Is the SUSY Algebra isomorphic for all Kahler Manifolds?

For a Kahler manifolds, 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. I am correct in understanding that one gets the same algebra for all Kahler manifolds?

edited May 16 2012 at 1:59

asked May 16 2012 at 1:23

Jean Delinez
1,537●6●14

In some sense, yes. The Kahler identities specify all the a priori nontrivial commutation relations. This gives us some abstract lie algebra referred to as the (2,2) SUSY algebra. The algebra of differential forms on the Kahler manifold need not be a faithful representation of this algebra. – Ryan Thorngren May 16 2012 at 1:35

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? – Jean Delinez May 16 2012 at 1:58

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. – Ryan Thorngren May 16 2012 at 2:42

ah ok - then that's good I suppose - thanks – Jean Delinez May 16 2012 at 3:16

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