0
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.