exactly parallels that of the previous order. Under a homogeneous modification, $$\overleftrightarrow{{{\mathfrak{T}}}_3}^{\pm} \mapsto \overleftrightarrow{{{\mathfrak{T}}}_3}^{\pm} + \delta\overleftrightarrow{{{\mathfrak{T}}}_3}^{\pm}.$$ In order for the symmetry constraints to remain satisfied, the $\delta {{\mathfrak{T}}}$’s must not contribute to any commutator of the algebra, both for commutators among the ${{\mathfrak{psu}}}(1|1)^2$ generators and for those with ${{\mathfrak{psu}}}(1,1|2)$ generators. Again, we have not found any non-trivial homogeneous solutions, or ruled them out. However, as for the $\mathcal{O}(g^2)$ solution, the successful checks of our solution with field theory computations implies that such a homogeneous contribution is not part of the field theory solution.

The two-loop dilatation operator
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From (\[eq:psU11\^2Alg\]) we can now compute ${\delta{\mathfrak{D}}}_4$ directly, $${\delta{\mathfrak{D}}}_4=2 {\left\{\overrightarrow{{{\mathfrak{T}}}}^+,\overleftarrow{{{\mathfrak{T}}}}^-\right\}}_4=2 {\left\{\overleftarrow{{{\mathfrak{T}}}}^+,\overrightarrow{{{\mathfrak{T}}}}^-\right\}}_4.$$ It follows that ${\delta{\mathfrak{D}}}_4$ is composed only of the ${{\mathfrak{T}}}_1$’s and ${{\mathfrak{h}}}$, the one-site harmonic number generator. After setting the similarity transformation ${{\mathfrak{y}}}$ to zero, without loss of generality, and using the vanishing of the squares of the ${{\mathfrak{T}}}$’s, we find $$\begin{aligned}
 {\delta{\mathfrak{D}}}_4&=& 2 {\left\{\overrightarrow{{{\mathfrak{T}}}_1}^+,{\left[\overleftarrow{{{\mathfrak{T}}}_1}^-,{\left\{\overleftarrow{{{\mathfrak{T}}}_1}^+,{\left[\overrightarrow{{{\mathfrak{T}}}_1}^-,{{\mathfrak{h}}}\right]}\right\}}\right]}\right\}}+ 2{\left\{\overleftarrow{{{\mathfrak{T}}}_1}^-,{\left[\overrightarrow{{{\mathfrak{T}}}_1}^+,{\left\{\overrightarrow{{{\mathfrak{T}}}_1}^-,{\left[\overleftarrow{{{\mathfrak{T}}}_1}^+,{{\mathfrak{h}}}\right]}\right\}}\right]}\right\}} \nonumber \\ &=&   2 {\left\{\overleftarrow{{{\mathfrak{T}}}_1}^+,{\left[\overrightarrow{{{\mathfrak{T}}}_1}^-,{\left\{\overrightarrow{{{\mathfrak{T}}}_1}^+,{\left[\overleftarrow{{{\mathfrak{T}}}_1}^-,{{\mathfrak{h}}}\right]}\right\}}\right]}\right\}}+ 2{\left\{\overrightarrow{{{\mathfrak{T}}}_1}^-,{\left[\overleftarrow{{{\mathfrak{T}}}_1}^+,{\left\{\overleftarrow{{{\mathfrak{T}}}_1}^-,{\left[\overrightarrow{{{\mathfrak{T}}}_1}^+,{{\mathfrak{h}}}\right]}\right\}}\right]}\right\}}. \nonumber \\ & &  \label{eq:D4solution} \end{aligned}$$ In this expression we have left out the coupling constant transformation parameterized by $\alpha$ in (\[eq:g3solution\]), which leads to $${\delta{\mathfrak{D}}}_4 \mapsto {\delta{\mathfrak{D}}}_4 + 2 \alpha \, {\delta{\mathfrak{D}}}_2.$$ However, to match field theory results 