main.tex

\documentclass{article}
\usepackage{graphicx, amsmath, amssymb, xcolor} % Required for inserting images

\title{Choosing Symmetry}
\author{Stephanie}
\date{January 2025}
\input{defs}
\begin{document}

\maketitle

\subsubsection*{Introducing Symmetry}
We have indeed chosen symmetry as this iteration's \emph{The Elementary Yard} theme.  Symmetry may be a concept one typically associates with beauty, balance, art, and design; yet there is "another side" to symmetry - the  mathematical side.  Much of the time "symmetry" is what makes math so beautiful!   


Graphs of cubic functions from calculus (the study of functions) have a natural symmetry to them.  And if you are not familiar with calculus, please still read on!, as we promise to gently introduce the concept of functions within this column.  

\begin{figure}
    \centering
    \includegraphics[width=0.5\linewidth]{Pics/Function_x^3.svg.png}
    \caption{The odd function $f(x) = x^3$  appears symmetric about the origin $(0,0).$}
    \label{fig:enter-label}
\end{figure}

\begin{itemize}
    \item \textbf{1)}  Draw some forms (artistic and mathematical) which you believe might constitute as symmetric.  
\end{itemize}
The mathematical study of symmetry is often attributed to the subject of \emph{abstract algebra}.   But before abstract algebra, and often after calculus, one typically studies \emph{linear algebra}.  Linear algebra studies vector spaces and linear transformations via matrices like the one below: 
\begin{center}
$\begin{bmatrix}
2 & \color{teal}3 & \color{magenta}6\\
\color{teal}3 & 4 & \color{orange}5\\
\color{magenta}6 & \color{orange}5 & 9
\end{bmatrix}$


\end{center} 
which we might say is a \emph{symmetric matrix}.
\begin{itemize}
   \item \textbf{2)} Refer to the glossary of Mathy Terms - can you write out another symmetric matrix or two? Congratulations if so - you'll be on your way to abstract algebra soon enough!
\end{itemize}
\subsubsection*{Sets and the Symmetric Group $S_n$}

One relevant object of study within abstract algebra is the symmetric group $S_n$, which define as the collection of all \emph{permutations of a set} .  

Let us have a set of 8 balls $B$ in green $g$, cyan $c$, orange $o$, pink $p$, indigo $i$, red $r$, yellow $y$, and white $w$.   We write $g, c, o, p, i, r, y, w \in B$ to say that these balls are in the set $B$.  
\begin{figure}
    \centering
    \includegraphics[width=0.5\linewidth]{Pics/Coloured_Balls.png}
    \caption{we have a set of balls $B = \{g, c, o, p, i, r, y, w\}$ where each ball is an \emph{element} in the set.}
    \label{fig:enter-label}
\end{figure}
A subset $A \subseteq B$ can be thought of as a combination of elements 
in a set.  We can even have the empty set $\varnothing$ consisting of 0 elements, 1-element subsets, 2-element subsets . . . even the whole set $B \subseteq B$ can be considered a subset of itself.  
\begin{itemize}
    \item \textbf{3)}  Can you list all possible subsets of our set of balls $B$?  
\end{itemize}
\begin{figure}
    \centering
    \includegraphics[width=0.5\linewidth]{Pics/bij.jpg}
    \caption{A bijection $f$ of points.}
    \label{fig:enter-label}
\end{figure}
Now suppose we have a set of points $P = \{a, b, c, d\}$.  We could apply a function $f$ to map $\{a, b, c, d\}$ to another 4 points in the set $\mathcal{P}$.  Specifically, we would apply $f(a), f(b), f(c), f(d)$  to give say points $l, m, n, k \in \mathcal{P}$.  Or, see that
\[f(a)= l,\]
\[f(b) = m.\]
\[f(c) = n,\]
\[f(d) = k,\]
for the points  $l, m, n, k \in \mathcal{P}$.  Specifically, we have a \emph{bijection} of sets via a bijective function $f$ where our points  $a, b, c, d \in P$ map \emph{onto} and are in \emph{one-to-one} with the points $l, m, n, k \in \mathcal{P}$.  
\begin{itemize}
    \item \textbf{4)} Allow $a \in P$ to map to any element $l, m, n, k \in \mathcal{P}$ via a bijective function $f$, and then continue to map  $b, c, d \in P$ to the remaining $l, m, n, k \in \mathcal{P}$.  How many total bijections can we define?  Feel free to draw it out!  Hint:  Your number of arrows should be your number of bijections.  
\end{itemize}

So here we have mapped the elements of one set to another via bijections.  But what if we were to map the elements of $P$ back to themselves?  Sounds strange, but certainly we could allow 
\[f(a)= a,\]
\[f(b) = b.\]
\[f(c) = c,\]
\[f(d) = d,\]
or 
\[f(a)= b,\]
\[f(b) = c.\]
\[f(c) = d,\]
\[f(d) = a,\]
 . . . and so on.  
 
 
 Well, why would we want to do that?  Why would we want to  map elements of a set back to themselves?  \emph {The bijections of a set to itself is the symmetric group $S_n$} - the relevant group we mentioned earlier.  At the end of the day, one way to think about symmetry is as a \emph{mapping of an object onto itself which preserves its original structure}. For our set $P$, the bijections to itself are symmetries that give rise to permutations.     

Mapping elements of a set back to themselves describes the $n!$ permutations on $n$ elements of the set.  
\begin{itemize}
    \item \textbf{5)}  Refer back to \textbf{3)}, where we found all the possible subsets of the set $B$.  How does a combination of elements differ from a permutation of elements?
\end{itemize}
We denote $n!$ as factorial of the non-negative integer $n$.  Specifically, 
\[n! = n \times (n - 1) \times (n - 2) \times (n - 3) \times . . .\times 3 \times 2 \times 1.\]
As for a concrete example, let $n = 6$ and we have 
\[6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720.\]
It follows that the symmetric group $S_6$ would have 720 elements.  
\subsubsection*{Choosing Combinations: Symmetry of the Binomial Coefficient}
We've decided that $n!$ represents the permutations on $n$ elements, but how can we calculate \emph{combinations}? 
\begin{itemize}
    \item \textbf{6)} Let's say we have a tray of 6 cupcakes each frosted a different color (perhaps pink, orange, yellow, white, green, and lavender) and your plan is to take 2 to-go.  How many different combinations of 2 cupcakes can you possibly take home with you?  Choose and calculate carefully!
\end{itemize}
\color{red} \textbf{Spoiler!} In other words you have 6 different cupcakes and are choosing 2, which we denote mathematically as ${}_6C_2$ or $\binom{6}{2}$.  We call  $\binom{n}{k} = \frac{n!}{k!(n - k)!}$ the \emph{binomial coefficient}. \color{black} 

You may interested to know that the binomial coefficient $\binom{n}{k}$ possesses a natural symmetry, which we urge you to explore in the final problem.  
\begin{itemize}
    \item \textbf{7)}  Back to balls!  Say we have 2 bins of 8 balls each.
    \begin{itemize}
      \item \textbf{a)}  Solve for $\binom{8}{3}$, choosing balls from bin $A$.  
      \item \textbf{b)}  Solve for $\binom{8}{5}$, choosing balls from bin $B$.  Do you notice anything about your answer?   
      \item \textbf{c)}  How can we explain any symmetries you might notice between choosing  balls from bin $A$ and bin $B$?  Thinking about other symmetries we have encountered throughout this column, can you infer how the symmetry of the binomial coefficient "works"?  
    \end{itemize}
\end{itemize}

\input{glossary}
\end{document}