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Consider $\mathcal{N}=1,d=4$ SUSY with $n$ chiral superfields $\Phi^i,$ Kaehler potential $K,$ superpotential $W$ and action ($\overline{\Phi}_i$ is complex conjugate of $\Phi^i$) $$ S= \int d^4x \left[ \int d^2\theta d^2\overline{\theta} K(\Phi,\overline{\Phi})+ \int d^2\theta W(\Phi) + \int d^2\overline{\theta} \overline{W}(\overline{\Phi})\right].$$

As in Weinberg (QFT, volume 3, page 89), requiring invariance of $K$ and $W$ under \begin{align}\delta\Phi^i &= i\epsilon Q^{i}_{~j}\Phi^j \\ \delta\overline{\Phi}_i &=- i\epsilon \overline{\Phi}_jQ_{~i}^{j}\end{align} with $\epsilon$ small positive constant real parameter and $Q$ hermitian matrix, one obtains some conditions on $K$ and $W,$ namely (denoting $K_i=\frac{\partial K}{\partial\Phi^i}$ and $K^i=\frac{\partial K}{\partial\overline{\Phi}_i},$ and the same for $W$) $$ K_i Q^i_{~j} \Phi^j = \overline{\Phi}_j Q^j_{~i} K^i $$ $$ W_i Q^i_{~j}\Phi^j=0=\overline{\Phi}_j Q^j_{~i} \overline{W}^i $$ and can then compute the Noether current associated to such transformations, promote $\epsilon$ to a full chiral superfield $\epsilon\Lambda$, etc.

The obtained invariance implies $\delta_{\epsilon}S=0$ for the above transformations.

QUESTION: is the reverse also true? namely, by requiring $\delta_{\epsilon}S=0$ (instead of the apparently stronger invariance condition on $K$ and $W$), does one obtain the same constraints as above on $K$ and $W?$

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1 Answer 1

The answer is yes; this can be shown by evaluating the variation of the action. The action consists of three terms, we will consider them separately:

$$\delta_\epsilon K(\Phi,\bar{\Phi})=\frac{\partial K(\Phi,\bar{\Phi})}{\partial\Phi^i}\delta_\epsilon \Phi^i+\frac{\partial K(\Phi,\bar{\Phi})}{\partial\bar{\Phi}_i}\delta_\epsilon\bar{\Phi}_i=i\epsilon K_i {Q^i}_j\Phi^j-i\epsilon K^i\bar{\Psi}_j{Q^j}_i,$$

$$\delta_\epsilon W(\Phi)=\frac{\partial W(\Phi)}{\partial \Phi^i}\delta_\epsilon \Phi^i=i\epsilon W_i {Q^i}_j\Phi^j$$

$$\delta_\epsilon \overline{W}(\bar{\Phi})=\frac{\partial \overline{W}(\bar{\Phi})}{\partial \bar{\Phi}_i}\delta_\epsilon \bar{\Phi}_i=-i\epsilon \overline{W}^i\bar{\Psi}_j{Q^j}_i,$$

where I have used the same conventions and abbreviations as you. In order for the variation of the action to vanish, each of the above expression has to be equal to zero separately. This leads to precisely the same constraints you have written down.

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thanks for the answer. the point lies precisely in i. getting convinced that the two contributions can be dealt with separately; ii. proving that vanishing of the action functional requires vanishing of the above pieces, which are kind of integrands, and for which I believe some kind of variational bicomplex argument is required –  jj_p Apr 13 '14 at 20:11

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