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where [itex](U,\varphi )[/itex] is some coordinate chart such that [itex]\varphi :U\rightarrow\mathbb{R}^{n}[/itex], with [itex]\varphi (p)= x= \lbrace x^{\mu}\rbrace[/itex].

Am I correct in assuming that this definition relies on the fact that the directional derivative of a function is independent of the curve one chooses to parametrise it by? If so, is this the correct way to prove it?

Let [itex]f:\mathcal{M}\rightarrow\mathbb{R}[/itex] be a differential function of class [itex]C^{k}[/itex] and let [itex]\gamma_{1}: (a,b)\rightarrow U[/itex] and [itex]\gamma_{2}: (a,b)\rightarrow U[/itex] be two curves, parametrised by [itex]t[/itex] and [itex]s[/itex], respectively, both passing through the point [itex]p\in U\subset\mathcal{M}[/itex] such that [itex]\gamma_{1} (0)=p= \gamma_{2} (0)[/itex]. Furthermore, suppose that $$\left(\varphi\circ\gamma_{1}\right)'(0)= \left(\varphi\circ\gamma_{2}\right)'(0)$$ (via the coordinate chart as defined above).

We have then, that the directional derivative of the function [itex]f[/itex] through the point [itex]p\in U\subset\mathcal{M}[/itex] is given by $$\frac{df}{dt}\Biggr\vert_{t=0}= \frac{d(f\circ\gamma_{1})}{dt}\Biggr\vert_{t=0} = \frac{\partial f (p)}{\partial x^{\mu}}\left(\varphi\circ\gamma_{1}\right)'(0) = \frac{\partial f (p)}{\partial x^{\mu}}\left(\varphi\circ\gamma_{2}\right)'(0) = \frac{d(f\circ\gamma_{2})}{ds}\Biggr\vert_{s=0}$$

As such, the directional derivative of [itex]f[/itex] at [itex]p\in U\subset\mathcal{M}[/itex] is independent of the curve it's parametrised by.

Given this we can define the a tangent vector [itex]\dot{q}[/itex] at a point [itex]q\in\mathcal{M}[/itex] as the equivalence class of curves passing though the point [itex]q\in\mathcal{M}[/itex] (as defined earlier). The tangent space to [itex]\mathcal{M}[/itex] at the point [itex]q\in\mathcal{M}[/itex] is then defined in the following manner $$\lbrace\dot{q}\rbrace = \lbrace [\gamma] \;\vert \quad\gamma (0)=q\rbrace$$ [itex]\dot{q}[/itex] then acts on functions [itex]f[/itex] (as defined earlier) to produce the directional derivative of [itex]f[/itex] at the point [itex]q[/itex] in the direction of [itex]\dot{q}[/itex] as follows $$\dot{q}[f] =\frac{d(f\circ\gamma)}{dt}\Biggr\vert_{t=0}$$

Would this be correct? (I'm deliberately using the notation [itex]\dot{q}[/itex] for the tangent vectors as I'm approaching it from a physicist's point of view, with the aim of motivating the phase space for Lagrangian dynamics, and explicitly showing why [itex]q[/itex] and [itex]\dot{q}[/itex] can be treated as independent variables in the Lagrangian).

From this, can one then prove that the definition of a tangent vector as an equivalence class of curves is independent of coordinate chart.

Suppose that [itex](U,\varphi )[/itex] and [itex](V, \psi )[/itex] are two coordinate charts such that [itex]U \cap V \neq\emptyset[/itex] and let [itex]p\in U \cap V[/itex]. Let [itex]\gamma_{1}[/itex] and [itex]\gamma_{2}[/itex] be two coordinate curves (as defined previously) such that [itex]\gamma_{1} (0)=p=\gamma_{2} (0)[/itex]. It follows from the chain rule, that $$\left(\psi\circ\gamma_{1}\right)^{\prime}(0)=\left((\psi\circ\varphi^{-1})\circ (\varphi\circ\gamma_{1})\right)^{\prime}(0) \qquad\qquad\qquad\qquad \\ = \left(\psi\circ\varphi^{-1}\right)^{\prime}(\varphi (p))\left(\varphi\circ\gamma_{1}\right)^{\prime} (0) \qquad \\ = \left(\psi\circ\varphi^{-1}\right) ^{\prime}(\varphi (p))\left(\varphi\circ\gamma_{2}\right)^{\prime} (0) \qquad \\ = \left((\psi\circ\varphi^{-1})\circ (\varphi\circ\gamma_{2})\right)^{\prime}(0) \qquad\quad \\ = \left(\psi\circ\gamma_{2}\right)^{\prime}(0)\qquad\qquad\qquad\qquad\quad$$

As such, if the equivalence relation holds in one coordinate chart [itex](U,\varphi )[/itex] then it holds in any other (as [itex](V, \psi )[/itex] was chosen arbitrarily, other than it overlap with [itex](U, \varphi )[/itex] in the neighbourhood of [itex]p\in \mathcal{M}[/itex]).

Would this be correct?

Apologies in advance for the long-windedness of this post, just keen to check my understanding.