## A local view of the global definition of a differential operator

Recall the global definition of a differential operator (of order *m*)

Also recall that differential operators are local, so that in particular, they induce (linear) operators between the spaces of sections over chart neighbourhoods in *X*. This allows us to reduce to the Euclidean case, and for the remainder of this discussion, we will assume (noting that the space of (smooth) sections of is naturally isomorphic to the space of (smooth) maps from , etc.)

**Proposition:** If *D* has the form of a “local differential operator” of order *m*, then

*Proof:*

Set as in the definition of a local differential operator. If then Leibniz’s differentiation rule yields are multi-indices, and sums of mulit-indices are taken term-wise, and .

We proceed by induction on *m*, the order of *D* (as a local D.O.). By the linearity of , it suffices to prove . Now, , which gives , establishing the base case. Now assume that all local differential operators of order at most are (global) differential operators of the same order. The Leibniz rule above gives:

Since —i.e., each in the sum has order at most , and so by the inductive hypothesis (and noticing that ), we find that QED.

Now we begin to prove the converse to the above proposition. For the purposes of the next lemma, we may relax the assumption that *D* is an operator between sections of vector bundles over Euclidean space.

**Lemma 1:** Let , then for every If every vanishes at a point .

*Proof:*

Again, by induction on *m*. If , and thus the base case is established. Now suppose the result for , and let . Now . By assumption, and so . On the other hand, by hypothesis, and so QED.

**Lemma 2: **Let Now, for every with the property that

The proof is left as an exercise. We give an outline of the inductive construction. To find , allow to act on the constant co-ordinate sections , and expand each in terms of the basis . Take to be the matrix thus determined—since *D* maps smooth vector fields to smooth vector fields, the co-ordinates of will be given by smooth functions of *x*. This gives a order approximation of *D*.

Now we take these sections and begin to multiply them by functions on *X* to obtain new sections—e.g., take where is the multi-index with only a single 1 in position *j*. In general, define the by for some multi-index . It is clear (or an exercise) that these matrices produce the desired result, and moreover, Lemma 1 shows that at most of these matrices will be non-zero (i.e., that the process terminates). Uniqueness is, of course, automatic.

**Theorem:** Differential operators are precisely local operators whose local form is that of a local differential operator.

*Proof:*

Because differential operators are local, we can immediately reduce to the Euclidean case. Showing “” was the content of the proposition, and so it suffices to show ““. Invoking the construction from lemma 2 (with *k = m*), let .

Consider the section . Recall Taylor’s Theorem: given , a polynomial of degree *m, *a neighbourhood , and for each multi-index such that for all . Choose take *m* to be the order of *D*, and apply Taylor’s Theorem to each individually to obtain on a neighbourhood . Take .

Now by locality, and so Hence, by construction, and so . On the other hand, , and so linearity reduces the question to showing .

Finally, are differential operators of order *m* (use the proposition to get this for ), and since smooth functions which all vanish at , we find by lemma 1, that , whence having been arbitrary. QED.

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