# Schuller GR Lecture 1 (topology)

###### tags: `physics` `note`

### Spacetime
Spacetime is a four-dimensional **topological manifold** with a **smooth atlas** carrying a **torsion-free connection** <u>compatible</u> with a **Lorentzian metric** and a **time orientation** satisfying the Einstein equations.

### Topological space
**motivation**: We want to establish some structure over a set to allow us talking about the continuity of maps, topology is such kind of structure.

**Definition**: Let $M$ be a set. A topology $\mathcal{O}$ is a subset of $\mathcal{P}(M)$ (the power set of $M$, a collection of all subset of $M$) satisfying
(1) $\phi \in \mathcal{O}$ and $M \in \mathcal{O}$
(2) if $u,v\in \mathcal{O}$, then $u\cap v\in \mathcal{O}$
(3) for any arbitrary index set $A$, $[u_{\alpha}\in \mathcal{O},\,\, \alpha\in A]\Rightarrow \bigcup\limits_{\alpha\in A}u_{\alpha}\in \mathcal{O}$

**Examples**:
1. $M=\{1,2,3\}$, $\mathcal{O}_1=\big\{\phi,\{1,2,3\}\big\}$ is a topology on $M$, while $\mathcal{O}_2=\big\{\phi,\{1\},\{2\},\{1,2,3\}\big\}$ is not.
2. Let $M$ be any set. The chaotic topology is $\mathcal{O}_{chaotic}:=\big\{\phi,M\big\}$, which is the topology with the fewest elements. And the discrete topology is $\mathcal{O}_{discrete}:=\mathcal{P}(M)$, which is the topology with the most elements.
3. $M=\mathbf{R}^d=\big\{p=(p_1,...,p_d)\,\big\vert\, p_i\in\mathbf{R}\big\}$
Soft ball: $B_r(p):=\{(q_1,...,q_d)\,\big\vert\, \sum(q_i-p_i)^2<r^2\}$ with $r\in\mathbf{R}^{+}$
define the standard topology: $u\in\mathcal{O}_{standard}\iff\forall \,p\in u,\exists r\in\mathbf{R}^{+}\rightarrow B_r(p)\subseteq u$

**Terminology**:
- $M$: set
- $\mathcal{O}$: topology
- $(M,\mathcal{O})$: topological space
- $u$ is an open set $\iff u\in\mathcal{O}$
- $A$ is a closed set $\iff$ $M\setminus A\in\mathcal{O}$
- $\phi$ is a closed and open set.

### Continuous map
Whether a map $f:M\rightarrow N$ between two sets $M$ and $N$ is contonuous depends on whuch topologies are chosen on $M$ and $N$.

**Definition**: Let $(M,\mathcal{O}_{M})$ and $(N,\mathcal{O}_{N})$ are topological spaces. Then a map
\begin{align} f:M&\rightarrow N \\m&\mapsto f(m)\end{align}
is called continuous (with respect to $\mathcal{O}_M$ and $\mathcal{O}_N$) if
\begin{equation}
\forall \,V\in\mathcal{O}_N,\,\,\text{preim}_f(V):=\big\{m\in M \,\big|\,f(m)\in V\big\}\in\mathcal{O_M}
\end{equation}
That is, a map is continuous if and only if the preimage of all open sets (in the range $N$) are open sets (in the domain $M$).
**Example**: Let $M=\{1,2\}$, with the topology $\mathcal{O}_M=\big\{\phi,\{1\},\{2\},\{1,2\} \big\}$, and $N=\{1,2\}$, with the topology $\mathcal{O}_N=\big\{\phi,\{1,2\}\big\}$. And a map $f:M\rightarrow N$ with $f(1)=2,\,\,f(2)=1$. Check that $f$ is continuous.
\begin{align}
&\text{preim}_f(\phi)=\phi\in\mathcal{O}_M\\
&\text{preim}_f(\{1,2\})=\{1,2\}\in\mathcal{O}_M
\end{align}
Thus, $f$ is continuous.
Now check the inverse of $f$. $f^{-1}:N\rightarrow M$
\begin{align}
&\text{preim}_f(\{1\})=\{2\}\notin\mathcal{O}_N
\end{align}
Thus, $f^{-1}$ is not continuous.

### Composition of continuous maps
If we have $M\xrightarrow{f} N\xrightarrow{g}P$, the composition $g\circ f$ is
\begin{align}
g\circ f:M&\rightarrow P\\
m&\mapsto (g\circ f)(m):=g\big(f(m)\big)
\end{align}

**Theorem**: If $f$, $g$ continuous, then $g\circ f$ continuous.
(proof): For any $V\in\mathcal{O}_P$
\begin{align}
\text{preim}_{g\circ f}(V)&=\big\{m\in M\,\big|\,(g\circ f)(m)\in V\big\}\\
&=\big\{m\in M\,\big|\,f(m)\in \text{preim}_g(V)\big\}\\
&=\text{preim}_f\big(\text{preim}_g(V)\big)\\
(g \text{ is continuous}\rightarrow)\because\,\text{preim}&_g(V)\in \mathcal{O}_N\\
(f \text{ is continuous}\rightarrow)\therefore\,\text{preim}&_f\big(\text{preim}_g(V)\big)\in \mathcal{O}_M\\
\Rightarrow \text{preim}&_{g\circ f}(V)\in \mathcal{O}_M
\end{align}

### Inheriting a topology
There are many useful ways to inherit a topology from some given topologiccal space(s). For our purpose, consider
\begin{equation}
S\subseteq M
\end{equation}
We try to construct a topology on $S$ from the topology $\mathcal{O}_M$ on $M$.
**Definition**(subset topology): $\mathcal{O}\big\rvert_S:=\big\{u\cap S\,\big|\, u\in\mathcal{O}_M\big\}\subseteq \mathcal{P}(S)$
(1) $\phi=\phi\cap S\Rightarrow \phi\in\mathcal{O}\big|_S$
　 $S=M\cap S \Rightarrow S\in \mathcal{O}\big|_S$
(2)If $A\in \mathcal{O}\big|_S$, $B\in \mathcal{O}\big|_S$ then $\exists \,\tilde{A}\in \mathcal{O}_M$ and $\tilde{B}\in \mathcal{O}_M$ such that $A=\tilde{A}\cap S$ and $B=\tilde{B}\cap S$.
Since $A\cap B=(\tilde{A}\cap S)\cap (\tilde{B}\cap S)=(\tilde{A}\cap \tilde{B})\cap S$, then we conclude that $A\cap B\in \mathcal{O}\big|_S$

Use this specific ways to inherit a topology from a superset, the continuity would be inherited, that is
\begin{equation}
S\subseteq M\xrightarrow{f}N
\end{equation}where f is continuous (with respect to $\mathcal{O}_M$ and $\mathcal{O}_N$), then the restriction of the map $f\big|_S:S\rightarrow N$ is also continuous.

**video link**:{%youtube 7G4SqIboeig %}