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Hidden supersymmetry, which is the classical(non-super) symmetry in the form of susy, acting on a non-Grassmann space (e.g., Grassmann space is $(t,x,\theta,\bar{\theta})$, corresponding non-Grassmann space is $(t,x)$)...see the review article on hidden supersymmetry arXiv:1004.5489.

In the language of Lie geometry, we can understand SUSY as a vector operating on a Grassmann space. Is there an analogous geometric explanation for hidden SUSY? Any comment is highly appreciated.

Manishearth
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Simon
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    Hi Simon - could you rephrase your question to focus on what you really want to ask, rather than inviting discussion? Though it's definitely a good topic to ask about. – David Z Mar 19 '13 at 01:04
  • @DavidZaslavsky he asks for a geometrical interpretation of hidden SUSY, as stated in the title. Why do you interpret the term "discussion" to the disfavor ot the OP? From the context it is clear that he is looking for an explanation instead of an unconstructive pointless discussion. Why interpreting the term in disvafor of the OP like this by language nitpicking? The word discussion could just be exchanged by the word explanation without changing the meaning of the question. It looks like you are looking for a pretext to close this good and interesting theoretical physics question. – Dilaton Mar 20 '13 at 12:00
  • Well, dear Dilaton, I do agree with David that I should rephrase my question. And you both are right, I would use the word 'explanation'. Thanks so much. – Simon Mar 20 '13 at 16:07
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    @Dilaton now that just doesn't make any sense. If I were looking for a pretext to close the question, why would I suggest to Simon how he could improve it? This isn't the first time you've completely misinterpreted what I write - please try to be careful about that. – David Z Mar 20 '13 at 18:01

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A large class of models possessing one dimensional supersymmetric quantum mechanics is described by the radial dynamics of the motion on rank-1 symmetric spaces (i.e., the Schroedinger operator is the radial Laplacian). Please see the following article by: Boya, Wehrhahn and Rivero. This result can be understood from the fact that the full Laplacian (acting on forms) on a Riemannian manifold is supersymmetric by construction:

$\Delta = d d^{*} + d^{*} d$

where $d$ and $d^{*}$ are the exterior differential and the co-differentials respectively.

David Bar Moshe
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