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In 2D (1,1) superconformal field theory, the invariant "distance" between two points $Z_1=(z_1,\theta_1)$ and $Z_1=(z_1,\theta_1)$ in superspace is $$Z_{12}=z_1-z_2-\theta_1\theta_2.$$ My question is how to compute the quantities when the $\theta$ appears in denominator or in square root, for example, how to expand $\frac{1}{Z_{12}}$ into component form? And how to expand \begin{equation} \sqrt{1-\frac{Z_{12}Z_{34}}{Z_{13}Z_{24}}} \end{equation} into component form?

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TL;DR: It is defined by expanding in the finite Taylor series of the $\theta$s.

More details:

  1. Recall that a supernumber $z=z_B+z_S$ has a body $z_B\in\mathbb{C}$, which is a complex number; and a soul $z_S$, which belongs to the ideal generated by Grassmann-odd generators.

  2. For any analytic function $f$, the supernumber $f(z)$ is defined via its formal Taylor series around the body $z_B$.

    In particular, for the reciprocal function $f(z)=z^{-1}$, this prescription only make sense if the body $z_B\neq 0$ is non-zero.

  3. Point 2 can be generalized, so that if supernumbers $z,w$ share the same body $z_B=w_B$, then $f(w)$ may be defined as a formal Taylor series around the other supernumber $z$.

    In particular for a superfield $\Phi(\theta)=\phi_0+{\cal O}(\theta)$ and its lowest component field $\phi_0$, they share the the same body $\Phi(\theta)_B=\phi_{0B}$, so that $f(\Phi(\theta))$ may be defined as a Taylor series (which happens to be finite!) around $\phi_0$.

  4. This generalizes in a straightforward way to several superfields.

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  • $\begingroup$ So the equation \begin{equation} \frac{1}{z_1-z_2-\theta_1\theta_2}=\frac{1}{z_1-z_2}\frac{1}{1-\frac{\theta_1\theta_2}{z_1-z_2}}=\frac{1}{z_1-z_2}(1+\frac{\theta_1\theta_2}{z_1-z_2}) \end{equation} is right? Since $(\theta_1\theta_2)^2=0$. $\endgroup$ – phys_student Oct 18 '19 at 7:33
  • $\begingroup$ $\uparrow$ Yes. $\endgroup$ – Qmechanic Oct 18 '19 at 7:58

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