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I'm studying the different definitions of the tangent space for abstract manifolds, and I'm struggling to prove that these abstract concepts reduce to the classical ones when dealing with submanifolds of $\mathbb{R}^n$. What I know:

  • The tangent space $T_pM$ can be defined as the space of all equivalence classes of curves $\gamma:(-\epsilon, \epsilon)\to M$ with $\gamma(0)=p$ under the equivalence relation $$ \gamma \sim \beta \quad\iff\quad (\mathbf{x}^{-1}\circ\gamma)'(0) = (\mathbf{x}^{-1}\circ\beta)'(0), $$ where $\mathbf{x}$ is a local chart of $M$ at $p$. This is easily shown to be independent of the chosen chart.

  • The tangent space $T_pM$ can also be defined as the space or all derivations $D$ which are linear functionals obeying the Leibniz rule, which act on the space of smooth functions defined locally around $p$.

  • I know how to prove that the latter two are equivalent (in the sense that the spaces are isomorphic), and related by $$ [\gamma]\mapsto \left(f \mapsto (f\circ\gamma)'(0)\right), $$ where the latter map can be thought of as the directional derivative with respect to $[\gamma]$

  • I also know that they are both isomorphic to $\mathbb{R}^n$.

  • For the the differential of a smooth map $\varphi:M\to N$, I use the definition using derivations derivations, i.e. $$\mathrm d\varphi_p(X)(f) = X(f\circ \varphi).$$

  • I know how to prove the chain rule and I also know that the differential $\mathrm d\varphi_p$ is an isomorphism whenever the map $\varphi$ is a diffeomorphism.

What I don't know how to prove: If $M$ is a submanifold of $\mathbb{R}^n$, the tangent space $T_pM$ is isomorphic to the space of all tangents $\dot\gamma(0)\in \mathbb{R}^n$ of curves $\gamma$ passing through $p$?

I can intuitively see why this should be the case given the information that I have, but I don't know how to write a proper proof of it.

One of the ideas is to show that two curves belong to the same equivalence class if and only if their tangent vectors at $p$ coincide, in which case I could do something as shown in this post. However, with my definition of the differential I do not directly see that $$ (\mathbf{x}^{-1}\circ\gamma)'(0) = \mathrm d\mathbf{x}^{-1}_p(\dot\gamma(0)). $$ That being said, I would guess that one can use the aforementioned isomorphisms to identify the differential with a map that operates in this way.

Another idea would be to prove that the directional derivative with respect to $[\gamma]$ actually coincides with the tangent vector $\dot\gamma(0)$, but I don't see how one can do this.

Could you help me with this?

EDIT: I have added a bounty. Let me know if my question is unclear or missing any of the details.

EDIT 2: After thinking about this a bit more, I don't think it is possible to show what I'm trying to show without first defining the tangent space and the differential for submanifolds of $\mathbb{R}^n$ in the natural way (i.e. using tangents to curves) and showing that the differential of a chart is an isomorphism, and then using that fact to show that all of the new notions reduce to the classical ones. I would love to be wrong regarding this though.

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    $\begingroup$ I agree. You should show that two curves are equivalent if they have the same tangent vector. I mean, that is essentially the definition of the equivalence relation. The relation you are worried about, $(\mathbf{x}^{-1} \circ \gamma)'(0) = \mathrm{d} \mathbf{x}^{-1}_p (\gamma'(0))$ is just the "chain rule". $\endgroup$ – Nick Feb 13 at 21:39

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