# Higher-Order Differential Operators as Vector Fields

On a $C^\infty$ manifold $M$, one can produce the tangent space $T_p M$ at a point by equivalence classes of tangents to smooth curves through the point $p$.

When realised this way, the tangent vectors are local derivations of functions defined at $p$. Globally,then, vector fields on $M$ act on global functions as differential operators. Adapted to some chart $(U, x^1, \ldots, x^n)$, a vector field $V$ looks like

$$\sum_{i=1}^{n} v^i \frac{\partial}{\partial x^i}$$

where the $v^i$ are functions on $U$. A lot of stuff comes from this, Lie theory, etc. All this is ok for me.

But what about, say, acceleration vectors of curves through a point? These have the same geometrical interpretation that velocity vectors do. I can construct a curve in some local coordinates with any arbitrary acceleration that I like. The resulting space has the same dimension as the tangent space, too. Presumably one could also extend such things to global objects. With some fooling around, these can be made to correspond (locally) to expressions of the form

$$\sum_{i=1}^{n} v^i \frac{\partial^2}{\partial^2 x^i}?$$

Going in the other direction, note that things of the above type close under the same Lie bracket as regular vector fields. Physically, acceleration fields look like forces, so one might expect such a thing to be physically meaningful -- yet I've never heard of them.

Obviously this all carries on to derivative vectors of all orders. Do these sit in some enveloping algebra for vector fields? Some multivector construction?

I would greatly appreciate any references, please, if you have them.

• Are you familiar with jets? It sounds like this is what you're trying to get at. Note that unless you have some geometric structure you can't express this kind of thing in terms of an algebraic construction on $T_p M$ - the acceleration lives in the double tangent bundle unless you have a connection. Aug 4, 2015 at 6:05
• @AnthonyCarapetis Not massively, but I'll look into it. Why is a connection necessary? I also have an inkling from personal searching that this might relate to Schouten-Nijenhuis algebras (?). Aug 4, 2015 at 6:13
• The velocity $\dot \gamma$ is a vector field, so in order to differentiate it to get another vector field we must take the covariant acceleration $\nabla_{\dot \gamma} \dot \gamma$. If you take the second derivative in coordinates then you will get different answers in different coordinate systems - for example curves that are straight (zero acceleration) in cartesian coordinates will not be straight in polar coordinates. Thus in the setting of smooth manifolds the double tangent bundle gets involved. After some more thought, I think the higher tangent bundles are all you're looking for. Aug 4, 2015 at 8:05
• I think the kind of correspondence you're looking for breaks down in higher order - at least the duality between jets of maps $\mathbb R \to M$ and those of maps $M \to \mathbb R$ does. I don't see a natural way to identify accelerations with 2nd order differential operators, even the "directional" ones (of the form $X^2$ for some vector field $X$). Google turned up this article which discusses the divergence of the iterated tangent bundle and the jet bundle at second order - it might be interesting. Aug 5, 2015 at 8:27
• Thanks for that link. I'll keep thinking about it. The non-uniqueness of solutions to second order DEs may also be an issue. Aug 6, 2015 at 2:21

Consider a moving particle on a manifold, it traces out a curve $$c$$. This is simply a map from the real line to the manifold and so we can take it's derivative directly via the tangent functor. Now if it were to return another map from the real line to the manifold we could take the derivative again and so obtain the acceleration. However this is not the case. What is returned is a tangent field along the curve $$c$$. We cannot derive this directly by the tangent functor. What we need to use is the covariant derivative. But this requires a connection.