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It is mentioned in first page of this paper by Seiberg and Komargodski that the Lagrangian in superspace of a $U(1)$ gauge SUSY theory with FI terms is not gauge invariant. However, the FI terms in superspace is $$ \xi \int d^4 \theta V $$ where $V$ is a vector superfield. Now if we do a gauge transformation on $V$, i.e., $$ V\longrightarrow V+ i (\Phi-\bar{\Phi}) $$ the FI term remains invariant since $$ \int d^4 \theta \Phi=\int d^4 \theta \bar{\Phi}=0. $$

So what is the source of the gauge non-invariance in superspace ?

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I think what they mean is FI-term is not gauge invariant under the full gauge symmetry of the theory, but under this remaining gauge freedom after WZ gauge, which is $U(1)$.

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  • $\begingroup$ well seen ... don't you cite the 2nd paragraph of the introduction from the 3rd sentence ? $\endgroup$ – user46925 Jan 22 '16 at 1:31
  • $\begingroup$ yes, but also this is well known that adding FI restricts YM theory to $G = U(1)$. $\endgroup$ – John Doe Jan 22 '16 at 6:10
  • $\begingroup$ it was to answer the last question ... But you said the main thing to the OP $\endgroup$ – user46925 Jan 22 '16 at 9:56
  • $\begingroup$ I understand that the supercurrent is gauge invariant only under the WZ gauge but the above FI term in superspace is invariant under the full gauge symmetry. Can't see the WZ restriction in superspace. $\endgroup$ – Axion Jan 22 '16 at 13:26
  • $\begingroup$ @Axion from where comes the exact form of the F terms you used ? from which model precisely ( and book ) ? $\endgroup$ – user46925 Jan 22 '16 at 17:10

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