the generators of ${{\mathfrak{psu}}}(1,1|2)$ and ${{\mathfrak{psu}}}(1|1)^2$ commute with each other.

Fields and states {#sec:States}
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The fields in this sector are derivatives ${\mathcal{D}}={\mathcal{D}}_{22}$ acting on the fermions $\overrightarrow{\psi}=\Psi_{42}$ and $\overleftarrow{\psi}=\dot{\Psi}^1_2$, or the bosons $\phi^\downarrow=\Phi_{34}$ and $\phi^\uparrow=\Phi_{24}$. We denote $k$ derivatives by a subscript $k$, $$\phi^\updownarrow_k\sim{\mathcal{D}}^k \phi^\updownarrow, \quad \overleftrightarrow{\psi_k}\sim{\mathcal{D}}^k \overleftrightarrow{\psi}, \qquad (k \geq 0).$$

The representation of the symmetry algebra acts on a spin chain. The states of the spin chain are tensor products $${|X_1X_2 \ldots X_n\rangle} \quad \mbox{where} \quad X_i \in \left\{\phi^\updownarrow_k, \, \overleftrightarrow{\psi}_k \right\}.$$ A generic state is a linear combination of these tensor products, with the cyclic identification $${|X_1 \ldots X_i X_{i+1} \ldots X_n\rangle}= (-1)^{(X_1\ldots X_i)(X_{i+1} \ldots X_n)} {|X_{i+1} \ldots X_n X_1 \ldots X_i\rangle}$$ $(-1)^{AB}$ is $-1$ if both $A$ and $B$ are fermionic, and $1$ otherwise.

The leading order representation {#sec:Reps}
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For the leading order representation of the symmetry algebra, at $\mathcal{O}(g^0)$, the generators have one-site to one-site vertices. That is, the generators’ action on the spin chain is a tensor product (given by the sum of its action on each individual site). Also, every time a fermionic generator passes a fermionic field, we must add a factor of $-1$. The non-vanishing actions for ${{\mathfrak{J}}}^{++}$, $\overrightarrow{{{\mathfrak{Q}}}}^{+\uparrow}$, 