and ${{\mathfrak{R}}}^{\uparrow\uparrow}$ are $$\begin{aligned}
 \label{eq:FieldGen}
{{\mathfrak{J}}}^{++}_0  {|\phi^\updownarrow_k\rangle}&=(k+1){|\phi^\updownarrow_{k+1}\rangle},&
{{\mathfrak{J}}}^{++}_0  {|\overleftrightarrow{\psi}_k\rangle}&=\sqrt{(k+1)(k+2)}{|\overleftrightarrow{\psi}_{k+1}\rangle},
\notag \\
\overrightarrow{{{\mathfrak{Q}}}_0}^{+\uparrow} {|\phi^\downarrow_k\rangle}&=\sqrt{k+1}{|\overrightarrow{\psi}_k\rangle},&
\overrightarrow{{{\mathfrak{Q}}}_0}^{+\uparrow} {|\overleftarrow{\psi}_k\rangle}&=-\sqrt{k+1}{|\phi^\uparrow_{k+1}\rangle}, \notag \\
{{\mathfrak{R}}}^{\uparrow\uparrow}{|\phi^\downarrow_k\rangle}&={|\phi^\uparrow_k\rangle}.\end{aligned}$$ The remaining parts of the leading order representation can be computed using hermitian conjugation and the algebra given in Appendix \[sec:psu112comm\]. To find hermitian conjugates, it is simplest to use the (tensor product of the) diagonal metric $$\langle \phi^\downarrow_m {|\phi^\downarrow_n\rangle} = \delta_{mn}, \quad \langle \phi^\uparrow_m {|\phi^\uparrow_n\rangle} = \delta_{mn}, \quad \langle{\overrightarrow{\psi}_m} {|\overrightarrow{\psi}_n\rangle} =  \delta_{mn}, \quad \langle \overleftarrow{\psi}_m {|\overleftarrow{\psi}_n\rangle} =  \delta_{mn}.$$ Then the condition for two operators, ${{\mathfrak{J}}}$ and ${{\mathfrak{J}}}^\dagger$, to be hermitian conjugates is $$\begin{gathered}
  {\langleX'_1 \ldots X'_m|}{{\mathfrak{J}}}{|X_1\ldots X_n\rangle}= {\langleX_1\ldots X_n|}{{\mathfrak{J}}}^\dagger {|X'_1 \ldots X'_m\rangle}. \end{gathered}$$

Constraints from Feynman rules
------------------------------

Before beginning to discuss quantum corrections to the symmetry algebra, we review the constraints from Feynman diagrams [@Beisert:2004ry]. Most simply, only connected interactions appear. For the planar theory, this implies that interactions will involve replacing a set of adjacent spins of the spin chain with a new set, not necessarily of the same length. In fact, power counting implies that the $\mathcal{O}(g^n)$ term of a generator is the sum of interactions involving a total of up to $(n+2)$ initial and final spin sites. Therefore, quantum corrections will deform the representation from the tensor 