Info from https://arxiv.org/pdf/quant-ph/9608006.pdf
Background information
Let $E$ be the group of possible errors on $n$ qubits.
$S'$ is a subgroup of $E$, which consists of the undetectable errors (bring codewords to codewords)
$S$ is an abelian subgroup of $S'$, which consists of errors which have no effect on the encoded space.
Note, every element of $S'$ commutes with $S$. In fact, $S'$ is the centralizer of $S$ in $E$.
Since $S$ is abelian, its elements can be simultaneously diagonalised, inducing a decomposition of $\mathbb{C}^{2^{n}}$ into orthogonal eigenspaces. We take the code $Q$ to be one of these eigenspaces.
To each eigenspace of $S$, there corresponds a homomorphism $\chi: S \rightarrow \mathbb{C}$, under which each element of $S$ is mapped to the corresponding eigenvalue. Then $\chi$ is a character of $S$ and $\chi(iI)=i$
My problem:
I am having a little trouble interpreting how the relationship between elements of the various subgroups affects the action on characters. The following bullet points are quotes from the text and my (most likely incorrect) interpretation of them:
- Statement: "Every element $e \in E$ normalises $S$ and so conjugation by $e$ induces an action on characters."
- My Interpretation: $eSe^{-1}=S$, so $\chi(eSe^{-1})=\chi(S)=${$\lambda$}
- Statement: "Since $S'$ commutes with $S$, elements of $S'$ induce the trivial action on characters"
- Interpretation: $s's=ss'$. If $ \chi(S)=${$\lambda$}$ \rightarrow \chi(Ss')=\chi(s'S)=${$\lambda$}
- Statement: "Any element outside of $S'$ negates the value of the character at each element of $S$ with which it anticommutes"
- Interpretation: $e \in E, e \notin S'$ $s \in S$, then if $es=-se$, and if $\chi(s)=\lambda_{i}$ then $\chi(es)=-\lambda_{i}$
Would somebody able to tell me where I have gone wrong with these interpretations?