[Math] Why are ‘differential operators on manifolds’ differential operators

differential-geometrymathematical physicssmooth-manifoldsvector-bundles

It is clear what is meant by a differential operator on $\mathbb{R}^n$ (or some open subset). However, it is not clear to me why differential operators on smooth manifolds are defined the way they are, involving sections of smooth sections of vector bundles.

Recall that if $E, F \to M$ are smooth vector bundles over a smooth manifold $M$, a differential operator $E \to F$ is a morphism between the sheaves of smooth sections of $E$ and $F$ such that over an affine chart $U$ trivialising both $E,F$, say with $E|_U \cong U \times \mathbb{R}^n$ and $F|_U \cong U \times \mathbb{R}^m$ , $C^{\infty}(U, \mathbb{R}^n) \cong \Gamma(U, E) \to \Gamma(U, F) \cong C^{\infty}(U, \mathbb{R}^m)$ is given by a classical differential operator.

Question: why is this is the right generalisation of differential operators to manifolds — and what is the (geometric/physical) significance of the vector bundles here?

After-thought:
Ah. I believe I was confused by the fact that differential operators over manifolds are labelled as differential operators on manifolds. Not being pedantic here. It's plain why one might want to consider how smooth assignments of directions vary in certain directions (that is, why one would want to partially differentiate sections of vector bundles). However, I was under the misconception that vector bundles were brought into the picture to facilitate the generalisation of partial derivatives directly on manifolds (which incidentally are just given by the theory over the trivial bundle). A bit silly on reflection.

Best Answer

Consider the following operators:

  1. Gradient - Acts on functions and return vector fields. On $M = \mathbb{R}^3$, it has the signature $\nabla \colon \Gamma(M,M \times \mathbb{R}) \rightarrow \Gamma(M,TM)$.
  2. Directional derivative of some vector field with respect to a fixed vector field $X$. Acts on vector fields and return vector fields. On $M = \mathbb{R}^3$, it has the signature $\nabla_X \colon \Gamma(M,TM) \rightarrow \Gamma(M,TM)$.
  3. Divergence - Acts on vector field and return functions. On $M = \mathbb{R}^3$, it has the signature $\mathrm{div} \colon \Gamma(M,TM) \rightarrow \Gamma(M,M \times \mathbb{R})$.

They are all classical first order differential operators with different domains and codomains that can be interpreted as acting on and returning sections of vector bundles. On $M = \mathbb{R}^3$, all the vector bundles involved (and in general) are trivial and so you can just pick some global isomorphism $TM \cong M \times \mathbb{R}^3$ and think of the operators as operators with signature $D \colon C^{\infty}(\mathbb{R}^3,\mathbb{R}^n) \rightarrow C^{\infty}(\mathbb{R}^3,\mathbb{R}^m)$ for appropriate $n$ and $m$ (they act on vector valued functions and return vector valued functions). However, working with arbitrary manifolds, the notion of a vector field which was before (identified with) just a tuple of smooth function (an element of $C^{\infty}(\mathbb{R}^3,\mathbb{R}^3)$) now generalizes to a section of a possibly non-trivial vector bundle $TM$ and so the operators now should be generalized to operators that act on sections of (possibly non-trivial) vector bundles. And indeed,

  1. Gradient is replaced/generalized with the differential of a function $d \colon \Gamma(M, M \times \mathbb{R}) \rightarrow \Gamma(M, T^{*}M)$ (function $\mapsto$ covector field) or, in the presence of a Riemannian metric, by a gradient operator $\nabla \colon \Gamma(M,M\times \mathbb{R}) \rightarrow \Gamma(M,TM)$ (function $\mapsto$ vector field).
  2. The directional derivative is generalized by the notion of a covariant derivative (also called an affine connection) denoted by $\nabla_X \colon \Gamma(M,TM) \rightarrow \Gamma(M,TM)$.
  3. Divergence is generalized in the presence of a Riemannian metric to an operator $\mathrm{div} \colon \Gamma(M,TM) \rightarrow \Gamma(M,M\times \mathbb{R})$.
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