Coordinate representation of component functions of a diffeomorphism

Question

Let $M$ be a smooth manifold of dimension $n$ and let $p \in M$. Choose a smooth chart $(U, \phi)$ around $p$ and then we have $\phi : U \to \widehat{U} = \phi[U]$ to be a diffeomorphism with $\phi(a) = (x^1(a), \dots, x^n(a))$ with $x^i : U \to \mathbb{R}$ being the component functions.

Now we have the coordinate repsentation of $x^j$ to be $\widehat{x^j} = x^j \circ \phi^{-1} : \widehat{U} \to \mathbb{R}$, but does $\widehat{x^j}(a^1, \dots, a^n) = a^j$ for all $(a^1, \dots, a^n) \in \widehat{U}$?

I was told that this is the case, but I'm having some trouble seeing how to show this since $\widehat{x^j}(a^1, \dots, a^n) = x^j(\phi^{-1}(a^1, \dots, a^n))$ and since $x^j$ could be any (smooth) function I don't see how I could show the above.

Answer 1

I believe that definitions might differ in different textbooks, but here's my understanding of the situation. I will try to be consistent with your notations as much as I can.

You have an open set $U\subseteq \mathbb{R}^n$ and a homeomorphism $\phi: U \to \widehat{U}=\phi[U]$ where $p\in\widehat{U} \subseteq M$ is an open set in $M$. Now, we have some standard functions on $\mathbb{R}^n$ that I'm going to call them "coordinate functions" that are defined as follows: $$x^i:U\to \mathbb{R}$$ $$x^i(u^1,\cdots,u^n)=u^i$$

I can define a similar set of coordinate functions on my manifold using $\phi$ and I'm going to call them $\widehat{x^i}$ by defining $\widehat{x^i}=x^i \circ \phi^{-1}$. That's all that there's to it. Nothing more.

Now if $p \in M$, then $M$ has some coordinates like $(a^1,\cdots,a^n)$ which is available after introducing $\phi$. More precisely, $\phi^{-1}(p)=(a^1,\cdots,a^n)$

In our notation, $\widehat{x^j}(p)=x^j\circ\phi^{-1}(p)=x^j(a^1,\cdots,a^n)=a^j$ which is what you wanted to show, I believe.

Your confusion probably stems from the fact that you're assuming that the points of the manifold $M$ are $n$-tuples, ignoring the fact that you need to first introduce a parametrization (i.e. a homeomorphism $\phi$) to refer to the points of $M$ like that. I hope everything is clear now.

Answer 2

So if I have a function from some set $X$ into $\mathbb{R}^k$, $f : X \to \mathbb{R}^k$ then the component functions $f^i : X \to \mathbb{R}$ satisfy $f(p) = (f^1(p), \dots, f^n(p))$ for all $p \in X$. I didn't realize at the time of posting the question but what this means is that the component functions $f^i$ are characterized by the fact that $(\pi_i \circ f)(p) = f^i(p)$ for all $p \in X$ and for all $ 1\leq i \leq k$. Note that $\pi_i$ here means the $i^{\text{th}}$ projection function from $\mathbb{R}^k$ to $\mathbb{R}$.

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Now translating the above to the question at hand, I have $x^i : U \to \mathbb{R}$ to be the component functions of the homeomorphism $\phi : U \to \phi[U] \subseteq \mathbb{R}^n$. Hence $(\pi_j \circ \phi)(p) = x^j(p)$ for all $p \in U$.

Now choose $(a^1, \dots, a^n) \in \widehat{U} = \phi[U]$. I will show that $\widehat{x^j}(a^1, \dots, a^n) = a^j$. Observe then that

\begin{align*} \widehat{x^j}(a^1, \dots, a^n) &= x^j(\phi^{-1}(a^1, \dots, a^n)) \\ &= (\pi_j \circ \phi)(\phi^{-1}(a^1, \dots, a^n)) \\ &= \pi_j(\phi(\phi^{-1}(a^1, \dots, a^n))) \\ &= \pi_j(a^1, \dots, a^n) \\ &= a^j \end{align*}

as desired.