# Showing that the $\mathcal{N=2}$ SUSY Effective Action is Duality-Invariant

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).

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 $${\rm Im}(\bar{z})~=~-{\rm Im}(z).\tag{B}$$