$Y_b$ run over all fields. For fixed $i$, only a finite number of the $c^{X_i}_{ab}$ will be non-zero.

Starting at this order, it is possible to construct interactions that vanish on closed spin chain states because of the cyclic identification. These correspond to gauge transformations in gauge theory. An example of a one-site to two-site gauge transformation satisfies, for all fields $X_i$ and a fixed field $Y$, $${{\mathfrak{T}}}{|X_i\rangle} = {|X_i Y\rangle} - (-1)^{Y X_i} {|Y X_i\rangle}.$$ In fact, the symmetry algebra can be satisfied at this order only up to gauge transformations [@Beisert:2004ry].

Only the ${{\mathfrak{psu}}}(1|1)^2$ generators receive $\mathcal{O}(g)$ corrections since only they change the length of the spin chain. The action of $\overrightarrow{{{\mathfrak{T}}}}_1^{+}$ on states composed of the $\overrightarrow{\psi}_k$ was derived in [@Beisert:2004ry], and it is straightforward to generalize to the full solution by requiring commutation with the ${{\mathfrak{psu}}}(1,1|2)$ algebra (up to gauge transformations).

$$\begin{aligned}
 \overrightarrow{{{\mathfrak{T}}}_1}^{+}{|\phi^\updownarrow_m\rangle} &=& \frac{1}{\sqrt{2}}\sum_{k=0}^{m-1} \frac{1}{\sqrt{k+1}}\left({|\overrightarrow{\psi_k}\phi^\updownarrow_{m-1-k}\rangle}-{|\phi^\updownarrow_{m-1-k}\overrightarrow{\psi_k}\rangle}\right), \notag \\  
\overrightarrow{{{\mathfrak{T}}}_1}^{+} {|\overrightarrow{\psi_m}\rangle} &=& \frac{1}{\sqrt{2}}\sum_{k=0}^{m-1} \sqrt{\frac{m+1}{(k+1)(m-k)}} {|\overrightarrow{\psi_k}\overrightarrow{\psi_m}_{-1-k}\rangle}, \notag \\ 
\overrightarrow{{{\mathfrak{T}}}_1}^{+}{|\overleftarrow{\psi_m}\rangle} &=& \frac{1}{\sqrt{2}}\sum_{k=0}^{m-1} \sqrt{\frac{m-k}{(k+1)(m+1)}}\left({|\overrightarrow{\psi_k}\overleftarrow{\psi_m}_{-1-k}\rangle}+{|\overleftarrow{\psi_m}_{-1-k}\overrightarrow{\psi_k}\rangle}\right) \notag \\
&-& \frac{1}{\sqrt{2(m+1)}}\sum_{k=0}^{m} \left({|\phi^\downarrow_k\phi^\uparrow_{m-k}\rangle}-{|\phi^\uparrow_k\phi^\downarrow_{m-k}\rangle}\right). \end{aligned}$$

Up to a single minus sign in the last line, $\overleftarrow{{{\mathfrak{T}}}_1}^{+}$ follows from switching $\overrightarrow{\psi_k}$ and $\overleftarrow{\psi_k}$ in the above expression for $\overrightarrow{{{\mathfrak{T}}}_1}^{+}$, $$\begin{aligned}
 \overleftarrow{{{\mathfrak{T}}}_1}^{+}{|\phi^\updownarrow_m\rangle} &=& \frac{1}{\sqrt{2}}\sum_{k=0}^{m-1} \frac{1}{\sqrt{k+1}}\left({|\overleftarrow{\psi_k}\phi^\updownarrow_{m-1-k}\rangle}-{|\phi^\updownarrow_{m-1-k}\overleftarrow{\psi_k}\rangle}\right), \notag \\  
