Horizontal lift of differential operator

Question

On a Riemannian manifold $M$, there is a canonical horizontal lift $X^{\mathrm{hor}}$ of vector fields $X$ to $TM$, which is characterized by the two properties that

1. $X^{\mathrm{hor}}$ is a horizontal vector
2. $d \pi (X^{\mathrm{hor}}) = X$.

I wonder how one would define the horizontal lift of higher order differential operators. For example they should satisfy the property $$ P^{\mathrm{hor}} (f u) = f P^{\mathrm{hor}} u$$ for any function $u \in C^\infty(TM$) and any $f$ of the form $f(X) = \tilde{f}(|X|)$ for some function $\tilde{f} \in C^\infty(\mathbb{R})$. One would like to say that $P^\mathrm{hor}$ maps functions that are "constant in horizontal directions" to zero but that doesn't make sense as the horizontal distribution is never integrable unless $M$ is flat (in which case, obviously, we have a nice definition of the horizontal lift of any differential operator).

The problem is clearly that the horizontal lift does not induce an algebra homomorphism on the algebra of differential operators because if two vector fields commute on $M$, then their horizontal lift usually does not commute. So there must be some kind of "quantization" going on.

More specifically: It is clear that there are probably many choices to lift differential operators, but it would be nice to list some properties in the line of those stated above that characterize a horizontal lift of any differential operator uniquely.

Answer 1

This is a sort of standard construction you can find in several places. I don't know where this was done first though...

OK: first you can extend your horizontal lift from vector fields to all (symmetric) contravariant tensor fields by requiring that the horizontal lift of a tensor product is the tensor product (now upstairs) of the horizontal lift. With other words, ${}^\mathrm{hor}$ is a algebra homomorphism from the tensor algebra on $M$ to the tensor algebra of $TM$.

The next step consist in lifting the connection itself. Since you are in a Riemannian situation, this is fairly easy and gives you a (Riemannian) connection on $TM$ (viewed as manifold, so the lifted connection differentiates sections of bundles over $TM$).

Finally, you can use this connection on $TM$ to "quantize" the lifted symmetric tensor fields on $TM$. This gives a global symbol calculus on $TM$ which matches the global symbol calculus on $M$ build out of the connection downstairs.

I don't have a reasonable reference for this at hand, but in my book (german, argghh...) you can find this kind of stuff in Exercise 5.16 and around...