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From Lee's book regarding smooth manifolds. He has the following proposition

Suppose $M$ is a smooth manifold with or without boundary and $p \in M$. Every $v \in T_pM$ is the velocity of some smooth curve in $M$

He then gives the following proof for the case without boundary.

First suppose that $p \in \operatorname{Int} M$ (which includes the case $\partial M = \emptyset $). Let $(U, \varphi)$ be a smooth coordinate chart centered at $p$, and write $v= \sum_{i}v^i \frac{\partial}{\partial x^i} \bigg|_{p}$ the coordinate basis. For sufficiently small $\varepsilon >0 $, let $\gamma :(-\varepsilon, \varepsilon) \to U$ be the curve whose coordinate representation is $$\gamma(t)= (tv^1, \dots,tv^n)$$ (Remember, this really means $\gamma(t)=\varphi^{-1}(tv^1, \dots tv^n)$.) This is a smooth curve with $\gamma(0)=p$, and the computation above shows that $\gamma'(0)=\sum_{i}v^i \frac{\partial}{\partial x^i} \bigg|_{\gamma(0)}=v$.

Few questions. I don't quite understand the note he gave that

Remember, this really means $\gamma(t)=\varphi^{-1}(tv^1, \dots tv^n)$.

the map $\gamma$ is defined from some interval to the manifold $M$, but why would this interval be contained in the space say $\Bbb R^n$ that's the chart's codomain? How does it make sense to say that $\gamma$ would be the inverse of $\varphi$?

Secondly how does he get the result

and the computation above shows that $\gamma'(0)=\sum_{i}v^i \frac{\partial}{\partial x^i} \bigg|_{\gamma(0)}=v$.

what computation does he mean? The way he defines $\gamma'$ is that $$\gamma'(0)=d\gamma \left(\frac{d}{dt} \bigg|_{t_0} \right) \in T_{\gamma(t_0)}M?$$

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There is some abuse of notation going on here.

[...] the map $\gamma$ is defined from some interval to the manifold $M$, but why would this interval be contained in the space say $\Bbb R^n$ that's the chart's codomain? How does it make sense to say that $\gamma$ would be the inverse of $\varphi$?

What exactly do you mean by "interval contained in the space that's the chart's codomain"? We're also not saying that $\gamma$ is the inverse of $\varphi$.

Lee is explaining that there is an abuse of notation in this definition. Recall the chart is a map $\varphi^{-1}: \mathbb{R}^n \to U$. Lee is writing $\gamma(t) = (tv^1, \dots, tv^n)$ but as you probably noticed, this doesn't make sense at first because the codomain of $\gamma$ is $M$ but he is giving an element in $\mathbb{R}^n$. Indeed, what he really means here is the definition $$\gamma(t) = \varphi^{-1}(tv^1, \dots, v^n) $$ which makes sense because $\varphi^{-1}:\mathbb{R}^n \to U \subseteq M$.

So basically, Lee is already writing out the map on the trivialized chart and giving you the definition on $\mathbb{R}^n$ which is essentially the same definition because $\varphi^{-1}$ is a diffeomorphism anyhow - it is also called the coordinate representation (p. 69). So he just omits $\varphi^{-1}$ from the notation because it can be cumbersome and distracting.

[...] what computation does he mean? The way he defines $\gamma'$ is that $$\gamma'(0)=d\gamma \left(\frac{d}{dt} \bigg|_{t_0} \right) \in T_{\gamma(t_0)}M?$$

He's referring to the computation one page earlier (p. 69): $\gamma'(t_0) = \frac{\mathrm{d} \gamma^i}{\mathrm{d} t}(t_0) \left. \frac{\partial}{\partial x^i} \right|_{\gamma(t_0)}$.

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  • $\begingroup$ What I meant is that is $\gamma$ mapping to the same space as the chart $\varphi$'s codomain i.e $\Bbb R^n$? I don't know what to call this space... and then we're picking the result from there with $\varphi^{-1}$ back to the manifold? Also I might need a refresher on the chain rule, but how is $$\frac{\mathrm{d} \gamma^i}{\mathrm{d} t}(t_0) \left. \frac{\partial}{\partial x^i} \right|_{\gamma(t_0)} = \sum_{i}v^i \frac{\partial}{\partial x^i} \bigg|_{\gamma(0)}?$$ Thanks for the answer btw! $\endgroup$ Commented Feb 10, 2022 at 19:53
  • $\begingroup$ @BernardLees There is no chain rule involved, it is $\frac{\mathrm{d}}{\mathrm{d}t}(tv^i) = v^i$. $\endgroup$
    – Qi Zhu
    Commented Feb 10, 2022 at 20:04
  • $\begingroup$ Is he using the Einstein summation convention in $\frac{\mathrm{d} \gamma^i}{\mathrm{d} t}(t_0) \left. \frac{\partial}{\partial x^i} \right|_{\gamma(t_0)}$? $\endgroup$ Commented Feb 10, 2022 at 20:07
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    $\begingroup$ @BernardLees Yes. :-) Note that 1. the indices $i$ would otherwise be undefined and 2. the derivative $\gamma'$ should depend on the partial derivative in all directions. $\endgroup$
    – Qi Zhu
    Commented Feb 11, 2022 at 6:32

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