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  Is the SUSY Algebra isomorphic for all Kähler Manifolds?

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For a Kähler manifold, the graded algebra generated by $\partial,\overline{\partial},\partial^*,\overline{\partial}^\ast$, the Lefschetz operator, and the dual Lefschetz operator, is called the supersymmetric algebra, or in physics parlance the $N=(2,2)$ SUSY algebra. Am I correct in understanding that one gets the same algebra for all Kähler manifolds?


This post imported from StackExchange MathOverflow at 2014-08-10 19:58 (UCT), posted by SE-user Jean Delinez

asked May 15, 2012 in Mathematics by Jean Delinez (25 points) [ revision history ]
edited Sep 15, 2014 by Dilaton
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@RyanThorngren: Probably bump functions are enough to show that the representation is faithful. But surely it's not enough to check just the generators --- you need to check all nontrivial brackets of generators.

This post imported from StackExchange MathOverflow at 2014-08-10 19:58 (UCT), posted by SE-user Theo Johnson-Freyd

As a physicist I know the following.

  1. Zumino shows that $(1,1)$ supersymmetry in the two-dimensional nonlinear sigma model get enhanced to $(2,2)$ supersymmetry for Kahler target spaces.
  2. By suitably twisting the $(2,2)$ model, various supersymmetry generators get mapped to operators such as $\bar{\partial}$ and so on. 

My sub-question is how and why  this physicist's proof is mathematically imprecise. In particular, what are the gaps that need to be filled in.

@RyanThorngren @40227 Any comments to my sub-question?

What statement are you interpreting this argument as a proof of?

Mostly statement 1 established the connection between supersymmetry and Kahlerity of the target space manifold.

Most recent comments show all comments
Thanks. This is what I was thinking. It was clear that the SUSY relations would be satisfied for any Kahler manifold, but it was not all clear to me that the representation would be faithful. Can you give me an example of a (projective) Kahler space where the rep is non-faithful?

This post imported from StackExchange MathOverflow at 2014-08-10 19:58 (UCT), posted by SE-user Jean Delinez
Actually, having thought about it some, I now think that if all we want is a representation of the lie algebra, then this representation will be faithful. All we need to show is that there are nonzero smooth functions which remain nonzero when we apply these operators. I think that a bump function does the trick in each case.

This post imported from StackExchange MathOverflow at 2014-08-10 19:58 (UCT), posted by SE-user Ryan Thorngren

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