= Solution
The original symmetry flips every $X$-basis bit, $s_i\mapsto s_i+1$, without changing any domain wall. Hence
$$
DG=D,
\qquad
D^\dagger D=I+G
$$
with the present normalization. On the dual side, periodic domain walls obey
$$
\prod_i\widetilde Z_{i+1/2}=1,
$$
so the original global symmetry becomes a constraint on the dual symmetry sector.
Periodic versus antiperiodic boundary conditions determine whether the product of dual domain walls is $+1$ or $-1$. Conversely, the original even and odd $G$ sectors correspond to choices of dual boundary twist. Keeping all sectors therefore requires summing over both symmetry charges and both boundary conditions; within each matched sector the intertwiner is invertible up to normalization.
If an operator $O$ is symmetric, $[O,G]=0$, it descends consistently to a dual operator satisfying $DO=\widetilde O D$. If it is nonsymmetric, it changes the global charge and cannot be represented by a local operator within one fixed dual boundary sector; its dual either changes the twist, acquires a disorder string, or is annihilated by the projected intertwiner.
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