The second weight of generalized Reed-Muller codes in most cases

Abstract

The second weight of the Generalized Reed-Muller code of length q ⁿ and order d over the finite field with q elements is now known for d < q and d > (n − 1)(q − 1). In this paper, we determine the second weight for the other values of d which are not multiples of q − 1 plus 1. For the special case d = a(q − 1) + 1 we give an estimate.

Appendix A: Proof of lemma 3.10

1.1 Preliminary remarks

Let us set

$$ P_1=\prod\limits_{i=1}^n (q-\alpha_i), $$

$$ P_2=(q-\gamma)\prod\limits_{i=a+2}^n (q-\alpha_i). $$

Hence we have to study the minimum value of P ₁ − P ₂. Note that in the particular case d = (n − 1)(q − 1) the value of a is n − 1 and P ₂ = (q − γ).

Lemma A.1

If we permute the first a elements α _i we don’t change the value of P₁ − P₂. When α_a + 1 < b, if we permute the last n − a − 1 elements we don’t change the value of P₁ − P₂. When α_a + 1 ≥ b, namely when γ = α_a + 1, if we permute α_a + 1 with one of the last n − a − 1 elements α _i such that α _i  ≥ b we don’t change the value of P₁ − P₂.

Proof

This can be seen directly on the formulas giving P ₁ and P ₂.□

Then, from now on, we will suppose that the sequences α are such that

$$ \alpha_1 \geq \alpha_2 \geq \cdots \geq \alpha_{a-1} \geq \alpha_a, $$

$$ \alpha_{a+2} \geq \alpha_{a+3} \geq \cdots \geq \alpha_{n} $$

if α _a + 1 < b, and that

$$ \alpha_1 \geq \alpha_2 \geq \cdots \geq \alpha_{a-1} \geq \alpha_a, $$

$$ \alpha_{a+1} \geq \alpha_{a+2} \geq \cdots \geq \alpha_{n} $$

if α _a + 1 ≥ b. In particular, when we transform a sequence, we always reorder the new obtained sequence in this way.

Lemma A.2

If we replace α _i by α _i  + 1 and if the new sequence is in V, then the new P₁ − P₂ is lower than the old one.

Proof

When i ≤ a the value of P ₁ decreases, the value of P ₂ is not modified. When i = a + 1 and α _a + 1 < b, P ₁ decreases, the value of P ₂ is not modified. When i ≥ a + 1 and α _a + 1 ≥ b, P ₂ and P ₁ decreases, then we must examine more precisely the behaviour of P ₁ − P ₂. The difference between the old value of P ₁ − P ₂ and the new one is

$$\begin{array}{lll} &&\prod\limits_{\begin{array}{c} j=1\\j\neq i \end{array}}^n \left ( q-\alpha_j\right) - \prod\limits_{\begin{array}{c} j=a+1\\j\neq i \end{array}}^n \left ( q-\alpha_j\right) \\ && \;\; = \left ( \prod\limits_{j=1}^a \left ( q-\alpha_j\right)-1 \right ) \prod\limits_{\begin{array}{c} j=a+1\\j\neq i \end{array}}^n \left ( q-\alpha_j\right). \end{array}$$

But, as α _a + 1 ≥ b, α _a  ≤ q − 2 and then

$$ \prod\limits_{j=1}^a \left ( q-\alpha_j\right) \geq 2. $$

We conclude that the new value is lower than the old one. It remains to study the case where α _a + 1 < b and i > a + 1. The difference between the old value of P ₁ − P ₂ and the new one is now

$$ \left ( \prod\limits_{j=1}^{a+1} \left ( q-\alpha_j\right)-(q-b) \right ) \prod\limits_{\begin{array}{c} j=a+2\\j\neq i \end{array}}^n \left ( q-\alpha_j\right). $$

But as (q − α _a + 1) > (q − b) we conclude that the new value of P ₁ − P ₂ is lower than the old one. □

Lemma A.3

The minimum in the equation (1) is reached for sequences α such that $\sum_{i=1}^n \alpha_i=K$.

Proof

It is sufficient to prove that if $\sum_{i=1}^n \alpha_i<K$ it is possible to add 1 to a well chosen α _i (and then increase the sum), and obtain a new sequence in V for which the new P ₁ − P ₂ is lower than the old one. Suppose that $\sum_{i=1}^n \alpha_i<K$.

• 1) Suppose that α _a + 1 ≥ b.

  • a) If (α ₁,...,α _a ) ≠ (q − 1,...,q − 1,q − 2) then there exists an index i such that 1 ≤ i ≤ a and such that α _i does not reach its maximal value. Then we can replace α _i by α _i  + 1. By Lemma A.2 we conclude that the new P ₁ − P ₂ is lower than the old one.

  • b) Now suppose that (α ₁,...,α _a ) = (q − 1,...,q − 1,q − 2).

    • α) If b = 0 then K = d + 1 = a(q − 1) + 1. But the sum of the first a elements is a(q − 1) − 1. So that α _a + 1 is at most 1 (this term exists because a ≤ n − 1). Then we can replace α _a + 1 by α _a + 1 + 1 because q ≥ 3. By Lemma A.2 we conclude that the new P ₁ − P ₂ is lower than the old one.

    • β) If b ≥ 1 then K = d + q − b = a(q − 1) + q. In this case we know that a ≤ n − 2. We have α _a + 1 + α _a + 2 + ...α _n  ≤ q then if α _a + 1 < q − 1 we can add 1 to this term, if α _a + 1 = q − 1 then α _a + 2 ≤ 1 and because q ≥ 3 it is possible to add 1 to to this term. By Lemma A.2 we conclude that the new P ₁ − P ₂ is lower than the old one.

• 2) Suppose that α _a + 1 < b. Then by Lemma A.2 if we replace α _a + 1 by α _a + 1 + 1, we obtain a new P ₁ − P ₂ lower than the old one.

From now on we will suppose that α is such that

$$ \sum\limits_{i=1}^n \alpha_i=K $$

Lemma A.4

Let 1 ≤ i ≤ a and a + 1 ≤ j ≤ n and suppose that α _j  > α _i . If we permute these two elements, and if we obtain a sequence which is in V, then for the new sequence the value of P₁ − P₂ is lower or equal to the old one.

Proof

Indeed P ₁ does not change, and P ₂ increases (if j > a + 1 or if j = a + 1 and α _j  > b) or does not change (if j = a + 1 and α _j  ≤ b).□

Lemma A.5

Suppose that 1 ≤ α _i  ≤ α _j  ≤ q − 2 and that we are in one of the following cases:

• (1) 1 ≤ j < i ≤ a;

• (2) a + 2 ≤ j < i ≤ n;

• (3) α_a + 1 ≥ b and a + 1 ≤ j < i ≤ n;

• (4) 1 ≤ j ≤ a and a + 2 ≤ i ≤ n.

Let us replace α _i by α _i  − 1 and α _j by α _j  + 1. If the new sequence is in V, the new value of P₁ − P₂ is lower than the old one.

Proof

• 1) Case 1 ≤ j < i ≤ a. The difference between the old value of P ₁ − P ₂ and the new value is

  $$ (\alpha_j-\alpha_i+1) \prod\limits_{k \neq i,j}(q-k) > 0. $$

• 2) Case a + 2 ≤ j < i ≤ n. The difference between the the old value of P ₁ − P ₂ and the new value is

  $$ (\alpha_j-\alpha_i+1) \left ( \prod\limits_{k=1}^{a+1} \left ( q-\alpha_k\right)-(q-\gamma) \right ) \prod\limits_{\begin{array}{c} k=a+2\\k\neq i,j \end{array}}^n \left ( q-\alpha_k\right) >0 . $$

  To verify that the previous expression is > 0 note that if α _a + 1 < b then γ = b and

  $$ \prod\limits_{k=1}^{a+1} \left ( q-\alpha_k\right) \geq (q-\alpha_{a+1}). $$

  Hence

  $$ \prod\limits_{k=1}^{a+1} \left ( q-\alpha_k\right)-(q-b)\geq (b-\alpha_{a+1})>0. $$

  If α _a + 1 ≥ b then γ = α _a + 1 and α _a  ≤ q − 2. Then

  $$ \prod\limits_{k=1}^{a+1} \left ( q-\alpha_k\right)-(q-\gamma) \geq 2(q-\alpha_{a+1}) - (q-\alpha_{a+1})=(q-\alpha_{a+1}) >0. $$

• 3) Case α _a + 1 ≥ b and a + 1 ≤ j < i ≤ n. The formula of the difference between the the old value of P ₁ − P ₂ and the new value is similar

  $$ (\alpha_j-\alpha_i+1) \left ( \prod\limits_{k=1}^{a} \left ( q-\alpha_k\right)-1 \right ) \prod\limits_{\begin{array}{c} k=a+1\\k\neq i,j \end{array}}^n \left ( q-\alpha_k\right) > 0. $$

  To verify that the previous expression is > 0 we have just to remark that α _a  ≤ q − 2, then $\prod_{k=1}^{a}\left ( q-\alpha_k\right)\geq 2$.

• 4) Case 1 ≤ j ≤ a and a + 2 ≤ i ≤ n. A simple computation shows that the difference between the old value of P ₁ − P2 and the new value is

  $$ (\alpha_j-\alpha_i+1)\prod\limits_{\begin{array}{c} k=1\\k\neq i,j \end{array}}^n \left ( q-\alpha_k\right)\quad +\quad (q-\gamma) \prod\limits_{\begin{array}{c} k=a+2\\\ k\neq i \end{array}}^n \left ( q-\alpha_k\right) >0 $$

1.2 A.2 The head of a best sequence

We give here the form of the first a terms of a sequence α for which P ₁ − P ₂ is minimum. We prove that α can be chosen such that one of the two following conditions holds:

• (1) (α ₁,...α _a − 1, α _a ) = (q − 1,...,q − 1,q − 2) and α _a + 1 ≥ b;

• (2) (α ₁,..., α _a ) = (q − 1,...,q − 1) and α _a + 1 < b;

• 1) Let us suppose first that there exists a sequence α such that α _a + 1 < b and for which P ₁ − P ₂ is minimum. We will prove that for such a sequences the first a terms can be set to q − 1. Suppose that there exists a j ≤ a such that α _j  < q − 1. We have

  $$ \sum\limits_{i=1}^{a+1} \alpha_i <a(q-1)+b=d, $$

  then α _a + 2 > 0. If α _a + 2 > α _j by Lemma A.4 we can permute the two terms to obtain a sequence which have a lower or equal P ₁ − P ₂. If α _a + 2 ≤ α _j , by Lemma A.5 the sequence obtained by replacing α _a + 2 by α _a + 2 − 1 and α _j by α _j  + 1 has a lower P ₁ − P ₂. So we have proved that we can increase the value of α _j and obtain a lower or equal P ₁ − P ₂.

• 2) Let us suppose now that there exists a sequence α, such that α _a + 1 ≥ b, for which P ₁ − P ₂ is minimum. We will prove that for such a sequence the first a − 1 terms can be set to q − 1 and α _a can be set to q − 2. Suppose that there exists a j ≤ a such that α _j  < q − 1 if j < a or or that α _j  < q − 2 if j = a.

  • a) If b = 0 then K = d + 1 = a(q − 1) + 1. But $\sum_{i=1}^{a} \alpha_i < a(q-1)-1$. Then α _a + 1 > 0. If α _a + 1 > α _j by Lemma A.4 we can permute the two terms to obtain a sequence which have a lower or equal P ₁ − P ₂. If α _a + 1 ≤ α _j , by Lemma A.5 the sequence obtained by replacing α _a + 1 by α _a + 1 − 1 and α _j by α _j  + 1 has a lower P ₁ − P ₂. So we have proved that we can increase the value of α _j and obtain a lower or equal P ₁ − P ₂.

  • b) If b > 0 then K = d + q − b = a(q − 1) + q. But $\sum_{i=1}^{a} \alpha_i < a(q-1)-1+$ and then $\sum_{i=1}^{a+1} \alpha_i < a(q-1)+q-2+$. Hence α _a + 2 > 0. With the same method than in the previous part 1) we prove that we can increase the value of α _j and obtain a lower or equal P ₁ − P ₂.

1.3 A.3 The tail of a best sequence

We give here the form of the terms α _i for i ≥ a + 1 of a sequence α for which P ₁ − P ₂ is minimum, assuming that the head is as in the previous subsection.

• 1) Let us suppose first that there exists a sequence α such that α _a + 1 < b and for which P ₁ − P ₂ is minimum. We have seen in the previous subsection that we can suppose that the first a terms are q − 1. We know that K = a(q − 1) + q. Then $\sum_{i=a+2}^{n} \alpha_i=q-\alpha_{a+1}$ using Lemma A.5 we can pack the terms α _i for i ≥ a + 2 in such a way that

  • a) if α _a + 1 = 0 then α _a + 2 = q − 1, α _a + 3 = 1 and α _i  = 0 for i > a + 3;

  • b) if b > α _a + 1 ≥ 1 then α _a + 2 = q − α _a + 1 and and α _i  = 0 for i > a + 2.

• 2) Let us suppose now that α _a + 1 ≥ b. We We have seen in the previous subsection that we can suppose that the a − 1 first a − 1 terms are q − 1 and α _a  = q − 2.

  • a) If b = 0 then K = a(q − 1) + 1. Then by Lemma A.5 we can pack the terms α _i for i ≥ a + 1 in such a way that α _a + 1 = 2 and α _i  = 0 for i > a + 1.

  • b) If b > 0 then K = a(q − 1) + q. Then by Lemma A.5 we can pack the terms α _i for i ≥ a + 1 in such a way that α _a + 1 = q − 1, α _a  + 2 = 2 and α _i  = 0 for i > a + 2.

1.4 A.4 The minimum value of P ₁ − P ₂

• 1) Case b=0. Then K = a(q − 1) + 1. The previous results give directly a sequence for which P ₁ − P ₂ is minimum:

  $$ \alpha_1=\cdots =\alpha_{a-1}= q-1, $$
  $$ \alpha_a=q-2, \quad \alpha_{a+1}=2, \quad \alpha_{a+2}=\cdots\alpha_n=0. $$

  For this sequence we have

  $$ P_1=2(q-2)q^{n-a-1} \quad P_2= (q-2)q^{n-a-1}, $$

  then the minimum value of P ₁ − P ₂ is

  $$ \mu = (q-2)q^{n-a-1}. $$

• 2) Case b=1. Then K = a(q − 1) + q and a ≤ n − 2.

  • a) Let us test first the assumption α _a + 1 = 0. The previous results give directly a sequence reaching the minimum of P ₁ − P ₂ under this assumption:

    $$ \alpha_1=\cdots =\alpha_{a}= q-1, $$
    $$ \alpha_{a+1}=0,\quad \alpha_{a+2}=q-1, \quad \alpha_{a+3}=1, \quad \alpha_{a+4}=\cdots\alpha_n=0. $$

    We remark that if a = n − 2 this case cannot occur because there is not enough room to contain all the α _i . For this sequence we have

    $$ P_1=q(q-1)q^{n-a-3},\quad P_2=(q-1)(q-1)q^{n-a-3}, $$

    so that the minimum of P ₁ − P ₂ under this assumption is

    $$ \mu_1=(q-1)q^{n-a-3}. $$

  • b) Now let us test the assumption α _a + 1 ≥ b = 1. The previous results give directly a sequence reaching the minimum of P ₁ − P ₂ under this assumption:

    $$ \alpha_1=\cdots =\alpha_{a-1}= q-1, $$
    $$ \alpha_{a}=q-2, \quad \alpha_{a+1}=q-1,\quad \alpha_{a+2}=2, \quad \alpha_{a+3}=\cdots\alpha_n=0. $$

    For this sequence we have

    $$ P_1=2(q-2)q^{n-a-2},\quad P_2=(q-2)q^{n-a-2}, $$

    so that the minimum of P ₁ − P ₂ under this assumption is

    $$ \mu_2=(q-2)q^{n-a-2}. $$

  • c) Conclusion on the case b = 1. Let us compare μ ₁ and μ ₂ (when a < n − 2)):

    $$ \mu_2-\mu_1=q^{n-a-1}-2q^{n-a-2}-q^{n-a-2}+q^{n-a-3}, $$
    $$ \mu_2-\mu_1=q^{n-a-2}(q-3)+q^{n-a-3}. $$

    But q ≥ 3, then μ ₂ > μ ₁. Hence the minimum value is μ ₁.

    Let us summarize the obtained result in the case b = 1:

    • if a < n − 2 then $\mu=\mu_1=(q-1)q^{n-a-3}$;

    • if a = n − 2 then $\mu=\mu_2=(q-2)q^{n-a-2}=q-2$.

• 3) Case 2 ≤ b < q − 1. Then K = a(q − 1) + q and a ≤ n − 2.

  • a) Test of the assumption α _a + 1 < b.

    • α) Test of the joint assumption α _a + 1 = 0. The previous results give directly a sequence reaching the minimum of P ₁ − P ₂ under this assumption:

      $$ \alpha_1=\cdots =\alpha_{a}= q-1, $$
      $$ \alpha_{a+1}=0,\quad \alpha_{a+2}=q-1, \quad \alpha_{a+3}=1, \quad \alpha_{a+4}=\cdots\alpha_n=0. $$

      This case cannot occur if a = n − 2. For this sequence we have

      $$ P_1=q(q-1)q^{n-a-3},\quad P_2=(q-b)(q-1)q^{n-a-3}. $$

      Then the minimum reached by P ₁ − P ₂ under this assumption is

      $$ \mu_1=b(q-1)q^{n-a-3}. $$

    • β) Test of the joint assumption α _a + 1 ≠ 0. The previous results shows that a sequence reaching the minimum of P ₁ − P ₂ under these assumptions is of the form

      $$ \alpha_1=\cdots =\alpha_{a}= q-1, $$
      $$ \alpha_{a+1}>0,\quad \alpha_{a+2}=q-\alpha_{a+1}, \quad \alpha_{a+3}=\cdots\alpha_n=0. $$

      For this sequence we have

      $$ P_1=(q-\alpha_{a+1})\alpha_{a+1}q^{n-a-2}, \quad P_2=(q-b)\alpha_{a+1}q^{n-a-2}, $$

      then

      $$ P_1-P_2=(b-\alpha_{a+1})\alpha_{a+1}q^{n-a-2}. $$

      The minimum of the quadratic polynomial (b − α _a + 1)α _a + 1 (with 1 ≤ α _a + 1 < b ≤ q − 2) is reached for α _a + 1 = 1 which gives for minimum of P ₁ − P ₂

      $$ \mu_2=(b-1)q^{n-a-2}. $$

  • b) Test of the assumption α _a + 1 ≥ b. The previous results shows that a sequence reaching the minimum of P ₁ − P ₂ under this assumption is

    $$ \alpha_1=\cdots =\alpha_{a-1}= q-1, $$
    $$ \alpha_{a}=q-2, \quad \alpha_{a+1}=q-1,\quad \alpha_{a+2}=2, \quad \alpha_{a+3}=\cdots\alpha_n=0. $$

    For this sequence we have

    $$ P_1=2(q-2)q^{n-a-2},\quad P_2=(q-2)q^{n-a-2}, $$

    so that the minimum of P ₁ − P ₂ under this assumption is

    $$ \mu_3=(q-2)q^{n-a-2}. $$

  • c) Conclusion of the case 2 ≤ b < q − 1. The minimum of P ₁ − P ₂ is

    $$ \mu =\min(\mu_1,\mu_2,\mu_3)=\mu_2=(b-1)q^{n-a-2}. $$

    Indeed, as q − 1 > b ≥ 2, we have q − 2 > b − 1 > 0, which prove that μ ₃ > μ ₂. To prove that μ ₁ > μ ₂ let us compute

    $$ \mu_1-\mu_2=b(q-1)q^{n-a-3}-(b-1)q^{n-a-2} = q^{n-a-2}-b\!q^{n-a-3}. $$

    But b < q, then μ ₁ − μ ₂ > 0.