New constant dimension subspace codes from block inserting constructions

Abstract

A basic problem of the constant dimension subspace coding is to determine the maximal possible size A_q(n,d,k) of a set of k-dimensional subspaces in $\mathbf {F}_{q}^{n}$ such that the subspace distance satisfies $\text {dis}(U,V) =2k-2 \dim (U \cap V) \geq d$ for any two different subspaces U and V in this set. We propose two constructions of constant dimension subspace codes that can insert flexibly into the generalized parallel linkage construction. In our constructions matrix blocks from small constant dimension codes and rank metric codes play important roles. Through a well-arranged combination for the matrix blocks, more than 120 new constant dimension subspace codes of distance 4, 6, 8 better than previously best known codes are constructed.

Appendix: A

In [10], the authors proposed a construction that composes several CDCs in Theorem 4. However, the result is incorrect since the minimal distance of the CDCs in construction is not satisfied ≥ d under some conditions.

More specifically, the proof of dis(W₃,W₄) ≥ d in Theorem 4 is reduced to $\dim (W_{3} + W_{4}) \geq k + \frac {d}{2}$, where W₃ is a subspace in the CDC B₁ and W₄ is a subspace in the CDC B₂. The authors claimed that $\dim (W_{3} + W_{4}) = \dim (W_{3} + W_{4}^{\prime }) = 2k - \dim (W_{3} \cap W_{4}^{\prime }) \geq k + \frac {d}{2}$, where $W_{4}^{\prime } = \text {rs} \left (\begin {array}{lllllll} I_{a_{2}} & | & M_{2} & | & \quad O_{2} & \\ O_{1} & |& M_{3} & | & O^{\prime } & | & M_{1}^{\prime } \end {array}\right )$, M₃ is a matrix with restricted rank $\leq a_{1} - \frac {d}{2}$ and $M_{1}^{\prime }$ is a matrix with unknown rank.

The problem is that $\dim (W_{3} + W_{4}^{\prime }) = 2k - \dim (W_{3} \cap W_{4}^{\prime })$ is not strictly proved in the paper. It is true if and only if $\dim (W_{3}) = \dim (W_{4}^{\prime })=k$. However, the Theorem 4 only gives that $\dim (W_{4}^{\prime }) = a_{2} + \text {rank} (M_{3} | O^{\prime } | M_{1}^{\prime }) \leq a_{2} + \text {rank} (M_{3}) + \text {rank} (M_{1}^{\prime }) \leq k - \frac {d}{2} + \text {rank} (M_{1}^{\prime })$. Thus $\dim (W_{4}^{\prime })=k$ is not satisfied if $\text {rank} (M_{3} | O^{\prime } | M_{1}^{\prime }) \neq a_{1}$.

We give a concrete example to show that the minimal distance is not satisfied in some cases. Using the same notations in Theorem 4 of [10], let n₁ = n₂ = 6,a₁ = 4,a₂ = 2,b₁ = b₂ = 2,d = 4,k = 6 and q = 2, then M₁ ∈ (4, 2, 2)₂ RMC, M₂ ∈ (2, 4, 2)₂ RMC and M₃ ∈ (4, 4, 2)₂ RRMC with rank ≤ 2. For simplification, we further assume M₁ and M₂ are zero matrices, and $M_{3} = \left (\begin {array}{llll} {1} {0} {0} {0} \\ {0} {0} {0} {0} \\ {0} {0} {0} {0} \\ {0} {1} {0} {0} \end {array}\right )$. From the construction, we have

$$ \begin{array}{@{}rcl@{}} W_{3} &=& \text{rs} (G_{3})= \text{rs} \left( \begin{array}{llllllllllll} 1 0 0 0 0 0 | 0 0 1 0 0 0 \\ 0 1 0 0 0 0 | 0 0 0 0 0 0 \\ 0 0 1 0 0 0 | 0 0 0 0 0 0 \\ 0 0 0 1 0 0 | 0 0 0 1 0 0 \\ 0 0 0 0 0 0 | 1 0 0 0 0 0 \\ 0 0 0 0 0 0 | 0 1 0 0 0 0 \end{array}\right) \in B_{1}, \text{and} \\ W_{4} &=& \text{rs} (G_{4}) = \text{rs} \left( \begin{array}{llllllllllll} 1 0 0 0 0 0 | 0 0 0 0 0 0 \\ 0 1 0 0 0 0 | 0 0 0 0 0 0 \\ 0 0 1 0 0 0 | 1 0 0 0 0 0 \\ 0 0 0 0 0 0 | 0 1 0 0 0 0 \\ 0 0 0 0 0 0 | 0 0 1 0 0 0 \\ 0 0 0 1 0 0 | 0 0 0 1 0 0 \end{array}\right) \in B_{2}. \end{array} $$

Using the same proof technique, $\dim (W_{3} + W_{4}) = \text {rank} \left (\begin {array}{ll} G_{3} \\ G_{4} \end {array}\right ) = \text {rank} \left (\begin {array}{ll} G_{3} \\ G_{4}^{\prime } \end {array}\right ) = \dim (W_{3} + W_{4}^{\prime }),$ where $W_{4}^{\prime } = \text {rs} (G_{4}^{\prime }) = \text {rs} \left (\begin {array}{llllllllllll} 1 0 0 0 0 0 | 0 0 0 0 0 0 \\ 0 1 0 0 0 0 | 0 0 0 0 0 0 \\ 0 0 1 0 0 0 | {0} 0 0 0 0 0 \\ 0 0 0 0 0 0 | 0 {0} 0 0 0 0 \\ 0 0 0 0 0 0 | 0 0 1 0 0 0 \\ 0 0 0 1 0 0 | 0 0 0 1 0 0 \end {array}\right )$ and the first (second) green zero is the result of subtracting the 9th (10th) row from the 5th (6th) row of $\left (\begin {array}{ll} G_{3} \\ G_{4} \end {array}\right )$.

It is easy to check that $\dim (W_{3}) = \dim (W_{4}) = 6$. According to the Theorem 4, dis(W₃,W₄) ≥ 4 should be satisfied, which implies that $\dim (W_{3} + W_{4})$ should be ≥ 8. However, $\dim (W_{4}^{\prime }) = 5$ and

$$ \begin{array}{@{}rcl@{}} \dim(W_{3} + W_{4}) &=& \dim(W_{3} + W_{4}^{\prime}) \\ & =& \dim(W_{3}) + \dim(W_{4}^{\prime}) - \dim(W_{3} \cap W_{4}^{\prime}) \\ & =& 6 + 5 - 4 = 7. \end{array} $$

Thus $\text {dis}(W_{3}, W_{4}) = 2\dim (W_{3} + W_{4}) - 2 \times 6 = 14 - 12 = 2$, which not meets the desired minimal distance.