# Definition of Differential operator

Definition 2.2, page 19 Let $$M$$ be a smooth manifold and $$E_i \rightarrow M$$ be two smooth vector bundles. A PDO $$P:\Gamma (M,E_0) \rightarrow \Gamma(M,E_1)$$ of order $$k$$ is a a linear map which satisfies the following properties:

1. $$P$$ is local in the sense that if $$s \in \Gamma(M,E_0)$$ vanishes on an open subset $$U \subseteq M$$, then so does $$Ps$$.

2. If $$x:U \rightarrow \Bbb R^n$$ is a chart, $$\phi_i: E_i \Big|_U \rightarrow U \times \Bbb K ^{p_i}$$ a trivialization, then the localizaed operator $$\phi_1 \circ P \circ \phi_0^{-1}$$ can be written as $$(\phi_1 \circ P \circ \phi_0^{-1}) (f)(y) = \sum_{|\alpha| \le k } A^{(\alpha)}(y) \frac{\partial^\alpha}{\partial x_\alpha} f(y)$$ for each $$f \in C^\infty(U, \Bbb K^{p_0})$$ where $$A^\alpha:U \rightarrow M_{p_1,p_0}(\Bbb K)$$. is a smooth function.

I returned to this definition after working on it for a while. I am confused now - why doesn't 2 => 1?

The problem is that without imposing (1) it does not make much sense to talk about (2). Statement (1) essentialy means that $$P$$ can be localized in a rather naive sense without problems. The point is that (1) implies that for any open subset $$U\subset M$$ and sections $$s_1,s_2$$ of $$E_0$$, the fact that $$s_1|_U=s_2|_U$$ implies $$P(s_1)|_U=P(s_2)|_U$$. Thus, there is a well defined operator $$P|_U:\Gamma(U,E_0)\to\Gamma(U,E_1)$$ defined by choosing extensions of locally defined sections, applying $$P$$ and restricting the result. The condition in (2) actually implicitly uses the restriction of $$P$$ to $$U$$ and (1) is needed for this restriction to be sufficiently closely related to $$P$$ itself to be useful. In the context you are studying, you can think about pseudodifferential operators as examples of operators for which localization is a much more subtle issue.

In fact, it is evan true that for linear operators (1) implies (2), by the so-called Peetre-theorem.