How to handle the differential of a vector field, ${\rm d}X:TM\to TTM$, in terms of (equivalence classes of) curves?

Solution 1:

It's unclear exactly what you're asking, but here are a few points to consider:

The tangent space of a manifold and the differential of a smooth map can be defined in a number of ways. Equivalence classes of curves is one which can be useful in some situations, but it tends to be cumbersome, because the vector space structure of tangent spaces is highly obfuscated, and choosing representatives introduces superfluous information. If you're looking for an alternative "intrinsic" definition, derivations are the standard choice.

If you want to "explicitly write down" the differential of a vector field $X$ on an $n$-manifold $M$, you can choose local coordinates on $M$ (and the induced coordinates on $TM$), and $dX$ will corresponds to the derivative of $X$ in the multivariable calculus sense, which is just a set of $n+n^2$ functions (the components of $X$ and all of their partial derivatives). When changing coordinates, however, these functions will not transform in a striaghtforward way, so if you with to formulate an intrinsic object which captures this information, there are a few options:

One option is to simply define a new object which obeys this transformation law. These are known as jets. The differential $dX$ is then equivalent to the jet $j^1(X)$, which is a section of the jet bundle $J^1(\pi_{TM})$. In fact, jects are closely related to the definition of the tangent space as equivalence classes of curves, and they are correspondingly difficult to work with.

Alternately, one can endow $TM$ with an extra piece of structure which allows us to "split" $dX$ into a 0th order part and a 1st order part in an intrinsic way. These are known as connections, and they can be formulated in a variety of ways.