Failure of duality for power weights over finite residue rings

Power weights over integer residue rings

For $m\ge4$, write $R_m=\mathbb Z/m\mathbb Z$ and let

$$w_{m,\ell}(x)=\min\{r,m-r\}^{\ell}, \qquad x\equiv r\pmod m,\quad 0\le r<m,$$

where $\ell\ge2$ is an integer. Extend this weight additively to vectors. For a linear code $C\le R_m^n$, put

$$W_C(y)=\sum_{c\in C}y^{w_{m,\ell}(c)}, \qquad A_j(C)=\#\{c\in C:w_{m,\ell}(c)=j\}.$$

Its dual is $C^\perp=\{z\in R_m^n:c\cdot z=0\text{ for every }c\in C\}$. Since all code pairs below have the same length, equality of these one-variable enumerators is equivalent to equality of the homogeneous enumerators in the original definition.

Theorem. For every $m\ge4$ and integer $\ell\ge2$, there are linear codes $C,D\le R_m^n$, for some $n$, such that

$$W_C=W_D, \qquad W_{C^\perp}\ne W_{D^\perp}.$$

Thus the power weight does not respect duality. This proves Conjecture 9.5 of Jay A. Wood, arXiv:2601.02608.

The argument first handles prime moduli using Wood's Theorem 2.8. A singular weight matrix is dealt with directly; in the nonsingular case, a power-sum inequality verifies the remaining hypothesis of that theorem. Explicit constructions cover $4,6,9$, and a divisor-lifting argument then covers every modulus. The divisor-reduction strategy was used by N. Abdelghany and J. A. Wood in their 2020 Lee-weight paper; the low-coefficient lifting argument needed here is proved below.

1. Two coefficients that survive lifting

For every modulus at least four, the only symbols of weight $1$ are $1$ and $-1$, and every other nonzero symbol has weight at least $2^\ell\ge4$. Consequently a vector of weight $1$ is a singleton with entry $\pm1$, and a vector of weight $2$ has exactly two entries, both in $\{1,-1\}$.

Suppose $d\mid m$, with $d\ge4$, and put $q=m/d$. The map

$$\iota:R_d^n\longrightarrow R_m^n, \qquad x\longmapsto qx,$$

is injective, and the image of an $R_d$-linear code is $R_m$-linear. Moreover,

$$w_{m,\ell}(qx)=q^\ell w_{d,\ell}(x).$$

Thus $W_C=W_D$ implies $W_{\iota(C)}=W_{\iota(D)}$. If $\pi:R_m^n\to R_d^n$ is coordinatewise reduction, then

$$\iota(C)^\perp=\pi^{-1}(C^\perp),$$

because $qc\cdot z=0\pmod m$ is equivalent to $c\cdot\pi(z)=0\pmod d$. Reduction gives a bijection between the ambient vectors of weight $j$ over these two rings for $j=1,2$: it preserves their support and their entries $\pm1$. Hence

$$A_j\bigl(\iota(C)^\perp\bigr)=A_j(C^\perp), \qquad j=1,2.$$

It therefore suffices to construct a pair distinguished by $A_1$ or $A_2$ for each prime $p\ge5$ and for $d=4,6,9$.

2. Every prime modulus

Wood's Example 9.4 already discusses the small primes $5$ and $7$. The argument here is uniform in the prime and does not assume that every power-weight matrix is nonsingular.

Fix a prime $p=2t+1\ge5$. The symmetry group of $w=w_{p,\ell}$ is $H=\{1,-1\}$, and the minimum positive weight is attained on exactly this orbit. Choose representatives $r_1,\ldots,r_t$ of the nonzero $H$-orbits and form the integer matrix

$$\mathcal W_{ij}=w(r_i r_j).$$

Every row and column has sum

$$S_\ell=\sum_{a=1}^t a^\ell.$$

The singular case

If $\mathcal W$ is singular, take a nonzero integral vector $v$ with $\mathcal Wv=0$. The constant column sum gives $\sum_i v_i=0$. Choose an integer $M>\max_i|v_i|$, and define multiplicities

$$\eta_i=M,\qquad \eta'_i=M+v_i.$$

Let $C$ and $D$ be the one-dimensional codes whose generator rows contain respectively $\eta_i$ and $\eta'_i$ copies of $r_i$. All multiplicities are positive, both generator maps are injective, and the codes have the same length. The identity $\mathcal W\eta=\mathcal W\eta'$ says that the corresponding scalar multiples of the two generator rows have equal weights, so $W_C=W_D$.

A dual vector of weight $2$ has two entries $\pm1$. For a generator row with no zero entries, these can cancel precisely when their two columns lie in the same $H$-orbit. There are two choices of signs for each pair. Therefore

$$A_2(C^\perp)=2\sum_i\binom{\eta_i}{2}, \qquad A_2(D^\perp)-A_2(C^\perp) =\sum_i v_i^2>0.$$

The nonsingular case

Now suppose $\mathcal W$ is nonsingular. We verify the hypotheses of Wood's Theorem 2.8. The symmetry and minimum-orbit hypotheses have already been checked, and nonsingularity is precisely nondegeneracy in that theorem.

Let $g$ be a primitive element of $\mathbb F_p$, write $a_j=w(g^j)$ for $j\in\mathbb Z/t\mathbb Z$, and put

$$c_h=\sum_{j=0}^{t-1}a_j a_{j+h}.$$

The values $a_j$ are a permutation of $1^\ell,\ldots,t^\ell$. Thus

$$c_0=S_{2\ell},\qquad c_h=c_{t-h}, \qquad \sum_{h=0}^{t-1}c_h=S_\ell^2.$$

If all the quantities in the last hypothesis of Wood's theorem were equal, then

$$pc_1=\cdots=pc_{\lfloor t/2\rfloor}=2S_\ell^2.$$

The preceding identities would imply

$$pS_{2\ell}=3S_\ell^2.$$

This is impossible. For real $x>0$, set $S_x=\sum_{a=1}^t a^x$. The function $F(x)=\log S_x$ is convex, since $F''(x)$ is the variance of $\log a$ under the probabilities $a^x/S_x$. Consequently

$$\frac{d}{dx}\log\frac{S_{2x}}{S_x^2} =2F'(2x)-2F'(x)\ge0.$$

At $x=2$, the usual power-sum formulas give

$$(2t+1)S_4-3S_2^2 =\frac{t(t+1)(2t+1)^2(t-1)(t+2)}{60}>0.$$

Since $\ell\ge2$, it follows that $pS_{2\ell}>3S_\ell^2$. Wood's theorem therefore applies. Its proof, using Proposition 4.8, constructs equal-length codes with equal weight enumerators whose dual $A_2$-coefficients differ. This completes the prime case, including the coefficient needed for lifting.

3. Modulus four

Failure for this modulus is already a consequence of Wood's classification; see Corollary 9.6 of arXiv:2404.07154v1, published in Contemporary Mathematics 826 (2025), 361–430. The following explicit pair also supplies the particular low-weight coefficient needed for lifting.

Put $a=2^\ell$ and $q=a/2-1$. Thus $a\ge4$ is even and $q\ge1$. Use length $n=a+q$. Let $C$ be generated over $R_4$ by the row containing $a$ copies of $1$, followed by $q$ copies of $2$. Let $D$ be generated by two rows: the first is $2$ on the first $a/2$ positions and zero elsewhere, while the second is $2$ on the next $a/2$ positions and zero elsewhere.

Both codes have four elements, and direct calculation gives

$$W_C(y)=W_D(y)=1+2y^{a^2/2}+y^{a^2}.$$

The generator row of $C$ has no zero columns. The two-row generator matrix of $D$ has exactly $q$ zero columns. Since a vector of weight $1$ is a singleton with entry $\pm1$, we obtain

$$A_1(C^\perp)=0, \qquad A_1(D^\perp)=2q=a-2>0.$$

4. Modulus six

Put

$$a=2^\ell,\qquad b=3^\ell,\qquad h=a+b-1, \qquad z=b-a-1.$$

Here $z>0$, since $3^\ell>2^\ell+1$. A generator row over $R_6$ with $u,v,w$ copies of $1,2,3$, respectively, has nonzero scalar-orbit weights

$$\begin{pmatrix} 1&a&b\\ a&a&0\\ b&0&b \end{pmatrix} \begin{pmatrix}u\\v\\w\end{pmatrix}.$$

The first two orbits, $\{\pm1\}$ and $\{\pm2\}$, have size two; the third is $\{3\}$.

Take the two multiplicity vectors

$$\eta=(2b-a,\,0,\,3a-2), \qquad \eta'=(2b+a,\,z,\,a-2).$$

Their entries are nonnegative integers, and each first coordinate is positive. Multiplication by the displayed matrix gives, respectively,

$$\begin{pmatrix} a(3b-1)\\ a(2b-a)\\ 2b(a+b-1) \end{pmatrix}, \qquad \begin{pmatrix} a(2b-a)\\ a(3b-1)\\ 2b(a+b-1) \end{pmatrix}.$$

Thus the two size-two orbit weights are interchanged and the third is unchanged. The second generator row is $z$ coordinates longer than the first. Append $z$ zero coordinates to the first row. The resulting cyclic codes $C,D$ have the same length, each has six elements, and $W_C=W_D$.

A weight-$1$ vector belongs to the dual precisely when its nonzero coordinate is a zero column of the generator row. Hence

$$A_1(C^\perp)=2z, \qquad A_1(D^\perp)=0.$$

5. Modulus nine

Put

$$a=2^\ell,\qquad b=3^\ell,\qquad c=4^\ell, \qquad S=1+a+c,\qquad L=S-2b.$$

Convexity of $x\mapsto x^\ell$ gives $2^\ell+4^\ell\ge2\cdot3^\ell$, so $L>0$.

First let $C_0$ be the cyclic code over $R_9$ generated by a row containing $b$ copies of each of $1,2,4$. It has length $3b$ and size nine. Multiplication by a unit permutes the three unit negation orbits, whose symbol weights are $1,a,c$. The six unit multiples of the row therefore have weight $bS$. Its two nonzero nonunit multiples, by $3$ and $6$, have weight $3b^2$. Thus

$$W_{C_0}(y)=1+6y^{bS}+2y^{3b^2}.$$

Next construct a two-dimensional ternary code $B$. Use the four projective column directions

$$(1,0),\quad (0,1),\quad (1,1),\quad (1,2)$$

with multiplicities $b,b,b,L$, respectively. Its length is $3b+L=S+b$. Every nonzero linear functional on $\mathbb F_3^2$ vanishes on exactly one of these four directions, and each direction is the kernel of exactly two nonzero functionals. Consequently its Hamming weight enumerator is

$$W_B^{\mathrm H}(y) =1+6y^{(S+b)-b}+2y^{(S+b)-L} =1+6y^S+2y^{3b}.$$

Embed $B$ in $R_9^{S+b}$ by multiplying all entries by $3$, and call the resulting $R_9$-linear code $D$. Each nonzero embedded symbol has power weight $b$, so

$$W_D(y)=1+6y^{bS}+2y^{3b^2}=W_{C_0}(y).$$

Append $L$ zero coordinates to $C_0$, obtaining $C$ of the same length as $D$. All projective-column multiplicities are positive, so $B$ has dimension two and neither its generator matrix nor that of $D$ has a zero column. Therefore

$$A_1(C^\perp)=2L>0, \qquad A_1(D^\perp)=0.$$

6. All remaining moduli

Every integer $m\ge4$ has a divisor that is either a prime $p\ge5$ or one of $4,6,9$. Indeed, if its only prime divisors are $2$ and $3$, then either $4\mid m$, or $9\mid m$, or $m=6$.

For such a divisor, the preceding sections provide equal-length codes with equal power-weight enumerators and different dual $A_1$ or $A_2$. Section 1 lifts that pair to $R_m$ and preserves the distinguishing coefficient. This proves the theorem for every stipulated modulus and exponent. $\square$