# How to motivate vectors as derivations?

In a manifold it's easy to motivate the definition of vectors as equivalence classes of curves. On the other hand the definition as derivations is harder to motivate. I know how to show that the space obtained with derivations is isomorphic to the space obtained with curves, so I know that defining with derivations works.

My doubt it is: how would I motivate the usage of derivations? This question is, how can it be intuitively visible that linearity and Leibniz rule alone are sufficient for assuring that the operator is a directional derivative?

I've seem some discussion about it on the "Applied Differential Geometry" book from William Burke where he tries to motivate this saying that a differential operator measures terms in the Taylor series and that Leibniz rule assures the operator is sensible only to first order terms. But I didn't get this idea yet.

## 2 Answers

The idea is, as always when dealing with local notions on manifolds, to pass to a chart and see what happens. On a chart (i.e. Euclidean space) you have an obvious correspondence between vectors and directional derivatives, in the sense that derivation in direction $v$ is given by $Df\cdot v$. If you write this down component-wise and lift it up to the manifold, you'll see the correspondence between tangent vectors and derivations on the manifold.

How can it be intuitively visible that linearity and Leibniz rule alone are sufficient for assuring that the operator is a directional derivative?

If we define a derivation by these conditions, we can prove that the set of derivations at a point is an $n$-dimensional vector space - this theorem should be found in any textbook covering abstract manifolds. Thus these conditions already give us a space of the correct size to be the space of directional derivatives - any extra conditions would either be superfluous, would decrease the dimension of the space or would make it fail to be a vector space at all.