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?



Lie super algebra presentation of the Kähler identities
Indeed there is. Apologies for tooting my own horn, but you can find it in [this paper](https://arxiv.org/abs/hep-th/9705161), cowritten with Chris Koehl and Bill Spence. Instead of repeating the explanation, I refer you to this [MathOverflow answer](https://mathoverflow.net/a/16215/394) 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.)


