For $(\mathfrak{g},[-,-])$ an ordinary Lie algebra let me say that a super-extension of it (maybe not the best terminology) is a super-Lie algebra $(\mathfrak{s}, [-,-]_{\mathfrak{s}})$ whose bosonic component is exactly $(\mathfrak{g}, [-,-])$, hence whose underlying super vector space is $\mathfrak{s} \simeq \underset{= \mathfrak{s}_{even}}{\underbrace{\mathfrak{g}}} \oplus \underset{= \mathfrak{s}_{odd}}{\underbrace{S}}$, with super Lie bracket $[-,-]_{\mathfrak{s}}$ restricting to $[-,-]$ if both arguments are in $\mathfrak{g}$.

From the even-even-odd component of the super-Jacobi identity it is easy to see that the even-odd component of $[-,-]_\mathfrak{s}$ is necessarily a Lie action $\rho$ of $\mathfrak{g}$ on $S$. Similarly from the even-odd-odd component of the super-Jacobi identity it is easy to see that the odd-odd-component of $[-,-]_\mathfrak{s}$ is necessarily a symmetric bilinear pairing $(-,-) : S \otimes S \to \mathfrak{g}$, which is equivariant with respect to this action.

What seems more subtle is to analyze the constraints imposed on this data by the odd-odd-odd component of the super-Jacobi identity. By taking all three arguments equal, one finds that it is necessary that $\rho_{(\psi,\psi)}(\psi) = 0$ for all $\psi \in S$. Also it is easy to see that for satisfying the super-Jacobi identity, it is sufficient that $\rho_{(\psi,\phi)} = 0$ is the zero-action, for all $\psi,\phi = 0$.

But is this necessary? What would be the general classification of super-extensions, in the above sense?

Specifically for the case that $\mathfrak{g} = \mathfrak{iso}(\mathbb{R}^{d-1,1}) \simeq \mathbb{R}^{d-1,1} \rtimes \mathfrak{so}(d-1,1)$ is the Poincaré Lie algebra, in some dimension $d$. Then it is easy to see that sufficient data for super-extensions $\mathfrak{s}$, in the above sense, are given by real $\mathfrak{so}(d-1,1)$-representations $S$ equipped with an $\mathfrak{so}$-equivariant bilinear symmetric pairing of the special form $(-,-) : S \otimes_{\mathbb{R}} S \to \mathbb{R}^{d-1,1} \hookrightarrow \mathfrak{iso}(\mathbb{R}^{d-1,1})$. Such are provided by real spin representations, and the result is the usual super Poincaré Lie algebras ("supersymmetry").

Now at this point, the existing literature points to the Haag–Łopuszański–Sohnius theorem. This states further conditions on a super-extension (e.g. that $P_a P^a$ remains a Casimir, and more) and then concludes that these are the only super-extensions satisfying this. Of course these extra conditions are well-motivated from expected behaviour of scattering matrices in field theories. But if we disregarded this and consider the purely mathematical problem of classifying all super-extensions -- in the above sense -- of the Poincaré Lie algebras: could there be more?

For $(\mathfrak{g},[-,-])$ an ordinary Lie algebra let me say that a super-extension of it (maybe not the best terminology) is a super-Lie algebra $(\mathfrak{s}, [-,-]_{\mathfrak{s}})$ whose bosonic component is exactly $(\mathfrak{g}, [-,-])$, hence whose underlying super vector space is $\mathfrak{s} \simeq \underset{= \mathfrak{s}_{even}}{\underbrace{\mathfrak{g}}} \oplus \underset{= \mathfrak{s}_{odd}}{\underbrace{S}}$, with super Lie bracket $[-,-]_{\mathfrak{s}}$ restricting to $[-,-]$ if both arguments are in $\mathfrak{g}$.

From the even-even-odd component of the super-Jacobi identity it is easy to see that the even-odd component of $[-,-]_\mathfrak{s}$ is necessarily a Lie action $\rho$ of $\mathfrak{g}$ on $S$. Similarly from the even-odd-odd component of the super-Jacobi identity it is easy to see that the odd-odd-component of $[-,-]_\mathfrak{s}$ is necessarily a symmetric bilinear pairing $(-,-) : S \otimes S \to \mathfrak{g}$, which is equivariant with respect to this action.

What seems more subtle is to analyze the constraints imposed on this data by the odd-odd-odd component of the super-Jacobi identity. By taking all three arguments equal, one finds that it is necessary that $\rho_{(\psi,\psi)}(\psi) = 0$ for all $\psi \in S$. Also it is easy to see that for satisfying the super-Jacobi identity, it is sufficient that $\rho_{(\psi,\phi)} = 0$ is the zero-action, for all $\psi,\phi \in S$.

But is this necessary? What would be the general classification of super-extensions, in the above sense?

Specifically for the case that $\mathfrak{g} = \mathfrak{iso}(\mathbb{R}^{d-1,1}) \simeq \mathbb{R}^{d-1,1} \rtimes \mathfrak{so}(d-1,1)$ is the Poincaré Lie algebra, in some dimension $d$. Then it is easy to see that sufficient data for super-extensions $\mathfrak{s}$, in the above sense, are given by real $\mathfrak{so}(d-1,1)$-representations $S$ equipped with an $\mathfrak{so}$-equivariant bilinear symmetric pairing of the special form $(-,-) : S \otimes_{\mathbb{R}} S \to \mathbb{R}^{d-1,1} \hookrightarrow \mathfrak{iso}(\mathbb{R}^{d-1,1})$. Such are provided by real spin representations, and the result is the usual super Poincaré Lie algebras ("supersymmetry").

Now at this point, the existing literature points to the Haag–Łopuszański–Sohnius theorem. This states further conditions on a super-extension (e.g. that $P_a P^a$ remains a Casimir, and more) and then concludes that these are the only super-extensions satisfying this. Of course these extra conditions are well-motivated from expected behaviour of scattering matrices in field theories. But if we disregarded this and consider the purely mathematical problem of classifying all super-extensions -- in the above sense -- of the Poincaré Lie algebras: could there be more?