main.tex

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\title{Prep Program Week 1 Notes}
\author {"Combinatorial Holonomy" REU}
\date{May 2024}

\begin{document}
\maketitle
\includegraphics[scale = .09]{pics/sweetsolids.jpg}
\tableofcontents
\clearpage
\section{Tetrahedron: $A_4$}
\includegraphics[scale = .33]{pics/Tetra.jpg}
\begin{itemize}
    \item V = 4
       \item F = 4
     \begin{itemize}
        \item 4, 3-fold axes from each vertex to opposite centroid, then \textbf{8 elements}, order 3 of the same conjugacy class (i.e. $a,b \in G$ are conjugate such that $b = gag^-1$ for all elements $g\in G$).
    \end{itemize}
    \item E = 6
    \begin{itemize}
        \item 3, 2-fold axes, then \textbf{3 elements}, order 2 of the same conjugacy class.
    \end{itemize}
\end{itemize}
Then \textbf{12 pure} rotational symmetries (which are orientation-preserving isometries) including $e$. \\\\
The rotation group acts faithfully as permutations of the 4 vertices; yet note that upon embedding 2 tetrahedra within a cube, the \emph{even} permutations of the diagonals of the cube preserve the structure of the tetrahedra while the odd permutations interchange them.  Thus we have the rotation group of the tetrahedron as isomorphic to $A_4$, with $A_4$ being a monomorphism into $S_4$.  We can also compute the order of 12 for $A_4$ with the formula $n!/2$ for any family $A_n$ of degree $n$.\\\\
Note $S_4$ as isomorphic to the tetrahedron's full isometry group (consisting of pure rotations, pure reflections, pure rotations with reflections, and $e$).
\subsection{Subgroup Structure of $A_4$}
\includegraphics[scale = .33]{pics/A4subs.png}
\begin{itemize}
    \item Trivial group \{()\} and the whole group.
    \item Cyclic Group $\mathbb{Z}_2$, containing 3 groups generated by rotations of order 2.
    \begin{itemize}
        \item \{(), (1,2), (3,4)\}
        \item \{(), (1,3), (2,4)\}
        \item \{(), (1,4), (2,3)\}
    \end{itemize}
    \item $A_3 \cong \mathbb{Z}_3$.
    \begin{itemize}
        \item \{(), (2,3,4), (2,4,3)\}
        \item \{(), (1,3,4), (1,3,4)\}
        \item \{(), (1,2,4), (1,4,2)\}
        \item \{(), (1,2,3), (1,3,2)\}
    \end{itemize}
    \item Klein Four-Group $V_4$: \{(), (1,2)(3,4), (1,3)(2,4), (1,4)(2,3)\}. 
    \end{itemize}
\subsection{Triangulated Surfaces Summary:}
   \begin{itemize}
       \item For a triangulated surface $S = (V,F)$:\\ Each $f\in F$ is of the form $f = \{v_1,v_2,v_3\}$ where for every distinct $v \in V$ contained in the set of faces $f \in F$ we can arrange these faces in a sequence $(f_1,f_2, . . .f_\delta)$ such that $f_i$ and $f_i+1$ are \textbf{adjacent} for $1 \leq i \leq \delta$.
       \item Adjacency:\\
       If $f_1 \cap f_2 = \{v_1,v_2\}$, $f_1$ and $f_2$ are adjacent.\\
       If $f_\delta$ is adjacent to $f_1$, then $v$ is an \emph{interior vertex} (or \emph{boundary vertex} otherwise).
       \item Each $e \in E$ contained in some $f \in F$ of $S$ is of the form $e = \{v_1,v_2,v_3\}$ where each face $f = \{v_1,v_2,v_3\}$ contains three edges $\{v_2,v_3\}$, $\{v_3,v_1\}$, and $\{v_1,v_2\}$.
       \begin{itemize}
           \item Boundary Edge: An $e \in E$ belongs to only one $f \in F$ for $S$.
           \item Interior Edge: An $e \in E$ belongs to two distinct $f \in F$ for $S$.
       \end{itemize}
       \item If all edges of $S$ are interior, then so are all vertices, and $S$ has \emph{no boundary}.
       \item \textbf{Connectedness}, in terms of:
       \begin{itemize}
           \item Vertices:\\
           For any two $v,v' \in V$, there exists a sequence $v = v_1,v_2, . . ., v_k = v' \in S$ such that $\{v_i, v_i+1$\} is an edge for all $1 \leq i < k$.
           \item Faces:\\
           For any two $f,f' \in F$, there exists a sequence $f = f_1,f_2, . . ., f_k = f' \in S$ such that $f_i$ is adjacent to $f_i+1$ for all $1 \leq i < k$.\\\\
           \textbf{Equivalence:} A surface is connected iff it is face-connected.            
       \end{itemize}
       \item \textbf{Automorphism Group}, Aut($S$):\\
       A bijection $\alpha: V \xrightarrow[]{} V$ of $S$ for  $\{v_1,v_2,v_3\} \in F$ is an automorphism iff  $\{\alpha(v_1),\alpha(v_2),\alpha(v_3)\} \in F$, the set of automorphisms of $S$ forming a group under composition.\\\\
      For any connected surface, an automorphism $\alpha \in $Aut$(S)$ which fixes all vertices of a single face is sufficient to completely determine the automorphism.\\\\
       A fixed face can be mapped at most 6 ways, or $|$Aut$(S)| \leq 6 |F|$, with $|$Aut$(S)| = 6 |F|$ being maximally symmetric.  
       
          \end{itemize}
\section{Cube: $S_4$}
\begin{tikzpicture}
\draw[green, thick] (0,0,0) -- (0,3,0) -- (3,3,0) -- 
(3,0,0) -- cycle;
\draw[green, thick] (0,0,3) -- (0,3,3) -- (3,3,3) -- (3,0,3) -- cycle;
\draw[green, thick] (0,0,0) -- (0,0,3);
\draw[green, thick] (0,3,0) -- (0,3,3);
\draw[green, thick] (3,0,0) -- (3,0,3);
\draw[green, thick] (3,3,0) -- (3,3,3);
\end{tikzpicture}
\begin{itemize}
    \item V = 8
    \begin{itemize}
        \item 4, 3-fold axes through opposite vertices, then \textbf{8 elements, order 3}.
    \end{itemize}
    \item F = 6
\begin{itemize}
    \item 3, 4-fold axes through opposite centroids
    \item We can break down these rotations to \textbf{3 elements, order 2} and \textbf{6 elements, order 4} (9 elements total). 
\end{itemize}
    \item E = 12
    \begin{itemize}
        \item 6, 2-fold axes through opposite edges, then \textbf{6 elements, order 2}. 
    \end{itemize}
\end{itemize}
Then \textbf{24 pure} rotational symmetries including $e$. Consider the \emph{group action} of this rotation group on pairs of opposite vertices of the cube (i.e. the four diagonals) and we have an injective homomorphism into $S_4$.  Then the rotation group of the cube is isomorphic to the permutation group $S_4$.  As the cube is dual to the octahedron, the octahedral group $O$ (alternatively known as the cube group) is also isomorphic to $S_4$. 
\subsection{Subgroup Structure of $S_4$}
\begin{itemize}
    \item Trivial group \{()\} and the whole group.
     \item $S_2$, or the trivial element and transposition, with 6 conjugate subgroups.
     \begin{itemize}
         \item $H_1$ = \{(), (1,2)\}.
     \end{itemize}
        \item Double transposition, with 3 conjugate subgroups:
        \begin{itemize}
            \item  $H_1$ = \{(),(1,2)(3,4)\}. 
        \end{itemize}
        \item $A_3$, or $A_n$ on subset size three, with 4 conjugates:
        \begin{itemize}
            \item $H_1$ = \{(), (1,2,3), (1,3,2)\}
        \end{itemize}
        \item $\mathbb{Z}_4$, or cyclic groups order 4, with 3 conjugate subgroups.
       \begin{itemize}
           \item $H_1$ = \{(), (1,2,3,4), (1,3)(2,4), (1,4,3,2)\}.
       \end{itemize}
        \item $V_4 \unlhd S_4$, including the trivial element and three double transpositions:
        \begin{itemize}
            \item \{(), (1,2)(3,4), (1,3)(2,4), (1,4)(2,3)\}.
        \end{itemize}
        \item Non-normal $V_4$, sending two element subsets within themselves, with 3 conjugates:
        \begin{itemize}
           \item $H_1$ = \{(),(1,2),(3,4),(1,2)(3,4)\}.
        \end{itemize}
       \item Maximal Subgroups, or $H \leq K \leq G$ where $H = K$ OR $K = G$:
    \begin{itemize}
        \item $A_4 \unlhd S_4$, the even permutations of $S_4$:\\\\
        \includegraphics[scale = .4]{pics/A4set.png}
        \item $S_3$, of order six fixing \{4\} and has 4 conjugates.
         \item $D_4$, with 3 conjugates.
      \end{itemize}
      \end{itemize}
\Large 30 subgroups!!!  11 conjugacy classes.\normalsize     
      
       



\section{The Fundamental Group: $\pi_1(X)$}
Intuitively: \emph{Start with a space (for example, a surface), and some point in it, and all the loops both starting and ending at this point—paths that start at this point, wander around and eventually return to the starting point. Two loops can be combined in an obvious way: travel along the first loop, then along the second. Two loops are considered equivalent if one can be deformed into the other without breaking. The set of all such loops with this method of combining and this equivalence between them is the fundamental group for that particular space.}
\subsection{Groups Review}
\begin{itemize}
    \item Homomorphism: Structure-preserving algebraic map $f: G \xrightarrow[]{} G'$ such that: 
    \begin{itemize}
        \item   $f (x \cdot y) = f(x) \cdot f(y)$
        \item $f(e) = e'$
        \item $f(x^{-1}) = f(x)^{-1}$
        \item monomorphisms (injective), epimorphisms (surjective), isomorphisms (bijective)
         \end{itemize}
   \item Kernel of $f$ is $f^{-1}(e')$, a subgroup of $G$.
   \item Normal Group $N \triangleleft G$: $gng^{-1} \in N$ for all $g \in G$ and $n \in N$
   \item Quotient Group: $G/H$
     
\end{itemize}
\subsection{The Simplest Homotopy Group}
\begin{itemize}
        \item  The fundamental group is a \textbf{homotopy invariant} of a topological space $X$.
        \item We often want to prove whether two given spaces are \emph{homeopmorphic}/or not.  Homeomorphic spaces (those which are homotopy equivalent) have isomorphic fundamental groups.  
        \item The fundamental group thus helps to distinguish  spaces, beyond that of conditions of \textbf{simple connectedness}.
        \item Any convex subspace of $\mathbb{R}^n$ has a trivial (one-element)
        fundamental group, and we can compare spaces by determining whether given spaces have a trivial fundamental group/or not (for example). 
\end{itemize}
As the \textbf{simplest homotopy group}, the fundamental group  $\pi_1(X,x_0)$ of the topological space $X$ is the group (or groupoid) of equivalence classes under homotopy of loops under the operation of path composition.  We should think of the parameter $t$ as time, the homotopy $F$ representing a continuous deforming from map $f$ to map $f'$ as $t$ travels from 0 to 1.
\begin{itemize}
    \item A \textbf{path} is a continuous map $f:[0,1] \xrightarrow[]{} X$ where $f(0) = x_0$ and $f(1) = x_1$.
    \begin{itemize}
        \item  A \textbf{loop} is a path such that $f(0) = f(1)$.
    \end{itemize}
       \item A \textbf{homotopy of paths} is a family of maps $f_t:[0,1] \xrightarrow[]{} X$ for $t \in [0,1]$.  The homotopy map $F:[0,1]^2 \xrightarrow{} X$ is also continous.
    \end{itemize}



\subsection{Homotopy of Paths}
 If $f$ and $f'$ are \emph{continuous maps} of space $X$ to space $Y$, $f$ is \textbf{homotopic} to $f'$ if there is a \emph{continuous map} $F: X \times I \xrightarrow{} Y $ where \[F(x,0) = f(x) \hspace{.66cm} \& \hspace{.66cm} F(x,1) = f'(x).\] 
Then $f \simeq f'$ is an equivalence relation.  If $f$ is a path in $X$,  we can denote its path-homotopic equivalence class as [$f$].\\\\
If two paths $f$ and $f'$ mapping the interval $I = [0,1]$  share the same initial point $x_0$ and final point $x_1$ and there is a continuous map: \[F: I \times I \xrightarrow[]{} X\] 
such that:

\begin{align*}
      F(s,0) = f(s) \hspace{.66cm}&\&\hspace{.66cm}F(s,1) = f'(s)\\
    F(0,t) = x_0 \hspace{.66cm}&\&\hspace{.66cm} F(1,t) = x_1 
      \end{align*}
for each $s\in I$ and $t\in I$.  Then we have a \textbf{path homotopy} $f \simeq_p f'$ as a stronger relation than mere homotopy.\\\\
For a convex subspace $A$ of $\mathbb{R}^n$, a straight-line homotopy moves point $f(x)$ to $g(x)$ via the straight line segment joining them.  Any two paths $f,g$ from $x_0$ to $x_1$ are path homotopic in $A$.

\section{Connectedness}
Topological space $X$ is \textbf{connected} if there does not exist a \textit{separation} of $X$, in which the union of a pair $U,V$ of non-empty disjoint open subsets is $X$ (i.e. $X$ cannot be separated into two "globs").  Connectedness helps to distinguish topological spaces!

A path is a continuous map $f:[0,1] \xrightarrow[]{} X$ where $f(0) = x_0$ and $f(1) = x_1$, a path-component being being an equivalence class of $X$ under the relation which makes $x_0$ equivalent to $x_1$. A space $X$ is path-connected if every pair of points of $X$ can be joined by a path in $X$. Then there is exactly one path-component. \\
\includegraphics[scale = .2]{pics/1920px-Path-connected_space.png}
The connectedness of intervals in $\mathbb{R}$ allows to show that:
\begin{itemize}
    \item A path-connected space $X$ is connected.
    \item Subsets  (intervals and rays) of $\mathbb{R}$ are connected iff they are path-connected.
    \item Open subsets of $\mathbb{R}^n$ are connected iff they are path-connected.
    \item Connectedness/path-connectedness are the same for finite topological spaces.
    \item Quotients of connected spaces are connected.
    \item The connected components of a space are disjoint unions of the path-connected components.
\end{itemize}
\section{Continuity}
\subsection{Generalized continuity of a function}
For topological spaces, a function $f: X \xrightarrow{} Y$ if for each open subset $V$ of $Y$, the set $f^{-1}(V)$ is an open subset of $X$.  Consider that $f$ is continuous relative to specific topologies on $X$ and $Y$ (i.e. the topologies specified for its domain and range).
\begin{itemize}
    \item If topology of the range space $Y$ is given by \textbf{basis} $\mathcal{B}$, show that the inverse image of every basis element is open to prove continuity of $f$.  As $V$ of $Y$ can be written as the union of basis elements, then $f^{-1}(V)$ is open if each set $f^{-1}(\mathcal{B}_\alpha)$ is open.  
\end{itemize}
If for a point $x \in X$ and each \textbf{neighborhood} $V$ of $f(x)$ there is a neighborhood $U$ of $X$ such that $f(U) \subset V$, then $f$ is continuous at point $x$.
\subsection{"Strong Continuity": Quotient Maps}
\includegraphics[scale = .5]{pics/quotient.png}
Let the quotient map $p: X \xrightarrow[]{} Y$ be a surjective map such that a subset $U$ of $Y$ is open in $Y$ iff $p^{-1}(U)$ is open in $X$.  Note that a subset $C$ of $X$ is \textbf{saturated} if $C$ contains every $p^{-1}(\{y\})$ that it intersects.  Then equivalently, $p$ is continuous and maps saturated open sets of $X$ to open sets of $Y$.  A quotient map is either open OR closed (or neither open nor closed).\\

If $X$ is a space and $A$ a set and if $p: X \xrightarrow[]{} A$ is surjective, then there exists exactly one toplogy $\mathcal{T}$ on $A$ relative to the quotient map $p$ known as the \textbf{quotient toplogy} induced $p$.  Then the topology $\mathcal{T}$ consists of subsets $U$ of $A$ such that $p^{-1}(U)$ is open $X$.\\

Let $X^*$ be a partition of the topological space $X$ into disjoint subsets whose union is $X$ and $p: X \xrightarrow[]{} X^*$ be the surjective map carrying each point of $X$ to each element of $X^*$ containing it.  $X^*$ is the \textbf{quotient space} in the quotient topology induced by $p$ in which elements of $X^*$ are equivalence classes in relation to $X$.  Then the open set of \emph{$X^*$ is a collection of equivalence classes whose union is an open set in $X$.}\\

Determining when a map out of a quotient space in continuous:
\begin{theorem}
    Let $p: X \xrightarrow[]{} Y$  be a quotient map.  Let $Z$ be a space and $g: X \xrightarrow[]{} Z$ be a map constant on each set $p^{-1}(\{y\})$ for $y \in Y$.
    Then $g$ induces a map $f: Y \xrightarrow[]{} Z$ such that $f \circ p = g$.
    Induced map $f$ is continuous iff $g$ is continuous, $f$ is a quotient map iff $g$ is a quotient map.  
\end{theorem}
\end{document}