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  Geometric origin of the Kahler potential for SUSY $\sigma$-models

+ 2 like - 0 dislike

I was reading Cecotti's book on Supersymmetric Field Theories, and there is a statement that confuses me. It is proven that if one considers a $\sigma$-model $\phi^i:~\Sigma\to\mathcal{M}$ with Lagrangian $$ \mathcal{L}=\frac{1}{2}g_{ij}\partial_\mu\phi^i\partial^\mu\phi^j+\frac{i}{2}h_{ij}\bar{\psi}^i\overline{\sigma}^\mu\partial_\mu\psi^j, $$ if we have $\mathcal{N}=1$, then $g=h$, and moreover, if one wants $\mathcal{N}>1$ supersymmetry, the target manifold must admit $\mathcal{N}-1$ parallel structures $I^a_{ij}$ satisfying the Clifford algebra $I^aI^b+I^bI^a=-2\delta^{ab}$, with $a=2,...,\mathcal{N}$ (a full explicit proof can also be found in Bagger, Supersymmetric Sigma Models, 1984).

In particular it means that for $\mathcal{N}=2$, $\mathcal{M}$ admits a complex structure $I^2=-1$, and a corollary is that $\mathcal{N}=2$ supersymmetric target manifolds must be Kähler.

This is confusing to me, because I would have thought that target manifolds of $\mathcal{N}=1$ must be Kähler. One of the first result in any SUSY course is that the most general SUSY kinetic term (for scalar multiplets) is $$ \mathcal{L}_\text{kin}=\frac{1}{2}K_{i\bar{\jmath}}\partial_\mu\phi^i\partial^\mu\bar{\phi}^\bar{\jmath}\qquad K_{i\bar{\jmath}}=\partial_i\overline{\partial}_\bar{\jmath}K $$ For some Kähler potential $K$.

Is the difference that in the second case, I assume already that I have complex scalars? In that case does that mean that despite the metric descending from a Kahler potential, the target space is not Kähler? If the target space of $\mathcal{N}=1$ is indeed Kähler, what is the complex structure?

This post imported from StackExchange Physics at 2016-12-22 17:33 (UTC), posted by SE-user Bulkilol

asked Dec 12, 2016 in Theoretical Physics by Bulkilol (65 points) [ revision history ]
edited Dec 22, 2016 by 40227

1 Answer

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It depends on the dimensionality of the space one works with. In $2D$, the kinetic term that you presented appears after the integration over 4 Grassmann coordinates in the expression $\int \mathrm{d} \ \sigma^2 \mathrm{d} \ \theta^4 K(\Phi, \overline{\Phi})$ which means $\cal{N}=2$. Perhaps it is the reason of your confusion?

This post imported from StackExchange Physics at 2016-12-22 17:33 (UTC), posted by SE-user Andrew Feldman
answered Dec 12, 2016 by Andrey Feldman (904 points) [ no revision ]
Indeed, I didn't realise that the result is valid only in 2 and 3 dimensions. However, I still have a question: If I start with the first Lagrangian in four dimensions and I stay completely agnostic about superspace. Is it possible to prove starting only from supersymmery transformations that there is a complex structure I^2?

This post imported from StackExchange Physics at 2016-12-22 17:33 (UTC), posted by SE-user Bulkilol
@Such a structure can be constructed from the covariant spinors on the target space, which existence is guaranteed by the presence of supersymmetry. Schematically, for the case of $\cal{N}=1$ and covariant spinor $\eta$, the Kahler form and the volume form have the form: $\omega_{\mu \nu}=\eta \Gamma_{\mu \nu} \eta$, $\Omega_{\mu \nu \rho}=\eta \Gamma_{\mu \nu \rho} \eta$. See, for example, the seminal paper "Vacuum configurations for superstrings" by Witten et al.

This post imported from StackExchange Physics at 2016-12-22 17:33 (UTC), posted by SE-user Andrew Feldman

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