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