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?
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asked May 16, 2012 at 1:23
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3$\begingroup$ 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. $\endgroup$– Ryan ThorngrenCommented May 16, 2012 at 1:35
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$\begingroup$ 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? $\endgroup$– Jean DelinezCommented May 16, 2012 at 1:58
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1$\begingroup$ 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. $\endgroup$– Ryan ThorngrenCommented May 16, 2012 at 2:42
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$\begingroup$ ah ok - then that's good I suppose - thanks $\endgroup$– Jean DelinezCommented May 16, 2012 at 3:16
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1$\begingroup$ @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. $\endgroup$– Theo Johnson-FreydCommented Jul 25, 2013 at 19:17
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