Understanding the "abuse of notation" in the differential of tangent vectors

Question

I am reading John Lee's Smooth Manifolds book, current looking at the bottom of Page 63 in which we are working out what the differential looks like in the special case that it's along the transition map between two charts.

Let $M$ be a smooth manifold and and $(U,\psi)$ and $(V,\phi)$ be two smooth coordinate charts with coordinate functions $(x^i)$ and $(\tilde x^i)$ respectively. It goes on to mention that we are engaging in a typical abuse of notation by writing the transition map as follows:

$$\phi\circ\psi^{-1}(x):=(\tilde x^1(x),...,\tilde x^n(x)).$$

It then mentions that here we are thinking of the $\tilde x^i$ in $\tilde x^i(x)$ as a coordinate function, but $x$ as representing a point.

I want to make sure I have understood what exactly about this is abuse of notation (my background is physics so I am no stranger to abuse of notation). I want to say that we are doing two things that are "questionable notation" here, the first is that (as the author mentions) before, the letter $x$ would have represented the coordinate functions on $U$ but now represent a point. And also that $\tilde x^j$ is no longer a function $$\tilde x^j:V\rightarrow \Bbb R^N$$ but instead the different function $$\tilde x^j:\psi(U)\rightarrow \phi(V).$$ Which leads on to me now looking at the final equation we end up with (which is much more recognisable to a physicist),

$$\frac{\partial \tilde x^{j}}{\partial x^{i}}(\psi(p)) \frac{\partial}{\partial \tilde x^{j}}|_{\psi(p)},$$

in which $\tilde x^k$ is now simultaneously two objects in the same expression. in the first "half" of the expression, $\frac{\partial \tilde x^j}{\partial x^i}(\psi(p))$, it is the transition map, and in the second "half", $\frac{\partial}{\partial \tilde x^j}\biggr|_{\psi(p)}$, it is the (strictly completely different) coordinate function on $V$.

My question is, is my reasoning above correct?

Thanks in advance :)

Answer 1

I am reading the book at the moment, so everything below may be completely wrong. But my interpretation is:

• $\varphi$ is a map from $U$ into $\mathbb{R}^n$, where each of the 'coordinate functions' denoted by $x^j$ in the expression $\varphi = (x^j)$ are the elements in the vector representation of $\varphi(p)$ for $p\in U$. If we denote the $j$-th projection by $\pi_j: \mathbb{R}^n\to \mathbb{R}$, which grabs the $j$-th coordinate,

  $$ \pi_j\circ \varphi = x^j: U\to\mathbb{R} $$

• The same thing holds for the second coordinate chart $(V,\psi)$. On an individual level, $\widetilde{x}^j$ is a map $V\to \mathbb{R}$

• The tangent vectors at $p$, with respect to the two coordinate charts are defined similarly.

• In the equation $\psi\circ\varphi^{-1}(x)=(\widetilde{x}^1,\ldots,\widetilde{x}^n)$, the $x$ on the left of the equality sign is the argument of
  $\varphi^{-1}$

$$\psi\circ(\varphi^{-1}\underbrace{(x)}_{\llap{\substack{x \in\varphi(U\cap V)}}}) =(\underbrace{\widetilde{x}^1}_{\rlap{\substack{\text{coordinate } \text{function on }U\cap V }}}\overbrace{(x)}^{\rlap{x\in U\cap V}},\ldots\ldots,\widetilde{x}^n(x)) $$

• Within the expansion on the right hand side, we think of each $\widetilde{x}^j(x)$ mapping points from the manifold $U\cap V$ directly into $\mathbb{R}^n$ without the intermediate $\varphi^{-1}$, here is where we also identify $x=\varphi^{-1}(x)$. The reasoning behind doing this is that $\psi$ is ultimately responsible for the coordinate representation of $x\in U\cap V$.

$$ d(\psi\circ\varphi^{-1})_{\varphi(p)}\biggl(\dfrac{\partial}{\partial x^i}\Bigr|_{\varphi(p)}\biggr)=\dfrac{\overbrace{\partial \widetilde{x}^j}^{\substack{\text{numerator as}\\\text{ the coordinate}\\\text{projection}\\ \text{of }\psi\circ\varphi^{-1}}}}{\underbrace{\partial x^i}_{{\substack{\text{denominator}\\\text{is the proper}\\\text{argument to}\\\psi\circ\varphi^{-1}}}}}(\varphi(p))\dfrac{\partial}{\partial \widetilde{x}^j}\Bigr|_{\psi(p)} $$