Quantum repetition subspace projector and a distance-one boundary error

From the standard pairing in [F1], $⟨e_i,e_j{⟩}_L={\sum }_{r<2}{\delta }_{ri}\overline{{\delta }_{rj}}={\delta }_{ij}$, so $e_0,e_1$ are orthonormal. Each physical factor is also $Q_2$ with this pairing; [F2] gives $⟨u_x,u_y{⟩}_P={\prod }_{k<3}{\delta }_{x_k{y}_k}$, and [F3] gives the ordered product basis. Thus $u_{000},u_{111}$ are orthonormal, and $V$ is isometric. For each basis vector $u_x$, [F5] gives $⟨e_i,V^{\ast }{u}_x{⟩}_L=⟨V{e}_i,u_x{⟩}_P$, which is $1$ for $(i,x)=(0,000)$ or $(1,111)$ and $0$ otherwise. Since $⟨e_i,y{⟩}_L=\overline{y_i}$, this computes $V^{\ast }{u}_{000}=e_0$, $V^{\ast }{u}_{111}=e_1$, and $V^{\ast }{u}_x=0$ for the other six basis strings. Thus $P=V{V}^{\ast }$ has exactly the stated diagonal entries and is the code projector in [L1] and [L2].

The map $E_X$ sends $u_{000}$ to $u_{100}$ and $u_{111}$ to $u_{011}$, both orthogonal to the code subspace by [F3]. Thus $P{E}_{X}P=0$, so $E_X$ is detectable with scalar $0$. By [F1] and [F4], its action on elementary tensors is $X\otimes I\otimes I$; hence it is supported on the first wire and has weight at most $1$.

The map $E_Z$ fixes $u_{000}$ and negates $u_{111}$. Thus $P{E}_{Z}P$ acts by eigenvalues $+1$ and $-1$ on the two-dimensional code subspace and is not a scalar multiple of $P$. By [F1] and [F4], it is $Z\otimes I\otimes I$, so it has weight at most $1$. It is non-detectable; every weight-zero operator is detectable by [L5]. Therefore the minimum non-detectable weight is $d=1$, and $\mathrm{wt}(E_Z)=d$.

By [L6], every operator of weight $<d=1$ is detectable; step 1.3 exhibits a non-detectable operator of weight exactly $d$. Thus the detection boundary is sharp, and no correction-map claim follows.