In supersymmetry or supergravity, textbooks always show that one can define a Kähler potential $K=K(\phi^i,(\phi^i)^\ast)$ and an holomorphic superpotential $W=W(\phi_i)$ such that the scalar potential is given by (up to some normalisation)

$$ V=\left|\frac{\partial W}{\partial \phi_i}\right|^2~~\text{(SUSY)}\qquad V=e^{-K}\left( |D_iW|^2-3|W|^2\right)~~\text{(SUGRA)}. $$

with $D_i=\partial_i+\partial_i K$ the Kähler covariant derivative.

Now, consider a $\sigma$-model $\Sigma\to \mathcal{M}$ $$ \mathcal{L}=\frac{1}{2}g_{ij}(\partial_\mu \phi^i)(\partial^\mu \phi^i)^\ast-V(\phi^i,(\phi^i)^\ast)+\text{(higher spins)} $$

If the sigma model is supersymmetric, it is a consequence of Berger theorem that $\mathcal{M}$ is a Kähler manifold (because SUSY restricts the holonomy of $\mathcal{M}$, see for instance Cecotti's wonderful book *Supersymmetric Field Theories*) and thus the metric is kähler and $g_{ij}=\partial_i\overline{\partial}_j K$ (locally), proving that there exists a Kähler potential (In the case of SUGRA, the target manifold is Hodge Kähler and thus the same result holds).

**The question now is:** Is there a way to prove only with similar geometric arguments that the scalar potentials take the form above? It is fairly easy to prove those result "brute force", but I'm looking for something more elegant.

Cecotti proves from Morse theory that in SQM we indeed must have a superpotential, because it is the only way (except if $\mathcal{M}$ has Killing vectors) to deform the superalgebra. But even if the same reasoning holds in higher dimensions, I fail to see way the SUGRA case should be different (and why it explicitely depends on the Kähler potential).

This post imported from StackExchange Physics at 2016-11-16 11:18 (UTC), posted by SE-user Bulkilol