& = D \kappa^a - \varrho^a_{\phantom{a} b} k^b - \frac{1}{\ell^2} \varrho^a h + \frac{1}{\ell^2} V^a \tau - \frac{1}{\ell} \bar{\varepsilon} \Gamma^{a0} \psi \, , \\
\delta V^a & = D \varrho^a - \varrho^a_{\phantom{a} b} V^b \,  , \\
\delta h & = d \tau - \varrho^a k_a + V^a \kappa_a + \bar{\varepsilon} \Gamma^0 \psi \, , \\
\delta \psi & = d \varepsilon + \frac{1}{4} \omega^{ab} \Gamma_{ab} \varepsilon - \frac{1}{4} \varrho^{ab} \Gamma_{ab} \psi - \frac{1}{2 \ell} \varrho^a \Gamma_a \psi + \frac{1}{2 \ell} V^a \Gamma_a \varepsilon \\
& = D \varepsilon - \frac{1}{4} \varrho^{ab} \Gamma_{ab} \psi - \frac{1}{2 \ell} \varrho^a \Gamma_a \psi + \frac{1}{2 \ell} V^a \Gamma_a \varepsilon \,  ,
\end{split}$$ and, restricting ourselves to supersymmetry, we have $$\label{susytr}
\begin{split}
& \delta \omega^{ab} = 0 \, , \quad
\delta k^a = - \frac{1}{\ell} \bar{\varepsilon} \Gamma^{a0} \psi \, , \quad
\delta V^a = 0 \, , \quad
\delta h = \bar{\varepsilon} \Gamma^0 \psi \, , \\
& \delta \psi = d \varepsilon + \frac{1}{4} \omega^{ab} \Gamma_{ab} \varepsilon + \frac{1}{2 \ell} V^a \Gamma_a \varepsilon = D \varepsilon + \frac{1}{2 \ell} V^a \Gamma_a \varepsilon \, .
\end{split}$$ Let us also mention, for the sake of completeness, that from the gauge variation of the curvature $2$-form 