# Expand superspace function into component form

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 $$$$\sqrt{1-\frac{Z_{12}Z_{34}}{Z_{13}Z_{24}}}$$$$ into component form?

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.

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