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I'm revisiting the definition for tangent spaces in Lee's Introduction to Smooth Manifolds and I'm trying to convince myself why we might define tangent vectors as derivations at a point $p\in M$:

Let $M$ be a smooth manifold, and let $p\in M$. A linear map $v:C^\infty(M)\to \mathbb{R}$ is called a derivation at $p$ if \begin{align*} v(fg) = f(p)vg + g(p)vf \end{align*} for all $f,g\in C^\infty(M)$.

So far, I know that if $M=\mathbb{R}^n$, then each derivation can be given as a directional derivative in some direction in $\mathbb{R}^n$. After reading the parts on the differential and its computation in coordinates, I'm still wondering why we would be interested in defining a tangent vector as a map that acts on functions on the manifold and the benefits from acting on smooth functions. The main reason that I can think of is that the collection of derivations at a point forms a vector space, which is we what want for a tangent space.

I have also looked at the approach of defining tangent vectors with equivalence classes of curves, but it seems that there's also an action on $f\in C^\infty(M)$ going on; we call curves $\gamma:J\to M$ the tangent vectors, and they have a directional-derivative-like operators that act on $f\in C^\infty(M)$ by \begin{align*} \left.\frac{d}{dt}(f\circ \gamma)(t)\right|_{t=0}. \end{align*} This seems really similar to how a vector in $\mathbb{R}^n$ defines its own directional derivative, but again, I'm not sure why the action on $f\in C^\infty(M)$ would be useful/significant.

Chen
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  • I saw this post a while ago, and I think this is where I got the "justification" that the space of derivations is a vector space. Looking back again, it seems that the directional derivative "encodes" some information about direction, and I guess the action on a scalar function $f\in C^\infty(M)$ somewhat gives us a "shape" or characterization of the manifold (if this makes sense). – Chen Aug 17 '20 at 03:08
  • When I first enconuntered derivations, I thought that they are simply a cleaner way of describing what is going on- the value of the directional derivative is independent of the curve you choose, it only depends on a choice of direction, so why bother with equivalence classes of curves if you can go straight to derivations, which as you say, encode all the information you want about a tangent vector. As I got to a more advanced level, it became obvious that derivations are the way to go, because you can bring out big algrbraic guns using them. cont. – Matematleta Aug 17 '20 at 03:21
  • The most abstract but most useful way to define the tangent space is as the dual space of a certain quotient space--the cotangent space. Here are some details. – Matematleta Aug 17 '20 at 03:24
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    There are several ways of thinking of tangent vectors in $\mathbb{R}^n$ which we could generalize to an arbitrary manifold: directional derivatives, velocities of curves, gradients of functions, and more. Each gives their own way to construct the tangent bundle, though all are ultimately isomorphic. The reason we so often choose to define tangent vectors as directional derivatives is that this definition is often easiest to work with. – Kajelad Aug 17 '20 at 03:52
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  • A complete treatment of all three definitions including isomorphisms connecting them, can be found in the first chapters of Jeffrey Lee's Manifolds and Differential Geometry. – Kelvin Lois Aug 17 '20 at 14:00
  • See also https://math.stackexchange.com/q/3766650 – Paul Frost Aug 17 '20 at 22:13

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