Puncturing maximum rank distance codes

Abstract

We investigate punctured maximum rank distance codes in cyclic models for bilinear forms of finite vector spaces. In each of these models, we consider an infinite family of linear maximum rank distance codes obtained by puncturing generalized twisted Gabidulin codes. We calculate the automorphism group of such codes, and we prove that this family contains many codes which are not equivalent to any generalized Gabidulin code. This solves a problem posed recently by Sheekey (Adv Math Commun 10:475–488, 2016).

1 Introduction

Let $M_{m,n}({\mathbb {F}}_{q})$, $m\le n$, be the rank metric space of all the $m\times n$ matrices with entries in the finite field ${\mathbb {F}}_{q}$ with q elements, $q=p^h$, p a prime. The distance between two matrices by definition is the rank of their difference. An (m, n, q; s)-rank distance code (also rank metric code) is any subset $\mathcal X$ of $M_{m,n}({\mathbb {F}}_{q})$ such that the distance between two of its distinct elements is at least s. An (m, n, q; s)-rank distance code is said to be linear if it is an ${\mathbb {F}}_{q}$-linear subspace of $M_{m,n}({\mathbb {F}}_{q})$.

It is known [9] that the size of an (m, n, q; s)-rank distance code $\mathcal X$ is bounded by the Singleton-like bound:

$$\begin{aligned} |\mathcal X| \le q^{n(m-s+1)}. \end{aligned}$$

When this bound is achieved, $\mathcal X$ is called an (m, n, q; s)-maximum rank distance code, or (m, n, q; s)-MRD code for short.

Although MRD codes are very interesting by their own and they caught the attention of many researchers in recent years [1, 8, 28, 29], such codes also have practical applications in error-correction for random network coding [15, 20, 31], space-time coding [32] and cryptography [14, 30].

Obviously, investigations of MRD codes can be carried out in any rank metric space isomorphic to $M_{m,n}({\mathbb {F}}_{q})$. In his pioneering paper [9], Ph. Delsarte constructed linear MRD codes for all the possible values of the parameters m, n, q and s by using the framework of bilinear forms on two finite-dimensional vector spaces over a finite field. Delsarte called such sets Singleton systems instead of maximum rank distance codes. Few years later, Gabidulin [13] independently constructed Delsarte’s linear MRD codes as evaluation codes of linearized polynomials over a finite field [21]. Although originally discovered by Delsarte, these codes are now called Gabidulin codes. In [19] Gabidulin’s construction was generalized to get different MRD codes. These codes are now known as Generalized Gabidulin codes. For $m=n$ a different construction of Delsarte’s MRD codes was given by Cooperstein [5] in the framework of the tensor product of a vector space over ${\mathbb {F}}_{q}$ by itself.

Recently, Sheekey [29] presented a new family of linear MRD codes by using linearized polynomials over ${\mathbb {F}}_{q^n}$. These codes are now known as generalized twisted Gabidulin codes. The equivalence classes of these codes were determined by Lunardon et al. [24]. In [27] a further generalization was considered giving new MRD codes when $m<n$; the authors call these codes generalized twisted Gabidulin codes as well. In this paper the term “generalized twisted Gabidulin code” will be used for codes defined in [29, Remark 8]. For different relations between linear MRD codes and linear sets see [7, 23, 29, Section 5], [6, Section 5]. To the extent of our knowledge, these are the only infinite families of linear MRD codes with $m<n$ appearing in the literature.

In [11] infinite families of nonlinear $(n,n,q;n-1)$-MRD codes, for $q\ge 3$ and $n\ge 3$ have been constructed. These families contain the nonlinear MRD codes provided by Cossidente et al. [6]. These codes have been afterward generalized in [10] by using a more geometric approach. A generalization of Sheekey’s example which yields additive but not ${\mathbb {F}}_{q}$-linear codes can be found in [26].

Let $\mathcal X$ be a rank distance code in $M_{n,n}({\mathbb {F}}_{q})$. For any given $m\times n$ matrix A over ${\mathbb {F}}_{q}$ of rank $m<n$, the set $A\mathcal X=\{AM: M\in \mathcal X\}$ is a rank distance code in $M_{m,n}({\mathbb {F}}_{q})$. The code $A\mathcal X$ is said to be obtained by puncturing$\mathcal X$withA and $A\mathcal X$ is called a punctured code. The reason of this definition is that if $A=(\mathrm{I}_m|\mathbf 0_{n-m})$, where $I_m$ and $\mathbf 0_{n-m}$ is the $m\times m$ identity and $m\times (n-m)$ null matrix, respectively, then the matrices of $A\mathcal X$ are obtained by deleting the last $n-m$ rows from the matrices in $\mathcal X$. Punctured rank metric codes have been studied before in [3, 25], but the equivalence problem among these codes have not been dealt with in these papers.

In [29, Remark 9] Sheekey posed the following problem:

Are the MRD codes obtained by puncturing generalized twisted Gabidulin codes equivalent to the codes obtained by puncturing generalized Gabidulin codes?

Here we investigate punctured codes and study the above problem in the framework of bilinear forms. We point out that the very recent preprint [34] deals with the same problem by using q-linearized polynomials. In [34] the authors investigate the middle nucleus and the right nucleus of punctured generalized twisted Gabidulin codes, for $m<n$. By exploiting these nuclei, they derive necessary conditions on the automorphisms of these codes which depend on certain restrictions for the parameters.

Let V and $V'$ be two vector spaces over ${\mathbb {F}}_{q}$ of dimensions m and n, respectively. Since the rank is invariant under matrix transposition, we may assume $m\le n$.

A bilinear form on V and $V'$ is a function $f:V\times V'\rightarrow {\mathbb {F}}_{q}$ that satisfies the identity

$$\begin{aligned} f\left( \sum _{i}{x_i v_i},\sum _{j}{x'_j v'_j}\right) =\sum _{i,j}{x_if\left( v_i,v'_j\right) x_j'}, \end{aligned}$$

for all scalars $x_i,x'_j\in {\mathbb {F}}_{q}$ and all vectors $v_i\in V$, $v'_j\in V'$. The set $\Omega _{m,n}=\Omega (V,V')$ of all bilinear forms on V and $V'$ is an mn-dimensional vector space over ${\mathbb {F}}_{q}$.

The left radical $\mathrm{Rad}(f)$ of any $f\in \Omega _{m,n}$ is by definition the subspace of V consisting of all vectors v satisfying $f(v,v')=0$ for every $v'\in V'$. The rank of f is the codimension of $\mathrm{Rad}(f)$, i.e.,

$$\begin{aligned} \mathrm{rank}(f)=m-\mathrm{dim}_{{\mathbb {F}}_{q}}(\mathrm{Rad}(f)). \end{aligned}$$

(1)

Then the ${\mathbb {F}}_{q}$-vector space $\Omega _{m,n}$ equipped with the above rank function is a rank metric space over ${\mathbb {F}}_{q}$.

Let $\{u_0,\ldots , u_{m-1}\}$ and $\{u'_0,\ldots , u'_{n-1}\}$ be a basis for V and $V'$, respectively. For any $f\in \Omega _{m,n}$, the $m\times n$${\mathbb {F}}_{q}$-matrix $M_f=(f(u_i,u'_j))$, is called the matrix offin the bases$\{u_0,\ldots , u_{m-1}\}$and$\{u'_0,\ldots , u'_{n-1}\}$. It turns out that the map

$$\begin{aligned} \begin{array}{rccc} \nu _{\{u_0,\ldots , u_{m-1};u'_0,\ldots , u'_{n-1}\}}:&{} \Omega _{m,n} &{} \rightarrow &{} M_{m,n}({\mathbb {F}}_{q})\\ &{} f &{} \mapsto &{} M_f \end{array} \end{aligned}$$

(2)

is an isomorphism of rank metric spaces with $\mathrm{rank}(f)=\mathrm{rank}(M_f)$.

Let $\Gamma \mathrm{L}(\Omega _{m,n})$ denote the general semilinear group of the mn-dimensional ${\mathbb {F}}_{q}$-vector space $\Omega _{m,n}$, that is, the group of all invertible semilinear transformations of $\Omega _{m,n}$. Let $\{w_1,\ldots ,w_{mn}\}$ be a basis for $\Omega _{m,n}$, and recall that $\mathrm{Aut}({\mathbb {F}}_{q})=\langle \phi _p\rangle$, where $\phi _p:{\mathbb {F}}_{q}\rightarrow {\mathbb {F}}_{q}$ is the Frobenius map $\lambda \mapsto \lambda ^p$. Using $\phi _p$, we define the map $\phi :\Omega _{m,n}\rightarrow \Omega _{m,n}$ by

$$\begin{aligned} \phi :\sum _{i}{\lambda _i w_i}\mapsto \sum _{i}{\lambda _i^p w_i}. \end{aligned}$$

Then $\phi$ is an invertible semilinear transformation of $\Omega _{m,n}$, and for $(a_{ij})\in \mathrm{GL}(mn,q)$ we have $(a_{ij})^\phi =(a_{ij}^p)$. Therefore $\phi$ normalizes the general linear group $\mathrm{GL}(mn,q)$ and we have $\Gamma \mathrm{L}(\Omega _{m,n})=\mathrm{GL}(\Omega _{m,n}) \rtimes \mathrm{Aut}({\mathbb {F}}_{q})$.

An automorphism of the rank metric space $\Omega _{m,n}$ is any transformation $\tau \in \Gamma \mathrm{L}(\Omega _{m,n})$ such that $\mathrm{rank}(f^\tau )=\mathrm{rank}(f)$, for all $f\in \Omega _{m,n}$. The automorphism group$\mathrm{Aut}(\Omega _{m,n})$ of $\Omega _{m,n}$ is the group of all automorphisms of $\Omega _{m,n}$, i.e.,

$$\begin{aligned} \mathrm{Aut}(\Omega _{m,n})=\{\tau \in \Gamma \mathrm{L}(\Omega _{m,n}):\mathrm{rank}(f^\tau )=\mathrm{rank}(f), \quad {\hbox {for all}}\ f\in \Omega _{m,n}\}. \end{aligned}$$

By [35, Theorem 3.4],

$$\begin{aligned} \mathrm{Aut}(\Omega _{m,n})=(\mathrm{GL}(V)\times \mathrm{GL}(V')) \rtimes \mathrm{Aut}({\mathbb {F}}_{q}) \quad \text{ for }\ m<n, \end{aligned}$$

and

$$\begin{aligned} \mathrm{Aut}(\Omega _{n,n})=(\mathrm{GL}(V')\times \mathrm{GL}(V'))\rtimes \langle \top \rangle \rtimes \mathrm{Aut}({\mathbb {F}}_{q})\quad \text{ for } m=n, \end{aligned}$$

where $\top$ is an involutorial operator. In details, any given $(g,g')\in \mathrm{GL}(V)\times \mathrm{GL}(V')$ defines the linear automorphism of $\Omega _{m,n}$ given by

$$\begin{aligned} f^{(g,g')}(v,v')=f(gv,g'v'), \end{aligned}$$

for any $f\in \Omega _{m,n}$. If A and B are the matrices of $g\in \mathrm{GL}(V)$ and $g'\in \mathrm{GL}(V')$ in the given bases for V and $V'$, then the matrix of $f^{(g,g')}$ is $A^tM_fB$, where t denotes transposition. Additionally, the semilinear transformation $\phi$ of $\Omega _{m,n}$ is the automorphism given by

$$\begin{aligned} f^{\phi }(v,v')=\left[ f\left( v^{\phi ^{-1}},{v'}^{\phi ^{-1}}\right) \right] ^p. \end{aligned}$$

If $M_f=(a_{ij})$ is the matrix of f in the given bases for V and $V'$, then the matrix of $f^{\phi }$ is $M_f^{\phi }=(a_{ij}^{p})$. Therefore $\phi$ normalizes the group $\mathrm{GL}(V)\times \mathrm{GL}(V')$. If $m<n$, the above automorphisms are all the elements in $\mathrm{Aut}(\Omega _{m,n})$.

If $m=n$, one may assume, and we do, $V'=V=\langle u_0,\ldots , u_{m-1}\rangle$. The involutorial operator $\top :\Omega _{n,n}\rightarrow \Omega _{n,n}$ is defined by setting

$$\begin{aligned} f^\top (v,v')=f(v',v). \end{aligned}$$

If $M_f=(a_{ij})$ is the matrix of f in the given bases for V and $V'$, then the matrix of $f^{\top }$ is the transpose matrix $M_f^t$ of $M_f$. The operator $\top$ acts on $\mathrm{GL}(V)\times \mathrm{GL}(V)$ by mapping $(g,g')$ to $(g',g)$.

For a given subset $\mathcal X$ of $\Omega _{m,n}$, the automorphism group of $\mathcal X$ is the subgroup of $\mathrm{Aut}(\Omega _{m,n})$ fixing $\mathcal X$. Two subsets $\mathcal X_1,\mathcal X_2$ of $\Omega _{m,n}$ are said to be equivalent if there exists $\varphi \in \mathrm{Aut}(\Omega _{m,n})$ such that $\mathcal X_2=\mathcal X_1^\varphi$.

The main tool we use in this paper is the kth cyclic model in $V(r,q^r)$ for an r-dimensional vector space V(r, q) over ${\mathbb {F}}_{q}$, where k is any positive integer such that $\mathrm{gcd}(r,k)=1$. This model generalizes the cyclic model introduced in [5, 12, 17], and it is studied in Sect. 2. In particular, the endomorphisms of the kth cyclic model are represented by $r\times r$$q^k$-circulant matrices over ${\mathbb {F}}_{q^r}$.

For any k such that $\mathrm{gcd}(m,k)=1=\mathrm{gcd}(n,k)$, the elements of $\Omega _{m,n}$ acting on the kth cyclic model of V and $V'$ are represented by $q^k$-circulant $m\times n$ matrices over ${\mathbb {F}}_{q^d}$, where $d=\mathrm{lcm}(m,n)$. We then have a description of the elements in $\mathrm{Aut}(\Omega _{m,n})$ in terms of $q^k$-circulant matrices.

In Sect. 3 we prove that the code obtained by puncturing an (n, n, q; s)-MRD code is an $(m,n,q;s+m-n)$-MRD code, where $n-s<m\le n$. In particular, the code in $\Omega _{m,n}$ obtained by puncturing a generalized Gabidulin code in $\Omega _{n,n}$ is a generalized Gabidulin code. Conversely, every generalized Gabidulin code in $\Omega _{m,n}$ can be obtained by puncturing a generalized Gabidulin code in $\Omega _{n,n}$.

By using the representation by $q^k$-circulant matrices of the elements of $\Omega _{m,n}$ acting on the kth cyclic model for V and $V'$, we calculate the automorphism group of some generalized Gabidulin code. In Sect. 3 we also construct an infinite family of MRD codes by puncturing generalized twisted Gabidulin codes [24, 29]. We calculate the automorphism group of these codes in Sect. 4. By using a recent result by Liebhold and Nebe [22], we prove in Sect. 5 that the above family contains many MRD codes which are inequivalent to the MRD codes obtained by puncturing generalized Gabidulin codes. This solves the problem posed by Sheekey in [29, Remark 9].

2 Cyclic models for bilinear forms on finite vector spaces

Let $V(r,q)=\langle u_0,\ldots , u_{r-1}\rangle _{{\mathbb {F}}_{q}}$, $r\ge 2$, be an r-dimensional vector space over the finite field ${\mathbb {F}}_{q}$. We denote the set of all linear transformations of V(r, q) by $\mathrm{End}(V(r,q))$.

Embed V(r, q) in $V(r,q^r)$ by extending the scalars. Concretely this can be done by defining $V(r,q^r)=\{\sum _{i=0}^{r-1}{\lambda _i u_i:\lambda _i \in {\mathbb {F}}_{q^r}}\}$.

Let $\xi :V(r,q^r)\rightarrow V(r,q^r)$ be the ${\mathbb {F}}_{q^r}$-semilinear transformation with associated automorphism $\delta :x\in {\mathbb {F}}_{q^r}\rightarrow x^q\in {\mathbb {F}}_{q^r}$ such that $\xi (u_i)=u_i$. Clearly, V(r, q) consists of all the vectors in $V(r,q^r)$ which are fixed by $\xi$.

In the paper [5], the cyclic model of V(r, q) was introduced by taking the eigenvectors $s_0,\ldots , s_{r-1}$ in $V(r,q^r)$ of a Singer cycle $\sigma$ of V(r, q); here a Singer cycle of V(r, q) is an element $\sigma$ of $\mathrm{GL}(V(r,q))$ of order $q^r-1$. The cyclic group $S=\langle \sigma \rangle$ is called a Singer cyclic group of$\mathrm{GL}(V(r,q))$ [18].

Since $s_0,\ldots , s_{r-1}$ have distinct eigenvalues in ${\mathbb {F}}_{q^r}$, they form a basis of the extension $V(r,q^r)$ of V(r, q).

In this basis the matrix of $\sigma$ is the diagonal matrix $\mathrm{diag}(w,w^q,\ldots ,w^{q^{r-1}})$, where w is a primitive element of ${\mathbb {F}}_{q^r}$ over ${\mathbb {F}}_{q}$ and $w^{q^i}$ is the eigenvalue of $s_i$. The action of the linear part $\ell _{\xi }$ of the ${\mathbb {F}}_{q^r}$-semilinear transformation $\xi$ is given by $\ell _{\xi }(s_i)=s_{i+1}$, where the indices are considered modulo r [5]. It follows that

$$\begin{aligned} V(r,q)=\left\{ \sum _{i=0}^{r-1}{x^{q^{i}}s_i}:x \in {\mathbb {F}}_{q^r}\right\} . \end{aligned}$$

(3)

We call $\{s_0,\ldots , s_{r-1}\}$ a Singer basis for V(r, q) and the representation (3) for V(r, q), or equivalently the set $\{(x,x^q,\ldots , x^{q^{r-1}}):x \in {\mathbb {F}}_{q^r}\}\subset {\mathbb {F}}_{q^r}^r$, is the cyclic model forV(r, q) [12, 17].

We point out that the ${\mathbb {F}}_{q^r}$-semilinear transformation $\phi :V(r,q^r)\rightarrow V(r,q^r)$ with associated automorphism the Frobenius map $\phi _p:x\in {\mathbb {F}}_{q^r}\rightarrow x^p\in {\mathbb {F}}_{q^r}$ such that $\phi (u_i)=u_i$ acts on the cyclic model (3) by mapping $xs_0+x^qs_1+\cdots + x^{q^{r-1}}s_{r-1}$ to $x^{pq^{r-1}}s_0+x^ps_1+\cdots + x^{pq^{r-2}}s_{r-1}$.

Let k be a positive integer such that $\mathrm{gcd}(k,r)=1$. Set $s^{(k)}_i=s_{ki \bmod r}$, for $i=0,\ldots , r-1$. For brevity, we use $[j]=q^{j}$ and $a^{[j]}=a^{q^{j}}$, for any $a\in {\mathbb {F}}_{q^r}$. It is clear that the exponent j is taken mod r because of the field size. Then we may write

$$\begin{aligned} V(r,q)=\left\{ \sum _{i=0}^{r-1}{x^{[ki]}s^{(k)}_i}:x \in {\mathbb {F}}_{q^r}\right\} . \end{aligned}$$

(4)

We call the representation (4) for V(r, q), or equivalently the set $\{(x,x^{[k]},\ldots , x^{[k(r-1)]}):x \in {\mathbb {F}}_{q^r}\}\subset {\mathbb {F}}_{q^r}^r$, the kth cyclic model forV(r, q).

It is easily seen that the linear part of the semilinear transformation $\xi ^k$ acts on the kth cyclic model for V(r, q) by mapping $s_i^{(k)}$ to $s_{i+1}^{(k)}$, with indices considered modulo r.

An $r\times r$$q^k$-circulantmatrix over ${\mathbb {F}}_{q^r}$ is a matrix of the form

$$\begin{aligned} D_{(a_0,a_1,\ldots ,a_{r-1})}^{(k)}= \left( \begin{array}{cccc} a_0 &{}\quad a_1 &{}\quad \cdots &{}\quad a_{r-1}\\ a_{r-1}^{[k]} &{}\quad a_{0}^{[k]} &{}\quad \cdots &{}\quad a_{r-2}^{[k]}\\ \vdots &{}\quad \vdots &{}\quad \ddots &{}\quad \vdots \\ a_{1}^{[k(r-1)]} &{}\quad a_{2}^{[k(r-1)]} &{}\quad \cdots &{}\quad a_0^{[k(r-1)]} \end{array}\right) \end{aligned}$$

with $a_i\in {\mathbb {F}}_{q^r}$. We say that the above matrix is generated by the array$(a_0,\ldots ,a_{r-1})$.

Let $\mathcal D_r^{(k)}({\mathbb {F}}_{q^r})$ denote the matrix algebra formed by all $r\times r$$q^k$-circulant matrices over ${\mathbb {F}}_{q^r}$ and $\mathcal B_r^{(k)}({\mathbb {F}}_{q^r})$ the set of all invertible $q^k$-circulant $r\times r$ matrices. When $k=1$, an $r\times r$q-circulant matrix over ${\mathbb {F}}_{q^r}$ is also known as a Dickson matrix, $\mathcal D_r({\mathbb {F}}_{q^r})=\mathcal D_r^{(1)}({\mathbb {F}}_{q^r})$ is the Dickson matrix algebra and $\mathcal B_r({\mathbb {F}}_{q^r})=\mathcal B_r^{(1)}({\mathbb {F}}_{q^r})$ is the Betti–Mathieu group [2, 4]. It is known that $\mathrm{End}(V(r,q))\simeq \mathcal D_r({\mathbb {F}}_{q^r})$ and $\mathcal B_r({\mathbb {F}}_{q^r})\simeq \mathrm{GL}(V(r,q))$ [21, 36].

Remark 2.1

In terms of matrix representation, the above isomorphisms are described as follows. Let $V(r,q)=\langle u_0,\ldots , u_{r-1}\rangle _{{\mathbb {F}}_{q}}$ and $\{s_0,\ldots , s_{r-1}\}$ a Singer basis for V(r, q) defined by the primitive element w of ${\mathbb {F}}_{q^r}$ over ${\mathbb {F}}_{q}$. Up to a change of the basis $\{u_0,\ldots , u_{r-1}\}$ in V(r, q), we may assume

$$\begin{aligned} u_i=w^{i}s_0+\cdots +w^{iq^{r-1}}s_{r-1}, \quad \mathrm {for}\ i=0,\ldots , r-1. \end{aligned}$$

Notice that $u_i\in \ V(r,q)$, for $i=0,\ldots , r-1$. The non-singular Moore matrix

$$\begin{aligned} E_{r}=\left( \begin{array}{cccc} 1 &{}\quad w &{}\quad \cdots &{}\quad w^{r-1}\\ 1 &{}\quad w^q &{}\quad \cdots &{}\quad w^{(r-1)q}\\ \vdots &{}\quad \vdots &{}\quad &{}\quad \vdots \\ 1 &{}\quad w^{q^{r-1}} &{}\quad \cdots &{}\quad w^{(r-1)q^{r-1}} \end{array}\right) \end{aligned}$$

(5)

is the matrix of the change of basis from $\{u_0,\ldots , u_{r-1}\}$ to $\{s_0,\ldots , s_{r-1}\}$. Therefore, the matrix map $D\in \mathcal D_{r}({\mathbb {F}}_{q^r})\rightarrow E_r^{-1}DE_r\in M_{r,r}({\mathbb {F}}_{q})$ realizes the above isomorphism.

Proposition 2.2

$\mathrm{End}(V(r,q))\simeq \mathcal D_r^{(k)}({\mathbb {F}}_{q^r})$ and $\mathrm{GL}(V(r,q))\simeq \mathcal B_r^{(k)}({\mathbb {F}}_{q^r})$.

Proof

For any $\mathbf a=(a_0,\ldots , a_{r-1})$ over ${\mathbb {F}}_{q^r}$, the $q^k$-circulant matrix $D_{\mathbf a}^{(k)}$ acts on the kth cyclic model (4) for V(r, q) by mapping $(x,x^{[k]},\ldots , x^{[k(r-1)]})$ to $(a_0x+a_1x^{[k]}+\cdots + a_{r-1}x^{[k(r-1)]},a_{r-1}^{[k]}x+a_0^{[k]}x^{[k]}+\cdots + a_{r-2}^{[k]}x^{[k(r-1)]},\ldots , a_{1}^{[k(r-1)]}x+a_2^{[k(r-1)]}x^{[k]}+\cdots + a_{0}^{[k(r-1)]}x^{[k(r-1)]})$, giving $D_{\mathbf a}^{(k)}$ is an endomorphism of (4). Let $D_\mathbf a, D_{\mathbf a'}\in \mathcal D_r^{(k)}({\mathbb {F}}_{q^r})$ such that $D_\mathbf a\mathbf x^t= D_{\mathbf a'}\mathbf x^t$, for every $\mathbf x=(x,x^{[k]},\ldots , x^{[k(r-1)]})$, $x \in {\mathbb {F}}_{q^r}$. Hence, $(a_0-a'_0)x+(a_1-a'_1)x^{[k]}+\cdots + (a_{r-1}-a'_{r-1})x^{[k(r-1)]}=0$, for all $x\in {\mathbb {F}}_{q^r}$. As the left hand side is a polynomial of degree at most $q^{r-1}$ with $q^r$ roots, we get $\mathbf a=\mathbf a'$. Therefore, matrices in $\mathcal D_r^{(k)}({\mathbb {F}}_{q^r})$ represent $q^{r^2}$ distinct endomorphisms of the kth cyclic model for V(r, q). As $q^{r^2}=|\mathrm{End}(V(r,q))|$, we get the result. $\square$

Remark 2.3

Let $K_r$ be the (permutation) matrix of the change of basis from $\{s_0^{(k)},\ldots ,{s_{r-1}}^{(k)}\}$ to $\{s_0,\ldots , s_{r-1}\}$. As $s^{(k)}_i=s_{ik \bmod r}$, for $i=0,\ldots , r-1$, then the ith column of $K_r$ is the array $(0,\ldots ,0,1,0,\ldots ,0)^t$ where 1 is in position $ik \bmod r$, for $i=0,\ldots ,r-1$. If $\tau \in \mathrm{End}(V(r,q))$ has $q^k$-circulant matrix $D_{(a_0,a_1,\ldots ,a_{r-1})}^{(k)}$ in the basis $\{s_0^{(k)},\ldots ,{s_{r-1}}^{(k)}\}$, then the matrix of $\tau$ in the Singer basis $\{s_0,\ldots ,s_{r-1}\}$ is the q-circulant matrix $D_{(b_0,\ldots ,b_{r-1})}=K_rD_{(a_0,a_1,\ldots ,a_{r-1})}^{(k)}K_r^{-1}$, for some array $(b_0,\ldots ,b_{r-1})$ over ${\mathbb {F}}_{q^r}$. Since $\mathrm{gcd}(k,r)=1$, we can write $1=lr+hk$, for some integers l, h, giving

$$\begin{aligned} b_i=a_{ih \bmod r}, \quad \mathrm {for}\ i=0,\ldots , r-1. \end{aligned}$$

Therefore, $\mathcal D_r^{(k)}({\mathbb {F}}_{q^r})=K_r^{-1}\mathcal D_r({\mathbb {F}}_{q^r})K_r$ and $\mathcal B_r^{(k)}({\mathbb {F}}_{q^r})=K_r^{-1}\mathcal B_r({\mathbb {F}}_{q^r})K_r$.

Remark 2.4

We explicitly describe the action of $\mathrm{Aut}({\mathbb {F}}_{q})$ on $V(r,q^r)$ in the Singer basis $\{s_0^{(k)},\ldots ,s_{r-1}^{(k)}\}$. By Remark 2.1 the invertible semilinear transformation $\phi$ of $V(r,q^r)$ defined by the Frobenius map $\phi _p:x\in {\mathbb {F}}_{q^r}\rightarrow x^p\in {\mathbb {F}}_{q^r}$ acts in the basis $\{s_0,\ldots ,s_{r-1}\}$ via the pair $(E_r(E_r^{-1})^p;\phi _p)$, where $E_r$ is the non-singular Moore matrix (5) and $(E_r^{-1})^p$ is the matrix obtained by $E_r^{-1}$ by applying $\phi _p$ to every entry. By Remark 2.3$\phi$ acts in the basis $\{s_0^{(k)},\ldots ,s_{r-1}^{(k)}\}$ via the pair $(K_r^{-1}E_r(E_r^{-1})^pK_r;\phi _p)$, since $K_r^p=K_r$.

Let $V=\langle u_0,\ldots , u_{m-1}\rangle _{{\mathbb {F}}_{q}}$ and $V'=\langle u'_0,\ldots , u'_{n-1}\rangle _{{\mathbb {F}}_{q}}$, with $m\le n$. If $m=n$ we take $V'=V=\langle u_0,\ldots , u_{m-1}\rangle _{{\mathbb {F}}_{q}}$. Let $\sigma$ and $\sigma '$ be Singer cycles of $\mathrm{GL}(V)$ and $\mathrm{GL}(V')$, respectively, with associated semilinear transformations $\xi$ and $\xi '$. Let $\{s_0,\ldots ,s_{m-1}\}$ and $\{s'_0,\ldots ,s'_{n-1}\}$ be a Singer basis for V and $V'$, defined by $\sigma$ and $\sigma '$, respectively. For any given positive integer k such that $\mathrm{gcd}(k,n)=\mathrm{gcd}(k,m)=1$, let $\{s^{(k)}_0,\ldots ,s^{(k)}_{m-1}\}$ and $\{s_0'^{(k)},\ldots ,s_{n-1}'^{(k)}\}$ be the bases of $V(m,q^m)$ and $V(n,q^n)$ defined as above. Therefore, we may consider $\Omega _{m,n}$ as the set of all bilinear forms acting on the kth cyclic model for V and $V'$. In addition, any element in $\mathrm{GL}(V)\times \mathrm{GL}(V')$ is represented by a pair $(A,B)\in \mathcal B_m^{(k)}({\mathbb {F}}_{q^m})\times \mathcal B_{n}^{(k)}({\mathbb {F}}_{q^n})$.

Set $e=\mathrm{gcd}(m,n)$ and $d=\mathrm{lcm}(m,n)$, the greatest common divisor and the least common multiple of m and n, respectively.

Let $\mathrm{Tr}_{q^d/q}$ denote the trace function from ${\mathbb {F}}_{q^d}$ onto ${\mathbb {F}}_{q}$:

$$\begin{aligned} \mathrm{Tr}_{q^d/q}:y \in {\mathbb {F}}_{q^d}\rightarrow \mathrm{Tr}_{q^d/q}(y)=\sum _{i=0}^{d-1}{y^{q^i}}\in {\mathbb {F}}_{q}. \end{aligned}$$

Since $\mathrm{gcd}(k,d)=1$, we may write $\mathrm{Tr}_{q^d/q}$ as

$$\begin{aligned} T^{(k)}:y \in {\mathbb {F}}_{q^d}\rightarrow T^{(k)}(y)=\sum _{i=0}^{d-1}{y^{[k]}}\in {\mathbb {F}}_{q}. \end{aligned}$$

For $0\le j\le e-1$ and a given $a\in {\mathbb {F}}_{q^d}$ and $v=xs^{(k)}_0+\cdots +x^{[k(m-1)]}s^{(k)}_{m-1}\in V$ and $v'=x's_0'^{(k)}+\cdots +{x'}^{[k(n-1)]}s_{n-1}'^{(k)}$, the map

$$\begin{aligned} f_{a,j}^{(k)}(v,v')=T^{(k)}(a x x'^{[kj]}) \end{aligned}$$

(6)

is a bilinear form on the kth cyclic model for V and $V'$. We set

$$\begin{aligned} \Omega _{j}^{(k)}=\{f_{a,j}^{(k)}:a\in {\mathbb {F}}_{q^d}\}, \quad \mathrm {for}\ 0\le j \le e-1. \end{aligned}$$

(7)

The following result gives the decomposition of $\Omega _{m,n}$ as sum of the subspaces $\Omega _{j}^{(k)}$.

Theorem 2.5

$$\begin{aligned} \Omega _{m,n}=\bigoplus _{j=0}^{e-1}{\Omega _{j}^{(k)}}. \end{aligned}$$

(8)

Proof

Let first assume $k=1$. For any e-tuple $\mathbf a=(a_0,\ldots ,a_{e-1})$ over ${\mathbb {F}}_{q^d}$ we define an $m\times n$ matrix $D_{\mathbf a}=D_{\mathbf a}^{(1)}=(d_{i,j})$ over ${\mathbb {F}}_{q^d}$ as follows. We will use indices from 0 for both rows and columns of $D_{{\mathbf {a}}}$. Let $d_{0,j}=a_j$, for $0\le j\le e-1$, and let $d_{i,j}=d_{i-1,j-1}$, where the row index is taken modulo m and the column index is taken modulo n. Notice that the above rule determines every entry of $D_{\mathbf a}$. In fact, $d_{i,j}=a_l^{q^s}$, where $l\equiv j-iv\pmod e$, $0\le l\le e-1$ and $s=\beta m+i$, where $\beta$ is the unique integer in $\{0,1,\ldots ,n/e-1\}$ such that $j-i \equiv l+\beta m \pmod n$.

Now let $f_{a,j}\in \Omega _{j}$. Then the matrix of $f_{a,j}$ in the Singer bases $\{s_0,\ldots , s_{m-1}\}$ and $\{s_0',\ldots , s_{n-1}'\}$ is the matrix obtained by applying the above construction to the array $\mathbf a=(0,\ldots ,0,a,0,\ldots ,0)$, with a in the jth position. It is now easy to see that the ${\mathbb {F}}_{q}$-spaces $\Omega _{j}$, for $j=0,\ldots ,e-1$ intersect trivially. By consideration on dimensions we may write $\Omega _{m,n}=\bigoplus _{j=0}^{e-1}{\Omega _{j}}$.

The kth cyclic model for $V'$ and V is obtained from the cyclic model by applying the changing of basis described in Remark 2.3. Therefore the ${\mathbb {F}}_{q}$-spaces $\Omega _{j}^{(k)}$, $k>1$, are pairwise skew and $\Omega _{m,n}=\bigoplus _{j=0}^{e-1}{\Omega _{j}^{(k)}}$. $\square$

Example 1

Let $m=2$, $n=6$ and $k=1$, so that $d=6$ and $e=2$. For any array $\mathbf a=(a_0,a_1)$ over $\mathbb F_{q^{6}}$, we have

$$\begin{aligned} D_\mathbf a=\left( \begin{array}{cccccc} a_0&{}\quad a_1&{}\quad a_0^{q^2}&{}\quad a_1^{q^2}&{}\quad a_0^{q^4}&{}\quad a_1^{q^4}\\ a_1^{q^5}&{}\quad a_0^q&{}\quad a_1^q&{}\quad a_0^{q^3}&{}\quad a_1^{q^3}&{}\quad a_0^{q^5} \end{array}\right) .\end{aligned}$$

Example 2

Let $m=4$, $n=6$ and $k=5$, so that $d=12$ and $e=2$. For any array $\mathbf a=(a_0,a_1)$ over $\mathbb F_{q^{12}}$, we have

$$\begin{aligned} D_{\mathbf a}^{(k)}=\left( \begin{array}{cccccc} a_0&{}\quad a_1&{}\quad a_0^{[8k]}&{}\quad a_1^{[8k]}&{}\quad a_0^{[4k]}&{}\quad a_1^{[4k]}\\ a_1^{[5k]}&{}\quad a_0^{[k]}&{}\quad a_1^{[k]}&{}\quad a_0^{[9k]}&{}\quad a_1^{[9k]}&{}\quad a_0^{[5k]}\\ a_0^{[6k]}&{}\quad a_1^{[6k]}&{}\quad a_0^{[2k]}&{}\quad a_1^{[2k]}&{}\quad a_0^{[10k]}&{}\quad a_1^{[10k]}\\ a_1^{[11k]}&{}\quad a_0^{[7k]}&{}\quad a_1^{[7k]}&{}\quad a_0^{[3k]}&{}\quad a_1^{[3k]}&{}\quad a_0^{[11k]} \end{array}\right) \\\nonumber \qquad = \left( \begin{array}{cccccc} a_0&{}\quad a_1&{}\quad a_0^{q^{4}}&{}\quad a_1^{q^4}&{}\quad a_0^{q^8}&{}\quad a_1^{q^8}\\ a_1^{q}&{}\quad a_0^{q^5}&{}\quad a_1^{q^5}&{}\quad a_0^{q^9}&{}\quad a_1^{q^9}&{}\quad a_0^q\\ a_0^{q^6}&{}\quad a_1^{q^6}&{}\quad a_0^{q^{10}}&{}\quad a_1^{q^{10}}&{}\quad a_0^{q^2}&{}\quad a_1^{q^2} \\ a_1^{q^7}&{}\quad a_0^{q^{11}}&{}\quad a_1^{q^{11}}&{}\quad a_0^{q^3}&{}\quad a_1^{q^3}&{}\quad a_0^{q^{7}} \end{array}\right) .\end{aligned}$$

We call a matrix of type $D_{\mathbf a}^{(k)}$ an $m\times n$$q^k$-circulantmatrix over ${\mathbb {F}}_{q^d}$, where $d=\mathrm{lcm}(m,n)$. We say that $D_\mathbf a^{(k)}$ is generated by the array$\mathbf a=(a_0,a_1,\ldots ,a_{e-1})$, where $e=\mathrm{gcd}(m,n)$. We will denote the set of all $m\times n$$q^k$-circulant matrices over ${\mathbb {F}}_{q^d}$ by $\mathcal D_{m,n}^{(k)}({\mathbb {F}}_{q^d})$.

The next result gives a description of $\Omega _{m,n}$ and $\mathrm{Aut}(\Omega _{m,n})$ in terms of $q^k$-circulant matrices.

Proposition 2.6

Let $m\le n$. Then $\Omega _{m,n}\simeq \mathcal D_{m,n}^{(k)}({\mathbb {F}}_{q^d})$.

If $m<n$, then

$$\begin{aligned} \mathrm{Aut}(\Omega _{m,n})\simeq \left( \mathcal B_m^{(k)}({\mathbb {F}}_{q^m})\times \mathcal B_n^{(k)}({\mathbb {F}}_{q^n})\right) \rtimes \mathrm{Aut}({\mathbb {F}}_{q}); \end{aligned}$$

if $m=n$, then

$$\begin{aligned} \mathrm{Aut}(\Omega _{n,n})\simeq \left( \mathcal B_n^{(k)}({\mathbb {F}}_{q^m})\times \mathcal B_n^{(k)}({\mathbb {F}}_{q^n})\right) \rtimes \langle \top \rangle \rtimes \mathrm{Aut}({\mathbb {F}}_{q}). \end{aligned}$$

Proof

For any $\mathbf a=(a_0,\ldots , a_{e-1})$ over ${\mathbb {F}}_{q^d}$ we consider the bilinear form $f_\mathbf a^{(k)}=f_{a_0,0}^{(k)}+\cdots +f_{a_{e-1},e-1}^{(k)}$. Straightforward calculation shows that the matrix of $f_\mathbf a^{(k)}$ in the bases $\{s^{(k)}_0,\ldots ,s^{(k)}_{m-1}\}$ and $\{s_0'^{(k)},\ldots ,s_{n-1}'^{(k)}\}$ is the $m\times n$$q^k$-circulant matrix $D_\mathbf a^{(k)}$ generated by $\mathbf a$. Now assume that $f_\mathbf a^{(k)}$ is the null bilinear form. Let $V=\langle u_0,\ldots , u_{m-1}\rangle _{{\mathbb {F}}_{q}}$ and $V'=\langle u_0',\ldots , u_{n-1}'\rangle _{{\mathbb {F}}_{q}}$. By Remarks 2.1 and 2.3 the matrix of $f_\mathbf a^{(k)}$ in the bases $\{u_0,\ldots , u_{m-1}\}$ and $\{u_0',\ldots , u_{n-1}'\}$ is $(K_m^{-1}E_m)^{t} D^{(k)}_\mathbf a(K_n^{-1}E_n)$, which is clearly the zero matrix. As $K_m^{-1}E_m$ and $K_n^{-1}E_n$ are both non-singular we get $D_\mathbf a^{(k)}$ is the zero matrix giving $\mathbf a$ is the zero array. Therefore, matrices in $\mathcal D_{m,n}^{(k)}({\mathbb {F}}_{q^r})$ represent $q^{de}=q^{mn}$ distinct bilinear forms acting on the kth cyclic models for V and $V'$. As $q^{mn}=|\Omega _{m,n}|$, we get $\Omega _{m,n}\simeq \mathcal D_{m,n}^{(k)}({\mathbb {F}}_{q^d})$.

To prove the second part of the Proposition we first note that Proposition 2.2 implies that the group of all ${\mathbb {F}}_{q}$-linear automorphisms of $\Omega _{m,n}$ is isomorphic to $(\mathcal B_m^{(k)}({\mathbb {F}}_{q^m})\times \mathcal B_n^{(k)}({\mathbb {F}}_{q^n}))$, if $m<n$, and to $(\mathcal B_m^{(k)}({\mathbb {F}}_{q^m})\times \mathcal B_n^{(k)}({\mathbb {F}}_{q^n}))\rtimes \langle \top \rangle$, if $m=n$.

If $D_{(a_0,\ldots , a_{e-1})}^{(k)}$ is the matrix of f in the bases $\{s_0,\ldots ,s_{m-1}\}$ and $\{s_0,\ldots ,s_{n-1}\}$ for V and $V'$ respectively, then $f^\phi$ is $D_{(a_0^p,\ldots , a_{e-1}^p)}^{(k)}$ by Remark 2.4. This concludes the proof. $\square$

Remark 2.7

The isomorphism $\nu =\nu _{\{s^{(k)}_0,\ldots ,s^{(k)}_{m-1};s_0'^{(k)},\ldots ,s_{n-1}'^{(k)}\}}:\Omega _{m,n}\rightarrow \mathcal D_{m,n}^{(k)}({\mathbb {F}}_{q^d})$ is described as follows. Let $V=\langle u_0,\ldots , u_{m-1}\rangle _{{\mathbb {F}}_{q}}$ and $V'=\langle u_0',\ldots , u_{n-1}'\rangle _{{\mathbb {F}}_{q}}$ and let $f\in \Omega _{m,n}$ with matrix $M_f$ over ${\mathbb {F}}_{q}$ in the bases $\{u_0,\ldots , u_{m-1}\}$ and $\{u'_0,\ldots , u'_{n-1}\}$ of V and $V'$. Since $\{u_0,\ldots , u_{m-1}\}$ is a basis for $V(m,q^d)$ and $\{u_0',\ldots , u_{n-1}'\}$ is a basis for $V(n,q^d)$, we can extend the action of f on $V\times V'$ to an action on $V(m,q^d)\times V(n,q^d)$ in the natural way. Let $f(s_0^{(k)},s_j'^{(k)})=a_{j}\in {\mathbb {F}}_{q^d}$, $j=0,\ldots , e-1$. By Remarks 2.1 and 2.3, the matrix of the change of basis from $\{u_0,\dots , u_{r-1}\}$ to $\{s_0^{(k)},\ldots ,s_0^{(k)}\}$ is $E_r^{-1}K_r$. Therefore, $\nu (f)=D_{\mathbf a}^{(k)}=(E_m^{-1}K_m)^{t} M_f (E_n^{-1}K_n)$, with $\mathbf a=(a_0,\ldots , a_{e-1})$. Since change of bases in $V(m,q^d)\times V(n,q^d)$ preserves the rank of bilinear forms, we have $\mathrm{rank}(f)=\mathrm{rank}(M_f)=\mathrm{rank}(D_\mathbf a^{(k)})$.

3 Puncturing generalized Gabidulin codes

Let $\mathcal X$ be a rank distance code in $M_{n,n}({\mathbb {F}}_{q})$ and A any given $m\times n$ matrix of rank m, $m<n$. It is clear that the set $A\mathcal X=\{AM: M\in \mathcal X\}$ is a rank distance code in $M_{m,n}({\mathbb {F}}_{q})$. We say that the code $A\mathcal X$, which we will denote by $\mathcal P_A(\mathcal X)$, is obtained by puncturing$\mathcal X$withA and $\mathcal P_A(\mathcal X)$ is known as a punctured code.

Theorem 3.1

(Sylvester’s rank inequality) [16, p. 66] Let A be an $m \times n$ matrix and M an $n\times n'$ matrix. Then

$$\begin{aligned} \mathrm{rank}(AM)\ge \mathrm{rank}(A)+\mathrm{rank}(M)-n. \end{aligned}$$

Theorem 3.2

(see also [3, Corollary 35]) Let $\mathcal X$ be an (n, n, q; s)-MRD code. Let A be any $m \times n$ matrix over ${\mathbb {F}}_{q}$ of rank m, with $n-s < m \le n$. Then the punctured code $\mathcal P_A(\mathcal X)$ is an $(m,n,q;s')$-MRD code, with $s'=s+m-n$.

Proof

We first show that the map $M\mapsto AM$ is injective. Assume $AM_1=AM_2$ for some distinct matrices $M_1, M_2 \in \mathcal X$. Then $A(M_1-M_2)=0$, giving $\mathrm{dim}(\ker A)\ge \mathrm{rank}\,(M_1-M_2)\ge s>0$, thus $\mathrm{rank}\, A=m-\mathrm{dim}(\ker A)<m$, a contradiction. Therefore, $|A \mathcal X|=|\mathcal X|=q^{n(n-s +1)}=q^{n(m-s'+1)}$.

By the Sylvester’s rank inequality, we have

$$\begin{aligned} \mathrm{rank}(AM_1-AM_2)\ge \mathrm{rank}(A)+\mathrm{rank}(M_1-M_2)-n \ge m+s-n=s'>0. \end{aligned}$$

It follows that $A\mathcal X$ is an $(m,n,q;s')$-MRD code. $\square$

Remark 3.3

Let B be matrix in $M_{m,n}({\mathbb {F}}_{q})$ of rank m. It is known that there exist $S\in \mathrm{GL}(m,q)$ and $T\in \mathrm{GL}(n,q)$ such that $B=SAT$ [16, p. 62]. Therefore

$$\begin{aligned} \mathcal P_B(\mathcal X)=\mathcal P_{SAT}(\mathcal X)=S\mathcal P_{A}(T\mathcal X), \end{aligned}$$

giving $\mathcal P_B(\mathcal X)$ is equivalent to the punctured code $\mathcal P_A(T\mathcal X)$. Note that $T\mathcal X$ is equivalent to $\mathcal X$.

We recall the construction of the generalized Gabidulin codes as given in [13]. For any positive integers t, k with $t\le n$ and $\mathrm{gcd}(k,n)=1$, set $\mathcal L_t^{(k)}({\mathbb {F}}_{q^n})$ to be the set of all $q^k$-polynomials over ${\mathbb {F}}_{q^n}$ of $q^k$-degree at most $t-1$, i.e.,

$$\begin{aligned} \mathcal L_t^{(k)}({\mathbb {F}}_{q^n})=\left\{ a_0+a_1x^{[k]}+\cdots +a_{t-1}x^{[k(t-1)]}:a_i \in {\mathbb {F}}_{q^n}\right\} . \end{aligned}$$

We note that by reordering the powers of x in any $f\in {\mathcal {L}}_t^{(k)}({\mathbb {F}}_{q^n})$ we actually find a q-polynomial. However, to study the generalized Gabidulin codes in terms of $q^k$-polynomials we need to keep the original order for the powers in f.

Let $g_0,\ldots , g_{m-1}\in {\mathbb {F}}_{q^n}$, $m\le n$, be linearly independent over ${\mathbb {F}}_{q}$. Let $G^{[k]}$ be the matrix

$$\begin{aligned} G^{[k]}=\left( \begin{array}{cccc} g_0 &{}\quad g_1 &{}\quad \cdots &{}\quad g_{m-1} \\ g_0^{[k]} &{}\quad g_1^{[k]} &{}\quad \cdots &{}\quad g_{m-1}^{[k]}\\ \cdots &{}\quad \cdots &{}\quad \cdots &{}\quad \cdots \\ g_0^{[(t-1)k]} &{}\quad g_1^{[(t-1)k]} &{}\quad \cdots &{}\quad g_{m-1}^{[(t-1)k]} \end{array}\right) . \end{aligned}$$

We consider the matrix $G^{[k]}$ as a generator matrix of a subset $\tilde{\mathcal G}_t^{(k)}$ of arrays over ${\mathbb {F}}_{q^n}$, i.e., $\tilde{\mathcal G}_t^{(k)} = \tilde{\mathcal G}_{(g_0,\ldots , g_{m-1});t}^{(k)}= \{(f(g_0),\ldots ,f(g_{m-1})) :f\in \mathcal L_t^{[k]}({\mathbb {F}}_{q^n})\}$.

Let $V'={\mathbb {F}}_{q^n}=\langle u'_0,\ldots , u'_{n-1}\rangle _{{\mathbb {F}}_{q}}$. The map

$$\begin{aligned} \begin{array}{rccl} \varepsilon =\varepsilon _{\{u'_0,\ldots , u'_{n-1}\}}: &{} {\mathbb {F}}_{q^n}&{} \longrightarrow &{} {\mathbb {F}}_{q}^n\\ &{} \sum {x_i u'_i} &{} \mapsto &{} (x_0,\ldots , x_{n-1})^t \end{array} \end{aligned}$$

maps the set $\tilde{\mathcal G}_t^{(k)}$ to the matrix set

$$\begin{aligned} \varepsilon \left( \tilde{\mathcal G}_t^{(k)}\right) =\left\{ (M^{(0)}\,M^{(1)}\,\ldots \,M^{(m-1)}):f\in \mathcal L^{(k)}_{t}({\mathbb {F}}_{q^n})\right\} \subseteq M_{n,m}({\mathbb {F}}_{q}), \end{aligned}$$

where $M^{(i)}=\varepsilon (f(g_i))$. Since the rank is invariant under matrix transposition and in this paper we consider matrix codes in $M_{m,n}({\mathbb {F}}_{q})$ with $m\le n$, we may take the matrix code $\mathcal G_t^{(k)}$ obtained by taking the transpose of the elements in $\varepsilon (\tilde{\mathcal G}_t^{(k)})$. Therefore $\mathcal G_t^{(k)}$ is a $(m,n,q;m-t+1)$-MRD code. These MRD codes are called generalized Gabidulin codes [19].

By Proposition 2.2, we may identify the elements in $\mathrm{End}(V')$ with elements in $\mathcal L^{(k)}_{n}({\mathbb {F}}_{q^n})$ via the map $D_{(a_0,\ldots ,a_{n-1})}\mapsto a_0+a_1x^{[k]}+\cdots +a_{n-1}x^{[k(n-1)]}$. Therefore, $\mathcal G_t^{(k)}$ consists of the matrices of the restriction over the subspace $V=\langle g_0,\ldots , g_{m-1}\rangle _{{\mathbb {F}}_{q}}$ of $V'={\mathbb {F}}_{q^n}$ of the endomorphisms (of $V'$) in ${\mathcal {L}}_{t}^{(k)}({\mathbb {F}}_{q^n})$. These matrices act on the set ${\mathbb {F}}_{q}^m$ of all row vectors as $v\mapsto vM$. As we are working in the framework of bilinear forms, we consider any matrix in $\mathcal G_t^{(k)}$ as a matrix of the restriction on $V\times V'$ of the bilinear form acting on $V'\times V'$ whose $n\times n$ matrix is the matrix in the basis $\{u'_0,\ldots , u'_{n-1}\}$ of an element in ${\mathcal {L}}_{t}^{(k)}({\mathbb {F}}_{q^n})$. By Proposition 2.6 the elements in $\mathcal G_t^{(k)}$ can be represented by $q^k$-circulant matrices over ${\mathbb {F}}_{q^d}$, where $d=\mathrm{gcd}(m,n)$.

The following result seems to be known, but we include a proof for the sake of completeness.

Theorem 3.4

Let $\mathcal G$ be any generalized Gabidulin $(n,n,q;n-t+1)$-code and let A be any given $m\times n$ matrix over ${\mathbb {F}}_{q}$ of rank m, with $t < m \le n$. Then the punctured code $\mathcal P_A(\mathcal G)$ is a generalized Gabidulin $(m,n,q;m-t+1)$-code. Conversely, every generalized Gabidulin $(m,n,q;m-t+1)$-code, with $1\le t\le m$, is obtained by puncturing a generalized Gabidulin $(n,n,q;n-t+1)$-code.

Proof

Let $V'={\mathbb {F}}_{q^n}=\langle u'_0,\ldots , u'_{n-1}\rangle _{{\mathbb {F}}_{q}}$. By the argument above, $\mathcal G$ is considered as the set of all bilinear forms acting on $V'\times V'$ whose matrix corresponds to a $q^k$-polynomial in $\mathcal L_{t}^{(k)}({\mathbb {F}}_{q^n})=\{a_0+a_1x^{[k]}+\cdots +a_{t-1}x^{[k(t-1)]}:a_i \in {\mathbb {F}}_{q^n}\}$.

The given matrix $A=(a_{ij})$ corresponds to the linear transformation

$$\begin{aligned} \begin{array}{rccl} \tau :&u'_i\mapsto & {} \sum _{j=0}^{n-1}{a_{ij}u'_j }, \end{array}\quad i=0,\ldots , m-1. \end{aligned}$$

As $\mathrm{rank}\, A=m$, the subspace $V=\langle \tau (u'_0),\ldots ,\tau (u'_{m-1})\rangle _{{\mathbb {F}}_{q}}$ is an m-dimensional subspace of $V'$. It follows that $\mathcal P_A(\mathcal G)$ consists of the matrices of the bilinear forms on $V\times V'$ in the bases $\{g_i=\tau (u_i): i=0,\ldots ,m-1\}$ and $\{u'_0,\ldots , u'_{n-1}\}$ of V and $V'$, respectively. Therefore $\mathcal P_A(\mathcal G)$ is the generalized Gabidulin code $\mathcal G_{(g_0,\ldots ,g_{m-1}),t}^{(k)}$. By Theorem 3.2$\mathcal G_{(g_0,\ldots ,g_{m-1}),t}^{(k)}$ is an $(m, n, q; m-t + 1)$-MRD code.

For the converse, let $\mathcal G_t^{(k)}=\mathcal G_{(g_0,\ldots ,g_{m-1});t}^{(k)}$ be a generalized Gabidulin code. Set $V=\langle g_0,\ldots , g_{m-1}\rangle _{{\mathbb {F}}_{q}}$ and extend $g_0,\ldots , g_{m-1}$ with $g_m,\ldots , g_{n-1}$ to form a basis of $V'={\mathbb {F}}_{q^n}=V(n,q)$. Then, elements in $\mathcal G_t^{(k)}$ are the restriction on $V\times V'$ of the bilinear forms acting on $V'\times V'$ whose matrix is the matrix of elements in $\mathcal L_{t}^{(k)}({\mathbb {F}}_{q^n})$ in the basis $g_0,\ldots , g_{n-1}$. The set $\overline{\mathcal G}_t^{(k)}=\mathcal G_{(g_0,\ldots , g_{n-1});t}^{(k)}$ of such bilinear forms is a generalized Gabidulin $(n,n,q;n-t+1)$-code. In addition, matrices in $\mathcal G_t^{(k)}$ are obtained from the matrices of $\overline{\mathcal G}_t^{(k)}$ by deleting the last $n-m$ rows, i.e., $\mathcal G_t^{(k)}=A\overline{\mathcal G}_t^{(k)}$ with $A=(I_m|O_{n-m})$. Therefore, the generalized Gabidulin code $\mathcal G_t^{(k)}$ is obtained by puncturing the generalized Gabidulin code $\overline{\mathcal G}_t^{(k)}$ with A. $\square$

From the proof of the previous result, we get the following description for the generalized Gabidulin codes.

Corollary 3.5

Let $g_0,\ldots , g_{m-1}\in {\mathbb {F}}_{q^n}, m\le n$, be linearly independent over ${\mathbb {F}}_{q}$, $V=\langle g_0,\ldots , g_{m-1}\rangle _{{\mathbb {F}}_{q}}$ and $V'={\mathbb {F}}_{q^n}$. Then the generalized Gabidulin code ${\mathcal {G}}_{(g_0,\ldots , g_{m-1});t}^{(k)}$ consists of the restriction on $V \times V'$ of the bilinear forms in ${\mathcal {L}}_{t}^{(k)}({\mathbb {F}}_{q^n})$.

Remark 3.6

Let $e=\mathrm{gcd}(m,n)$ and $d=\mathrm{lcm}(m,n)$. From the arguments contained in Sect. 2 there exists a (Singer) basis of $V=\langle g_0,\ldots ,g_{m-1}\rangle _{{\mathbb {F}}_{q}}$ and a (Singer) basis of $V'={\mathbb {F}}_{q^n}$ such that the elements in $\mathcal G_{(g_0,\ldots ,g_{m-1});t}^{(k)}$ may be represented as $m\times n$$q^k$-circulant matrices over ${\mathbb {F}}_{q^d}$.

Remark 3.7

By the isomorphism $\Omega _{m,n}\simeq \mathcal D_{m,n}^{(1)}({\mathbb {F}}_{q^d})$ stated by Proposition 2.6, the Gabidulin code $\mathcal G_{(g_0,\ldots ,g_{m-1}),t}^{(1)}$ is actually the Delsarte code defined by (6.1) in [9] with $V=\langle g_0,\ldots , g_{m-1}\rangle _{{\mathbb {F}}_{q}}$.

In the rest of the paper, m will be a divisor of n. We set $r=n/m$. Let $V=\langle u_0,\ldots , u_{m-1}\rangle _{{\mathbb {F}}_{q}}$ and $V'=\langle u'_0,\ldots , u'_{n-1}\rangle _{{\mathbb {F}}_{q}}$ be two vector spaces over ${\mathbb {F}}_{q}$ of dimension m and n, respectively. If $m=n$ we take $V'=V=\langle u_0,\ldots , u_{n-1}\rangle _{{\mathbb {F}}_{q}}$.

In the light of the isomorphism $\nu _{\{u_0,\ldots , u_{m-1};u'_0,\ldots , u'_{n-1}\}}$ described by (2), every bilinear form acting on $V\times V'$ may be identified with an $m\times n$ matrix over ${\mathbb {F}}_{q}$. In other words, if we assume V is an m-dimensional subspace of $V'$ after a vector-space isomorphism, then the bilinear forms in $\Omega _{m,n}$ are the restrictions on $V\times V'$ of the bilinear form in $\Omega _{n,n}$. Thus, $\Omega _{m,n}$ is the puncturing of $\Omega _{n,n}$ by a suitable $m\times n$ matrix of rank m.

In this paper we work with cyclic models for vector spaces over ${{\mathbb {F}}_{q}}$. Let $\{s'_0,\ldots , s'_{n- 1}\}$ be a Singer basis for $V'$. We note that not all m-dimensional subspaces of $V'$ can be represented with a cyclic model over ${\mathbb {F}}_{q^m}$. Therefore, we need to choose suitable vectors $\{s_0,\ldots , s_{m-1}\}$ in $V'$ such that the projection of the vectors in the cyclic model for $V'$ on the ${\mathbb {F}}_{q}^{m}$-span of $\{s_0,\ldots ,s_{m-1}\}$ is an m-dimensional subspace over ${\mathbb {F}}_{q}$ represented cyclically. This is what we do in the rest of this section.

Let $\sigma '$ be a Singer cycle of $V'$ with associated primitive element $w'$. Let $\{s_0',\ldots ,s_{n-1}'\}$ be the Singer basis for $V'$ defined by $\sigma '$. Note that $s_i'\in V(n,q^n)$, for $i=0\ldots , n-1$. Set $s_i=\sum _{j=0}^{r-1} s_{i+jm}'$ for $i=0,1,\ldots , m-1$, $\sigma =\sigma '^{(q^{n}-1)/(q^m-1)}$ and $w=w'^{(q^{n}-1)/(q^m-1)}$. Then $\sigma$ has order $q^m-1$ and $\omega$ is a primitive element of ${\mathbb {F}}_{q^m}$ over ${\mathbb {F}}_{q}$. It is easily seen that $s_{i}$ is an eigenvector for $\sigma$ with eigenvalue $w^{q^i}$, for $i=0,\ldots ,m-1$. Since m divides n, the ${\mathbb {F}}_{q^m}$-span $V(m,q^m)$ of $\{s_0,\ldots , s_{m-1}\}$ is contained in $V(n,q^n)$. Let $\xi$ be the semilinear transformation on $V(m,q^m)$ whose linear part is defined by $\ell _{\xi }(s_i)=s_{i+1}$, where the indices are considered modulo m, and whose companion automorphism is $\delta :x\in {\mathbb {F}}_{q^m}\mapsto x^q\in {\mathbb {F}}_{q^m}$. Since the subset $\{xs_0+\cdots +x^{q^{m-1}}s_{m-1}:x \in {\mathbb {F}}_{q^m}\}$ of $V(m,q^m)$ is fixed pointwise by $\xi$, it is a cyclic model for an m-dimensional vector space V over ${\mathbb {F}}_{q}$. By Proposition 2.6, every bilinear form on $V\times V'$ can be represented by $m\times n$q-circulant matrices over ${\mathbb {F}}_{q^n}$.

Lemma 3.8

Let k be a positive integer such that $\mathrm{gcd}(k,n)=1$. Then $s_i^{(k)}=\sum _{j=0}^{r-1} s_{i+jm}'^{(k)}$, for $i=0,1,\ldots , m-1$.

Proof

For $i=0,1,\ldots , m-1$ we have $s_i^{(k)}=s_{ki\bmod m}=\sum _{j=0}^{r-1} s_{(ki\bmod m)+jm}'$. On the other hand, $s_{i+jm}'^{(k)}=s_{k(i+jm)\bmod n}$. Therefore we need to prove

$$\begin{aligned} \{(ki \bmod m)+jm:j=0,\ldots , r-1\}=\{k(i+lm)\bmod n:l=0,\ldots , r-1\}. \end{aligned}$$

Let $ki=tm+s$ with $0\le s \le m-1$. For each $j\in \{0,1,\ldots ,r-1\}$ we need to find $l\in \{0,1,\ldots ,r-1\}$ such that $s+jm \equiv (tm+s+klm) \bmod n$. This is equivalent to

$$\begin{aligned} j \equiv (t+ kl)\bmod r, \end{aligned}$$

(9)

as $r=n/m$. Since $\mathrm{gcd}(k,n)=1$, we also have $\mathrm{gcd}(k,r)=1$. Let $k^{-1}$ denote the inverse of k modulo r. With $l=k^{-1}(j-t)\bmod r$ Eq. (9) is satisfied and

$$\begin{aligned} \{k^{-1}(j-t)\bmod r :j \in \{0,1,\ldots ,r-1\}\}=\{0,1,\ldots ,r-1\}. \end{aligned}$$

Hence the assertion is proved. $\square$

The above lemma implies the following result.

Proposition 3.9

Let k be a positive integer such that $\mathrm{gcd}(k,n)=1$ and m a divisor of n. Let $\{s_0',\ldots ,s_{n-1}'\}$ be a Singer basis for $V'$. Set $s_i^{(k)}=\sum _{j=0}^{r-1} s_{i+jm}'^{(k)}$, for $i=0,1,\ldots , m-1$. Then the ${\mathbb {F}}_{q}$-subspace $\left\{ \sum _{i=0}^{m-1}{x^{[ki]}s_i^{(k)}} :x \in {\mathbb {F}}_{q^m}\right\}$ of $V(n,q^n)$ is a kth cyclic model for V.

Proof

By Remark 2.3 the kth cyclic model for V is obtained from the cyclic model $V=\{\sum _{i=0}^{m-1}{x^{q^i}s_i} :x \in {\mathbb {F}}_{q^m}\}$ by applying the change of basis, say $\kappa$, from $\{s_0,\ldots ,s_{m-1}\}$ with $s_i=\sum _{j=0}^{r-1} s_{i+jm}'$ for $i=0,1,\ldots , m-1$, to $\{s_0^{(k)},\ldots ,s_{m-1}^{(k)}\}$. We recall that $\kappa$ is represented by the permutation matrix $K_{m}^{-1}$. By Lemma 3.8, $s_i^{(k)}=\sum _{j=0}^{r-1} s_{i+jm}'^{(k)}$, for $i=0,1,\ldots , m-1$. This implies that the image under $\kappa$ of the cyclic model $\{\sum _{i=0}^{m-1}{x^{q^i}s_i} :x \in {\mathbb {F}}_{q^m}\}$ is $\left\{ \sum _{i=0}^{m-1}{x^{[ki]}s_i^{(k)}} :x \in {\mathbb {F}}_{q^m}\right\}$. $\square$

Let f be any given bilinear form acting on the kth cyclic model of $V'$ with $q^k$-circulant matrix $D_\mathbf a^{(k)}$ in the basis $\{s_0'^{(k)},\ldots , s_{n-1}'^{(k)}\}$. As the matrix of the coordinates of the vectors $s_0^{(k)},\ldots , s_{m-1}^{(k)}$ in this basis is the $1\times r$ block matrix $A=(I_m \mid I_m \mid \ldots \mid I_m)$, the restriction $f|_{V\times V'}$ of f on $V\times V'$ has matrix $\overline{D}=AD_\mathbf a^{(k)}$ in the bases $\{s_0^{(k)},\ldots , s_{m-1}^{(k)}\}$ and $\{s_0'^{(k)},\ldots , s_{n-1}'^{(k)}\}$.

To make notation easier, we index the rows and columns of an $m\times n$ matrix M by elements in $\{0,\ldots ,m-1\}$ and $\{0,\ldots , n-1\}$. Further, $M_{(i)}$ and $M^{(j)}$ will denote the ith row and jth column of M, respectively.

For $i=0,\ldots ,m-1$, we have $A_{(i)}=(0,\ldots ,0,1,0,\ldots ,0,1,0\ldots ,0,1,0\ldots )$, with 1 at position $i+km$, for $k=0,\ldots , r-1$, and 0 elsewhere. Let $D_\mathbf a^{(k)}$ be generated by the array $\mathbf a=(a_0,\ldots ,a_{n-1})$. Then, $[D_\mathbf a^{(k)}]^{(j)}=(a_j,a_{j-1}^{[k]},\ldots ,a_{j+1}^{[k(n-1)]})^t$, for $j=0,\ldots ,n-1$. Therefore, the (i, j)-entry of $\overline{D}$ is

$$\begin{aligned} a_{j-i}^{[ki]}+a_{j-(i+m)}^{[k(i+m)]}+\cdots +a_{j-(i+(r-1)m)}^{[k(i+(r-1)m)]}=\sum _{h=0}^{r-1}a_{j-(i+hm)}^{[k(i+hm)]}, \end{aligned}$$

where indices are taken modulo n. It turns out that $\overline{D}$ is the $m\times n$$q^k$-circulant matrix over ${\mathbb {F}}_{q^n}$ generated by the array $(\sum _{h=0}^{r-1}a_{-hm}^{[khm]},\sum _{h=0}^{r-1}a_{1-hm}^{[khm]},\ldots , \sum _{h=0}^{r-1}a_{m-1-hm}^{[khm]})$. In particular, if $\mathbf a=(a_0,\ldots ,a_{t-1},0,\ldots ,0)$ for some $t\in \{1,\ldots , m\}$, then $\overline{D}$ is generated by the m-array $(a_0,\ldots ,a_{t-1},0,\ldots ,0)$ giving $\overline{D}$ to be the matrix of a bilinear form in the rank distance code

$$\begin{aligned} \Phi _{m,n,t}^{(k)}=\bigoplus _{j=0}^{t-1}{\Omega _{j}^{(k)}}, \end{aligned}$$

(10)

where the ${\mathbb {F}}_{q}$-subspaces $\Omega _{j}^{(k)}$ are given in (7). Note that the dimension of $\Phi _{m,n,t}$ over ${\mathbb {F}}_{q}$ is nt. The above arguments together with Theorem 3.4 prove the following result.

Theorem 3.10

Let $m>1$ be any divisor of n. For any $t\in \{1,\ldots , m\}$, the subset $\Phi _{m,n,t}^{(k)}$ of $\Omega _{m,n}$ is a generalized Gabidulin $(m,n,q;m-t+1)$-code.

We now describe the generalized twisted Gabidulin codes as provided by Sheekey in the recent paper [29] by using the framework of $q^k$-polynomials over ${\mathbb {F}}_{q^n}$. In [24] the equivalence between different generalized twisted Gabidulin codes was addressed.

Let $\mathrm{N}_{q^n/q}$ denote the norm map from ${\mathbb {F}}_{q^n}$ onto ${\mathbb {F}}_{q}$:

$$\begin{aligned} \mathrm{N}_{q^n/q} :y\in {\mathbb {F}}_{q^n}\mapsto \mathrm{N}_{q^n/q}(y)=\prod _{j=0}^{n-1}{y^{q^j}}\in {\mathbb {F}}_{q}. \end{aligned}$$

Theorem 3.11

[24, 29] For any $t\in \{1,\ldots ,n-1\}$ and $\mu \in {{\mathbb {F}}_{{q^n}}^\times }$ with $\mathrm{N}_{q^n/q}(\mu )\ne (-1)^{nt}$, define the subset

$$\begin{aligned} \Gamma _{n,n,t,\mu ,s}^{(k)}=\left\{ f_{a,0}^{(k)}+f_{\mu a^{q^{sk}}, t}^{(k)}:a\in {\mathbb {F}}_{q^n}\right\} \end{aligned}$$

of $\Omega _{n,n}$ and put

$$\begin{aligned} \mathcal H_{n,n,t,\mu ,s}^{(k)}=\Gamma _{n,n,t,\mu ,s}^{(k)}\oplus \Omega _{1}^{(k)}\oplus \cdots \oplus \Omega _{t-1}^{(k)}. \end{aligned}$$

Then $\mathcal H_{n,n,t,\mu ,s}^{(k)}$ is an $(n,n,q;n-t+1)$-MRD code which is not equivalent to $\Phi _{n,n,t}^{(k)}$ if $t \ne 1,n-1$.

Let A be the $1\times r$ block matrix $(I_m \mid I_m \mid \ldots \mid I_m)$. By Theorem 3.2, for any $t\in \{1,\ldots ,m-1\}$, the punctured code $\mathcal P_A(\mathcal H_{n,n,t,\mu ,s}^{(k)})$ is an MRD code with the same parameters as the code $\Phi _{m,n,t}^{(k)}$ defined by (10). We denote such $(m,n,q;m-t+1)$-MRD code by $\mathcal H_{m,n,t,\mu ,s}^{(k)}$. By using the decomposition (8) of $\Omega _{m,n}^{(k)}$ with $e=m$, we get

$$\begin{aligned} \mathcal H_{m,n,t,\mu ,s}^{(k)}=\Gamma _{m,n,t,\mu ,s}^{(k)} \oplus \Omega _{1}^{(k)}\oplus \cdots \oplus \Omega _{t-1}^{(k)} \end{aligned}$$

(11)

where $\Gamma _{m,n,t,\mu ,s}^{(k)}=\{f_{a,0}^{(k)}+f_{\mu a^{q^{sk}}, t}^{(k)}:a\in {\mathbb {F}}_{q^n}\}$ and $f_{a,j}^{(k)}\in \Omega _{j}^{(k)}$ is defined by (6). It turns out that $\mathcal H_{m,n,t,\mu ,s}^{(k)}$ is a linear MRD code of dimension nt.

Remark 3.12

It is easy to see that the MRD code $\mathcal H_{m,n,t,\mu ,s}^{(k)}$ has the same parameters as the $(m,rm/2,q;m-1)$-MRD codes provided in [7] only for $t=2$ and $n=m$.

4 The automorphism group of some punctured generalized Gabidulin codes

Very recently, Liebhold and Nebe calculated the group of all linear automorphisms of any generalized Gabidulin code. Here we give the finite fields version of Theorem 4.9 in [22].

Theorem 4.1

[22] Let $\mathcal G_t^{(k)}=\mathcal G_{(g_0,\ldots , g_{m-1}),t}^{(k)}$ be a generalized Gabidulin code. Let ${\mathbb {F}}_{q}\le \mathbb {F}_{q^{m'}} \le {\mathbb {F}}_{q^n}$ be the maximal subfield of ${\mathbb {F}}_{q^n}$ such that $\langle g_0,\ldots , g_{m-1}\rangle _{\mathbb {F}_q}$ is an $\mathbb {F}_{q^{m'}}$-subspace. Let $\mathbb {F}_{q^d}$ be the minimal subfield of ${\mathbb {F}}_{q^n}$ such that $\langle g_0,\ldots , g_{m-1}\rangle _{{\mathbb {F}}_{q}}$ is contained in a one-dimensional $\mathbb {F}_{q^d}$-subspace. Then there is a subgroup G of $\mathrm{Gal}(\mathbb {F}_{q^d}/{\mathbb {F}}_{q})$ such that the group of all linear automorphisms of $\mathcal G_t^{(k)}$ is isomorphic to

$$\begin{aligned} \left( {\mathbb {F}_{{q^{m'}}}^\times } \times \mathrm{GL}(n/d,q^d) \right) \rtimes G. \end{aligned}$$

Remark 4.2

The generalized Gabidulin code $\Phi _{m,n,t}^{(k)}$ defined by (10) corresponds to the case $\langle g_0,\ldots , g_{m-1}\rangle _{{\mathbb {F}}_{q}}={\mathbb {F}}_{q^m}$ in the previous theorem.

In spite of Theorem 4.1, we believe that it is useful to have an explicit description of the full automorphism group of an MRD code to compare MRD codes among each other; see [29, 33].

Let $S'=\langle \sigma '\rangle$ be a Singer cyclic group of $\mathrm{GL}(V')$ with associated semilinear transformation $\xi '$ as described in Sect. 2. In the Singer basis $\{s_0'^{(k)},\ldots ,s_{n-1}'^{(k)}\}$, the matrix of $\sigma '$ is the diagonal matrix $\mathrm{diag}(w,w^{[k]},\ldots ,w^{[k(n-1)]})$. Therefore, $(\sigma '^{i},\sigma '^{j})$ acts on $\Omega _{n,n}$ by mapping the bilinear form with $q^k$-circulant matrix $D_{(a_0,\ldots , a_{n-1})}^{(k)}$ to the bilinear form with matrix $D_{(w^ia_0w^j,\ldots , w^ia_{n-1}w^{j[k(n-1)]})}^{(k)}$. The matrix of the linear part $\ell _{\xi '}$ of $\xi '$ is the permutation matrix $D_{(0,\ldots ,0,1)}^{(k)}$. Then $(\ell _{\xi '},\ell _{\xi '})$ acts on $\Omega _{n,n}$ by mapping the bilinear form with $q^k$-circulant $D_{(a_0,\ldots , a_{n-1})}^{(k)}$ to the bilinear form with matrix $D_{(a_0^{[k]},\ldots ,a_{n-1}^{[k]})}^{(k)}$. Set $\bar{\ell }=(\ell _{\xi '},\ell _{\xi '})$ and $\bar{C}$ be the cyclic subgroup of $\mathrm{Aut}(\Omega _{n,n})$ generated by $\bar{\ell }$. It turns out that any element in $(S'\times S')\rtimes \bar{C}\rtimes \mathrm{Aut}({\mathbb {F}}_{q})$ fixes every component $\Omega _{j}^{(k)}$ of $\Omega _{n,n}$.

In the paper [29], Sheekey gave a complete description of the automorphism group of the MRD codes $\Phi _{n,n,t}^{(k)}$ and $\mathcal H_{n,n,t,\mu ,s}^{(k)}$, for any $t\in \{1,\ldots ,n-2\}$.

Theorem 4.3

[29] Let $q=p^h$, p a prime.

• (i) For any given $t\in \{1,\ldots ,n-2\}$, the automorphism group of $\Phi _{n,n,t}^{(k)}$ is the semidirect product $(S'\times S')\rtimes \bar{C}\rtimes \mathrm{Aut}({\mathbb {F}}_{q})$.

• (ii) For any given $t\in \{1,\ldots ,n-2\}$, the automorphism group of $\mathcal H_{n,n,t,\mu ,s}^{(k)}$ is the subgroup of $(S'\times S')\rtimes \bar{C} \rtimes \mathrm{Aut}({\mathbb {F}}_{q})$ whose elements correspond to the triples $((D_{\mathbf a},D_{\mathbf b});\bar{\ell }^i; p^e)$, with $\mathbf a=(a,0,\ldots ,0)$, $\mathbf b=(b,0,\ldots ,0)$ and $a^{q^s-1}b^{q^s-q^t}=\mu ^{p^eq^i-1}$.

We now give more information on the automorphism group of the MRD code $\mathcal H_{n,n,t,\mu ,s}^{(k)}$.

Corollary 4.4

$|\mathrm{Aut}(\mathcal H_{n,n,t,\mu ,s}^{(k)})|=l(q^n-1)(q^{gcd(n,s,t)}-1)$, for some divisor l of nh. If $\mathbb F_{p^d}$ is the smallest subfield of ${\mathbb {F}}_{q^n}$ which contains $\mu$, then $\frac{nh}{d}\, |\, l$. In particular, if $\mu \in \mathbb F_{p}^\times$ then $l=nh$.

Proof

Let $G_{n,k}=\{x^{q^k-1} :x \in {{\mathbb {F}}_{{q^n}}^\times }\}$. Note that $G_{n,k}$ is a subgroup of the multiplicative group ${{\mathbb {F}}_{{q^n}}^\times }$. To calculate $|G_{n,k}|$ it is enough to observe that $x^{q^k-1}=y^{q^k-1}$ for some $x,y\in {{\mathbb F}}_{{q^n}}^\times$ if and only if $x/y \in {{\mathbb {F}}_{{q^k}}^\times },$ and hence $x/y \in {{\mathbb {F}}_{{q^{\mathrm {gcd}(n,k)}}}^\times }$. It follows that $|G_{n,k}|=\frac{q^n-1}{q^{\mathrm{gcd}(n,k)}-1}$. Also, the size of the subgroup $G_{n,k} \cap G_{n,j}$ is

$$\begin{aligned} c_{n,k,j}=\mathrm{gcd}\left( \frac{q^n-1}{q^{\mathrm{gcd}(n,k)}-1},\frac{q^n-1}{q^{\mathrm{gcd}(n,j)}-1}\right) . \end{aligned}$$

Note that for each $c\in {\mathbb F_{{q^n}}^\times }$ we have $|G_{n,k} \cap cG_{n,j}|\in \{0,c_{n,k,j}\}$.

Let $d_{n,k,j}$ denote the number of pairs $(x,y)\in {\mathbb {F}}_{{q^n}}^\times \times {\mathbb {F}}_{{q^n}}^\times$ such that

$$\begin{aligned} x^{q^k-1}y^{q^j-1}=1. \end{aligned}$$

If $x^{q^k-1}y^{q^j-1}=1$, then $x^{q^k-1}=(1/y)^{q^j-1}$ and hence $x^{q^k-1} \in G_{n,k} \cap G_{n,j}$. It follows that we can choose $x_0=x^{q^k-1}$ in $c_{n,k,j}$ different ways, and it uniquely defines $y_0=y^{q^j-1}$. We have

$$\begin{aligned} \left| \left\{ x :x^{q^k-1}=x_0\right\} \right| =\left| \left\{ x :x^{q^k-1}=1\right\} \right| =q^{\mathrm{gcd}(n,k)}-1 \end{aligned}$$

and

$$\begin{aligned} \left| \left\{ y :y^{q^j-1}=y_0\right\} \right| =\left| \left\{ y :y^{q^j-1}=1\right\} \right| =q^{\mathrm{gcd}(n,j)}-1, \end{aligned}$$

thus

$$\begin{aligned} d_{n,k,j}=(q^n-1)(q^{\mathrm{gcd}(n,k,j)}-1). \end{aligned}$$

If $x^{q^k-1}y^{q^j-1}=c$, then $x^{q^k-1}=c (1/y)^{q^j-1}$, thus $x^{q^k-1} \in G_{n,k} \cap cG_{n,j}$. By the above arguments, we get that for each $c\in {\mathbb F_{{q^n}}^\times }$, the number of pairs $(x,y)\in {\mathbb F_{{q^n}}^\times }\times {\mathbb F_{{q^n}}^\times }$ such that $x^{q^k-1}y^{q^j-1}=c$ is either 0, or $d_{n,k,j}$.

Now, let $\mu$ be a given element in ${\mathbb F_{{q^n}}^\times }, q = p^h$ and let $\mathcal H$ denote the set of integers r, such that

$$\begin{aligned} x^{q^k-1}y^{q^j-1}=\mu ^{p^r-1} \end{aligned}$$

has a solution in ${\mathbb F_{{q^n}}^\times } \times {\mathbb F_{{q^n}}^\times }$. By the above arguments, we have $0\in \mathcal H$. If $x_0^{q^k-1}y_0^{q^j-1}=\mu ^{p^r-1}$ and $x_1^{q^k-1}y_1^{q^j-1}=\mu ^{p^s-1}$, then

$$\begin{aligned} \left( x_0^{p^s}\right) ^{q^k-1}\left( y_0^{p^s}\right) ^{q^j-1}=\mu ^{p^{r+s}-p^s} \end{aligned}$$

and

$$\begin{aligned} \left( x_0^{p^s}x_1\right) ^{q^k-1}\left( y_0^{p^s}y_1\right) ^{q^j-1}=\mu ^{p^{r+s}-1}, \end{aligned}$$

thus $r,s \in \mathcal H$ yields $r+s \in \mathcal H$ giving $\mathcal H$ is an additive subgroup in $\mathbb {Z}_{nh}$. Therefore, $l=|\mathcal H|$ divides nh.

By the above arguments, the number of triples $(a,b,i)\in {\mathbb F_{{q^n}}^\times } \times {\mathbb F_{{q^n}}^\times } \times \mathbb {Z}_{nh}$ such that $a^{q^s-1}b^{q^s-q^t}=\mu ^{p^{i}-1}$ is $l(q^n-1)(q^{gcd(n,s,t)}-1)$, for some divisor l of nh. If $\mathbb F_{p^d}$ is the smallest subfield of ${\mathbb {F}}_{q^n}$ which contains $\mu$, then $\mathcal H$ contains the additive subgroup of $\mathbb {Z}_{nh}$ generated by d, giving $\frac{nh}{d}\, |\, l$.

If $\mu \in {\mathbb F_{{p}}^\times }$, then $\mu ^{p^i-1}=1$ for all $i\in {\mathbb Z}_{nh}$. $\square$

Let $m>1$ be any divisor of n and k any positive integer such that $\mathrm{gcd}(k,n)=1$. Let $S=\langle \sigma \rangle$ and $S'=\langle \sigma ' \rangle$ be Singer cyclic groups of $\mathrm{GL}(V)$ and $\mathrm{GL}(V')$, respectively, and $\{s_0,\ldots , s_{m-1}\}$ and $\{s_0',\ldots , s_{n-1}'\}$ the Singer bases defined by $\sigma$ and $\sigma '$.

Set $\ell =(\ell _\xi ,\ell _{\xi '})$ and let C be the cyclic subgroup of $\mathrm{Aut}(\Omega _{m,n})$ generated by $\ell$. Then $C\simeq \mathrm{Aut}({\mathbb {F}}_{q^m}/{\mathbb {F}}_{q})$.

Remark 4.5

To make notation easier, in all the arguments used in the actual and in the next section we assume $k=1$. We put $\Omega _{j}=\Omega _{j}^{(1)}$ and $\Phi _{m,n,t-1}=\Phi _{m,n,t-1}^{(1)}$. By Lemma 3.8, the same arguments work perfectly well for any k with $\mathrm{gcd}(k,n)=1$ if the cyclic model of vector spaces and q-circulant matrices involved are replaced by the kth cyclic model and $q^{k}$-circulant matrices. The details are left to the reader.

Both the Singer cyclic groups $S=\langle \sigma \rangle$ and $S'=\langle \sigma ' \rangle$, as well as the cyclic group C, fix every component $\Omega _{j}$ of $\Omega _{m,n}$ giving that every element in $(S\times S')\rtimes C$ is an automorphism of $\Phi _{m,n,t}$, for any $t\in \{1,\ldots ,m\}$.

Theorem 4.6

Let $\overline{T}'$ be the subset of $\mathrm{End}(V')$ whose elements correspond to q-circulant matrices defined by an array of type

$$\begin{aligned} (c_0,\underbrace{0,\ldots ,0}_{m-1\mathrm { \ times}},c_{1},\underbrace{0,\ldots ,0}_{m-1\mathrm { \ times}}, c_{r-1},\underbrace{0,\ldots ,0}_{m-1\mathrm { \ times}}) \end{aligned}$$

over ${\mathbb {F}}_{q^n}$, and set $T'=\overline{T}'\cap \mathrm{GL}(V')$. Then, for any given $t\in \{1,\ldots ,m-1\}$, the automorphism group of $\Phi _{m,n,t}$ is the semidirect product $(S\times T')\rtimes C\rtimes \mathrm{Aut}({\mathbb {F}}_{q})$.

Proof

Straightforward calculations show that the given group is a subgroup of $\mathrm{Aut}(\Phi _{m,n,t})$.

Let $\varphi =(A,B;\theta )\in \mathrm{Aut}(\Phi _{m,n,t})$, with $A\in {\hbox {GL}}(V), B\in {\hbox {GL}}(V')$ and $\theta \in {\hbox {Aut}}({\mathbb {F}}_{q})$. As $\Phi _{m,n,t}$ is fixed by the semilinear automorphism $\phi$ defined by the Frobenius map $x\mapsto x^p$, we may assume $\theta =\mathbf 1$. We identify the elements A and B with their Dickson matrices in $\mathcal B_{m}({\mathbb {F}}_{q^m})$ and $\mathcal B_{n}({\mathbb {F}}_{q^n})$, respectively. To suit our present needs, we set $A^t=D_{(a_0,a_1,\ldots ,a_{m-1})}$ and $B=D_{(b_0,b_1,\ldots ,b_{n-1})}^t$.

Let f be any given element in $\Phi _{m,n,t}$ with q-circulant matrix $D_{\mathbf a}$. Then, the Dickson matrix of $f^\varphi$ is defined by the m-tuple formed by the first m entries of $(A^t D_{\mathbf a}B)_{(0)}$.

The lth entry, with $0\le l\le m-1$, in $(A^t D_{\mathbf a}B)_{(0)}$ is given by the inner product $(A^t D_{\mathbf a})_{(0)}\cdot B^{(l)}$, with $B^{(l)}=(b_{-l}^{q^l},b_{-l+1}^{q^l},\ldots ,b_{-l+n-1}^{q^{l}})$ where subscripts are taken modulo n.

We recall that $M^{(i)}$ and $M_{(j)}$ denotes the ith column and the jth row of M, respectively.

Let $f=f_{\alpha ,j}$ with $\alpha \in {\mathbb {F}}_{q^n}$ and $0\le j\le m-1$. Then $D_{\mathbf a}$ has only n nonzero entries, one in each column. More precisely, the nonzero entry of the hth column of $D_{\mathbf a}$ is $\alpha ^{q^{h-j}}$ at position $(h-j) \bmod m$. It follows that the hth entry of

$$\begin{aligned} (A^t D_{\mathbf a})_{(0)}=\left( A^t_{(0)}D_{\mathbf a}^{(0)},A^t_{(0)}D_{\mathbf a}^{(1)},\ldots ,A^t_{(0)}D_{\mathbf a}^{(n-1)}\right) \end{aligned}$$

is $a_{h-j}\alpha ^{q^{h-j}}$, where the subscript $h-j$ is taken modulo m. Hence, the lth entry of $(A^tD_{\mathbf a}B)_{(0)}$ is

$$\begin{aligned} \sum _{h=0}^{n-1}a_{h-j}\alpha ^{q^{h-j}}b_{h-l}^{q^l}= \sum _{h=0}^{m-1}a_{h-j}\sum _{k=0}^{r-1}{\alpha ^{q^{h-j+km}}b_{h-l+km}^{q^l}}. \end{aligned}$$

(12)

Since we are assuming that $\varphi$ fixes $\Phi _{m,n,t}=\bigoplus _{j=0}^{t-1}{\Omega _{j}}$, we must have (by putting $h-j=i$)

$$\begin{aligned} \sum _{i=0}^{m-1}a_{i}\sum _{k=0}^{r-1}{\alpha ^{q^{i+km}}b_{i+j-l+km}^{q^l}}=0 \end{aligned}$$

(13)

for $0\le j\le t-1$, $t\le l\le m-1$.

Since Eq. (13) holds for all $\alpha \in {\mathbb {F}}_{q^n}$, we get

$$\begin{aligned} a_ib_{i+j-l}= a_ib_{i+j-l+m}= \cdots = a_ib_{i+j-l+(r-1)m}=0, \end{aligned}$$

for $0\le i \le m-1$ and $t\le l\le m-1$.

As $A\in \mathcal B_m({\mathbb {F}}_{q^m})$, $(a_0,a_1,\ldots ,a_{m-1})\ne (0,\ldots ,0)$. By applying a suitable element in the cyclic subgroup C of $\mathrm{Aut}(\Omega _{m,n})$ generated by $\ell$, we may assume $a_0\ne 0$. Therefore, we get

$$\begin{aligned} b_{j-l}= b_{j-l+m}=\cdots =b_{j-l+(r-1)m}=0, \end{aligned}$$

for $0\le j\le t-1$ and $t\le l\le m-1$. By considering subscripts modulo n, we see that the possible nonzero entries in $(b_0,\ldots , b_{n-1})$ are those in position km, with $0\le k\le r-1$, with at least one of them nonzero.

In $B^{(l)}$, the only nonzero entries are $b_{km}^{q^l}$ in $(l+km)$th positions, for $0\ \le k\le r-1$. Then the expression (12) for the lth entry in $(A^tD_{\mathbf a}B)_{(0)}$, reduces to $a_{l-j}\sum _{k=0}^{r-1}{\alpha ^{q^{l-j+km}}b_{km}^{q^l}}$, which must be zero for $0\le j\le t-1$, $t\le l\le m-1$ and all $\alpha \in {\mathbb {F}}_{q^n}$. In addition, $1\le l-j\le m-1$ gives $(a_0,a_1,\ldots ,a_{m-1})=(a_0,0,\ldots ,0)$, $a_0\ne 0$, and therefore $\mathrm{Aut}(\Phi _{m,n,t})$ has the prescribed form. $\square$

Remark 4.7

Statement (i) in Theorem 4.3 is obtained by taking $m=n$ in the previous Theorem.

Remark 4.8

By Remark 4.5, Theorem 4.6 provides also the description of the automorphism group of the generalized Gabidulin code $\Phi _{m,n,t}^{(k)}$. We notice that the codes $\Phi _{m,n,t}^{(k)}$ are defined in different cyclic models for $\Omega _{m,n}$, for different values of k.

Proposition 4.9

Let $\overline{T}'$ be defined as in Theorem 4.6. Then $\overline{T}'\simeq \mathrm{End}(V(n/m,q^m))$ and $T'=\overline{T}'\cap \mathrm{GL}(V')\simeq \mathrm{GL}(n/m,q^m)$.

Proof

Set $r=n/m$. By Proposition 2.2 we have $\mathrm{End}(V(r,q^m))\cong \mathcal D_r({\mathbb {F}}_{q^n})$, where $\mathcal D_r({\mathbb {F}}_{q^n})$ is the Dickson matrix algebra of all the $q^m$-circulant $r\times r$ matrices acting on the cyclic model $W=\{(x,x^{q^m},\ldots , x^{q^{n-m}}):x \in {\mathbb {F}}_{q^n}\}$ for $V(r,q^m)$. Both W and $V'=\{(x,x^{q},\ldots , x^{q^{n-1}}):x \in {\mathbb {F}}_{q^n}\}$ are n-dimensional vector spaces over ${\mathbb {F}}_{q}$ and the map

$$\begin{aligned} \begin{array}{cccc} \tau : &{} W &{} \longrightarrow &{}V'\\ &{} (x,x^{q^m},\ldots , x^{q^{n-m}}) &{}\mapsto &{} (x,x^{q},\ldots , x^{q^{n-1}}) \end{array} \end{aligned}$$

is an isomorphism of vector spaces.

A straightforward computation shows that $\tau$ induces the group isomorphism

$$\begin{aligned} \begin{array}{cccc} \bar{\tau }: &{} \mathcal D_r({\mathbb {F}}_{q^n}) &{} \longrightarrow &{}T'\\ &{} D_{(c_0,c_1,\ldots ,c_{r-1})} &{}\mapsto &{} D_{(c_0,0,\ldots 0,c_1,0,\ldots ,0,c_{r-1},0,\ldots ,0)} \end{array}. \end{aligned}$$

This is enough to get the result. $\square$

Corollary 4.10

For any given $t\in \{1,\ldots ,m-1\}$,

$$\begin{aligned} \mathrm{Aut}\left( \Phi _{m,n,t}\right) \simeq \left( {\mathbb F_{{q^m}}^\times }\times \mathrm{GL}(n/m,q^m)\right) \rtimes \mathrm{Aut}({\mathbb {F}}_{q^m}/{\mathbb {F}}_{q})\rtimes \mathrm{Aut}({\mathbb {F}}_{q}). \end{aligned}$$

Proof

Since the Singer cyclic group S of $\mathrm{GL}(V)$ is isomorphic to the multiplicative group ${\mathbb F_{{q^m}}^\times }$ and the cyclic group C generated by $\ell$ is isomorphic to $\mathrm{Aut}({\mathbb {F}}_{q^m}/{\mathbb {F}}_{q})$, the result follows from Theorem 4.6 and Proposition 4.9. $\square$

5 The automorphism group of some punctured generalized twisted Gabidulin code

The following result gives information on the geometry of the punctured code $\mathcal H_{m,n,t,\mu ,s}^{(k)}$, and it will be used to calculate the automorphism group of this MRD code. We apply arguments similar to those used by Shekeey [29]. As we did in the previous section, we consider only the case $k=1$ to make notation easier. The arguments below work perfectly well in the general case. We put $\Omega _{j}=\Omega _{j}^{(1)}$, $\Phi _{m,n,t-1}=\Phi _{m,n,t-1}^{(1)}$ and $\mathcal H_{m,n,t,\mu ,s}=\mathcal H_{m,n,t,\mu ,s}^{(1)}$.

Theorem 5.1

Let $m>1$ be any divisor of n and $\mu \in {\mathbb F_{{q^n}}^\times }$ such that $\mathrm{N}_{q^n/q}(\mu )\ne (-1)^{nt}$. For any given $t\in \{1,\ldots ,m-2\}$ and $s\not \equiv 0,\pm 1,\pm 2 \pmod m$, $\bigoplus _{j=1}^{t-1}\Omega _{j}$ is the unique subspace of $\mathcal H_{m,n,t,\mu ,s}$ which is equivalent to $\Phi _{m,n,t-1}$.

Proof

Let $\varphi =(A,B;\theta )\in \mathrm{Aut}(\Omega _{m,n})$ such that $\Phi _{m,n,t-1}^\varphi$ is contained in $\mathcal H_{m,n,t,\mu ,s}$. Here, $\theta =\phi ^e$. As every component $\Omega _j$ is fixed by the semilinear automorphism $\phi$, we may assume $\theta =\mathbf 1$. We identify the elements A and B with their Dickson matrices in $\mathcal B_{m}({\mathbb {F}}_{q^m})$ and $\mathcal B_{n}({\mathbb {F}}_{q^n})$, respectively. To suit our present needs, we set $A^t=D_{(a_0,a_1,\ldots ,a_{m-1})}$ and $B=D_{(b_0,b_1,\ldots ,b_{n-1})}^t$.

Let $f=f_{\alpha ,j}$ with $\alpha \in {\mathbb {F}}_{q^n}$ and $0\le j\le t-2$. Let $D_{\mathbf a}$ be the Dickson matrix of f in the Singer bases $s_0,\ldots , s_{m-1}$ and $s_0',\ldots , s_{n-1}'$. Set $r=n/m$. By arguing as in the proof of Theorem 4.6, we get that the lth entry in $(A^tD_{\mathbf a}B)_{(0)}$ is given by

$$\begin{aligned} \sum _{h=0}^{m-1}{a_{h-j} \sum _{k=0}^{r-1}{\alpha ^{q^{h-j+km}}b_{h-l+km}^{q^l}}}, \end{aligned}$$

(14)

where the indices of the entries of A and B are taken modulo m and n, respectively. Since we are assuming that $\Phi _{m,n,t-1}^\varphi$ is contained in $\mathcal H_{m,n,t,\mu ,s}$, we must have (after substituting $h-j$ with i)

$$\begin{aligned} \sum _{i=0}^{m-1}{a_i\left( \sum _{k=0}^{r-1}{\alpha ^{q^{i+km}}b_{i+j-l+km}^{q^l}}\right) }=0, \end{aligned}$$

for $0\le j\le t-2$, $t+1\le l\le m-1$ and all $\alpha \in {\mathbb {F}}_{q^n}$. Therefore,

$$\begin{aligned} a_{i}b_{i+j-l+km} = 0,\quad \mathrm {for}\ 0\le i\le m-1. \end{aligned}$$

As $B\in \mathcal B_{n}({\mathbb {F}}_{q^n})$, some of the $b_i$’s are nonzero. On the other hand, the cyclic group $C=\langle \ell \rangle$ fixes every component $\Omega _{j}$. Hence, we can assume $b_0\ne 0$ and get $a_{l-j}=0$, for $0\le j\le t-2$ and $t+1\le l\le m-1$, i.e.,

$$\begin{aligned} a_{l} =a_{l-1}=\cdots = a_{l-t+2}= 0 \end{aligned}$$

for $t+1\le l\le m-1$, giving $(a_0,\ldots ,a_{m-1})=(a_0,a_1,a_2,0\ldots ,0)$. Whenever $a_i\ne 0$, we get

$$\begin{aligned} b_{i+j-l+km} = 0, \end{aligned}$$

for $0\le j\le t-2$, $0\le k\le r-1$, i.e.,

$$\begin{aligned} b_{i+km+1}=\cdots =b_{i+(k+1)m-3}=0 \end{aligned}$$

(15)

for $i=0,1,2$ and $0\le k\le r-1$ since $j-l$ can take all integers from $\{1-m, 2-m, \ldots , -4,-3\}$. We now compare the 0th and tth entries of $(A^tD_\mathbf aB)_{(0)}$. From (14) we can see that the 0th entry of $(A D_{\mathbf a} B^t)_{(0)}$ is

$$\begin{aligned} \sum _{i=0,1,2}{a_i\left( \sum _{k=0}^{r-1}{\alpha ^{q^{i+km}}b_{i+j+km}}\right) } \end{aligned}$$

and the tth entry is

$$\begin{aligned} \sum _{i=0,1,2}{a_i\left( \sum _{k=0}^{r-1}{\alpha ^{q^{i+km}}b_{i+j-t+km}^{q^t}}\right) }. \end{aligned}$$

Since we are assuming that $\Phi _{m,n,t-1}^\varphi$ is contained in $\mathcal H_{m,n,t,\mu ,s}$, we must have

$$\begin{aligned} \mu \left[ \sum _{i=0,1,2}{a_i\left( \sum _{k=0}^{r-1}{\alpha ^{q^{i+km}}b_{i+j+km}}\right) }\right] ^{q^s}=\sum _{i=0,1,2}{a_i\left( \sum _{k=0}^{r-1}{\alpha ^{q^{i+km}}b_{i+j-t+km}^{q^t}}\right) }, \end{aligned}$$

for all $\alpha \in {\mathbb {F}}_{q^n}$, i.e.,

$$\begin{aligned} \sum _{i=0,1,2}{a_i\left( \sum _{k=0}^{r-1}{\alpha ^{q^{i+km}}b_{i+j-t+km}^{q^t}}\right) }-\mu \sum _{i=0,1,2}{a_i^{q^s}\left( \sum _{k=0}^{r-1}{\alpha ^{q^{i+km+s}}b_{i+j+km}^{q^s}}\right) }=0, \end{aligned}$$

for all $\alpha \in {\mathbb {F}}_{q^n}$. Since $s\ne \pm i+km$, for $i=0,1,2$ and $0\le k\le r-1$, we get

$$\begin{aligned} \{km+i :i=0,1,2,\, 0\le k \le r-1\} \cap \{km+s+i :i=0,1,2,\, 0\le k \le r-1\}=\emptyset \end{aligned}$$

and hence

$$\begin{aligned} \left\{ \begin{array}{l} \mu a_ib_{i+j+km}=0\\ a_ib_{i+j-t+km}=0, \end{array} \right. \end{aligned}$$

for $0\le j\le t-2$ and $0\le k\le r-1$. Thus, whenever $a_i\ne 0$, we get

$$\begin{aligned} \left\{ \begin{array}{l} b_{i+j+km}=0\\ b_{i+j-t+km}=0. \end{array} \right. \end{aligned}$$

The first equation with $j=0$, the second with $j=t-2$ and (15) give

$$\begin{aligned} b_{i+km}=\cdots =b_{i+(k+1)m-2}=0, \end{aligned}$$

for $i=0,1,2$ and $0\le k\le r-1$.

For $a_0\ne 0$ we get

$$\begin{aligned} b_{km}=\cdots =b_{(k+1)m-2}=0, \end{aligned}$$

for $a_1\ne 0$ we get

$$\begin{aligned} b_{km+1}=\cdots =b_{(k+1)m-1}=0 \end{aligned}$$

and for $a_2\ne 0$ we get

$$\begin{aligned} b_{km+2}=\cdots =b_{(k+1)m}=0, \end{aligned}$$

with $0\le k\le r-1$.

Hence, just one of the $a_i$’s is nonzero. By choosing a suitable element in C we can assume $a_0\ne 0$ so that

$$\begin{aligned} (b_0,\ldots ,b_{n-1})=(0,\ldots ,0,b_{m-1},0,\ldots ,0,b_{2m-1},0,\ldots ,0,b_{n-1}). \end{aligned}$$

By recalling that $B=D_{(b_0,\ldots ,b_{n-1})}^t$, we can see that the action of $\varphi =(A,B;\mathbf 1)$ on $\Omega _j^{(1)}$ is the following:

$$\begin{aligned} \begin{array}{lcl} f_{\alpha ,j}^\varphi \left( v,v'\right) &{} = &{} f_{\alpha ,j}\left( a_0 x,b_{n-1}^q{x'}^q+b_{\left( r-1\right) m-1}^{q^{m+1}}{x'}^{q^{m+1}}+\cdots +b_{m-1}^{q^{\left( r-1\right) m+1}}{x'}^{q^{\left( r-1\right) m+1}}\right) \\ &{} = &{} \mathrm{Tr}\left( \alpha a_0 x\left( b_{n-1}{x'}+b_{\left( r-1\right) m-1}^{q^{m}}{x'}^{q^{m}}+\cdots +b_{m-1}^{q^{\left( r-1\right) m}}{x'}^{q^{\left( r-1\right) m}}\right) ^{q^{j+1}}\right) \end{array} \end{aligned}$$

giving $f_{\alpha ,j}^\varphi \in \Omega _{j+1}$. By consideration on dimensions we have

$$\begin{aligned} \Omega _{j}^\varphi =\Omega _{j+1} \end{aligned}$$

giving that $\Phi _{m,n,t-1}^\varphi =\bigoplus _{j=1}^{t-1}\Omega _{j}$ is the unique subspace of $\mathcal H_{m,n,t,\mu ,s}$ which is equivalent to $\Phi _{m,n,t-1}$. $\square$

We are now in position to calculate the automorphism group of the MRD code $\mathcal H_{m,n,t,\mu ,s}$.

Theorem 5.2

Let $q=p^h$, p a prime. Let $m>1$ be any divisor of n and $\mu \in {\mathbb F_{{q^n}}^\times }$ such that $\mathrm{N}_{q^n/q}(\mu )\ne (-1)^{nt}$. Set $r=n/m$. For any given $t\in \{1,\ldots ,m-2\}$ and $s\not \equiv 0,\pm 1,\pm 2\pmod m$, the automorphism group of $\mathcal H_{m,n,t,\mu ,s}$ is the subgroup of $(S\times T')\rtimes C\rtimes \mathrm{Aut}({\mathbb {F}}_{q})$ whose elements correspond to triples $((D_{\mathbf a},D_{\mathbf b}); \ell ^i;\phi ^e)$, where $\mathbf a=(a,0,\ldots ,0)$, with $a\in {\mathbb {F}}_{q^m}$, and $\mathbf b=(b_0,\underbrace{0,\ldots ,0}_{m-1\mathrm { \ times}},b_{m},\underbrace{0,\ldots ,0}_{m-1\mathrm { \ times}}, b_{(r-1)m},\underbrace{0,\ldots ,0}_{m-1\mathrm { \ times}})$, with $b_{lm}\in {\mathbb {F}}_{q^n}$ such that

$$\begin{aligned} \mu a^{q^s-1} b_{lm}^{(q^s-q^t)q^{-lm}} = \mu ^{p^eq^{i-lm}}, \end{aligned}$$

for $0\le l\le r-1$ whenever $b_{lm}$ is nonzero.

Proof

From Theorem 5.1 every automorphism of $\mathcal H_{m,n,t,\mu ,s}$ must fix $\bigoplus _{j=1}^{t-1}\Omega _{j}$ giving $\mathrm{Aut}(\mathcal H_{m,n,t,\mu ,s})$ is a subgroup of $\mathrm{Aut}(\bigoplus _{j=1}^{t-1}\Omega _{j})$ which in turn is conjugate to $\mathrm{Aut}(\Phi _{m,n,t-1})$. By Theorem 4.6, $\mathrm{Aut}(\Phi _{m,n,t-1})=(S \times T')\rtimes C\rtimes \mathrm{Aut}({\mathbb {F}}_{q})$, and it is easy to see that this group fixes every component $\Omega _{j}$. Let $\varphi =((A,B),\ell ^i;\theta )\in \mathrm{Aut}(\mathcal H_{m,n,t,\mu ,s})$ with $\theta =\phi ^e$. As $\varphi$ fixes every component $\Omega _{j}$, then $\varphi$ must fix $\Gamma _{m,n,t,\mu ,s}$. In addition, $\ell ^i$ maps $\Gamma _{m,n,t,\mu ,s}$ to $\Gamma _{m,n,t,\mu ^{q^i},s}$, thus the above condition holds if and only if $\varphi =((A,B),\mathbf 1;\theta )$ maps $\Gamma _{m,n,t,\mu ^{q^i},s}$ to $\Gamma _{m,n,t,\mu ,s}$.

Let $D_{(a,0,\ldots ,0)}$ and $D_{(b_0,\ldots ,0,b_m,0,\ldots ,0,b_{(r-1)m},0,\ldots ,0)}^t$ be the q-circulant matrix of A and B, respectively. Let $f=f_{\alpha ,0}+f_{\mu ^{q^i}\alpha ^{q^s},t}$ be any bilinear form in $\Gamma _{m,n,t,\mu ^{q^i},s}$. Then, $f^{ \varphi }$ is the bilinear form defined by

$$\begin{aligned} f^{ \varphi }\left( v,v'\right)= & {} f_{\alpha ^{p^e},0}\left( ax,b_0x'+b_m {x'}^{q^m}+\cdots +b_{\left( r-1\right) m}{x'}^{q^{\left( r-1\right) m}}\right) \\&+f_{\mu ^{p^eq^i}\alpha ^{q^s p^e},t}\left( ax,b_0x'+b_m {x'}^{q^m}+\cdots +b_{\left( r-1\right) m}{x'}^{q^{\left( r-1\right) m}}\right) \\= & {} \mathrm{Tr}\left( \alpha ^{p^e}ax\left( b_0x'+b_m {x'}^{q^m}+\cdots +b_{\left( r-1\right) m}{x'}^{q^{\left( r-1\right) m}}\right) \right) \\&+\, \mathrm{Tr}\left( \mu ^{p^eq^i}\alpha ^{q^sp^e}ax\left( b_0x'+b_m {x'}^{q^m}+\cdots +b_{\left( r-1\right) m}{x'}^{q^{\left( r-1\right) m}}\right) ^{q^t}\right) \\= & {} \mathrm{Tr}\left( a\left( \alpha ^{p^e}b_0+\alpha ^{p^eq^{\left( r-1\right) m}}b_m^{q^{\left( r-1\right) m}}+\cdots +\alpha ^{p^eq^{m}}b_{\left( r-1\right) m}^{q^{m}}\right) xx'\right) \\&+\, \mathrm{Tr}\left( a\left( \mu ^{p^eq^i}\alpha ^{p^eq^s}b_0^{q^t}+\mu ^{p^eq^{\left( r-1\right) m+i}}\alpha ^{p^eq^{\left( r-1\right) m+s}}b_m^{q^{\left( r-1\right) m+t}}\right. \right. \\&\left. \left. +\cdots + \mu ^{p^eq^{m+i}}\alpha ^{p^eq^{m+s}}b_{\left( r-1\right) m}^{q^{m+t}}\right) x{x'}^{q^t}\right) . \end{aligned}$$

For $f^{ \varphi }$ to lie in $\Gamma _{m,n,t,\mu ,s}$ we must have

$$\begin{aligned} \begin{array}{l} a^{q^s}\mu \left( \alpha ^{p^eq^s}b_0^{q^s}+\alpha ^{p^eq^{\left( r-1\right) m+s}}b_m^{q^{\left( r-1\right) m+s}}+\cdots +\alpha ^{p^eq^{m+s}}b_{\left( r-1\right) m}^{q^{m+s}}\right) \\ \quad =a\left( \mu ^{p^eq^i}\alpha ^{p^eq^s}b_0^{q^t}+\mu ^{p^eq^{\left( r-1\right) m+i}}\alpha ^{p^eq^{\left( r-1\right) m+s}}b_m^{q^{\left( r-1\right) m+t}}+\cdots +\mu ^{p^eq^{m+i}}\alpha ^{p^eq^{m+s}}b_{\left( r-1\right) m}^{q^{m+t}}\right) \end{array} \end{aligned}$$

for all $\alpha \in {\mathbb {F}}_{q^n}$. This yields

$$\begin{aligned} \begin{array}{lcl} \mu a^{q^s} b_{lm}^{q^{(r-l)m+s}}= & {} a\mu ^{p^eq^{(r-l)m+i}} b_{lm}^{q^{(r-l)m+t}} \end{array} \end{aligned}$$

giving

$$\begin{aligned} \begin{array}{lcl} \mu a^{q^s-1} b_{lm}^{(q^s-q^t)q^{-lm}}= & {} \mu ^{p^eq^{i-lm}}, \end{array} \end{aligned}$$

(16)

for $0\le l\le r-1$ whenever $b_{lm}$ is nonzero. $\square$

Remark 5.3

Statement ii) in Theorem 4.3 is obtained by taking $m=n$ in the previous Theorem.

Remark 5.4

By Lemma 3.8, the arguments used in the proof of Theorems 5.2 work perfectly well for any k with $\mathrm{gcd}(k,n)=1$ if the cyclic model of vector spaces and q-circulant matrices involved are replaced by the kth cyclic model and $q^{k}$-circulant matrices. This implies that the automorphism group of the punctured code $\mathcal H_{m,n,t,\mu ,s}^{(k)}$ is the subgroup of $(S\times T')\rtimes C\rtimes \mathrm{Aut}({\mathbb {F}}_{q})$ whose elements correspond to triples $((D_{\mathbf a}^{(k)},D_{\mathbf b}^{(k)}); \ell ^i;\phi ^e)$, where $\mathbf a=(a,0,\ldots ,0)$, with $a\in {\mathbb {F}}_{q^m}$, and $\mathbf b=(b_0,\underbrace{0,\ldots ,0}_{m-1\mathrm { \ times}},b_{m},\underbrace{0,\ldots ,0}_{m-1\mathrm { \ times}}, b_{(r-1)m},\underbrace{0,\ldots ,0}_{m-1\mathrm { \ times}})$ with $b_{lm}\in {\mathbb {F}}_{q^n}$ such that

$$\begin{aligned} \mu a^{q^{sk}-1} b_{lm}^{(q^{sk}-q^t)q^{-lm}} = \mu ^{p^eq^{i-lm}}, \end{aligned}$$

(17)

for $0\le l\le r-1$ whenever $b_{lm}$ is nonzero.

Theorem 5.5

Let $m>1$ be any divisor of n and $\mu \in {\mathbb F_{{q^n}}^\times }$, such that $\mathrm{N}_{q^n/q}(\mu )\ne (-1)^{nt}$. For any given $t\in \{1,\ldots ,m-2\}$, the punctured code $\mathcal H_{m,n,t,\mu ,s}^{(k)}$, with $s\not \equiv 0,\pm 1,\pm 2\pmod m$ and $\mathrm{gcd}(n,sk-t)<m$ is not equivalent to any generalized Gabidulin code.

Proof

In Sect. 3 we have seen that $\mathcal H_{m,n,t,\mu ,s}^{(k)}$ is an MRD $(m,n,q;m-t+1)$-code. Therefore $\mathcal H_{m,n,t,\mu ,s}^{(k)}$ has the same parameters as any generalized Gabidulin code $\mathcal G_{(g_0,\ldots ,g_{m-1});t}^{(j)}$, with $g_0,\ldots ,g_{m-1}\in {\mathbb {F}}_{q^n}$ linearly independent over ${\mathbb {F}}_{q}$.

By Theorem 4.1 the subgroup $L_{\mathcal G}$ of all linear automorphisms of $\mathcal G_{(g_0,\ldots ,g_{m-1});t}^{(j)}$ is isomorphic to

$$\begin{aligned} \left( {\mathbb {F}_{{q^{m'}}}^\times }\times \mathrm{GL}(n/d,q^d) \right) \rtimes G, \end{aligned}$$

for some divisors $m'$ and d of n and a subgroup G of $\mathrm{Aut}({\mathbb {F}}_{q^d}/{\mathbb {F}}_{q})$. Note that m divides d. We represent the elements of $L_\mathcal G$ by pairs of type $((a,A);\varphi )$, with $a\in {\mathbb {F}_{{q^{m'}}}^\times }, A\in \mathrm{GL}(n/d,q^d)$ and $\varphi \in G$. In particular the subgroup $\{(1,A);{ id})\}$ of $L_\mathcal G$ is isomorphic to $\mathrm{GL}(n/d,q^d)$.

By way of contradiction, assume that $\mathcal H_{m,n,t,\mu ,s}^{(k)}$ is equivalent to $\mathcal G_{(g_0,\ldots ,g_{m-1});t}^{(j)}$, for some $g_0,\ldots ,g_{m-1}\in {\mathbb {F}}_{q^n}$ linearly independent over ${\mathbb {F}}_{q}$. Then, $\mathrm{Aut}(\mathcal H_{m,n,t,\mu ,s}^{(k)})$ must be isomorphic to $\mathrm{Aut}(\mathcal G_{(g_0,\ldots ,g_{m-1});t}^{(j)})$. In particular the subgroup $L_\mathcal H$ of all linear automorphisms of $\mathcal H_{m,n,t,\mu ,s}^{(k)}$ must be isomorphic to $L_\mathcal G$.

Set $r=n/m$. By Theorem 5.2 and Remark 5.4, $L_\mathcal H$ is the subgroup of $(S\times T')\rtimes C$ whose elements correspond to pairs $((D_{\mathbf a}^{(k)},D_{\mathbf b}^{(k)}); \ell ^i)$, where $\mathbf a=(a,0,\ldots ,0)$, with $a\in {\mathbb {F}}_{q^m}$, and $\mathbf b=(b_0,0,\ldots ,0,b_{m},0,\ldots ,0, b_{(r-1)m},0,\ldots ,0)$, with $b_{lm}\in {\mathbb {F}}_{q^n}$ satisfying Eq. (17) with $e=0$.

For $i=0$ and $\mathbf a=(1,0,\ldots ,0)$, the pairs $((I_m,D_{\mathbf b}^{(k)}), { id})$, with $b_{lm}\in {\mathbb {F}}_{q^n}$ such that

$$\begin{aligned} \mu b_{lm}^{(q^{sk}-q^t)q^{-lm}} = \mu ^{q^{-lm}}, \end{aligned}$$

(18)

for $0\le l\le r-1$ whenever $b_{lm}$ is nonzero, form a subgroup B of $L_\mathcal H$ which should be isomorphic to $\mathrm{GL}(n/d,q^d)$. By raising to the $q^{lm}$th power both sides of Eq. (18), it becomes

$$\begin{aligned} b_{lm}^{q^t(q^{sk-t}-1)}=\mu ^{1-q^{lm}}. \end{aligned}$$

(19)

It is clear that the elements $b_{lm}\in {\mathbb {F}}_{q^n}$ that satisfy Eq. (19) corresponds to the solutions of

$$\begin{aligned} X^{q^{sk-t}-1}=\mu ^{1-q^{lm}}. \end{aligned}$$

(20)

over ${\mathbb {F}}_{q^n}$. If this equation has no solution in ${\mathbb {F}}_{q^n}$ then $b_{lm}=0$.

Let $x,y\in {\mathbb F_{{q^n}}^\times }$ be distinct solutions of Eq. (20). Then $(x/y)^{q^{sk-t}-1}=1$, or equivalently $x/y\in {\mathbb F_{{q^c}}^\times },$ with $c=\mathrm{gcd}(n,sk-t)$. Thus the solutions of Eq. (20) over ${\mathbb {F}}_{q^n}$ are exactly the elements in $\{\lambda x:x \in {\mathbb F_{{q^c}}^\times \}}$ where $\lambda \in {\mathbb {F}}_{q^n}$ is a fixed solution of (20). Therefore the number of solutions of (20) is either 0 or $q^c-1$. In any case this number is strictly less than $q^m$. It follows that

$$\begin{aligned} |B|\le q^{cr}-1<q^{n}-1=|\mathrm{GL}(1,q^n)|\le |\mathrm{GL}(n/d,q^d)|. \end{aligned}$$

From the above inequality it follows that the subgroup B is not isomorphic to $\mathrm{GL}(n/d,q^d)$. This contradicts the assumption that $\mathcal H_{m,n,t,\mu ,s}^{(k)}$ is equivalent to $\mathcal G_{(g_0,\ldots ,g_{m-1});t}^{(j)}$. The result then follows. $\square$

Corollary 5.6

Let $m>1$ be any divisor of n and $\mu ,\nu \in {\mathbb F_{{q^n}}^\times }$ such that $\mathrm{N}_{q^n/q}(\mu )\ne (-1)^{nt}\ne \mathrm{N}_{q^n/q}(\nu )$. For any given $t\in \{1,\ldots ,m-2\}$ and $s\not \equiv 0,\pm 1,\pm 2\pmod m$ the punctured code $\mathcal H_{m,n,t,\mu ,s}^{(k)}$ is equivalent to $\mathcal H_{m,n,t,\nu ,u}^{(k)}$ if and only if there exist $j\in \{0,\ldots , r-1\}$, an integer i, a nonzero element $a\in {\mathbb {F}}_{q^m}$ and $b_{hm}\in {\mathbb {F}}_{q^n}$, $h=0,\ldots , r-1$, such that $a\mu ^{p^eq^{(r-h)m+i}} b_{hm}^{q^{(r-h)m+t}} = a^{q^u} \nu b_{(h-j)m}^{q^{(j-h)m+u}}$, for $0\le h\le r-1$ (with indices of b considered modulo n) and the $q^k$-circulant matrix $D_{\mathbf b}$ defined by $\mathbf b=(b_0,\underbrace{0,\ldots ,0}_{m-1\mathrm { \ times}},b_{m},\underbrace{0,\ldots ,0}_{m-1\mathrm { \ times}}, b_{(r-1)m},\underbrace{0,\ldots ,0}_{m-1\mathrm { \ times}})$ is non-singular.

Proof

We argue with $k=1$. By Theorem 5.1, $\mathcal H_{m,n,t,\mu ,s}$ and $\mathcal H_{m,n,t,\nu ,u}$ contain a unique subspace equivalent to $\Phi _{m,n,t-1}$. Therefore, any isomorphism from $\mathcal H_{m,n,t,\mu ,s}$ to $\mathcal H_{m,n,t,\nu ,u}$ is in $\mathrm{Aut}(\Phi _{m,n,t-1})$. By using similar arguments as in the proof of Theorem 5.2, we may consider isomorphisms of type $((A,B),\mathbf 1;\theta )$. Let $f=f_{\alpha ,0}+f_{\mu ^{q^i}\alpha ^{q^s},t}$ be any bilinear form in $\Gamma _{m,n,t,\mu ^{q^i},s}$. Then, $f^{ \varphi }$ lies in $\Gamma _{m,n,t,\nu ,u}$ if and only if

$$\begin{aligned}&a^{q^u}\nu \left( \alpha ^{p^eq^u}b_0^{q^u}+\alpha ^{p^eq^{(r-1)m+u}}b_m^{q^{(r-1)m+u}}+\cdots +\alpha ^{p^eq^{m+u}}b_{(r-1)m}^{q^{m+u}}\right) \\&\quad =a\left( \mu ^{p^eq^i}\alpha ^{p^eq^s}b_0^{q^t}+\mu ^{p^eq^{(r-1)m+i}}\alpha ^{p^eq^{(r-1)m+s}}b_m^{q^{(r-1)m+t}}\right. \\&\qquad \left. +\cdots +\mu ^{p^eq^{m+i}}\alpha ^{p^eq^{m+s}}b_{(r-1)m}^{q^{m+t}}\right) \end{aligned}$$

for all $\alpha \in {\mathbb {F}}_{q^n}$. This yields $s=jm+u$ for some $j\in \{0,\ldots , r-1\}$ giving $a\mu ^{p^eq^{(r-h)m+i}} b_{hm}^{q^{(r-h)m+t}} = a^{q^u} \nu b_{(h-j)m}^{q^{(j-h)m+u}}$, for $0\le h\le r-1$. Straightforward calculations show that the latter conditions imply that $\mathcal H_{m,n,t,\mu ,s}$ is equivalent to $\mathcal H_{m,n,t,\nu ,u}$.

$\square$