}\ |\ Q_{B})$ be a totally unimodular matrix with $Q_{B}$ as a column permutation of the identity matrix. Let $x$ be linearly dependent on the rows of $Q.$ Let $x_{T},x_{B}$ denote the vectors $x|_{T},x|_{B}$ respectively. Then

[TABLE]

Proof.

We have $x_{T}^{T}=x_{B}^{T}Q_{T},$ so that $||x_{T}||^{2}=x_{T}^{T}x_{T}=x_{B}^{T}Q_{T}Q_{T}^{T}x_{B}.$ Now the maximum magnitude of an entry in $Q_{T}Q_{T}^{T}$ must be less or equal to $|T|$ since entries of $Q_{T}$ are $0,\pm{1}.$ It follows that the maximum magnitude eigenvalue $\lambda_{max}$ of $Q_{T}Q_{T}^{T}$ must satisfy $|\lambda_{max}|\leq|T|\times|B|.$

Therefore, using Cauchy$-$Schwarz inequality,

$||x_{T}||^{2}=||x_{B}^{T}Q_{T}Q_{T}^{T}x_{B}||\leq||x_{B}^{T}||\times||Q_{T}Q_{T}^{T}x_{B}||\leq|\lambda_{max}|||x_{B}||^{2}\leq|T|\times|B|\times||x_{B}||^{2}.$ The result follows. ∎

The next result states that a vector, of the kind in the above lemma, that is linked to a given vector, is easy to find.

Lemma 20**.**

Let \mbox{\cal V}_{SP} be a regular vector space on $S\uplus P.$

Let x_{P}\in\mbox{\cal V}_{SP}\circ P. Then there exists (x_{S},x_{P})\in\mbox{\cal V}_{SP} such that $||x_{S}||\leq\sqrt{|S|\times|P|}||x_{P}||.$ Moreover, if a standard representative matrix is available for \mbox{\cal V}_{SP}, built according to the priority sequence $(P,S),$ then $x_{S}$ can be computed in $\tilde{O}(mnlog(M))$ time, where $m=r({\cal V}_{SP}\circ P),$ $log(M)$ is the maximum bit size of an entry in $x_{P}.$ and $n=|S|-r({\cal V}_{SP}\times S)=r({\cal V}_{SP}^{\perp}\circ S).$ The maximum bit size of an entry in $x_{S}$ is $O(log(mM)).$

Proof.

The standard representative matrix with respect the base $B$ of \mbox{\cal M}({\cal V}_{SP}), since it is picked according to the priority sequence $(P,S),$ must have the form

[TABLE]

We know, by Lemma 17, that this matrix is totally unimodular so that all its entries are $0,\pm{1}.$ Now let x_{P}\in\mbox{\cal V}_{SP}\circ P. We know, from the discussion in Subsection 3.2.2, that $x_{P}^{T}=\lambda^{T}(I_{2P_{1}}\ |\ Q_{2P_{2}}).$ Therefore, $\lambda^{T}=x_{P_{1}}^{T}.$ Let us pick $(x_{S},x_{P})$ as

[TABLE]

This yields $x_{S}^{T}=x_{P_{1}}^{T}({\mathbf{0}}_{2S_{1}}\ \vdots\ |Q_{2S_{2}}).$

By Lemma 19 , we know that

$||x_{S}||\leq\sqrt{(|S_{2}|\times|P_{1}|)}\times||x_{P_{1}}||\leq\sqrt{(|S|\times|P|)}\times||x_{P}||.$

It is clear that $x_{S}$ can be computed in $O((|P_{1}|\times|S_{2}|)log(M))$ time, where $log(M)$ is the maximum bit size of an entry in $x_{P}.$ Therefore, since $m=r({\cal V}_{SP}\circ P)=|P_{1}|,n=|S_{2}|=|S|-r({\cal V}_{SP}\times S)=r({\cal V}_{SP}^{\perp}\circ S),$ the computation can be done in $O(mnlog(M))$ time.

Since $x_{S}^{T}=x_{P_{1}}^{T}({\mathbf{0}}_{2S_{1}}\ \vdots\ Q_{2S_{2}}),$ $Q_{2S_{2}}$ has $0,\pm 1$ entries and $m$ rows, and maximum bit size of an entry in $x_{P}$ is $O(log(M)),$ we must have that the maximum bit size of an entry in $x_{S}$ is $O(log(mM)).$

∎

7.2 Approximate successive minima bases

An important problem associated with number lattices is the computation of approximately shortest vectors of various kinds in the lattice. Even the approximate (with respect to a factor polyomial in the dimension of the lattice) versions of these problems are known to be hard ([9],[1]). In this subsection, we examine what can be inferred from short vectors of a number lattice about short vectors of a linked number lattice, when the linking is through a regular vector space.

Definition 21**.**

Let $L_{S}$ be a number lattice with basis matrix $B_{S}$ of rank $n.$ For $i=1,\cdots m,$ we define the $i^{th}$ successive minimum as $\lambda_{i}(L_{S})=inf\{r,dim(span(L_{S}\cap\overline{B}(0,r)))\geq i\},$ where $\overline{B}(0,r)\equiv\{x\in\Re^{|S|},||x||\leq r\}$ is the closed ball of radius $r$ around $0_{S}.$

It is clear from the definition of the succesive minima that if $i\leq j$ then $\lambda_{i}(L_{S})\leq\lambda_{j}(L_{S}),$ and that there exists a basis matrix for $L_{S}$ with $i^{th}$ row having length $\lambda_{i}(L_{S}).$

Definition 22**.**

*We say $B_{S}$ is an SM-basis matrix for $L_{S}$ iff the $i^{th}$ row of $B_{S}$ has length $\lambda_{i}(L_{S}).$

Let $\mathbf{\alpha}\equiv(\alpha_{1},\cdots,\alpha_{n}).$ We say $B_{S}$ is an $\alpha$SM-basis matrix for $L_{S}$ iff the $i^{th}$ row of $B_{S}$ has length $\leq\alpha_{i}\lambda_{i}(L_{S}).$ If we wish to not explicitly mention $\alpha,$ $\alpha$SM-basis matrices are also referred to as reduced basis matrices.*

Remark 5**.**

It is immediate from the definition that if $B_{S}$ is an SM-basis matrix for $L_{S},$ then its $i^{th}$ row has length no more than the length of its $j^{th}$ row, where $i\leq j.$ However, this is not necessarily true in the case of an $\alpha$SM-basis matrix.

The following lemma is an immediate consequence of the definition of an SM-basis matrix.

Lemma 23**.**

Let $\hat{B}_{S}$ be an SM-basis matrix for the number lattice $L_{S},$ and let $B_{S}$ be any basis matrix of $L_{S}.$ Then there exists a matrix $B^{\prime}_{S},$ obtained by permuting the rows of $B_{S}$ such that the length of the $i^{th}$ row of $B^{\prime}_{S}$ is less or equal to that of the $i^{th}$ row of $\hat{B}_{S}.$

We now present a way of relating lengths of vectors in the number lattice parts of two invertibly linked generalized number lattices. We need the following routine lemma whose proof is relegated to the appendix.

Lemma 24**.**

Let $\nu(\cdot),\mu(\cdot)$ be permutations of the set $\{1,\cdots,n\}.$ Then, for every $j\in\{1,\cdots,n\}$ there exists an $i\in\{1,\cdots,n\}$ such that $\nu(i)\leq j\leq\mu(i).$

Theorem 25**.**

Let ${\cal V}_{SP}$ be a regular vector space, and let ${\cal V}_{P}\equiv{\cal V}_{SP}\times P,{\cal V}_{S}\equiv{\cal V}_{SP}\times S.$

Let ${\cal K}_{P},{\cal K}_{S}$ be generalized number lattices with ${\cal K}_{P}\subseteq{\cal V}_{SP}\circ P,{\cal K}_{S}\equiv{\cal V}_{SP}\leftrightarrow{\cal K}_{P}.$ Let ${\cal K}_{P}={\cal V}_{P}+L_{P},$ where ${\cal V}_{P},L_{P},$ are orthogonal.

${\cal K}_{S}={\cal V}_{S}+L_{S},$* where ${\cal V}_{S},L_{S},$ are orthogonal. Further $L_{S}\subseteq{\cal V}_{SP}\leftrightarrow L_{P}$ and $L_{P}\subseteq{\cal V}_{SP}\leftrightarrow L_{S}.$* 2. 2.

Let $x_{P}\in L_{P},x_{P}\neq 0_{P}.$ Then, there is a unique ${x}_{S}\in L_{S}$ such that $({x}_{S},x_{P})\in{\cal V}_{SP}.$ Further, $x_{S}\neq 0_{S},$ $||{x}_{S}||\leq\sqrt{(|S|\times|P|)}\times||x_{P}||.$ Similarly, if $x_{S}\in L_{S},x_{S}\neq 0_{S},$ then, there is a unique ${x}_{P}\in L_{P}$ such that $({x}_{S},x_{P})\in{\cal V}_{SP}.$ Further, $x_{P}\neq 0_{P},$ $||{x}_{P}||\leq\sqrt{(|S|\times|P|)}\times||x_{S}||.$ 3. 3.

Let $B_{P}$ be a basis matrix for $L_{P},$ with rows $x_{1P},\cdots,x_{nP}.$ Let ${x}_{iS},i=1,\cdots,n,$ be the unique vector in $L_{S}$ such that $({x}_{iS},x_{iP})\in{\cal V}_{SP},i=1,\cdots,n,$ and let $B_{S}$ be the basis matrix for $L_{S},$ with ${x}_{iS},i=1,\cdots n,$ as the $i^{th}$ row. Let the rows of $B_{P},B_{S}$ be permuted in order of increasing length to yield the matrices $B^{\prime}_{P},B"_{S}.$ Let $B^{\prime}_{P}$ have rows ${x}^{\prime}_{1P},\cdots,{x}^{\prime}_{nP}$ and let $B"_{S}$ have rows ${x}"_{1S},\cdots,{x}"_{nS}.$ Then,

[TABLE] 4. 4.

*Let $\hat{B}_{P}$ be an *$\alpha$SM- basis matrix for $L_{P}$ with rows $\hat{x}_{1P},\cdots,\hat{x}_{nP},$ and with $||\hat{x}_{iP}||\leq||\hat{x}_{jP}||,i\leq j.$

Let $\hat{B}_{S}$ be any basis matrix for $L_{S}$ with rows $\hat{x}_{1S},\cdots,\hat{x}_{nS},$ and with $||\hat{x}_{iS}||\leq||\hat{x}_{jS}||,i\leq j.$ Then,

[TABLE] 5. 5.

*Let $\hat{B}_{P}$ be an *$\alpha$*SM- basis matrix for $L_{P}$ with rows $\hat{x}_{1P},\cdots,\hat{x}_{nP},$ and with $||\hat{x}_{iP}||\leq||\hat{x}_{jP}||,i\leq j.$ Let ${x}_{iS},i=1,\cdots,n,$ be the unique vector in $L_{S}$ such that $({x}_{iS},\hat{x}_{iP})\in{\cal V}_{SP},i=1,\cdots,n,$ and let $B_{S}$ be the matrix with ${x}_{iS},i=1,\cdots,n,$ as the $i^{th}$ row. Then $B_{S}$ is a *$\beta$*SM-basis matrix for *$L_{S},$

where $\mathbf{\beta}\equiv(\beta_{1},\cdots,\beta_{n})=(\alpha_{1}(|S|\times|P|),\cdots,\alpha_{n}(|S|\times|P|)).$

Proof.

1. From Theorem 9, we know that there exists $L^{\prime}_{S}\subseteq{\cal K}_{S}$ such that ${\cal K}_{S}={\cal V}_{S}+L^{\prime}_{S},L^{\prime}_{S}\cap{\cal V}_{S}={\mathbf{0}}_{S},L_{P}\subseteq{\cal V}_{SP}\leftrightarrow L^{\prime}_{S}.$ Let $L_{S}$ be the projection of $L^{\prime}_{S}$ onto ${\cal V}_{S}^{\perp}.$ Then ${\cal V}_{S},L_{S}$ are orthogonal and $L_{S}+{\cal V}_{S}=L^{\prime}_{S}+{\cal V}_{S}={\cal K}_{S}.$

Next, since $0_{P}\in L_{P},$ we must have that ${\cal V}_{SP}\times S\subseteq{\cal V}_{SP}\leftrightarrow L_{P}$ and it is given that $L^{\prime}_{S}\subseteq{\cal V}_{SP}\leftrightarrow L_{P}$ so that ${\cal V}_{SP}\times S+L^{\prime}_{S}\subseteq{\cal V}_{SP}\leftrightarrow L_{P}.$ Therefore, $L_{S}\subseteq{\cal V}_{SP}\times S+L^{\prime}_{S}\subseteq{\cal V}_{SP}\leftrightarrow L_{P}.$

Let $x_{P}\in{\cal V}_{SP}\leftrightarrow L^{\prime}_{S}.$ Then there exists $x^{\prime}_{S}\in L^{\prime}_{S}$ such that $(x^{\prime}_{S},x_{P})\in{\cal V}_{SP}.$ For any $x"_{S}\in{\cal V}_{SP}\times S,$ we have $(x"_{S},0_{P})\in{\cal V}_{SP}$ so that $(x^{\prime}_{S}+x"_{S},x_{P})\in{\cal V}_{SP}.$ But $L^{\prime}_{S}\subseteq L_{S}+{\cal V}_{SP}\times S$ which means that $x"_{S}$ can be chosen so that $x^{\prime}_{S}+x"_{S}\in L_{S}.$ Thus $x_{P}\in{\cal V}_{SP}\leftrightarrow L_{S}.$ But ${\cal V}_{SP}\leftrightarrow L^{\prime}_{S}\supseteq L_{P}.$ Therefore, ${\cal V}_{SP}\leftrightarrow L_{S}\supseteq L_{P}.$

Thus, ${\cal K}_{S}={\cal V}_{S}+L_{S},$ with ${\cal V}_{S},L_{S}$ orthogonal, with ${\cal V}_{SP}\leftrightarrow L_{P}\supseteq L_{S},$ and with ${\cal V}_{SP}\leftrightarrow L_{S}\supseteq L_{P}.$

2. Let $x_{P}\in L_{P},x_{P}\neq 0_{P}.$ Since ${\cal V}_{SP}\circ P\supseteq{\cal K}_{P}\supseteq L_{P},$ we must have that $x_{P}\in{\cal V}_{SP}\circ P.$ From Lemma 20, it is clear that there exists a vector ${x}^{\prime}_{S}$ such that $({x}^{\prime}_{S},x_{P})\in{\cal V}_{SP}$ and such that $||{x}^{\prime}_{S}||\leq\sqrt{(|S|\times|P|)}\times||x_{P}||.$ Since ${\cal V}_{SP}\leftrightarrow{\cal K}_{P}={\cal K}_{S},$ we must have that ${x}^{\prime}_{S}\in{\cal K}_{S}.$ Since {\cal K}_{S}=\mbox{\cal V}_{S}+L_{S}, and \mbox{\cal V}_{S},L_{S} are orthogonal to each other, we have ${x}^{\prime}_{S}=x"_{S}+{x}_{S},{x"}_{S}\in{\cal V}_{S},{x}_{S}\in L_{S},$ $||{x}_{S}||\leq||x^{\prime}_{S}||\leq\sqrt{(|S|\times|P|)}\times||x_{P}||.$ Now $x"_{S}\in{\cal V}_{S}={\cal V}_{SP}\times S,$ so that $(x"_{S},0_{P})\in{\cal V}_{SP}.$ Therefore $(({x^{\prime}}_{S},x_{P})-(x"_{S},0_{P}))=({x}_{S},x_{P})\in{\cal V}_{SP}.$

By Theorem 9, part 3(a), ${x}_{S}$ is the unique vector in $L_{S}$ such that $({x}_{S},x_{P})\in{\cal V}_{SP}$ and since $x_{P}$ is nonzero, so is ${x}_{S}.$

The above argument clearly works with $S,P$ interchanged so that the result is also true with $S,P$ interchanged.

3. Let $\mu(\cdot),\nu(\cdot)$ be permutations of $\{1,\cdots,n\}$ such that $x_{iP}=x^{\prime}_{\mu(i)P},x_{iS}=x"_{\nu(i)S},i=1,\cdots,n.$ From the previous part of the theorem we have that

[TABLE]

Let $j\in\{1,\cdots,n\}.$ Then, by Lemma 24, there exists $i\in\{1,\cdots,n\}$ such that $\nu(i)\leq j\leq\mu(i).$ We then have

[TABLE]

Interchanging $S,P$ and $\mu(\cdot),\nu(\cdot)$ in the above argument we get $||x"_{jS}||\leq\sqrt{(|S|\times|P|)}||x^{\prime}_{jP}||.$

4. Let $\hat{B}_{S}$ be any basis matrix for $L_{S}$ with rows $\hat{x}_{1S},\cdots,\hat{x}_{nS},$ and with $||\hat{x}_{iS}||\leq||\hat{x}_{jS}||,i\leq j.$ Let $x_{iP},i=1,\cdots,n$ be the unique vectors in $L_{P}$ such that $(\hat{x}_{iS},{x}_{iP})\in{\cal V}_{SP}.$ Let $x"_{iP}\in\{x_{1P},\cdots,x_{nP}\},i=1,\cdots,n$ be such that $||x"_{jP}||\leq||x"_{iP}||,j\leq i.$ Let $B"_{P}$ be the matrix with its $i^{th}$ row vector as $x"_{iP},i=1,\cdots,n.$ We then have, by the previous part, since $\hat{B}_{P}$ is an $\alpha$SM basis matrix,

[TABLE]

5. Let $\tilde{B}_{S}$ be an SM-basis matrix for $L_{S}$ with rows $\tilde{x}_{1S},\cdots,\tilde{x}_{nS}.$ We have

[TABLE]

∎

Remark 6**.**

In Theorem 25 we work with ${\cal V}_{P}={\cal V}_{SP}\times P.$ Our primary interest is in linking number lattices. So if ${\cal K}_{P}={\cal V}_{P}+L_{P}$ and ${\cal V}_{P}\supseteq{\cal V}_{SP}\times P,L_{P}\cap{\cal V}_{P}={\mathbf{0}}_{P}$ and $L_{P},{\cal V}_{SP}\times P$ are orthogonal, we redefine our generalized number lattice on $P$ to be $\hat{{\cal K}}_{P}\equiv{\cal V}_{SP}\times P+L_{P}$ and work with $\hat{{\cal K}}_{P}$ in place of ${\cal K}_{P}.$ Theorem 25 would then be immediately applicable.

8 Approximate successive minima bases of dual number lattices

We show in this section that from certain approximate successive minima bases of lattices, such bases can be built for dual number lattices, efficiently.

8.1 LLL- reduced bases

A special class of $\alpha SM-basis$ matrices that can be built in polynomial time is the LLL- reduced basis. We give the definition and important properties of this basis matrix below.

Definition 26**.**

Let $B_{P}\equiv(b_{1},\cdots,b_{m})^{T},$ be an $(m\times n)$ basis matrix of number lattice $L_{P}.$ We say $b^{*}_{1},\cdots,b^{*}_{m}$ is the Gram-Schmidt orthogonalization of $b_{1},\cdots,b_{m}$ iff

$b_{1}^{*}=b_{1};$** 2. 2.

$b_{i}^{*}=b_{i}-\Sigma_{j=1}^{i-1}\alpha_{ij}b_{j}^{*},$* where $\alpha_{ij}=\frac{<b_{i},b_{j}^{*}>}{<b_{j},b_{j}^{*}>}.$*

It can be seen that $b^{*}_{1},\cdots,b^{*}_{m}$ are orthogonal vectors and have the same span as $b_{1},\cdots,b_{m}.$

We define $B_{P}^{*}\equiv(b^{*}_{1},\cdots,b^{*}_{m})^{T}.$

It is clear that $B_{P}=KB_{P}^{*},$ where $K$ is a lower triangular matrix with $K_{ii}=1,K_{ij}=\alpha_{ij},j\leq i\leq m.$

Note that, if $\hat{B}_{P}\equiv\hat{K}\hat{B}_{P}^{*},$ where $\hat{K}$ is a lower triangular matrix with $1^{\prime}$s along the diagonal and the rows of $\hat{B}_{P}^{*}$ are orthogonal to each other, then the rows $\hat{b}_{i}^{*},i=1,\cdots m$ of $\hat{B}_{P}^{*}$ constitute the Gram-Schmidt orthogonalization of the rows $\hat{b}_{i},i=1,\cdots m,$ of $\hat{B}_{P}.$

Let us rewrite $B_{P}=KB_{P}^{*}$ as $B_{P}=KD(D^{-1}B_{P}^{*}),$ where $D$ is a diagonal matrix with

$D_{ii}=||b^{*}_{i}||,i=1,\cdots,m.$ Observe that the rows of $D^{-1}B_{P}^{*}$ have length $1$ and are orthogonal to each other. Let $F\equiv KD.$

Definition 27**.**

We say $B_{P}$ is LLL-reduced, or, equivalently, satisfies the size reduction and the Lovasz condition with respect to its Gram-Schmidt orthogonalization, with $\delta\in(\frac{1}{4},1),$ iff

(size reduction) $|\alpha_{ij}|\leq\frac{1}{2},1\leq i\leq m;$

• 2.

(Lovasz condition) $(\alpha_{(i+1)i})^{2}||b^{*}_{i}||^{2}+||b^{*}_{i+1}||^{2}\geq\ \delta||b^{*}_{i}||^{2},i.e.,|F_{(i+1)i}|^{2}+|F_{(i+1)(i+1)}|^{2}\geq\ \delta(|F_{ii}|^{2}).$

The following is an important property of LLL-reduced bases (Proposition 1.12 of [11]).

Theorem 28**.**

*For $\delta=\frac{3}{4},$ an LLL reduced basis $B_{P}\equiv(b_{1},\cdots,b_{m})^{T}$ of $L_{P},$ satisfies

$<b_{j},b_{j}>\ \leq 2^{m-1}(\lambda_{j}(L_{P}))^{2}.$ Therefore, an LLL-reduced basis for $L_{P}$ is an $\alpha$SM-basis matrix with $\alpha\equiv(\alpha_{1},\cdots,\alpha_{m}),\alpha_{i}=2^{\frac{m-1}{2}},i\leq m.$*

LLL-reduced bases can be built in polynomial time from an integral basis of $L_{P}.$ The worst case complexity of the LLL-algorithm is $\tilde{O}(m^{4}n(log(\hat{M}))^{2}),$ where $m$ is the dimension of $L_{P},$ $n=|P|,$ $\hat{M}$ is the maximum norm of a row in the original integral basis of $L_{P}$ ([11]). For comparison purposes we use the (simply stated) complexity of the original LLL-algorithm. Faster algorithms are available ([22]).

We will show below, that once we have an LLL-reduced basis of $L_{P},$ reduced bases, as good as the LLL-basis, of lattices linked to $L_{P},$ can be found more efficiently (Theorem 31). This would be true even if one compares with the algorithms in [22].

8.2 LLL-reduced basis of dual from primal

Let ${\cal K}_{P}=L_{P}+{\cal V}_{P},$ where $L_{P}$ is an integral lattice orthogonal to ${\cal V}_{P}.$ Let ${\cal V}_{P}$ have $B_{P}\equiv(b_{1},\cdots,b_{m})^{T}$ as a basis matrix. Let ${\cal K}_{P}^{d}=L_{P}^{(2)}+{\cal V}_{P}^{(2)},$ where $L_{P}^{(2)},{\cal V}_{P}^{(2)}$ are orthogonal. By Theorem 1, we know that ${\cal V}_{P}^{(2)}=(L_{P}+{\cal V}_{P})^{\perp}$ and $L_{P}^{(2)}$ has $B_{P}^{(2)}$ as a basis matrix, where $row(B_{P}^{(2)})=row(B_{P})$ and $B_{P}(B_{P}^{(2)})^{T}=\ identity\ matrix\ I_{m}.$

It may be noted that a basis matrix of $L_{P}^{(2)}$ would in general be rational rather than integral. However, for construction of LLL-reduced basis, it can be treated as an integral matrix multiplying every entry by $det(B_{P}B_{P}^{T}).$ This integer has size $\tilde{O}(\hat{M}^{m}),$ i.e., bit size $\tilde{O}(mlog(\hat{M})),$ where $m$ is the dimension of $L_{P}$ and $\hat{M}$ is the maximum size of $||B_{P}(i)||.$

The resulting matrix would have maximum norm $M^{\prime}$ of a row of bit length $log(M^{\prime})=\tilde{O}(mlog(\hat{M})).$

The LLL-reduced basis for such a matrix can be computed, if one proceeds from a basis of $L_{P}^{(2)},$ in $\tilde{O}(m^{4}n(log(M^{\prime}))^{2})=\tilde{O}(m^{6}n(log(\hat{M}))^{2})$ where $m$ is the dimension of $L_{P},$ $n=|P|,$ $\hat{M}$ is the maximum size of $||B_{P}(i)||.$

There are two components to the LLL-algorithm. The final reduced basis has to satisfy the size reduction condition and the Lovasz condition. The latter is more involved. If it is not satisfied we have to swap the successive bases which violate it and begin all over again. We will show that given an LLL-reduced basis for $L_{P},$ one can build one for $L_{P}^{(2)}$ without having to verify the Lovasz condition. The method given below is not new. Although not explicitly stated in the lecture on dual lattices in [23], it is essentially immediate from ideas there.

We have $B_{P}=F(D^{-1}B_{P}^{*}).$

Let $B_{P}^{{(2+)}}\equiv(K^{-1})^{T}D^{-1}(D^{-1}B_{P}^{*})=(F^{-1})^{T}(D^{-1}B_{P}^{*}).$

It is clear that $B_{P}(B_{P}^{{(2+)}})^{T}=I_{m},$ since $(D^{-1}B_{P}^{*})(D^{-1}B_{P}^{*})^{T}=I_{m},$ and further, that $row(B_{P})=row(B_{P}^{{(2+)}}).$

Let $B_{P}^{{(2)}}$ be the matrix obtained from $B_{P}^{{(2+)}}$ by reversing the order of its rows. Let $E,H,G,$ be obtained respectively by reversing the order of the columns and the order of the rows of $D^{-1},(K^{-1})^{T},(F^{-1})^{T}.$ For computation of $G$ from $F$ one may use the algorithm due to Dixon ([5]) which has complexity $\tilde{O}(m^{4}log(\hat{M})),$ where $\hat{M}$ is the maximum size of $||B_{P}(i)||.$ (Faster algorithms are available ([26]) but the expressions are not simple to state.)

Let $T_{k}$ denote the $k\times k$ matrix obtained by reversing the order of rows of $I_{k}.$ Note that $C^{\prime}=T_{k}C$ has the rows of $C$ in reverse order. In particular this means that $T_{k}^{2}=I_{k}.$

We then have

[TABLE]

It is clear that $B_{P}^{{(2)}}=GEB_{P}^{*}=HE^{2}B_{P}^{*}.$

Since the rows of $E^{2}B_{P}^{*}$ are orthogonal to each other and $H$ is a lower triangular matrix with $1^{\prime}$s along the diagonal, we have the following lemma.

Lemma 29**.**

The Gram-schmidt orthogonalization of $B_{P}^{{(2)}}$ is $E^{2}B_{P}^{*}.$ The Lovasz condition for $B_{P}^{{(2)}}$ with respect to its Gram-schmidt orthogonalization is $|G_{(i+1)i}|^{2}+|G_{(i+1)(i+1)}|^{2}\geq\ \delta(|G_{ii}|^{2}).$

We now have

Theorem 30**.**

If $B_{P}$ satisfies the size reduction condition with respect to its Gram-schmidt orthogonalization $B_{P}^{*},$ then $B_{P}^{{(2)}}$ satisfies the size reduction condition for the entries $(i+1,i),i=1,\cdots,m-1,$ with respect to its Gram-schmidt orthogonalization $E^{2}B_{P}^{*}.$ 2. 2.

If $B_{P}$ satisfies the Lovasz condition with respect to its Gram-schmidt orthogonalization $B_{P}^{*},$ then $B_{P}^{{(2)}}$ satisfies the Lovasz condition with respect to its Gram-schmidt orthogonalization $E^{2}B_{P}^{*}.$

The proof is relegated to F.

While the matrix $B_{P}^{{(2)}}=HE^{2}B_{P}^{*},$ satisfies the Lovasz condition, it may not satisfy the size reduction condition, since the entries $H_{ij},i>j+1$ are not necessarily of magnitude less or equal to $\frac{1}{2}.$ We therefore require one iteration, without swapping, of the LLL-algorithm, for performing the size reduction. The matrix, if scaled by $det(B_{P}B_{P}^{T})$ would become integral. The maximum norm $M^{\prime}$ of a row of this matrix would be $\tilde{O}((\hat{M})^{m}),$ where $\hat{M}$ is the maximum norm of a row of $B_{P}.$ If no swapping is involved, the size reduction alone takes $\tilde{O}(m^{2}nlog(M^{\prime}))=\tilde{O}(m^{3}nlog(\hat{M}))$ time, where $n=|P|$ ([22],[23]). The final LLL-reduced matrix is obtained by scaling down the result of the size reduction by $det(B_{P}B_{P}^{T}).$ We saw earlier, that computation of $G$ from $F$ can be done in $\tilde{O}(m^{4}log(\hat{M}))$ time. Let $M$ be the maximum magnitude of an entry in $L_{P}.$ Then it is clear that $log(\hat{M})=O(log(|P|M))=O(log(nM)).$ The multiplication $GEB_{P}^{*}$ can be done in $\tilde{O}((m^{\theta}n)log(\hat{M}))=\tilde{O}((m^{\theta}n)log(n{M}))$ time. Since $\theta<3,$ the overall computation of $B_{P}^{(2)"}$ from $B_{P}$ can be performed in $\tilde{O}((m^{4}+m^{3}n)log(n{M}))$ time.

The above is a substantial improvement over computing using the LLL-algorithm directly on the basis $B_{P}^{{(2)}}$ of $L^{(2)}_{P}$ (complexity $\tilde{O}(m^{6}n(log(n{M}))^{2})$). We summarize this discussion in part 1 of Theorem 31.

9 $\alpha$SM-basis matrices for number lattices through

linking and dualization

We summarize in this section our results on approximate SM-bases of number lattices related through linking by regular vector spaces and by dualization.

Theorem 31**.**

Let ${\cal V}_{SP}$ be a regular vector space, ${\cal K}_{P}$ be a generalized number lattice with ${\cal K}_{P}={\cal V}_{P}+L_{P},$ where $L_{P}$ is an integral lattice orthogonal to ${\cal V}_{P}.$ Let ${\cal V}_{SP}\times P={\cal V}_{P}$ and let ${\cal V}_{SP}\circ P=span({\cal K}_{P}).$

Let ${\cal K}_{S}\equiv{\cal V}_{SP}\leftrightarrow{\cal K}_{P},$ with ${\cal K}_{S}={\cal V}_{S}+L_{S},$ where ${\cal V}_{S},L_{S},$ are orthogonal.

Let ${\cal K}_{P}^{d}={\cal V}_{P}^{(2)}+L_{P}^{(2)},$ where ${\cal V}_{P}^{(2)},L_{P}^{(2)},$ are orthogonal. Let $B_{P}$ be an LLL-reduced basis for $L_{P}$ available with its Gram-Schmidt orthogonalization. Let $m\equiv dim(L_{P}),n\equiv|P|,$ and let $M$ be the maximum size of an entry in $B_{P}.$

*Let standard representative matrix $Q_{SP}$ of ${\cal V}_{SP},$ which is built according to the priority $(P,S),$ be available. We then have the following. *

*If $L_{P}$ is integral, an *$\alpha$SM-basis matrix $B_{P}^{(2)"}$ for $L_{P}^{(2)},$ where $\alpha\equiv(\alpha_{1},\cdots,\alpha_{m}),\alpha_{i}=2^{\frac{m-1}{2}},i=1,\cdots,m$ can be built in $\tilde{O}((m^{4}+m^{3}n)log(n{M}))$ time. Further, the maximum bit size of an entry in $B_{P}^{(2)"}$ would be $\tilde{O}(mlog(nM)).$ 2. 2.

*A *$\beta$*SM-basis matrix $B_{S}$ for $L_{S},$ where $\beta\equiv(\beta_{1},\cdots,\beta_{m}),\beta_{i}=(|S|\times|P|)2^{\frac{m-1}{2}},i=1,\cdots,m$ can be built in time *

$\tilde{O}(r^{4}log(|S|))+\tilde{O}(r^{2}mlog((|S|\times|P|)M))+O((r({\cal V}_{SP}^{\perp}\circ S))(r({\cal V}_{SP}\circ P))mlog(M)),$

where $r\equiv min(r({\cal V}_{SP}\times S),r({\cal V}_{SP}^{\perp}\circ S)).$ Further, the maximum bit size of an entry in $B_{S}$ would be $\tilde{O}(log((|S|\times|P|)M)).$ 3. 3.

*If $L_{P}$ is an integral matrix, and an LLL-reduced basis for $L_{P}^{(2)}$ has already been computed as in part 1 above, a *$\beta$SM-basis matrix $B^{(2)}_{S}$ for $L_{S}^{(2)},$ where $\beta\equiv(\beta_{1},\cdots,\beta_{m}),\beta_{i}=(|S|\times|P|)2^{\frac{m-1}{2}},i=1,\cdots,m,$ can be built in time

${O}((r({\cal V}_{SP}))^{2}|S\uplus P|)+\tilde{O}(\hat{r}^{4}log(|S|))+\tilde{O}(\hat{r}^{2}m^{2}log(M)+\hat{r}^{2}mlog(|S|\times|P|))$

$+O((r({\cal V}_{SP}\circ S)(r({\cal V}_{SP}^{\perp}\circ P)m^{2}log(M)),$

where $\hat{r}\equiv min(r({\cal V}_{SP}^{\perp}\times S),r({\cal V}_{SP}\circ S)).$ Further, the maximum bit size of an entry in $B^{(2)}_{S}$ would be $\tilde{O}(mlog(M)+log((|S|\times|P|))).$

We need the following lemma for the proof of Theorem 31.

Lemma 32**.**

Let ${\cal V}_{SP},{\cal K}_{P},L_{P},$ be as in the statement of Theorem 31. Let $Q_{SP}$ be a standard representative matrix of a regular vector space ${\cal V}_{SP}$ built according to the priority sequence $(P,S).$ Let $x_{P}\in L_{P}.$ Then,

to build a standard representative matrix $Q^{\prime}_{SP}$ of ${\cal V}_{SP}$ according to the priority sequence $(S,P)$ takes ${O}((r({\cal V}_{SP}))^{2}|S\uplus P|)$ time; 2. 2.

to find a vector $x_{S}$ such that $(x_{S},x_{P})\in{\cal V}_{SP},$ $||x_{S}||\leq\sqrt{(|S|\times|P|)}\times||x_{P}||$ and such that maximum bit length of an entry in $x_{S}$ is $O(log(|P|M)),$ where $log(M)$ is the maximum bit length of an entry in $x_{P}$ takes $O((r({\cal V}_{SP}^{\perp}\circ S))(r({\cal V}_{SP}\circ P))log(M))$ time. 3. 3.

to project $x_{S}$ onto ${\cal V}_{S},$ where ${\cal V}_{S}$ stands for ${\cal V}_{SP}\times S$ or ${\cal V}_{SP}^{\perp}\circ S$ takes

$\tilde{O}((r^{4}log(|S|))+\tilde{O}((r^{2}(log((|S|\times|P|)M)),$* time where $r\equiv min(r({\cal V}_{SP}\times S),r({\cal V}_{SP}^{\perp}\circ S)).$ The projections $x^{\prime}_{S},x"_{S}$ respectively of $x_{S}$ onto ${\cal V}_{S},{\cal V}_{S}^{\perp}$ have maximum bit size $O(log((|S|\times|P|)M)).$*

Proof.

1. This follows from Lemma 18.

2. This follows from Lemma 20.

3. Note that a standard representative matrix, say $Q_{S},$ of ${\cal V}_{SP}\times S$ is visible in the standard representative matrix $Q_{SP}$ of ${\cal V}_{SP}.$ and a standard representative matrix, say $Q^{\prime}_{S},$ of ${\cal V}_{SP}\circ S$ is visible in the standard representative matrix $Q^{\prime}_{SP}$ of ${\cal V}_{SP}.$

Let $x^{\prime}_{S},x"_{S}$ be projections of $x_{S}$ onto ${\cal V}_{SP}\times S,$ $({\cal V}_{SP}\times S)^{\perp},$ respectively. Since $Q_{SP}$ is totally unimodular, so is $Q_{S}.$ We then have,

[TABLE]

The maximum magnitude of entries in $Q_{S}Q_{S}^{T}$ is $|S|.$ Therefore the time for inversion ([5]) is

$\tilde{O}((r({\cal V}_{SP}\times S))^{4}log(|S|).$

From Lemma 20, the maximum bit size of an entry in $x_{S}$ is $O(log(|P|M)).$ The maximum bit size of an entry in $x^{\prime}_{S},x"_{S}$ can be seen to be $O(log((|S|\times|P|)M)).$ The multiplication time can be seen to be $\tilde{O}((r({\cal V}_{SP}\times S))^{2}log((|S|\times|P|)M)).$

Instead of computing $x^{\prime}_{S}$ first, we could have computed $x"_{S}.$ We would then have had to interpret $Q_{S}$ as a standard representative matrix for ${\cal V}_{SP}^{\perp}\circ S=({\cal V}_{SP}\times S)^{\perp}.$ If $Q^{\prime}_{S}=(I|K)$ is a standard representative matrix of ${\cal V}_{SP}\times S,$ then $Q"_{S}\equiv(-K^{T}|I)$ is a standard representative matrix for $({\cal V}_{SP}\times S)^{\perp}={\cal V}_{SP}^{\perp}\circ S.$ Thus $Q"_{S}$ is a totally unimodular matrix. Therefore, the above calculation holds for computing $x"_{S}$ first with $Q_{S}$ denoting a standard representative matrix of ${\cal V}_{SP}^{\perp}\circ S.$

Therefore the time for inversion can be taken to be $\tilde{O}(r^{4}log(|S|)),$ and the multiplication time can be seen to be $\tilde{O}(r^{2}log((|S|\times|P|)M)),$ where $r\equiv min(r({\cal V}_{SP}\times S),r({\cal V}_{SP}^{\perp}\circ S)).$

∎

Proof of Theorem 31.

1. This is clear from the discussion at the end of Subsection 8.2.

2. Given $x_{P}\in L_{P},$ to find $x_{S}$ such that $(x_{S},x_{P})\in{\cal V}_{SP}$ takes $\tilde{O}((r({\cal V}_{SP}^{\perp}\circ S))(r({\cal V}_{SP}\circ P))log(max|x_{P}(i)|)=\tilde{O}((r({\cal V}_{SP}^{\perp}\circ S)))(r({\cal V}_{SP}\circ P)log(M))$ time, by Lemma 20. This has to be done as many times as there are rows of $B_{P}.$ To project $x_{S}$ onto ${\cal V}_{SP}\times S$ takes $\tilde{O}((r({\cal V}_{SP}\times S))^{4}log(|S|))+\tilde{O}((r({\cal V}_{SP}\times S))^{2}log((|S|\times|P|)M))$ time. The first term, corresponds to the inversion which has to be done only once. The multiplication has to be done as many times as there are rows of $B_{P}.$ Therefore, to compute $B_{S}$ takes time

[TABLE]

If the projection of $x_{S}$ had been performed onto $({\cal V}_{SP}\times S)^{\perp}$ the computation time would be given by the above expression with $r({\cal V}_{SP}\times S)$ being replaced by $r(({\cal V}_{SP}\times S)^{\perp})=r({\cal V}_{SP}^{\perp}\circ S).$

We have, {\cal K}_{P}^{d}=\mbox{\cal V}^{(2)}_{P}+L^{(2)}_{P}, with \mbox{\cal V}^{(2)}_{P},L^{(2)}_{P}, orthogonal and {\cal K}_{S}^{d}=\mbox{\cal V}^{(2)}_{S}+L^{(2)}_{S}, with \mbox{\cal V}^{(2)}_{S},L^{(2)}_{S}, orthogonal. We are given that ${\cal V}_{SP}\circ P=span({\cal K}_{P}).$ Further, \mbox{\cal V}^{(2)}_{P}=({\cal V}_{P}+L_{P})^{\perp}=({\cal K}_{P})^{\perp}=({\cal V}_{SP}\circ P)^{\perp}={\cal V}_{SP}^{\perp}\times P and $span(L_{P})=span(L^{(2)}_{P}).$ By Theorem 1, $span({\cal K}_{P}^{d})={\cal V}_{P}^{\perp}=({\cal V}_{SP}\times P)^{\perp}={\cal V}_{SP}^{\perp}\circ P.$ We have, by IDT (Theorem 11), ${\cal K}_{S}^{d}=({\cal V}_{SP}\leftrightarrow{\cal K}_{P})^{d}=({\cal V}_{SP})^{\perp}\leftrightarrow{\cal K}_{P}^{d}.$

Since ({\cal V}_{SP})^{\perp}\leftrightarrow{\cal K}_{P}^{d}={\cal K}_{S}^{d},{\cal V}_{SP}^{\perp}\times P=\mbox{\cal V}^{(2)}_{P},{\cal V}_{SP}^{\perp}\circ P\supseteq{\cal K}_{P}^{d}, Theorem 25 is applicable.

To build a standard representative matrix $Q^{\prime}_{SP}$ of ${\cal V}_{SP}$ according to the priority sequence $(S,P)$ from $Q_{SP}$ takes ${O}((r({\cal V}_{SP}))^{2}|S\uplus P|)$ time. A standard representative matrix of ${\cal V}_{SP}^{\perp}$ according to the priority sequence $(P,S)$ immediately becomes available. Within this matrix standard representative matrices are visible for ${\cal V}_{SP}^{\perp}\times S,({\cal V}_{SP}^{\perp}\times S)^{\perp}={\cal V}_{SP}\circ S.$ The maximum bit size of an entry of $L^{(2)}_{P}$ is $\tilde{O}(mlog(M)).$ Therefore, by the computation in part 2 above, using ${\cal V}_{SP}^{\perp}$ in place of ${\cal V}_{SP},$ $(mlog(M)+log(|S|\times|P|))$ in place of $log(M)$ for maximum bit size, and taking $r\equiv min(r({\cal V}_{SP}\circ S),r({\cal V}_{SP}^{\perp}\times S)),$ the result follows. $\Box\,$

Remark 7**.**

We do not have a general method for building a reduced basis for the dual lattice from a general reduced basis of the primal lattice. For the LLL-reduced basis, fortunately we have an efficient method for building a reduced basis for the dual. If $L_{P}$ is a number lattice on $P$ and ${\cal V}_{SP}$ is a vector space on $S\uplus P,$ satisfying appropriate conditions relative to $L_{P},$ we can efficiently build a reduced basis for $L_{S},$ the number lattice part of ${\cal V}_{SP}\leftrightarrow L_{P},$ from a reduced basis for $L_{P}.$ However, there appears no easy way of building the reduced basis of the dual of $L_{S},$ starting from an LLL-reduced basis of $L_{P},$ except through the use of the implicit duality theorem as in part 3 of Theorem 31.

10 Closest and shortest vectors

In Section 7, we studied the case where a vector in a number lattice is uniquely lnked to that in another. In this section we examine the situation where the vector linked to a given vector is not unique. We examine questions of the kind ‘what is a shortest vector linked to a given vector?’ We show that the answer is related to the notion of a ‘closest vector’ defined below.

Let $L_{P}\subseteq{\mathbb{Q}}^{|P|}$ and let $x_{P}\in{\mathbb{Q}}^{|P|}.$ We say $\hat{x}_{P}\in L_{P}$ is closest in $L_{P}$ to $x_{P}$ iff whenever $x^{\prime}_{P}\in L_{P},$ we have $||\hat{x}_{P}-x_{P}||\leq||{x^{\prime}}_{P}-x_{P}||.$ We say $\hat{x}_{P}\in L_{P}$ is $\alpha-$closest in $L_{P},\alpha\in{\mathbb{Q}},$ to $x_{P}$ iff whenever $x^{\prime}_{P}\in L_{P},$ we have $||\hat{x}_{P}-x_{P}||\leq\alpha||{x^{\prime}}_{P}-x_{P}||.$

The closest vector problem is to find a vector closest in $L_{P}$ to a given vector $x_{P}.$ This problem is known to be NP-Hard ([1]). However, there is a polynomial time algorithm available ([2]) for finding an $\alpha-$closest vector in $L_{P}$ to a given vector, for $\alpha=2^{\frac{m}{2}}.$ This algorithm is given below.

Let $B_{P}$ be an LLL-reduced basis matrix for $L_{P}.$ Let $B_{P}=(b_{1P},\cdots,b_{mP})^{T}=KB^{*}_{P}=K(b^{*}_{1P},\cdots,b^{*}_{mP})^{T},$ where $B^{*}_{P}$ is the Gram-Schmitt orthogonalization for $B_{P}.$ Let $x_{P}\in{\mathbb{Q}}^{|P|}.$ Then the vector

[TABLE]

is $2^{\frac{m}{2}}-$closest in $L_{P}$ to $x_{P}.$

We now examine some natural questions about short vectors which are related to the closest vector problem.

Question 1. Suppose $L_{PQ}$ is a number lattice with $(x_{P},x_{Q})\in L_{PQ}.$ What is a shortest vector in $L_{PQ}$ whose restriction to $P$ is $x_{P}?$

We reformulate this question in terms of closest vectors in the lemma below.

Lemma 33**.**

Let $L_{PQ}$ be a number lattice with $(x_{P},x_{Q})\in L_{PQ}.$

A vector $\hat{x}_{Q}$ is a closest vector to $x_{Q}$ in $L_{PQ}\times Q$ iff

$(x_{P},x_{Q}-\hat{x}_{Q})$* is the shortest vector in $L_{PQ}$ whose restriction to $P$ is $x_{P},$ i.e.,*

$(x_{P},{x}_{Q}-\hat{x}_{Q})\in L_{PQ}$* and whenever $(x_{P},x^{\prime}_{Q})\in L_{PQ},||x_{Q}-\hat{x}_{Q}||\leq||x^{\prime}_{Q}||.$* 2. 2.

*The vector $\hat{x}_{Q}$ is an *$\alpha-$*closest vector to $x_{Q}$ in $L_{PQ}\times Q,$ iff *

$(x_{P},{x}_{Q}-\hat{x}_{Q})\in L_{PQ}$* and whenever $(x_{P},x^{\prime}_{Q})\in L_{PQ},||x_{Q}-\hat{x}_{Q}||\leq\alpha||x^{\prime}_{Q}||.$*

Proof.

Let $(x_{P},x_{Q})\in L_{PQ}.$ Then $(x_{P},x_{Q}-x"_{Q})\in L_{PQ}$ iff $x"_{Q}\in L_{PQ}\times Q.$

We have that $\hat{x}_{Q}$ is a closest vector to $x_{Q}$ in $L_{PQ}\times Q,$

iff $\langle(x_{Q}-x"_{Q}),(x_{Q}-x"_{Q})\rangle\geq\langle(x_{Q}-\hat{x}_{Q})),(x_{Q}-\hat{x}_{Q})\rangle,$ whenever $x"_{Q}\in L_{PQ}\times Q,$ i.e.,

iff $\langle(x_{P},x_{Q}-x"_{Q}),(x_{P},x_{Q}-x"_{Q})\rangle\geq\langle(x_{P},x_{Q}-\hat{x}_{Q}),(x_{P},x_{Q}-\hat{x}_{Q})\rangle,$ whenever $(x_{P},x_{Q}-x"_{Q})\in L_{PQ}.$

The proof of the second part is similar. ∎

Question 2. Let $L_{P}$ be a number lattice. Let $L^{\prime}_{P}$ be the projection of $L_{P}$ onto a vector space ${\cal V}_{P}.$ Given a vector $x^{\prime}_{P}\in L^{\prime}_{P}$ what is a shortest vector in $L_{P}$ whose projection onto ${\cal V}_{P}$ is $x^{\prime}_{P}?$

We can reduce this to Question 1 above as follows.

Let $B_{P}$ be a basis matrix for $L_{P},$ with ${x}_{iP},i=1,\cdots,m,$ as the $i^{th}$ row. Resolve ${x}_{iP}$ as ${x}_{iP}={x^{\prime}}_{iP}+{x"}_{iP},i=1,\cdots,m,{x^{\prime}}_{iP}\in{\cal V}_{P},{x"}_{iP}\in{\cal V}_{P}^{\perp}.$ Let $P^{\prime},P"$ be disjoint copies of $P$ and let $\hat{x}_{P^{\prime}},\hat{x}_{P"}$ denote copies respectively on $P^{\prime},P"$ of \hat{x}_{P}\in\mbox{\cal F}_{P}. Let $B_{P^{\prime}P"}$ be the matrix with $({x^{\prime}}_{iP^{\prime}},{x"}_{iP"}),i=1,\cdots m,$ as the $i^{th}$ row. Let $L_{P^{\prime}P"}$ be the number lattice on $P^{\prime}\cup P"$ generated by the rows of $B_{P^{\prime}P"}.$ It is clear that $x_{P}\in L_{P}$ iff $({x^{\prime}}_{P^{\prime}},{x"}_{P"})\in L_{PP"},$ where ${x^{\prime}}_{P},{x"}_{P"}$ are respectively the projections of $x_{P}$ onto ${\cal V}_{P},{\cal V}_{P}^{\perp}.$ We have that $||x_{P}||^{2}=||{x^{\prime}}_{P^{\prime}}||^{2}+||{x"}_{P"}||^{2}.$ Further since the rows of $B_{P}$ are linearly independent, so are the rows of $B_{P^{\prime}P"}.$ Therefore, $B_{P^{\prime}P"}$ is a basis matrix for $L_{P^{\prime}P"}.$

The projection $L^{\prime}_{P}$ of $L_{P}$ onto ${\cal V}_{P}$ has the copy $L_{P^{\prime}P"}\circ P^{\prime}.$ We have seen that the lengths of corresponding vectors in $L_{P},$ $L_{P^{\prime}P"}$ are the same. Therefore, Question 2 can be rephrased as ‘given $x^{\prime}_{P^{\prime}}\in L_{P^{\prime}P"}\circ P^{\prime}$ what is a shortest vector $({x^{\prime}}_{P^{\prime}},{x"}_{P"})\in L_{P^{\prime}P"}?,$’ which is Question 1.

We next examine what information about $L_{P},$ in terms of short vectors, can be garnered from ${\cal K}_{S}\equiv{\cal V}_{SP}\leftrightarrow L_{P}.$

We need a preliminary lemma before we phrase the next question.

Lemma 34**.**

Let ${\cal V}_{SP}$ be a vector space, $L_{P}\subseteq{\cal V}_{SP}\circ P,$ be a number lattice, with $span(L_{P})\supseteq{\cal V}_{SP}\times P.$ Let ${\cal K}_{S}\equiv{\cal V}_{SP}\leftrightarrow L_{P}.$ Then ${\cal K}_{S}={\cal V}_{SP}\times S+L_{S},$ where ${\cal V}_{SP}\times S,L_{S},$ are orthogonal and $L_{S}$ is a number lattice.

Proof.

Let $L^{\prime}_{P}$ be the projection of $L_{P}$ onto $({\cal V}_{SP}\times P)^{\perp}.$ Let ${\cal K}_{P}\equiv L^{\prime}_{P}+{\cal V}_{SP}\times P.$ Since $span(L_{P})\supseteq{\cal V}_{SP}\times P,$ we have that $span(L_{P})=span(L^{\prime}_{P}+{\cal V}_{SP}\times P)=span({\cal K}_{P}),$ and that $L_{P}+{\cal V}_{SP}\times P=L^{\prime}_{P}+{\cal V}_{SP}\times P={\cal K}_{P}.$ It is clear that ${\cal V}_{SP}\leftrightarrow{\cal K}_{P}\supseteq{\cal V}_{SP}\leftrightarrow L_{P}.$ Let $x_{S}\in{\cal V}_{SP}\leftrightarrow{\cal K}_{P}.$ Then there exists $\hat{x}_{P}\in{\cal K}_{P}$ such that $(x_{S},\hat{x}_{P})\in{\cal V}_{SP}.$ Now ${\cal K}_{P}=L_{P}+{\cal V}_{SP}\times P$ so that $\hat{x}_{P}=x_{P}+x"_{P},x_{P}\in L_{P},x"_{P}\in{\cal V}_{SP}\times P.$ Therefore $(x_{S},\hat{x}_{P})+(0_{S},x"_{P})=(x_{S},x_{P})\in{\cal V}_{SP}.$ Thus, ${\cal V}_{SP}\leftrightarrow L_{P}\supseteq{\cal V}_{SP}\leftrightarrow{\cal K}_{P}$ and therefore ${\cal V}_{SP}\leftrightarrow L_{P}={\cal V}_{SP}\leftrightarrow{\cal K}_{P}.$ From Theorem 25, part 1, it follows that ${\cal K}_{S}={\cal V}_{SP}\times S+L_{S},$ where ${\cal V}_{SP}\times S,L_{S},$ are orthogonal and $L_{S}$ is a number lattice. ∎

Question 3. Let ${\cal V}_{SP}$ be a regular vector space, $L_{P}\subseteq{\cal V}_{SP}\circ P,$ be a number lattice, with $span(L_{P})\supseteq{\cal V}_{SP}\times P.$ Let ${\cal K}_{P}=L_{P}+{\cal V}_{SP}\times P,$ with $L_{P},{\cal V}_{SP}\times P,$ not necessarily orthogonal. Let ${\cal K}_{S}\equiv{\cal V}_{SP}\leftrightarrow L_{P}.$ Given $x_{S}\in L_{S},$ what is a shortest vector $x_{P}\in L_{P}$ such that $(x_{S},x_{P})\in{\cal V}_{SP}?$

First let $L^{\prime}_{P}$ be the projection of $L_{P}$ onto $({\cal V}_{SP}\times P)^{\perp}.$ Let ${\cal K}_{P}\equiv L^{\prime}_{P}+{\cal V}_{SP}\times P.$ It is clear that ${\cal K}_{P}=L_{P}+{\cal V}_{SP}\times P.$ By Lemma 34, ${\cal K}_{S}\equiv{\cal V}_{SP}\leftrightarrow{\cal K}_{P}={\cal V}_{SP}\leftrightarrow L_{P}$ and we can write ${\cal K}_{S}={\cal V}_{SP}\times S+L_{S},$ where $L_{S}$ is orthogonal to ${\cal V}_{SP}\times S.$ Since ${\cal V}_{SP}\circ S\supseteq{\cal K}_{S},{\cal V}_{SP}\times S\subseteq{\cal K}_{S},$ by Theorem 9, we have that ${\cal V}_{SP}\leftrightarrow{\cal K}_{S}={\cal K}_{P}.$ By Theorem 25, if $x_{S}\in L_{S}$ there is a unique vector $x^{\prime}_{P}\in L^{\prime}_{P}$ such that $(x_{S},x^{\prime}_{P})\in{\cal V}_{SP}.$ Further, since $(x_{S},x^{\prime}_{P})\in{\cal V}_{SP},$ a vector $(x_{S},x_{P})\in{\cal V}_{SP}$ iff $x_{P}-x^{\prime}_{P}\in{\cal V}_{SP}\times P.$

Question 3. now reduces to Question 2. We find the unique $x^{\prime}_{P}\in L^{\prime}_{P}$ such that $(x_{S},x^{\prime}_{P})\in{\cal V}_{SP}.$ Next we find a shortest vector $x_{P}\in L_{P}$ whose projection onto $({\cal V}_{SP}\times P)^{\perp}$ is $x^{\prime}_{P}.$ Note that $x_{P}-x^{\prime}_{P}\in{\cal V}_{SP}\times P,$ so that $(x_{S},x_{P})\in{\cal V}_{SP}.$ Therefore $x_{P}$ is a shortest vector in $L_{P},$ such that $(x_{S},x_{P})\in{\cal V}_{SP}.$

Remark 8**.**

Questions 1, 2 and 3 can be rephrased in terms of ‘$\alpha-$’ closest rather than ‘shortest’ and the answers are similar.

11 Conclusion

We have shown that ideas that have proved useful in the study of electrical networks, viz., implicit inversion theorem (IIT) (Theorem 8) and implicit duality theorem (IDT) (Theorem 11), which are basic to implicit linear algebra, can be used with profit to study number lattices linked through regular vector spaces and through dualization.

We have built new self dual number lattices from old by using IDT, in a manner analogous to building new reciprocal electrical networks from old.

Using IIT, we have related properties of number lattices invertibly linked through a regular vector space and using IDT, to that of duals of such lattices. We have shown that reduced bases for such number lattices can be built efficiently starting from such a basis for one of them.

We have shown that the short vector problem under certain additional restrictions can be solved by solving an appropriate closest vector problem.

Appendix A Variations on the problem of finding basis for a number lattice

1. Consider the problem of finding the basis for the number lattice $\hat{L}_{S}$ of integral solutions to the equation

[TABLE]

where $A_{S}$ has full row rank and rows of $A_{S}$ generate the rows of $B.$ The SL-algorithm finds a unimodular matrix

[TABLE]

We claim that $(R_{21}\ \vdots\ R_{22})$ is a basis matrix for the number lattice $\hat{L}_{S}.$

To see this, first it is clear that rows of $(R_{21}\ \vdots\ R_{22})$ are in $\hat{L}_{S}.$

Next, since $R$ is unimodular, we can write any integral solution to Equation 8 as

[TABLE]

If $\lambda_{1}^{T}$ is nonzero

[TABLE]

since $\hat{A}_{S}$ has full row rank. Therefore $\lambda_{1}^{T}=0$ and any integral solution to Equation 8 can be written as

[TABLE]

Thus, $(R_{21}\ \vdots\ R_{22})$ is a basis matrix for the number lattice $\hat{L}_{S}.$

2. The problem of finding a basis for the collection of all integral vectors contained in a vector space \mbox{\cal V}_{S} is equivalent to finding a basis matrix for the integral solutions to the equation $x^{T}Q_{S}^{T}=0,$ where $Q_{S}$ is a basis matrix for ${\cal V}_{S}^{\perp}.$

3. Finding a basis for the intersection of a number lattice with a vector space can be handled similarly. Let $C_{S}$ be a basis matrix for number lattice $L_{S}.$ Let ${\cal V}_{S},{\cal V}_{S}^{\perp}$ be complementary orthogonal vector spaces with representative matrices $B_{S},Q_{S}$ respectively. A vector $x_{S}^{T}$ belongs to $L_{S}\cap{\cal V}_{S},$ iff $x_{S}^{T}=\lambda^{T}C_{S},\lambda\ \mbox{integral}$ and $x_{S}^{T}Q_{S}^{T}=0.$ We first find a basis matrix $\hat{C}$ for integral solutions to $\lambda^{T}C_{S}Q_{S}^{T}=0$ and then compute $\hat{C}C_{S}.$ We claim that this matrix is a basis matrix for $L_{S}\cap{\cal V}_{S}.$

The rows of $\hat{C}C_{S}$ belong to $L_{S}\cap{\cal V}_{S},$ since they are linear combinations of rows of $C_{S}$ orthogonal to ${\cal V}_{S}^{\perp}.$ To see the converse, any vector $x_{S}^{T}$ in $L_{S}\cap{\cal V}_{S},$ must satisfy $x_{S}^{T}=\lambda^{T}C_{S},\ \lambda\ \mbox{integral and}\ \lambda^{T}C_{S}Q_{S}^{T}=0.$ Any integral vector in the solution space of $\lambda^{T}C_{S}Q_{S}^{T}=0,$ can be written as $\sigma^{T}\hat{C},\ \sigma\ \mbox{integral}.$ Therefore any vector $x_{S}^{T}$ in $L_{S}\cap{\cal V}_{S},$ can be written $\sigma^{T}\hat{C}C_{S},\ \sigma\ \mbox{integral}.$

Appendix B Proof of Lemma 7

1. Suppose \mbox{{\cal K}{SP}}\leftrightarrow\mbox{{\cal K}{PQ}}=\mbox{{\cal K}{SQ}} and {\bf 0}_{PQ}\in\mbox{{\cal K}{PQ}}.

It is clear from the definition of the matched composition operation that \mbox{{\cal K}{SP}}\circ S\supseteq\mbox{{\cal K}{SQ}}\circ S.

Since {\bf 0}_{PQ}\in\mbox{{\cal K}{PQ}}, if \mbox{{\bf f}{S}}\oplus{\bf 0}_{P}\in\mbox{{\cal K}{SP}}, we must have that \mbox{{\bf f}{S}}\oplus{\bf 0}_{Q}\in\mbox{{\cal K}{SQ}}. Thus, \mbox{{\cal K}{SP}}\times S\subseteq\mbox{{\cal K}_{SQ}}\times S.

2. On the other hand suppose \mbox{{\cal K}{SP}}\times S\subseteq\mbox{{\cal K}{SQ}}\times S and \mbox{{\cal K}{SP}}\circ S\supseteq\mbox{{\cal K}{SQ}}\circ S. Let \hat{{\cal K}}_{PQ}\equiv\mbox{{\cal K}{SP}}\leftrightarrow\mbox{{\cal K}{SQ}}, i.e., $\hat{{\cal K}}_{PQ}$ is the collection of all vectors \mbox{{\bf f}{P}}\oplus\mbox{{\bf f}{Q}} s.t. for some vector \mbox{{\bf f}{S}}, \mbox{{\bf f}{S}}\oplus\mbox{{\bf f}{Q}}\in\mbox{{\cal K}{SQ}}, \mbox{{\bf f}{S}}\oplus\mbox{{\bf f}{P}}\in\mbox{{\cal K}_{SP}}.

Since ${\cal K}_{SP}$ is closed under subtraction, it contains the zero vector, the negative of every vector in it and is closed under addition.

Since {\bf 0}_{S}\oplus{\bf 0}_{P}\in\mbox{{\cal K}{SP}}, we must have that {\bf 0}_{S}\in\mbox{{\cal K}{SP}}\times S and therefore {\bf 0}_{S}\in\mbox{{\cal K}{SQ}}\times S. It follows that {\bf 0}_{S}\oplus{\bf 0}_{Q}\in\mbox{{\cal K}{SQ}}.

Hence, by definition of $\hat{{\cal K}}_{PQ},{\bf 0}_{PQ}\in\hat{{\cal K}}_{PQ}.$ Further, since both \mbox{{\cal K}{SP}},\mbox{{\cal K}{SQ}}, are closed under addition, so is $\hat{{\cal K}}_{PQ}$ since \hat{{\cal K}}_{PQ}=\mbox{{\cal K}{SP}}\leftrightarrow\mbox{{\cal K}{SQ}}.

We will now show that \mbox{{\cal K}{SP}}\leftrightarrow\hat{{\cal K}}_{PQ}=\mbox{{\cal K}{SQ}}.

Let \mbox{{\bf f}{S}}\oplus\mbox{{\bf f}{Q}}\in\mbox{{\cal K}{SQ}}. Since \mbox{{\cal K}{SP}}\circ S\supseteq\mbox{{\cal K}{SQ}}\circ S, we must have that \mbox{{\bf f}{S}}\oplus\mbox{{\bf f}{P}}\in\mbox{{\cal K}{SP}}, for some \mbox{{\bf f}{P}}. By the definition of $\hat{{\cal K}}_{PQ},$ we have that \mbox{{\bf f}{P}}\oplus\mbox{{\bf f}_{Q}}\in\hat{{\cal K}}_{PQ}.

Hence, \mbox{{\bf f}{S}}\oplus\mbox{{\bf f}{Q}}\in\mbox{{\cal K}_{SP}}\leftrightarrow\hat{{\cal K}}_{PQ}.

Thus, \mbox{{\cal K}{SP}}\leftrightarrow\hat{{\cal K}}_{PQ}\supseteq\mbox{{\cal K}{SQ}}.

Next, let \mbox{{\bf f}{S}}\oplus\mbox{{\bf f}{Q}}\in\mbox{{\cal K}{SP}}\leftrightarrow\hat{{\cal K}}_{PQ}, i.e., for some \mbox{{\bf f}{P}},\mbox{{\bf f}{S}}\oplus\mbox{{\bf f}{P}}\in\mbox{{\cal K}{SP}} and \mbox{{\bf f}{P}}\oplus\mbox{{\bf f}_{Q}}\in\hat{{\cal K}}_{PQ}.

We know, by the definition of $\hat{{\cal K}}_{PQ},$ that there exists \mbox{{\bf f}{S}}^{\prime}\oplus\mbox{{\bf f}{Q}}\in\mbox{{\cal K}{SQ}} s.t. \mbox{{\bf f}{S}}^{\prime}\oplus\mbox{{\bf f}{P}}\in\mbox{{\cal K}{SP}}.

Since ${\cal K}_{SP}$ is closed under subtraction, we must have, (\mbox{{\bf f}{S}}-\mbox{{\bf f}{S}}^{\prime})\oplus{\bf 0}_{P}\in\mbox{{\cal K}_{SP}}.

Hence, \mbox{{\bf f}{S}}-\mbox{{\bf f}{S}}^{\prime}\in\mbox{{\cal K}{SP}}\times S\subseteq\mbox{{\cal K}{SQ}}\times S.

Hence (\mbox{{\bf f}{S}}-\mbox{{\bf f}{S}}^{\prime})\oplus{\bf 0}_{Q}\in\mbox{{\cal K}_{SQ}}.

Since ${\cal K}_{SQ}$ is closed under addition and \mbox{{\bf f}{S}}^{\prime}\oplus\mbox{{\bf f}{Q}}\in\mbox{{\cal K}_{SQ}},

it follows that (\mbox{{\bf f}{S}}-\mbox{{\bf f}{S}}^{\prime})\oplus{\bf 0}_{Q}+\mbox{{\bf f}{S}}^{\prime}\oplus\mbox{{\bf f}{Q}}=\mbox{{\bf f}{S}}\oplus\mbox{{\bf f}{Q}} also belongs to ${\cal K}_{SQ}$.

Thus, \mbox{{\cal K}{SP}}\leftrightarrow\hat{{\cal K}}_{PQ}\subseteq\mbox{{\cal K}{SQ}}.

3. From parts 1 and 2 above, the equation \mbox{{\cal K}{SP}}\leftrightarrow\mbox{{\cal K}{PQ}}=\mbox{{\cal K}{SQ}} can be satisfied by some $\hat{{\cal K}}_{PQ}$ if and only if \mbox{{\cal K}{SP}}\circ S\supseteq\mbox{{\cal K}{SQ}}\circ S and \mbox{{\cal K}{SP}}\times S\subseteq\mbox{{\cal K}_{SQ}}\times S.

Next, let $\hat{{\cal K}}_{PQ}$ satisfy the equation \mbox{{\cal K}{SP}}\leftrightarrow\hat{{\cal K}}_{PQ}=\mbox{{\cal K}{SQ}} and be closed under addition.

From part 2, we know that if $\hat{{\cal K}}_{PQ}$ satisfies \mbox{{\cal K}{SP}}\circ P\supseteq\hat{{\cal K}}_{PQ}\circ P and \mbox{{\cal K}{SP}}\times P\subseteq\hat{{\cal K}}_{PQ}\times P,

then \mbox{{\cal K}{SP}}\leftrightarrow(\mbox{{\cal K}{SP}}\leftrightarrow\hat{{\cal K}}_{PQ})=\hat{{\cal K}}_{PQ}. But $\hat{{\cal K}}_{PQ}$ satisfies \mbox{{\cal K}{SP}}\leftrightarrow\hat{{\cal K}}_{PQ}=\mbox{{\cal K}{SQ}} and satisfies \mbox{{\cal K}{SP}}\circ P\supseteq\hat{{\cal K}}_{PQ}\circ P and \mbox{{\cal K}{SP}}\times P\subseteq\hat{{\cal K}}_{PQ}\times P,

It follows that for any such $\hat{{\cal K}}_{PQ},$ we have \mbox{{\cal K}{SP}}\leftrightarrow\mbox{{\cal K}{SQ}}=\hat{{\cal K}}_{PQ}.

This proves that \hat{{\cal K}}_{PQ}\equiv\mbox{{\cal K}{SP}}\leftrightarrow\mbox{{\cal K}{SQ}} is the only solution to the equation \mbox{{\cal K}{SP}}\leftrightarrow\mbox{{\cal K}{PQ}}=\mbox{{\cal K}{SQ}}, under the condition \mbox{{\cal K}{SP}}\circ P\supseteq\mbox{{\cal K}{PQ}}\circ P and \mbox{{\cal K}{SP}}\times P\subseteq\mbox{{\cal K}_{PQ}}\times P. $\Box\,$

Appendix C Diagrams of expressions of linkages

A regular expression of linkages is best represented by means of a diagram, where nodes correspond to individual linkages, and edges correspond to index sets common to two linkages.

Example 35**.**

Consider the regular expression

[TABLE]

Figure 1 (a) shows the diagram for this expression. (Individual linkages are represented only by the subscript. For instance ${(S_{5}(-A)VC)}$ represents ${\cal K}_{(S_{5}(-A)VC)}.$) Let $N$ be a subset of the nodes of the diagram. The subexpression of $\cal E$ on $N$ will have as diagram the subgraph with nodes $N$ and edges with both endpoints in $N.$

We need use only $`\leftrightarrow$’ or only $`\rightleftharpoons$’ since the other operation can be handled by changing the sign of one of the index sets as in ${\cal K}_{(S_{5}(-A)VC)}.$ We will make edges corresponding to $`\leftrightarrow$’ bold and those corresponding to $`\rightleftharpoons$’ dotted. In the present expression we have used only $`\leftrightarrow$’.

Suppose in the subexpression ${\cal K}_{LR}\leftrightarrow{\cal K}_{RM}$ we have ${\cal K}_{LR}\circ R\supseteq{\cal K}_{RM}\circ R,{\cal K}_{LR}\times R\subseteq{\cal K}_{RM}\times R.$ We then make the edge corresponding to $R$ directed from ${\cal K}_{LR}$ to ${\cal K}_{RM}.$ In the present expression, we have taken ${\cal K}_{S_{1}QRT},{\cal K}_{S_{3}RPV},$ to satisfy ${\cal K}_{S_{1}QRT}\circ R\supseteq{\cal K}_{S_{3}RPV}\circ R,{\cal K}_{S_{1}QRT}\times R\subseteq{\cal K}_{S_{3}RPV}\times R,$ so that in the diagram there is a directed edge labelled $R,$ from the node ${S_{1}QRT}$ to the node ${S_{3}RPV}.$

Let us define the dual to the above expression to be

[TABLE]

The diagram for this dual expression is shown in Figure 1 (b). It is identical to that of the primal except that the a node ${\cal K}_{S}$ has been replaced by ${\cal K}^{d}_{S}$ and a bold edge corresponding to an index set $P$ has been replaced by a dotted edge corresponding to $P.$ Further, directed edges retain their direction in the diagram of the dual. This is because $({\cal K}_{LR}\leftrightarrow{\cal K}_{RM})^{d}\ =\ {\cal K}^{d}_{LR}\rightleftharpoons{\cal K}^{d}_{RM}$ and ${\cal K}_{LR}\circ R\supseteq{\cal K}_{RM}\circ R,{\cal K}_{LR}\times R\subseteq{\cal K}_{RM}\times R$ is equivalent to ${\cal K}^{d}_{LR}\times R\subseteq{\cal K}^{d}_{RM}\times R,{\cal K}^{d}_{LR}\circ R\supseteq{\cal K}^{d}_{RM}\circ R.$

Let the above ‘primal’ expression have the evaluation ${\cal K}_{S_{1}S_{2}S_{3}S_{4}S_{5}S_{6}}.$

From IDT (Theorem 11), it follows that the dual expression evaluates to ${\cal K}^{d}_{S_{1}S_{2}S_{3}S_{4}S_{5}S_{6}}.$

Finally, if every one of the linkages in the expression is a full dimensional number lattice, by Theorem 12, every subexpression will evaluate to a full dimensional number lattice.

Let ${\cal K}_{LR},{\cal K}_{RM}$ be as above with a directed bold edge from ${\cal K}_{LR}$ to ${\cal K}_{RM},$ in the diagram of the expression.

Let us divide the nodes of the diagram into two sets $N_{1},N_{2}$ with ${\cal K}_{LR}\in N_{1}$ and ${\cal K}_{RM}\in N_{2},$ where nodes in $N_{1}$ have no edges between them. The subexpression on $N_{1}$ will have the form ${\cal K}_{GHLR}\equiv\oplus_{i,j}\{{\cal K}_{G_{i}H_{j}}\}\oplus{\cal K}_{LR},$ where the $G_{i},H_{j},L,R,M$ are mutually disjoint, no index set occurs more than once and $G\equiv\uplus G_{i},H\equiv\uplus H_{j}.$ Let the subexpression on $N_{2}$ evaluate to ${\cal K}_{CHRM}.$

The original expression can be simplified to ${\cal K}_{GHLR}\leftrightarrow{\cal K}_{CHRM}.$ The diagram of this reduced expression will have only two nodes but many edges corresponding to the $H_{j}$ and $R.$

We will show that in the reduced diagram, there will be a directed edge corresponding to $R,$ from node ${\cal K}_{GHLR}$ to node ${\cal K}_{CHRM},$ in addition to the edges corresponding to the $H_{j}.$

Firstly, it can be shown that ${\cal K}_{LR}\circ R\supseteq{\cal K}_{CHRM}\circ R$ and ${\cal K}_{LR}\times R\subseteq{\cal K}_{CHRM}\times R$ as follows.

By Theorem 8, we have that ${\cal K}_{YZ}\circ Z\supseteq({\cal K}_{XY}\leftrightarrow{\cal K}_{YZ})\circ Z$ and ${\cal K}_{YZ}\times Z\subseteq({\cal K}_{XY}\leftrightarrow{\cal K}_{YZ})\times Z.$ By repeated application of this result, if necessary, we can infer that ${\cal K}_{RM}\circ R\supseteq{\cal K}_{CHRM}\circ R$ and ${\cal K}_{RM}\times R\subseteq{\cal K}_{CHRM}\times R.$ But we have ${\cal K}_{LR}\circ R\supseteq{\cal K}_{RM}\circ R$ and ${\cal K}_{LR}\times R\subseteq{\cal K}_{RM}\times R.$ Thus, ${\cal K}_{LR}\circ R\supseteq{\cal K}_{CHRM}\circ R$ and ${\cal K}_{LR}\times R\subseteq{\cal K}_{CHRM}\times R.$

Next, since ${\cal K}_{GHLR}=\oplus_{i,j}\{{\cal K}_{G_{i}H_{j}}\}\oplus{\cal K}_{LR},$ we must have ${\cal K}_{GHLR}\circ R={\cal K}_{LR}\circ R$ and ${\cal K}_{GHLR}\times R={\cal K}_{LR}\times R.$ It follows that ${\cal K}_{GHLR}\circ R={\cal K}_{LR}\circ R\supseteq{\cal K}_{CHRM}\circ R$ and ${\cal K}_{GHLR}\times R={\cal K}_{LR}\times R\subseteq{\cal K}_{CHRM}\times R.$ Therefore, in the reduced diagram there will be a directed edge corresponding to $R,$ from node ${\cal K}_{GHLR}$ to node ${\cal K}_{CHRM}.$

The dual expression will have a corresponding reduced expression ${\cal K}^{d}_{GHLR}\rightleftharpoons{\cal K}^{d}_{CHRM}.$ Here again the arrow for the edge corresponding to $R$ will be from ${\cal K}^{d}_{GHLR}$ to ${\cal K}^{d}_{CHRM}.$

If the directed edge is from $N_{2}$ to $N_{1},$ in general, the diagram of the dual reduced expression may only have an undirected edge.

In the case of the expression

[TABLE]

we can take $N_{1},$ for instance, to be $N_{1}\equiv{\cal K}_{S_{1}QRT}\oplus{\cal K}_{S_{5}(-A)VC},$ since there is no edge between the nodes.

We then have the reduced expression

[TABLE]

[TABLE]

Figure 2(a) shows the diagram of this reduced expression. Here, the nodes have been split into $N_{1},N_{2},$ with $N_{1}$ to the right. We have

$N_{2}\equiv{\cal K}_{S_{2}S_{3}S_{4}S_{6}ACTQRV}=({\cal K}_{S_{2}QPW}\leftrightarrow{\cal K}_{S_{3}RPV}\leftrightarrow{\cal K}_{S_{4}ATB}\leftrightarrow{\cal K}_{S_{6}CBW}).$

The internal edges (which appear in Figure 1), labelled $B,P,W$ involving nodes of $N_{2}$ have been deleted, there are undirected edges labelled $T,Q,V,C,A,$ and a directed edge from $N_{1}$ to $N_{2},$ labelled $R$ corresponding to the directed edge from ${\cal K}_{S_{1}QRT}$ to ${\cal K}_{S_{3}RPV}$ in Fig 1(a). Its meaning here is that

[TABLE]

Figure 2(b), shows the diagram of the dual of the reduced expression. It is identical to the diagram in Figure 2(a), except that we have to interpret the node, say $\alpha\beta\gamma$ as ${\cal K}^{d}_{\alpha\beta\gamma},$ the dotted edges as corresponding to ‘$\rightleftharpoons$’ and the directed dotted edge from $N_{1}$ to $N_{2},$ labelled $R$ as corresponding to the directed dotted edge from ${\cal K}^{d}_{S_{1}QRT}$ to ${\cal K}^{d}_{S_{3}RPV}$ in Fig 1(b). Its meaning here is that

[TABLE]

These ideas extend to the case where the set of nodes of the diagram is partitioned into $\{N_{1},\cdots,N_{k}\},$ if there are internal edges, only the block of nodes $N_{k}$ contains them, and finally, the directed edges are either between the node sets $N_{1},\cdots,N_{k-1},$ or into $N_{k}.$

Appendix D Proof of Lemma 18

It is sufficient to prove the result when the priority sequences for $B_{1},B_{2},$ are of the form

$(e_{1},\cdots,e_{i-1},e,e^{\prime},e_{i+2},\cdots e_{|S|})$ for $B_{1},$ and of the form $(e_{1},\cdots,e_{i-1},e^{\prime},e,e_{i+2},\cdots e_{|S|})$ for $B_{2}.$ Without loss of generality we can assume the standard representative matrix of ${\cal V}_{S}$ with respect to the base $B_{1}$ can be written as

[TABLE]

where $k_{2}=\pm 1$ and the order of columns is $B_{1}\cap B_{2},e,e^{\prime},(S-B_{1})\cap(S-B_{2}).$ To convert this into a standard representative matrix of ${\cal V}_{S}$ with respect to $B_{2}$ we have to add or subtract the last row from the earlier rows in order to convert the entries in the column $e^{\prime}$ in these earlier rows from $k_{1}$ to the ${\mathbf{0}}$ column. This results in a standard representative matrix and therefore must be totally unimodular. So the entries in the columns corresponding to $e,(S-B_{1})\cap(S-B_{2})$ in this new matrix must be $0,\pm 1$ and therefore all numbers encountered in this computation are $0,\pm 1.$ This computation is $O(|B_{1}\cap B_{2}||S|)$ when $|B_{1}-B_{2}|=1$ and in general has to be repeated $|B_{1}-B_{2}|$ times so that the computation is finally $O(|B_{1}|^{2}|S|)=O(m^{2}|S|).$ $\Box\,$

Appendix E Proof of Lemma 24

It is sufficient to prove the following statement for the pair of permutations, $(id(\cdot),\rho(\cdot)),\rho\equiv\mu\nu^{-1}.$

Statement Let $S\equiv\{t,\cdots,t+n\},t\in\mathbb{Z},$ and let $\rho(\cdot)$ act on $S.$ Let $S^{\prime}\equiv\{j,i\leq j\leq\rho(i),i\in S\}.$ Then, $S^{\prime}=S.$

Note that if the statement is true for $t=t^{\prime}$ it is also true for $t=t^{\prime}+s,s\in\mathbb{Z}.$

The statement is clearly true for $n=1.$ Let the statement be true for $n\leq k.$

Now, let $n=k+1.$ Let $T_{r}\equiv\{t,\cdots,t+r\}\subseteq S.$ We have $T_{0}\subseteq S^{\prime}.$

We will show that if $0\leq r\leq k,$ then $T_{r+1}\equiv\{t,\cdots,t+r+1\}\subseteq S^{\prime}.$

If $\rho(T_{r})=T_{r},$ then $\rho(S-T_{r})=S-T_{r}.$ So the statement is true since it is true for $n\leq k$ and we have $0\leq r\leq k,0\leq k-r\leq k.$ (We are in this case dealing with two permutations on sets of smaller size.)

Suppose $\rho(T_{r})\neq T_{r}.$ Then there exists $m,0\leq m\leq r$ such that $\rho(t+m)>t+r.$ Therefore, $t+m<t+r+1\leq\rho(t+m),$ so that $t+r+1\in S^{\prime},$ i.e., $T_{r+1}\equiv\{t,\cdots,t+r+1\}\subseteq S.$

We conclude that $S=T_{k+1}\subseteq S^{\prime}.$

Thus the statement is true for n=k+1. $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \Box\,$

Appendix F LLL-reduced basis for dual number lattice

The proof of Theorem 30 is based on the following simple lemma which can be verified by direct calculation.

Lemma 36**.**

*Let $F$ be a lower triangular matrix with

$F_{ii}=||b^{*}_{i}||,i=1,\cdots,m\ \mbox{and}\ F_{(i+1)i}=\alpha_{(i+1)i}||b^{*}_{i}||,i=1,\cdots,m-1.$ Then, $G\equiv T_{m}(F^{-1})^{T}T_{m}$ is a lower triangular matrix satisfying*

[TABLE]

*Proof of Theorem 30

*1. This is clear from Lemma 36.

2. The Lovasz condition for $B_{S}$ is

[TABLE]

Dividing the above inequality throughout by $||b^{*}_{i}||^{2}\times||b^{*}_{i+1}||^{2},$ we get

[TABLE]

Therefore,

[TABLE]

$i.e.,|G_{(i+1)i}|^{2}+|G_{(i+1)(i+1)}|^{2}\geq\delta(|G_{ii}|^{2}),i=1,\cdots,m-1,$ which proves the required result. $\Box\,$