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The effective action of the $\mathcal{N}=2$ supersymmetric $SU(2)$ gauge theory contains the following term;

$$Im\int d^{4}xd^{2}\theta d^{2}\bar{\theta}\Phi^{\dagger}\mathcal{F}'(\Phi)$$

Where $\Phi$ is a superfield, $\mathcal{F}$ is some (holomorphic), function and $\mathcal{F}'(\Phi)=\frac{d\mathcal{F}}{d\Phi}$.

Define $$\Phi_{D}=\mathcal{F}'(\Phi)\quad \text{and}\quad \mathcal{F}_{D}'(\Phi_{D})=-\Phi.$$

According to this paper, we have;

$$ Im\int d^{4}xd^{2}\theta d^{2}\bar{\theta}\Phi^{\dagger}\mathcal{F}'(\Phi) = Im\int d^{4}xd^{2}\theta d^{2}\bar{\theta}\Phi^{\dagger}_{D}\mathcal{F}_{D}'(\Phi_{D})\tag{9.7} $$ So that this term is invariant under the transformation.

Plugging in the definitions of $\Phi_{D}$ and $\mathcal{F}_{D}$, we have;

$$ Im\int d^{4}xd^{2}\theta d^{2}\bar{\theta}\Phi^{\dagger}\mathcal{F}'(\Phi) = Im\int d^{4}xd^{2}\theta d^{2}\bar{\theta}(-\mathcal{F}_{D}'(\Phi_{D}))^{\dagger}\Phi_{D} $$

So I need to show that (up to a total derivative), $-(\mathcal{F}_{D}'(\Phi_{D}))^{\dagger}\Phi_{D}=\Phi^{\dagger}\mathcal{F}'(\Phi).$

I feel like this should be immediate, but I just can't seem to make it work.

I have tried integrating by parts, as well as the following formal manipulation;

$$ -\left(\frac{d\mathcal{F}_{D}}{d\Phi_{D}}(\Phi_{D})\right)^{\dagger}\Phi_{D} = -\Phi_{D}^{\dagger}\left(\frac{d\mathcal{F}_{D}}{d\Phi_{D}}\right)^{\dagger}\Phi_{D} = \Phi_{D}^{\dagger}\left(\frac{d\mathcal{F}_{D}}{d\Phi_{D}}\right)\Phi_{D} = \Phi_{D}^{\dagger}\mathcal{F}'_{D}(\Phi_{D}). $$ Integrating by parts didn't go anywhere, and I have been unable to justify either of the first two equalities above (I am including them only for completeness as far as my attempts).

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  1. Eq. (9.7) follows from the involutive properties of the Legendre transformation $$F(\phi)\quad\longrightarrow \quad-F_D(\phi_D)~:=~\phi\phi_D - F(\phi),\tag{A}$$ which is spelled out in eq. (9.6). Note the extra minus in the definition (A).

  2. The minus disappears again because $${\rm Im}(\bar{z})~=~-{\rm Im}(z).\tag{B}$$

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