= Supersymmetric Liouville equation
{title2=$iD_-D_+\Phi=e^\Phi$}
In <two-dimensional N=(1,1) superspace>, one convention for the supersymmetric Liouville equation is $iD_-D_+\Phi=e^\Phi$. With the ordered expansion $\Phi=\phi+i\theta^-\psi_++i\theta^+\psi_-+i\theta^-\theta^+F$ and <left Grassmann derivatives>, its left side is
$$
F+i\theta^-\partial_-\psi_- -i\theta^+\partial_+\psi_+ -i\theta^-\theta^+\partial_-\partial_+\phi.
$$
Writing $N=\Phi-\phi$, <Grassmann parity> gives $N^2=2\theta^-\theta^+\psi_+\psi_-$ and $N^3=0$. Hence
$$
e^\Phi=e^\phi[1+i\theta^-\psi_++i\theta^+\psi_-+\theta^-\theta^+(iF+\psi_+\psi_-)].
$$
Comparing coefficients eliminates the <auxiliary field> as $F=e^\phi$ and gives $\partial_-\psi_-=e^\phi\psi_+$, $\partial_+\psi_+=-e^\phi\psi_-$, $\partial_-\partial_+\phi=-e^{2\phi}+ie^\phi\psi_+\psi_-$. The fermion bilinear sign depends on the stated ordering and derivative convention.
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