Tangent Space is naturally isomorphic to $\{p\} \times \mathbb{R}^n$?

Question

I am studying differential geometry of curves and surfaces and trying to make sense of the definition of the tangent space. I have seen this before, studying manifolds:

The tangent space to a manifold $M$ is a space of derivations $D: C^{\infty}_p(M) \rightarrow \mathbb{R}$ that satifies $D(fg) = D(f)g(p) + f(p)D(g)$ and is denoted by $T_pM$. But some books of differential geometry of curves and surfaces defines it to be $\{p\} \times \mathbb{R}^2$.

This makes sense because they're cleary isomorphic, considering a two dimensional surface immersed in $\mathbb{R}^3$. But is it a natural isomorphism? If so, does it require to be an immersed surface or this result holds for manifolds in general? If it is not natural, why the books of differential geometry of surfaces defines this way?

Thanks in advance!

Answer 1

In finite dimensions, there is a canonical isomorphism: Every tangent vector $v\in T_pM$ can be seen as a directional derivative $D_v$, which is a derivation in your above sense.

To make this more precise: A tangent vector $v\in T_pM$ to a manifold $M$ at a point $p\in M$ is defined as an equivalence class $[\gamma_v]$ of curves: Two curves $\gamma_1,\gamma_2\in C^\infty((-\varepsilon,\varepsilon),M)$ with $\gamma_1(0)=\gamma_2(0)=p$ are tangentially equivalent, if $$\left.\frac{d}{dt}\right|_{t=0} \Phi(\gamma_1(t)) = \left.\frac{d}{dt}\right|_{t=0}\Phi(\gamma_2(t))$$ with respect to some chart $(\Phi,U)$ with $p\in U$ and $\Phi : U\rightarrow\mathbb{R}^n$.

Then, for any smooth function $f\in C^\infty(M,\mathbb{R})$, its directional derivative $D_v f$ into the direction $v\in T_pM$ is defined by $$D_v f := \left.\frac{d}{dt}\right|_{t=0}f(\gamma_v(t))$$ for some curve $\gamma_v:(-\varepsilon,\varepsilon)\rightarrow M$ that represents $v$ in the above sense.