In case you have the Kahler and super- potential [itex]K,W[/itex]: [itex] K(T,S,C) = -log (S +S^{*}) -3 log ( T+ T^{*} - C C^{*}) [/itex] [itex] W(T,S,C)= C^{3} + d e^{-aS} +b [/itex] with [itex]T,S,C[/itex] chiral super fields, [itex]b,d[/itex] complex numbers and [itex]a>0[/itex]. I tried to calculate the local F-terms arising from this. The local F-terms for the i-th chiral superfield are given by: [itex] F_{i}= D_{i}W = K_{i}W + W_{i}[/itex] where in the rhs the index i denotes the derivative wrt to the i-th field. eg [itex]W_{S}=\frac{\partial W}{\partial S}[/itex] However I'm having a slight problem with the particular derivative. See what I mean...taking it: [itex] F_{S}= K_{S} W + W_{S} = - \frac{C^{3} + d e^{-aS} +b}{S+S^{*}} -d a e^{-aS}[/itex] correct? On the other hand, if I try to work with the covariant derivative wrt to the conjugate fields: [itex] F^{*}_{S}= D_{S^{*}} W^{*} = K_{S^{*}} W^{*} + W_{S^{*}} [/itex] I don't get the complex conjugate of the above. Because in this case [itex]W_{S^{*}}=0[/itex] and so: [itex] F^{*}_{S}= - \frac{(C^{3} + d e^{-aS} +b)^{*}}{S+S^{*}}[/itex] what's the problem?
Ah found the mistake.... again by writing in LaTeX it became obvious- In the F* equation I needed the W* derivative as the second term... However In the case of [itex]F[/itex] let's say... How can I see if its module squared is simultaneously zero or not? [itex] |F_{T}|^{2}= \frac{9}{(T+T^{*} - CC^{*})^{2}} |C^{3}+ d e^{-aS} +b |^{2} [/itex] [itex] |F_{S}|^{2}= | \frac{C^{3} + d e^{-aS} +b}{S+S^{*}} + d a e^{-aS}|^{2} [/itex] [itex] |F_{C}|^{2}= | \frac{3 C^{*} (C^{3}+d e^{-aS} +b)}{T+T^{*}-CC^{*}} +3 C^{2}|^{2} [/itex]