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

What is the Noether charge associated with Kahler invariance of supergravity (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.

• 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. Nov 14 '13 at 18:15
• @Trimok thank you for the reference. I looked it up and it was informative, but did not directly relate to the question. Nov 18 '13 at 10:48

## 1 Answer

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.