## A global definition of a differential operator

** Background **

and are smooth vector bundles over a smooth manifold . We have seen how to define differential operators locally, that is, over . How should we define *global* differential operators?

** Definitions **

Let us write . We define a Leibniz bracket by which acts on a section by . That is,

This is similar to the Lie bracket of two vector fields, .

We set and inductively define differential operators of order at most by requiring that provided for each .

Recall that for a section , its *support* is the closure of .

We would like to compare differential operators to local operators, those operators which only use local information in the following sense: An operator is local if, for each , .

Lemma 1A differential operator is local.

*Proof:* The only obvious way to prove this lemma is by induction. It is clearly true for : if is a section, then .

Now suppose that this is true for . Let , be arbitrarily chosen. Let be any open set for which . By Urysohn’s lemma, there is some smooth function with and . In particular, . (For brevity’s sake, call such an a “support function.”) Then since , we observe that

The support of the sum of two sections is the union of the supports of the sections since the sum is zero exactly when both sections are zero. Therefore,

The support of the product of two sections is the intersection of the supports of the sections, since the product is zero exactly when at least one of the sections is zero. Thus .

Additionally, by the inductive assumption. So

This containment holds for arbitrary containing . Therefore, the support of is contained in the intersection of the closures of these open sets:

One might naively wonder if this proves anything substantial: Are there ever any operators that aren’t local? In fact, yes; here’s an example. Let , , so that the sections of and both constitute smooth functions on . Let be a function with support on the northern hemisphere of , i.e., . Define an operator by

We see that cannot be local. Consider, say, a function with support only on the southern hemisphere of , so that . But by examining the integral , it’s clear that is not a subset of .

This gives intuition to the word “local:” a local operator acting on a section of determines based only on the behavior of in a neighborhood of , rather than on any global information. This corollary justifies the intuition: If on a neighborhood in two sections of agree, i.e., , then, restricted to , .

*Proof:* Put . Then and . Therefore, , so on we have that .

Since we’ve seen that differential operators are local, the natural next question is whether there are any non-differential local operators. This is answered by a theorem of Peetre, proved in the 1960s:

Theorem 2All local operators are differential operators.

The next question is whether in local coordinates differential operators can be represented in the form

The answer is “yes”:

Theorem 3All differential operators are, locally, differential operators.

## Leave a Reply