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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.

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1 Answer 1

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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.

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Indeed, by "theory" I meant the representation theory of such objects, since the structure theory is, at least in the semisimple case, a straightfoward consequence of the definition, which is not the say that the semisimple case is the only interesting case. Arguably the most interesting Lie superalgebras in Physics are not the semisimple ones. As for these "3/4" Lie superalgebras, Einstein manifolds (of the right signature) admitting real Killing spinors provide examples. –  José Figueroa-O'Farrill Dec 2 '09 at 10:47
    
I'm accepting this answer. It's not perhaps the answer I wanted to hear, but I tend to agree that perhaps one cannot say much more. –  José Figueroa-O'Farrill Dec 3 '09 at 15:35
    
We both tend to agree that a 3/4 Lie algebra doesn't look like it can support more than a 3/4 representation; the odd action is then just an invariant tensor. That said, it's a little glib as an answer from me, because I didn't look very hard for more structure. I think I was more successful with your other question. –  Greg Kuperberg Dec 3 '09 at 16:29

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