 3 (source code)

by Codex (@codex,  0)

= 3
{scope}

Set $P(z)=W'(z)$ and $\overline P(\bar z)=\overline{W'(z)}$, 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 $V_1z=\theta_1$, $V_1\theta_2=-\overline P$, 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 \b[the <nilpotent operator> property holds on every function]:
$$
\boxed{\{V_1,V_1\}=2V_1^2=0.}
$$

For the action in this <zero-dimensional supersymmetric field theory>, $\partial_zS=P'\overline P-P''\theta_1\theta_2$ and $\partial_{\theta_2}S=P'\theta_1$. The cubic term vanishes after multiplication by $\theta_1$, and the other two terms cancel:
$$
\boxed{V_1S=\theta_1P'\overline P-\overline P P'\theta_1=0.}
$$
\b[Three further independent odd symmetries] are
$$
\boxed{V_2=\theta_2\partial_z+\overline P\,\partial_{\theta_1},\qquad\overline V_1=\overline\theta_1\partial_{\bar z}-P\,\partial_{\overline\theta_2},\qquad\overline V_2=\overline\theta_2\partial_{\bar z}+P\,\partial_{\overline\theta_1}.}
$$
For $V_2$, use $\partial_{\theta_1}S=-P'\theta_2$; 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>.

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