\\  + & \sum_{\substack{0 \leq m  \\ 0 \leq k < m}} & \frac{m-k}{\sqrt{2(k+1)(m+1)}}\,\mbox{Tr}\,{\left\{\overrightarrow{\psi_k},\overleftarrow{\psi_m}_{-1-k}\right\}} \check{\overleftarrow{\psi_m}} \nonumber \\ + & \sum_{\substack{0 \leq m \\ 0 \leq k \leq m}} & \frac{1}{\sqrt{2(m+1)}}\,\mbox{Tr}\,{\left[\phi_k^\downarrow,\phi_{m-k}^\uparrow\right]} \check{\overleftarrow{\psi_m}}. \end{aligned}$$

There is a similar expression for $\overleftarrow{{{\mathfrak{T}}}_1}^+$, which can be read from (\[eq:leftT1+\]). For the hermitian conjugates, the ${{\mathfrak{T}}}_1^-$, simply perform the switch $$X_i \leftrightarrow \check{X}_i, \quad \forall X_i.$$ Substituting these expressions into the expressions for ${{\mathfrak{x}}}$, gives its non-planar version. Then the expressions given for ${{\mathfrak{T}}}_3$ and ${\delta{\mathfrak{D}}}_4$ become non-planar. Because the proof that the planar solution satisfies the symmetry constraints is independent of planarity, the non-planar generalization still satisfies the symmetry constraints.

While we do not have a proof that this is the correct non-planar solution, our solution accurately includes wrapping interactions, which can be thought of as special cases of non-planar interactions. Wrapping interactions apply to two-site states, for which the planar solution and the non-planar generalization are equivalent. Since the ${{\mathfrak{T}}}_1$’s map one site to two sites or vice-versa, the action of ${\delta{\mathfrak{D}}}_4$ is well defined even on two-site states. Since acting with the ${{\mathfrak{T}}}^+$’s on two-site states yields three-site states, two-site states are in the same multiplets as three-site states. Therefore, adding special 