Exercise 1.20 (Basis of regular coordinate balls)

Question

extbf{Proposition 1.19.} Every smooth manifold has a countable basis of regular coordinate balls.
ewlie

	extbf{Exercise 1.20.} Prove Proposition 1.19.

Elu Thingol · Accepted Answer

The difference in the proof of these two assertions (Lemma 1.10 and Proposition 1.19) is difficult to spot at first sight. In the definition of precompact coordinate balls in Lemma 1.10, we did not require the coordinate chart to smoothly map the set in question to a ball centered at $0$ in $\mathbb{R}^n$. In Proposition 1.19, we will have to adjust for this restriction.

\begin{proof}
    Let $M$ be a 	extit{smooth} $n$-manifold. First, we consider the special case in which $M$ can be covered by a single 	extit{smooth} chart. Suppose $\varphi:M 	o \widetilde{U}\,$, $\widetilde{U} \subseteq \mathbb{R}^n$, is a global coordinate map, and let $\mathcal{B}$ be the collection of all open balls $B_r(x) \subseteq \mathbb{R}^n$ such that $r$ is rational, $x$ has rational coordinates and $B_{r^{\prime}}(x) \subseteq \widetilde{U}$ for some $r^{\prime}>r$. 
    %\begin{equation*}
    %    \mathcal{B}=\left\{ B_r(x) \subseteq \mathbb{R}^n \mid r \in \mathbb{Q}, x \in \mathbb{Q}^night\}
    %\end{equation*}
    Each such ball is precompact in $\widetilde{U}$, and it is easy to check that $\mathcal{B}$ is a countable basis for the topology of $\widetilde{U}$. Because $\varphi$ is a 	extit{diffeomorphism}, it follows that the collection of sets of the form $\varphi^{-1}\left(B_{r}(x)ight)$ for $B_{r}(x) \in \mathcal{B}$ is a countable basis for the topology of $M$. 	extit{Now consider another coordinate chart $\psi_x(p) := \phi(p) - \phi(x)$. Then for each ball $B:=\varphi^{-1}\left(B_{r}(x)ight)$, the smooth coordinate map $\psi_x: B^{\prime}	o\widetilde{U}$, $B^{\prime} := \varphi^{-1}\left(B_{r^{\prime}}(x)ight)$, satisfies
    \begin{equation*}
        \psi_x(B) = B_{r}(0), \quad \psi_x(\overline{B}) = \overline{\psi_x(B)} = \overline{B_{r}(0)}, \quad \psi_x(B^{\prime}) = B_{r^{\prime}}(0).
    \end{equation*}
    Thus, such collection consists only of regular coordinate balls}, with the restrictions of $\psi_x$ as coordinate maps.
ewline

Now let $M$ be an arbitrary smooth $n$-manifold. The generalization then follows in the exact same way as in the proof of Lemma 1.10.

% Previous attempt:
    %	extit{for each ball $B_{r^{\prime}}(x)$, the function $\psi_x(p) = \varphi(p) - \varphi(x)$ is a diffeomorphism as well.} If follows that the collection of sets 	extit{of the form $\psi_x^{-1}(B)$ for $B \in \mathcal{B}$} is a countable basis for the topology of $M$. 	extit{Furthermore, for each of these balls $B:=\psi_x^{-1}(B_{r}(x))$, the restriction $\varphi:B^{\prime}	o\widetilde{U}$, $B^{\prime}:=\psi_x^{-1}(B_{r^{\prime}}(x))$ satisfies:
    %\begin{equation*}
    %    \varphi(B) = B_{r^{\prime}}(0), \quad \varphi(\overline{B}) = \overline{\varphi_x(B)} = \overline{B_{r^{\prime}}(0)}, \quad \varphi(B^{\prime}) = B_{r^{\prime}}(0).
    %\end{equation*}
    %Thus, such collection consists only of regular coordinate balls}, with the restrictions of $\psi$ as coordinate maps.    
\end{proof}

Exercise 1.20 (Basis of regular coordinate balls)

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Exercise 1.20 (Basis of regular coordinate balls)

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