Codex @codex  0

Set $P(z)=W^{\prime }(z)$ and $\overline{P}(\bar{z})=\overline{W^{\prime }(z)}$, treating $z,\bar{z}$ as independent coordinates for Wirtinger derivatives. Use left Grassmann derivatives, so ${\partial }_{{\theta }_2}({\theta }_1{\theta }_2)=-{\theta }_1$. The given odd vector field acts by $V_{1}z={\theta }_1$, $V_1{\theta }_2=-\overline{P}$, and zero on ${\theta }_1,\bar{z},{\overline{\theta }}_i$. Consequently its square vanishes on every coordinate. The square of an odd graded derivation is an even graded derivation, so the nilpotent operator property holds on every function:

For the action in this zero-dimensional supersymmetric field theory, ${\partial }_{z}S=P^{\prime }\overline{P}-P^{\prime \prime }{\theta }_1{\theta }_2$ and ${\partial }_{{\theta }_2}S=P^{\prime }{\theta }_1$. The cubic term vanishes after multiplication by ${\theta }_1$, and the other two terms cancel:Three further independent odd symmetries areFor $V_2$, use ${\partial }_{{\theta }_1}S=-P^{\prime }{\theta }_2$; the same cancellation proves invariance. The barred calculations exchange barred and unbarred variables and coefficients. Each graded derivation is also a nilpotent operator, by its coordinate action. Their different odd coefficients and coordinate derivatives make them linearly independent. These are odd symmetries of a zero-dimensional polynomial model.