3/4-Lie superalgebras: how much of a theory can one develop?

Question

Let $\mathfrak{s} = \mathfrak{s}_0 \oplus \mathfrak{s}_1$ be a real Lie superalgebra. (The ground field does not matter much, but at least one formula will not work as written if the characteristic is 2 or 3.) Recall that this means that there is a bilinear 2-graded bracket $[-,-]$ with three components

(a) $\mathfrak{s}_0 \times \mathfrak{s}_0 \to \mathfrak{s}_0$ (skewsymmetric)

(b) $\mathfrak{s}_0 \times \mathfrak{s}_1 \to \mathfrak{s}_1$

(c) $\mathfrak{s}_1 \times \mathfrak{s}_1 \to \mathfrak{s}_0$ (symmetric)

satisfying the Jacobi identity, which splits into 4 components, which can be paraphrased as

(1) $\mathfrak{s}_0$ is a Lie algebra under (a)

(2) $\mathfrak{s}_1$ is an $\mathfrak{s}_0$-module under (b)

(3) the map in (c) is $\mathfrak{s}_0$-equivariant

(4) $[[x,x],x] = 0$ for all $x \in \mathfrak{s}_1$

The fact that the first three components can be written using words, whereas the fourth is easiest via a formula, suggests that they should perhaps be treated differently.

Indeed, over time I have come across many examples of superalgebras where the first three components of the Jacobi identity are satisfied but not the fourth. I'd like to call them 3/4-Lie superalgebras. I would like to know how far can this notion be pushed and in particular how much of the theory of Lie superalgebras still works in the 3/4 case.

To motivate this seemingly random question, let me end by pointing out one generic example where they arise. There are others, but they are lengthier to describe.

Let $\mathfrak{g}$ be a metric Lie algebra; that is, a Lie algebra with an ad-invariant inner product $(-,-)$ and let $V$ be a symplectic $\mathfrak{g}$-module; that is, one possessing a $\mathfrak{g}$-invariant symplectic form $\langle-,-\rangle$. Now let $\mathfrak{s} = \mathfrak{g} \oplus V$. Then maps (a) and (b) are obvious: given by the Lie bracket on $\mathfrak{g}$ and the action of $\mathfrak{g}$ on $V$, respectively. Map (c) is the transpose of map (b) using the inner products of both $\mathfrak{g}$ and $V$; in other words, if $x,y \in V$ then $[x,y] \in \mathfrak{g}$ is defined by $$([x,y],a) = \langle a\cdot x,y\rangle$$ for all $a \in \mathfrak{g}$.

Then it is easy to see that $[x,y] = [y,x]$ and that (1)-(3) are satisfied, whereas in general (4) is not satisfied and instead defines a subclass of symplectic $\mathfrak{g}$-modules.

Answer 1

At least in the semisimple case, it doesn't seem like there can be a theory of 3/4-Lie superalgebras other than a fairly predictable set of examples within Cartan-Weyl theory. I don't know how to do all of the relevant calculations, but it is easy to see roughly how they would go.

Suppose that the even part $s_0$ is a semisimple Lie algebra $g$. Then $s_1$ is can be a self-dual representation $V$ of $g$. Unless possibly $g$ is $E_6$, I don't think that $V$ can be anything other than a self-dual representation. So then the form is a symmetric element of $\mathrm{Inv}(g \otimes V \otimes V)$, which is a vector space whose dimension can be computed. The dimension is at most the rank of $g$ when $V$ is irreducible. Indeed it is usually the rank of $g$, when the highest weight of $V$ is far away from the walls of the Weyl chamber. The dimension of the symmetric part is smaller, but it is often non-zero. (This is the part that I don't know how to compute off the top of my head.)

So inevitably there will be a classification of these 3/4 Lie algebras using these branching rules. Besides the classification, the only theory that springs to mind is representations of these 3/4 Lie algebras. A representation $W$ would first off be a representation of $g$ (and I suppose a super vector space?). Then there would be a $g$-invariant map $V \otimes W \to W$ which represents the action of $V$. I don't see a rationale for imposing restrictions on this map other than $g$-invariance, for otherwise $g \oplus V$ would not be the adjoint representation of itself. So once again, there is some Cartan-Weyl classification that says what $W$ can be, and I'm not sure what more you could say.