# Rank-Metric Lattices

Giuseppe Cotardo G. C. is supported by the Irish Research Council, grant n.
GOIPG/2018/2534. School of Mathematics and Statistics, University College
Dublin, Ireland  Alberto Ravagnani A. R. is supported by the Dutch Research
Council through grants VI.Vidi.203.045, OCENW.KLEIN.539, and by the Royal
Academy of Arts and Sciences of the Netherlands. Department of Mathematics and
Computer Science, Eindhoven University of Technology, the Netherlands

###### Abstract

We introduce the class of rank-metric geometric lattices and initiate the
study of their structural properties. Rank-metric lattices can be seen as the
$q$-analogues of higher-weight Dowling lattices, defined by Dowling himself in
1971. We fully characterize the supersolvable rank-metric lattices and compute
their characteristic polynomials. We then concentrate on the smallest rank-
metric lattice whose characteristic polynomial we cannot compute, and provide
a formula for it under a polynomiality assumption on its Whitney numbers of
the first kind. The proof relies on computational results and on the theory of
vector rank-metric codes, which we review in this paper from the perspective
of rank-metric lattices. More precisely, we introduce the notion of lattice-
rank weights of a rank-metric code and investigate their properties as
combinatorial invariants and as code distinguishers for inequivalent codes.

## Introduction

This paper initiates the study of rank-metric lattices and of their connection
with the invariant theory of rank-metric codes. We define a rank-metric
lattice as the combinatorial subgeometry of $\smash{\mathbb{F}_{q^{m}}^{n}}$
whose atoms are the projective points corresponding to vectors having small
rank (defined later).

Rank-metric lattices can be seen as the $q$-analogues of a family of lattices
that have been studied since the seventies, namely higher-weight Dowling
lattices. These were introduced by Dowling himself [21, 22] in connection with
the main problem of classical coding theory, i.e., computing the largest
dimension of an error-correcting code in the Hamming metric with given length
and correction capability. This question connects with the famous conjecture
by Segre on the largest size of an arc in a projective space over a finite
field [34]. Higher-weight Dowling lattices have been studied by various
authors [21, 22, 8, 9, 28, 10, 24, 38, 33, 11], yet the techniques for
computing their characteristic polynomial are to date unknown.

Computing the characteristic polynomials of rank-metric lattices is equivalent
to counting/estimating the number of rank-metric codes with given length and
correction capability, a line of research that is becoming increasingly
important in contemporary coding theory; see e.g. [2, 16, 25, 31].

The first member of the family of rank-metric lattices is precisely the
$q$-analogue of the first member of the family of higher-weight Dowling
lattices. Indeed, the former is (isomorphic to) the lattice of subspaces of
$\mathbb{F}_{q}^{n}$, and the latter is the lattice of subsets of
$\\{1,\ldots,n\\}$. It is natural to ask to which extent this analogy holds.
In this paper, we show that divergences in the behavior of $q$-analogous
lattices already appear when examining the second family members. For example,
while the second higher-Weight Dowling lattices is supersolvable (it is known
as the $q$-analogue of the partition lattice), the second rank-metric lattice
is not. In fact, in Section 3 of this paper we provide a complete
characterization of the modular elements of rank-metric lattices, and
determine all the supersolvable ones. The proof techniques follow to some
extent Bonin’s approach [9], but there are important technical differences in
the arguments, which almost never translate from higher-weight Dowling
lattices to rank-metric lattices (we will explain why).

In Section 4 we compute the characteristic polynomial of some rank-metric
lattices, and establish general properties of their roots by combining the
characterization of modular elements mentioned above with Stanley’s modular
factorization theorem [35]. The smallest rank-metric lattices whose
characteristic polynomial we cannot compute is the one generated by vectors in
$\smash{\mathbb{F}_{q^{4}}^{4}}$ of rank at most $2$. Using a computational
approach based on coding theory, in Section 5 we provide a closed formula for
its characteristic polynomial under the assumption that its second Whitney
number is a polynomial in $q$. The formula we obtain is compatible with all
the other properties of rank-metric lattices we establish throughout the
paper.

In Section 6 we consider applications of rank-metric lattices to the invariant
theory of (vector) rank-metric codes. We show how rank-metric lattices can be
used to define new distinguishers for this class of codes. We start from the
observation that the elements of the first rank-metric lattice are exactly the
optimal rank-metric anticodes defined in [32] in connection with a class of
invariants for (vector) rank-metric codes, namely the generalized vector rank
weights; see [29]. In this paper, we extend the definition of [32] to “higher
weight” rank-metric lattices. We then classify some families of rank-metric
codes that show interesting extremality properties with respect to these
notions and study the concepts lattice weight distribution and lattice
binomial moments in this context.

##  1 Posets and Lattices

In this section we recall some definitions and results on posets and lattices
that we will need throughout the paper. A standard reference for this part is
[37, Chapter 3]. The reader who is already familiar with lattice theory can
skip this section.

In the sequel, $(\mathscr{P},\leq)$ denotes a finite poset. We say that
$(\mathscr{P},\leq)$ is trivial if the cardinality of $\mathscr{P}$ is $1$. We
sometimes abuse notation and simply write $\mathscr{P}$ for
$(\mathscr{P},\leq)$. Moreover, for $s,t\in\mathscr{P}$ we write $s<t$ for
$s\leq t$ and $s\neq t$.

######  Definition 1.1.

Elements $s,t\in\mathscr{P}$ are comparable if $s\leq t$ or $t\leq s$. We say
that $t$ covers $s$ (or that $s$ is covered by $t$) if $s<t$ and there is no
element $u\in\mathscr{P}$ such that $s<u<t$. The notation is
$s\mathrel{<\kern-6.00006pt\raise 0.06458pt\hbox{$\cdot$}}t$.

######  Definition 1.2.

The join of $s,t\in\mathscr{P}$, should it exist, is the element
$u\in\mathscr{P}$ that satisfies $s\leq u$, $t\leq u$, and $u\leq v$ for all
$v\in\mathscr{P}$ with $s\leq v$ and $t\leq v$. Dually, the meet of
$s,t\in\mathscr{P}$, should it exist, is the element $u\in\mathscr{P}$ that
satisfies $u\leq s$, $u\leq t$, and $v\leq u$ for all $v\in\mathscr{P}$ with
$v\leq s$ and $v\leq t$. The poset $\mathscr{P}$ has minimum element
$0\in\mathscr{P}$ if $0\leq s$ for all $s\in\mathscr{P}$. Dually, it has
maximum element $1\in\mathscr{P}$ if $s\leq 1$ for all $s\in\mathscr{P}$.

######  Notation 1.3.

It is easy to check that the join (resp., the meet) of $s,t\in\mathscr{P}$,
should it exist, is unique and we denote it by $s\vee t$ (resp., by $s\wedge
t$). The minimum (resp., the maximum) element of $\mathscr{P}$, should it
exist, is unique and it is denoted by $0_{\mathscr{P}}$ (resp., by
$1_{\mathscr{P}}$), or by $0$ (resp., $1$) if the poset is clear from context.

######  Definition 1.4.

The Möbius function of $\mathscr{P}$ is defined recursively by
$\mu_{\mathscr{P}}(s,t)=1$ if $s=t$, $\mu_{\mathscr{P}}(s,t)=-\sum_{s\leq
u<t}\mu_{\mathscr{P}}(s,u)$ if $s<t$, and $\mu_{\mathscr{P}}(s,t)=0$
otherwise. We sometimes write $\mu_{\mathscr{P}}(t)$ for
$\mu_{\mathscr{P}}(0,t)$, when $\mathscr{P}$ has minimum element $0$.

The following result is a cornerstone of combinatorics.

######  Proposition 1.5 (Möbius Inversion Formula).

Let $\mathbb{K}$ be a field and let $f:\mathscr{P}\longrightarrow\mathbb{K}$
be a function. Define $g:\mathscr{P}\longrightarrow\mathbb{K}$ by
$g(t)=\sum_{s\leq t}f(s)$ for all $t\in\mathscr{P}$. Then

$f(t)=\sum_{s\leq t}\mu_{\mathscr{P}}(s,t)\,g(s)\quad\mbox{for all
$t\in\mathscr{P}$}.$

Chains play a crucial role in determining the structure of a poset. They are
defined as follows.

######  Definition 1.6.

A chain in $\mathscr{P}$ is a non-empty subset $C\subseteq\mathscr{P}$ such
that every $s,t\in C$ are comparable. We say that $C$ is maximal if there is
no chain in $\mathscr{P}$ that strictly contains $C$. We say that
$\mathscr{P}$ is graded if all maximal chains in $\mathscr{P}$ have the same
cardinality.

We will also need the following concepts.

######  Definition 1.7.

Suppose that $\mathscr{P}$ is graded and with minimum element $0$. The rank
function $\rho_{\mathscr{P}}:\mathscr{P}\to\mathbb{Z}$ of $\mathscr{P}$ is
uniquely defined by $\rho_{\mathscr{P}}(0)=0$ and
$\rho_{\mathscr{P}}(t)=\rho_{\mathscr{P}}(s)+1$ whenever $s,t\in\mathscr{P}$
and $s\mathrel{<\kern-6.00006pt\raise 0.06458pt\hbox{$\cdot$}}t$. In this case
we call $\rho_{\mathscr{P}}(s)$ the rank of $s$ and the elements
$s\in\mathscr{P}$ with $\rho_{\mathscr{P}}(s)=1$ are called atoms of
$\mathscr{P}$. We denote the set of atoms of $\mathscr{P}$ by
$\textup{At}(\mathscr{P})\subseteq\mathscr{P}$. We call $\mathscr{P}$
semimodular if it is graded and

$\rho_{\mathscr{P}}(s\wedge t)+\rho_{\mathscr{P}}(s\vee
t)\leq\rho_{\mathscr{P}}(s)+\rho_{\mathscr{P}}(t)\quad\mbox{for all
$s,t\in\mathscr{P}$.}$ (1)

An element $t\in\mathscr{P}$ is called modular if equality holds in (1) for
all $s\in\mathscr{P}$.

In this paper, we focus on the following special family of posets, called
lattices. Standard examples of posets (such as the poset of subsets of a
finite set and the poset of linear subspaces of a space over a finite field)
are in fact lattices.

######  Definition 1.8.

A finite lattice $(\mathscr{L},\leq)$ is a finite poset where every
$s,t\in\mathscr{L}$ have a join and a meet. In this case, join and meet can be
seen as commutative and associative operations
$\vee,\wedge:\mathscr{L}\times\mathscr{L}\to\mathscr{L}$. In particular, the
join (resp., the meet) of a non-empty subset $S\subseteq\mathscr{L}$ is well-
defined as the join (resp., the meet) of its elements and denoted by $\vee S$
(resp., by $\wedge S$). Furthermore, $\mathscr{L}$ has a minimum and maximum
element ($0_{\mathscr{L}}$ and $1_{\mathscr{L}}$, resp.). Finally, if
$\mathscr{L}$ is graded then the rank of $\mathscr{L}$ is
$\textup{rk}(\mathscr{L})=\rho_{\mathscr{L}}(1_{\mathscr{L}})$.

One of the most studied combinatorial invariants of a lattice is its
characteristic polynomial, defined as follows.

######  Definition 1.9.

Let $\mathscr{L}$ be a finite graded lattice with rank function $\rho$ and
Möbius function $\mu$. The characteristic polynomial of $\mathscr{L}$ is the
element of $\mathbb{Z}[\lambda]$ defined as

$\chi(\mathscr{L};\lambda):=\sum_{s\in\mathscr{L}}\mu(s)\lambda^{\textup{rk}(\mathscr{L})-\rho(s)}=\sum_{i=0}^{\textup{rk}(\mathscr{L})}w_{i}(\mathscr{L})\lambda^{\textup{rk}(\mathscr{L})-i},$

where

$w_{i}(\mathscr{L})=\sum_{\begin{subarray}{c}s\in\mathscr{L}\\\
\rho(s)=i\end{subarray}}\mu(s)$

is the $i$-th Whitney number of the first kind of $\mathscr{L}$. The $i$-th
Whitney number of the second kind of $\mathscr{L}$, denoted by
$W_{i}(\mathscr{L})$, is the number of elements of $\mathscr{L}$ of rank $i$.

Computing the characteristic polynomial of a lattice is a standard problem in
combinatorics. For some lattices, this problem is more “approachable” than for
others.

######  Definition 1.10.

A geometric lattice is a finite lattice $(\mathscr{L},\leq)$ which is
semimodular and atomistic, i.e., every element $s\in\mathscr{L}$ is the join
of a set of atoms of $\mathscr{L}$.

In [35, 36], Stanley identified a class of geometric lattices whose
characteristic polynomial has a precise combinatorial significance. They can
be defined as follows.

######  Definition 1.11.

A geometric lattice $\mathscr{L}$ is called supersolvable if $\mathscr{L}$ has
a maximal chain made of modular elements.

The following result is known as Stanley’s modular factorization Theorem. It
gives a sufficient condition for the characteristic polynomial of a geometric
lattices to have a factor.

######  Theorem 1.12 (see [35, Theorem 2]).

Let $\mathscr{L}$ be a geometric lattice of rank $n$ and let $t\in\mathscr{L}$
be a modular element. We have

$\chi(\mathscr{L};\lambda)=\chi([0,t];\lambda)\sum_{\begin{subarray}{c}x\in\mathscr{L}\\\
x\wedge
t=0\end{subarray}}\mu_{\mathscr{L}}(x)\lambda^{n-\rho_{\mathscr{L}}(x)-\rho_{\mathscr{L}}(t)}.$

As a corollary of the previous theorem we finally obtain Stanley’s
decomposition result for the characteristic polynomial of a geometric
supersolvable lattice. The corollary shows that the polynomial splits into
linear factors, whose roots have a precise combinatorial significance; see
also [36, Section 3].

######  Corollary 1.13.

Let $\mathscr{L}$ be a supersolvable geometric lattice of rank $n$ and let
$0=t_{0}\mathrel{<\kern-6.00006pt\raise
0.06458pt\hbox{$\cdot$}}t_{1}\mathrel{<\kern-6.00006pt\raise
0.06458pt\hbox{$\cdot$}}\cdots\mathrel{<\kern-6.00006pt\raise
0.06458pt\hbox{$\cdot$}}t_{n}=1$ be a maximal chain of modular elements of
$\mathscr{L}$. We have

$\chi(\mathscr{L};\lambda)=\prod_{i=1}^{n}(\lambda-|\\{a\in\textup{At}(\mathscr{L})\mid
a\leq t_{i},\,a\not\leq t_{i-1}\\}|).$

Corollary 1.13 and its generalizations (see e.g. [26, 7]) are standard tools
for computing the characteristic polynomial of geometric lattices.
Unfortunately, not all geometric lattices are supersolvable, hence their
characteristic polynomial has to be computed using other approaches. This will
be the case of rank-metric lattices, which we introduce in the next section.

##  2 Rank-Metric Lattices

In this section we define the main combinatorial objects studied in this
paper, namely rank-metric lattices. These can be seen as the $q$-analogue of
the higher-weights Dowling lattices, which were introduced in 1971 by Dowling
[21] and later investigated and generalized by several authors; see below for
further details on this.

We will also illustrate the connection between rank-metric lattices and the
problem of computing the number of rank-metric codes of given dimension and
minimum distance lower bounded by a given integer $d$. The latter is a central
problem in contemporary coding theory.

######  Notation 2.1.

Throughout this paper, $q$ is a prime power and $n$, $m$ are integers with
$n,m\geq 2$. We denote by $\mathbb{F}_{q}$ the finite field with $q$ elements
and by $\mathbb{F}_{q^{m}}$ its extension of degree $m$. We let
$\\{e_{1},\ldots,e_{n}\\}$ be the standard basis of $\mathbb{F}_{q}^{n}$. With
a small abuse of notation we denote by $\\{e_{1},\ldots,e_{n}\\}$ also the
standard basis of $\mathbb{F}_{q^{m}}^{n}$. All dimensions in the sequel are
computed over $\mathbb{F}_{q^{m}}$, unless otherwise stated. We denote by
$\leq$ the inclusion of linear spaces, to emphasize the order structure.

We start by defining the rank of a vector with entries from the field
extension $\mathbb{F}_{q^{m}}$.

######  Definition 2.2.

The rank (weight) of a vector $v\in\mathbb{F}_{q^{m}}$ is the dimension over
$\mathbb{F}_{q}$ of the $\mathbb{F}_{q}$-span of its entries. We denote it by
$\textup{rk}(v)$.

The rank of a vector is closely related to that of a matrix. To see this, let
$\Gamma=\\{\gamma_{1},\ldots,\gamma_{m}\\}$ be an ordered
$\mathbb{F}_{q}$-basis of $\mathbb{F}_{q^{m}}$. For a vector
$v\in\mathbb{F}_{q^{m}}^{n}$ we let $\Gamma(v)$ be the $n\times m$ matrix with
entries in $\mathbb{F}_{q}$ defined by

$v_{i}=:\sum_{j=1}^{m}\Gamma(v)_{i,j}\gamma_{j}\quad\mbox{for all $i$}.$

By definition, the map
$\mathbb{F}_{q^{m}}^{n}\longrightarrow\mathbb{F}_{q}^{n\times m}$ given by
$v\longmapsto\Gamma(v)$ is an $\mathbb{F}_{q}$-isomorphism and we have
$\textup{rk}(v)=\textup{rk}(\Gamma(v))$ for all bases $\Gamma$ and for all
$v\in\mathbb{F}_{q^{m}}^{n}$.

The following properties of the vector rank will be needed repeatedly. They
easily follow from the connection between vector and matrix rank explained
above.

######  Proposition 2.3.

The following hold.

  1. 1.

$\textup{rk}(v)\geq 0$ for all $v\in\mathbb{F}_{q^{m}}^{n}$, with equality if
and only if $v=0$.

  2. 2.

$\textup{rk}(v)=\textup{rk}(\lambda v)$ for all $v\in\mathbb{F}_{q^{m}}^{n}$
and $\lambda\in\mathbb{F}_{q^{m}}\setminus\\{0\\}$.

  3. 3.

$\textup{rk}(v+w)\leq\textup{rk}(v)+\textup{rk}(w)$ for all
$v,w\in\mathbb{F}_{q^{m}}^{n}$ (triangular inequality).

  4. 4.

$\textup{rk}(v+w)\geq\textup{rk}(v)-\textup{rk}(w)$ for all
$v,w\in\mathbb{F}_{q^{m}}^{n}$.

We are now ready to introduce the class rank-metric lattices, whose elements
are the $\mathbb{F}_{q^{m}}$-linear subspaces of $\mathbb{F}_{q^{m}}^{n}$
having a basis of vectors with rank bounded from above, ordered by inclusion.

######  Definition 2.4.

Let $i\in\\{1,\ldots n\\}$. The rank-metric lattice (RML in short)
$\mathscr{L}_{i}(n,m;q)$ associated with the $4$-tuple $(i,n,m,q)$ is the
geometric sublattice of $\mathscr{L}(\mathbb{F}_{q^{m}}^{n})$ whose atoms are
the $1$-dimensional $\mathbb{F}_{q^{m}}$-subspaces of $\mathbb{F}_{q^{m}}^{n}$
generated by the vectors of rank at most $i$, that is,

$\displaystyle\mathscr{L}_{i}(n,m;q):=\\{\left<v_{1},\ldots,v_{\ell}\right>_{\mathbb{F}_{q^{m}}}:\ell\geq
1,\,v_{1},\ldots,v_{\ell}\in T_{i}(n,m;q)\\},$

where $T_{i}(n,m;q):=\\{v\in\mathbb{F}_{q^{m}}^{n}:\textup{rk}(v)\leq i\\}$.
The order is given by the inclusion of $\mathbb{F}_{q^{m}}$-linear subspaces
of $\mathbb{F}_{q^{m}}^{n}$.

We will need the following observations, which we will use without explicitly
referring to them.

######  Remark 2.5.

It follows from the definitions that $\mathscr{L}_{i}(n,m;q)$ is a geometric
lattice of rank $n$, where the rank function is given by the
$\mathbb{F}_{q^{m}}$-dimension. Furthermore, for any
$X,Y\in\mathscr{L}_{i}(n,m;q)$ we have $X\vee_{i}Y=X+Y$ (the ordinary sum of
linear subspaces) and $X\wedge_{i}Y=\left<x\in X\cap Y:\textup{rk}(x)\leq
i\right>$, where $\vee_{i}$ and $\wedge_{i}$ denote the operations of join and
meet in $\mathscr{L}_{i}(n,m;q)$, respectively. By definition we have that
$\mathscr{L}_{0}(n,m;q)$ is the trivial lattice $\\{0\\}$ and that
$\mathscr{L}_{n}(n,m;q)=\\{X\leq\mathbb{F}_{q^{m}}^{n}\\}$ is the lattice of
$\mathbb{F}_{q^{m}}$-subspaces of $\mathbb{F}_{q^{m}}^{n}$.

As already mentioned, rank-metric lattices can be seen as the $q$-analogues
(or also rank-metric analogues) of a well-studied class of lattices, known
under the name of higher-weight Dowling lattices. These combinatorial
geometries were introduced by Dowling in connection with a fundamental problem
in coding theory; see [21, 22]. Their theory and connections with the Critical
Problem by Crapo and Rota was further explored by Zaslavsky [38], Bonin [8,
9], Kung [28], Brini [10], Games [24], Britz [11], and more recently by
Ravagnani [33]. The techniques for computing the characteristic polynomials of
these lattices for arbitrary parameters are currently unknown.

######  Definition 2.6.

Let $Q$ be a prime power. The Hamming weight of a vector
$v\in\mathbb{F}_{Q}^{n}$ is
$\textup{wt}^{\textup{H}}(v):=|\\{i\in[n]:v_{i}\neq 0\\}|$. For
$i\in\\{1,\ldots,n\\}$, the higher-weight Dowling lattice (HWDL in short)
$\mathscr{H}_{i}(n;Q)$ associated to the 3-tuple $(i,n,Q)$ is the geometric
sublattice of $\mathscr{L}(\mathbb{F}_{Q}^{n})$ whose atoms are the
$1$-dimensional $\mathbb{F}_{Q}$-subspaces of $\mathbb{F}_{Q}^{n}$ generated
by vectors of Hamming weight at most $i$, that is

$\displaystyle\mathscr{H}^{\textup{H}}_{i}(n;Q):=\\{\left<v_{1},\ldots,v_{n}\right>_{\mathbb{F}_{Q}}:\ell\geq
1,\,v_{1},\ldots,v_{\ell}\in T_{i}^{\textup{H}}(n;Q)\\},$

where
$T_{i}^{\textup{H}}(n;Q):=\\{v\in\mathbb{F}_{Q}^{n}:\textup{wt}^{\textup{H}}(v)\leq
i\\}$.

Computing the characteristic polynomial (or equivalently the Whitney numbers
of the first kind) of higher-weight Dowling lattices would allow to solve a
long-standing open problem in coding theory and finite geometry, namely the
famous Segre’s conjecture, known as the MDS Conjecture in the context of
coding theory. More generally, computing the characteristic polynomial of
higher-weight Dowling lattices would allow one to give formulas for the number
of Hamming-metric codes having given dimension and minimum distance bounded
from below by an integer $d$. The $q$-analogy between higher-weight Dowling
lattices and rank-metric lattices “lifts” to an analogy between the
corresponding questions about codes, as we now briefly illustrate.

######  Definition 2.7.

A (rank-metric) code is an $\mathbb{F}_{q^{m}}$-linear subspace
$C\leq\mathbb{F}_{q^{m}}^{n}$. The minimum (rank) distance of $C$ is
$d(C):=\min\\{\textup{rk}(v):v\in\mathbb{F}_{q^{m}}^{n}\mid v\neq 0\\}$, where
the zero code $\\{0\\}$ has minimum distance $\min\\{n,m\\}+1$ by definition.
In this context, $n$ is called the length of $C$. The dual of $C$ is the code
$C^{\perp}:=\\{v\in\mathbb{F}_{q^{m}}^{n}:c\cdot v=0\textup{ for all }c\in
C\\}$, where $\cdot$ denotes the standard inner product of vectors.

The equivalence between the problem of counting rank-metric codes with given
dimension and minimum distance lower bounded by $d$ and that of computing the
Whitney numbers of rank-metric lattices is expressed by Theorem 2.10 below
(the general version of the result is [33, Theorem 3.1]). In order to state
it, we will need the following symbols and the notion of Gaussian binomial
coefficients; see [1].

######  Notation 2.8.

We let $w_{j}(i,n,m;q)$ be the $j$-th Whitney number of the second kind of the
lattice $\mathscr{L}_{i}(n,m;q)$. We denote by $\alpha_{k}(i,n,m;q)$ the
number of $k$-dimensional rank-metric codes $C\leq\mathbb{F}_{q^{m}}^{n}$ with
$d(C)\geq i+1$. The $j$-th Whitney number of the first kind of
$\mathscr{L}_{i}(n,m;q)$ is denoted by $W_{j}(i,n,m;q)$.

######  Definition 2.9.

Let $a,b$ be integers and let $Q$ be a prime power. The $Q$-binomial
coefficients of $a$ and $b$ is

$\begin{bmatrix}{a}\\\ {b}\end{bmatrix}_{Q}=\begin{cases}0&\textup{ if
}b<0\textup{ or }0\leq a<b,\\\\[4.30554pt] 1&\textup{ if }b=0\textup{ and
}a\geq 0,\\\\[4.30554pt]
\displaystyle\prod_{i=0}^{b-1}\frac{Q^{a-i}-1}{Q^{i+1}-1}&\textup{ if
}b>0\textup{ and }a\geq b.\\\\[4.30554pt] \end{cases}$

We can now state the rank-metric instance of [33, Theorem 3.1].

######  Theorem 2.10.

The following hold for any $j\in\\{0,\ldots,n\\}$.

  1. (1)

$\displaystyle
w_{j}(i,n,m;q)=\sum_{k=0}^{j}\alpha_{k}(i,n,m;q)\begin{bmatrix}{n-k}\\\
{j-k}\end{bmatrix}_{q^{m}}(-1)^{j-k}q^{m\binom{j-k}{2}}$.

  2. (2)

$\displaystyle\alpha_{j}(i,n,m;q)=\sum_{k=0}^{j}w_{k}(i,n,m;q)\begin{bmatrix}{n-k}\\\
{j-k}\end{bmatrix}_{q^{m}}$.

A very natural question is for which values of $j$ we have
$\alpha_{j}(i,n,m;q)=0$. A partial answer to this question is given by the
following result, which follows from [20, Theorem 5.4].

######  Theorem 2.11 (rank-metric Singleton bound).

Let $1\leq k\leq n$ and $1\leq d\leq\min\\{n,m\\}$ be integers. Let
$C\leq\mathbb{F}_{q^{m}}^{n}$ be a code of dimension $k\geq 1$ and minimum
distance at least $d$. We have

$mk\leq\max\\{n,m\\}(\min\\{n,m\\}-d+1).$

###### Proof.

Let $\Gamma$ be a basis of $\mathbb{F}_{q^{m}}$ over $\mathbb{F}_{q}$. Then
$\\{\Gamma(v)\,:\,v\in C\\}$ is an $\mathbb{F}_{q}$ linear space of matrices,
of the same cardinality as $C$, in which every non-zero matrix has rank at
least $d$. Therefore by [20, Theorem 5.4] we have that
$|C|\leq\min\\{q^{m(n-d+1)},q^{n(m-d+1)}\\}$. The desired bound follows from
the fact that $|C|=q^{mk}$. ∎

We conclude this section with the following lemma summarizing some well-known
properties of Gaussian binomial coefficients. We will need them in various
proofs later.

######  Lemma 2.12.

Let $a,b,c$ be integers, let $Q$ be a prime power, and let $x,y$ be rational
numbers. The following hold.

  1. 1.

$\displaystyle\begin{bmatrix}{a}\\\ {b}\end{bmatrix}_{Q}\begin{bmatrix}{b}\\\
{c}\end{bmatrix}_{Q}=\begin{bmatrix}{a}\\\
{c}\end{bmatrix}_{Q}\begin{bmatrix}{a-c}\\\ {a-b}\end{bmatrix}_{Q}$.

  2. 2.

$\displaystyle\begin{bmatrix}{a+b}\\\
{c}\end{bmatrix}_{Q}=\sum_{j=0}^{c}q^{j(b-c+j)}\begin{bmatrix}{a}\\\
{j}\end{bmatrix}_{Q}\begin{bmatrix}{b}\\\
{c-j}\end{bmatrix}_{Q}=\sum_{j=0}^{c}Q^{(c-j)(a-j)}\begin{bmatrix}{a}\\\
{j}\end{bmatrix}_{Q}\begin{bmatrix}{b}\\\ {c-j}\end{bmatrix}_{Q}$.

  3. 3.

$\displaystyle\sum_{j=0}^{c}\begin{bmatrix}{c}\\\
{j}\end{bmatrix}_{Q}(-1)^{j}Q^{\binom{j}{2}}x^{c-j}y^{j}=\begin{cases}0&\textup{
if }c=0,\\\ \displaystyle\prod_{j=0}^{c-1}(x-Q^{j}y)&\textup{ if }c\geq
1.\end{cases}$

##  3 Supersolvability

As Stanley’s modular factorization theorem shows, the characteristic
polynomial of supersolvable geometric lattices is particularly easy to compute
and completely determined by the combinatorial structure of the lattice’s
atoms; see Theorem 1.12. It is therefore very natural to ask which rank-metric
lattices are supersolvable, a question that we answer completely in this
section.

Continuing the analogy with the Hamming-metric case, we will first state which
higher-weight Dowling lattices are supersolvable. These were fully
characterized by Bonin in [9] and are very few, as the following result shows.

######  Theorem 3.1 (see [9]).

The only non-trivial supersolvable higher-weight Dowling lattices are
$\mathscr{H}_{1}(n;q)$, $\mathscr{H}_{2}(n;q)$, $\mathscr{H}_{n-1}(n;q)$ for
$n\geq 2$, and $\mathscr{H}_{n}(n;q)$.

The rank-metric analogue of Theorem 3.1, to the proof of which this section is
entirely devoted, is the following.

######  Theorem 3.2.

The lattice $\mathscr{L}_{i}(n,m;q)$ is supersolvable if and only if one of
the following holds:

  1. 1.

$i\in\\{1,n-1,n\\}$,

  2. 2.

$m\in\\{2,\ldots,i\\}$.

Note Theorem 3.2 shows that the parameter $q$ does not play any role in
determining whether or not $\mathscr{L}_{i}(n,m;q)$ is supersolvable.

######  Remark 3.3.

The approaches developed in [9, 22] extends only in part from the Hamming to
the rank metric, because of technical differences between the two metrics that
arise in the proofs. In fact, while the second higher-weight Dowling lattice
$\mathscr{H}_{2}(n;q)$ is supersolvable (it is known as the $q$-analogue of
the partition lattice), the second rank-metric lattice
$\mathscr{L}_{2}(n,m;q)$ is not supersolvable, as we will prove in this
section. Our main result is a complete characterization of the modular
elements of $\mathscr{L}_{i}(n,m;q)$, from which Theorem 3.2 follows easily.
The proof of the characterization is splitted across a number of preliminary
results, which we will combine at the end of the section.

We start with the rank-metric analogues of [9, Lemmas 1.2 and 1.4]. Their
proofs extend easily from the Hamming-metric case.

######  Lemma 3.4.

Let $X\in\mathscr{L}_{i}(n,m;q)$. Then $X$ is modular if and only if
$X\wedge_{n}Y=X\wedge_{i}Y$ for all $Y\in\mathscr{L}_{i}(n,m;q)$.

###### Proof.

Recall that, by definition of modularity, $X$ is a modular element of
$\mathscr{L}_{i}(n,m;q)$ if and only if, for all $Y\in\mathscr{L}_{i}(n,m;q)$,

$\dim(X\wedge_{i}Y)=\dim(X)+\dim(Y)-\dim(X\vee_{i}Y).$ (2)

Since $X\vee_{i}Y=X\vee_{n}Y$, we have that the right hand side of (2) is
$X\wedge_{n}Y$. It follows that $X$ is modular in $\mathscr{L}_{1}(n,m;q)$ if
and only if $\dim(X\wedge_{i}Y)=\dim(X\wedge_{n}Y)$. The statement now follows
from the fact that $X\wedge_{i}Y\leq X\wedge_{n}Y$. ∎

######  Lemma 3.5.

Let $X$ be a modular element of $\mathscr{L}_{i}(n,m;q)$ and let $x\in
X\setminus T_{i}(n,m;q)$. If $y\in T_{i}(n,m;q)$ and $x+y\in T_{i}(n,m;q)$,
then $y\in X$.

###### Proof.

We have $\textup{rk}(x)\geq i+1$, $\textup{rk}(y)\leq i$, and
$\textup{rk}(x+y)\leq i$. Define the lattice element
$Y:=\left<x+y,y\right>_{\mathbb{F}_{q^{m}}}\in\mathscr{L}_{i}(n,m;q)$ and
assume toward a contradiction that $y\notin X$. Since $y\in Y\setminus X$, we
have $X\wedge_{n}Y=\left<x\right>_{\mathbb{F}_{q^{m}}}$. Since $x\notin
T_{i}(n,m;q)$, we have $X\wedge_{i}Y=\\{0\\}$. Therefore $X$ is not modular by
Lemma 3.4. ∎

We will establish Theorem 3.2 in various steps and by treating the cases
$i=1$, $i=2$, and $i\geq 3$ separately. While the general proof strategy is
similar to that of [9], the technical details are different and heavily rely
on the properties of the rank weight.

######  Proposition 3.6.

$\mathscr{L}_{1}(n,m;q)$ is isomorphic to the lattice of
$\mathbb{F}_{q}$-subspaces of $\mathbb{F}_{q}^{n}$. In particular,
$\mathscr{L}_{1}(n,m;q)$ is a modular lattice.

###### Proof.

We will construct a poset isomorphism
$f:(\mathscr{L},\leq)\longrightarrow\mathscr{L}_{1}(n,m;q)$, where
$(\mathscr{L},\leq)$ is the lattice of $\mathbb{F}_{q}$-subspaces of
$\mathbb{F}_{q}^{n}$ ordered by inclusion. Define
$f(U):=\left<U\right>_{\mathbb{F}_{q^{m}}}=U\otimes_{\mathbb{F}_{q}}\mathbb{F}_{q^{m}}$
for all $U\in\mathscr{L}$. It is easy to see that $f$ is order preserving and
injective. Therefore to conclude the proof it suffices to prove that $f$ is
surjective. To see this, let $X\in\mathscr{L}_{1}(n,m;q)$ and let
$\\{v_{1},\ldots,v_{j}\\}$ be an $\mathbb{F}_{q^{m}}$-basis of $X$ made of
vectors of rank 1. Write $v_{\ell}=\lambda_{\ell}w_{\ell}^{\prime}$ for
$\ell\in\\{1,\ldots,j\\}$, where
$\lambda_{\ell}\in\mathbb{F}_{q^{m}}\setminus\\{0\\}$ and
$w_{\ell}\in\mathbb{F}_{q}^{n}$. Then
$X=U\otimes_{\mathbb{F}_{q}}\mathbb{F}_{q^{m}}=f(U)$, where
$U=\left<w_{1},\ldots,w_{j}\right>_{\mathbb{F}_{q}}$, concluding the proof. ∎

######  Remark 3.7.

The proof of Proposition 3.6 shows that the elements of
$\mathscr{L}_{1}(n,m;q)$ are precisely those subspaces of $\mathbb{F}_{q^{m}}$
having a basis made of vectors with entries in $\mathbb{F}_{q}$. These are the
optimal rank-metric anticodes; see [32, Section 4].

The following lemma summarizes some properties of the rank of a vector that
will be crucial in our approach. These are mainly inherited from the
correspondence between vector rank and matrix rank illustrated in Section 2.
They will be applied in several instances throughout this section, without
referring to them explicitly, to avoid making the exposition too cumbersome.

######  Lemma 3.8.

Let $v\in\mathbb{F}_{q^{m}}^{n}$ be a non-zero vector. Then $\textup{rk}(v)$
is the smallest $r$ for which there exist rank-1 vectors
$v_{1},\ldots,v_{r}\in\mathbb{F}_{q^{m}}^{n}$ with $v=v_{1}+\ldots+v_{r}$.
Furthermore, any such rank-1 vectors $v_{1},\ldots,v_{r}$ are linearly
independent over $\mathbb{F}_{q^{m}}$. Finally, for every
$S\subseteq\\{1,\ldots,r\\}$ we have that $\sum_{j\in S}v_{j}$ has rank $|S|$.

###### Proof.

The first part of the statement follows from the fact that $v\mapsto\Gamma(v)$
is a rank-preserving $\mathbb{F}_{q}$-linear isomorphism (for any basis
$\Gamma$).

For the second part, suppose towards a contradiction that the rank-$1$ vectors
$v_{1},\ldots,v_{r}$ are not linearly independent over $\mathbb{F}_{q^{m}}$
and let $\lambda_{1},\ldots,\lambda_{r-1}\in\mathbb{F}_{q^{m}}$ such that
(without loss of generality)
$v_{r}=\lambda_{1}v_{1}+\cdots+\lambda_{r-1}v_{r-1}$. We then have

$v=v_{1}+\cdots+v_{r}=(1+\lambda_{1})v_{1}+\cdots+(1+\lambda_{r-1})v_{r-1},$

which implies $\textup{rk}(v)\leq r-1$ by the triangular inequality; see
Proposition 2.3. This is a contradiction.

In order to prove the last part of the statement, let
$S\subseteq\\{1,\ldots,r\\}$ and $w:=\sum_{j\in S}v_{j}$. Suppose towards a
contradiction that $\textup{rk}(w)<|S|$. Define the vector $u:=\sum_{j\notin
S}v_{j}$ and observe that $\textup{rk}(u)\leq r-|S|$, since it is the sum of
$r-|S|$ rank-$1$ vectors. By the triangle inequality we have

$r=\textup{rk}(v)=\textup{rk}(w+u)\leq\textup{rk}(w)+\textup{rk}(u)<|S|+r-|S|=r,$

which is a contradiction. ∎

The following two lemmas are devoted to the case $i=2$. Note that the proof
techniques do not apply to the Hamming-metric case. In fact, the Hamming-
metric analogues of the following two lemmas are false; see [22].

######  Lemma 3.9.

Let $X$ be a modular element of $\mathscr{L}_{2}(n,m;q)$. If $X$ contains an
element of rank $3$, then $X=\mathbb{F}_{q^{m}}^{n}$.

###### Proof.

Suppose that $X$ has an element of rank $3$, say $x$. By Lemma 3.8, there
exist $x_{1},x_{2},x_{3}\in\mathbb{F}_{q^{m}}^{n}$ of rank $1$, linearly
independent over $\mathbb{F}_{q^{m}}$, with $x=x_{1}+x_{2}+x_{3}$. Define
$y:=x_{1}+x_{2}$ and note that $x-y=x_{3}$. We have $y,x_{3}\in T_{2}(n,m;q)$,
which implies that $y,x_{3}\in X$, by Lemma 3.5. Analogously, $x_{1},x_{2}\in
X$. If $n=3$, then this immediately shows that $X=\mathbb{F}_{q^{m}}^{3}$,
because $x_{1}$, $x_{2}$ and $x_{3}$ are linearly independent over
$\mathbb{F}_{q^{m}}$.

We assume $n\geq 4$ in the remainder of the proof. We complete
$\\{x_{1},x_{2},x_{3}\\}$ to a basis of $\mathbb{F}_{q^{m}}^{n}$ by selecting
vectors $x_{4},\ldots,x_{n}\in\mathbb{F}_{q^{m}}^{n}$ of rank $1$ (observe
that is always possible because the rank-$1$ vectors of
$\mathbb{F}_{q^{m}}^{n}$ span $\mathbb{F}_{q^{m}}^{n}$). For all
$j\in\\{1,\ldots,n\\}$, let $\lambda_{j}\in\mathbb{F}_{q^{m}}\setminus\\{0\\}$
such that $x_{j}=\lambda_{j}y_{j}$, for some
$y_{j}\in\mathbb{F}_{q}^{n}\setminus\\{0\\}$. Note that the $\lambda_{j}$’s
exist because $\textup{rk}(x_{j})=1$ for all $j\in\\{1,\ldots,n\\}$. Define
the vector $z:=x_{1}+x_{2}-\lambda_{1}y_{4}$, so that

$z=x_{1}+x_{2}-\lambda_{1}y_{4}=\lambda_{1}(y_{1}-y_{4})+x_{2},\qquad
x-z=x_{3}+\lambda_{1}y_{4}.$

It follows that $z,x-z\in T_{2}(n,m;q)$, since they are sum of at most two
vectors of rank $1$ as $y_{1}-y_{4}\in\mathbb{F}_{q}^{n}$, and therefore $z\in
X$ by Lemma 3.5. Note that we can re-write $x_{4}$ as

$x_{4}=\lambda_{4}y_{4}=\lambda_{1}^{-1}\lambda_{4}(y-z),$

which is a linear combination of elements of $X$. Therefore $x_{4}\in X$ as
well. Analogously, one can prove that $x_{5},\ldots,x_{n}\in X$, which in turn
implies that $X=\mathbb{F}_{q^{m}}^{n}$. ∎

######  Lemma 3.10.

Let $X\in\mathscr{L}_{2}(n,m;q)\setminus\\{\mathbb{F}_{q^{m}}^{n}\\}$. If $X$
contains an element of rank $4$, then $X$ is not modular.

###### Proof.

Suppose that $X$ has an element of rank $4$, say $x$. By Lemma 3.8, there
exist $x_{1},x_{2},x_{3},x_{4}\in\mathbb{F}_{q^{m}}^{n}$ of rank $1$, linearly
independent over $\mathbb{F}_{q^{m}}$, that sum to $x$. We examine two cases.

  * •

Suppose that $x_{j}\in X$ for some $j\in\\{1,2,3,4\\}$. We have that $x-x_{j}$
is an element of $X$ of rank $3$ and therefore $X$ cannot be modular by Lemma
3.9.

  * •

Suppose $x_{1},x_{2},x_{3},x_{4}\notin X$ and assume towards a contradiction
that $X$ is modular. Define $y:=x_{1}+x_{2}$ and note that $y,x-y\in
T_{2}(n,m;q)$. Therefore $y\in X$ by Lemma 3.5. For all $j\in\\{1,2,3,4\\}$,
take $\lambda_{j}\in\mathbb{F}_{q^{m}}\setminus\\{0\\}$ and
$y_{j}\in\mathbb{F}_{q}^{n}\setminus\\{0\\}$ such that
$x_{j}=\lambda_{j}y_{j}$, and define

$z_{1}:=x_{1}+x_{2}+\lambda_{3}y_{1},\qquad
z_{2}:=x_{3}+x_{4}-\lambda_{3}y_{1},\qquad
Y:=\left<z_{1},z_{2}\right>_{\mathbb{F}_{q^{m}}}.$

Observe that

$z_{1}=(\lambda_{1}+\lambda_{3})y_{1}+x_{2},\qquad
z_{2}=\lambda_{3}(y_{3}-y_{1})+x_{4}.$

Since $y_{1}$, $x_{2}$, $y_{3}-y_{1}$ and $x_{4}$ all have rank at most 1, we
have $Y\in\mathscr{L}_{2}(n,m;q)$. We have $z_{1}\notin X$, otherwise
$z_{1}-y=\lambda_{3}y_{1}\in X$ and hence $\lambda_{1}y_{1}=x_{1}\in X$ (since
$X$ is a linear space), which is false by assumption. Since $x=z_{1}+z_{2}\in
Y$ and $Y$ has dimension at most $2$, we have
$X\wedge_{n}Y=\left<x\right>_{\mathbb{F}_{q}}$. On the other hand, $x\notin
T_{2}(n,m;q)$ because $\textup{rk}(x)=4$, and therefore
$X\wedge_{2}Y=\left<0\right>_{\mathbb{F}_{q}}$. This implies that $X$ is not
modular by Lemma 3.4. ∎

We next characterize the modular elements of the lattice
$\mathscr{L}_{i}(n,m;q)$ for $i\geq 3$. Again, the proofs of the following
four lemmas heavily use the properties of the rank weight stated in Lemma 3.8.
We start with two results that, technically, only require $i\geq 2$, but that
we will need in the case $i\geq 3$ only.

######  Lemma 3.11.

Let $i\geq 2$, $j\in\\{1,\ldots,i\\}$, and let $X$ be a modular element of
$\mathscr{L}_{i}(n,m;q)$. Let $x_{1},\ldots,x_{i+j}\in\mathbb{F}_{q^{m}}^{n}$
be of rank $1$. Suppose that $x:=x_{1}+\ldots+x_{i+j}\in X$ and that
$\textup{rk}(x)\geq i+1$. Then $x_{s}-x_{p}\in X$ for all
$s,p\in\\{1,\ldots,i+j\\}$.

###### Proof.

The result is clear if $s=p$. We henceforth assume $s\neq p$, say
$(s,p)=(1,2)$ without loss of generality. Define
$y:=x_{1}+x_{3}+\cdots+x_{i+1}$ and $z:=x_{2}+x_{3}+\cdots+x_{i+1}$. Observe
that $y,z\in X$ by Lemma 3.5, since $x\in X\setminus T_{i}(n,m;q)$ and
$z,y,x-y,x-z\in T_{i}(n,m;q)$, as they are sum of at most $i$ vectors of rank
$1$ (we are using $j\leq i$ here). Finally, we have that $y-z=x_{1}-x_{2}\in
X$ because it is sum of elements of $X$. ∎

######  Lemma 3.12.

Let $i\geq 2$ and let $X$ be a modular element of $\mathscr{L}_{i}(n,m;q)$. If
$X$ contains an element of rank strictly bigger than $i$, then it must contain
an element of rank either $i+1$ or $i+2$.

###### Proof.

Since $X$ contains an element of rank strictly bigger than $i$ and the sum of
two vectors of rank $\leq i$ has rank $\leq 2i$, $X$ must contain an element
$x$ of rank $i+j$, for some $1\leq j\leq i$. If $j\in\\{1,2\\}$ then we are
done. Now suppose $j\geq 3$ and write $x=x_{1}+\ldots+x_{i+j}$, where
$x_{1},\ldots x_{i+j}\in\mathbb{F}_{q^{m}}^{n}$ have rank 1 and are linearly
independent over $\mathbb{F}_{q^{m}}$. Let $y:=x_{1}+\ldots+x_{i}$ and observe
$y,x-y\in T_{i}(n,m;q)$ by Lemma 3.8. Therefore by Lemma 3.5 we have that
$y\in X$. Let $z_{1}:=x_{i+1}-x_{i+3}$ and $z_{2}:=x_{i+2}-x_{i+3}$. By Lemma
3.11 we have $z_{1},z_{2}\in X$, since $X$ is modular. Now observe that

$z:=y+z_{1}+z_{2}=x_{1}+\cdots+x_{i}+x_{i+1}+x_{i+2}-2x_{i+3}$

has rank $i+1$, $i+2$, or $i+3$, since by Proposition 2.3 we have

$i+1=\textup{rk}(x_{1}+\cdots+x_{i+2})-\textup{rk}(2x_{i+3})\leq\textup{rk}(z)\leq
i+3,$

where the last inequality from the fact that $z$ is a sum of at most $i+3$
vectors of rank $1$. If $z$ has rank $i+1$ or $i+2$ then we are done. If $z$
has rank $i+3$, then $y+z_{1}\in X$ must have rank $i+1$ or $i+2$, since
$z_{2}$ has rank $1$ or $2$ (we are using the triangular inequality again).
Therefore in any case we have exhibited a vector of rank $i+1$ or $i+2$. ∎

In the proofs of the following two lemmas we need the assumption $i\geq 3$.
For completeness of the exposition we will stress when the assumption is
technically used.

######  Lemma 3.13.

Let $i\geq 3$ and let $X$ be a modular element of $\mathscr{L}_{i}(n,m;q)$. If
$X$ contains an element of rank $i+2$, then it contains an element of rank
$i+1$.

###### Proof.

Let $x\in X$ be of rank $i+2$ and write $x:=x_{1}+\ldots+x_{i+2}$ for some
rank-$1$ elements $x_{1},\ldots,x_{i+2}\in\mathbb{F}_{q^{m}}^{n}$ that are
linearly independent over $\mathbb{F}_{q^{m}}$. Define $y:=x_{1}+\ldots+x_{i}$
and $z:=x_{1}+\cdots+x_{i-1}$. Notice that $y,z,x-y,x-z\in T_{i}(n,m;q)$ by
Lemma 3.8 (here we use $i\geq 3$). Therefore, by Lemma 3.5, $y,z,x-y$ and
$x-z$ are elements of $X$. Finally, we have that
$x-y+z=x_{1}+\ldots+x_{i-1}+x_{i+1}+x_{i+2}$ is an element of $X$ of rank
$i+1$ by Lemma 3.8. ∎

######  Lemma 3.14.

Let $i\geq 3$ and let $X$ be a modular element of $\mathscr{L}_{i}(n,m;q)$. If
$X$ contains an element of rank $i+1$, then $X=\mathbb{F}_{q^{m}}^{n}$.

###### Proof.

Suppose that $X$ contains an element of rank $i+1$, say $x$. Write
$x=x_{1}+\ldots+x_{i+1}$ for some
$x_{1},\ldots,x_{i+1}\in\mathbb{F}_{q^{m}}^{n}$ of rank $1$ and linearly
independent over $\mathbb{F}_{q^{m}}$. Note that for all
$j\in\\{1,\ldots,i+1\\}$ we have $x-x_{j}\in T_{i}(n,m;q)$. Therefore by Lemma
3.5 we have $x_{j}\in X$ for all $j$. We now complete
$\\{x_{1},\ldots,x_{i+1}\\}$ to a basis of $\mathbb{F}_{q^{m}}^{n}$ by
selecting vectors $x_{i+2},\ldots,x_{n}\in\mathbb{F}_{q^{m}}^{n}$ of rank $1$
(this is always possible because the rank-$1$ vectors of
$\mathbb{F}_{q^{m}}^{n}$ span $\mathbb{F}_{q^{m}}^{n}$). Thus
$\left<x_{1},\ldots,x_{i+1},x_{i+2},\ldots,x_{n}\right>_{\mathbb{F}_{q^{m}}}=\mathbb{F}_{q^{m}}^{n}$.
For $s\in\\{i+2,\ldots,n\\}$, define the vector $z_{s}:=x_{1}+x_{2}-x_{s}$ and
notice that $z_{s},x-z_{s}\in T_{i}(n,m;q)$ by construction (here is where the
assumption $i\geq 3$ is technically needed). Lemma 3.5 implies that $z_{s}\in
X$ and therefore $x_{s}\in X$, for all $s\in\\{i+2,\ldots,n\\}$, since
$x_{1},x_{2}\in X$ by the first part of the proof and $X$ is a linear space.
It follows that $X$ contains a basis for $\mathbb{F}_{q^{m}}^{n}$ and
therefore $X=\mathbb{F}_{q^{m}}^{n}$, as claimed. ∎

We can finally characterize the modular elements of any lattice
$\mathscr{L}_{i}(n,m;q)$ for $2\leq i\leq n$. The following is the analogue of
[9, Theorem 1.3]. Note that in the rank-metric context (and in sharp contrast
to the Hamming-metric case) it also holds for $i=2$.

######  Theorem 3.15.

For all $i\geq 2$, the modular elements of $\mathscr{L}_{i}(n,m;q)$ are the
elements of the set $\\{X\leq\mathbb{F}_{q^{m}}^{n}:X\subseteq
T_{i}(n,m;q)\\}\cup\\{\mathbb{F}_{q^{m}}^{n}\\}$.

###### Proof.

First of all we note that if $X\subseteq T_{i}(n,m;q)$, then
$X\wedge_{i}Y\in\mathscr{L}_{i}(n,m;q)$ and therefore
$X\wedge_{i}Y=X\wedge_{n}Y$, for all $Y\in\mathscr{L}_{i}(n,m;q)$. It follows
that $X$ is modular in $\mathscr{L}_{i}(n,m;q)$ by Lemma 3.4. Moreover,
$\mathbb{F}_{q^{m}}^{n}$ is trivially modular.

In order to prove the other implication, suppose that $X$ is a modular element
of $\mathscr{L}_{i}(n,m;q)$ that contains a vector of rank strictly greater
than $i$. By combining Lemmas 3.9 and 3.10 in the case $i=2$ and Lemmas 3.12,
3.13 and 3.14 in the case $i\geq 3$, we conclude that
$X=\mathbb{F}_{q^{m}}^{n}$. This establishes the theorem. ∎

We are finally ready to establish Theorem 3.2.

###### Proof of Theorem 3.2.

The supersolvability of the lattice $\mathscr{L}_{1}(n,m;q)$ immediately
follows from the fact that $\mathscr{L}_{1}(n,m;q)$ is a modular lattice by
Proposition 3.6. We assume $i\geq 2$ in the remainder of the proof. Let
$\\{e_{1},\ldots,e_{n}\\}$ be the canonical basis of $\mathbb{F}_{q^{m}}^{n}$.
Consider the following maximal chain in $\mathscr{L}_{i}(n,m;q)$:

$\left<0\right>_{\mathbb{F}_{q^{m}}}\mathrel{<\kern-6.00006pt\raise
0.06458pt\hbox{$\cdot$}}\left<e_{1}\right>_{\mathbb{F}_{q^{m}}}\mathrel{<\kern-6.00006pt\raise
0.06458pt\hbox{$\cdot$}}\left<e_{1},e_{2}\right>_{\mathbb{F}_{q^{m}}}\mathrel{<\kern-6.00006pt\raise
0.06458pt\hbox{$\cdot$}}\cdots\mathrel{<\kern-6.00006pt\raise
0.06458pt\hbox{$\cdot$}}\left<e_{1},e_{2},\ldots,e_{n}\right>_{\mathbb{F}_{q^{m}}}.$
(3)

For $1\leq k\leq n$, all the elements of
$\left<e_{1},e_{2},\ldots,e_{k}\right>_{\mathbb{F}_{q^{m}}}$ have rank at most
$k$ by the triangular inequality. Moreover, Theorem 3.15 implies that if
$i\in\\{n-1,n\\}$ or $m\in\\{2,\ldots,i\\}$, then all the element of the chain
in (3) are modular. This shows that, by definition, $\mathscr{L}_{i}(n,m;q)$
is supersolvable in those cases.

We conclude the proof by showing that $\mathscr{L}_{i}(n,m;q)$ cannot be
supersolvable if $m\geq i+1$ and $i\in\\{2,\ldots,n-2\\}$. We will do this by
showing a much stronger fact: All the elements of $\mathscr{L}_{i}(n,m;q)$ of
rank (i.e., dimension) $i+1$ cannot be modular. To see this, let
$X\in\mathscr{L}_{i}(n,m;q)$ have dimension $i+1$ and let
$\smash{G\leq\mathbb{F}_{q^{m}}^{(i+1)\times n}}$ be the full-rank matrix in
reduced row-echelon form whose rows are a basis for $X$. Let
$\Gamma:=\\{\gamma_{1},\ldots,\gamma_{i+1}\\}$ be $\mathbb{F}_{q}$-linearly
independent elements of $\mathbb{F}_{q^{m}}$ (they exist because $m\geq i+1$).
One can easily check that $\sum_{s=1}^{i+1}\gamma_{s}G_{s}$ has rank at least
$i+1$. Therefore $X$ cannot be modular by Theorem 3.15. ∎

Table 1 illustrates Theorem 3.2 visually. For a fixed $n\geq 3$, it shows the
“supersolvability region” of $\mathscr{L}_{i}(n,m;q)$.

| $m$
---|---
|  $2$ |  $3$ |  $4$ |  $\cdots$ |  $j$ |  $j+1$ |  $\cdots$
$i$ |  $1$ |  |  |  |  |  |  |
|  $2$ |  |  |  |  |  |  |
|  $3$ |  |  |  |  |  |  |
|  $\vdots$ |  |  |  |  |  |  |
|  $j$ |  |  |  |  |  |  |
|  $\vdots$ |  |  |  |  |  |  |
|  $n-1$ |  |  |  |  |  |  |
|  $n$ |  |  |  |  |  |  |
Table 1: The “region” of supersolvable rank-metric lattices for $n\geq 3$.

##  4 The Characteristic Polynomial of $\mathscr{L}_{i}(n,m;q)$

In this section we derive explicit formulas for the Whitney numbers of the
first and second kind of some rank-metric lattices, including the
supersolvable ones we characterized in Theorem 3.2. We also compute the
critical exponent (defined later) of rank-metric lattices for some parameter
sets. We start with the case $i=1$.

######  Proposition 4.1.

The following hold.

  1. (1)

$w_{j}(1,n,m;q)=(-1)^{j}q^{\binom{j}{2}}\begin{bmatrix}{n}\\\
{j}\end{bmatrix}_{q}$ for all $j\in\\{0,\ldots,n\\}$.

  2. (2)

$W_{j}(1,n,m;q)=\begin{bmatrix}{n}\\\ {j}\end{bmatrix}_{q}$ for all
$j\in\\{0,\ldots,n\\}$.

  3. (3)

$\displaystyle\chi(\mathscr{L}_{1}(n,m;q);\lambda)=\sum_{j=0}^{n}(-1)^{j}q^{\binom{j}{2}}\begin{bmatrix}{n}\\\
{j}\end{bmatrix}_{q}\lambda^{n-j}=\prod_{j=0}^{n-1}(\lambda-q^{j}).$

###### Proof.

The first two formulas follow from Proposition 3.6 and the definitions of
Whitney numbers of the first and second kind. The third formula follows from
the definition of characteristic polynomial and Lemma 2.12. ∎

We now treat the case $i=n$. The following result easily follows from the
definitions and its proof is therefore omitted.

######  Proposition 4.2.

The following hold.

  1. (1)

$w_{j}(n,n,m;q)=(-1)^{j}q^{m\binom{j}{2}}\begin{bmatrix}{n}\\\
{j}\end{bmatrix}_{q^{m}}$ for all $j\in\\{0,\ldots,n\\}$.

  2. (2)

$W_{j}(n,n,m;q)=\begin{bmatrix}{n}\\\ {j}\end{bmatrix}_{q^{m}}$ for all
$j\in\\{0,\ldots,n\\}$.

  3. (3)

$\displaystyle\chi(\mathscr{L}_{n}(n,m;q);\lambda)=\sum_{j=0}^{n}(-1)^{j}q^{m\binom{j}{2}}\begin{bmatrix}{n}\\\
{j}\end{bmatrix}_{q^{m}}\lambda^{n-j}=\prod_{j=0}^{n-1}(\lambda-q^{mj}).$

It is interesting to observe the very close “structural” analogy between
Propositions 4.1 and 4.2, where the only difference is the presence of the
factor $m$.

Since the rank of a vector $v\in\mathbb{F}_{q^{m}}^{n}$ cannot exceed $m$, the
next result follows from Proposition 4.2.

######  Corollary 4.3.

Suppose that $m\in\\{2,\ldots,i\\}$. The following hold.

  1. (1)

$w_{j}(i,n,m;q)=(-1)^{j}q^{m\binom{j}{2}}\begin{bmatrix}{n}\\\
{j}\end{bmatrix}_{q^{m}}$ for all $j\in\\{0,\ldots,n\\}$.

  2. (2)

$W_{j}(i,n,m;q)=\begin{bmatrix}{n}\\\ {j}\end{bmatrix}_{q^{m}}$ for all
$j\in\\{0,\ldots,n\\}$.

  3. (3)

$\displaystyle\chi(\mathscr{L}_{i}(n,m;q);\lambda)=\sum_{j=0}^{n}(-1)^{j}q^{m\binom{j}{2}}\begin{bmatrix}{n}\\\
{j}\end{bmatrix}_{q^{m}}\lambda^{n-j}=\prod_{j=0}^{n-1}(\lambda-q^{mj}).$

In the next result we study the case $i=n-1$ under the assumption that $m\geq
n$. We will use Stanley’s modular factorization theorem.

######  Theorem 4.4.

If $n\leq m$, then

$\chi(\mathscr{L}_{n-1}(n,m;q);\lambda)=\left(\lambda-q^{(n-1)m}+\prod_{s=1}^{n-1}(q^{m}-q^{s})\right)\prod_{j=0}^{n-2}(\lambda-q^{mj}).$

###### Proof.

By Corollary 1.13 and the proof of Theorem 3.2, we have

$\chi(\mathscr{L}_{n-1}(n,m;q);\lambda)\\\
=\prod_{j=1}^{n}\left(\lambda-\left|\left\\{A\in\textup{At}(\mathscr{L}_{i}(n,m;q)):A\leq\left<e_{1},\ldots,e_{j}\right>_{\mathbb{F}_{q^{m}}},\,A\not\leq\left<e_{1},\ldots,e_{j-1}\right>_{\mathbb{F}_{q^{m}}}\right\\}\right|\right),$
(4)

were we set the span over $\mathbb{F}_{q^{m}}$ of the empty set to be
$\\{0\\}$. Therefore it remains to compute the quantities

$a_{j}:=\left|\left\\{A\in\textup{At}(\mathscr{L}_{n-1}(n,m;q)):A\leq\left<e_{1},\ldots,e_{j}\right>_{\mathbb{F}_{q^{m}}},A\not\leq\left<e_{1},\ldots,e_{j-1}\right>_{\mathbb{F}_{q^{m}}}\right\\}\right|$

for $j\in\\{1,\ldots,n\\}$. Observe that the atoms of $\mathscr{L}_{i}(n,m,q)$
below $\left<e_{1},\ldots,e_{j}\right>_{\mathbb{F}_{q^{m}}}$ but not below
$\left<e_{1},\ldots,e_{j-1}\right>_{\mathbb{F}_{q^{m}}}$ are the
$1$-dimensional subspace of $\mathbb{F}_{q^{m}}^{n}$ generated by a vector of
the form $(\star,\cdots,\star,1,0,\cdots,0)$, where the $1$ is in position $j$
and the $\star$’s are arbitrary elements of $\mathbb{F}_{q^{m}}$. It follows
that $a_{j}=q^{(j-1)m}$, for all $j\in\\{1,\ldots,n-1\\}$. Moreover,

$a_{n}=q^{(n-1)m}-\left|\left\\{\left<x\right>_{\mathbb{F}_{q^{m}}}\leq\mathbb{F}_{q^{m}}^{n}:\textup{rk}(x)=n\right\\}\right|=q^{(n-1)m}-\prod_{s=1}^{n-1}(q^{m}-q^{s}).$

Substituting these expressions into (4) gives the desired result. ∎

We now extract from Theorem 4.4 some more explicit information about the
Whitney numbers of $\mathscr{L}_{n-1}(n,m;q)$.

######  Corollary 4.5.

The following hold for $n\leq m$.

  1. (1)

$w_{0}(n-1,n,m;q)=1$.

  2. (2)

$\displaystyle
w_{j}(n-1,n,m;q)=(-1)^{j}\left(q^{m\binom{j}{2}}\begin{bmatrix}{n}\\\
{j}\end{bmatrix}_{q^{m}}-q^{m\binom{j-1}{2}}\begin{bmatrix}{n-1}\\\
{j-1}\end{bmatrix}_{q^{m}}\prod_{s=1}^{n-1}(q^{m}-q^{s})\right)$ for all
$j\in\\{1,\ldots,n\\}$.

  3. (3)

$W_{0}(n-1,n,m;q)=1$.

  4. (4)

$\displaystyle W_{1}(n-1,n,m;q)=\begin{bmatrix}{n}\\\
{1}\end{bmatrix}_{q^{m}}-\prod_{s=1}^{n-1}(q^{m}-q^{s})$.

  5. (5)

$W_{j}(n-1,n,m;q)=\begin{bmatrix}{n}\\\ {j}\end{bmatrix}_{q^{m}}$ for all
$j\in\\{2,\ldots,n\\}$.

###### Proof.

Parts (1) and (3) follow from the definitions. Part (2) follows from Theorem
4.4, since by Lemma 2.12 we have

$\chi(\mathscr{L}_{n-1}(n,m;q);\lambda)\\\ \begin{aligned}
&=\prod_{j=0}^{n-1}(\lambda-q^{mj})+\prod_{s=1}^{n-1}(q^{m}-q^{s})\prod_{j=0}^{n-2}(\lambda-q^{mj})\\\
&=\sum_{j=0}^{n}(-1)^{j}q^{m\binom{j}{2}}\begin{bmatrix}{n}\\\
{j}\end{bmatrix}_{q^{m}}\lambda^{n-j}+\prod_{s=1}^{n-1}(q^{m}-q^{s})\sum_{j=1}^{n}(-1)^{j-1}q^{m\binom{j-1}{2}}\begin{bmatrix}{n-1}\\\
{j-1}\end{bmatrix}_{q^{m}}\lambda^{n-j}.\end{aligned}$

By the definition of Whitney numbers of the second kind we have

$W_{1}(n-1,n,m;q)=\begin{bmatrix}{n}\\\
{1}\end{bmatrix}_{q^{m}}-\left|\left\\{\left<x\right>_{\mathbb{F}_{q^{m}}}\leq\mathbb{F}_{q^{m}}^{n}:\textup{rk}(x)=n\right\\}\right|,$

which implies (4). Finally, (5) follows from the observation that any subspace
of $\mathbb{F}_{q^{m}}^{n}$ of dimension $j\geq 2$ has a basis of vectors of
rank $\leq n-1$. To see this, let $X\leq\mathbb{F}_{q^{m}}^{n}$ be of
dimension $j\in\\{2,\ldots,n\\}$ and let
$\smash{G\in\mathbb{F}_{q^{m}}^{j\times n}}$ be the matrix in reduced row-
echelon form whose rows form a basis of $X$. Every row of $X$ has at least
$j-1$ zeroes. Therefore its rank is at most $n-j+1\geq n-1$. ∎

The following result is the $q$-analogue of [9, Theorem 3.3]. It computes some
of the roots of the characteristic polynomial of rank-metric lattices, which
is a standard question in lattice theory.

######  Proposition 4.6.

For all $i\in\\{1,\ldots,n\\}$, the characteristic polynomial of
$\mathscr{L}_{i}(n,m;q)$ has the integers $1,q^{m},\ldots,q^{(i-1)m}$ among
its roots.

###### Proof.

Let $\smash{U_{i}:=\left<e_{1},\ldots,e_{i}\right>_{\mathbb{F}_{q^{m}}}}$. All
the elements of $U_{i}$ have rank less or equal than $i$. Therefore $U_{i}$ is
a modular element of $\mathscr{L}_{i}(n,m;q)$ by Theorem 3.15. As an
application of Theorem 1.12 we obtain

$\chi(\mathscr{L}_{i}(n,m;q);\lambda)=\chi([0,U_{i}];\lambda)\sum_{\begin{subarray}{c}X\in\mathscr{L}_{i}(n,m;q)\\\
X\wedge U_{i}=0\end{subarray}}\mu(0,X)\,\lambda^{n-i-\dim(X)},$ (5)

where $\mu$ is the Möbius function of $\mathscr{L}_{i}(n,m;q)$. The desired
result now follows from the fact that the interval $[0,U_{i}]$ is the lattice
of subspaces of $U_{i}$, whose characteristic polynomial is

$\chi([0,U_{i}];\lambda)=\prod_{s=0}^{i-1}\left(\lambda-q^{sm}\right);$

see e.g. [37]. Combining this with (5) concludes the proof. ∎

The following example shows that converse of Corollary 1.13 does not hold in
general. Recall that, by Proposition 3.6, the rank-metric lattice
$\mathscr{L}_{2}(n,m;q)$ is not supersolvable.

######  Example 4.7.

Let $q=2$, $n=4$, and $m=3$. We have $\alpha_{0}(2,4,3;2)=1$. Observe that the
maximum rank of an element of $\mathbb{F}_{2^{3}}^{4}$ is $3$ and that

$\alpha_{1}(2,4,3;2)=\left|\left\\{\left<v\right>_{\mathbb{F}_{2^{3}}}:v\in\mathbb{F}_{2^{3}}^{4},\,\textup{rk}(v)=3\right\\}\right|=\begin{bmatrix}{4}\\\
{3}\end{bmatrix}_{2}(2^{3}-2)(2^{3}-2^{2})=360.$

We also have

$\alpha_{2}(2,4,3;2)=\left|\left\\{C\leq\mathbb{F}_{2^{3}}^{4}:\dim(C)=2,\,d(C)\geq
3\right\\}\right|.$

On the other hand, by the rank-metric Singleton bound of Theorem 2.11 we have
$6\leq 4(3-d+1)$ for any $2$-dimensional code of $\mathbb{F}_{2^{3}}^{4}$ of
minimum distance $d$. This implies $d\leq 2$ and therefore
$\alpha_{2}(2,4,3;2)=0$. A similar argument using again the rank-metric
Singleton bound of Theorem 2.11 shows that
$\alpha_{3}(2,4,3;2)=\alpha_{4}(2,4,3;2)=0$. Finally, by applying Theorem 2.10
we obtain

$(w_{j}(2,4,3;2):j\in\\{0,\ldots,4\\})=(1,-225,11680,-89280,77824).$

Therefore, by definition, we have
$\chi(\mathscr{L}_{2}(4,3;2);\lambda)=\lambda^{4}-225\lambda^{3}+11680\lambda^{2}-89280\lambda+77824$,
whose factorization is $(\lambda-1)(\lambda-8)(\lambda-64)(\lambda-152)$.
However, $\mathscr{L}_{2}(4,3;2)$ is not supersolvable by Theorem 3.2.

######  Remark 4.8.

The linear factor $\lambda-64=\lambda-2^{2\cdot 3}$ in the previous example
might suggest that the characteristic polynomial of $\mathscr{L}_{2}(4,3;2)$
can be computed as in the proof of Theorem 4.4 via the chain

$0\mathrel{<\kern-6.00006pt\raise 0.06458pt\hbox{$\cdot$}}\langle
e_{1}\rangle_{\mathbb{F}_{2^{3}}}\mathrel{<\kern-6.00006pt\raise
0.06458pt\hbox{$\cdot$}}\langle
e_{1},e_{2}\rangle_{\mathbb{F}_{2^{3}}}\mathrel{<\kern-6.00006pt\raise
0.06458pt\hbox{$\cdot$}}\langle
e_{1},e_{2},e_{3}\rangle_{\mathbb{F}_{2^{3}}}\mathrel{<\kern-6.00006pt\raise
0.06458pt\hbox{$\cdot$}}\mathbb{F}_{2^{3}}^{4}.$

However, the proof of Lemma 3.9 shows that $\langle
e_{1},e_{2},e_{3}\rangle_{\mathbb{F}_{2^{3}}}$ is not modular. To see this in
a more direct way, let $1,\alpha,\beta\in\mathbb{F}_{2^{3}}$ be linearly
independent over $\mathbb{F}_{2}$. Define $x=e_{1}+\alpha e_{2}+\beta e_{3}$
and $z=e_{1}+e_{4}+\alpha e_{2}$. Let $U=\langle
z,x+z\rangle_{\mathbb{F}_{2^{3}}}$. It is easy to check that
$X\in\mathscr{L}_{2}(4,3;2)$. Moreover, $\langle
e_{1},e_{2},e_{3}\rangle_{\mathbb{F}_{2^{3}}}\cap U=\langle x\rangle$, because
$z\in U\setminus\langle e_{1},e_{2},e_{3}\rangle_{\mathbb{F}_{2^{3}}}$. This
contradicts the definition of modularity for $\langle
e_{1},e_{2},e_{3}\rangle_{\mathbb{F}_{2^{3}}}$.

We now turn to the critical exponent of rank-metric lattices, which is one of
the most important invariant of a combinatorial geometry. It was defined by
Crapo and Rota in [17] and it measures the largest dimension of a linear space
that distinguishes, i.e., avoids, a given collection of projective points.
Computing the critical exponent of a combinatorial geometry is a hard problem
in general, known under the name of Critical Problem; see [17].

In this paper, we give the definition of critical exponent directly in the
context of rank-metric lattices.

######  Definition 4.9.

The critical exponent of $\mathscr{L}_{i}(n,m;q)$ is

$\textup{crit}(\mathscr{L}_{i}(n,m;q))=n-\max\\{k\in\mathbb{Z}_{\geq
0}:\alpha_{k}(i,n,m;q)\neq 0\\},$

where $\alpha_{k}(i,n,m;q)$ is defined in Notation 2.8.

In the next result we explicitly compute the value of the critical exponent
for some parameter sets.

######  Theorem 4.10.

  1. (1)

If $n\leq m$, then $\textup{crit}(\mathscr{L}_{i}(n,m;q))=i$.

  2. (2)

If $n>m$, then $\textup{crit}(\mathscr{L}_{i}(n,m;q))\geq ni/m$. Furthermore,
$\textup{crit}(\mathscr{L}_{i}(n,m;q))=ni/m$ if $m\mid ni$ and one of the
following holds:

    1. (a)

$n=tm$ and $i=m-k$ for some nonnegative integers $t,k$;

    2. (b)

$n=mk/2$ for some nonnegative integer $k$, $m$ even, and $i\in\\{2,m-2\\}$;

    3. (c)

$n=8$, $m=6$, $i=3$ and $q=2^{h}$ with $h\equiv 1\mod 6$;

    4. (d)

$n=8$, $m=4$, $i=3$ and $q=2^{h}$ for some odd nonnegative integer $h$.

###### Proof.

We start observing that, by definition, we have

$\max\\{k\in\mathbb{Z}_{>0}:\alpha_{k}(i,n,m;q)\neq 0\\}\\\ \begin{aligned}
&=\max\\{k\in\mathbb{Z}_{>0}:\exists\,C\leq\mathbb{F}_{q^{m}}^{n}\textup{ s.t.
}\dim(C)=k\textup{ and }d(C)\geq i+1\\}\\\
&\leq\frac{\max\\{n,m\\}(\min\\{n,m\\}-i)}{m},\end{aligned}$ (6)

where the latter inequality follows from the rank-metric Singleton bound; see
Theorem 2.11. Therefore, by the definition of critical exponent, we have

$\textup{crit}(\mathscr{L}_{i}(n,m;q))\geq
n-\frac{\max\\{n,m\\}(\min\\{n,m\\}-i)}{m},$ (7)

where we have equality in (7) if and only if there exists a code
$C\leq\mathbb{F}_{q^{m}}^{n}$ of dimension
$\max\\{n,m\\}(\min\\{n,m\\}-i)/{m}$ and minimum distance at least $i+1$.
Again by the rank-metric Singleton bound (Theorem 2.11), this happens if and
only if there exists a code $C\leq\mathbb{F}_{q^{m}}^{n}$ of dimension
$\max\\{n,m\\}(\min\\{n,m\\}-i)/{m}$ and minimum distance exactly $i+1$. In
the remainder of the proof we describe some parameter sets for which the
existence of such a code has been established.

If $m\geq n$ then the existence of a code $C\leq\mathbb{F}_{q^{m}}^{n}$ of
dimension $n-i$ and minimum distance $i+1$ has been established by Delsarte
[20] for all $i$. If $m<n$, the existence is only known for some parameters,
which have been conveniently collected in [30, Remark 2.3] and to which we
refer (see also [4, 18, 5]). These correspond precisely to the parameters
listed in the statement. ∎

By combining Theorem 4.10 with [33, Corollary 3.4], we obtain a recursion for
the Whitney numbers of the first kind of $\mathscr{L}_{i}(n,m;q)$ in some
parameter ranges.

######  Proposition 4.11.

Let $j\in\\{0,\ldots,n\\}$. We have

$w_{j}(i,n,m;q)=-\sum_{s=0}^{j-1}w_{s}(i,n,m;q)\begin{bmatrix}{n-s}\\\
{j-s}\end{bmatrix}_{q^{m}}$ (8)

if $n\leq m$ and $n-i<j$, or if $n>m$ and $n(m-i)/m<j$.

##  5 On the Whitney Numbers of $\mathscr{L}_{2}(4,4;q)$

In this section we concentrate on the “smallest” non-supersolvable rank-metric
lattices whose characteristic polynomials we cannot compute by applying the
results of Section 4, namely, lattices of the form $\mathscr{L}_{2}(4,4;q)$.
We give explicit formulas for their characteristic polynomials, under the
assumption that their second Whitney numbers of the first kind are a
polynomial in $q$. This technical assumption will allow us to follow a
computational approach based on rank-metric codes, Theorem 2.10, and Lagrange
interpolation.

######  Notation 5.1.

In this section, $n=m=4$. We denote by $M(q)$ the number of $2$-dimensional
MRD codes $C\leq\mathbb{F}_{q^{4}}^{4}$ to simplify the notation.

We start by observing the following.

######  Proposition 5.2.

We have:

  1. (1)

$\displaystyle w_{2}(2,4,4;q)=q^{4}\begin{bmatrix}{4}\\\
{2}\end{bmatrix}_{q^{4}}-\begin{bmatrix}{3}\\\
{1}\end{bmatrix}_{q^{4}}\,\sum_{j=3}^{4}\begin{bmatrix}{4}\\\
{j}\end{bmatrix}_{q}\,\prod_{s=1}^{j-1}(q^{4}-q^{s})+M(q)$;

  2. (2)

$\displaystyle W_{2}(2,4,4;q)=\begin{bmatrix}{4}\\\
{2}\end{bmatrix}_{q^{4}}-M(q)$.

Moreover, $w_{2}(2,4,4;q)$ is a polynomial in $q$ if and only if $M(q)$ is a
polynomial in $q$.

###### Proof.

By Theorem 2.10, we have

$w_{2}(2,4,4;q)=\sum_{k=0}^{2}\alpha_{k}(2,4,4;q)\begin{bmatrix}{4-k}\\\
{2-k}\end{bmatrix}_{q^{4}}(-1)^{2-k}\,q^{4\binom{2-k}{2}}.$ (9)

Furthermore:

  * •

$\alpha_{0}(2,4,4;q)=1$;

  * •

$\displaystyle\alpha_{1}(2,4,4;q)=\sum_{j=3}^{4}\begin{bmatrix}{4}\\\
{j}\end{bmatrix}_{q}\,\prod_{s=1}^{j-1}(q^{4}-q^{s})$;

  * •

$\alpha_{2}(2,4,4;q)=M(q)$.

These expressions combined with (9) give the first part of the statement. For
the second part, note that by Theorem 2.11 we have $d(C)\leq 3$ for all
$C\leq\mathbb{F}_{q^{4}}^{4}$ of dimension 2. Therefore

$W_{2}(2,4,4;q)=\begin{bmatrix}{4}\\\
{2}\end{bmatrix}_{q^{4}}-|\\{C\leq\mathbb{F}_{q^{4}}^{4}:d(C)>2\\}|=\begin{bmatrix}{4}\\\
{2}\end{bmatrix}_{q^{4}}-M(q).$

The last part of the statement follows from Theorem 2.10 and the fact that
$\alpha_{3}(2,4,4;q)=\alpha_{4}(2,4,4;q)=0$ by Theorem 2.11. ∎

The following observation will be crucial in our approach.

######  Lemma 5.3.

Let $u,v\in\mathbb{F}_{2^{4}}^{4}$ be vectors of rank 4. The number of
2-dimensional MRD codes $C\leq\mathbb{F}_{2^{4}}^{4}$ containing $u$ is the
same as the number of 2-dimensional MRD codes $C\leq\mathbb{F}_{2^{4}}^{4}$
containing $v$.

###### Proof.

Since the entries of $u$ and $v$ both span $\mathbb{F}_{q^{4}}$ over
$\mathbb{F}_{2}$, for all $j\in\\{1,\ldots,i\\}$ there exist
$\smash{a_{1,j},\ldots a_{4,j}\in\mathbb{F}_{2}}$ with
$\smash{v_{j}=\sum_{i=1}^{4}u_{i}a_{i,j}}$. Let $A$ be the matrix whose
$(i,j)$ entry is $a_{i,j}$. Then $A$ has size $4\times 4$ and invertible.
Moreover, $v=u\cdot A$. Therefore the mapping
$\mathbb{F}_{q^{4}}^{4}\to\mathbb{F}_{q^{4}}^{4}$ given by $x\mapsto x\cdot A$
is rank-preserving and thus induces a bijection between the 2-dimensional MRD
codes $C\leq\mathbb{F}_{q^{4}}^{4}$ containing $u$ and the 2-dimensional MRD
codes $C\leq\mathbb{F}_{q^{4}}^{4}$ containing $v$. This establishes the
lemma. ∎

######  Notation 5.4.

We let $\smash{\hat{M}(q)}$ be the quantity defined by the previous lemma,
that is, $\smash{\hat{M}(q)}$ is the number of $2$-dimensional MRD codes in
$\smash{\mathbb{F}_{q^{4}}^{4}}$ containing a given vector of rank $4$.

The quantities $M(q)$ and $\hat{M}(q)$ are closely connected to each other, as
the following result shows.

######  Proposition 5.5.

We have

$\hat{M}(q)=\frac{M(q)(q^{3}-q^{2}-q-1)}{q^{5}(q-1)^{3}(q+1)(q^{2}+q+1)}.$

###### Proof.

We count the elements of the set

$\Sigma:=\\{(C,v):C\leq\mathbb{F}_{q^{4}}^{4}\textup{ is
MRD},\,\dim(C)=2,\,v\in\mathbb{F}_{q^{4}}^{4},\,\textup{rk}(v)=2\\}$ (10)

in two ways. Since the number of rank-4 vectors in $\mathbb{F}_{q^{4}}^{4}$ is
$(q^{4}-1)(q^{4}-q)(q^{4}-q^{2})(q^{4}-q^{3})$, by Lemma 5.3 we have

$|\Sigma|=\hat{M}(q)(q^{4}-1)(q^{4}-q)(q^{4}-q^{2})(q^{4}-q^{3}).$ (11)

Now fix a $2$-dimensional MRD code $C$ in $\smash{\mathbb{F}_{2^{4}}^{4}}$ and
recall for every such $C$ the number of rank-4 vectors $v\in C$ is the same
and is computed, for example, in [19, Corollary 28]. Therefore,

$\displaystyle|\Sigma|$ $\displaystyle=M(q)\left(q^{8}-1-\begin{bmatrix}{4}\\\
{1}\end{bmatrix}_{q}(q^{4}-1)\right)$ (12)
$\displaystyle=M(q)\,q\,(q-1)\,(q+1)\,(q^{2}+1)\,(q^{3}-q^{2}-q-1).$

The statement follows from comparing (11) with (12). ∎

We now start working under the assumption that $w_{2}(2,4,3;q)$ is a
polynomial in $q$. By Proposition 5.2, this is equivalent to assuming that
$M(q)$ is a polynomial in $q$.

The values of $\hat{M}(q)$ for some small values of $q$ (we managed up to
$q=9$) can be computed using algebra software combined with an argument that
we outline very briefly. We let $\alpha$ be a primitive element of
$\mathbb{F}_{q^{4}}$ and $v:=(1,\alpha,\alpha^{2},\alpha^{3})$ be a vector of
rank $4$. Then $\smash{\hat{M}(q)}$ is the number of 2-dimensional MRD codes
$\smash{C\leq\mathbb{F}_{q^{4}}^{4}}$ containing $v$. We consider the code
whose generator matrix is

$\begin{pmatrix}1&\alpha&\alpha^{2}&\alpha^{3}\\\ 0&1&A&B\end{pmatrix}$

for some $A,B\in\mathbb{F}_{q^{4}}^{4}$ such that $\\{1,A,B\\}$ is a set of
elements that are linearly independent over $\mathbb{F}_{q}$. Observe that the
matrices

$\begin{pmatrix}1&\alpha&\alpha^{2}&\alpha^{3}\\\
0&1&A&B\end{pmatrix},\qquad\begin{pmatrix}1&\alpha&\alpha^{2}&\alpha^{3}\\\
0&1&A^{\prime}&B^{\prime}\end{pmatrix}$

generate the same code if and only if $A=A^{\prime}$ and $B=B^{\prime}$.
Therefore to compute $\hat{M}(q)$ it suffices to count all the pairs $(A,B)$
such that, for all non-zero $\gamma\in\mathbb{F}_{q^{4}}$,

$(1,\alpha,\alpha^{2},\alpha^{3})+(0,\gamma,\gamma A,\gamma B)$

has rank 3 or 4. This is equivalent to saying that, for all
$i\in\\{1,\ldots,q^{4}-1\\},$

$(1,\alpha,\alpha^{2},\alpha^{3})+\alpha^{i}(0,1,A,B)$

has rank 3 or 4. Using the matrix representation of vectors one can finally
show that $\hat{M}(q)$ is the number of
$(A_{1},A_{2},A_{3},A_{4},B_{1},B_{2},B_{3},B_{4})\in\mathbb{F}_{q}^{8}$ such
that

$I+\begin{pmatrix}0&0&0&0\\\ 1&0&0&0\\\ A_{1}&A_{2}&A_{3}&A_{4}\\\
B_{1}&B_{2}&B_{3}&B_{4}\end{pmatrix}\cdot M^{i}$

has rank 3 or 4 for all $i\in\\{1,\ldots,q^{4}-1\\}$, where $M$ is the
companion matrix of minimal polynomial of $\alpha$ over $\mathbb{F}_{q}$. By
combining this computational approach with Proposition 5.5 we obtain the
following result.

######  Lemma 5.6.

The values of $M(q)$ for $q\in\\{2,3,4,5,7,8\\}$ are the following:

  * •

$M(2)=1344$,

  * •

$M(3)=6368544$,

  * •

$M(4)=998645760$,

  * •

$M(5)=43710000000$,

  * •

$M(7)=11599543859904$,

  * •

$M(8)=103734668427264$.

Our next step is to use the previous values to interpolate $M(q)$ under the
polynomiality assumption. The putative degree of $M(q)$ is too large (up to
16, as we will explain later) for this to be possible. We therefore proceed by
constructing a polynomial of smaller degree which we can interpolate with the
values computed in Lemma 5.6. We start by recalling the following folklore
result from elementary algebra.

######  Lemma 5.7.

Let $P(x),Q(x)\in\mathbb{Z}[x]$ be polynomials. If $\\{a\in\mathbb{Z}\mid
P(a)/Q(a)\in\mathbb{Z}\\}=+\infty$, then $Q(x)$ divides $P(x)$ in
$\mathbb{Z}[x]$.

By combining Lemma 5.7 with Proposition 5.5, we deduce that if $M(q)$ is a
polynomial in $q$, then $\hat{M}(q)$ is a polynomial in $q$ as well. Under
these assumptions, write

$M(q)=\frac{\hat{M}(q)\,q^{5}(q-1)^{3}(q+1)(q^{2}+q+1)}{q^{3}-q^{2}-q-1}$

and observe that $q^{5}(q^{3}-1)(q^{2}-1)(q-1)$ and $q^{3}-q^{2}-q-1$ are
coprime in $\mathbb{Z}[q]$, which implies that $q^{3}-q^{2}-q-1$ divides
$\hat{M}(q)$ as a polynomial. Therefore we may write
$\hat{M}(q)=Z(q)(q^{3}-q^{2}-q-1)$, for some $Z(q)\in\mathbb{Z}[q]$ of degree
at most 5. The bound on the degree follows from the fact that

$M(q)\leq\begin{bmatrix}{4}\\\ {2}\end{bmatrix}_{q^{4}}=q^{16}+\mbox{lower
order terms in $q$},$

which gives $\deg(M(q))\leq 16$ and thus $\deg(\hat{M}(q))\leq 8$. By
definition, we have

$Z(q)=\frac{M(q)}{q^{5}(q^{3}-1)(q^{2}-1)(q-1)}.$

Since $Z(q)$ has degree at most 5, it can be interpolated using the 6 values
for $M(q)$ computed in Lemma 5.6. The passages are elementary and the final
outcome is

$Z(q)=\frac{1}{2}\,(q^{5}-q^{4}-q^{3}-q^{2}),$

which in turn gives

$M(q)=\frac{1}{2}\,q^{7}(q^{3}-1)(q^{2}-1)(q-1)(q^{3}-q^{2}-q-1).$ (13)

Finally, by applying Proposition 5.2 and Theorem 2.11, after lengthy
computations one obtains the following result.

######  Theorem 5.8.

If $w_{2}(2,4,4;q)$ is a polynomial in $q$, then

$\displaystyle w_{0}(2,4,4;q)$ $\displaystyle=1,$ $\displaystyle
w_{1}(2,4,4;q)$ $\displaystyle=-\left(q^{8}+q^{7}+2q^{6}-q^{3}+1\right),$
$\displaystyle w_{2}(2,4,4;q)$
$\displaystyle=\frac{1}{2}\,\left(q^{16}+3q^{14}+q^{13}+3q^{12}+q^{11}+q^{10}-q^{9}+q^{7}+4q^{6}-2q^{3}\right),$
$\displaystyle w_{3}(2,4,4;q)$
$\displaystyle=-\frac{1}{2}\left(q^{20}+3q^{18}+q^{17}+2q^{16}-q^{15}+3q^{12}+2q^{11}+q^{10}-q^{9}-2q^{8}-q^{7}\right),$
$\displaystyle w_{4}(2,4,4;q)$
$\displaystyle=\frac{1}{2}\left(q^{20}+3q^{18}+q^{17}+q^{16}-q^{15}-3q^{14}-q^{13}+q^{11}\right).$

In particular, we have

$\chi(\mathscr{L}_{2}(4,4;q);\lambda)=(\lambda-1)(\lambda-q^{4})\Biggl{(}\lambda^{2}-(q^{8}+q^{7}+2q^{6}-q^{4}-q^{3})\lambda\\\
+\frac{1}{2}\left(q^{16}+3q^{14}+q^{13}+q^{12}-q^{11}-3q^{10}-q^{9}+q^{7}\right)\Biggr{)}.$

######  Remark 5.9.

It is interesting to observe that the final formula of Theorem 5.8 is
compatible with the modularity of $\langle
e_{1},e_{2}\rangle_{\mathbb{F}_{q^{4}}}$ in $\mathscr{L}_{2}(4,4;q)$, which
implies that $(\lambda-1)(\lambda-q^{4})$ is always a factor of
$\chi(\mathscr{L}_{2}(4,4;q);\lambda)$ by Theorem 1.12. We could also
computationally obtain the value of $M(9)$ and it coincides with the
prediction made by Equation (13).

As an application of Theorem 5.8, we compute the characteristic polynomial of
$\mathscr{L}_{2}(4,4;q)$ for some values of $q$, under the assumption that its
second Whitney number of the first kind is a polynomial in $q$. The final
results are summarized in Table LABEL:tab:valueschi.

Table 2: The characteristic polynomial of $\mathscr{L}_{2}(4,4;q)$, under the assumption that its second Whitney number of the first kind is a polynomial in $q$. We display the unique factorization of the putative polynomial. $q$ |  $\chi(\mathscr{L}_{2}(4,4;q);\lambda)$
---|---
$2$ |  $(\lambda-1)(\lambda-16)(\lambda^{2}-488\lambda+60736)$
$3$ |  $(\lambda-1)(\lambda-81)(\lambda^{2}-10098\lambda+29574801)$
$4$ |  $(\lambda-1)(\lambda-256)(\lambda^{2}-89792\lambda+2588286976)$
$5$ |  $(\lambda-1)(\lambda-625)(\lambda^{2}-499250\lambda+86141640625)$
$7$ |  $(\lambda-1)(\lambda-2401)(\lambda^{2}-6820898\lambda+17687732901601)$
$8$ |  $(\lambda-1)(\lambda-4096)(\lambda^{2}-19394048\lambda+147637824126976)$
$9$ |  $(\lambda-1)(\lambda-6561)(\lambda^{2}-48885282\lambda+962216318765601)$
|

##  6 Lattice-Rank Weights of Rank-Metric Codes

In this section we show an application of rank-metric lattices to coding
theory. We introduce new structural invariants (sometimes called
distinguishers) for rank-metric codes based on rank-metric lattices. We call
these invariants lattice-rank weights and show that they extend the
generalized vector rank weights introduced in [32]. In our approach, the
elements of the rank-metric lattice $\mathscr{L}_{i}(n,m;q)$ play the same
role that optimal anticodes play in [32]. In fact, the elements of
$\mathscr{L}_{1}(n,m;q)$ are precisely the optimal anticodes of [32].

In the second part of this section we then define the notions of lattice
binomial moments and lattice rank weight distribution that are naturally
associated with the invariants we introduce. We derive derive the
corresponding MacWilliams-type identities and discuss some families of rank-
metric codes that are extremal with respect to these new notions.

######  Notation 6.1.

Throughout this section we work with fixed $n,m\geq 2$, which we generally
don’t remember in the notation to improve readability. We also fix a code
$C\leq\mathbb{F}_{q^{m}}^{n}$ of dimension $k$ and minimum distance $d$. All
dimensions are computed over $\mathbb{F}_{q^{m}}$, unless otherwise stated.

We recall the notion of equivalent codes.

######  Definition 6.2.

We say that the code $C,D\leq\mathbb{F}_{q^{m}}^{n}$ are equivalent if there
exists an $\mathbb{F}_{q^{m}}$-linear map
$\varphi:\mathbb{F}_{q^{m}}\rightarrow\mathbb{F}_{q^{m}}$ such that
$\textup{rk}(v)=\textup{rk}(\varphi(v))$ for all
$\smash{v\in\mathbb{F}_{q^{m}}^{n}}$ and $\varphi(C)=D$.

Studying the invariants of codes is a natural way of investigating their
structure. Several invariants are particularly relevant for applications due
to their property of distinguishing between inequivalent codes. The latter
problem is central not only in coding theory but also in some related research
areas. For example, code distinguishers provide an attack tool in code-based
cryptography. We propose the following invariants.

######  Definition 6.3.

For $i\in\\{1,\ldots,n\\}$ and $j\in\\{1,\ldots,k\\}$, the $(i,j)$-th lattice-
rank weight of $C$ is the number

$\ell_{j}^{(i)}(C):=\min\\{\dim(X):X\in\mathscr{L}_{i}(n,m;q),\,\dim(C\cap
X)\geq j\\}.$

The $(i,j)$-th dual lattice-rank weight of $C$ is the number

$\hat{\ell}_{j}^{(i)}(C):=\min\\{\dim(X^{\perp}):X\in\mathscr{L}_{i}(n,m;q),\,\dim(C\cap
X)\geq j\\}.$

######  Remark 6.4.

If $i\in\\{1,n\\}$ or $m\in\\{2,\ldots,i\\}$, then the dual of an element of
$\mathscr{L}_{i}(n,m;q)$ is still an element of $\mathscr{L}_{i}(n,m;q)$.
However, this property does not hold in general. Examples can be easily found.

For ease of notation, we write $\smash{\ell_{j}^{(i)}}$,
$\smash{\ell_{j}^{(i)\perp}}$, $\smash{\hat{\ell}_{j}^{(i)}}$ and
$\smash{\ell_{j}^{(i)\perp}}$ instead of $\smash{\ell_{j}^{(i)}(C)}$,
$\smash{\ell_{j}^{(i)}(C^{\perp})}$, $\smash{\hat{\ell}_{j}^{(i)}(C)}$ and
$\smash{\hat{\ell}_{j}^{(i)}(C^{\perp})}$ respectively.

We also recall the following definition of generalized rank-weights proposed
in [29] and later characterized via anticodes in [32].

######  Definition 6.5.

For $j\in\\{1,\ldots,k\\}$, the $j$-th generalized rank weight of $C$ is

$m_{j}(C):=\min\\{\dim(A):A\in\mathcal{A},\,\dim(A\cap C)\geq j\\},$

where $\mathcal{A}$ denotes the set of $\mathbb{F}_{q^{m}}$-subspaces of
$\mathbb{F}_{q^{m}}^{n}$ that have a basis made of vectors with entries in
$\mathbb{F}_{q}$.

Again, we write $m_{j}$ instead of $m_{j}(C)$. We have the following result
which, in particular, shows that the lattice rank weights extend the
generalized rank weights as a code invariant.

######  Theorem 6.6.

The following hold.

  1. (1)

$\ell_{j}^{(1)}=m_{j}$ for all $j\in\\{1,\ldots,k\\}$.

  2. (2)

$\ell_{1}^{(i)}=\left\lceil\frac{d}{i}\right\rceil$ for all
$i\in\\{1,\ldots,n\\}$.

  3. (3)

$\ell_{j}^{(i)}\leq n$ for all $i\in\\{1,\ldots,n\\}$ and
$j\in\\{1,\ldots,k\\}$.

  4. (4)

$\ell_{j}^{(i)}<\ell_{j+1}^{(i)}$ for all $i\in\\{1,\ldots,n\\}$ and
$j\in\\{1,\ldots,k-1\\}$.

  5. (5)

$\ell_{j}^{(i+1)}\leq\ell_{j}^{(i)}$ for all $i\in\\{1,\ldots,n-1\\}$ and
$j\in\\{1,\ldots,k\\}$.

  6. (6)

$\ell_{j}^{(i)}\leq n-k+j$ for all $i\in\\{1,\ldots,n\\}$ and
$j\in\\{1,\ldots,k\\}$.

  7. (7)

$\ell_{j}^{(i)}\geq\left\lceil\frac{d}{i}\right\rceil+j-1$ for all
$i\in\\{1,\ldots,n\\}$ and $j\in\\{1,\ldots,k\\}$.

###### Proof.

We establish the various properties separately.

  1. (1)

Proposition 3.6 implies that the elements of $\mathscr{L}_{1}(n,m;q)$ are
precisely the $\mathbb{F}_{q^{m}}$-subspaces of $\mathbb{F}_{q^{m}}^{n}$ with
a basis made of vectors with entries in $\mathbb{F}_{q}$. The statement
follows from Remark 3.7 and Definition 6.5.

  2. (2)

Every $v\in C$ can be expressed as the sum of no less than
$\left\lceil\textup{rk}(v)/i\right\rceil$ element of rank less or equal then
$i$. Hence we have

$\displaystyle\ell_{1}^{(i)}$
$\displaystyle=\min\\{\dim(X):X\in\mathscr{L}_{i}(n,m;q)\mid\dim(C\cap X)\geq
1\\}$
$\displaystyle=\min\left\\{\left\lceil\frac{\textup{rk}(v)}{i}\right\rceil:v\in
C\right\\}=\left\lceil\frac{d}{i}\right\rceil.$

  3. (3)

The property follows from the fact that $\dim(X)\leq n$ for any
$X\in\mathscr{L}_{i}(n,m;q)$.

  4. (4)

Let $X\in\mathscr{L}_{i}(n,m;q)$ such that $\dim(X\cap C)\geq j+1$ and
$\smash{\dim(X)=\ell_{j}^{(i)}}$. Take an element $Y\in\mathscr{L}_{i}(n,m;q)$
such that $Y\leq X$ and $\smash{\dim(Y)=\dim(X)-1=\ell_{j}^{(i)}-1}$. We will
prove that $\dim(Y\cap C)\geq j$. Note that, since $Y\leq X$, we have $Y\cap
C=Y\cap(X\cap C)$ and $Y+(X\cap C)\leq X$. Therefore,

$\displaystyle\dim(Y\cap C)$ $\displaystyle=\dim(Y\cap(X\cap C))$
$\displaystyle=\dim(Y)+\dim(X\cap C)-\dim(Y+(X\cap C))$
$\displaystyle\geq\dim(Y)+\dim(X\cap C)-\dim(X)$
$\displaystyle=\ell_{j}^{(i)}-1+j+1-\ell_{j}^{(i)},$

which implies $\dim(Y\cap C)\geq j$ and concludes the proof.

  5. (5)

The property follows from the definition of $(i,j)$-th lattice-rank weight.

  6. (6)

By (3) and (4) we have an increasing chain
$\smash{\ell_{j}^{(i)}<\ell_{j+1}^{(i)}<\cdots<\ell_{k}^{(i)}\leq n}$. It
follows that $\smash{\ell_{j}^{(i)}+k-j\leq n}$.

  7. (7)

By (2) and (4) we have the decreasing chain
$\smash{\ell_{j}^{(i)}>\ell_{j-1}^{(i)}>\cdots>\ell_{1}^{(i)}=\left\lceil\frac{d}{i}\right\rceil}$.
It follows that $\smash{\ell_{j}^{(i)}\geq\ell_{1}^{(i)}+j-1}$.

This concludes the proof. ∎

######  Remark 6.7.

Observe that, by part (2) of Theorem 6.6, we have that
$\smash{\ell_{1}^{(i)}=1}$ for all $i\geq d$. This shows that
$\smash{\ell_{1}^{(i)}=\ell_{1}^{(i+1)}}$ for $i\geq d$ and therefore the
bound in part (5) of Theorem 6.6 is sometimes met with equality if $j=1$.
Notice that the same bound can be met with equality also for $j\neq 1$.
Consider, for example,
$\smash{C:=\left<e_{1},e_{2}\right>_{\mathbb{F}_{q^{m}}}\leq\mathbb{F}_{q^{m}}^{n}}$.
It is not difficult to check that $\smash{\ell_{1}^{(i)}=1}$ and
$\smash{\ell_{2}^{(i)}=2}$ for all $i\in\\{1,\ldots,n\\}$, which shows that
$\smash{\ell_{2}^{(i)}=\ell_{2}^{(i+1)}}$ for all $i\in\\{1,\ldots,n\\}$.

In the following example we show that lattice-rank weights can distinguish
between inequivalent codes when the “fundamental” codes parameters (length,
dimension, minimum distance, and generalized weights) don’t.

######  Example 6.8.

Let $\alpha$ be the primitive element of $\mathbb{F}_{2^{4}}$ with
$\alpha^{4}+\alpha+1=0$ and consider the codes

$C:=\left<(1,0,\alpha^{5},\alpha),(0,1,\alpha^{14},\alpha^{13})\right>,\qquad
D:=\left<(1,0,\alpha^{6},\alpha^{10}),(0,1,\alpha^{7},\alpha^{3})\right>,$

where all the spans in this example are over $\mathbb{F}_{2^{4}}$. The
following hold.

  1. (1)

$\textup{rk}((1,0,\alpha^{5},\alpha))=\textup{rk}((0,1,\alpha^{14},\alpha^{13}))=\textup{rk}((1,0,\alpha^{6},\alpha^{10}))=\textup{rk}((0,1,\alpha^{7},\alpha^{3}))=3$

  2. (2)

$\ell_{1}^{(1)}(C)=2$. This value is attained, for example, by the space
$\left<(1,1,0,0),(0,0,1,1)\right>$ of $\mathscr{L}_{1}(4,4;2)$, whose
intersection with $C$ is $\left<(1,1,\alpha^{12},\alpha^{12})\right>$.

  3. (3)

$\ell_{1}^{(1)}(D)=2$. This value is attained, for example, by the space
$\left<(1,0,1,1),(0,1,0,1)\right>$ of $\mathscr{L}_{1}(4,4;2)$, whose
intersection with $C$ is $\left<(1,\alpha^{6},1,\alpha^{13})\right>$.

  4. (4)

$\ell_{2}^{(1)}(C)=\ell_{2}^{(1)}(D)=4$. This means that the only element of
$\mathscr{L}_{1}(4,4;2)$ that intersects $C$ and $D$ in a subspace of
dimension at lest $2$ is the full space $\mathbb{F}_{2^{4}}^{4}$.

Therefore, by Theorem 6.6, $C$ and $D$ have the same “fundamental” code
parameters. On the other hand, the following hold.

  1. (5)

$\ell_{1}^{(2)}(C)=1$. This value is attained, for example, by the element
$\left<(1,\alpha^{6},0,1)\right>$ of $\mathscr{L}_{2}(4,4;2)$.

  2. (6)

$\ell_{1}^{(2)}(D)=1$. This value is attained, for example, by the element
$\left<(1,\alpha^{6},1,\alpha^{13})\right>$ of $\mathscr{L}_{2}(4,4;2)$.

  3. (7)

$\ell_{2}^{(2)}(C)=2$. This is because $C\in\mathscr{L}_{2}(4,4;2)$, since
$C=\left<(1,1,\alpha^{12},\alpha^{12}),(1,\alpha^{6},0,1)\right>$.

  4. (8)

$\ell_{2}^{(2)}(D)=3$. This is because $D\notin\mathscr{L}_{2}(4,4;2)$ and the
smallest element of $\mathscr{L}_{2}(4,4;2)$ containing $D$ is
$\left<(1,0,0,\alpha^{11}),(0,1,0,\alpha^{14}),(0,0,1,\alpha^{8})\right>$.

This allows us to conclude that $C$ and $D$ are inequivalent codes, even
though they share the same generalized rank weights.

We now introduce the notions of weight distribution and binomial moments
associated to the lattice-rank weights of a code. We show that partial
information about the former provides partial information about the latter.
These two notions capture structural properties of the underlying code that
have been investigated before in the coding theory literature; see, for
example, [23, 6, 14, 13]. In particular, the binomial moments have been used
to define and study the zeta function of codes; see [23] for the Hamming
metric and [6, 15] for the rank metric.

In the context outlined above, we also derive the MacWilliams identities for
rank-metric codes associated to the lattice-rank weights. These identities
provide a connection between the invariants of a code and those of its dual.

######  Definition 6.9.

Let $i\in\\{1,\ldots,n\\}$ and $j\in\\{1,\ldots,k\\}$. The $(i,j)$-th lattice-
rank distribution of $C$ is define as the integer vector whose $(u+1)$-th
component, for $u\in\\{0,\ldots,n\\}$, is

$A_{j,u}^{(i)}(C):=\sum_{\scriptsize\begin{matrix}X\in\mathscr{L}_{i}(n,m;q)\\\
\dim(X)=u\end{matrix}}A_{j,X}^{(i)}(C),$

where, for $X\in\mathscr{L}_{i}(n,m;q)$,

$A_{j,X}^{(i)}(C):=|\\{D\leq C\cap X:\dim(D)=j\textup{ and
}\nexists\;Y\in\mathscr{L}_{i}(n,m;q)\textup{ s.t. }D\leq Y<X\\}|.$

######  Definition 6.10.

Let $i\in\\{1,\ldots,n\\}$, $j\in\\{1,\ldots,k\\}$ and $u\in\\{0,\ldots,n\\}$.
The $(i,j,u)$-th lattice binomial moment of $C$ is defined as

$B_{j,u}^{(i)}(C):=\sum_{\begin{subarray}{c}X\in\mathscr{L}_{i}(n,m;q)\\\
\dim(X)=u\end{subarray}}B_{j,X}(C),$

where

$B_{j,X}(C)=\begin{bmatrix}{\dim(C\cap X)}\\\
{j}\end{bmatrix}_{q^{m}}\quad\mbox{for all $X\in\mathscr{L}_{i}(n,m;q)$.}$

To further simplify the notation, in the sequel we will write
$\smash{A_{j,u}^{(i)}}$ and $\smash{B_{j,u}^{(i)}}$ instead of
$\smash{A_{j,u}^{(i)}(C)}$ and $\smash{B_{j,u}^{(i)}(C)}$, respectively. The
following result shows that $\smash{B_{j,u}^{(i)}}$ is fully determined by the
code parameters, for $u$ in some intervals.

######  Proposition 6.11.

We have

$B_{j,u}^{(i)}=\begin{cases}0&\textup{ if }u<\ell_{j}^{(i)},\\\
\begin{bmatrix}{k+u-n}\\\
{j}\end{bmatrix}_{q^{m}}|\\{X\in\mathscr{L}_{i}(n,m;q):\dim(X)=u\\}|&\textup{
if }u>n-\hat{\ell}_{1}^{(i)\perp}.\end{cases}$

###### Proof.

If $u<\ell_{j}^{(i)}$, then for all $X\in\mathscr{L}_{i}(n,m;q)$ such that
$\dim(X)=u$ we have

$\begin{bmatrix}{\dim(C\cap X)}\\\ {j}\end{bmatrix}_{q^{m}}=0.$

On the other hand, if $\smash{n-u<\hat{\ell}_{1}^{(i)\perp}}$ then
$C^{\perp}\cap X^{\perp}=\\{0\\}$ and the following holds:

$\displaystyle B_{j,u}^{(i)}(C)$
$\displaystyle=\sum_{\begin{subarray}{c}X\in\mathscr{L}_{i}(n,m;q)\\\
\dim(X)=u\end{subarray}}\begin{bmatrix}{\dim(C\cap X)}\\\
{j}\end{bmatrix}_{q^{m}}$
$\displaystyle=\sum_{\begin{subarray}{c}X\in\mathscr{L}_{i}(n,m;q)\\\
\dim(X)=u\end{subarray}}\begin{bmatrix}{n-\dim(C^{\perp}+X^{\perp})}\\\
{j}\end{bmatrix}_{q^{m}}$
$\displaystyle=\sum_{\begin{subarray}{c}X\in\mathscr{L}_{i}(n,m;q)\\\
\dim(X)=u\end{subarray}}\begin{bmatrix}{\dim(C^{\perp}\cap
X^{\perp})+k+u-n}\\\ {j}\end{bmatrix}_{q^{m}}$
$\displaystyle=\sum_{\begin{subarray}{c}X\in\mathscr{L}_{i}(n,m;q)\\\
\dim(X)=u\end{subarray}}\begin{bmatrix}{k+u-n}\\\ {j}\end{bmatrix}_{q^{m}}$
$\displaystyle=\begin{bmatrix}{k+u-n}\\\
{j}\end{bmatrix}_{q^{m}}|\\{X\in\mathscr{L}_{i}(n,m;q):\dim(X)=u\\}|.$

This concludes the proof. ∎

######  Remark 6.12.

We have that $B_{j,u}^{(i)}=A_{j,u}^{(i)}=0$ for all $u<\ell_{j}^{(i)}$.

The following result shows the relation between the invariants we just
defined. Its proof is similar to the one of [13, Theorem 6.4] and therefore we
omit it.

######  Theorem 6.13.

Let $i\in\\{1,\ldots,n\\}$, $j\in\\{1,\ldots,k\\}$ and
$u,v\in\\{0,\ldots,n\\}$. Define

$f_{i}(u,v)=|\\{(X,Y)\in\mathscr{L}_{i}(n,m;q)\times\mathscr{L}_{i}(n,m;q):Y\leq
X,\,\dim(X)=u,\,\dim(Y)=v\\}|$

and denote by $\mu^{(i)}$ the Möbius function of the lattice
$\mathscr{L}_{i}(n,m;q)$. The following hold.

  1. (1)

$\displaystyle B_{j,u}^{(i)}=\sum_{v=0}^{u}A_{j,v}^{(i)}\,f_{i}(u,v)$.

  2. (2)

$\displaystyle
A_{j,v}^{(i)}=\sum_{u=0}^{v}\mu^{(i)}(u,v)B_{j,u}^{(i)}\,f_{i}(v,u)$.

The following can be seen as the MacWilliam identities for lattice-rank
weights. We derive them passing through the lattice binomial moments. Using
Theorem 6.13, an analogous result can be derived for the lattice-rank
distribution. MacWilliam-types identities in this form have already been
established in the coding theory literature in several contexts. The proof of
this result is similar to the one of [13, Theorem 6.7] and we therefore omit
it.

######  Theorem 6.14.

The following holds for all $i,j\in\\{1,\ldots,k\\}$ and
$u\in\\{1,\ldots,n\\}$.

$B_{j,u}^{(i)}=\sum_{p=0}^{i}q^{mp(k+u-n-i+p)}\begin{bmatrix}{k+u-n}\\\
{i-p}\end{bmatrix}_{q^{m}}\sum_{\begin{subarray}{c}X\in\mathscr{L}_{i}(n,m;q)\\\
\dim(X)=u\end{subarray}}\begin{bmatrix}{\dim(C^{\perp}\cap X^{\perp})}\\\
{j}\end{bmatrix}_{q^{m}}$

We conclude this section by characterizing some classes of rank-metric codes
that exhibit interesting rigidity properties with respect to lattice-rank
weights. The following is the vector-analogue of the family of $j$-TBMD codes;
see [14, Definition 4.1] for matrix codes and [13, Definition 6.9] for higher-
order tensors. The latter are a special class of tensor rank-metric whose
tensor binomial moments are fully determined by the code parameters.

######  Definition 6.15.

Let $i\in\\{1,\ldots,n\\}$ and $j\in\\{1,\ldots,k\\}$. We say that $C$ is a
$(i,j)$-LBMD code (short for lattice binomial moment determined) code if

$n-\ell_{j}^{(i)}-\hat{\ell}_{1}^{(i)\perp}<0.$

As an immediate consequence of Proposition 6.11 and Theorem 6.13, we have that
the lattice binomial moments and the lattice-rank distribution of $C$
associated with the lattice $\mathscr{L}_{i}(n,m;q)$ are fully determined by
its code parameters. The following two results are straighforward consequences
of Theorem 6.6 and Proposition 6.11.

######  Proposition 6.16.

Let $i\in\\{1,\ldots,n\\}$ and $j\in\\{1,\ldots,k\\}$. If $C$ is $(i,j)$-LBMD
then $C$ is $(i-1,j)$-LBMD and $(i,j+1)$-LBMD.

######  Proposition 6.17.

Let $i\in\\{1,\ldots,n\\}$ and $j\in\\{1,\ldots,k\\}$. If $C$ is $(i,j)$-LBMD,
then the $(i,j)$-th lattice rank distribution and binomial moments are fully
determined by $n$, $k$, $i$, and $j$.

We now turn to the lattice-analogues of MTR codes, which were introduced in
[15].

######  Definition 6.18.

We say that $C\leq\mathbb{F}_{q^{m}}^{n}$ is perfect if
$\smash{C=\left<0\right>_{\mathbb{F}_{q^{m}}}}$ or if $C$ has a basis made of
rank-$1$ vectors. Equivalently, $C$ is perfect if
$C\in\mathscr{L}_{1}(n,m;q)$.

In the linear algebra literature, perfect spaces are often called Frobenius-
closed.

######  Remark 6.19.

The concept of perfect space was first introduced in [3], for matrix spaces,
and later in [13], for higher order tensors. As observed in [12, Proposition
14.45] and [15, Proposition 3.4], there is a strong connection between perfect
spaces and the tensor rank of a tensor code. More precisely, the tensor rank
can be defined as the smallest dimension of a perfect space containing a
tensor code. As a consequence, the value of $\smash{\ell_{k}^{(1)}}$ can be
seen as the vectorial analogue of the tensor rank of a tensor code the bound
of part (7) of Theorem 6.6 is the corresponding Kruskal-type bound; see [27].

In Remark 6.19, we observed that Property (7) of Theorem 6.6 can be regarded
as vector analogue of the tensor-rank bound. Matrix rank-metric codes meeting
the Kruskal’s bound with equality, namely the MTR codes, are of great
interested in coding theory for their properties of optimality in terms of
storage and encoding; see [15, Table 1]. We introduce the vector analogue of
this family of codes.

######  Definition 6.20.

Let $i\in\\{1,\ldots,n\\}$. We say that the code $C$ is $i$-lattice-optimal if
it attains the bound of part (7) of Theorem 6.6 for $j=k$, i.e., if

$\ell_{k}^{(i)}=\ell_{1}^{(i)}+k-1=\left\lceil\frac{d}{i}\right\rceil+k-1.$

In the next result we show that $C$ is $i$-lattice-optimal if and only if $C$
attains the bound of part (7) of Theorem 6.6 for $i$ and for all
$j\in\\{1,\ldots,k\\}$. In particular, the lattice-rank weights of such a $C$
indexed by $i$ are fully determined by $\smash{\ell_{1}^{(i)}}$.

######  Proposition 6.21.

The following statements are equivalent for all $i\in\\{1,\ldots,n\\}$.

  1. (1)

$C$ is a $i$-lattice-optimal code.

  2. (2)

$\ell_{j}^{(i)}=\ell_{1}^{(i)}+j-1$ for all $j\in\\{1,\ldots,k\\}$.

###### Proof.

If $\ell_{j}^{(i)}=\ell_{1}^{(i)}+j-1$ for all $j\in\\{1,\ldots,k\\}$ then, in
particular, $\ell_{k}^{(i)}=\ell_{1}^{(i)}+k-1$, which implies that $C$ is
$i$-lattice-optimal. On the other hand, let $C$ be $i$-lattice-optimal. By
parts (2) and (4) of Theorem 6.6 we have an increasing chain

$\ell_{1}^{(i)}<\ell_{2}^{(i)}<\cdots<\ell_{k}^{(i)}=\ell_{1}^{(i)}+k-1.$

This implies the statement. ∎

As a consequence of Theorem 6.6 and [32, Corollary 19], we have that property
(6) of Theorem 6.6 can be seen as a generalization of the Singleton-type bound
for rank-metric codes (Theorem 2.11), which can be recovered for $i=1$ and
$j=1$. This motivates the following definition.

######  Definition 6.22.

Let $i\in\\{1,\ldots,n\\}$ and $j\in\\{1,\ldots,k\\}$. We say that $C$ is
$(i,j)$-LMRD (short for lattice maximum rank distance) if $C$ meets the bound
of property (6) of Theorem 6.6 with equality, i.e., if

$\ell_{j}^{(i)}=n-k+j.$

The following result can be seen as the vector-analogue of [14, Theorem 4.15].

######  Proposition 6.23.

$C$ is MRD if and only if $C$ is $(1,1)$-LBMD.

###### Proof.

Recall that by Remark 6.4 we have that if
$\\{X^{\perp}:X\in\mathscr{L}_{1}(n,m;q)\\}=\mathscr{L}_{1}(n,m;q)$. If $C$ is
MRD then $C^{\perp}$ is MRD as well [20] and we have

$n-d-d^{\perp}=n-(n-k+1)-(k+1)=-2<0,$

which implies that $C$ is $(1,1)$-LBMD. On the other hand, if $C$ is
$(1,1)$-LBMD then we have $n<d+d^{\perp}$. Therefore, we must have
$d+d^{\perp}=n+2$ and $C$ is MRD, for example, by [19, Corollary 18]. ∎

We conclude this work with an example that illustrates some notion of
optimality defined in this section.

######  Example 6.24.

Let $q=2$ and $n=m=4$. Consider
$C:=\left<(1,0,\alpha^{2},\alpha^{7}),(0,1,\alpha^{6},\alpha^{2})\right>$,
where the span is over $\mathbb{F}_{2^{4}}$ and $\alpha$ is a primitive
element of $\mathbb{F}_{2^{4}}$ with $\alpha^{4}+\alpha+1=0$. We have
$C^{\perp}=\left<(1,0,\alpha^{6},\alpha^{10}),(0,1,\alpha^{11},\alpha^{6})\right>$.
Recall that by Remark 6.4 we have that
$\\{X^{\perp}:X\in\mathscr{L}_{1}(n,m;q)\\}=\mathscr{L}_{1}(n,m;q)$. The
following hold.

  1. (1)

$\ell_{1}^{(1)}(C)=\ell_{1}^{(1)}(C^{\perp})=3$, which implies that
$d(C)=d(C^{\perp})=3$ by Theorem 6.6. This value is attained, for example, by
the element $\left<(1,0,1,0),(0,1,0,0),(0,0,0,1)\right>$ of
$\mathscr{L}_{1}(4,4;2)$, whose intersections with $C$ and $C^{\perp}$ are the
span of $(1,\alpha^{2},1,\alpha^{3})$ and $(1,\alpha^{2},1,\alpha)$,
respectively.

  2. (2)

$\ell_{2}^{(1)}(C)=4$. This means that the only element of
$\mathscr{L}_{1}(4,4;2)$ containing $C$ is the full space
$\smash{\mathbb{F}_{2^{4}}^{4}}$.

Therefore, we have

$\displaystyle 4-\ell_{1}^{(1)}(C)-\ell_{1}^{(1)}(C^{\perp})=-2<0,\qquad
4-\ell_{2}^{(1)}(C)-\ell_{1}^{(1)}(C^{\perp})=-3<0.$

It follows that $C$ is $(1,1)$-LBMD and $(1,2)$-LBMD. Moreover, one can easily
check that $C$ is also MRD. Note that this is in line with Proposition 6.23.

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