1. In defining
the infinitesimal cotangent Lie algebroid, the underlying graded algebra is defined as
$$ (T^\ast_{inf} \mathfrak{a})^\ast_\bullet
\;:=\;
\mathfrak{a}^\ast_\bullet \oplus Der(CE(\mathfrak{a}))_\bullet $$
I understand this is a degree-by-degree direct sum. What is in principle the grading on ##Der(CE(\mathfrak{a}))_\bullet##, such that in the next example for the action Lie algebroid, ##\frac{\partial}{\partial c^\alpha}## and ##\frac{\partial}{\partial \phi^a}## have degree ##-1## and ##0## respectively?
2. I am not sure I got why
## C^\infty(X)/ (\frac{\partial S}{\partial \phi^a}) \simeq C^\infty(X_{dS=0}) ## when ##X## is a superpoint.
My understanding of the given explanation is the following. As ##X## is a superpoint, each generator ##f## in ##C^\infty(X) := \mathbb{R} \oplus V## where ##V## is a finite dimensional vector space that is a nilpotent ideal, and we could write ##f = f_0 + f_1a_1 + ... + f_na_n## where ##f_i \in \mathbb{R}## and ##a_i## are the basis vectors of ##V##. Thus as long as ##f## vanishes where ##\frac{\partial S}{\partial \psi^a}## vanishes, we could generate ##f## from ##\frac{\partial S}{\partial \psi^a}## by scaling term by term. Such ##f## is zero in the quotient, and so the quotient is exactly the algebra of functions on ##X_{dS=0}##. But
(a) Is ##X## being a superpoint a necessary condition? If ##X## is an usual manifold, we could also do the scaling point by point.
(b) We are not actually allowed to talk about "points" in a superpoint. Are we,
again, secretly using a 1-1 correspondence between ##\{ X\rightarrow \mathbb{R}^1 \}## and ## \{X_{even} \rightarrow \mathbb{R}^1 \}## with ##C^\infty(X_{even})## generated from generators of ##C^\infty(X) ## regarded in even degree?
