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In this post, the answer proves that there is a bijection between the following definitions for the tangent space

Suppose $M$ is a smooth manifold (not necessarily a submanifold of $\mathbb{R} ^n$) and $p \in M$. Define \begin{align*} C_p (M) := \left\{ \gamma : (-\epsilon , \epsilon ) \rightarrow M \text{ differentiable with } \gamma (0) = p \right\}. \end{align*} Define an equivalence relation on $C_p (M)$ by \begin{align*} \alpha \sim \beta \iff (\phi \circ \alpha )' (0) = (\phi \circ \beta )' (0) \text{ for some chart } \phi. \end{align*} The tangent space $T_p M$ at $p$ is defined by $T_p M = C_p (M) / \sim $.

and

Let $p \in U \subseteq M$, where $U$ is open. The tangent space of $M$ at $p$ is the vector space $$ T_p M = \{ \, \text{derivations on $U$ at $p$} \, \} \equiv \text{Der}_p(C^\infty(U)). $$ The subscript $p$ tells us the point at which we are taking the tangent space.

where

A derivation on an open subset $U \subseteq M$ at $p \in U$ is a linear map $X: C^\infty(U) \to \mathbb{R}$ satisfying the Leibniz rule $$ X(fg) = f(p) X(g) + g(p) X(f). $$

I have no problems with the proof. However, I am aware that we can attach a vector space structure to each space, with the first being

For each $(h,U)$ with $p \in U$, $T_p M$ is a vector space with operations $$ v + w : = h_* ^{-1} (h_* (v) + h_* (w)) \quad \text{and} \quad \lambda v : = h_* ^{-1} (\lambda h_* (v)), $$ with $h_* : [\gamma ]_{~} \mapsto (h \circ \gamma )' (0)$.

and for the second being the structure being inherited by the dual space in which it is contained.

Question: Are these vector spaces isomorphic? I tried to prove it but to no avail and I am wondering if this is even true.

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    They're finite dimensional vector spaces of the same dimension, so they are automatically isomorphic. – Christian Sykes Feb 07 '25 at 19:02
  • https://math.stackexchange.com/questions/200926/how-are-tangent-spaces-defined-through-equivalence-classes-of-curves?rq=1 – Volk Feb 07 '25 at 20:25
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    All the proofs will give you a linear bijection. – Moishe Kohan Feb 07 '25 at 23:47
  • @ChristianSykes, true but the isomorphism here is more than that. It is a so-called natural isomorphism, which means the isomorphism can be defined in a basis-independent way. This is not true for any two vector spaces with the same dimension. – Deane Feb 09 '25 at 04:20

1 Answers1

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Yeah, those spaces are isomorphic, as already suggested in the comments. However, there is no only a bijection coming from the fact that they have the same dimension. Indeed, you can take the natural map constructed in the answer of the link mentioned by you and prove that the map is not only bijective, but a linear map once you fix the structures of vector space that you defined. Let me know if you need help to do this.