# Proposition 3.23 from Lee's Introduction to smooth manifolds

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 $$\gamma'(0)=d\gamma \left(\frac{d}{dt} \bigg|_{t_0} \right) \in T_{\gamma(t_0)}M?$$

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): $$\gamma'(t_0) = \frac{\mathrm{d} \gamma^i}{\mathrm{d} t}(t_0) \left. \frac{\partial}{\partial x^i} \right|_{\gamma(t_0)}$$.

• 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! Feb 10, 2022 at 19:53
• @BernardLees There is no chain rule involved, it is $\frac{\mathrm{d}}{\mathrm{d}t}(tv^i) = v^i$. Feb 10, 2022 at 20:04
• Is he using the Einstein summation convention in $\frac{\mathrm{d} \gamma^i}{\mathrm{d} t}(t_0) \left. \frac{\partial}{\partial x^i} \right|_{\gamma(t_0)}$? Feb 10, 2022 at 20:07
• @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. Feb 11, 2022 at 6:32