invariant tensor , which yields $$\label{invadscarrollsuperF}
\begin{split}
& \langle J_{ab} J_{cd} \rangle = \alpha_0 \left( \delta_{ad} \delta_{bc} - \delta_{ac} \delta_{bd} \right) \, , \quad 
\langle J_{ab} H \rangle = \alpha_1 \epsilon_{ab} \, , \quad 
\langle K_a P_b \rangle = - \alpha_1 \epsilon_{ab} \, , \quad
\langle P_a P_b \rangle = 0 \, , \\
& \langle Q_\alpha Q_\beta \rangle = 2 \alpha_1  C_{\alpha \beta} \, .
\end{split}$$ Then, applying the $\ell \rightarrow \infty$ limit to the CS action we get $$\label{CSC}
\begin{split}
I^{\text{super-Carroll}}_{CS} & = \frac{k}{4 \pi} \int_\mathcal{M} \Bigg \lbrace \frac{\alpha_0}{2} \left( \omega^a_{\phantom{a} b} \mathcal{R}^b_{\phantom{b} a} \right) + \alpha_1 \left( \epsilon_{ab} \mathcal{R}^{ab} h - 2 \epsilon_{ab} \mathcal{K}^a V^b + 2 \bar{\psi} \Psi \right)  \\
& - d \left( \frac{\alpha_1}{2} \epsilon_{ab} \omega^{ab} h - \alpha_1 \epsilon_{ab} k^a V^b \right) \Bigg \rbrace \, ,
\end{split}$$ which is now written in terms of the super-Carroll curvatures appearing in (they should not be confused with the super-AdS Carroll ones). Observe that the same action can be derived by using the non-vanishing components in and the connection $1$-form for the Carroll superalgebra in the general expression . Furthermore, we can see that the exotic term, proportional to $\alpha_0$, now reduces purely to the so-called Lorentz 