6
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Trimok Nov 14 '13 at 18:15
  • $\begingroup$ @Trimok thank you for the reference. I looked it up and it was informative, but did not directly relate to the question. $\endgroup$ – Neuneck Nov 18 '13 at 10:48
3
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.