Are the coefficients of the weight enumerator polynomial of a stabilizer code always integers?

Question

Consider an $ [\![n,k]\!] $ stabilizer code. Define the weight enumerator polynomial $ A(x) $ of the code as $$ A(x):=A_0+A_1x+\dots+A_nx^n $$
where $$ A_j:=\frac{1}{(2^k)^2} \sum_{p \in P_n,\,\mathrm{wt}(p)=j} |\mathrm{tr}(p \Pi)|^2. $$ Here $ \Pi $ is the projector onto the code subspace. Is it the case that the $ A_j $ are always integers?

Answer 1

TL;DR: For the stabilizer codes, $A_j$ is the number of stabilizer operators with Hamming weight $j$ and thus a non-negative integer.

---

Suppose $g_1,\dots,g_{n-k}$ are generators of the stabilizer group $S$ of a $[\![n,k]\!]$ stabilizer code. Define $S^w:=\{s\in S\,|\,\mathrm{wt}(s)=w\}$ and $P_n^w:=\{p\in P_n\,|\,\mathrm{wt}(p)=w\}$. Then $$ \begin{align} A_j &= \frac{1}{4^k}\sum_{p\in P_n^j}[\mathrm{tr}(p\Pi)]^2\tag1\\ &= \frac{1}{4^k}\sum_{p\in P_n^j}\left[\mathrm{tr}\left(p\prod_{i=1}^{n-k}\frac{I+g_i}{2}\right)\right]^2\tag2\\ &= \frac{1}{4^k}\sum_{p\in P_n^j}\left[\frac{1}{2^{n-k}}\mathrm{tr}\left(p\sum_{b_1=0}^1\dots\sum_{b_{n-k}=0}^1g_1^{b_1}\dots g_{n-k}^{b_{n-k}}\right)\right]^2\tag3\\ &= \frac{1}{4^n}\sum_{p\in P_n^j}\left[\mathrm{tr}\left(p\sum_{s\in S}s\right)\right]^2\tag4\\ &= \frac{1}{4^n}\sum_{p\in P_n^j}\left(\sum_{s\in S}\mathrm{tr}(ps)\right)^2\tag5\\ &= \frac{1}{4^n}\left[\sum_{p\in P_n^j\setminus S}\left(\sum_{s\in S}\mathrm{tr}\left(ps\right)\right)^2+\sum_{p\in S^j}\left(\sum_{s\in S}\mathrm{tr}\left(ps\right)\right)^2\right]\tag6\\ &= \frac{1}{4^n}\left(0+\sum_{p\in S^j}\left(2^n\right)^2\right)\tag7\\ &=|S_j|\tag8 \end{align} $$ in analogy with the classical case. This gives us another way to see that $A_0+\dots+A_n=2^{n-k}=|S|$.