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  What's the Noether charge associated with Kaehler invariance of SuGra?

+ 6 like - 0 dislike

The title basically sais it all:

What is the Nother charge associated with Kaehler invariance of SuGra?

As the question is rather tangential to what I need to do, I have not tried explicitly calculating it myself, but I'm sure that I'm not the first one to wonder...

This post imported from StackExchange Physics at 2014-03-07 13:18 (UCT), posted by SE-user Neuneck
asked Nov 13, 2013 in Theoretical Physics by Neuneck (110 points) [ no revision ]
Found this Chapter 17A.2, p 378, in Freedman/ Van Proeyen Supergravity book. This is about Kahler-Hodge manifolds (apparently linked to matter-coupled N=1 supergravity),"Kahler charges" apply to fermions only, and are (like magnetic monopoles), multiple of a minimum charge.

This post imported from StackExchange Physics at 2014-03-07 13:18 (UCT), posted by SE-user Trimok
@Trimok thank you for the reference. I looked it up and it was informative, but did not directly relate to the question.

This post imported from StackExchange Physics at 2014-03-07 13:18 (UCT), posted by SE-user Neuneck

1 Answer

+ 4 like - 0 dislike

Answer: There is none.

The issue at hand is that the Kaehler invariance is just that - an invariance, not a continuous symmetry of the fields. Most prominently the superpotential must transform as $$ \mathcal W \to \mathcal W e^{-h} $$ A general superpotential that leads to consistent theories is $$ \mathcal W =\frac{1}{2} m_{ab} \phi^a \phi^b + \frac{1}{3} Y_{abc} \phi^a \phi^b \phi^c $$ with at least one of the $m_{ab}$ and $Y_{abc}$ non-zero. From this is is obvious, that no transformation of the fields $\phi^a$ exists, such that $\mathcal W \to \mathcal W e^{-h}$ without redefining the couplings.

Thus, there is a Kaehler invariance, which involves a redefinition of the couplings and has its value on its own (e.g. on non-simply connected internal spaces, the Kaehler potential might only be defined locally, with definitions on different charts being equal up to Kaehler transformations $\mathcal K' = \mathcal K + f(\phi) + \bar f(\bar\phi)$), but this is not a symmetry in the sense of Noether's theorem.

This post imported from StackExchange Physics at 2014-03-07 13:18 (UCT), posted by SE-user Neuneck
answered Nov 18, 2013 by Neuneck (110 points) [ no revision ]

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