I understand that a tangent vector, tangent to some point $p$ on some $n$-dimensional manifold $\mathcal{M}$ can defined in terms of an equivalence class of curves $[\gamma]$ (where the curves are defined as $\gamma:(a,b)\rightarrow U\subset\mathcal{M}$, passing through said point, such that $\gamma (0)= p$), under the equivalence relation $$\gamma_{1} \sim \gamma_{2} \iff \left(\varphi\circ\gamma_{1}\right)'(0)= \left(\varphi\circ\gamma_{2}\right)'(0)$$ where $(U,\varphi )$ is some coordinate chart such that $\varphi :U\rightarrow\mathbb{R}^{n}$, with $\varphi (p)= x= \lbrace x^{\mu}\rbrace$.

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 $f:\mathcal{M}\rightarrow\mathbb{R}$ be a differential function of class $C^{k}$ and let $\gamma_{1}:(a,b)\rightarrow U$ and $\gamma_{2}:(a,b)\rightarrow U$ be two curves, parametrised by $t$ and $s$, respectively, both passing through the point $p\in U\subset\mathcal{M}$ such that $\gamma_{1} (0)=p= \gamma_{2} (0)$. 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 $f$ through the point $p\in U\subset\mathcal{M}$ 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 $f$ at $p\in U\subset\mathcal{M}$ is independent of the curve it's parametrised by.

Given this we can define the a tangent vector $\dot{q}$ at a point $q\in\mathcal{M}$ as the equivalence class of curves passing though the point $q\in\mathcal{M}$ (as defined earlier). The tangent space to $\mathcal{M}$ at the point $q\in\mathcal{M}$ is then defined in the following manner $$\lbrace\dot{q}\rbrace = \lbrace [\gamma] \;\vert \quad\gamma (0)=q\rbrace$$ $\dot{q}$ then acts on functions $f$ (as defined earlier) to produce the directional derivative of $f$ at the point $q$ in the direction of $\dot{q}$ 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 $\dot{q}$ 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 $q$ and $\dot{q}$ 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 $(U,\varphi )$ and $(V, \psi )$ are two coordinate charts such that $U \cap V \neq\emptyset$ and let $p\in U \cap V$. Let $\gamma_{1}$ and $\gamma_{2}$ be two coordinate curves (as defined previously) such that $\gamma_{1} (0)=p=\gamma_{2} (0)$. 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 $(U,\varphi )$ then it holds in any other (as $(V, \psi )$ was chosen arbitrarily, other than it overlap with $(U, \varphi )$ in the neighbourhood of $p\in \mathcal{M}$).

Would this be correct?

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


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