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?

  • $\begingroup$ You may be interested in Huybrechts' Complex Geometry: An Introduction, section 3, appendix B titled 'SUSY for Kähler Manifolds'. $\endgroup$ May 2, 2019 at 12:17

2 Answers 2


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.)

  • $\begingroup$ Great, thanks for the link: So if my quick reading of your old answer is correct, then together with the anti-commutation relations between the partial differential operators and their adjoints, the super Lie algebra is $\frak{so}(3,1)$? $\endgroup$ May 1, 2019 at 20:06
  • $\begingroup$ I'm not sure what you mean by $\mathfrak{so}(3,1)$. For me that is a Lie algebra, isomorphic to the Lie algebra of Lorentz transformations on 4-dimensional Minkowski spacetime. I believe that the Lie superalgebra in question is the centraliser of a spatial translation in the $N=1$ 4-dimensional Poincaré superalgebra. $\endgroup$ May 1, 2019 at 20:09
  • $\begingroup$ Great, thanks a lot for your help! $\endgroup$ May 1, 2019 at 20:40

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 $\partial,\overline{\partial}, \partial^*,\overline{\partial}^*$.

Is the SUSY Algebra isomorphic for all Kähler Manifolds?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.