Let $$ \cdots \longrightarrow E_n \overset{\pi_n}\longrightarrow E_{n-1} \longrightarrow \cdots \longrightarrow E_1\overset{\pi_1}\longrightarrow E_0$$ be a sequence of fibered manifolds (surjective submersions). Define $\Lambda^0:=\bigcup_{n\geq 0}C^{\infty}(E_n)/\sim$, where $f \sim g$ iff $f$ is the pullback of $g$ by the sequence or viceversa. $\Lambda^0$ is an $\mathbb R$-algebra and every $C^{\infty}(E_n)$ is embedded on it.
I want to study the algebra of derivations $D$: $$ D:\Lambda^0 \longrightarrow \Lambda^0$$ $$ D \text{ is }\mathbb R \text{-linear}$$ $$ D(fg)=D(f)g+gD(f)$$
The restriction to $C^{\infty}(E_n)$ gives a derivation over $[\cdot]$: $$D_n:=D|_{C^{\infty}(E_n)}:C^{\infty}(E_n)\longrightarrow \Lambda^0$$ $$ D_n(f):=D([f]) $$ $$ D_n(fg)=D_n(f)[g]+[f]D_n(g)$$
- Can we affirm that there is a number $k(n)\geq n$ such that $\text{Im}(D_n)\subset C^{\infty}(E_{k(n)})$? If so, can we say something about $k(n)$?
- Example: Let $X_n:E_n\longrightarrow TE_{n-1}$ a sequence of vector fields over the projection $\pi_n:E_n\longrightarrow E_{n-1}$. As $(X_n)_{\xi}\in T_{\pi_n(\xi)}E_{n-1}$, $X_n$ is equivalent to a derivation along $\pi_n^*$: $$X_n: C^{\infty}(E_{n-1})\longrightarrow C^{\infty}(E_{n})$$ $$ X_n(f)(\xi):=(X_n)_{\xi}(f) $$ $$ X_n(fg)=X_n(f)\pi_n^*g+\pi_n^*f X_n(g) $$ The derivation: $$ D:\Lambda^0 \longrightarrow \Lambda^0$$ $$ D[f]:=[X_nf], \;\; f\in C^{\infty}(E_{n-1})$$ is well defined iff $\pi_{n+1}^*(X_nf)=X_{n+1}(\pi_n^*f)$, iff $d\pi_n\circ X_{n+1}=X_n\circ \pi_{n+1}$ ($X_n$ are related by the sequence).
- Is every derivation $D$ of the form $D_n=X_{n}$, where $X_n$ are vector fields $X_n:E_{k(n)}\longrightarrow TE_{n-1}$ along $E_{k(n)}\longrightarrow E_n$, related by the sequence?