Regarding first order differential operator and derivative endomorphism

Question

I am reading "Differential operators and actions of Lie algebroids" by Kosmann-Schwarzbach and Mackenzie.

There is some confusion regarding the terminology.

Let $E\rightarrow M$ be a vector bundle.

A first order differential operator on $\Gamma(M,E)$ is an $\mathbb{R}$-linear map $D:\Gamma(M,E)\rightarrow \Gamma(M,E)$ for which there exists a vector field $D_M$ of $M$ such that $D(f\alpha)=fD(\alpha)+D_M(f)\alpha$ for all $f\in C^\infty(M)$ and $\alpha\in \Gamma(M,E)$.

A derivative endomorphism of $\Gamma(M,E)$ is an $\mathbb{R}$-linear map $D:\Gamma(M,E)\rightarrow \Gamma(M,E)$ for which there exists an $\mathbb{R}$-linear endomorphism $D_M$ of $C^\infty(M)$ such that $D(f\alpha)=fD(\alpha)+D_M(f)\alpha$ for all $f\in C^\infty(M)$ and $\alpha\in \Gamma(M,E)$.

There is some inconsistency with the terminology used in "General theory of Lie groupoids and Lie algebroids" but that is not my question here.

The paper says that if we start with a derivative endomorphism, then "it follows that $D$ is a first order differential operator." I do not understand how is this straightforward or if it is correct at all.

If I take any first order differential operator $D$, then, ignoring that $D_M$ (which is a vector field) has a derivation property, we can just focus on the $\mathbb{R}$-linearity and conclude that $D$ is a derivative endomorphism of $\Gamma(M,E)$. So, any first order differential operator should be a derivative endomorphism. But, the same paper says "a first-order differential operator is a derivative endomorphism of $\Gamma(M,E)$ if and only if it has scalar symbol."

What is that I am misunderstanding here?

The terminology is interchanged in the Mackenzie's book "General theory of Lie groupoids and Lie algebroids". What is called here as first order differential operator is called as derivative endomorphism there and vice versa.

I also think the name "derivative endomorphism" should be used when $D_M$ is a vector field, a derivation of $C^\infty(M)$.

Answer 1

Substituting $f=f_1f_2$ in the definition of a derivative endomorphism immediately implies that $D_M$ is a derivation, using the fact that $g_1ψ=g_2ψ$ for all vector fields $ψ$ implies $g_1=g_2$, where $g_1$ and $g_2$ are smooth functions.

Thus, the two notions coincide. An important remark: the first property in the original post does not define first-order differential operators, but rather singles out a special class of first-order differential operators that the authors want to study.