## I Introduction

We consider low rank lattice codes for transmission over a noisy erasure channel as illustrated in Fig. 1. In this figure

information symbols are mapped by an encoder/modulator to a vector

, where is a rank- lattice in . The output of the additive noise channel is , where is a noise vector independent of and with independent components. Components of are then erased by an erasure network, whose outputs are obtained by retaining only those symbols of indexed by subsets in a given sub-collection of subsets; thus coincides with is the positions identified by . As an example, with , and ,. A decoder estimates the source symbols based on

with a probability of error denoted

. The objective is to minimize for each by designing a single codebook which satisfies a power constraint , wheredenotes expectation with respect to a uniform distribution on the codebook. Here we consider as our sub-collection

,*all -subsets*of . Our paper is organized as follows. Prior work and lattice background is in Sec. II. Two performance bounds are presented in Sec. III, constructions for codes in dimension are presented and compared to the derived bounds in Sec. IV. A summary is in Sec. LABEL:sec:summary.

We use the acronym w.l.o.g to mean ‘without loss of generality’.

## Ii Prior Work and Review of Lattice Terminology

The problem considered here may be viewed as a code design problem for a special case of the *compound channel*, see e.g. [blackwell1959capacity], [csiszar1991capacity]. This work was motivated by a study on cross layer coding that appeared in [courtade2011optimal]. For prior contributions on the Gaussian erasure channel, please refer to [ozccelikkale2014unitary] and the references therein.We now develop notation and some basic definitions for low rank lattices in .
Let be a collection of orthonormal column vectors in and let denote the associated orthonormal matrix. Let denote a generator matrix of full rank for a lattice . We will refer to as the *mother* lattice. Let

(1) |

is a rank- lattice in . Let denote the Gram matrix of ( is the transpose operator).

The *determinant* of a lattice is defined in terms of the determinant of its Gram matrix by
Let denote the radius of the largest inscribed sphere in a Voronoi cell of . The packing density of is defined in terms of the volume of a unit-radius Euclidean ball in by

(2) |

We denote by , the largest packing density that can be acheived by any lattice in . The problem of finding lattices that maximize the packing density is a classical problem in number theory and geometry, with several excellent references [SPLAG], [gruber1987geometry].

The following definitions are from [gruber1987geometry]. A body captures the notion of a solid subset of , specifically, is a *body* if it has nonempty interior and is contained in the closure of its interior [gruber1987geometry].
A body is said to be *centrally symmetric* if , where . A closed body with the property that for any , the point for every is called a *star body*. While convex bodies are star bodies, the converse is not true. A simple example, and one directly relevant to us is the star body formed by the union of centrally symmetric ellipsoids in . A lattice is said to be *admissible* for , or -admissible, if no non-zero point in lies in . The greatest lower bound of over all admissible lattices is called the lattice constant of , denoted (which is set to is there are no admissible lattices). A -admissible lattice with is said to be a *critical* lattice for .

A lattice is said to be a *packing lattice* for a body if the sets and are disjoint for all non-zero . It is known, Thm.1, Ch. 3, Sec 20 [gruber1987geometry], that is a packing lattice for centrally symmetric, convex body if and only if it is admissible for . Thus, for a convex body, the problem of finding a packing lattice for is equivalent to that of finding an admissible lattice for . The connection between packings and admissibility for non-convex bodies is messier. The distinction arises because for a centrally symmetric body , is a lattice packing of if and only if it is admissible for , where denotes the set sum or Minkowski sum. If the centrally symmetric body is also convex, then and thus packing problems and admissibility problems are closely related. On the other hand, if is centrally symmetric but non-convex, in order to solve a packing problem for one must solve an admissibility problem for , and this set may not be as easily described as .

In our application, we need to index body by subset in a given sub-collection of subsets and our problem is one of packing , which is non-convex. While admissibility for non convex centrally symmetric body says nothing in general about packings for , it turns out that our design problem is equivalent to finding a critical lattice for because the decoder knows . Thus it is possible to draw on the theory of admissible lattices for star bodies. This theory provides several key ingredients to help find a solution to this problem. Most notably, in the chapter on Mahler’s compactness theorem [cassels2012introduction], Theorem VII states that *every critical lattice for a bounded star body has linearly independent points on the boundary of .*

## Iii Bounds

Let . For , let be the lattice obtained be retaining only those coordinates that are in or equivalently is the projection of into the subspace where is the th unit vector in . For any k-subset , we denote by the submatrix obtained by extracting from the rows identified by . The generator matrix for is and its Gram matrix .

Define the (packing volume) contraction ratio

(3) |

let and let .

### Iii-a Determinant Upper Bound

We will use symbols ,

to denote the arithmetic mean and geometric mean, respectively, of the real numbers

over some index set . When a -dim mother lattice is set in using a basis , the projections on the subsets , cannot all be simultaneously good. There are two important factors that measure the ‘goodness’ of the projections—the packing density and the scale of the lattices . The following theorem develops one of two bounds presented in this paper.###### Theorem 1.

(Determinant Bound) Given a mother lattice and orthonormal basis , let and be respectively, the geometric mean of the volume contraction ratios and packing densities of the child lattices , taken over all -subsets of . Then

(4) |

Equality holds if and only if all child lattices have equal determinants.

###### Proof.

The packing densities of the mother lattice and child lattice are related by the following identity

(5) |

Compute the geometric mean of both sides over the collection of -subsets to get

(6) |

From the arithmetic-geometric mean inequality it follows that

(7) |

and equality holds if and only if is a constant with respect to . However

(8) | |||||

where in (a) we have used the Cauchy-Binet formula, see e.g. [horn2012matrix]. The remainder of the proof follows directly. ∎

The following corollary is immediate.

###### Corollary 1.

Given mother lattice and orthonormal basis , let be the minimum volume contraction ratio of the child lattices , taken over all -subsets of , and the geometric mean of the packing densities. Then

(9) |

Equality holds in the left inequality iff all the contraction ratios are equal and all the child lattices have equal determinants. Equality holds in the right inequality iff all child lattices achieve the optimal packing density in dimension .

### Iii-B Trace Upper Bound

###### Theorem 2.

(Trace Bound) For an code, the compaction ratio is bounded as

(10) |

Equality holds if the shortest vector of each lattice is the image of the shortest vector in .

###### Proof.

Upon summing over all -subsets we obtain

(11) | |||||

By definition the smallest packing radius of any child lattice satisfies

(12) |

for any non-zero and any -subset . Upon averaging over subsets we obtain the upper bound

(13) | |||||

Equality holds if is the shortest vector in for all . Thus

(14) |

and (10) follows immediately. ∎

## Iv Analysis of Some Codes

We construct for for various values of and various mother lattices . Numerical results for the , are presented in Fig. 2, in which is plotted as a function of the packing density of the mother lattice. We have plotted the determinant bound using both the optimal and the cubic lattice for the child lattices. We have also plotted the trace bound.

Observe that in the case there is a significant gap between the best possible construction and the upper bounds. In the case performance close to the determinant bound is achieved by setting the mother lattice to be the cubic lattice. Also with the cubic lattice as the mother lattice, since the trace bound is lower than the determinant bound, this is proof that it is impossible to simultaneously achieve the packing density of when the mother lattice is the cubic lattice.

(a) | |

(b) |

Geometrically, the ability to decode correctly, post-erasure, with iid Gaussian noise is determined by the largest noise sphere which can be packed by the projected (or child) lattice in each of the subspaces . When each noise sphere is projected back onto the subspace spanned by the columns of , a noise sphere is transformed into an ellipsoid. To see this consider the noise sphere in subspace . Setting , this leads to the noise ellipsoid . The packing of the noise ellipsoids , (recall that is the half the length of the shortest non-zero vector in ), by the mother lattice is shown in the six panels in Fig. 3(a), one for each subspace. Fig. 3(b) shows the star body . This illustrates the star body, which is the union of the six ellipses, two of which are circles. Also shown are points of the lattice , which in this case is an admissible lattice for this star body and illustrates also the interpretation as a code design problem for the compound channel. Observe that is *simultaneously* good as a packing for all six erasure configurations, i.e. for each of the bodies . Also, is *simultaneously* critical for five of the six bodies (notice that one circle does not touch any of the lattice points).

### Iv-a

Let , . We obtain . The trace and determinant upper bounds yield . Hence this construction is optimal.

### Iv-B

With , and

we obtain six child lattices with Gram matrices

(15) |

All six child lattices are similar to . The first one has shortest vector of square length and all the others have square length . This code achieves , .

The trace upper bound is , regardless of the mother lattice, while the determinant upper bound depends on the mother lattice, and is and for (hexagonal lattice) and , resp. Thus the determinant bound is tighter than the trace bound but greater than . This construction does not meet the trace bound or the determinant bound with equality. However with , the following theorem shows this to be the best possible. A computer-based search has also failed to reveal any improvements in .

###### Theorem 3.

Let be a matrix with orthonormal columns, so that . Let . Let where is a 2-subset of . Let be the length of the shortest non-zero vector in . Then for at least one of the six 2-subsets of .

###### Proof.

For this proof we will write

(16) |

Our proof is by contradiction. Suppose that the shortest non-zero vector in all has square length . Now , in at most one position , else the shortest would be smaller than . The same is true for , and . Hence there exists one position , w.l.o.g. , such that , and . It follows that (i) and must be of opposite sign and (ii) . (i) is true because if , are of the same sign and then or . Assuming , we have which contradicts the hypothesis that . (ii) is true for if not or , and by the same argument as in (i) cannot exceed .

Since , it follows that for positions . In at least one of these positions, say , and . Now and must be of the same sign and , by a proof similar to that used before. Thus for , and . Again for , both and cannot hold for the same , hence, either and or and . We assume the first case. The proof for the other case is similar.

We have already proved that and are of the same sign. Assume they are both positive (if not reverse signs of all elements of ). Further, assume that and consider positions (if , then the same proof applies but for positions ). We now break up the proof into two cases:

Case 1: ( and of the same sign): Either (a) or (b) . If (a) then and . Thus , hence .

If (b) then let (). Thus and it follows that . Since it follows that and that and thus . Thus . But then and it follows that .

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