commit 3dd4ca66c4dc14dda970a937a9db053f3cf5bb70
parent 77ea854fcc4709ce5310c3d2200975dd69a7a602
Author: miksa <milutin@popovic.xyz>
Date: Wed, 10 Mar 2021 18:10:58 +0100
finished week 4 ;)
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| A | week4.pdf | | | 0 | |
| M | week4.tex | | | 506 | ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++- |
2 files changed, 505 insertions(+), 1 deletion(-)
diff --git a/week4.pdf b/week4.pdf
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diff --git a/week4.tex b/week4.tex
@@ -14,12 +14,18 @@
\usepackage[colorlinks=true,naturalnames=true,plainpages=false,pdfpagelabels=true]{hyperref}
\usepackage[parfill]{parskip}
+\usepackage{tikz}
+\usetikzlibrary{patterns,decorations.pathmorphing,positioning}
+
\theoremstyle{definition}
\newtheorem{definition}{Definition}
\theoremstyle{definition}
\newtheorem{question}{Question}
+\theoremstyle{definition}
+\newtheorem{example}{Example}
+
\theoremstyle{theorem}
\newtheorem{theorem}{Theorem}
@@ -41,10 +47,508 @@
\maketitle
\tableofcontents
+\section{Characters}
+\begin{definition}
+ The characters $\chi _D$ of a group representation $D$, are the \textit{traces} of the linear operators
+ of of the representation or their matrix elements.
+ \begin{equation}
+ \chi _D (g) \equiv \text{Tr}(D(g)) = \sum _i (D(g))_{ii}
+ \end{equation}
+\end{definition}
+Advantages of the characters are:
+\begin{itemize}
+ \item $\text{Tr}(AB) = \text{Tr}(BA)$ so $\text{Tr}(D(g^{-1}g_1g)) = \text{Tr}(D(g_1))$
+ \item equivalent representations have the \textit{same} characters
+ \item characters are different for inequivalent irreducible representation,
+ say $D_a$ and $D_b$ then there is a orthogonality relation up to $N$
+ (number of elements in the group):
+ \begin{align*}
+ \frac{1}{N} \sum _{g\in G} \chi _{D_a}(g)^*\chi_{D_b}(g) = \delta _{ab}
+ \end{align*}
+\end{itemize}
+Furthermore characters are a \textit{complete} basis for functions that are constant on the conjugacy
+ class. Suppose $F(g_1)$ is such a function. We can expand this function in terms of matrix elements
+ of irreducible representations.
+ \begin{align*}
+ F(g_1) &= \sum_{a,j,l}\frac{1}{n_a} c_{jk}^{a} (D_a(g_1))_{jk} \\
+ &= \sum_{a,j,l}\frac{1}{n_a} c_{jk}^{a} (D_a(g^{-1}g_1g))_{jk} \\
+ &= \sum_{a,j,k,g,l,m}\frac{1}{n_a} c_{jk}^{a} (D_a(g^{-1}))_{jl}(D_a(g_1))_{lm}(D_a(g))_{mk} \\
+ &= \sum_{a,j,l}\frac{1}{n_a} c_{jk}^{a} (D_a(g_1))_{lm} \delta _{jk} \delta _{lm} \\
+ &= \sum_{a,j,l}\frac{1}{n_a} c_{jj}^{a} (D_a(g_1))_{ll} \\
+ &= \sum_{a,j,l}\frac{1}{n_a} c_{jj}^{a} \chi _a (g_1)
+ \end{align*}
+where $n_a$ denotes the dimension of the representation $D_a$.
+\newline
+
+
+We can use characters to find out how many irreducible representations appear in a reducible one and
+decompose it to it's irreducible components. We define the projection operator onto the subspace that
+transforms under the representation of $a$, where $D$ is a arbitrary representation.
+\begin{equation}
+ P_a = \frac{n_a}{N} \sum _{g\in G} \chi _{D_a}(g)^* D(g)
+\end{equation}
+This gives us the projection to the original basis.
+
\section{Spring System in a Equilateral Triangle}
+Consider three masses on the edges of a equilateral triangle connected by springs.
+The system has 6 degrees of freedom, the $x, y$ coordinates of the three masses.
+
+\begin{figure}[h!]
+ \centering
+ \begin{tikzpicture}[
+ mass/.style = {draw,circle, minimum size=0.6cm, inner sep=0pt, thick},
+ spring/.style = {decorate,decoration={zigzag, pre length=1cm,post length=1cm,segment length=5pt}},
+ ]
+ \node[mass] (m1) at (2,3) {$1$};
+ \node[mass] (m2) at (-2,3) {$2$};
+ \node[mass] (m3) at (0,0) {$3$};
+ \draw[spring] (m1) -- node[above] {} (m2);
+ \draw[spring] (m2) -- node[above] {} (m3);
+ \draw[spring] (m3) -- node[above] {} (m1);
+ \end{tikzpicture}
+ \caption{Spring System Equilateral Triangle (not equilateral in the picture)\label{drawing}}
+\end{figure}
+
+First we will see what we can find just by looking at the symmetries of the system.
\subsection{Group Theoretical Approach}
+The 6 degrees of freedom means that the system can be described with a 6 dimensional space.
+This is a tensor product of a 2 dimensional space of $x$ and $y$ coordinates and a 3 dimensional
+space of the masses (blocks).
+\begin{align*}
+ (x_1, y_1, x_2, y_2,x_3, y_3)
+\end{align*}
+The 3 dimensional space has $S_3$ symmetry (Group of all permutations of a three-element set),
+it can be represented with $D_3$ the dihedral group.
+\begin{align*}
+ D_3(e) =
+ \begin{pmatrix}
+ 1 & 0 & 0 \\
+ 0 & 1 & 0 \\
+ 0 & 0 & 1 \\
+ \end{pmatrix} \;\;\;\;
+ D_3(a_1) = \begin{pmatrix}
+ 0 & 0 & 1 \\
+ 1 & 0 & 0 \\
+ 0 & 1 & 0 \\
+ \end{pmatrix} \;\;\;\;
+ D_3(a_2) =
+ \begin{pmatrix}
+ 0 & 1 & 0 \\
+ 0 & 0 & 1 \\
+ 1 & 0 & 0 \\
+ \end{pmatrix} \;\;\;\; \\
+ D_3(a_3) =
+ \begin{pmatrix}
+ 0 & 1 & 0 \\
+ 1 & 0 & 0 \\
+ 0 & 0 & 1 \\
+ \end{pmatrix} \;\;\;\;
+ D_3(a_4) = \begin{pmatrix}
+ 1 & 0 & 0 \\
+ 0 & 0 & 1 \\
+ 0 & 1 & 0 \\
+ \end{pmatrix} \;\;\;\;
+ D_3(a_5) =
+ \begin{pmatrix}
+ 0 & 0 & 1 \\
+ 0 & 1 & 0 \\
+ 1 & 0 & 0 \\
+ \end{pmatrix} \;\;\;\;
+\end{align*}
+The 2 dimensional space also transforms under $S_3$, under a representation $D_2$
+\begin{align*}
+ D_2(e) =
+ \begin{pmatrix}
+ 1 & 0 \\
+ 0 & 1 \\
+ \end{pmatrix} \;\;\;\;
+ D_2(a_1) = \begin{pmatrix}
+ -\frac{1}{2} & -\frac{\sqrt{3}}{2} \\
+ \frac{\sqrt{3}}{2} & -\frac{1}{2} \\
+ \end{pmatrix} \;\;\;\;
+ D_2(a_2) =
+ \begin{pmatrix}
+ -\frac{1}{2} & \frac{\sqrt{3}}{2} \\
+ -\frac{\sqrt{3}}{2} & -\frac{1}{2} \\
+ \end{pmatrix} \;\;\;\; \\
+ D_2(a_3) =
+ \begin{pmatrix}
+ -1 & 0 \\
+ 0 & 1 \\
+ \end{pmatrix} \;\;\;\;
+ D_2(a_4) = \begin{pmatrix}
+ \frac{1}{2} & -\frac{\sqrt{3}}{2} \\
+ \frac{\sqrt{3}}{2} & -\frac{1}{2} \\
+ \end{pmatrix} \;\;\;\;
+ D_2(a_5) =
+ \begin{pmatrix}
+ \frac{1}{2} & -\frac{\sqrt{3}}{2} \\
+ -\frac{\sqrt{3}}{2} & -\frac{1}{2} \\
+ \end{pmatrix} \;\;\;\;
+\end{align*}
+Then $D_6$ is the tensor product of $D_3$ and $D_2$:
+\begin{equation}
+ (D_6(g))_{i\mu k\nu}=(D_3(g))_{ij}(D_2(g)_{\mu \nu}
+\end{equation}
+For $a_2$ we would than have block matrices instead of $1$ in $D_3$:
+\begin{align*}
+ D_6(a_2) =
+ \begin{pmatrix}
+ 0 & D_2(a_2) & 0 \\
+ 0 & 0 & D_2(a_2) \\
+ D_2(a_2) & 0 & 0 \\
+ \end{pmatrix}
+\end{align*}
+All other elements follow accordingly.
+
+Now $S_3$ has two 1 dimensional irreducible representations which are trivial because they map to the
+identity and one, 2 dimensional irreducible representation $D_2$ bellow is a character table of these
+representations
+\begin{table}[h!]
+ \begin{center}
+ \caption{Character table of $S_3$}
+ \label{tab:table1}
+ \begin{tabular}{l|l|l|l}
+ $S_3$ & $e$ & $\{a_1, a_2\}$ & $\{a_3, a_4, a_5\}$ \\
+ \hline
+ $\chi _0$ & 1 & 1 & 1 \\
+ \hline
+ $\chi _1$ & 1 & 1 & -1 \\
+ \hline
+ $\chi _2$ & 2 & -1 & 0 \\
+ \hline
+ $\chi _3$ & 3 & 0 & 1 \\
+ \hline
+ $\chi _6$ & 6 & 0 & 0 \\
+ \end{tabular}
+ \end{center}
+\end{table}
+
+To find the normal modes of the oscillation around equilibrium we project $D_6(g)$ to $D_0$, $D_1$ and $D_2$
+which are the irreducible representatives of $S_3$.
+\newline
+We start with $D_0$:
+\begin{align}
+ P_0 &= \frac{1}{6} \sum _{g\in G} \chi _{0}(g)^* D_6(g) \nonumber \\
+ &=\frac{1}{6}
+ \begin{pmatrix}
+ D_2(e)+D_2(a_4) & D_2(a_2)+D_2(a_3) & D_2(a_1)+D_2(a_5) \\
+ D_2(a_1)+D_2(a_3) & D_2(e)+D_2(a_5) & D_2(a_2)+D_2(a_4)\\
+ D_2(a_1)+D_2(a_5) & D_2(a_1)+D_2(a_4) & D_2(e)+D_2(a_3) \\
+ \end{pmatrix} \nonumber\\
+ &=
+ \begin{pmatrix}
+ \frac{1}{2} \\
+ \frac{\sqrt{3}}{6} \\
+ -\frac{1}{2} \\
+ \frac{\sqrt{3}}{6} \\
+ 0 \\
+ \frac{1}{\sqrt{3}} \\
+ \end{pmatrix}
+ \begin{pmatrix}
+ \frac{1}{2} & \frac{\sqrt{3}}{6} & -\frac{1}{2} &\frac{\sqrt{3}}{6} & 0 & \frac{1}{\sqrt{3}}
+ \end{pmatrix}
+ \label{eig: 1}
+\end{align}
+For $D_1$ we get:
+\begin{align}
+ P_1 &= \frac{1}{6} \sum _{g\in G} \chi _{1}(g)^* D_6(g) \nonumber \\
+ &=\frac{1}{6}
+ \begin{pmatrix}
+ D_2(e)-D_2(a_4) & D_2(a_2)-D_2(a_3) & D_2(a_1)-D_2(a_5) \\
+ D_2(a_1)-D_2(a_3) & D_2(e)-D_2(a_5) & D_2(a_2)-D_2(a_4)\\
+ D_2(a_1)-D_2(a_5) & D_2(a_1)-D_2(a_4) & D_2(e)-D_2(a_3) \\
+ \end{pmatrix} \nonumber\\
+ &=
+ \begin{pmatrix}
+ -\frac{\sqrt{3}}{6} \\
+ \frac{1}{2} \\
+ -\frac{\sqrt{3}}{6} \\
+ -\frac{1}{2} \\
+ \frac{1}{\sqrt{3}} \\
+ 0
+ \end{pmatrix}
+ \begin{pmatrix}
+ -\frac{\sqrt{3}}{6} & \frac{1}{2} & -\frac{\sqrt{3}}{6} &-\frac{1}{2} & \frac{1}{\sqrt{3}} & 0
+ \end{pmatrix}
+ \label{eig: 2}
+\end{align}
+And for $D_2$ we get:
+\begin{align}
+ P_2 &= \frac{2}{6} \sum _{g\in G} \chi _{2}(g)^* D_6(g) \nonumber \\
+ &=\frac{2}{6}
+ \begin{pmatrix}
+ 2D_2(e) & -D_2(a_2) & -D_2(a_1) \\
+ -D_2(a_1) & 2D_2(e) & -D_2(a_2)\\
+ -D_2(a_1) & -D_2(a_1) & 2D_2(e) \\
+ \end{pmatrix} \nonumber
+\end{align}
+And the nontrivial modes provided by $P_2$ can be calculated by including translation in $x$ and $y$
+direction $T_x$ and $T_y$:
+\begin{align*}
+ T_x = \frac{1}{3}
+\begin{pmatrix}
+ 1 & 0 & 1 & 0 & 1 & 0 \\
+ 0 & 0 & 0 & 0 & 0 & 0 \\
+ 1 & 0 & 1 & 0 & 1 & 0 \\
+ 0 & 0 & 0 & 0 & 0 & 0 \\
+ 1 & 0 & 1 & 0 & 1 & 0 \\
+ 0 & 0 & 0 & 0 & 0 & 0 \\
+\end{pmatrix}
+ \;\;\;\;
+ T_y = \frac{1}{3}
+\begin{pmatrix}
+ 0 & 0 & 0 & 0 & 0 & 0 \\
+ 0 & 1 & 0 & 1 & 0 & 1 \\
+ 0 & 0 & 0 & 0 & 0 & 0 \\
+ 0 & 1 & 0 & 1 & 0 & 1 \\
+ 0 & 0 & 0 & 0 & 0 & 0 \\
+ 0 & 1 & 0 & 1 & 0 & 1 \\
+\end{pmatrix}
+\end{align*}
+To see the modes we move mass $3$ in the $y$ direction which is the vector $\begin{pmatrix}0&0&0&0&0&1\end{pmatrix}$ and its mode can be get by appying it to $P_2 -T_x -T_y$:
+\begin{equation}
+ \begin{pmatrix}
+ \frac{\sqrt{3}}{6}&-\frac{1}{6}& -\frac{\sqrt{3}}{6} & -\frac{1}{6} & 0 & \frac{1}{3}
+ \end{pmatrix}
+ \label{eig: 3}
+\end{equation}
+
+The three vectors in Equasions \ref{eig: 1}, \ref{eig: 2} and \ref{eig: 3} are one of the normal modes of
+the system, at the same time they are the eigenvectors of the motion equation of the
+system which will be introduced in the next chapter.
+Further more the corresponding modes are equaual to the following motions.
+Note there are three more nodes mabye I will get later into them, but they can be calculated like the
+mode from Eq. \ref{eig: 3}.
+
+\begin{figure}[h]
+ \begin{subfigure}[b]{0.32\textwidth}
+ \centering
+ \resizebox{\linewidth}{!}{
+ \begin{tikzpicture}[
+ mass/.style = {draw,circle, minimum size=0.6cm, inner sep=0pt, thick},
+ spring/.style = {decorate,decoration={zigzag, pre length=1cm,post length=1cm,segment length=5pt}},
+ arrow/.style = {thick, ->, shorten >= 2pt, shorten <=2pt}
+ ]
+ \node[mass] (m1) at (2,3) {$1$};
+ \node[mass] (m2) at (-2,3) {$2$};
+ \node[mass] (m3) at (0,0) {$3$};
+
+ \draw[spring] (m1) -- node[above] {} (m2);
+ \draw[spring] (m2) -- node[above] {} (m3);
+ \draw[spring] (m3) -- node[above] {} (m1);
+
+ \draw[arrow] (m1) -- (2.5, 3.5);
+ \draw[arrow] (m2) -- (-2.5, 3.5);
+ \draw[arrow] (m3) -- (0, -0.7);
+
+ \end{tikzpicture}
+ }
+ \caption{Motion given by \ref{eig: 1}\\Breathing Mode}
+ \end{subfigure}
+ \begin{subfigure}[b]{0.32\textwidth}
+ \centering
+ \resizebox{\linewidth}{!}{
+ \begin{tikzpicture}[
+ mass/.style = {draw,circle, minimum size=0.6cm, inner sep=0pt, thick},
+ spring/.style = {decorate,decoration={zigzag, pre length=1cm,post length=1cm,segment length=5pt}},
+ arrow/.style = {thick, ->, shorten >= 2pt, shorten <=2pt}
+ ]
+ \node[mass] (m1) at (2,3) {$1$};
+ \node[mass] (m2) at (-2,3) {$2$};
+ \node[mass] (m3) at (0,0) {$3$};
+
+ \draw[spring] (m1) -- node[above] {} (m2);
+ \draw[spring] (m2) -- node[above] {} (m3);
+ \draw[spring] (m3) -- node[above] {} (m1);
+
+ \draw[arrow] (m1) -- (1.5, 3.5);
+ \draw[arrow] (m2) -- (-2.5, 2.5);
+ \draw[arrow] (m3) -- (0.7, 0);
+ \end{tikzpicture}
+ }
+ \caption{Motion given by \ref{eig: 2}\\ Rotation}
+ \end{subfigure}
+ \begin{subfigure}[b]{0.32\textwidth}
+ \centering
+ \resizebox{\linewidth}{!}{
+ \begin{tikzpicture}[
+ mass/.style = {draw,circle, minimum size=0.6cm, inner sep=0pt, thick},
+ spring/.style = {decorate,decoration={zigzag, pre length=1cm,post length=1cm,segment length=5pt}},
+ arrow/.style = {thick, ->, shorten >= 2pt, shorten <=2pt}
+ ]
+ \node[mass] (m1) at (2,3) {$1$};
+ \node[mass] (m2) at (-2,3) {$2$};
+ \node[mass] (m3) at (0,0) {$3$};
+
+ \draw[spring] (m1) -- node[above] {} (m2);
+ \draw[spring] (m2) -- node[above] {} (m3);
+ \draw[spring] (m3) -- node[above] {} (m1);
+
+ \draw[arrow] (m1) -- (2.5, 2.5);
+ \draw[arrow] (m2) -- (-2.5, 2.5);
+ \draw[arrow] (m3) -- (0, 0.7);
+ \end{tikzpicture}
+ }
+ \caption{Motion given by \ref{eig: 3}}
+ \end{subfigure}
+\end{figure}
+
+
+
+
\subsection{Physical Approach}
+The physical approach would be to construct the lagrangian $\mathfrak{L} = T - V$. Were $T$ is
+simply the kinetic energy of the system in $\eta = (x_1, y_1, x_2, y_2, x_3, y_3)$ coordinates and
+for simplicity we set all masses to $m$.
+\begin{align*}
+ T = \frac{m}{2} \dot{\eta} _i \dot{\eta} ^i \;\;\;\; \text{for\footnote{Einstein Summation Convention}}
+ \;\; i = 1,\dots,6
+\end{align*}
+For $V$ the potential energy we have three springs and \textit{small oscillations around equilibrium},
+two of them are the offset of the one to the angle
+of $\theta = \pm \frac{\pi}{3}$, which is $\begin{pmatrix} cos(\theta) \\ sin(\theta)\end{pmatrix} =
+ \begin{pmatrix} \frac{1}{2} \\ \pm \frac{\sqrt{3}}{2}\end{pmatrix}$. Then $V$ is:
+\begin{align*}
+ V &= \frac{k}{2} U^i_{\; j} \eta _i \eta ^j \\
+ &= \frac{k}{2} \bigg((x_1 - x_2)^2 +(\frac{1}{2}(x_2-x_3 + \frac{\sqrt{3}}{2} (y_2 - y_3))^2
+ +(\frac{1}{2}(x_1-x_3) + \frac{\sqrt{3}}{2}(y_1 - y_3))^2 \bigg)
+\end{align*}
+Where $U$ is:
+\begin{align*}
+ U = \frac{1}{4}
+ \begin{pmatrix}
+ 5 & \sqrt{3} & -4 & 0 & -1 & -\sqrt{3} \\
+ \sqrt{3} & 3 & 0 & 0 & -\sqrt{3} & -3 \\
+ -4 & 0 & 5 & -\sqrt{3} & -1 & \sqrt{3} \\
+ 0 & 0 & -\sqrt{3} & 3 & \sqrt{3} & -3 \\
+ -1 & -\sqrt{3} & -1 & \sqrt{3} & 2 & 0 \\
+ -\sqrt{3} & -3 & \sqrt{3} & -3 & 0 & 6
+ \end{pmatrix}
+\end{align*}
+
+The Euler-Lagrange equations then give us :
+\begin{equation}
+ m\ddot{\eta}^i = -k U^i_{\; j} \eta ^j
+\end{equation}
+With the exponential ansatz $\eta = \eta _0 e^{i\omega t}$ we get
+\begin{equation}
+ U^i_{\; j}\eta ^j = \lambda \eta ^i \;\;\;\;\;\;\; \text{where} \;\; \lambda = \frac{m\omega^2}{k}
+\end{equation}
+The normal modes are the eigenvalues of $U$, which were calculated only using symmetry in Equations
+\ref{eig: 1}, \ref{eig: 2} and \ref{eig: 3}. With some more character theory we can extract even more
+information explicitly on the eigenvalues. To sum it up there are four different eigenvalues, which means
+that some modes have the same frequency, to find the frequencies (eigenvalues) go through a calculation in
+diagonal coordinates of $U$ and $D(g)$ (we know all traces of $D(g)$) the trace is then invariant and the
+sum of all eigenvectors, it will give 3 equation with four unknown but one is the trivial oscillation
+with $\lambda = 0$ making the system solvable.
+
\section{Noncommutative geometric Finite Spaces}
-\subsection{The Metric on Finite Discrete Spaces}
+\subsection{Metric on Finite Discrete Spaces}
+Let $X$ be a \textit{finite discrete space}, described by a structure space $\hat{A}$ of
+a matrix algebra $A$. To describe distance between two points in X (as we would in a metric space)
+we use an array $\{d_{ij}\}_{i, j \in X}$ of \textit{real nonnegative} entries on $X$
+such that
+\begin{itemize}
+ \item $d_{ij} = d_{ji}$ Symmetric
+ \item $d_{ij} \leq d_{ik} d_{kj}$ Triangle Inequality
+ \item $d_{ij} = 0$ for $i=j$ (the same element)
+\end{itemize}
+
+\begin{example}
+ The \textit{discrete metric} on a discrete space X is $d_{ij}=1$ for $i\neq j$ and $d_{ij}=0$
+ for $i = j$
+ \newline
+ Properties of the discrete metric \url{https://en.wikipedia.org/wiki/Discrete_space#Properties}
+\end{example}
+
+The commutative case, where $A$ is assumed commutative can desrcibe the metric on $X$ in terms of
+algebraic data. The result is the following theorem can be proved.
+\begin{theorem}
+ Let $d_{ij}$ be a metric on $X$ a finite discrete space with $N$ points, $A = \mathbb{C}^N$
+ with elements $a = (a(i))_{i=1}^N$ such that $\hat{A} \simeq X$. Then there exists a
+ representation $\pi$ of $A$ on a finite-dimensional inner product space $H$ and a symmetric
+ operator $D$ on $H$ such that
+ \begin{equation}
+ d_{ij} = \sup_{a\in A}\{|a(i)-a(j)| : ||[D, \pi(a)]|| \leq 1\}
+ \end{equation}
+\end{theorem}
+
+\begin{proof}
+ Claim that this would follow from the equality:
+ \begin{equation}
+ ||[D, \pi(a)]|| = \max_{k\neq l} \big\{\frac{1}{d_{kl}}|a(k) - a(l)|\big\}
+ \label{induction}
+ \end{equation}
+ This can be proved with induction. Set $N=2$ then $H=\mathbb{C}^2$, $\pi:A\rightarrow L(H)$ and
+ a hermitian matrix $D$.
+ \begin{align}
+ \pi(a) =
+ \begin{pmatrix}
+ a(1) & 0 \\
+ 0 & a(2)
+ \end{pmatrix}
+ \;\;\;\;
+ D =
+ \begin{pmatrix}
+ 0 & (d_{12})^{-1} \\
+ (d_{21})^{-1} & 0
+ \end{pmatrix}
+ \end{align}
+ Then:
+ \begin{align}
+ ||[D, \pi(a)]|| = (d_{12})^{-1} | a(1) - a(2)|
+ \end{align}
+ Suppose this holds for $N$ with $\pi_N$, $H_N = \mathbb{C}^N$ and $D_N$.
+ Then it holds for $N+1$ with $H_{N+1} = H_{N} \oplus \bigoplus_{i=1}^N H_N^i$ and
+ \begin{align}
+ \pi_{N+1}(a(1),\dots,a(N+1)) = \pi_N(a(1),\dots,a(N))
+ \oplus
+ \begin{pmatrix}
+ a(1) & 0 \\
+ 0 & a(N+1)
+ \end{pmatrix}
+ \oplus \cdots \oplus
+ \begin{pmatrix}
+ a(N) & 0 \\
+ 0 1 & a(N+1)
+ \end{pmatrix}
+ \end{align}
+ And $D_{N+1}$:
+ \begin{align}
+ D_{N+1} = D_N
+ \oplus
+ \begin{pmatrix}
+ 0 & (d_{1(N+1)})^-1 \\
+ (d_{1(N+1)})^-1 & 0
+ \end{pmatrix}
+ \oplus \cdots \oplus
+ \begin{pmatrix}
+ 0 & (d_{N(N+1)})^-1 \\
+ (d_{N(N+1)})^-1 & 0
+ \end{pmatrix}
+ \end{align}
+ From this follows \ref{induction}.
+ Then we can continue the proof, we set for fixed $i, j$, $a(k) = d_{ik}$, which gives
+ $|a(i) - a(j)| = d_{ij}$
+ \begin{align}
+ \Rightarrow \frac{1}{d_{kl}} | a(k) - a(l) | = \frac{1}{d_{kl}} | d_{ik} - d_{il} | \leq 1
+ \end{align}
+\end{proof}
+
+The translation of the metric on $X$ into algebraic data assumes commutativity in $A$, but this can be
+extended to noncommutative matrix algebra, with the following metric on a structure space $\hat{A}$
+of a matrix algebra $M_{n_i}(\mathbb{C}$
+\begin{equation}
+ d_{ij} = \sup_{a\in A}\big\{|\text{Tr}(a(i)) - \text{Tr}((a(j))|: ||[D, a]|| \leq 1\big\}
+\end{equation}
+This is special case of the Connes' distance formula on a structure space of $A$.
+
+\begin{definition}
+ A \textit{finite spectral triple} is a tripe $(A, H, D)$, where $A$ is a unital $*$-algebra,
+ faithfully represented on a finite-dimensional Hilbert space $H$, with a symmetric operator
+ $D: H \rightarrow H$.
+\end{definition}
+$A$ is automatically a matrix algebra.
\end{document}
