Lie super algebra presentation of the Kähler identities

Question

For any Kähler manifold $(M,h)$, with Lefschetz operators $L$ and $\Lambda$, and counting operator $H$, we have the following the well-known Kähler-Hodge identities:

\begin{align*} [\partial,L] = 0, && [\overline{\partial},L] = 0, & & [\partial^*,\Lambda] = 0, && [\overline{\partial}^*, \Lambda] = 0, \\ [L,\partial^*] = i\overline{\partial}, & & [L,\overline{\partial}^*] = - i\partial, & & [\Lambda,\partial] = i\overline{\partial}^*, & & [\Lambda,\overline{\partial}] = - i\partial^*, \end{align*} In physics literature this is often referred to a "supersymmetric algebra". Does there exist a more mathematical understanding of this object, perhaps as a Lie super algebra?

Answer 1

Indeed there is. Apologies for tooting my own horn, but you can find it in this paper, cowritten with Chris Koehl and Bill Spence. Instead of repeating the explanation, I refer you to this MathOverflow answer from a decade ago.

Added after the comment

From the point of view espoused in that paper, this is what remains when one takes a supersymmetric field theory in 4-dimensional Minkowski spacetime and dimensionally reduces to a line. Taking that line to be spacelike, the supersymmetry algebra breaks to the centraliser of the three-dimensional complement of that line. Schematically, the Poincaré superalgebra is $$\mathfrak{sp}= \mathfrak{so}(3,1) \oplus S \oplus V $$ where $V$ is the 4-dimensional real vector representation of $\mathfrak{so}(3,1)$ and $S$ is the 4-dimensional real spinorial representation of $\mathfrak{so}(3,1)$. Both $V$ and $S$ are irreducible representations. The Lie superalgebra is $\mathbb{Z}$-graded with $\mathfrak{so}(3,1)$ in degree $0$, $S$ in degree $-1$ and $V$ in degree $-2$. The only non-obvious bracket is $$ [S,S] = V $$ Now pick a spacelike vector $v \in V$. Then the centraliser of $v^\perp$ in $\mathfrak{sp}$ is the Lie subalgebra $$\mathfrak{so}(2,1) \oplus S \oplus V $$

Geometrically, $v^\perp$ acts trivially, $v$ is central and acts like the Laplacian, the $\mathfrak{so}(2,1) \cong \mathfrak{sl}(2,\mathbb{R})$ subalgebra is spanned by $L,\Lambda,H$ and $S$ is spanned by $\partial$, $\partial^*$, $\bar\partial$ and $\bar\partial^*$.

(You may have to complexify everything, by the way.)

Answer 2

I did not know these identities but after a small search, i think that some relations are missing from your post:
In: http://mathworld.wolfram.com/KaehlerIdentities.html
some additional relations (see eqs (17)-(19) and the LHS of eqs (10)-(13); the RHS do not seem to be independent relations, they can probably be extracted from the rest) are included. If you take into account these additional relations, then your algebra is a Lie superalgebra (or: $\mathbb{Z}_2$-graded lie algebra), with a basis of the even subspace consisting of $L$, $\Lambda$, $H$ and a basis of the odd subspace consisting of $\partial,\overline{\partial}, \partial^*,\overline{\partial}^*$.

Related:
Is the SUSY Algebra isomorphic for all Kähler Manifolds?