Chain rule for the superderivative

Question

A one dimensional complex supermanifold $X$ is locally described by an ordinary complex coordinate $z$ and an anticommuting coordinate $\theta$, $\theta^2 = 0$.

The superderivative is the square root of the derivative in the following sense: Take the vector field $D_{\theta} = \partial_{\theta} + \theta\partial_z$ so that $\frac12[D_{\theta},D_{\theta}]= \partial_z$, where $[ \ , \ ]$ denotes the super-commutator.

It is stated (see for example pg. 4 Equation 2.9 in https://arxiv.org/pdf/1209.2459.pdf) that under a superanalytic change of coordinates $(z, \theta) \mapsto (\overline{z}(z, \theta), \overline{\theta}(z, \theta))$ the superderivative $D_{\theta}$ transforms as $$ D_{\theta} = (D_{\theta} \overline{z} + \overline{\theta} D_{\theta} \overline{\theta} )\frac{\partial}{\partial_{\overline{z}}} + D_{\theta} \overline{\theta} D_{\overline{\theta}} $$

My Question: When I use the chain rule I find that: $$D_{\theta} = (D_{\theta} \overline{z})\frac{\partial}{\partial_{\overline{z}}} + D_{\theta} \overline{\theta} D_{\overline{\theta}}. $$

In particular, I am missing the term $\overline{\theta} D_{\theta} \overline{\theta} $ term. I cannot figure out how to reproduce the given transformation $D_{\theta}$ using the ordinary chain rule from calculus.

How does one use the chain rule on $D_{\theta}$ to produce the term $\overline{\theta} D_{\theta} \overline{\theta} $?

Answer 1

Direct application of definitions and chain rule gives $$ D_\theta = (D_\theta \hat{\theta}) \partial_{\hat{\theta}}+(D_\theta \hat{z}) \partial_{\hat{z}}. $$ Then eliminate $\partial_{\hat{\theta}}$ in favour of $D_{\hat{\theta}}$: $$ D_\theta = (D_\theta \hat{\theta}) (D_{\hat{\theta}}-\hat{\theta}\partial_{\hat{z}})+(D_\theta \hat{z}) \partial_{\hat{z}}. $$ Rearranging terms, we get $$ D_\theta =(D_\theta \hat{\theta})D_{\hat{\theta}} +(D_\theta \hat{z}-\hat{\theta}D_\theta \hat{\theta}) \partial_{\hat{z}} $$ as claimed in (2.9) of the given reference (note the sign).