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I have already constructed the tangent space to a manifold, denoted $T_pM$, and I have a good basis for it $\{\hat e_{(\mu)}\}$. (I followed the method of equivalence classes of curves tangent at $p$.) I have also shown that the space of directional derivative operators at $p$ forms a vector space, which I denote $D_p$. To show that the $\{\hat e_{(\mu)}\}$ basis for the tangent space may be properly identified with the $\{\partial_\mu\}$ partial derivative basis for the space of directional derivatives, I need to show that $T_pM$ and $D_p$ are isomorphic vector spaces. An isomorphism $\phi$ between vector spaces is a bijection (an injection and a surjection):

$$ \phi:T_pM\to D_p\qquad\text{such that}\qquad V^\mu\hat e_{(\mu)}\mapsto W^\mu\partial_\mu ~~. $$

However, it seems like I need to act the directional derivatives on a function before I can demonstrate this. For instance, proving injectivity requires showing that $\phi(A)=\phi(B)$ implies $A=B$, but it appears that I will only be able to do this if I act the directional derivative on a function $f$ to obtain a value. If I proceed from there, then I will only have proven the isomorphism of $T_pM$ and $D_pf$. Usually one says, "Since $f$ was arbitrary, the result for $D_pf$ holds for the operator space $D_p$ itself," but that seems like a non-rigorous, heuristic argument to me, or hand-waving, essentially. Therefore, I request input on how one might rigorously demonstrate the isomorphism of the tangent space with the given vector space of operators. If there is language differentiating the operator space $D_p$ from the space of operators acting on functions $D_pf$, that might also be useful.

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In the following answer I will assume you are familiar with germs of functions and derivations at a point. I will provide an argument starting with the definition of tangent space of a manifolds in terms of derivations and germs. I'll finally prove a theorem about the dimension of a tangent space defined in such a way and the proof relies on showing what you're asking. Nonetheless, if you are not interested in the argument being made in term of germs, as you used a different approach, just read the proof of the theorem, that construct explicitly the isomorphism \eqref{B} you're looking for.

Let $M$ be a differentiable manifold. Given $p\in M$, consider the real vector space of all derivations at a point p $$\mathrm{Der}(p)=\{v:C^{\infty}_p\rightarrow\mathbb{R}\text{ derivation}\}=:\mathrm{T}_pM\tag{A}\label{A}$$ where $C^{\infty}_p$ is the set of germs at $P$.

We'll define \eqref{A} to be the tangent space at $P$.

So this is how one defines the tangent space for a general smooth manifold. The idea now is the following, we want to use the geometric idea of tangent space we have in $\mathbb{R}^n$, which coincides with $\mathbb{R}^n$ itself of course, and use the properties of smooth manifolds to show how these are related. First we'll need a lemma, though.

Lemma

Let $x_0\in\mathbb{R}^n, f\in C^{\infty}_{x_0}$ a germ at $p$, then the following holds $$\exists g_i\in C^{\infty}_{x_0}: f(x)=f(x_0)+g_i(x^i-x^i_0)\qquad g_i(x_0)=\frac{\partial f}{\partial x^i}\bigg\lvert_{x_0}$$ in a sufficiently small neighborhood of $x_0$ (summation over repeated indices understood). This is basically a version of Taylor's formula for germs, let me know if you want me to prove it.

Note that the assumption that $f\in C^{\infty}_{x_0}$ is crucial as it follows that $g_i\in C^{\infty}_{x_0}$. Were $f\in C^{k}_{x_0}$ for some $k>0$, we'd end up with $g_i\in C^{k-1}_{x_0}$, which is trouble for the derivation that follows. The intuitive reason is that $C^k$ derivations "don't know" how to act on $C^{k-1}$ derivations in general, so our results hold for the smooth case. Nonetheless, there are other ways to define tangent spaces for $C^k$ manifolds. Moving on, let's prove the theorem.

Theorem

Let $M$ be a differentiable manifold, $p\in M$, then $\dim\mathrm{T}_p M=\dim M$

The idea of the proof is to show that for $M=\mathbb{R}^n$ and $x_0\in\mathbb{R}^n$, $\mathrm{T}_{x_0}\mathbb{R}^n$ coincides with the geometric notion of tangent space (which is $\mathbb{R}^n$ itself). Then we'll use the fact that arbitrary manifolds are locally diffeomorphic to $\mathbb{R}^n$.

Proof

Consider $M=\mathbb{R}^n$ and $U\subset\mathbb{R}^n$ open. Let $x_0\in\mathbb U$. Consider the map $$\iota:\mathbb{R}^n\rightarrow\mathrm{T}_{x_0}\mathbb{R}^n\qquad \iota(v)=v^i\frac{\partial}{\partial x^i}\bigg\lvert_{x_0}\quad\forall v=(v^1, v^2... v^n)\in\mathbb{R}^n\tag{B}\label{B}$$ where the domain $\mathbb{R}^n$ is understood as the geometric notion of tangent space to $\mathbb{R}^n$ at $x_0$, which is $\mathbb{R}^n$ itself. We'll proceed by proving that \eqref{B} is a vector space isomorphism (it's obviously a linear map):

  1. Injectivity. Consider the coordinate functions $$x^i: U\rightarrow\mathbb{R} \qquad x=(x^1,... x^n)\mapsto x_i\quad i=1,...n.\tag{C}\label{C}$$ It suffices to prove that for a non-zero vector $v=(v^1, v^2... v^n)\in\mathbb{R}^n$, the image is non-zero (a linear map $f:V\rightarrow W$ is injective iff $f^{-1}(0_W)=0_V$). In that case, consider $v=(v^1, v^2... v^n)\neq0$, then $$\iota(v)=v^i\frac{\partial}{\partial x^i}\bigg\lvert_{x_0}.$$ Now, a derivation is the zero derivation iff it is zero on every germ. But if we apply it to the coordinate functions \eqref{C}, very clearly:

$$\iota(v)(x^j)=v^i\frac{\partial}{\partial x^i}\bigg\lvert_{x_0}(x^j)=v^i\delta^j_i=v^j \quad j=1,...n.$$ Given that $v\neq0$, $\exists j\in{1,...n}$ such that $v^j\neq 0$ and thus the derivation $\iota(v)$ is non-zero and we have proved injectivity.

  1. Surjectivity. Now consider a derivation $D\in \mathrm{T}_{x_0}\mathbb{R}^n$, we will show that $\exists u\in\mathbb{R}^n$ such that $\iota(u)=D$. Consider again the coordinate functions \eqref{C} and define $$a^i:=D(x^i).\tag{D}\label{D}$$ Clearly $a=(a^1,...a^n)\in\mathbb{R}^n$ Using the lemma above, locally for an arbitrary germ $f\in C^{\infty}_{x_0}$, $f(x)=f(x_0)+g_i(x)(x^i-x^i_0)$ using linearity and Leibniz rule \begin{align} D(f)&=D(f(x_0)+g_i(x)(x^i-x^i_0))=\underbrace{D((f(x_0))}_{=0}+D(g_i(x)(x^i-x^i_0))=\\ &=D(g_i(x))\underbrace{(x^i_0-x^i_0)}_{=0}+g_i(x_0)D(x^i-x^i_0)=g_i(x_0)D(x^i)-g_i(x_0)\underbrace{D(x^i_0)}_{=0}=\\ &=\frac{\partial f}{\partial x^i}\bigg\lvert_P D(x_i)=a^i\frac{\partial f}{\partial x^i}\bigg\lvert_P=a^i\frac{\partial}{\partial x^i}\bigg\lvert_P(f)=\iota(a)(f) \end{align} where I have used also the fact that derivations are zero acting on constant germs and are evaluated at $x_0$ and the second part of the lemma for $g_i$. So, $D(f)=\iota(a)(f)$ for an arbitrary $f\in C^\infty_{x_0}$, which means that $D=\iota(a)$ and thus $w=a$ is the vector we were seeking.

So we've proved that $\mathbb{R}^n\simeq\mathrm{T}_{x_0}\mathbb{R}^n$. Note that this isomorphism is natural, since $\mathbb{R}^n$ has a canonical basis.

To conclude the proof, we consider a generic manifold $M$. Let $(U, \varphi)\quad \varphi:U\overset{\sim}{\rightarrow}\varphi(U)$ be a local chart and $p\in U$. In particular $\varphi$ is a diffeomorphism and thus its differential at $p$ $$d\varphi_p: \mathrm{T}_p M\rightarrow \mathrm{T}_{\varphi(p)}\varphi(U)=\mathrm{T}_{\varphi(p)}\mathbb{R}^n.$$ is a vector space isomorphism, so $\mathrm{T}_p M\overset{d\varphi_p}{\simeq}\mathrm{T}_{\varphi(p)}\mathbb{R}^n\simeq\mathbb{R}^n$, which proves the theorem. To conclude, note the the isomorphism is not natural as it depends on the choice of the chart. More practically, each chart induces a different basis of the tangent space which is identified with the canonical basis of $\mathbb{R}^n$ via its differential.

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  • $\begingroup$ Thank you. You appear mostly to have followed the method in Tu. I will closely study your reply, and follow up with any questions. This post has been flagged for removal twice already though, so I might send you a direct message if the post disappears. Thanks again. I think I was missing the fact that two vector spaces are isomorphic if they have the same dimension, which is glaringly obvious now. $\endgroup$ Feb 12 at 22:58
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    $\begingroup$ I don't doubt Tu's book has a similar proof, this is a pretty standard theorem in differential geometry; it may be even more similar to Lee's. Anyhow, I followed the derivation I learned in a course, which used a book not translated in English. I don't think there is a way to send direct messages here. You may ask to migrate it, I guess. $\endgroup$ Feb 12 at 23:10

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