Supersymmetry invariants On page 158 of Fields, the supersymmetry algebra is represented in terms of the action on supercoordinates as
$$\delta \theta^\alpha = \epsilon^\alpha$$
$$\delta\bar{\theta}^{\dot{\alpha}} = \bar{\epsilon}^{\dot{\alpha}}$$
$$\delta x^{\alpha\dot{\beta}} = \frac{1}{2}i(\epsilon^\alpha \bar{\theta}^{\dot{\beta}} + \bar{\epsilon}^{\dot{\beta}}\theta^\alpha)$$
Further down on the page, the following statement is made:

In the supersymmetric case the infinitesimal invariants under the q's (and therefore p) are
$$d\theta^\alpha, \qquad d\bar{\theta}^{\dot{\alpha}}, \qquad dx^{\alpha\dot{\beta}} + \frac{1}{2}i(d\theta^\alpha)\bar{\theta}^{\dot{\beta}} + \frac{1}{2}i(d\bar{\theta}^{\dot{\beta}})\theta^\alpha$$

I get a sense that these should follow from the definition of the supersymmetry transformations, but what exactly does the quoted statement mean?
Am I correct in inferring that $d\theta^\alpha = \theta'^\alpha - \theta^\alpha$ where $\theta'^\alpha$ is $\theta^\alpha$ + SUSY variation of $\theta^\alpha$ by an infinitesimal SUSY parameter (more precisely by an infinitesimal scalar times the supersymmetry parameter $\epsilon$)? Thinking of the SUSY variation $\delta_\epsilon$ as something similar to a BRST variation, is this statement about nilpotency of the algebra in form notation or something?
 A: No, you are not corrected in your interpretation of the statement. The statement is that theses differentials, or $1$-forms, are supersymetric in the sense that:
$d(\theta^{\alpha}+\delta_{\epsilon}\theta)=d(\theta^{\alpha}+\epsilon^{\alpha})=0$
and the same for the other differentials. These $d$ is not the same as $\delta_{\epsilon}$. It is just a exterior derivative acting on "smooth functions".
No, the supersymmetry is not a BRST symmetry. The only thing in common between these two is the fact that they are both fermionic symmetries, just that. The supersymmetry is not nilpotent, quite the contrary, it obeys the following algebra:
$$
\{Q_\alpha,Q_{\dot\beta}\}\propto P_{\alpha\dot\beta}
$$
as you can check using the formulas above and the fact that $\delta_{\epsilon}=\epsilon^{\alpha}Q_\alpha+\bar{\epsilon}^{\dot\beta}Q_{\dot\beta}$, and $P_{\alpha\dot\beta}$ being the generator of translations of $x_{\alpha\dot\beta}\rightarrow x_{\alpha\dot\beta}+\Lambda_{m}\sigma^{m}_{\alpha\dot\beta}$.
