====== 2020-01-22, Wednesday ======

Definition of topological manifold, examples, coordinate chart and smoothly compatible coordinate charts. 

==== Topological Manifold ====

A topological manifold $M$ of dimension $n$ is a topological space such that  
  - $M$ is a Hausdorff space, for every pair of distinct points $x, y \in M$, there are disjoint open subsets $U, V \subset M$ such that $x \in U, y \in V$.
  - $M$ is second countable. There exists a countable basis for the topology of $M$. ((Recall a collection of open subsets in M is a topological basis, if all other open set can be written as union or finite intersections of the basis elements.))
  - $M$ is locally Euclidean of dimension $n$: each point of $M$ has a neighborhood that is homeormophic to an open subset of $\R^n$. 

[[https://en.wikipedia.org/wiki/Non-Hausdorff_manifold#Line_with_two_origins | A line with two origins]] is an example of non-Hausdorff space but is locally Euclidean. 

==== Coordinate Chart ====
Let $M$ be a topological n-manifold. A //coordinate chart// on M is a pair $(U, \varphi)$, where $U$ is ane open subset of $M$ and $\varphi: U \to \hat U$ is a homeomorphism from $U$ to an open subset $\hat U = \varphi(U) \subset \R^n$. Let $p \in M$, if $U$ is a neighborhood of $p$ and $\varphi(p) = 0$, we say $(U, \varphi)$ is //centered at // $p$. 

==== Example: $S^n$ (or just $S^1$) ====
We follow the Example 1.4 in Lee for $S^n$. 

To show that $S^n$ is an $n$-manifold, we only need to show it is locally Euclidean, since $S^n$ as a topological subspace of $\R^{n+1}$, it is automatically Hausdorff and second-countable. 

In each of the $n+1$ axis directions in $\R^{n+1}$, we can cut the sphere $S^n$ in half, and get 
$$ U_i^+ = \{ \vec x \in S^n \mid x_i > 0 \} \quad  U_i^- = \{ \vec x \in S^n \mid x_i < 0 \} $$
These gives $2n+2$ open sets on $S^n$. Clearly, they cover $S^n$. 

We consider maps
$$\gdef\B{\mathbb B} \pi_i^\pm: U_i^\pm \to \B^n, \quad (x_1, \cdots, x_{n+1}) \mapsto (x_1, \cdots, \hat x_i, \cdots, x_{n+1}) $$
where $\B^n \subset \R^n$ is the unit open ball in $\R^n$, and hat means omit that variable. We can check $\pi_i^+$ is a homeomorphism from $U_i^+$. Thus $(U_i^\pm, \pi_i^\pm)$ is a coordinate patch, for each $i=1, \cdots, n+1$. 

==== Smooth structure ====
Two charts $(U, \varphi)$ and $(V, \psi)$ are said to be //smoothly compatible// if either $U \cap V = \emptyset$, or $\psi \circ \varphi^{-1}: \varphi(U\cap V) \to \psi(U \cap V)$ is a diffeomorphism. 

**?** Let $f: A \to B$ be a homeomorphism of open subsets in $\R^n$. If $f$ is smooth, is it automatically true that $f^{-1}$ is smooth? 


$$\gdef\acal{\mathcal A}$$
An ** atlas for $M$ ** is a collection of charts whose domains cover $M$. An atlas $\acal$ is smooth, if any two charts in it are smoothly compatible with each other. 


==== Further Reading ====

See this note about difference between topological structure and differential structure. 
https://www3.nd.edu/~lnicolae/FYsem2003.pdf