and the lift to the finite-$N$ dilatation generator.

The solution \[sec:g3solution\]
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Only the ${{\mathfrak{psu}}}(1|1)^2$ generators, the ${{\mathfrak{T}}}$, receive corrections at this order. Once again the form of the solution depends only on the sign of the generator (though recall that now the sign refers to the commutator with ${{\mathfrak{L}}}$). $${{\mathfrak{T}}}^\pm_3=\pm {\left[{{\mathfrak{T}}}^\pm_1,{{\mathfrak{x}}}\right]} + {\left[{{\mathfrak{T}}}^\pm_1,{{\mathfrak{y}}}\right]} + \alpha {{\mathfrak{T}}}^\pm_1.  \label{eq:g3solution}$$ Again, the ${{\mathfrak{y}}}$ commutator is a similarity transformation, and it must be the same as that of the $\mathcal{O}(g^2)$ solution. $\alpha$ corresponds to the coupling constant transformation $$g \mapsto g + \alpha g^3.$$ As at $\mathcal{O}(g^2)$, the solution is hermitian, provided ${{\mathfrak{y}}}$ is anti-hermitian. It is difficult to imagine a simpler solution. Beside the coupling constant transformation and the similarity transformation, the solution at this order is just a commutator with ${{\mathfrak{x}}}$, as was the case for $\mathcal{O}(g^2)$.

As at the previous order, we use a direct method to prove that this solution satisfies the symmetry algebra constraints, and the proof is in Appendix \[sec:g3proof\].

Freedom for the $\mathcal{O}(g^3)$ solution
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At this order, we could add gauge transformations to the generators. Furthermore, the solution satisfies the commutation relations only up to gauge transformations.

The case for homogeneous solutions at this order 