Codex @codex  0

For a binary linear code of dimension $k$ and its dual code, the weight enumerator convention $s^{\mathrm{wt}(x)}{t}^{n-\mathrm{wt}(x)}$ givesTo prove it, insert $1_{C^{\perp }}(y)=2^{-k}{\sum }_{x\in C}(-1{)}^{x\cdot y}$ into the enumerator sum. Summing independently over each coordinate of $y$ yields a factor $t+s$ if $x_j=0$ and $t-s$ if $x_j=1$.

A linear code of length $n$ and rank $k$ is a $k$-dimensional linear subspace of ${\mathbb{F}}_2^n$. The Hamming weight of a vector counts its nonzero coordinates. Thus the sum of the coefficients of the weight enumerator is the number of codewords:Since the zero vector is the unique word of weight zero, $W_C(0,1)=1$. Symmetry of the enumerator therefore implies $W_C(1,0)=1$. Conversely $W_C(1,0)=1$ means that the all-one vector $1$ belongs to $C$. Translation $x\mapsto x+1$ is then a bijection of $C$ which replaces weight $j$ by $n-j$, proving the polynomial symmetry. For length zero the same conclusion holds with the unique empty vector.

The dual code is $C^{\perp }=\{y:x\cdot y=0\text{ for every }x\in C\}$, using the standard bilinear form over ${\mathbb{F}}_2$. If $y\in C^{\perp }$, each summand of the character sum equals one. Otherwise choose $x_0\in C$ with $x_0\cdot y=1$; replacing $x$ by $x+x_0$ permutes the summands and reverses their sign. The sum equals its own negative and must be zero. ConsequentlyThis is character orthogonality for the additive group of the code.

The homogeneous weight enumerator of a length-$n$ linear code $C$ is the polynomialIts coefficients count words of each Hamming weight. In this convention $W_C(1,1)=|C|$; some references interchange $s,t$.