& = d h + V^a k_a - \frac{1}{2} \bar{\psi} \Gamma^0 \psi   = \mathfrak{H} - \frac{1}{2} \bar{\psi} \Gamma^0 \psi \,   , \\
\Psi & = d \psi + \frac{1}{4} \omega^{ab} \Gamma_{ab} \psi + \frac{1}{2 \ell} V^a \Gamma_a \psi \,  .
\end{split}$$ Let us also mention that, considering the Bianchi identity $\nabla F =0$ ($\nabla = d + [A, \cdot]$), we obtain $$\label{bianchdscarrollsuper}
\begin{split}
d \mathcal{R}^{ab} & = \frac{2}{\ell^2} R^a V^b \,  , \\
D \mathcal{K}^a  & = \mathcal{R}^a_{\phantom{a} b} k^b + \frac{1}{\ell^2} R^a h - \frac{1}{\ell^2} V^a \mathcal{H} - \frac{1}{\ell} \bar{\psi} \Gamma^{a0} \Psi \, , \\
D R^a & = \mathcal{R}^a_{\phantom{a} b} V^b \,   , \\
d \mathcal{H} & = R^a k_a - V^a \mathcal{K}_a + \bar{\psi} \Gamma^0 \Psi \,  , \\
D \Psi & = \frac{1}{4} \mathcal{R}^{ab} \Gamma_{ab} \psi + \frac{1}{2 \ell} R^a \Gamma_a \psi - \frac{1}{2 \ell} V^a \Gamma_a \Psi \,  ,
\end{split}$$ where $D=d+\omega$.

Now, in order to construct a CS action (that is an action of the form ) invariant under the super-AdS Carroll group, we require the connection $1$-form given in and the corresponding non-vanishing components of the invariant tensor.

Concerning the invariant tensor, which is the fundamental ingredient for the construction of a CS 