Orbit Structure of Grassmannian G2,m and a Decoder for Grassmann Code C(2, m)

Abstract

In this article, we consider decoding Grassmann codes, linear codes associated to the Grassmannian and its embedding in a projective space. We look at the orbit structure of Grassmannian arising from the multiplicative group ${\mathbb{F}}_{q^m}^{*}$ in $G{L}_m\left(q\right)$. We project the corresponding Grassmann code onto these orbits to obtain a subcode of a $q$–ary Reed-Solomon code. We prove that some of these projections contain an information set of the parent Grassmann code. By improving the decoding capacity of Peterson’s decoding algorithm for the projected subcodes, we prove that one can correct up to $\lfloor (d-1)/2\rfloor$ errors for Grassmann code, where $d$ is the minimum distance of Grassmann code.

I. Introduction

Let $q$ be a prime power and ${\mathbb{F}}_q$ be a finite field with $q$ elements. Let $\ell$ and $m$ be positive integers satisfying $\ell \le m$. The Grassmann variety $G_{\ell ,m}$ is an algebraic variety over ${\mathbb{F}}_q$ whose points are $\ell$-dimensional subspaces of an $m$ dimensional vector space $V$ over ${\mathbb{F}}_q$. It is customary to assume $V={\mathbb{F}}_q^m$ although $V$ can be any ${\mathbb{F}}_q$-linear vector space of dimension $m$. To every such algebraic variety, one can associate a natural linear code, thinking the variety as a projective system [20]. The linear code associated to the Grassmanian $G_{\ell ,m}$, in this way, is known as the Grassmann code and is denoted by $C(\ell ,m)$. Ryan [16], [17] initiated the study of Grassmann codes $C(\ell ,m)$ over the binary field. Later, Nogin [14] continued the study of Grassmann code $C(\ell ,m)$ over a general finite field. They proved that the Grassmann code $C(\ell ,m)$ is an ${\left[n,k,d\right]}_q$ code where

$$n={\left[\begin{array}{c}m\\ \ell \end{array}\right]}_q,k=\left(\begin{array}{c}m\\ \ell \end{array}\right),\text{and}d=q^{\ell (m-\ell )}.$$ (1)

Here ${\left[\begin{array}{c}m\\ \ell \end{array}\right]}_q$ is the Gaussian binomial coefficient given by

$${\left[\begin{array}{c}m\\ \ell \end{array}\right]}_q:=\frac{\left(q^m-1\right)\left(q^{m-1}-1\right)\cdots \left(q^{m-\ell +1}-1\right)}{\left(q^{\ell }-1\right)\left(q^{\ell -1}-1\right)\cdots (q-1)}.$$

Mathematicians have been studying different aspects of Grassmann codes since they were discovered. For example, the weight spectrum of Grassmann codes $C\left(2,m\right)$, $C\left(3,6\right)$ was computed by Nogin [14], [15]. Kaipa-Pillai [11] computed the weight spectrum of the code. $C\left(3,7\right)$. Some of the initial and terminal generalized Hamming weights of $C(\ell ,m)$ are also known [4], [6], [14]. The automorphism group of $C(\ell ,m)$ is quite large and fully determined [5]. Further, the structure of the minimum weight codewords of the dual Grassmann code $C(\ell ,m{)}^{\perp }$ is well understood [1]. There is was proved that the support of any minimum weight codeword of $C(\ell ,m{)}^{\perp }$ are three points of a line of the Grassmannian and conversely, any three points of a line of the Grassmannian $G_{\ell ,m}$ is the support a minimum weight codeword of $C(\ell ,m{)}^{\perp }$.

Apart from Grassmann codes, Schubert codes, the codes associated to Schubert varieties in Grassmannians, has been a topic of equal interest among researchers. The study of Schubert codes was initiated by Ghorpade-Lachaud [4] and a conjecture about the minimum distance of Schubert codes was proposed. The conjecture is known as the MDC for Schubert codes. After several attempts and proving the MDC in many different cases [3], [9], [8], the MDC was settled in affirmative sense [21], [7]. Different aspects of Grassmann and Schubert codes have been studied by several mathematicians but the the decoding problem for these codes have not been explored in much detail. So far, no effective decoding algorithm for Grassmann code or Schubert code is known. In a recent work, the second named author together with P. Beelen proposed a decoding algorithm for Grassmann codes [2]. In this article, paths in Grassmannians were used to construct certain parity checks for Grassmann codes and a majority logic decoder was proposed using these parity checks. But unfortunately, the proposed majority voting algorithm is not effective, since the proposed decoder can correct approximately $d/2^{\ell +1}$ errors. In other words, even in the simplest cases, the decoder could not correct up to $\lfloor (d-1)/2\rfloor$ errors. Further, the second named author extended the majority voting decoder to Schubert codes corresponding to Schubert varieties in Grassmannian $G_{2,m}$ [18]. Interestingly, in some cases, the proposed decoder for Schubert code is effective but in most of the cases it is not. Therefore, the problem of proposing an effective decoder for Grassmann and Schubert code is still open, even in the simplest case such as codes associated to Grassmannian $G_{2,m}$ and Schubert varieties in $G_{2,m}$.

In this article, we study the decoding problem for Grassmann code $C\left(2,m\right)$ We consider the action of the cyclic group ${\mathbb{F}}_{{\mathrm{q}}^{\mathrm{*}}}^{\mathrm{*}}$ onto $G_{2,m}$, thinking $G_{2,m}$ as ordered pair of elements in ${\mathbb{F}}_{q^m}$, and study the orbits of this action. We see that the projection of Grassmann code $C\left(2,m\right)$ onto these orbits can be thought of as a subcode of certain Reed-Solomon code. Moreover, most of these projected subcodes contains an information set of $C\left(2,m\right)$. We use such subcodes and use Peterson’s decoding algorithm to correct the errors. As a consequence, we are able to correct $\lfloor d-1/2\rfloor$ errors for the Grassmann code $C\left(2,m\right)$ where d is the minimum distance of the code.

II. Preliminaries

In this section, we recall the definition of Grassmann variety and the construction of Grassmann codes. As earlier, positive integers $\ell ,m$ satisfying $\ell \le m$ and a finite field ${\mathbb{F}}_q$ with $q$ elements are fixed throughout the article. Define the set $\mathbb{I}(\ell ,m)$ by

$$\mathbb{I}(\ell ,m):=\left\{\left({\alpha }_1,{\alpha }_2,\dots ,{\alpha }_{\ell }\right):1\le {\alpha }_1<{\alpha }_2<\cdots <{\alpha }_{\ell }\le m\right\}$$

and fix a linear order on $I(\ell ,m)$. Let $V$ be an $m$-dimensional vector space over ${\mathbb{F}}_q$. The Grassmannian $G_{\ell }\left(V\right)$ of all $\ell$-planes of vector space $V$ is defined by

$$G_{\ell }\left(V\right):=\{P:P\subset V\mathrm{is}\mathrm{a}\mathrm{linear}\mathrm{subspace}\mathrm{and}\dim P=\ell \}.$$

The Grassmannian $G_{\ell }\left(V\right)$ can be embedded into a projective space via the Plucker map. More precisely, fix an ordered basis $ℬ$ of $V$. For $P\in G_{\ell }\left(V\right)$, let $M_P$ be an $\ell \times m$ matrix whose rows are coordinates of some basis of $P$ with respect to $ℬ$. The Plucker map $\mathrm{P}\mathrm{l}$ is

$$\text{Pl}:G_{\ell }(V)\to {\mathbb{P}}^{\left(\begin{array}{c}m\\ \ell \end{array}\right)-1};P\mapsto {\left[{\left(M_P\right)}_{\alpha }\right]}_{\alpha \in \mathbb{I}(\ell ,m)}$$ (2)

where ${\left(M_P\right)}_{\alpha }$ denotes the $\ell \times \ell$ minor of $M_P$ corresponding to columns of $M_P$ indexed by tuple $\alpha$. The image of the Plucker map is given by the zero set of a bunch of quadratic polynomials and hence defined a projective algebraic variety, known as Grassmann variety. For a detail study on Grassmann varieties we refer to [10], [12]. It is known that if $V$ and $V^{\prime }$ are two vector spaces of dimension $m$ over ${\mathbb{F}}_q$, then there exists an automorphism of ${\mathbb{P}}^{\left(\begin{array}{c}m\\ \ell \end{array}\right)-1}$ mapping $G_{\ell }\left(V\right)$ to $G_{\ell }\left(V^{\prime }\right)$. Therefore, for the rest of the article, we denote by $G_{\ell ,m}$, the Grassmannian of all $\ell$-planes of ${\mathbb{F}}_q^m$. To every projective algebraic variety, one can associate a linear code using the language of projective systems [20, Ch.1]. More precisely, each nondegenerate subset $𝒫$ of a projective space ${\mathbb{P}}^{k-1}$ over ${\mathbb{F}}_q$ corresponds to a unique linear code. Further, the minimum distance and the generalized Hamming weights of the corresponding code can be studied from hyperplane sections of $𝒫$ with linear subspaces of ${\mathbb{P}}^{k-1}$. Therefore, there is a linear code which corresponds to the Grassmannian $G_{\ell ,m}$. The code associated to $G_{\ell ,m}$ in this way is known as the Grassmann code and is denoted by $C(\ell ,m)$. To get into more details, we would like to recall the construction of Grassmann code.

Let $\underset{\_}{\mathbf{X}}=\left(X_{ij}\right)$ be an $\ell \times m$ matrix of $\ell m$ indeterminates $X_{ij}$ over ${\mathbb{F}}_q$. For every $\alpha \in \mathbb{I}(\ell ,m)$, let ${\mathbf{X}}_{\alpha }$ denote the $\ell \times \ell$ minor of $\mathbf{X}$ corresponding to columns labeled by $\alpha$. Let ${\mathbb{F}}_q{\left[{\mathbf{X}}_{\alpha }\right]}_{\alpha \in \mathbb{I}(\ell ,m)}$ be the ${\mathbb{F}}_q$ linear space spanned by all minors ${\mathbf{X}}_{\alpha }$. For each $P\in G_{\ell ,m}$, let $M_P$ be a matrix corresponding to point $P$ as in equation (2), and represent the Grassmannian $G_{\ell ,\mathrm{m}}$ as $G_{\ell ,\mathrm{m}}\left(\mathbb{A}\right)=\left\{M_{P_1},\dots ,M_{P_n}\right\}$ be a set of such matrices corresponding to each point $P_t\in G_{\ell ,m}$ in some fixed order, where $n=\left|G_{\ell ,m}\right|$. Consider the evaluation map

$$\begin{gathered} \mathrm{Ev}:{\mathbb{F}}_q{\left[{\mathbf{X}}_{\alpha }\right]}_{\alpha \in \mathbb{I}(\ell ,m)}⟶{\mathbb{F}}_q^n \\ f\left({\mathbf{X}}_{\alpha }\right)\mapsto \left(f\left(M_{P_1}\right),\dots ,f\left(M_{P_n}\right)\right). \end{gathered}$$

The image of the evaluation map Ev is known as the Grassmann code and is denoted by $C(\ell ,m)$. The Grassmann code $C(\ell ,m)$ is an $[n,k,d{]}_q$ linear code where $n,k$ and $d$ are given by equation (1). Clearly, the codewords of Grassmann codes $C(\ell ,m)$ are indexed by points of $G_{\ell ,m}$. Therefore, we may use points $P_1,\dots ,P_n\in G_{\ell ,m}$ as an indexing set for coordinates of codewords in $C(\ell ,m)$.

In this article, we only study the Grassmann code $C(2,m)$. We write a $2\times m$ generic matrix as

$$\underset{\_}{\mathbf{X}}=\left[\begin{array}{cccc}{X}_1& X_2& \cdots & X_m\\ Y_1& Y_2& \cdots & Y_m\end{array}\right]$$

and write the first row of the indeterminate matrix $\underset{\_}{\mathbf{X}}$ as $\mathbf{X}$ and the second row of $\underset{\_}{\mathbf{X}}$ as $\mathbf{Y}$. In the next section, we will study the orbit structure of Grassmannian $G_{2,m}$ under the natural action induced by the cyclic group ${\mathbb{F}}_{q^m}^{\mathrm{*}}$ but before we get into the orbit structure, we recall the definition of the trace function of field extensions.

Let ${\mathbb{F}}_{q^m}$ be the field extension of ${\mathbb{F}}_q$ of degree $m$.

Definition 1:

The trace function of ${\mathbb{F}}_{q^m}$ over ${\mathbb{F}}_q$ is defined and denoted by

$$\mathrm{Tr}\left(x\right)=x+x^q+x^{q^2}+\cdots +x^{q^{m-1}}.$$

Note that ${\mathbb{F}}_{q^m}$ is an $m$ dimensional vector space over ${\mathbb{F}}_q$ and Tr is an ${\mathbb{F}}_q$ linear map. Trace functions are play a key role in our decoder.

III. Orbit Structure of Grassmannian ${\mathit{G}}_{\mathbf{2},\mathit{m}}$

In this section, we study the natural action of the cyclic group of ${\mathbb{F}}_{q^m}^{\mathrm{*}}$ on Grassmannian $G_{2,\mathrm{m}}$. Our goal is to understand the orbits of $G_{2,m}$ under this action and the behavior of the projection of the code $C(2,m)$ onto these orbits. Before going into further details, we shall fix some notations that we will be using throughout the article. As fixed in the last section, let ${\mathbb{F}}_{q^m}$ be the field extension of ${\mathbb{F}}_q$ of degree $m$. We know that ${\mathbb{F}}_{q^m}$ is an $m$-dimensional ${\mathbb{F}}_q$-vector space and hence ${\mathbb{F}}_q^m$ and ${\mathbb{F}}_{q^m}$ are isomorphic as an ${\mathbb{F}}_q$ space. We fix an isomorphism between ${\mathbb{F}}_q^m$ and ${\mathbb{F}}_{q^m}$. Keeping this in mind, we may think of $G_{2,m}$ as a subset of ${\mathbb{F}}_{q^m}^{\mathrm{*}}\times {\mathbb{F}}_{q^m}^{\mathrm{*}}$ consisting of tuples $⟨\alpha ,\beta {⟩}_{{\mathbb{F}}_q}$, where $\alpha ,\beta \in {\mathbb{F}}_{q^m}^{\mathrm{*}}$ span a two dimensional subspace over ${\mathbb{F}}_q$. In other words, we think ${\mathbb{F}}_{q^m}^{\mathrm{*}}\times {\mathbb{R}}_{q^m}^{\mathrm{*}}$ as the set of $2\times m$ matrices over ${\mathbb{F}}_q$ of rank one and two and $G_{2,m}$ is, up to equivalence, the set of those matrices that are of rank two. For $\alpha \in {\mathbb{F}}_{q^m}$, we denote by $\left({\alpha }_1,\dots ,{\alpha }_m\right)\in {\mathbb{F}}_q^m$, the coordinates of $\alpha$ and vice-versa. Furthermore, we treat the subspace span by tuple $⟨\alpha ,\beta ⟩$ same as the point of $G_{2,m}$ spanned by coordinates $\left({\alpha }_1,\dots ,{\alpha }_m\right)$ and $\left({\beta }_1,\dots ,{\beta }_m\right)$. Having said that, we recall the following trivial lemma.

Lemma 2:

For every $\alpha \in {\mathbb{F}}_{q^m}^{*}$, the map given by $X\mapsto {\mathrm{Tr}}_{{\mathbb{F}}_{q^m}/{\mathbb{F}}_q}(\alpha X)$ is a nonzero ${\mathbb{F}}_q$-linear functional of $V={\mathbb{F}}_{q^m}$. Note that, the trace map is ${\mathbb{F}}_q$-linear in both $X$ and $\alpha$. Therefore, for every $X\in {\mathbb{F}}_{q^m}$ and $X_i\in {\mathbb{F}}_q$, there exist some $\alpha \in {\mathbb{F}}_{q^m}$ such that ${Tr}_{{\mathbb{F}}_{q}m/{\mathbb{F}}_q}\left(\alpha X\right)=X_i$. Furthermore, if $X_i\in {\mathbb{F}}_q$ is a coordinate of $X\in {\mathbb{F}}_{q^m}$ (via isomorphism treating it in ${\mathbb{F}}_q^m$), then there exist some $\alpha \in {\mathbb{F}}_{q^m}^{\mathrm{*}}$ such that ${Tr}_{{\mathbb{F}}_{q}m/{\mathbb{F}}_q}\left(\alpha X\right)=X_i$. This plays a very important role. The next lemma is an immediate consequence of Lemma 2.

Lemma 3:

Let $\alpha ,\beta \in {\mathbb{F}}_{q^m}^{\mathrm{*}}$ and let $W={\mathbb{F}}_{q^m}\times {\mathbb{F}}_{q^m}$ be the ${\mathbb{F}}_q$ vector space. Then functions of the form

$$f_{\alpha ,\beta }(X,Y)=\mathrm{T}\mathrm{r}\left(\alpha X\right)\mathrm{T}\mathrm{r}\left(\beta Y\right)-\mathrm{T}\mathrm{r}\left(\alpha Y\right)\mathrm{T}\mathrm{r}\left(\beta X\right)$$

are determinantal functions on $W$ as a vector space over ${\mathbb{F}}_q$.

What we mean is that the function $f_{\alpha ,\beta }(X,Y)$ is an alternating bilinear map on $W$. Therefore, 2×2 minors $X_i{Y}_j-X_j{Y}_i$ of the $2\times m$ matrix $\mathbf{X}$, can be written as linear combination of functions $f_{\alpha ,\beta }(X,Y)$ for $\alpha ,\beta \in {\mathbb{F}}_{q^m}^{\mathrm{*}}$. In other words, one can think of Grassmann code as evaluation functions ${\sum }_{\alpha ,\beta } f_{\alpha ,\beta }(X,Y)$ on Grassmannian $G_{2,m}$ as a subset of ${\mathbb{F}}_{q^m}^{\mathrm{*}}\times {\mathbb{F}}_{q^m}^{\mathrm{*}}$.

Now we are ready to look at the natural group action of ${\mathbb{F}}_{q^m}^{\mathrm{*}}$ on Grassmannian $G_{2,m}$. Let $\gamma$ be a generator of ${\mathbb{F}}_{q^m}^{\mathrm{*}}$. The natural action of ${\mathbb{F}}_{q^m}^{\mathrm{*}}$ on $G_{2,m}$ is defined by

$${\mathbb{F}}_{q^m}^{\mathrm{*}}\times G_{2,m}\to G_{2,m}\mathrm{defined}\mathrm{by}(\gamma ,⟨\alpha ,\beta {⟩}_{{\mathbb{F}}_q})\mapsto ⟨\gamma \alpha ,\gamma \beta {⟩}_{{\mathbb{F}}_q}.$$ (3)

Let ${𝒪}_1,\dots ,{𝒪}_r$ be the orbits of this action. Therefore, if $⟨\alpha ,\beta {⟩}_{{\mathbb{F}}_q}\in {𝒪}_j$ is some element of orbit ${𝒪}_j$ then ${𝒪}_j=\left\{{⟨{\gamma }^i\alpha ,{\gamma }^i\beta ⟩}_{{\mathbb{F}}_q}:1\le i\le q^m-1\right\}$. Since $\gamma$ generates ${\mathbb{F}}_{q^m}^{\mathrm{*}}$ and for any $⟨\alpha ,\beta {⟩}_{{\mathbb{F}}_q}\in G_{2,m}$, we have $\alpha \in {\mathbb{F}}_{q^m}^{\mathrm{*}}$ and hence we may assume that each orbit has one element of the form $⟨1,\delta {⟩}_{{\mathbb{F}}_q}$. We denote the orbit containing $⟨1,\delta {⟩}_{{\mathbb{F}}_q}$ as ${𝒪}_{\delta }$ and we call the element $⟨1,\delta {⟩}_{{\mathbb{F}}_q}$ an orbit representative of ${𝒪}_{\delta }$.

Example 4:

Assume $m=4$ and $q=2$. Let $\gamma$ be such that ${\gamma }^4=\gamma +1$. In this case ${\mathbb{F}}_{16}={\mathbb{F}}_2\left[\gamma \right]$ and $⟨\gamma ⟩={\mathbb{F}}_{16}^{\mathrm{*}}$. Note that if $⟨1,\alpha {⟩}_{F_q}$ is an orbit representative of the orbit $O_{\alpha }$, then $⟨1,\alpha {⟩}_{{\mathbb{F}}_q}=\{\mathrm{0,1},\alpha ,1+\alpha \}$. By remowing 0 we may just write this set as $\{1,\alpha ,\alpha +1\}$. Further, as $\gamma$ is a generator of the field ${\mathbb{F}}_{16}$, we have $\alpha ={\gamma }^i$ and $1+\alpha ={\gamma }^j$ for some $i$ and $j$. As the subspace $⟨1,\alpha {⟩}_{{\mathbb{F}}_q}$ is of dimension 2, we get $\alpha \ne \mathrm{0,1}$. This leaves $16-2=14$ possibilities for $\alpha$. Further, each of those $\alpha$ and $\alpha +1$ generates the same space. Therefore, there are 7 different spaces of the form $⟨1,\alpha ⟩$. These are $\left\{1,{\gamma }^1,{\gamma }^4\right\},$ $\left\{1,{\gamma }^2,{\gamma }^8\right\},$ $\left\{1,{\gamma }^{14},{\gamma }^3\right\},$ $\left\{1,{\gamma }^5,{\gamma }^{10}\right\},\left\{1,{\gamma }^{13},{\gamma }^6\right\},$ $\left\{1,{\gamma }^7,{\gamma }^9\right\}$ and $\left\{1,{\gamma }^{11},{\gamma }^{12}\right\}$. Since the action of $\gamma$ maps $\left\{1,{\gamma }^i,{\gamma }^j\right\}$ to $\left\{\gamma ,{\gamma }^{i+1},{\gamma }^{j+1}\right\}$. A direct computation shows that there are three orbits, namely: The orbits $O_{\gamma },$ $O_{{\gamma }^2}$ and $O_{\gamma 5}$. The orbit $O_{\gamma }$ has 15 elements and contains orbit representatives $\left\{1,\gamma ,{\gamma }^4\right\},$ $\left\{1,{\gamma }^3,{\gamma }^{14}\right\}$ and $\left\{1,{\gamma }^{11},{\gamma }^{12}\right\}$. The orbit $O_{{\gamma }^2}$ also has 15 elements and it contains orbit representatives $\left\{1,{\gamma }^2,{\gamma }^8\right\}\left\{1,{\gamma }^6,{\gamma }^{13}\right\}$ and $\left\{1,{\gamma }^7,{\gamma }^9\right\}$. Finally the orbit $O_{{\gamma }^5}$ has 5 elements and it has only one orbit representative, namely $\left\{1,{\gamma }^5,{\gamma }^{10}\right\}$. Note that this gives in total 35 spaces, i.e. full Grassmannian $G_{\mathrm{2,4}}$.

The Grassmannian $G_{\mathrm{2,5}}$ is represented as elements of ${\mathbb{F}}_{32}$. Here, the field ${\mathbb{F}}_{32}$ has 31 nonzero elements. If $⟨1,\alpha ⟩{\mathbb{F}}_q$ is an orbit representative of the orbit $O_{\alpha }$, then there are 30 choices for $\alpha$. Further, since $\alpha$ and $1+\alpha$ generate the same space, we have only 15 choices of the subspace $⟨1,\alpha {)}_{{\mathbb{F}}_q}$. Furthermore, if $\gamma$ is a generator of ${\mathbb{F}}_{32}^{\mathrm{*}}$, then the action of $\gamma$ does not fix any elements of $G_{\mathrm{2,5}}$ Thus, all orbits will have size 31 and hence there are 5 orbits of size 31. Likewise, for the Grassmannian $G_{\mathrm{2,7}}$ represented as elements of ${\mathbb{F}}_{128}$ has 21 orbits of size 127.

In the next two lemmas we’ll understand why in the case of $m=5$ and $m=7$, the orbits structure of $G_{2,m}$ are quite uniform.

Lemma 5:

Let $P=⟨1,\alpha ⟩{\mathbb{F}}_q$ be a two dimensional ${\mathbb{F}}_q$-linear subspace of ${\mathbb{F}}_{q^m}$. Suppose that $\beta \in {\mathbb{F}}_{q^m}\backslash {\mathbb{F}}_q$. Then $\beta P=P$ if and only if $\alpha \in {\mathbb{F}}_{q^2}$, i.e. $P={\mathbb{F}}_{q^2}$.

Proof:

Let $P=⟨1,\alpha {)}_{{\mathbb{F}}_q}=\left\{a+b\alpha \mid a,b\in {\mathbb{F}}_q\right\}$ be as in the hypothesis and let $\beta \in {\mathbb{F}}_{q^m}\backslash {\mathbb{F}}_q$ such that $\beta P=P$. Then, as $\beta \in \beta P$ and $\beta P=P$, we have $\beta =a+b\alpha$ for some $a,b\in {\mathbb{F}}_q$. Since $\beta \notin {\mathbb{F}}_q$ we have $b\ne 0$. Also, as $1\in \beta P$ there exist $c,d\in {\mathbb{F}}_q$ such that $1=(c+d\alpha )\beta$. Putting the value of $\beta$, we get $1=(c+d\alpha )(a+b\alpha )$. Therefore, $1=bd{\alpha }^2+(ad+bc)\alpha +cd$ and hence $\alpha$ satisfies a polynomial equation over ${\mathbb{F}}_q$ of degree 2. It follows that $\alpha \in {\mathbb{F}}_{q^2}$. For the reverse implication, note that if $\alpha \in {\mathbb{F}}_{q^2}\backslash {\mathbb{F}}_q$, then $P={\mathbb{F}}_{q^2}$ and in this case, clearly $\alpha P=P$. ■

In the next lemma, we consider $G_{2,m}$ when $m$ is odd and count the number of orbits in $G_{2,m}$ under the action defined in equation (3) and compute the size of each orbit.

Lemma 6:

If $m$ is an odd integer, then there are $\frac{q^{m-1}-1}{q^2-1}$ orbits and the size of each orbit is $\frac{q^m-1}{q-1}$.

Proof:

The proof is a simple consequence of the orbit-stabilizer theorem and Lemma 5. Since $m$ is odd, there does not exist any $\alpha \in {\mathbb{F}}_{q^m}\backslash {\mathbb{F}}_q$ such that $\alpha \in {\mathbb{F}}_{q^2}$. Therefore, from lemma 5, we know that for any $P\in G_{2,m}$, the stabilizer of $P$ has size $(q-1)$, namely elements of ${\mathbb{F}}_q^{\mathrm{*}}$. Now from the orbit-stabilizer theorem, we get that the orbit of each of $P$ is of size $\frac{q^m-1}{q-1}$. Further, as $G_{2,m}$ is the disjoint union of orbits of $P\in G_{2,m}$, the number of orbits is $\left|G_{2,m}\right|/\left|O_{\delta }\right|$, where $O_{\delta }$ is an arbitrary orbit. As a result, we get the total number of orbits in $G_{2,m}$ are $\frac{q^{m-1}-1}{q^2-1}$. ■

This lemma justifies the nature of orbits of $G_{2,m}$ in cases $m=5,7$ that we discussed in the Example 4. The next lemma counts the size of orbits and total number of orbits in $G_{2,m}$ when $m$ is even.

Lemma 7:

If $m$ is even, then there are $q\frac{q^{m-2}-1}{q^2-1}$ orbits of size $\frac{q^m-1}{q-1}$ and exactly one orbit of size $\frac{q^m-1}{q^2-1}$.

Proof:

Let $P\in G_{2,m}$ be an arbitrary element of the Grassmannian. If $P={\mathbb{F}}_{q^2}^{\mathrm{*}}$ then we have $\beta P=P$ if and only if $\beta \in {\mathbb{F}}_{q^2}^{\mathrm{*}}$. In other words, the stabilizer of $P$ in this case is of size $q^2-1$ and hence from orbit-stabilizer theorem we get that the orbit of $P$ in this case is of size $\frac{q^m-1}{q^2-1}$. On the other hand, if $P\ne {\mathbb{F}}_{q^2}$, then from lemma 5 we know that in this case we will have $\beta P=P$ if and only if $\beta \in {\mathbb{F}}_q^{\mathrm{*}}$. In other words, in this case the stabilizer of $P$ is of size $q-1$ and hence the orbits of $P$ is of size $\frac{q^m-1}{q-1}$. Now as the cardinality of $G_{2,m}$ is $\frac{\left(q^m-1\right)\left(q^{m-1}-1\right)}{\left(q^2-1\right)(q-1)}$ and if there are $r$ orbits of size $\frac{q^m-1}{q-1}$, then we have

$$\frac{\left(q^m-1\right)\left(q^{m-1}-1\right)}{\left(q^2-1\right)(q-1)}=r\frac{q^m-1}{q-1}+\frac{q^m-1}{q^2-1}.$$

Solving for $r$ gives there are $q\frac{q^{m-2}-1}{q^2-1}$ orbits of size $\frac{q^m-1}{q-1}$. ■

Now we count the number of orbit representatives a particular orbit might have. For convenience we use the notation $⟨1,\alpha ⟩$ to represent ${\mathbb{F}}_q$-subspace spanned by $1,\alpha$.

Lemma 8:

Let $\alpha ,\beta \in {\mathbb{F}}_{q^m}\backslash {\mathbb{F}}_q$. Then ${𝒪}_{\alpha }={𝒪}_{\beta }$ if and only if

$$\beta =\frac{a\alpha +b}{c\alpha +d},a,b,c,d\in {\mathbb{F}}_q$$

where $ad\ne bc$.

Proof:

Assume that $\alpha ,\beta \in {\mathbb{F}}_{q^m}\backslash {\mathbb{F}}_q$. Note that this is equivalent to ${\mathrm{d}\mathrm{i}\mathrm{m}}_{{\mathbb{F}}_q}⟨1,\alpha ⟩={\mathrm{d}\mathrm{i}\mathrm{m}\mathrm{p}}_{{\mathbb{F}}_q}⟨1,\beta ⟩=2$.

Suppose that ${𝒪}_{\alpha }={𝒪}_{\beta }$. Then there exists a nonzero $\gamma \in {\mathbb{F}}_{q^m}$ such that $⟨1,\alpha ⟩=\gamma ·⟨1,\beta ⟩=⟨\gamma ,\gamma \beta ⟩$. Since $\gamma ,\gamma \beta \in ⟨1,\alpha ⟩$, it follows that there exist $a,b\in {\mathbb{F}}_q$ such that $\gamma \beta =a\alpha +b$ and there exist $c,d\in {\mathbb{F}}_q$ such that $\gamma =c\alpha +d$. Therefore $(c\alpha +d)\beta =a\alpha +b$ which implies $\beta =\frac{a\alpha +b}{c\alpha +d}$. Note that if $ad=bc$ then $\beta \in {\mathbb{F}}_q$ which is a contradiction of the hypothesis.

Conversely, assume that $=\frac{a\alpha +b}{c\alpha +d},a,b,c,d\in {\mathbb{F}}_q$. Consider the action of $c\alpha +d\in {\mathbb{F}}_{q^m}$ on $⟨1,\beta ⟩$. Note that if $ad=bc$ then $\beta \in {\mathbb{F}}_q$ which is a contradiction of the hypothesis. Then

$$(c\alpha +d)\cdot ⟨1,\beta ⟩=⟨c\alpha +d,(c\alpha +d)\beta ⟩=⟨c\alpha +d,a\alpha +b⟩=⟨1,\alpha ⟩.$$

This completes the proof of the lemma. ■

Lemma 9:

Suppose that $\alpha \notin {\mathbb{F}}_{q^2}$. If

$$\frac{a_1\alpha +b_1}{c_1\alpha +d_1}=\frac{a_2\alpha +b_2}{c_2\alpha +d_2}$$

where $a_1{d}_1\ne b_1{d}_1,a_2{d}_2\ne b_2{d}_2$, then there exists a nonzero $ϵ\in {\mathbb{F}}_q$ such that $a_1=ϵa_2,b_1=ϵb_2,c_1=ϵc_2,d_1=ϵd_2$.

Proof:

Suppose

$$\frac{a_1\alpha +b_1}{c_1\alpha +d_2}=\frac{a_2\alpha +b_2}{c_2\alpha +d_2}$$

where $a_1{d}_1\ne b_1{c}_1,a_2{d}_2\ne b_2{c}_2$. Then

$$\left(a_1\alpha +b_1\right)\left(c_2\alpha +d_2\right)=\left(a_2\alpha +b_2\right)\left(c_1\alpha +d_1\right).$$

The polynomial

$$R\left(X\right)=\left(a_{1}X+b_1\right)\left(c_{2}X+d_2\right)-\left(a_{2}X+b_2\right)\left(c_{1}X+d_1\right)$$

is a polynomial of degree at most two with $\alpha$ as a root. If $\alpha$ is a root of $R\left(X\right)$ then either $\alpha \in {\mathbb{F}}_{q^2}$ or $R\left(X\right)$ is the zero polynomial. But as $\alpha \notin {\mathbb{F}}_{q^2}$, we get $R\left(X\right)=0$. This implies

$$\left(a_{1}X+b_1\right)\left(c_{2}X+d_2\right)-\left(a_{2}X+b_2\right)\left(c_{1}X+d_1\right)=0.$$

Consequently, either $c_{1}X+d_1$ or $a_{2}X+b_2$ must be a nonzero ${\mathbb{F}}_q$ multiple of $a_{1}X+b_1$. Now, if $c_{1}X+d_1=ϵ\left(a_{1}X+b_1\right)$, then $a_1{d}_1=b_1{c}_1$ which contradicts the hypothesis of the Lemma. Therefore, we get $a_{2}X+b_2=ϵ\left(a_{1}X+b_1\right)$ and hence

$$\left(a_{1}X+b_1\right)\left(c_{2}X+d_2\right)-ϵ\left(a_{1}X+b_1\right)\left(c_{1}X+d_1\right)=0.$$

Thus

$$\left(a_{1}X+b_1\right)\left(c_{2}X+d_2-ϵ\left(c_{1}X+d_1\right)\right)=0$$

which implies $c_{2}X+d_2=ϵ\left(c_{1}X+d_1\right)$. This establishes the proof. ■

Now we are ready to count exact number of orbit representatives of an orbit $O$ of $G_{2,m}$ of the form $⟨1,\alpha ⟩$ provided $\alpha \notin {\mathbb{F}}_{q^2}$.

Lemma 10:

Let ${𝒪}_{\alpha }$ be an orbit under the action of ${\mathbb{F}}_{q^m}^{\mathrm{*}}$ on $G_{2,m}$ and let $\alpha \notin {\mathbb{F}}_{q^2}$. Then there are $q^3-q$ different elements $\beta \in {\mathbb{F}}_{q^m}$ such that $⟨1,\beta ⟩\in {𝒪}_{\alpha }$.

Proof:

We shall count the number of distinct elements of the form $\frac{a\alpha +b}{c\alpha +d}$ where $ad\ne bc$. Lemma 9 implies that $\frac{a\alpha +b}{c\alpha +d}$ is represented by $q-1$ distinct multiples of $(a,b,c,d)$. There are $\left(q^2-1\right)\left(q^2-q\right)$ different quadruples in ${\mathbb{F}}_q^4$ such that $ad\ne bc$ Therefore there are $q^3-q=\frac{\left(q^2-1\right)\left(q^2-q\right)}{q-1}$ distinct multiples defining different $\frac{a\alpha +b}{c\alpha +d}$ quotients. ■

IV. Evaluation of the determinant function on Orbits of ${\mathit{G}}_{\mathbf{2},\mathit{m}}$

Our bound and decoding algorithm of Grassmann codes hinges on the fact that the Grassmann code $C(2,m)$ is a subcode of quadratic forms in $2m$ variables, namely one variable for each entry of the $2\times m$ generic matrix. But in this section we will think this code in a slightly different way. We know that reordering the points of Grassmannian only gives an equivalent code. Therefore, we first fix an order on orbits and then order points in each orbits. We may think $C(2,m)$ as evaluation of determinantal functions on representatives of points in each orbit in fixed order. Also, we have seen that the determinantal functions can be written as a ${\mathbb{F}}_q$-linear combinations of functions $f_{\alpha ,\beta }(X,Y)$ where $\alpha ,\beta \in {\mathbb{F}}_{q^m}$. Recall that, the orbit of ${\langle 1,\delta \rangle }_{{\mathbb{F}}_q}$ is the set ${𝒪}_{\delta }=\left\{{({\gamma }^i,{\gamma }^i\delta }_{{\mathbb{F}}_q}\mid 0\le i\le q^m-1\right\}$ where $\gamma \in {\mathbb{F}}_{q^m}^{\mathrm{*}}$ is a generator of the multiplicative cyclic group ${\mathbb{F}}_{q^m}^{\mathrm{*}}$. Note that, when determinantal functions $f_{\alpha ,\beta }(X,Y)$ is evaluated on an arbitrary point ${⟨{\gamma }^i,{\gamma }^i\delta ⟩}_{{\mathbb{F}}_q}$ of the orbit ${𝒪}_{\delta }$, it gives

$$f_{\alpha ,\beta }\left({\gamma }^d,{\gamma }^1\delta \right)=\mathrm{T}\mathrm{r}\left(\alpha {\gamma }^1\right)\mathrm{T}\mathrm{r}\left(\beta {\gamma }^1\delta \right)-\mathrm{T}\mathrm{r}\left(\alpha {\gamma }^1\delta \right)\mathrm{T}\mathrm{r}\left(\beta {\gamma }^1\right).$$

Since $\alpha ,$ $\beta$ and $\delta$ are fixed, we may think $f_{\alpha ,\beta }(X,Y)$ as polynomial in one variable when it is evaluated on orbit ${𝒪}_{\delta }$. Hence, we consider polynomials

$$f_{\alpha ,\beta }(T,\delta T)=\mathrm{T}\mathrm{r}\left(\alpha T\right)\mathrm{T}\mathrm{r}\left(\beta \delta T\right)-\mathrm{T}\mathrm{r}\left(\alpha \delta T\right)\mathrm{T}\mathrm{r}\left(\beta T\right)$$

where $,$ $\beta \in {\mathbb{F}}_{q^m}$. It implies that the evaluation of determinantal functions $f_{\alpha ,\beta }(X,Y)$ on orbits ${𝒪}_{\delta }$ is also given by the evaluation of polynomials $f_{\alpha ,\beta }(T,\delta T)$ on certain elements of ${\mathbb{F}}_{q^m}^{\mathrm{*}}$. This evaluation also gives a linear code which we denote by $C^{\delta }(2,m)$. We now try to understand the nature of code $C^{\delta }(2,m)$ for the orbit ${𝒪}_{\delta }$.

Let $N=\frac{q^m-1}{q-1}$. We have seen that all orbits (except possibly one) in $G_{2,m}$ are of size $N$. Let ${𝒪}_{\delta }$ be such an orbit and let $P_1,\dots ,P_N$ are representative points of ${𝒪}_{\delta }$. As we discussed, each of these $P_i$ is of the form ${⟨{\gamma }^{j_i},{\gamma }^{j_i}\delta ⟩}_{{\mathbb{F}}_q}$ for some $0\le j_i\le q^m-1$. Once $\delta$ is fixed, we may think the point $P_i$ as ${\gamma }^{j_i}\in {\mathbb{F}}_{q^m}^{*}$. Therefore, for the rest of the draft an orbit ${𝒪}_{\delta }$ of size $N$ and representative points $P_1\dots ,P_N$ as above is fixed. Let $L_{\delta }\left({\mathbb{F}}_q\right)$ be the ${\mathbb{F}}_q$ linear span of the set $\left\{f_{\alpha ,\beta }\left(T,\delta T\right)\left.:\alpha ,\beta \in {\mathbb{F}}_{q^m}\right\}\right.$ and let Ev be the evaluation map obtained by evaluating functions of $L_{\delta }\left({\mathbb{F}}_q\right)$ on points $P_1,\dots ,P_N$. The image of Ev is an ${\mathbb{F}}_q$-linear code. Further, it is not hard to see that

$$\mathrm{Im}\left(\mathrm{E}\mathrm{v}\right)=C^{\delta }\left(2,m\right).$$

Remark 11:

Note that, the determinantal function $f_{\alpha ,\beta }(T,\delta T)$ depends on $\alpha ,\beta \in {\mathbb{F}}_{q^m}$, therefore when we think a function in $L_{\delta }\left({\mathbb{F}}_q\right)$ as a polynomial in $T$, the coefficients are in the field ${\mathbb{F}}_{q^m}$. But since determinantal functions $f_{\alpha ,\beta }(T,\delta T)$ are defined by Trace function, the evaluation of $f_{\alpha ,\beta }(T,\delta \mathrm{\Gamma })$ on points $P_i$ are in the field ${\mathbb{F}}_q$ and hence the code $C^{\delta }(2,m)$ is a code over the field ${\mathbb{F}}_q$.

We have the following trivial lemma.

Lemma 12:

The code $C^{\delta }(2,m)$ is a projection of the code $C(2,m)$ onto the coordinates in the orbit ${𝒪}_{\delta }$.

Proof:

This is a simple consequence of the fact that the code $C(2,m)$ is the evaluation of the space $L_{\delta }\left({\mathrm{F}}_q\right)$ on points of $G_{2,m}$ ■

Lemma 13:

If $\alpha ,\beta \in {\mathbb{R}}_{q^m}$ is such that $f_{\alpha ,\beta }(T,\delta T)$ is a nonzero polynomial. Then $q+1\le \mathrm{d}\mathrm{e}\mathrm{g}\left(f_{\alpha ,\beta }(T,\delta T)\right)\le q^{m-1}+q^{m-2}$.

Proof:

We simply expand the determinantal function $f_{\alpha ,\beta }(T,\delta T)$ using the trace function $\text{Tr}\left(T\right)=\sum _{i=0}^{m-1}{T}^{q^i}$. Note that,

$$\begin{gathered} f_{\alpha ,\beta }(T,\delta T)=\mathrm{Tr}(\alpha T)\mathrm{Tr}(\beta \delta T)-\mathrm{Tr}(\alpha \delta T)\mathrm{Tr}(\beta T) \\ =\left(\sum _{i=0}^{m-1}{\left(\alpha T\right)}^{q^i}\right)\left(\sum _{j=0}^{m-1}{\left(\beta \delta T\right)}^{q^j}\right) \\ -\left(\sum _{i=0}^{m-1}{\left(\beta T\right)}^{q^i}\right)\left(\sum _{j=0}^{m-1}{\left(\alpha \delta T\right)}^{q^j}\right) \\ =\left(\sum _{i=0}^{m-1}\sum _{j=0}^{m-1}{(\alpha T)}^{q^i}{(\beta \delta T)}^{q^j}\right) \\ -\left(\sum _{i=0}^{m-1}\sum _{j=0}^{m-1}{\left(\beta T\right)}^{q^i}{\left(\alpha \delta T\right)}^{q^j}\right) \\ =\sum _{i=0}^{m-1}\sum _{j=0}^{m-1}{\left(\alpha T\right)}^{q^i}{\left(\beta \delta T\right)}^{q^j}-{\left(\beta T\right)}^{q^i}{\left(\alpha \delta T\right)}^{q^j} \\ =\sum _{i=0}^{m-1}\sum _{j=0}^{m-1}\left({\alpha }^{q^i}{\beta }^{q^j}-{\beta }^{q^i}{\alpha }^{q^j}\right){\delta }^{q^j}{T}^{q^i+q^j} \\ =\sum _{i=0}^{m-1}\sum _{j=0,j\ne 1}^{m-1}\left({\alpha }^{q^i}{\beta }^{q^j}-{\beta }^{q^i}{\alpha }^{q^j}\right){\delta }^{q^j}{T}^{q^i+q^j} \\ =\sum _{i=0}^{m-2}\sum _{j\ge i}^{m-1}\left({\alpha }^{q^i}{\beta }^{q^j}-{\beta }^{q^i}{\alpha }^{q^j}\right)\left({\delta }^{q^j}-{\delta }^{q^i}\right)T^{q^i+q^j} \end{gathered}$$

where we used ${\alpha }^{q^i}{\beta }^{q^j}-{\beta }^{q^i}{\alpha }^{q^j}=0$ for $i=j$. Clearly, the degree of this polynomial is at least $q+1$ and at most $q^{m-1}+q^{m-2}$. ■

The next corollary is an immediate consequence of Lemma 13.

Corollary 14:

Suppose that the function $f_{\alpha ,\beta }(T,\delta T)$ is not identically zero over ${\mathbb{F}}_{q^m}^{\mathrm{*}}$. Then $f_{\alpha ,\beta }(T,\delta T)$ can have at most $q^{m-1}+q^{m-2}-q-1$ many zeros in ${\mathbb{F}}_{q^m}^{\mathrm{*}}\cdot$

Note that the polynomial $f_{\alpha ,\beta }(T,\delta T)$ is divisible by $T^{q+1}$. Next, we determine the dimension of the code $C^{\delta }(2,m)$. To do so, we first determine a spanning set for the vector space $L_{\delta }\left({\mathbb{F}}_q\right)$. We determine it in the next two lemmas.

Lemma 15:

Let $\delta$ be a fixed nonzero element of ${\mathbb{F}}_{q^m}$. Suppose that $\delta$ is contained in the field ${\mathbb{F}}_{q^d}$ where $d\mid m,d>1$. Then

$$L_{\delta }\left({\mathbb{F}}_q\right)\subseteq ⟨\left\{\left({\delta }^{q^j}-{\delta }^{q^i}\right)T^{q^i+q^j},0\le i<j\le m-1,dVj-i\right\}⟩{\mathbb{F}}_{q^m}$$

Proof:

It is enough to prove that for each $\alpha ,\beta \in {\mathbb{F}}_{q^m}$, the determinantal function $f_{\alpha ,\beta }(T,\delta T)$ can be written as a ${\mathbb{F}}_{q^m}$-linear combination of monomials in the set

$$\left\{\left({\delta }^{q^j}-{\delta }^{q^i}\right)T^{q^i+q^j},0\le i<j\le m-1,d\nmid (j-i)\right\}.$$

Lemma 13 states that

$$f_{\alpha ,\beta }(T,\delta T)=\sum _{i=0}^{m-2} \sum _{j\ge i}^{m-1} \left({\alpha }^{q^i}{\beta }^{q^j}-{\beta }^{q^i}{\alpha }^{q^j}\right)\left({\delta }^{q^j}-{\delta }^{q^i}\right)T^{q^i+q^j}.$$

The condition on $\delta$ implies that ${\delta }^{q^d}=\delta$. For any term of the form $X^{q^i+q^j}$ where $j=nd+i$, we obtain that the

$$\delta q^j-\delta q^i=\delta q^{nd+i}-{\delta }^{q^i}={\left({\delta }^{q^{nd}}\right)}^{q^i}-{\delta }^{q^i}={\delta }^{q^i}-{\delta }^i=0.$$

Therefore, the expansion of $f_{\alpha ,\beta }(T,\delta T)$ has no terms of the form $T^{q^i+q^j}$ where $d\mid j-i$. ■

In the next lemma we prove that both the spaces discussed in the last lemma are the same.

Lemma 16:

Let $\delta$ be a fixed nonzero element of ${\mathbb{F}}_{q^m}$. Suppose that $\delta$ is contained in the field ${\mathbb{F}}_{q^d}$ where $d\mid m,d>1$. Then

$$L_{\delta }\left({\mathbb{F}}_q\right)={⟨\left\{\left({\delta }^{q^j}-{\delta }^{q^i}\right)T{q}^i+q^j,0\le i<j\le m-1,d\nmid (j-i)\right\}⟩}_{{\mathbb{F}}_{q^m}}$$

Proof:

In the view of Lemma 15, we only have to show that for every $0\le s<r\le m-1$, satisfying $dVr-s$, there exist $\alpha ,\beta \in {\mathbb{F}}_{q^m}$ such that $\left({\delta }^{q^r}-{\delta }^{q^s}\right)T^{q^s+q^r}$ can be written as an ${\mathbb{F}}_q$-linear combination of some $f_{\alpha ,\beta }(T,\delta T)$. Let $\gamma ,{\gamma }^q,\dots ,{\gamma }^{q^{m-1}}$ be a normal basis for ${\mathbb{F}}_{q^m}$ over ${\mathbb{F}}_q$. This implies that the matrix given by ${\left({\gamma }^{q^{i+j}}\right)}_{0\le i,j\le m-1}$ nonsingular. Thus for any $0\le s\le m-1$ there exists $b_{s,1},b_{s,2},\dots ,b_{s,m-1}\in {\mathbb{F}}_q$ such that

$$\sum _{i=0}^{m-1} b_{s,i}\left({\gamma }^{q^i},{\gamma }^{q^{i+1}},\dots ,{\gamma }^{q^{i+m-1}}\right)=e_s$$

where $e_s$ is the standard basis vector with a 1 in $s^{\text{th}}$ position and zeroes everywhere else. As the vector $\left({\gamma }^{q^i},{\gamma }^{q^{i+1}},\dots ,{\gamma }^{q^{i-1}}\right)$ is the coefficient vector for $\mathrm{T}\mathrm{r}\left({\gamma }^{q^i}T\right)$, (omitting monomials not of the form $T^{q^j}$ ), taking dot product of both sides with vector $\left(T,T^q,\dots ,T^{q^{m-1}}\right)$, we get

$$\sum _{i=0}^{m-1}{b}_{s,i}\text{Tr}\left({\gamma }^{q^i}T\right)=T^{q^s}.$$ (4)

Thus, for each fixed $\beta \in {\mathbb{F}}_{q^m}$, we have

$$\begin{gathered} \sum _{i=0}^{m-1} b_{s,i}{f}_{{\gamma }^{q^i},\beta }(T,\delta \beta T) \\ =\sum _{i=0}^{m-1} b_{s,i}\left(\mathrm{T}\mathrm{r}\left({\gamma }^{q^i}T\right)\mathrm{T}\mathrm{r}\left(\delta \beta T\right)-\mathrm{T}\mathrm{r}\left({\gamma }^{q^i}\delta T\right)\mathrm{T}\mathrm{r}\left(\beta T\right)\right) \\ =\sum _{i=0}^{m-1} b_{s,i}\mathrm{T}\mathrm{r}\left({\gamma }^{q^i}T\right)\mathrm{T}\mathrm{r}\left(\delta \beta T\right)-\sum _{i=0}^{m-1} b_{s,i}\mathrm{T}\mathrm{r}\left({\gamma }^{q^i}\delta T\right)\mathrm{T}\mathrm{r}\left(\beta T\right) \\ =T^{q^s}\mathrm{T}\mathrm{r}\left(\delta \beta T\right)-(\delta T{)}^{q^s}\mathrm{T}\mathrm{r}(\beta T). \end{gathered}$$

Now, taking $b_{r,1},b_{r,2},\dots ,b_{r,m-1}$ as in equation (4) and consider the linear combination

$$\begin{gathered} \sum _{i=0}^{m-1} b_{r,i}\left(T^{q^s}\mathrm{T}\mathrm{r}\left(\delta {\gamma }^{q^i}T\right)-(\delta T{)}^{q^s}\mathrm{T}\mathrm{r}\left({\gamma }^{q^i}T\right)\right) \\ =\sum _{i=0}^{m-1} b_{r,i}{T}^{q^s}\mathrm{T}\mathrm{r}\left(\delta {\gamma }^{q^i}T\right)-\sum _{i=0}^{m-1} b_{r,i}(\delta T{)}^{q^s}\mathrm{T}\mathrm{r}\left({\gamma }^{q^i}T\right) \\ =T^{q^s}\sum _{i=0}^{m-1} b_{r,i}\mathrm{T}\mathrm{r}\left(\delta {\gamma }^{q^i}T\right)-(\delta T{)}^{q^s}\sum _{i=0}^{m-1} b_{r,i}\mathrm{T}\mathrm{r}\left({\gamma }^{q^i}T\right) \\ =T^{q^s}(\delta T{)}^{q^r}-(\delta T{)}^{q^s}{T}^{q^r} \\ =\left({\delta }^{q^r}-{\delta }^{q^s}\right)T^{q^s+q^r}. \end{gathered}$$

In other words, for any $0\le r<s\le m-1$ with $d\nmid r-s$, monomials $\left({\delta }^{q^r}-{\delta }^{q^s}\right)T^{q^s+q^r}$ can be written as linear combination of $f_{\alpha ,\beta }(T,\delta T)$ for some $\alpha ,\beta \in {\mathbb{F}}_{q^m}$. This completes the proof of the Lemma. ■

We have now found a simple basis for the space $L_{\delta }\left({\mathbb{F}}_q\right)$. Note that the evaluation map Ev is an injective map therefore, the code $C^{\delta }(2,m)$ and the space $L_{\delta }\left({\mathbb{F}}_q\right)$ are of same dimension. We have the following corollary.

Corollary 17:

Let $\delta \in {\mathbb{F}}_{q^m}^{*}$. Let $1<d\le m$ be a positive integer such that $d\mid m$ and $\delta \in {\mathbb{P}}_{q^d}$. Then the code $C^{\delta }(2,m)$ has dimension $\left(\begin{array}{c}m\\ 2\end{array}\right)-\left(\begin{array}{c}\frac{m}{d}\\ 2\end{array}\right)d$. Consequently, if $m$ is prime, then $\mathrm{d}\mathrm{i}\mathrm{m}{C}^{\delta }(2,m)=\mathrm{d}\mathrm{i}\mathrm{m}C(2,m)$.

Proof:

The kernel of evaluation map ${\mathrm{Ev}}_L:{\mathbb{F}}_{q^m}\left[T\right]\to {\mathbb{F}}_{q^m}^N$ is an ideal of the form $⟨\mathrm{\Pi }\gamma \in L(T-\gamma )⟩$. The polynomial $\mathrm{\Pi }\gamma \in L(T-\gamma )$ has degree $N$. Therefore the evaluation map on ${\mathrm{E}\mathrm{v}}_L:{\mathbb{F}}_{q^m}\left[T\right]\to {\mathbb{F}}_{q^m}^N$. As the monomials in $L_{\delta }\left({\mathbb{F}}_q\right)$ all have degrees strictly smaller than $N$, it follows that the evaluation map ${\mathrm{E}\mathrm{v}:L}_{\delta }\left({\mathbb{F}}_q\right)\to {\mathbb{F}}_{q^m}^N$ is injective. Therefore,

$$\begin{gathered} {\dim }_{{\mathbb{F}}_q}{C}^{\delta }(2,m)={\mathrm{d}\mathrm{i}\mathrm{m}}_{{\mathbb{F}}_q}{L}_{\delta }\left({\mathbb{F}}_q\right)={\mathrm{d}\mathrm{i}\mathrm{m}}_{{\mathbb{F}}_{q^m}}{L}_{\delta }\left({\mathbb{F}}_q\right) \\ =\left(\begin{array}{c}m\\ 2\end{array}\right)-\left(\begin{array}{c}\frac{m}{d}\\ 2\end{array}\right)d. \end{gathered}$$

■

V. Decoding Grassmann code $\mathit{C}(\mathbf{2},\mathit{m})$

In this section, we propose a decoding algorithm for the Grassmann code $C(2,m)$. The minimum distance of $C(2,m)$ is $d=q^{2(m-2)}$. Ideally, one strives for a decoding algorithm capable of correcting up to $\lfloor (d-1)/2\rfloor$ errors. In a recent joint work [2], the second named author proposed a decoding algorithm for Grassmann codes $C(\ell ,m)$ but unfortunately, the proposed algorithms decoding capability is far from the optimal decoding capability of the code $C(\ell ,m)$. For example, asymptotically, the proposed decoder can correct only up to $⌊(d-1)/2^{\ell +1}⌋$ errors for $C(\ell ,m)$. In other words, for $C(2,m)$ the proposed algorithm can asymptotically correct around $\lfloor (d-1)/8\rfloor$ errors. Here we propose a list decoder for Grassmann code $C(2,m)$ that can correct up to $\lfloor (d-1)/2\rfloor$ errors for $C(2,m)$.

We propose an algorithm in this section that can comect up to $\lfloor (d-1)/2\rfloor$ errors for $C(2,m)$. We have seen that for most of the orbits ${𝒪}_{\delta }$ in $G_{2,m}$, codes $C^{\delta }(2,m)$ are of same dimension as the Grassmann code $C(2,m)$. Therefore, such $C^{\delta }(2,m)$ contains an information set of $C(2,m)$. Also, from Lemma 13 we know that the space $L_{\delta }\left({\mathbb{R}}_q\right)$ contains polynomials of degree bounded between $q+1$ and $q^{m-1}+q^{m-2}$ and the points of orbits ${𝒪}_{\delta }$ can be thought of as a subset of ${\mathbb{F}}_{q^m}^{\mathrm{*}}$, therefore we may think the code $C^{\delta }(2,m)$ as a subcode of a $\left[N,q^{m-1}+q^{m-2}-q\right]$ - RS code, where $N=\left|{𝒪}_{\delta }\right|$. Our decoding algorithm is based on these projections of $C(2,m)$. Our decoder uses Peterson’s decoding algorithm to obtain an information set of $C(2,m)$ and with the information set obtain the correct codeword of $C(2,m)$. But before we dive into details, we recall the following well known result for Reed-Solomon codes [13].

Proposition 18 (Petersen’s decoding algorithm):

An $[n,k,d=n-k+1]$ Reed–Solomon code can decode $⌊\frac{n-k+1}{2}⌋$ errors with complexity $𝒪\left(n^3\right)$.

The following remark could be useful in understanding the technical lemmas that will be given later in this section.

Remark 19:

As earlier, let ${𝒪}_{\delta }$ be an orbit in $G_{2,m}$ with cardinality $\left|{𝒪}_{\delta }\right|=N$, where $N=\frac{q^m-1}{q-1}$. We have seen that we may think the points of ${𝒪}_{\delta }$ as ${\gamma }^i$ for some $i$, as ${\gamma }^i$ represents point ${\langle {\gamma }^i,{\gamma }^i\delta \rangle }_{{\mathbb{F}}_q}$ of ${𝒪}_{\delta }$. Under this identification, ${𝒪}_{\delta }=\left\{1,\gamma ,\dots ,{\gamma }^{N-1}\right\}$. Without loss of generality we may assume that the coordinates of $C^{\delta }(2,m)$ are indexed on the set ${𝒪}_{\delta }$ in this fixed order. The restriction of $C(2,m)$ onto ${𝒪}_{\delta }$ is the code obtained by evaluating the functions $T^{q+1},T^{q^2+1},T^{q^2+q},\dots ,T^{q^{m-1}+q^{m-2}}$ on ${𝒪}_{\delta }$. We can decode it as a codeword of the Reed–Solomon code obtained by evaluating the monomials $T^j,q+1\le j\le q^{m-1}+q^{m-2}$ on ${𝒪}_{\delta }$. However, the sparsity of the monomials $T^{q^i+q^j},0\le i<j\le m-1$ is useful in for decoding. Note that if $f$ is spanned by the monomials $T^{q^i}+q^{\mathrm{j}}$, then

$$f\left({\gamma }^{jN}T\right)={\gamma }^{2jN}f\left(T\right)$$ (5)

Since ${\mathbb{F}}_{q^m}^{\mathrm{*}}={𝒪}_{\delta }\cup {\gamma }^N{𝒪}_{\delta }\cup {\gamma }^{2N}{𝒪}_{\delta }\cdots \cup {\gamma }^{(q-2)N}{𝒪}_{\delta }$ we can decode $\mathrm{E}\mathrm{v}\left(f\right)$ on ${\mathrm{F}}_{q^m}^{\mathrm{*}}$ instead by extending the values from ${𝒪}_{\delta }$ to ${\mathbb{F}}_{q^m}^{\mathrm{*}}$. For example, let $N=\left|{𝒪}_s\right|$ and $r\in {\mathbb{F}}_q^N$ be the received word. We extend this word to $r_{\text{ext}}\in {\mathbb{F}}_q^{q^m}-1$ as follows:

$$r_{ext}=(r\left|{\gamma }^{2N}r\right|{\gamma }^{4N}r|\cdots |{\gamma }^{(2(q-2))N}r).$$

Note that we used equation (5) to extend $r$ to $r_{\text{ext}}$. The word $r_{\text{ext}}$ is comprised of $q-1$ multiples of $r$. The errors positions in $r$ are replicated in each of the $q-1$ copies of $r$. Identifying in this way, we may change the problem of decoding $r$ as a codeword of $C^{\delta }(2,m)$, to a problem of decoding $r_{ext}$ as a codeword of Reed–Solomon code generated by evaluating the same monomials $T^{q^i+q^j},0\le i<j\le m-1$ over ${\mathbb{F}}_{q^m}^{\mathrm{*}}$. This certainly increases our decoding capacity.

Definition 20:

Let $C$ be a linear code. We say a procedure is a list decoder of size l correcting $\tau$ errors if given a received word $r\in {\mathbb{F}}_q^n$ the algorithm outputs a list $L$ of codewords such that $L$ contains all codewords at distance $\tau$ or less from $r$.

The classical Guruswami–Sudan algorithm list decoder can correct more than $\frac{n-k+1}{2}$ errors for a $[n,k,d]$ Reed–Solomon code. It can correct up to $n-1-\sqrt{n(k-1)}$ errors with a polynomial list size. We propose a slightly different list decoder. Instead of performing polynomial interpolation to find codewords close to the received word, we decode all possible values of the highest degree term. For example: take the [15, 5, 11] Reed–Solomon code obtained by evaluating the polynomials of degree 4 or less on the 15 nonzero points of ${\mathbb{F}}_{16}$. The code can correct 5 errors. That is, given $r=c+e$ where $c=\mathrm{E}\mathrm{v}\left(f\right)$, $deg\left(f\right)\le 4$ and $wt\left(e\right)\le 5$ we can determine $f\left(X\right)$ such that $\mathrm{E}\mathrm{v}\left(f\right)=c$. We are going to decode a subcode of the [15, 5, 11] Reed–Solomon code, $C_{\text{sub}}$. This subcode is the code obtained by the evaluations of $1,X,X^2,X^4$. The monomial $X^3$ is not part of $C_{\text{sub}}$. Since $C_{sub}$ is a subcode of a [15, 5, 11] code we can correct 5 errors. We propose a slightly expensive way of correcting 6 errors. Suppose that the evaluation of $f\left(X\right)=1+X^2+X^4$ was sent and 6 errors occurred. We propose that to decode 6 errors, instead of decoding with a classical list decoder, we try to decode the 16 combinations $r-a\mathrm{E}\mathrm{v}\left(X^4\right)$ for $a\in {\mathbb{F}}_{16}$. We know that when $a=1$ the evaluation $c-a\mathrm{E}\mathrm{v}\left(X^4\right)$ corresponds to a codeword of the [15, 3, 13] Reed–Solomon code where we can correct 6 errors instead of 5. Because the monomial $X^3$ is not part of $C_{\text{sub}}$ we know the coefficient corresponding to $X^3$ is zero. To decode in the subcode [15, 3, 13] Reed–Solomon subcode spanned by $1,X$ and $X^2$ we need only to check all 16 values of the coefficient of $X^4$ instead of all 256 values of both the coefficient of $X^4$ and $X^3$. Our code of interest, the code $C^{\delta }(2,m)$ is obtained by evaluating the monomials $T^3,T^5,T^6,T^9,T^{10},T^{12}$. Since we know our codeword is the evaluation of a polynomial with no term of the form $T^{11}$, decoding all possibilities for the coefficient of $T^{12}$ will contain a received word from a polynomial of degree 10 or less. This leads to an improvement of the decoding radius at the expense of decoding each codeword 16 times.

By correcting 6 errors, we mean that if 6 errors or less occur, then we can always find a closest codeword to the received word. It may be counter intuitive to correct 6 errors with a [15, 5, 11] Reed–Solomon code. However, if we compare our received word to all 16⁵ possible codewords, then we surely may be able to find a closest codeword to our received word and decode 6 errors. In our case we need not go as far as checking all possible coefficients. Since the projection $C(2,m{)}^{{𝒪}_{\delta }}$ is a subcode of a Reed–Solomon code generated by the evaluation of a sparse set of monomials, one needs to check only the possibilities for the coefficient of the highest term to obtain a significant improvement of the decoding radius. Having to check all possibilities for all coefficients would lead to an exponential complexity decoder. In our case, we need only to check one, which makes our decoder have polynomial complexity.

In the following algorithm we denote by RSDecoder $(L,k,r)$ the output of decoding the received word $r$ as a received codeword from a Reed–Solomon of dimension $k$ evaluated at $L$. For the next lemma we assume that ${𝒪}_{\delta }$ is an orbit of $G_{2,m}$ with $\left|{𝒪}_{\delta }\right|=N$ and ${𝒪}_{\delta }=\left\{1,\gamma ,\dots ,{\gamma }^{N-1}\right\}$ as in remark 19. We also fix the order.

Algorithm 1.

Input: a received word $r=c+e$ where $c\in C^{\delta }(2,m)$.

ListRS ← [ ].

for $a\in {\mathbb{F}}_{q^m}$ do

$r_a\leftarrow \left(r_i-a{\alpha }_i^{q^{m-1}}+q^{m-2}\right)$

$r_{ext}\leftarrow \left(r_a\left|{\gamma }^{2N}{r}_a\right|{\gamma }^{4N}{r}_a|\cdots |{\gamma }^{2(q-2)N}{r}_a\right)$.

$c^{\prime }\leftarrow RSDecoder\left({\mathbb{F}}_{q^m}^{\mathrm{*}},q^{m-1}+q^{m-3}-q-1,r_{ext}\right)$.

Append $c$ to ${\mathit{List}}_{\mathit{RS}}$

end for

Lemma 21:

We can list–decode up to $\tau =\frac{q^{m-1}-\left(q^{m-1}+q^{m-2}-q-1\right)}{2(q-1)}$ errors for the code $C^{\delta }(2,m)$ with complexity $𝒪\left(q^m{\left(q^{m-1}\right)}^3\right)$ and list size $q^m$.

Proof:

The proof is little technical. Let $c\in C^{\delta }(2,m)$ be a codeword transmitted through a noisy channel and let $r=c+e$ be the received word with error $e$ satisfying ${\mathrm{w}\mathrm{t}}_H\left(e\right)\le \tau$. The codeword $c\in C^{\delta }(2,m)$ means there exist a polynomial

$$f\left(T\right)=\sum _{0\le i<j\le m-1} a_{i,j}{T}^{q^i+q^j}$$

such that $f\left({\gamma }^i\right)=c_{i+1}$, where $c=\left(c_1,\dots ,c_N\right)$. For each $a\in {\mathbb{F}}_{q^m}$, let $s^a\in {\mathbb{F}}_q^N$ be the word defined by $\left(a{\gamma }^i\left(q^{m-1}+q^{m-2}\right)\right)0\le i\le N-1$. Note that $s_i^a$’s are the evaluation of monomial $a{T}^{q^{m-1}+q^{m-2}}$ on ${\gamma }^i$, i.e, on points of ${𝒪}_{\delta }$. For every $a\in {\mathbb{F}}_{q^m}$, we extend $r^a=r-s^a$ to $r_{\text{ext}}^a$ and decode $r_{\text{ext}}^a$ as a codeword from a Reed–Solomon code spanned by $T^s$ for $q+1\le s\le q^{m-1}+q^{m-3}$. This is possible due to equation (5) and the fact that the corresponding Reed-Solomon code can correct up to $(q-1)\tau$ errors. Note that for the unique value of $a=a_{i,j}$ for $i=q^{m-2}$ and $j=q^{m-1}$, the word $r_{ext}^a$ decode as

$$g_a\left(T\right)=\sum _{0\le 1<j\le m-\mathrm{1,1}\ne m-2} a_{i,j}{T}^{q^i+q^j}.$$

Consequently, we get a list of $q^m$ polynomials, namely, $\left\{g_a\left(T\right)+a{T}^{q^{m-2}+q^{m-1}}\right\}$. Among this list, there will be exactly one $a\in {\mathbb{F}}_{q^m}$, for which the polynomial $g_a\left(T\right)+a{T}^{q^{m-2}}+q^{m-1}$ is the sent codeword. ■

In the next lemma we express the number of elements in a field extension not lying in any proper subfield of the extension field. This number counts some the orbits of the Grassmannian which contain an information set. These orbits will be useful for decoding $C(2,m)$.

Proposition 22:

Let $m\ge 4$ be a positive integer and let ${\mathbb{F}}_{q^m}$ be a finite field with $q^m$ elements. If $m$ is a prime or $m\ge 8$, then there are at least $q^m-q^{m-3}+1$ elements $\alpha \in {\mathbb{F}}_{q^m}$ such that ${\mathbb{F}}_q\left[\alpha \right]$ is not any proper subfield of ${\mathbb{F}}_{q^m}$.

Proof:

When $m$ is prime, the statement follows trivially. Now assume $m\ge 8$ and $p_1,\dots ,p_s$ are all distinct primes ( in descending order) dividing $m$. Let $m_i=m/p_i$. Let $F$ be any proper subfield of ${\mathbb{F}}_{q^m}$ satisfying ${\mathbb{F}}_q\subseteq F⊊{\mathbb{F}}_{q^m}$. Then $F$ is contained in a unique subfield of order $m_i$ for some $1\le i\le s$. Therefore, to establish the proposition, we only need to show that

$$q^{m_1}+\cdots +q^{m_s}\le q^{m-3}.$$

But this will follow as

$$\begin{gathered} q^{m_1}+\cdots +q^{m_s}<q^{m_1}+\cdots +q+1 \\ =\frac{q^{m_1+1}-1}{q-1} \\ <q^{m_1+1}. \end{gathered}$$

Therefore, we now only need to show that $m_1+1\le m-3$. Now, 2 is the smallest prime that can divide $m$, we get $m_1\le m/2$. Thus, we have

$$\begin{gathered} m_1+1\le m/2+1\le m-3 \\ \iff m\ge 8. \end{gathered}$$

This completes the proof of the Proposition. ■

The next lemma is an important step in proposing the decoder for Grassmann code $C(2,m)$. In this lemma, we give the error distribution and prove that if the transmitted codeword is not too corrupt, there will always be an orbit ${𝒪}_{\delta }$ of $G_{2,m}$ with an information set such that the projection the received word onto this orbit is also not too corrupt and we can apply Reed-Solomon decoder. Now suppose $c\in C(2,m)$ is the transmitted codeword and $r$ is the received codeword with error $e$

Lemma 23:

In the above setting, if less than $\frac{q^{2m-4}}{2}$ errors occurred, i.e., ${\mathrm{w}\mathrm{t}}_H\left(e\right)\le \frac{q^{2m-4}}{2}$, then there exists an orbit ${𝒪}_{\delta }$ of $G_{2,m}$ of size $N=\frac{q^m-1}{q-1}$ which contains at most $\frac{q^m-1-\left(q^{m-1}+q^{m-3}-q-1\right)}{2(q-1)}$ errors and ${\mathbb{F}}_q\left[\delta \right]$ is not contained in any proper subfield of ${\mathbb{F}}_{q^m}$.

Proof:

The proof is divided into two parts. First, we prove the lemma for the case $m\ge 8$ or $m$ is a prime and then deal with the cases $m=\mathrm{4,6}$ separately. So let us assume that $m\ge 8$ or $m$ is a prime. In that case, Proposition 22 guarantees that there are at least $q^m-q^{m-3}$ elements such that ${\mathbb{F}}_q\left[\delta \right]$ does not lie in any proper subfield of ${\mathbb{F}}_{q^m}$. In particular, there will be at least $q^m-q^{m-3}$ elements that will lie in orbits of size $N$. Also, from lemma 10, we know that each such orbit contains $q^3-q$ elements from these $q^m-q^{m-3}$ elements. If there are less than $q^{2(m-2)}/2$ errors, the average error per orbit with information set is

$$\frac{q^{2m-4}}{2}\frac{q^3-q}{q^m-q^{m-3}}$$

It is easy to verify now that

$$\frac{q^{2m-4}}{2}\frac{q^3-q}{q^m-q^{m-3}}\le \frac{q^m-1-\left(q^{m-1}+q^{m-3}-q-1\right)}{2\left(q-1\right)}.$$

Next, if $m=4$, then there are $q^4-q^2$ elements in ${\mathbb{F}}_{q^4}$ that does not lie in ${\mathbb{F}}_{q^2}$. In particular, these points lie on orbits of size $N$ and each such orbit contains exactly $q^3-q$ elements. Moreover, these orbits contains an information set. Note that,

$$\frac{q^4}{2}\frac{q^3-q}{q^4-q^2}\le \frac{q^4-1-\left(q^3+q-q-1\right)}{2\left(q-1\right)}.$$

Thus, the average error per orbit with information set is less than the error correction capacity.

The case $m=6$ is similar to the case $m=4$. There are $q^6-q^3-q^2+q$ elements such that ${\mathbb{F}}_q\left[\delta \right]$ is not contained in a proper subfield of ${\mathbb{F}}_{q^6}$. Further each such orbit contains exactly $q^3-q$ elements among these. Thus there are $\frac{q^6-q^2-q^2+q}{q^3-q}$ orbits of size $N$ and these orbits contains an information set. If there are less than $q^8/2$ errors, the average error per orbit with information set is

$$\frac{q^8}{2}\frac{q^3-q}{q^6-q^3-q^2-q}.$$

Now it is straightforward to verify that

$$\frac{q^8}{2}\frac{q^3-q}{q^6-q^3-q^2+q}\le \frac{q^6-1-\left(q^5+q^3-q-1\right)}{2(q-1)}.$$

This completes the proof of the lemma. ■

Now we are ready to prove the main result of this article. In the next theorem, we give a decoding algorithm for the Grassmann code $C(2,m)$ which decodes up to $\lfloor (d-1)/2\rfloor$ errors. It applies Algorithm 1 at each orbit ${𝒪}_{\delta }$ with an information set and recovers a codeword from $C(2,m)$ using the codeword recovered from $C^{\delta }(2,m)$. It returns the closest such codeword to the received word.

Algorithm 2 works is by outputting the closest codeword from a long list of candidates. The algorithm takes the received word $r$ and looks at the subword $r_{{𝒪}_{\delta }}$ given by the positions of $r$ on each orbit ${𝒪}_{\delta }$ with an information set. At each orbit ${𝒪}_{\delta }$ there are $q^m$ candidates (one for each possible leading term). If any of the $q^m$ candidates decodes successfully we extend it to a codeword of $C(2,m)$. If this candidate codeword is the first codeword found or is closest to $r$ than the previous candidate codeword, then it is stored as the current candidate codeword $c_{out}$. This process continues checking all orbits ${𝒪}_{\delta }$ with an information set. At the end, the algorithm outputs the candidate codeword $c_{out}$.

If less than $\frac{D}{2}$ errors occurred the algorithm is guaranteed to work. Lemma 23 implies there is a clean orbit ${𝒪}_c$ with few errors. When we decode the $q^m$ candidates on this clean orbit we are guaranteed that one of the $q^m$ candidates will decode $r_{{𝒪}_c}$ correctly as a codeword of the corresponding Reed–Solomon code. Since ${𝒪}_c$ contains an information set of $C(2,m)$, that candidate codeword $c_{\text{clean}}$ is at distance $<\frac{D}{2}$ to $r$. No other candidate codeword is closer to the received word $r$ to a codeword with the same information set as $c_{\text{clean}}$. Therefore $\frac{D}{2}$ errors are corrected. We state the algorithm and prove its correctness as follows.

Theorem 24:

Let $m\ge 4$ and let $C(2,m)$ be the corresponding Grassmann code. Let $d=q^{2(m-2)}$. Suppose $r=c+e$ where $c\in C(2,m)$. If ${\mathrm{w}\mathrm{t}}_H\left(e\right)<\frac{d}{2}$, then Algorithm 2 returns $c$.

Proof:

Let $c$ be the sent codeword of $C(2,m)$ and $r=c+e$ where ${\mathrm{w}\mathrm{t}}_H\left(e\right)<\frac{q^{2\mathrm{m}\mathrm{m}-4}}{2}$. Algorithm 2 projects $r$ onto all orbits ${𝒪}_{\delta }$ with an information set. Since ${\mathrm{w}\mathrm{t}}_H\left(e\right)<\frac{d}{2}$, Lemma 23 implies there is an orbit ${𝒪}_{\delta }$ where the projection $r\mid O_{\delta }$ has less than $\frac{q^m-1-\left(q^{m-1}+q^{m-3}-q-1\right)}{2(q-1)}$ errors. Therefore there is one codeword in the output of ${𝒪}_{\delta }$ which contains the same information set as the sent codeword $c$. Since ${\text{wt}}_H\left(e\right)<\frac{d}{2}$, $c$ is closest to $r$ than any other codeword. Therefore once the algorithm loops through ${𝒪}_{\delta }$, $c_{out}$ will be set to $c$ and it will remain unchanged. Therefore $c$ is the output of the algorithm. ■

Algorithm 2.

Input: a received word $r=c+e$ where $c\in C(2,m)$.

$c_{\mathit{out}}\leftarrow \left(\right)$

for ${𝒪}_{\delta }$ in the set of orbits containing an information set. do

Codewords ← Algorithm 1$\left(r^{{𝒪}_{\delta }}\right)$

for $c^{{𝒪}_{\delta }}\in$ Codewords do

c′ ← extend the codeword $c^{{𝒪}_{\delta }}$ to a codeword in $C(2,m)$ with the same information bits as $c^{{𝒪}_{\delta }}$.

if $d\left(c^{\prime },r\right)<d\left(c_{\mathit{out}},r\right)$ then

$c_{\mathit{out}}\leftarrow c^{\prime }$

end if

end for

end for

Return $c_{\mathit{out}}$.

Note that the length of the Grassmann code $C(2,m)$ is ${\left[\begin{array}{c}m\\ 2\end{array}\right]}_q=\frac{\left(q^m-1\right)\left(q^{m-1}-1\right)}{\left(q^2-1\right)(q-1)}$ which is of the order $q^{2(m-2)}$.

Theorem 25:

Algorithm 2 decodes a codeword of $C(2,m)$ with complexity $𝒪\left(q^{5m}\right)=𝒪\left(n^{\frac{5}{2}}\right)$, where $n$ is the length of the code $C(2,m)$.

Proof:

The algorithm decodes a codeword of a Reed–Solomon code of length $n=q^m-1$ for each orbit ${𝒪}_{\delta }$ with an information set. If $m$ is odd, there are $\frac{q^{m-1}-1}{q^2-1}$ orbits of size $\frac{q^m-1}{q-1}$. If $m$ is even there are $q\frac{q^{m-2}-1}{q^2-1}$ of size $\frac{q^m-1}{q-1}$. In either case, the number of orbits on which we project and decode is $𝒪\left(q^{m-3}\right)$. The Reed–Solomon decoding algorithm has complexity $𝒪\left(q^{3m}\right)$. Therefore the complexity of the decoding algorithm on each orbit is $𝒪\left(q^{4m}\right)$. As there are $𝒪\left(q^{m-3}\right)$ orbits, the overall complexity is $𝒪\left(q^{5m}\right)$. ■

VI. Decoding Example

In this section we decode errors for the $C\left(\mathrm{2,4}\right)$ binary Grassmann code.

A. Decoding $C\left(\mathrm{2,4}\right)$

The code $C\left(\mathrm{2,4}\right)$ is a [35, 6, 16] code. We aim to correct 7 errors. As mentioned in Example 4 $G_{\mathrm{2,4}}$ has two orbits of size 15 and one orbit of size 5. If ${\mathbb{F}}_{16}={\mathbb{F}}_2\left[\gamma \right]$ where ${\gamma }^4=\gamma +1$, then the two orbits of size 15 are the ones containing $\langle 1,\gamma \rangle$ and $\langle 1,{\gamma }^2\rangle$ respectively. The orbit of size 5 is the one containing the space $⟨1,{\gamma }^5⟩$. On each orbit of size 15, the code $C^{\delta }(2,4)$ is a subcode of a Reed–Solomon code obtained by evaluating the monomials $T^3,T^5,T^6,T^9,T^{10},T^{12}$ on the 15 nonzero points of ${\mathbb{F}}_{16}^{*}$. Note that the Reed–Solomon code spanned by $T^3,T^4,\dots ,T^{10},T^{11},T^{12}$ has distance $15-(12-3)=6$. We shall correct 3 errors by decoding as a codeword of the Reed–Solomon code spanned by $T^3,T^4,\dots ,T^{10}$ instead. This means that we need to decode 16 possibilities, one for each possible coefficient of $T^{12}$.

Let ${𝒪}_1$ be the orbit of $⟨1,\gamma ⟩,$ ${𝒪}_2$ be the orbit of $⟨1,{\gamma }^2⟩$, and ${𝒪}_3$ be the orbit of $⟨1,{\gamma }^5⟩$. Suppose that the zero codeword was sent and that seven errors occurred; three errors in ${𝒪}_1$ and four errors in ${𝒪}_2$. The orbit ${𝒪}_3$ does not contain an information set and is ignored by our algorithm. Let $r$ be the received word. We shall assume that $r$ restricted to each orbit is

$$r_1:=r^{{𝒪}_1}=(\mathrm{1,1},\mathrm{1,0},\mathrm{0,0},\mathrm{0,0},\mathrm{0,0},\mathrm{0,0},\mathrm{0,0},0)$$

and

$$r_2:=r^{{𝒪}_2}=\left(\mathrm{1,1},\mathrm{1,1},\mathrm{0,0},\mathrm{0,0},\mathrm{0,0},\mathrm{0,0},\mathrm{0,0},0\right).$$

Denote by ${\mathrm{E}\mathrm{v}}_L\left(T^{12}\right)={\left({\gamma }^{12i}\right)}_{i=0}^{12}$ the evaluation vector of the monomial $T^{12}$.

Our decoding algorithm is as follows. First try to decode on ${𝒪}_1$. For each $a\in {\mathbb{F}}_{16}$ let $s^a={\left(a{\gamma }^{12i}\right)}_{i=0}^{12}$ and $r^a=r_1-s^a$, then decode $r^a$ as a Reed–Solomon codeword spanned by the evaluation of $T^3,T^4,\dots ,T^{10}$. In this way we consider all possibilities for the coefficient of $T^{12}$. We have implemented this decoder in SAGEMATH [19]. We get 16 possible codewords.

$•0$

$\begin{array}{l}•a{T}^{12}+a{T}^{10}+a^2{T}^9+T^8+(a^3+a^2+1)T^6+a^2{T}^5\\ +a{T}^4+(a+1)T^3\end{array}$

$\begin{array}{l}•a^2{T}^{12}+(a^2+a)T^{10}+(a+1)T^9+a^2{T}^8+a{T}^6\\ +(a^2+a+1)T^5+a{T}^4+(a^2+1)T^3\end{array}$

$\begin{array}{l}•a^3{T}^{12}+a^2{T}^{10}+(a^3+a^2)T^9+(a^3+a^2+a+1)T^8\\ +(a^2+a)T^6+(a+1)T^5+(a^2+a+1)T^4\\ +(a^3+a^2+a+1)T^3\end{array}$

$\begin{array}{l}•(a+1)T^{12}+(a^3+a^2+a)T^{10}+(a^2+1)T^9+a^3{T}^8\\ +(a^3+a+1)T^5+(a^2+a+1)T^4+a{T}^3\end{array}$

$\begin{array}{l}•(a^2+a)T^{12}+a^2{T}^{10}+(a^2+a+1)T^9\\ +(a^3+a+1)T^8+a^3{T}^6+(a+1)T^5+a{T}^4\\ +(a^2+a)T^3\end{array}$

$\begin{array}{l}•(a^3+a^2)T^{12}+a{T}^{10}+(a^3+a^2+a+1)T^9+a{T}^8\\ +(a^3+1)T^6+a^2{T}^5+(a^3+a^2+1)T^4+(a^3+a)T^3\end{array}$

$\begin{array}{l}•(a^3+a+1)T^{12}+(a^3+a)T^{10}+(a^3+1)T^9\\ +(a+1)T^8+a^2{T}^6+a^3{T}^5+(a+1)T^4\\ +(a^3+a^2+1)T^3\end{array}$

$\begin{array}{l}•(a^2+1)T^{12}+(a^3+a)T^{10}+a{T}^9+(a^2+1)T^8\\ +(a^3+a+1)T^6+a^3{T}^5+(a^3+a)T^4+a^2{T}^3\end{array}$

$\begin{array}{l}•(a^3+a)T^{12}+(a^2+a)T^{10}+a^3{T}^9+(a^3+a^2)T^8\\ +(a^3+a)T^6+(a^2+a+1)T^5+(a^2+1)T^4\\ +(a^3+a^2)T^3\end{array}$

$\begin{array}{l}•(a^2+a+1)T^{12}+a^3{T}^{10}+(a^2+a)T^9\\ +(a^3+a^2+a+1)T^8+(a^3+a)T^6+(a^3+a^2)T^5\\ +(a^3+a^2+a)T^4+(a^2+a+1)T^3\end{array}$

$\begin{array}{l}•(a^3+a^2+a)T^{12}+(a^3+a+1)T^9\\ +(a^3+a^2+a+1)T^8+a{T}^6+(a^3+a^2+1)T^4\\ +(a^3+1)T^3\end{array}$

$\begin{array}{l}•(a^3+a^2+a+1)T^{12}+(a^3+a^2)T^{10}\\ +(a^3+a)T^9+(a^3+a^2+1)T^8+(a+1)T^6+(a^3+a^2+a+1)T^5\\ +a{T}^4+a^3{T}^3\end{array}$

$\begin{array}{l}•(a^3+a^2+1)T^{12}+(a^3+a^2+a)T^{10}\\ +(a^3+a^2+a)T^9+(a^2+1)T^8+(a+1)T^6\\ +(a^3+a+1)T^5+(a^2+a)T^4+(a^3+a+1)T^3\end{array}$

$\begin{array}{l}•(a^3+1)T^{12}+a^3{T}^{10}+(a^3+a^2+1)T^9\\ +(a^2+1)T^8+(a+1)T^6+(a^3+a^2)T^5+(a^2+1)T^4\\ +(a^3+a^2+a)T^3\end{array}$

$\begin{array}{l}•T^{12}+(a^3+a^2)T^{10}+T^9+(a^3+a^2)T^8+(a^2+a)T^6\\ +(a^3+a^2+a+1)T^5+T^4+T^3\end{array}$

It seems there are 16 possible codewords for the received word. However, only the zero codeword is in the span of $T^3,T^5,T^6,T^9,T^{10},T^{12}$. All other polynomials contain $T^8$ and $T^4$, which implies they do not correspond to a codeword of $C^{\delta }\left(\mathrm{2,4}\right)$ nor $C\left(\mathrm{2,4}\right)$.

Now we decode on ${𝒪}_2$. For each $a\in {\mathbb{F}}_{16}$ let $r^a=r_2-s^a$. We decode $r^a$ as a Reed–Solomon codeword spanned by the evaluation of $T^3,T^4,\dots ,T^{10}$. We get 16 possible codewords:

$\begin{array}{l}•(a+1)T^{10}+(a^2+a)T^8+a^3{T}^6+(a^2+1)T^5\\ +(a^3+a^2+a+1)T^4\end{array}$

$\begin{array}{l}•a{T}^{12}+a{T}^{10}+a^2{T}^9+T^8+(a^3+a^2+1)T^6+a^2{T}^5\\ +a{T}^4+(a+1)T^3\end{array}$

$\begin{array}{l}•a^2{T}^{12}+a^2{T}^{10}+(a+1)T^9+(a^3+a^2)T^8+T^6\\ +(a+1)T^5+a^2{T}^4+(a^2+1)T^3\end{array}$

$\begin{array}{l}•a^3{T}^{12}+a^3{T}^{10}+(a^3+a^2)T^9+(a^2+a)T^8\\ +(a^3+a^2)T^6+(a^3+a^2)T^5+(a^3+a^2+a)T^4\\ +(a^3+a^2+a+1)T^3\end{array}$

$\begin{array}{l}•(a+1)T^{12}+(a+1)T^{10}+(a^2+1)T^9\\ +(a^3+a^2+1)T^8+(a^3+a)T^6+(a^2+1)T^5+T^4\\ +a{T}^3\end{array}$

$\begin{array}{l}•(a^2+a)T^{12}+(a^2+a)T^{10}+(a^2+a+1)T^9\\ +(a^3+a^2+1)T^8+(a^3+1)T^6+(a^2+a+1)T^5\\ +T^4+(a^2+a)T^3\end{array}$

$\begin{array}{l}•(a^3+a^2)T^{12}+(a^3+a^2)T^{10}+(a^3+a^2+a+1)T^9\\ +(a^3+a)T^8+(a+1)T^6+(a^3+a^2+a+1)T^5\\ +(a^2+a+1)T^4+(a^3+a)T^3\end{array}$

$\begin{array}{l}•(a^3+a+1)T^{12}+(a^3+a+1)T^{10}+(a^3+1)T^9\\ +(a^3+a+1)T^8+(a^3+a^2+a+1)T^6\\ +(a^3+1)T^5+(a^3+a^2+1)T^4+(a^3+a^2+1)T^3\end{array}$

$\begin{array}{l}•(a^2+1)T^{12}+(a^2+1)T^{10}+a{T}^9+(a^2+a+1)T^6\\ +a{T}^5+(a+1)T^4+a^2{T}^3\end{array}$

$\begin{array}{l}•(a^3+a)T^{12}+(a^3+a)T^{10}+a^3{T}^9+(a^2+a+1)T^8+\\ (a^3+a^2)T^6+a^3{T}^5+(a^3+a^2+1)T^4+(a^3+a^2)T^3\end{array}$

$\begin{array}{l}•(a^2+a+1)T^{12}+(a^2+a+1)T^{10}+(a^2+a)T^9\\ +T^8+(a^3+1)T^6+(a^2+a)T^5+(a^3+a^2+1)T^4\\ +(a^2+a+1)T^3\end{array}$

$\begin{array}{l}•(a^3+a^2+a)T^{12}+(a^3+a^2+a)T^{10}\\ +(a^3+a+1)T^9+(a^3+a+1)T^8+(a^2+a)T^6\\ +(a^3+a+1)T^5+(a+1)T^4+(a^3+1)T^3\end{array}$

$\begin{array}{l}•(a^3+a^2+a+1)T^{12}+(a^3+a^2+a+1)T^{10}\\ +(a^3+a)T^9+(a^2+a+1)T^8+a{T}^6+(a^3+a)T^5\\ +(a^3+a^2)T^4+a^3{T}^3\end{array}$

$\begin{array}{l}•(a^3+a^2+1)T^{12}+(a^3+a^2+1)T^{10}\\ +(a^3+a^2+a)T^9+(a^2+a)T^8+T^6\\ +(a^3+a^2+a)T^5+(a+1)T^4+(a^3+a+1)T^3\end{array}$

$\begin{array}{l}•(a^3+1)T^{12}+(a^3+1)T^{10}+(a^3+a^2+1)T^9\\ +(a^3+a)T^8+(a^3+1)T^6+(a^3+a^2+1)T^5\\ +(a^2+1)T^4+(a^3+a^2+a)T^3\end{array}$

$•T^{12}+T^{10}+T^9+(a^3+a^2)T^8+T^6+T^5+a^3{T}^4+T^3$

It seems there are 16 more possible codewords for the received word. However all polynomials contain $T^8$ and $T^4$, which implies they do not correspond to a codeword of $C\left(\mathrm{2,4}\right)$. Therefore we recover 0 as the sent codeword.

B. Decoding a nontrivial codeword

Now we shall decode the codeword given by $\mathrm{E}\mathrm{v}({det}_{\left[12\right]}\left(X\right))$. We recall that ${\mathbb{F}}_{16}={\mathbb{F}}_2\left[a\right]$ where $a^4=a+1$. We take the powers $1,a,a^2,a^3$ as a basis of ${\mathbb{F}}_{16}$ over ${\mathbb{F}}_2$. Its dual basis is given by $a^3+1,a^2,a,1$. If the vector space $W=⟨\mathbf{x},\mathbf{y}⟩$ where $\mathbf{x}=x_0+x_{1}a+x_2{a}^2+x_3{a}^3$ and $\mathbf{y}=y_0+y_{1}a+y_2{a}^2+y_3{a}^3$ then the function

$$\begin{gathered} de{t}_{[12\}}\left(W\right)={\text{Tr}}_{{\mathbb{F}}_{16}/{\mathbb{F}}_2}\left(\left(a^3+1\right)\mathbf{x}\right){\text{Tr}}_{{\mathbb{F}}_{16}/{\mathbb{F}}_2}\left(a^2\mathbf{y}\right) \\ -{\text{Tr}}_{{\mathbb{F}}_{16}/{\mathbb{F}}_2}\left(a^2\mathbf{x}\right){\text{Tr}}_{{\mathbb{F}}_{16}/{\mathbb{F}}_2}\left(\left(a^3+1\right)\mathbf{y}\right) \end{gathered}$$

On the orbit ${𝒪}_1,$ $W=⟨\mathbf{x},a\mathbf{x}⟩$ and

$$\begin{gathered} de{t}_{\left\{12\right\}}\left(W\right)={\text{Tr}}_{{\mathbb{F}}_{16}/{\mathbb{F}}_2}\left(\left(a^3+1\right)\mathbf{x}\right){\text{Tr}}_{{\mathbb{F}}_{16}/{\mathbb{F}}_2}\left(a^3\mathbf{x}\right) \\ -{\text{Tr}}_{{\mathbb{F}}_{16}/{\mathbb{F}}_2}\left(a^2\mathbf{x}\right){\text{Tr}}_{{\mathbb{F}}_{16}/{\mathbb{F}}_2}\left(\left(a^4+a\right)\mathbf{x}\right) \end{gathered}$$

$$\begin{gathered} de{t}_{\left\{12\right\}}\left(W\right)={\text{Tr}}_{{\mathbb{F}}_{16}/{\mathbb{F}}_2}\left(\left(a^3+1\right)\mathbf{x}\right){\text{Tr}}_{{\mathbb{F}}_{16}/{\mathbb{F}}_2}\left(a^3\mathbf{x}\right) \\ -{\text{Tr}}_{{\mathbb{F}}_{16}/{\mathbb{F}}_2}\left(a^2\mathbf{x}\right){\text{Tr}}_{{\mathbb{F}}_{16}/{\mathbb{F}}_2}\left(\left(a^4+a\right)\mathbf{x}\right) \end{gathered}$$

$$\begin{array}{l}de{t}_{\left\{12\right\}}(W)=a{\mathbf{x}}^{12}+\left(a^2+a+1\right){\mathbf{x}}^{10}+a^2{\mathbf{x}}^9\\ +\left(a^2+1\right){\mathbf{x}}^6+\left(a^2+a\right){\mathbf{x}}^5+(a+1){\mathbf{x}}^3.\end{array}$$

On the orbit ${𝒪}_2,W=⟨\mathbf{x},a^2\mathbf{x}⟩$ and

$$\begin{gathered} {det}_{\left\{12\right\}}\left(W\right)={\text{Tr}}_{{\mathbb{F}}_{16}/{\mathbb{F}}_2}\left(\left(a^3+1\right)\mathbf{x}\right){\text{Tr}}_{{\mathbb{F}}_{16}/{\mathbb{F}}_2}\left(a^4\mathbf{x}\right) \\ +{\text{Tr}}_{{\mathbb{F}}_{16}/{\mathbb{F}}_2}\left(a^2\mathbf{x}\right){\text{Tr}}_{{\mathbb{F}}_{16}/{\mathbb{F}}_2}\left(\left(a^5+a^2\right)\mathbf{x}\right). \end{gathered}$$

$$\begin{gathered} {det}_{\left\{12\right\}}(W)=\left(a^3+a^2\right){\mathbf{x}}^{12}+\left(a^2+a+1\right){\mathbf{x}}^{10} \\ +\left(a^3+a^{2}a+1\right){\mathbf{x}}^9+a^3{\mathbf{x}}^6+\left(a^2+a\right){\mathbf{x}}^5 \\ +\left(a^3+a\right){\mathbf{x}}^3. \end{gathered}$$

Suppose $c=\mathrm{Ev}({det}_{\left\{12\right\}}\left(X\right))$ is the intended codeword. The codeword restricted to ${𝒪}_1$ is equal to (1, 0, 0, 0, 1, 0, 0, 1, 1, 1, 1, 1, 1, 0, 0). The codeword restricted to ${𝒪}_2$ equals: (0, 0, 0, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1). As in the previous case, we shall assume there are three errors on ${𝒪}_1$ and four errors in ${𝒪}_2$.

When we attempt to decode (0, 1, 1, 0, 1, 0, 0, 1, 1, 1, 1, 1, 1, 0, 0) on the orbit ${𝒪}_1$ we get the following candidate codewords:

$•(a^2+1)x^{10}+x^8+a^3{x}^6+a{x}^5+a{x}^4$

$\begin{array}{l}•a{x}^{12}+(a^2+a+1)x^{10}+a^2{x}^9+(a^2+1)x^6\\ +(a^2+a)x^5+(a+1)x^3\end{array}$

$\begin{array}{l}•a^2{x}^{12}+(a+1)x^{10}+(a+1)x^9+(a^3+a+1)x^8\\ +(a^3+a^2+1)x^6+(a^2+1)x^5+a{x}^4+(a^2+1)x^3\end{array}$

$\begin{array}{l}•a^3{x}^{12}+x^{10}+(a^3+a^2)x^9+(a^3+a^2)x^8\\ +(a^3+a^2+a+1)x^6+x^5+(a^2+1)x^4\\ +(a^3+a^2+a+1)x^3\end{array}$

$\begin{array}{l}•(a+1)x^{12}+(a^3+a+1)x^{10}+(a^2+1)x^9\\ +(a^3+a^2)x^8+(a+1)x^6+(a^3+1)x^5+x^4+a{x}^3\end{array}$

$\begin{array}{l}•(a^2+a)x^{12}+x^{10}+(a^2+a+1)x^9+a^2{x}^8\\ +(a^2+a+1)x^6+x^5+a{x}^4+(a^2+a)x^3\end{array}$

$\begin{array}{l}•(a^3+a^2)x^{12}+(a^2+a+1)x^{10}\\ +(a^3+a^2+a+1)x^9+(a^3+a^2+a+1)x^8\\ +(a^2+a+1)x^6+(a^2+a)x^5+(a^3+a^2+1)x^4\\ +(a^3+a)x^3\end{array}$

$\begin{array}{l}•(a^3+a+1)x^{12}+(a^3+a^2+a+1)x^{10}\\ +(a^3+1)x^9+(a^2+1)x^8+(a^2+a)x^6+(a^3+a)x^5\\ +(a^2+1)x^4+(a^3+a^2+1)x^3\end{array}$

$\begin{array}{l}•(a^2+1)x^{12}+(a^3+a^2+a+1)x^{10}+a{x}^9\\ +(a^3+a^2+a+1)x^8+(a^3+a^2+a+1)x^6\\ +(a^3+a)x^5+(a^3+a^2+a)x^4+a^2{x}^3\end{array}$

$\begin{array}{l}•(a^3+a)x^{12}+(a+1)x^{10}+a^3{x}^9\\ +(a^3+a^2+a+1)x^8+(a+1)x^6+(a^2+1)x^5\\ +(a^2+a+1)x^4+(a^3+a^2)x^3\end{array}$

$\begin{array}{l}•(a^2+a+1)x^{12}+(a^3+a^2+1)x^{10}+(a^2+a)x^9\\ +(a^2+1)x^8+(a^3+a^2+a)x^6+(a^3+a^2+a)x^5\\ +(a^3+a)x^4+(a^2+a+1)x^3\end{array}$

$\begin{array}{l}•(a^3+a^2+a)x^{12}+(a^2+1)x^{10}+(a^3+a+1)x^9\\ +a{x}^8+(a^3+a^2)x^6+a{x}^3\\ +(a^3+a^2+1)x^4+(a^3+1)x^3\end{array}$

$\begin{array}{l}•(a^3+a^2+a+1)x^{12}+(a^3+1)x^{10}+(a^3+a)x^9\\ +(a^2+1)x^8+(a^2+a)x^6+(a^3+a^2+1)x^5\\ +(a^2+a)x^4+a^3{x}^3\end{array}$

$\begin{array}{l}•(a^3+a^2+1)x^{12}+(a^3+a+1)x^{10}\\ +(a^3+a^2+a)x^9+(a^3+a^2+1)x^8+(a^2+a)x^6\\ +(a^3+1)x^5+a{x}^4+(a^3+a+1)x^3\end{array}$

$\begin{array}{l}•(a^3+1)x^{12}+(a^3+a^2+1)x^{10}+(a^3+a^2+1)x^9\\ +(a+1)x^8+x^6+(a^3+a^2+a)x^5+(a+1)x^4\\ +(a^3+a^2+a)x^3\end{array}$

$\begin{array}{l}•x^{12}+(a^3+1)x^{10}+x^9+a^3{x}^8+(a^2+1)x^6\\ +(a^3+a^2+1)x^5+(a^2+a+1)x^4+x^3\end{array}$

We remark that there is a candidate codeword for each possible leading monomial $a_{12}{T}^{12}$ since we attempt to decode for all possibilities of that leading term. Note that the only polynomial spanned by $T^3,T^6,T^6,T^9,T^{10},T^{12}$ is the second entry of the list, corresponding to ${\mathrm{T}\mathrm{r}}_{{\mathbb{F}}_{16}/{\mathbb{F}}_2}((a^3+1)\mathbf{x}){\mathrm{T}\mathrm{r}}_{{\mathbb{F}}_{16}/{\mathbb{F}}_2}(a^3\mathbf{x})-{\mathrm{T}\mathrm{r}}_{{\mathbb{F}}_{16}/{\mathbb{F}}_2}(a^2\mathbf{x}){\mathrm{T}\mathrm{r}}_{{\mathrm{F}}_{16}/{\mathbb{F}}_2}((a^4+a)\mathbf{x})$. In this way, the original codeword is part of the list of possible codewords given by the decoding algorithm. We may recover the codeword by evaluating $a{T}^{12}+(a^2+a+1)T^{10}+a^2{T}^9+(a^2+1)T^6+(a^2+a)T^5+(a+1)T^3$ on ${𝒪}_1$ and extending the entries on an information set to the remaining positions of the codeword. One can also recover the function from the coefficients of the polynomial. Now we also consider what happens with ${𝒪}_2$ which has 4 errors. When we decode (1, 1, 1, 1, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1) on the orbit ${𝒪}_2$ we get the following possible codewords:

$\begin{array}{l}•(a+1)T^{10}+(a^2+a)T^8+a^3{T}^6+(a^2+1)T^5\\ +(a^3+a^2+a+1)T^4\end{array}$

$\begin{array}{l}•a{T}^{12}+a{T}^{10}+a^2{T}^9+T^8+(a^3+a^2+1)T^6+a^2{T}^5\\ +a{T}^4+(a+1)T^3\end{array}$

$\begin{array}{l}•a^2{T}^{12}+a^2{T}^{10}+(a+1)T^9+(a^3+a^2)T^8+T^6\\ +(a+1)T^5+a^2{T}^4+(a^2+1)T^3\end{array}$

$\begin{array}{l}•a^3{T}^{12}+a^3{T}^{10}+(a^3+a^2)T^9+(a^2+a)T^8\\ +(a^3+a^2)T^6+(a^3+a^2)T^5+(a^3+a^2+a)T^4\\ +(a^3+a^2+a+1)T^3\end{array}$

$\begin{array}{l}•(a+1)T^{12}+(a+1)T^{10}+(a^2+1)T^9\\ +(a^3+a^2+1)T^8+(a^3+a)T^6+(a^2+1)T^5\\ +T^4+a{T}^3\end{array}$

$\begin{array}{l}•(a^2+a)T^{12}+(a^2+a)T^{10}+(a^2+a+1)T^9\\ +(a^3+a^2+1)T^8+(a^3+1)T^6+(a^2+a+1)T^5\\ +T^4+(a^2+a)T^3\end{array}$

$\begin{array}{l}•(a^3+a^2)T^{12}+(a^3+a^2)T^{10}+(a^3+a^2+a+1)T^9\\ +(a^3+a)T^8+(a+1)T^6+(a^3+a^2+a+1)T^5\\ +(a^2+a+1)T^4+(a^3+a)T^3\end{array}$

$\begin{array}{l}•(a^3+a+1)T^{12}+(a^3+a+1)T^{10}+(a^3+1)T^9\\ +(a^3+a+1)T^8+(a^3+a^2+a+1)T^6+(a^3+1)T^5\\ +(a^3+a^2+1)T^4+(a^3+a^2+1)T^3\end{array}$

$\begin{array}{l}•(a^2+1)T^{12}+(a^2+1)T^{10}+a{T}^9+(a^2+a+1)T^6\\ +a{T}^5+(a+1)T^4+a^2{T}^3\end{array}$

$\begin{array}{l}•(a^3+a)T^{12}+(a^3+a)T^{10}+a^3{T}^9+(a^2+a+1)T^8\\ +(a^3+a^2)T^6+a^3{T}^5+(a^3+a^2+1)T^4\\ +(a^3+a^2)T^3\end{array}$

$\begin{array}{l}•(a^2+a+1)T^{12}+(a^2+a+1)T^{10}+(a^2+a)T^9\\ +T^8+(a^3+1)T^6+(a^2+a)T^5+(a^3+a^2+1)T^4\\ +(a^2+a+1)T^3\end{array}$

$\begin{array}{l}•(a^3+a^2+a)T^{12}+(a^3+a^2+a)T^{10}\\ +(a^3+a+1)T^9+(a^3+a+1)T^8+(a^2+a)T^6\\ +(a^3+a+1)T^5+(a+1)T^4+(a^3+1)T^3\end{array}$

$\begin{array}{l}•(a^3+a^2+a+1)T^{12}+(a^3+a^2+a+1)T^{10}\\ +(a^3+a)T^9+(a^2+a+1)T^8+a{T}^6+(a^3+a)T^5\\ +(a^3+a^2)T^4+a^3{T}^3\end{array}$

$\begin{array}{l}•(a^3+a^2+1)T^{12}+(a^3+a^2+1)T^{10}\\ +(a^3+a^2+a)T^9+(a^2+a)T^8+T^6\\ +(a^3+a^2+a)T^5+(a+1)T^4+(a^3+a+1)T^3\end{array}$

$\begin{array}{l}•(a^3+1)T^{12}+(a^3+1)T^{10}+(a^3+a^2+1)T^9\\ +(a^3+a)T^8+(a^3+1)T^6+(a^3+a^2+1)T^5\\ +(a^2+1)T^4+(a^3+a^2+a)T^3\end{array}$

$•T^{12}+T^{10}+T^9+(a^3+a^2)T^8+T^6+T^5+a^3{T}^4+T^3$

None of the possible codewords given by ${𝒪}_2$ can correspond to a codeword of the projection ${\mathrm{C}}^{{𝒪}_2}\left(\mathrm{2,4}\right)$ code nor a codeword of the $C\left(\mathrm{2,4}\right)$ Grassmann code since their evaluation polynomials contain the monomial $T^8$. Out of the 32 possible codewords, only one is a possible codeword of the Grassmann code. In this way the original codeword is recovered.

C. Several codeword candidates

It may be possible that each orbit offers different candidates for the corresponding codeword. As before, let us assume that we are decoding 7 errors when decoding $C\left(\mathrm{2,4}\right)$. Suppose $c=\mathrm{E}\mathrm{v}\left(0\right)$ is the intended codeword. However we shall assume all seven enrors occur on ${𝒪}_2$. In particular the received word restricted to ${𝒪}_1$ is $r_{{𝒪}_1}\text{=(0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0)}$. The received word restricted to ${𝒪}_2$ is $r_{{𝒪}_2}=(1,0,0,0,0,1,0,1,1,0,0,1,1,0,1)$. Note that the restriction of the received word on ${𝒪}_2$ is only one position away from the restriction of $\mathrm{E}\mathrm{v}\left({det}_{12}\left(X\right)\right)$ on ${𝒪}_2$. In this case we get a single candidate codeword from each orbit. The candidate codeword from ${𝒪}_1$ is the zero codeword. The candidate codeword from ${𝒪}_2$ is the evaluation of $\left(a^3+a^2\right)T^{12}+\left(a^2+a+1\right)T^{10}+\left(a^3+a^2+a+1\right)T^9+$ $a^3{T}^6+\left(a^2+a\right)T^5+\left(a^3+a\right)T^3$, which corresponds to the codeword given by $\mathrm{E}\mathrm{v}\left({det}_{12}\left(X\right)\right.$). The distance of Ev(0) to the received word is 7. The distance of the received word to $\mathrm{E}\mathrm{v}\left({det}_{12}\left(X\right)\right)$ is 9. Since the closest codeword is Ev(0) the algorithm outputs Ev(0) thereby correcting the 7 errors correctly.