Tangent vectors: arrows vs. derivatives

Question

I have a very hard time accepting the differential-geometric definition of a vector as a derivative operator, $$v = v^{\mu} \partial_{\mu}.$$ I want to make sure that the following line of reasoning is correct - if it is, then I will finally be at peace with the aforementioned definition.

(Geometrical) vectors are most naturally introduced as displacements in an affine space $\mathbb{A}$. E.g. a geometrical vector $\vec{v} \in \mathbb{V}$ (a vector space) can be thought of as a difference of two points (positions) $\mathcal{P, Q} \in \mathbb{A}$, so that $$\vec{v} = \mathcal{Q}-\mathcal{P}.$$ We can think of $\vec{v}$ as a good old fashioned arrow starting at $\mathcal{P}$ and ending at $\mathcal{Q}$. Of course, we cannot play this game outside of $\mathbb{A}$, but this is still valid locally in any manifold $\mathcal{M}$ since manifolds are locally Euclid (more specific, affine) spaces. If $\gamma:\mathbb{R} \to \mathcal{M}$ is a curve in $\mathcal{M}$ with $\gamma(0) = \mathcal{P}$, we can subtract $\mathcal{P}$ from its infinitesimally close neighbor, which lies on $\gamma$, to get a tangent vector$$\vec{v} = \gamma'(0).$$

First question: Am I correct to assume that $\gamma'(0)$ can be thought of as an arrow lying in the tangent space $T_{\mathcal{P}}\mathcal{M}$, not necessarily ending anywhere in $\mathcal{M}$? Furthermore, it should be obvious from the definition that this is a pure geometrical object and not a (differential) operator?

Second question: Is the operator-vector $v = v^{\mu} \partial_\mu$ just an object isomorphic to $\vec{v}$ and is therefore also known as the tangent vector, but whose character is more algebraic and less geometric? In other words, although $\vec{v}$ (an arrow) and $v$ (a derivative) are isomorphic, they are clearly different objects?

Third question: If the answer to the last question is yes, given some coordinates $\{x^\mu\}$, we should be able to write $$\vec{v} = v^{\mu} \vec{e}_\mu,$$ where now $\vec{e}_\mu$ is a geometric vector (arrow) basis and not a differential operator $\partial_\mu$. How is $\vec{e}_\mu$ related to $\gamma$?

Answer 1

You're certainly on the right track. Yes, $\vec v$ is a vector in the tangent space $T_PM$, almost never a vector joining $P\in M$ to a point $Q\in M$.

The main philosophy is to think of tangent directions $\vec v$ as equivalent to directional derivatives $D_{\vec v}$. Knowing $D_{\vec v}f(P)$ for all smooth functions $f$ (defined on a neighborhood of $P$, say) tells you what $\vec v$ must be. Given a coordinate system $x^\mu$ and a representation $\vec v = \sum v^\mu \vec e_\mu$, you can think of $\vec e_\mu$ as the tangent vectors to the $x^\mu$-coordinate curves (just like we think of $\vec e_j$ in Euclidean space as the tangent vectors to the $x_j$-axis).

EDIT: In particular, note that $D_{\vec v} f(P) = df_P(\vec v) =\left(\sum v^\mu\vec e_\mu\right)(f)$, or, if you insist, $=\left(\sum v^\mu\partial_\mu\right)(f)$.

Answer 2

Tangent vector are two-face objects. But further than displacements one should think them also directions.

In this way ones generalizes to manifolds the ideas learned in vector calculus on $\Bbb{R}^n$ about how vectors determine positions but also directions to which one can measure variation of several kind of functions: scalar, vectorial, tensorial.

Answer 3

There are as far as I know 2 ways to define tangent vectors for a manifold (and I guess a third for submanifolds of $\mathbb R^n$). They are as equivalence classes of curves with the same derivative and as derivations. For submanifolds you can just think about the hyperplane tangent to the manifold as your tangent space.

The reason that you might expect to think about a tangent vector as a derivative is that in $\mathbb R^n$ there is a natural directional derivative corresponding to any vector, which is the directional derivative in that direction, scaled by the norm of the vector.

For your first question I would say you are pretty much right - the tangent space is just some vector space so the vectors are not in $M$ technically. Whether it is just a pure geometrical object is another thing, you could view it as a derivative by defining its operation on a real-valued function $f$ by $v(f)=(f\circ\gamma)'(0)$.

Your second question I would view as mostly a question of technicalities more so than anything else. To be clear, when you write $\gamma'(0)$, that is really an equivalence class of curves with the same derivative at your base point, so yes, that is a different object to a derivative operator. The thing is that really the tangent space is a vector space, so you could take any other vector space of the same dimension and say those object are your tangent vectors.

For you last question, you have $\gamma'(0)=v^\mu\vec e_\mu$.

Answer 4

The point is that the definition of a manifold is abstract, i.e. manifolds are not by definition embedded in some $\mathbb R^n$. This allows us to be more flexible when proving statements, i.e. we do not require to provide an embedding each time we want to show that something is a manifold (although we do know they can all be embedded). The point of this directional derivative definition is essentially the same ; if you embed the manifold, your isomorphism of your manifold with its embedded version will map the abstract "directional derivative" tangent vector to the vector "based at $p$ pointing to $q$" in $T_pM$. But this last idea is somehow an embedded picture, so it is no surprise that the abstract picture is, well, more abstract.

Hope that helps,