# Why is $\text{dim}_{\mathbb{C}}Q(\chi) = 2^{n-(n-k)}$?

I am confused about a point in the proof of Theorem 27.3.6 which claims that $$\text{dim}_{\mathbb{C}}Q(\chi) = 2^{n-(n-k)}$$ in a "Concise Encyclopedia of Coding Theory" (page 663/664). The details are as follows:

Theorem 27.3.6: Let $$C$$ be an $$(n-k)-$$dimensional self-orthogonal subspace of $$\mathbb{F}_{2}^{n-k}$$ under the symplectic inner product. Let $$d := w_{Q} (C^{\perp_{s}}$$\ $$C) = \min$${$$wt_{Q}(v) \in > C^{\perp_{s}}$$\ $$C$$}. Then there is an $$[[n,k,d]]$$-qubit stabilizer code $$Q$$.

Useful definitions:

• A quantum operator $$E$$ on $$\mathbb{C}^{2^{n}}$$ is of the form $$i^{\lambda}X(a)Z(b)$$
• $$\epsilon_{n}$$ is the set of error operators $$E$$
• $$\overline{\epsilon_{n}}=\epsilon_{n} / Z(\epsilon_{n})$$
• $$C:=\overline{G} \in \overline{\epsilon}_{n}$$ Please find the relevant section of the proof of this theorem attached below. Please note that $$Q(\chi) = L_{\chi}(C^{2^{n}})$$ is the subspace $$Q(\chi) =$${$$|v\rangle \in \mathbb{C}^{2^{n}}|g|v\rangle = \chi(g) |v\rangle \text{ } \forall g \in G$$}.

Proof

Here is proposition 27.3.5

I don't understand how they came to the conclusion that $$\text{dim}_{\mathbb{C}}Q(\chi) = 2^{n-(n-k)}$$. I understand that the dimension of $$C$$, which we lift to $$\overline{G}$$ is $$n-k$$. And I understand that $$\text{dim}_{\mathbb{C}}Q(\chi)$$ is the same for all $$\chi \in \hat{G}$$. But I can't figure out why the dimension of $$Q(\chi)$$ is $$2^{2-(n-k)}$$.

Any help or hints would be greatly appreciated! Perhaps I am missing something simple.

The codespace is the simultaneous $$+1$$-eigenspace of the independent commuting Pauli generators $$G_1,\ldots, G_{(n-k)}$$ which act on a vector space of dimension $$2^n$$ (i.e. bit strings). The dimension of the $$+1$$-eigenspace of a single $$G_i$$ is $$2^n/2=2^{n-1}$$. This can be seen via Lagrange's theorem, as each generator induces a surjective group homomorphism $$\mathbb{Z}_2^n\to \mathbb{Z}_2$$, that sends each codeword in $$\{0,1\}^n$$ to $$\pm1$$, (these are the characters $$f_\alpha(z)=(-1)^{\langle \alpha,z\rangle}$$ of the abelian subgroup generated by $$G_1,\ldots, G_{(n-k)}$$). Since all the generators commute and are independent (no generator is a product of the other generators) it follows that the dimension of the simultaneous $$+1$$-eigenspace is $$2^n/2^{n-k}=2^{n-(n-k)}=2^k$$, that is the dimension of the space is halved by each of the $$n-k$$ generators (see Why do stabilizer cut the Hilbert space into two halves?).