MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The notion of "supersymmetry" that I am aware of proceeds as follows. One fixes a spacetime $\mathbb R^n$ and signature; I will write $\mathrm{SO}(n)$ for the corresponding group of orthogonal transformations, as if I had chosen Euclidean signature, but of course this story also works with $\mathrm{SO}(n-1,1)$, .... In any case, $\mathrm{SO}(n)$ has a double cover $\mathrm{Spin}(n)$, which is usually the universal cover. The usual symmetry of usual spacetime is the group of rigid transformations $\mathrm{ISO}(n) = \mathrm{SO}(n) \ltimes \mathbb R^n$, and it is convenient to work instead with the double cover $\mathrm{ISpin}(n) = \mathrm{Spin}(n) \ltimes \mathbb R^n$. I realize that these are not quite the standard names, but no matter; for example, you may replace the "I" in $\mathrm{SISO}$ with a "U", and the "O" with a "Y", if such changes make you feel better.

Now, suppose that you choose for yourself some real spin representation $\mathbb S$ of $\mathrm{Spin}(n)$ and a symmetric $\mathrm{Spin}(n)$-equivariant pairing $\Gamma: \mathbb S \otimes \mathbb S \to \mathbb R^n$. Exactly what flavor of choices this requires depends on the value of $n$ mod 8, and on the signature, and perhaps other things. As usual, "spin representation" means it extends to the appropriate clifford algebra.

Let $\pi \mathbb S$ denote the supervector space which is the "parity reversal" of $\mathbb S$. Then $\Gamma$ makes sense as an antisymmetric pairing on $\pi\mathbb S$. Indeed, it defines a nonabelian super Lie algebra structure on the supervector space $\pi\mathbb S \oplus \mathbb R^n$, whose only nontrivial bracket is $\Gamma : (\pi\mathbb S)^{\wedge 2} \to \mathbb R^n$. Lacking a good notation, I will call this Lie algebra $\mathfrak t$. It is a central extension of $\pi\mathbb S$ by $\mathbb R^n$, and fits into a nontrivial short exact sequence $\mathbb R^n \to \mathfrak t \to \pi\mathbb S$. Exponentiating of finite-dimensional super Lie algebras is trivial if the non-super part can be exponentiated, and so we end up with a nonabelian super translation group $T$, also given as a central extension, now of super groups, of $\pi \mathbb S$ by $\mathbb R^n$. By definition, everything is $\mathrm{Spin}(n)$-equivariant. The super isometry group is the extension $\mathrm{SISO}(n,\mathbb S) = \mathrm{Spin}(n) \ltimes T$.

My question is: What happens when flat space $\mathbb R^n$ is replaced by some other geometry? Simple cases include the round sphere, and de Sitter and anti de Sitter spaces. More complicated examples include other symmetric or homogeneous spaces for other Lie groups. Any "$G_2$ supersymmetry" or "$E_8$ supersymmetry" out there?

The motivation for my question comes from an observation that I learned from Dyson's excellent paper Birds and Frogs: $\mathrm{ISO}(n)$ is simply a limit as some "curvature" parameter goes to $\infty$ of $\mathrm{SO}(n+1)$, which is in the technical sense a more simple group. Note that there will no longer be a well-defined "super translation group", as the round sphere is not (usually) a group, but perhaps there is still be a simple deformation of $\mathrm{SISO}(n)$, acting on some super homogeneous space?

share|cite|improve this question
up vote 3 down vote accepted

Homogeneous superspaces generalizing the "Super-Poincaré\Lorentz = Superspace" exist. These superspaces are given as coset spaces of super-Lie groups $G/H$. Please see for example, the following article by: Schunck and Wainwright containing an explicit construction of the supersphere $S^{2|2}$ as the supercoset space $UOSP(1|2)/U(1)$. There are also treatments of the super-Ads case, please see for example: N. Alonso-Alberca, E.Lozano-Tellechea, T. Ortin .

In super-coset spaces, the standard construction of the (super) frame bundle and (super) $H-$ connection applies. However, In the contrast to the nonsupersymmetric case, where the algebra of $G$ is realized as an algebra of isometries of the Killing-Cartan metric; this metric does not have a natural generalization to supermanifolds, but we can still talk about isometries leaving the super-frame fields and the super-$H$ connection invariant up to a gauge transformation. The action of these super-isometries on functions (superfields) or more generally sections of super-vector bundles can be realized by means of differential operators.

There is another type of generalization of "flat supersymmetry", to an arbitrary spin manifold $(M, g)$, (which can be applied to homogeneous spaces having spin structures) by means of the construction of a super-cotangent bundle $T^{*}M \oplus \pi TM$, please,see for example the following work by: J.P. Michel. The super-cotangent bundle is a symplectic supermanifold and the generators of the supersymmetry algebra can obtained as a quantization of the corresponding symbols realized in the Poisson algebra of the super-cotangent bundle. The Dirac operator plays the role of the supersymmetry generator (it Poisson squares to the Laplacian). These supermanifold describe the phase spaces of a spinning particle moving on $(M, g)$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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