Basis for $T_x \mathbb{R}$

Question

Consider the manifold $\mathbb{R}$ with coordinate chart $(\mathbb{R}, \psi)$ where $\psi(x)= x^3$.

I am looking at the tangent space of $\mathbb{R}$ with respect to the chart $\psi$.

Lee states that for any $x \in \mathbb{R}$, $\frac{\partial}{\partial x^1}$ will form a basis for $T_x \mathbb{R}$. In this case mentioned above I think $x^1 = x^3$.

However, I am not sure how to interpret $\frac{\partial}{\partial x^3}$ as a basis for the tangent space. In particular, what does it mean to take the derivative with respect to the coordinate function $x^i$?

Is there a geometric interpretation of this?

Answer 1

Remark: If $c:(-\epsilon, \epsilon) \to M$ and $t$ is the parameter (i.e coordinate function for $\mathbb{R}$) then we define;

$$c_*\left(\frac{d}{dt}\Bigr|_{p}\right) = c'(p): p \in (-\epsilon, \epsilon)$$

1. $\psi: \mathbb{R} \to \mathbb{R}$

2. Let $x,y$ be the coordinate function of $\mathbb{R}$ in the domain, range respectively

3. Then $\dfrac{d}{dx}\Bigr|_p, \dfrac{d}{dy}\Bigr|_{\psi(p)}$ form basis for the tangent space of $\mathbb{R}$ at their respective points

4. The differential $\psi_*: T_p \mathbb{R} \to T_{\psi(p)} \mathbb{R}$ and so;

$$\psi_*\left(\frac{d}{dx}\Bigr|_p\right) = \lambda \cdot \frac{d}{dy}\Bigr|_{\psi(p)}$$

5. Apply $y$ to both sides:

$$\psi_*\left(\frac{d}{dx}\Bigr|_p\right)(y) = \frac{d}{dx}\Bigr|_p(y \circ \psi) = \lambda$$

6. Here by definition $y \circ \psi = \psi$ and so;

$$\frac{d}{dx}\Bigr|_p(\psi) = 3p^2 = \lambda$$

7. Thus $\psi_* = 3p^2 \dfrac{d}{dy}\Bigr|_{\psi(p)}$ and by definition $\psi_* = \psi'$