Is there a coordinate-free definition of a differential operator?

Question

A differential operator is normally defined to be a sum of partial derivatives with respect to a given basis, i.e. something that looks like $\sum_{\vert {\alpha} \vert \leq n} a_{\alpha}(x) \cdot \partial^{\alpha}$. Differential operators of interest tend to be naturally seen in a coordinate-free way -- for example, the laplacian is the divergence of the gradient. Is there a coordiante-free way to define a general differential operator?

Answer 1

We can define them without explicit coordinates using smooth vector fields (sections of the tangent bundle). If $M$ is a manifold, then we can (a bit loosely) define differential operators of order, at most, $m$ to be finite linear combinations of, at most, $m$ vector fields.

If you want to be more explicit and careful on the construction and expression, first define $\text{Diff}^0(M)=C^\infty(M)$. Next, define $\text{Diff}^1(M)$ to be operators $P:C^\infty(M)\rightarrow C^\infty(M)$ of the form $P=V+f,$ where $V$ is a smooth vector field, and $f\in C^\infty(M).$ Finally, define $\text{Diff}^m(M)$ to be operators $P:C^\infty(M)\rightarrow C^\infty(M)$ of the form $$P=\sum\limits_{k=1}^K P_{k1}P_{k2}\cdots P_{kN_k},$$ where each $P_{kj}\in \text{Diff}^1(M)$, $K\in\mathbb{N}$, and $N_k\leq m.$

Answer 2

There is a characterization due to Peetre which ensures that $P:\mathcal{D}(\Omega) \rightarrow \mathcal{D}'(\Omega)$ is a differential operator if, and only if $\hbox{supp }(Pu) \subset \hbox{supp } u$, where $\Omega$ is a open subset of $\mathbb{R}^n$. See: Peetre, J. Une Caractérisation Abstraite des Opérateurs Différentiels and Peetre, J. Réctification a L'article - Une Caractérisation Abstraite des Opérateurs Différentiels Théorème 2.

There are several generalizations of this result for manifolds and fiber bundles.