week4.tex (20498B)
1 \documentclass[a4paper]{article} 2 3 \usepackage[T1]{fontenc} 4 \usepackage[utf8]{inputenc} 5 6 \usepackage{mathptmx} 7 8 \usepackage{subcaption} 9 \usepackage[shortlabels]{enumitem} 10 \usepackage{amsmath,amssymb} 11 \usepackage{amsthm} 12 \usepackage{bbm} 13 \usepackage{graphicx} 14 \usepackage[colorlinks=true,naturalnames=true,plainpages=false,pdfpagelabels=true]{hyperref} 15 \usepackage[parfill]{parskip} 16 17 \usepackage{tikz} 18 \usetikzlibrary{patterns,decorations.pathmorphing,positioning} 19 20 \theoremstyle{definition} 21 \newtheorem{definition}{Definition} 22 23 \theoremstyle{definition} 24 \newtheorem{question}{Question} 25 26 \theoremstyle{definition} 27 \newtheorem{example}{Example} 28 29 \theoremstyle{theorem} 30 \newtheorem{theorem}{Theorem} 31 32 \theoremstyle{theorem} 33 \newtheorem{exercise}{Exercise} 34 35 \theoremstyle{definition} 36 \newtheorem{solution}{Solution} 37 38 \newtheorem*{idea}{Proof Idea} 39 40 \title{University of Vienna\\ Faculty of Physics\\ \vspace{1.25cm} 41 Notes on\\ Noncommutative Geometry and Particle Phyiscs} 42 \author{Milutin Popovic \\ Supervisor: Dr. Lisa 43 Glaser} 44 \date{Week 4: 05.03 - 12.03} 45 46 \begin{document} 47 48 \maketitle 49 \tableofcontents 50 \newpage 51 52 \section{Characters} 53 \begin{definition} 54 The characters $\chi _D$ of a group representation $D$, are the \textit{traces} of the linear operators 55 of of the representation or their matrix elements. 56 \begin{equation} 57 \chi _D (g) \equiv \text{Tr}(D(g)) = \sum _i (D(g))_{ii} 58 \end{equation} 59 \end{definition} 60 Advantages of the characters are: 61 \begin{itemize} 62 \item $\text{Tr}(AB) = \text{Tr}(BA)$ so $\text{Tr}(D(g^{-1}g_1g)) = \text{Tr}(D(g_1))$ 63 \item equivalent representations have the \textit{same} characters 64 \item characters are different for inequivalent irreducible representation, 65 say $D_a$ and $D_b$ then there is a orthogonality relation up to $N$ 66 (number of elements in the group): 67 \begin{align} 68 \frac{1}{N} \sum _{g\in G} \chi _{D_a}(g)^*\chi_{D_b}(g) = \delta _{ab} 69 \end{align} 70 \end{itemize} 71 Furthermore characters are a \textit{complete} basis for functions that are constant on the conjugacy 72 class. Suppose $F(g_1)$ is such a function. We can expand this function in terms of matrix elements 73 of irreducible representations. 74 \begin{align} 75 F(g_1) &= \sum_{a,j,l}\frac{1}{n_a} c_{jk}^{a} (D_a(g_1))_{jk} \\ 76 &= \sum_{a,j,l}\frac{1}{n_a} c_{jk}^{a} (D_a(g^{-1}g_1g))_{jk} \\ 77 &= \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} \\ 78 &= \sum_{a,j,l}\frac{1}{n_a} c_{jk}^{a} (D_a(g_1))_{lm} \delta _{jk} \delta _{lm} \\ 79 &= \sum_{a,j,l}\frac{1}{n_a} c_{jj}^{a} (D_a(g_1))_{ll} \\ 80 &= \sum_{a,j,l}\frac{1}{n_a} c_{jj}^{a} \chi _a (g_1) 81 \end{align} 82 where $n_a$ denotes the dimension of the representation $D_a$. 83 \newline 84 85 86 We can use characters to find out how many irreducible representations appear in a reducible one and 87 decompose it to it's irreducible components. We define the projection operator onto the subspace that 88 transforms under the representation of $a$, where $D$ is a arbitrary representation. 89 \begin{equation} 90 P_a = \frac{n_a}{N} \sum _{g\in G} \chi _{D_a}(g)^* D(g) 91 \end{equation} 92 This gives us the projection to the original basis. 93 94 \section{Spring System in a Equilateral Triangle} 95 Consider three masses on the edges of a equilateral triangle connected by springs. 96 The system has 6 degrees of freedom, the $x, y$ coordinates of the three masses. 97 98 \begin{figure}[h!] 99 \centering 100 \begin{tikzpicture}[ 101 mass/.style = {draw,circle, minimum size=0.6cm, inner sep=0pt, thick}, 102 spring/.style = {decorate,decoration={zigzag, pre length=1cm,post length=1cm,segment length=5pt}}, 103 ] 104 \node[mass] (m1) at (2,3) {$1$}; 105 \node[mass] (m2) at (-2,3) {$2$}; 106 \node[mass] (m3) at (0,0) {$3$}; 107 \draw[spring] (m1) -- node[above] {} (m2); 108 \draw[spring] (m2) -- node[above] {} (m3); 109 \draw[spring] (m3) -- node[above] {} (m1); 110 \end{tikzpicture} 111 \caption{Spring System Equilateral Triangle (not equilateral in the picture)\label{drawing}} 112 \end{figure} 113 114 First we will see what we can find just by looking at the symmetries of the system. 115 \subsection{Group Theoretical Approach} 116 The 6 degrees of freedom means that the system can be described with a 6 dimensional space. 117 This is a tensor product of a 2 dimensional space of $x$ and $y$ coordinates and a 3 dimensional 118 space of the masses (blocks). 119 \begin{align} 120 (x_1, y_1, x_2, y_2,x_3, y_3) 121 \end{align} 122 The 3 dimensional space has $S_3$ symmetry (Group of all permutations of a three-element set), 123 it can be represented with $D_3$ the dihedral group. 124 \begin{align} 125 D_3(e) = 126 \begin{pmatrix} 127 1 & 0 & 0 \\ 128 0 & 1 & 0 \\ 129 0 & 0 & 1 \\ 130 \end{pmatrix} \;\;\;\; 131 D_3(a_1) = \begin{pmatrix} 132 0 & 0 & 1 \\ 133 1 & 0 & 0 \\ 134 0 & 1 & 0 \\ 135 \end{pmatrix} \;\;\;\; 136 D_3(a_2) = 137 \begin{pmatrix} 138 0 & 1 & 0 \\ 139 0 & 0 & 1 \\ 140 1 & 0 & 0 \\ 141 \end{pmatrix} \;\;\;\; \\ 142 D_3(a_3) = 143 \begin{pmatrix} 144 0 & 1 & 0 \\ 145 1 & 0 & 0 \\ 146 0 & 0 & 1 \\ 147 \end{pmatrix} \;\;\;\; 148 D_3(a_4) = \begin{pmatrix} 149 1 & 0 & 0 \\ 150 0 & 0 & 1 \\ 151 0 & 1 & 0 \\ 152 \end{pmatrix} \;\;\;\; 153 D_3(a_5) = 154 \begin{pmatrix} 155 0 & 0 & 1 \\ 156 0 & 1 & 0 \\ 157 1 & 0 & 0 \\ 158 \end{pmatrix} \;\;\;\; 159 \end{align} 160 The 2 dimensional space also transforms under $S_3$, under a representation $D_2$ 161 \begin{align} 162 D_2(e) = 163 \begin{pmatrix} 164 1 & 0 \\ 165 0 & 1 \\ 166 \end{pmatrix} \;\;\;\; 167 D_2(a_1) = \begin{pmatrix} 168 -\frac{1}{2} & -\frac{\sqrt{3}}{2} \\ 169 \frac{\sqrt{3}}{2} & -\frac{1}{2} \\ 170 \end{pmatrix} \;\;\;\; 171 D_2(a_2) = 172 \begin{pmatrix} 173 -\frac{1}{2} & \frac{\sqrt{3}}{2} \\ 174 -\frac{\sqrt{3}}{2} & -\frac{1}{2} \\ 175 \end{pmatrix} \;\;\;\; \\ 176 D_2(a_3) = 177 \begin{pmatrix} 178 -1 & 0 \\ 179 0 & 1 \\ 180 \end{pmatrix} \;\;\;\; 181 D_2(a_4) = \begin{pmatrix} 182 \frac{1}{2} & -\frac{\sqrt{3}}{2} \\ 183 \frac{\sqrt{3}}{2} & -\frac{1}{2} \\ 184 \end{pmatrix} \;\;\;\; 185 D_2(a_5) = 186 \begin{pmatrix} 187 \frac{1}{2} & -\frac{\sqrt{3}}{2} \\ 188 -\frac{\sqrt{3}}{2} & -\frac{1}{2} \\ 189 \end{pmatrix} \;\;\;\; 190 \end{align} 191 Then $D_6$ is the tensor product of $D_3$ and $D_2$: 192 \begin{equation} 193 (D_6(g))_{i\mu k\nu}=(D_3(g))_{ij}(D_2(g)_{\mu \nu} 194 \end{equation} 195 For $a_2$ we would than have block matrices instead of $1$ in $D_3$: 196 \begin{align} 197 D_6(a_2) = 198 \begin{pmatrix} 199 0 & D_2(a_2) & 0 \\ 200 0 & 0 & D_2(a_2) \\ 201 D_2(a_2) & 0 & 0 \\ 202 \end{pmatrix} 203 \end{align} 204 All other elements follow accordingly. 205 206 Now $S_3$ has two 1 dimensional irreducible representations which are trivial because they map to the 207 identity and one, 2 dimensional irreducible representation $D_2$ bellow is a character table of these 208 representations 209 \begin{table}[h!] 210 \begin{center} 211 \caption{Character table of $S_3$} 212 \label{tab:table1} 213 \begin{tabular}{l|l|l|l} 214 $S_3$ & $e$ & $\{a_1, a_2\}$ & $\{a_3, a_4, a_5\}$ \\ 215 \hline 216 $\chi _0$ & 1 & 1 & 1 \\ 217 \hline 218 $\chi _1$ & 1 & 1 & -1 \\ 219 \hline 220 $\chi _2$ & 2 & -1 & 0 \\ 221 \hline 222 $\chi _3$ & 3 & 0 & 1 \\ 223 \hline 224 $\chi _6$ & 6 & 0 & 0 \\ 225 \end{tabular} 226 \end{center} 227 \end{table} 228 229 To find the normal modes of the oscillation around equilibrium we project $D_6(g)$ to $D_0$, $D_1$ and $D_2$ 230 which are the irreducible representatives of $S_3$. 231 \newline 232 We start with $D_0$: 233 \begin{align} 234 P_0 &= \frac{1}{6} \sum _{g\in G} \chi _{0}(g)^* D_6(g) \nonumber \\ 235 &=\frac{1}{6} 236 \begin{pmatrix} 237 D_2(e)+D_2(a_4) & D_2(a_2)+D_2(a_3) & D_2(a_1)+D_2(a_5) \\ 238 D_2(a_1)+D_2(a_3) & D_2(e)+D_2(a_5) & D_2(a_2)+D_2(a_4)\\ 239 D_2(a_1)+D_2(a_5) & D_2(a_1)+D_2(a_4) & D_2(e)+D_2(a_3) \\ 240 \end{pmatrix} \nonumber\\ 241 &= 242 \begin{pmatrix} 243 \frac{1}{2} \\ 244 \frac{\sqrt{3}}{6} \\ 245 -\frac{1}{2} \\ 246 \frac{\sqrt{3}}{6} \\ 247 0 \\ 248 \frac{1}{\sqrt{3}} \\ 249 \end{pmatrix} 250 \begin{pmatrix} 251 \frac{1}{2} & \frac{\sqrt{3}}{6} & -\frac{1}{2} &\frac{\sqrt{3}}{6} & 0 & \frac{1}{\sqrt{3}} 252 \end{pmatrix} 253 \label{eig: 1} 254 \end{align} 255 For $D_1$ we get: 256 \begin{align} 257 P_1 &= \frac{1}{6} \sum _{g\in G} \chi _{1}(g)^* D_6(g) \nonumber \\ 258 &=\frac{1}{6} 259 \begin{pmatrix} 260 D_2(e)-D_2(a_4) & D_2(a_2)-D_2(a_3) & D_2(a_1)-D_2(a_5) \\ 261 D_2(a_1)-D_2(a_3) & D_2(e)-D_2(a_5) & D_2(a_2)-D_2(a_4)\\ 262 D_2(a_1)-D_2(a_5) & D_2(a_1)-D_2(a_4) & D_2(e)-D_2(a_3) \\ 263 \end{pmatrix} \nonumber\\ 264 &= 265 \begin{pmatrix} 266 -\frac{\sqrt{3}}{6} \\ 267 \frac{1}{2} \\ 268 -\frac{\sqrt{3}}{6} \\ 269 -\frac{1}{2} \\ 270 \frac{1}{\sqrt{3}} \\ 271 0 272 \end{pmatrix} 273 \begin{pmatrix} 274 -\frac{\sqrt{3}}{6} & \frac{1}{2} & -\frac{\sqrt{3}}{6} &-\frac{1}{2} & \frac{1}{\sqrt{3}} & 0 275 \end{pmatrix} 276 \label{eig: 2} 277 \end{align} 278 And for $D_2$ we get: 279 \begin{align} 280 P_2 &= \frac{2}{6} \sum _{g\in G} \chi _{2}(g)^* D_6(g) \nonumber \\ 281 &=\frac{2}{6} 282 \begin{pmatrix} 283 2D_2(e) & -D_2(a_2) & -D_2(a_1) \\ 284 -D_2(a_1) & 2D_2(e) & -D_2(a_2)\\ 285 -D_2(a_1) & -D_2(a_1) & 2D_2(e) \\ 286 \end{pmatrix} \nonumber 287 \end{align} 288 And the nontrivial modes provided by $P_2$ can be calculated by including translation in $x$ and $y$ 289 direction $T_x$ and $T_y$: 290 \begin{align} 291 T_x = \frac{1}{3} 292 \begin{pmatrix} 293 1 & 0 & 1 & 0 & 1 & 0 \\ 294 0 & 0 & 0 & 0 & 0 & 0 \\ 295 1 & 0 & 1 & 0 & 1 & 0 \\ 296 0 & 0 & 0 & 0 & 0 & 0 \\ 297 1 & 0 & 1 & 0 & 1 & 0 \\ 298 0 & 0 & 0 & 0 & 0 & 0 \\ 299 \end{pmatrix} 300 \;\;\;\; 301 T_y = \frac{1}{3} 302 \begin{pmatrix} 303 0 & 0 & 0 & 0 & 0 & 0 \\ 304 0 & 1 & 0 & 1 & 0 & 1 \\ 305 0 & 0 & 0 & 0 & 0 & 0 \\ 306 0 & 1 & 0 & 1 & 0 & 1 \\ 307 0 & 0 & 0 & 0 & 0 & 0 \\ 308 0 & 1 & 0 & 1 & 0 & 1 \\ 309 \end{pmatrix} 310 \end{align} 311 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 $P_2 -T_x -T_y$: 312 \begin{equation} 313 \begin{pmatrix} 314 \frac{\sqrt{3}}{6}&-\frac{1}{6}& -\frac{\sqrt{3}}{6} & -\frac{1}{6} & 0 & \frac{1}{3} 315 \end{pmatrix} 316 \label{eig: 3} 317 \end{equation} 318 319 The three vectors in Equasions \ref{eig: 1}, \ref{eig: 2} and \ref{eig: 3} are one of the normal modes of 320 the system, at the same time they are the eigenvectors of the motion equation of the 321 system which will be introduced in the next chapter. 322 Further more the corresponding modes are equaual to the following motions. 323 Note there are three more nodes mabye I will get later into them, but they can be calculated like the 324 mode from Eq. \ref{eig: 3}. 325 326 \begin{figure}[h] 327 \begin{subfigure}[b]{0.32\textwidth} 328 \centering 329 \resizebox{\linewidth}{!}{ 330 \begin{tikzpicture}[ 331 mass/.style = {draw,circle, minimum size=0.6cm, inner sep=0pt, thick}, 332 spring/.style = {decorate,decoration={zigzag, pre length=1cm,post length=1cm,segment length=5pt}}, 333 arrow/.style = {thick, ->, shorten >= 2pt, shorten <=2pt} 334 ] 335 \node[mass] (m1) at (2,3) {$1$}; 336 \node[mass] (m2) at (-2,3) {$2$}; 337 \node[mass] (m3) at (0,0) {$3$}; 338 339 \draw[spring] (m1) -- node[above] {} (m2); 340 \draw[spring] (m2) -- node[above] {} (m3); 341 \draw[spring] (m3) -- node[above] {} (m1); 342 343 \draw[arrow] (m1) -- (2.5, 3.5); 344 \draw[arrow] (m2) -- (-2.5, 3.5); 345 \draw[arrow] (m3) -- (0, -0.7); 346 347 \end{tikzpicture} 348 } 349 \caption{Motion given by \ref{eig: 1}\\Breathing Mode} 350 \end{subfigure} 351 \begin{subfigure}[b]{0.32\textwidth} 352 \centering 353 \resizebox{\linewidth}{!}{ 354 \begin{tikzpicture}[ 355 mass/.style = {draw,circle, minimum size=0.6cm, inner sep=0pt, thick}, 356 spring/.style = {decorate,decoration={zigzag, pre length=1cm,post length=1cm,segment length=5pt}}, 357 arrow/.style = {thick, ->, shorten >= 2pt, shorten <=2pt} 358 ] 359 \node[mass] (m1) at (2,3) {$1$}; 360 \node[mass] (m2) at (-2,3) {$2$}; 361 \node[mass] (m3) at (0,0) {$3$}; 362 363 \draw[spring] (m1) -- node[above] {} (m2); 364 \draw[spring] (m2) -- node[above] {} (m3); 365 \draw[spring] (m3) -- node[above] {} (m1); 366 367 \draw[arrow] (m1) -- (1.5, 3.5); 368 \draw[arrow] (m2) -- (-2.5, 2.5); 369 \draw[arrow] (m3) -- (0.7, 0); 370 \end{tikzpicture} 371 } 372 \caption{Motion given by \ref{eig: 2}\\ Rotation} 373 \end{subfigure} 374 \begin{subfigure}[b]{0.32\textwidth} 375 \centering 376 \resizebox{\linewidth}{!}{ 377 \begin{tikzpicture}[ 378 mass/.style = {draw,circle, minimum size=0.6cm, inner sep=0pt, thick}, 379 spring/.style = {decorate,decoration={zigzag, pre length=1cm,post length=1cm,segment length=5pt}}, 380 arrow/.style = {thick, ->, shorten >= 2pt, shorten <=2pt} 381 ] 382 \node[mass] (m1) at (2,3) {$1$}; 383 \node[mass] (m2) at (-2,3) {$2$}; 384 \node[mass] (m3) at (0,0) {$3$}; 385 386 \draw[spring] (m1) -- node[above] {} (m2); 387 \draw[spring] (m2) -- node[above] {} (m3); 388 \draw[spring] (m3) -- node[above] {} (m1); 389 390 \draw[arrow] (m1) -- (2.5, 2.5); 391 \draw[arrow] (m2) -- (-2.5, 2.5); 392 \draw[arrow] (m3) -- (0, 0.7); 393 \end{tikzpicture} 394 } 395 \caption{Motion given by \ref{eig: 3}} 396 \end{subfigure} 397 \end{figure} 398 399 400 401 402 \subsection{Physical Approach} 403 The physical approach would be to construct the lagrangian $\mathfrak{L} = T - V$. Were $T$ is 404 simply the kinetic energy of the system in $\eta = (x_1, y_1, x_2, y_2, x_3, y_3)$ coordinates and 405 for simplicity we set all masses to $m$. 406 \begin{align} 407 T = \frac{m}{2} \dot{\eta} _i \dot{\eta} ^i \;\;\;\; \text{for\footnote{Einstein Summation Convention}} 408 \;\; i = 1,\dots,6 409 \end{align} 410 For $V$ the potential energy we have three springs and \textit{small oscillations around equilibrium}, 411 two of them are the offset of the one to the angle 412 of $\theta = \pm \frac{\pi}{3}$, which is $\begin{pmatrix} cos(\theta) \\ sin(\theta)\end{pmatrix} = 413 \begin{pmatrix} \frac{1}{2} \\ \pm \frac{\sqrt{3}}{2}\end{pmatrix}$. Then $V$ is: 414 \begin{align} 415 V &= \frac{k}{2} U^i_{\; j} \eta _i \eta ^j \\ 416 &= \frac{k}{2} \bigg((x_1 - x_2)^2 +(\frac{1}{2}(x_2-x_3 + \frac{\sqrt{3}}{2} (y_2 - y_3))^2 417 +(\frac{1}{2}(x_1-x_3) + \frac{\sqrt{3}}{2}(y_1 - y_3))^2 \bigg) 418 \end{align} 419 Where $U$ is: 420 \begin{align} 421 U = \frac{1}{4} 422 \begin{pmatrix} 423 5 & \sqrt{3} & -4 & 0 & -1 & -\sqrt{3} \\ 424 \sqrt{3} & 3 & 0 & 0 & -\sqrt{3} & -3 \\ 425 -4 & 0 & 5 & -\sqrt{3} & -1 & \sqrt{3} \\ 426 0 & 0 & -\sqrt{3} & 3 & \sqrt{3} & -3 \\ 427 -1 & -\sqrt{3} & -1 & \sqrt{3} & 2 & 0 \\ 428 -\sqrt{3} & -3 & \sqrt{3} & -3 & 0 & 6 429 \end{pmatrix} 430 \end{align} 431 432 The Euler-Lagrange equations then give us : 433 \begin{equation} 434 m\ddot{\eta}^i = -k U^i_{\; j} \eta ^j 435 \end{equation} 436 With the exponential ansatz $\eta = \eta _0 e^{i\omega t}$ we get 437 \begin{equation} 438 U^i_{\; j}\eta ^j = \lambda \eta ^i \;\;\;\;\;\;\; \text{where} \;\; \lambda = \frac{m\omega^2}{k} 439 \end{equation} 440 The normal modes are the eigenvalues of $U$, which were calculated only using symmetry in Equations 441 \ref{eig: 1}, \ref{eig: 2} and \ref{eig: 3}. With some more character theory we can extract even more 442 information explicitly on the eigenvalues. To sum it up there are four different eigenvalues, which means 443 that some modes have the same frequency, to find the frequencies (eigenvalues) go through a calculation in 444 diagonal coordinates of $U$ and $D(g)$ (we know all traces of $D(g)$) the trace is then invariant and the 445 sum of all eigenvectors, it will give 3 equation with four unknown but one is the trivial oscillation 446 with $\lambda = 0$ making the system solvable. 447 448 \section{Noncommutative geometric Finite Spaces} 449 \subsection{Metric on Finite Discrete Spaces} 450 Let $X$ be a \textit{finite discrete space}, described by a structure space $\hat{A}$ of 451 a matrix algebra $A$. To describe distance between two points in X (as we would in a metric space) 452 we use an array $\{d_{ij}\}_{i, j \in X}$ of \textit{real nonnegative} entries on $X$ 453 such that 454 \begin{itemize} 455 \item $d_{ij} = d_{ji}$ Symmetric 456 \item $d_{ij} \leq d_{ik} d_{kj}$ Triangle Inequality 457 \item $d_{ij} = 0$ for $i=j$ (the same element) 458 \end{itemize} 459 460 \begin{example} 461 The \textit{discrete metric} on a discrete space X is $d_{ij}=1$ for $i\neq j$ and $d_{ij}=0$ 462 for $i = j$ 463 \newline 464 Properties of the discrete metric \url{https://en.wikipedia.org/wiki/Discrete_space#Properties} 465 \end{example} 466 467 The commutative case, where $A$ is assumed commutative can desrcibe the metric on $X$ in terms of 468 algebraic data. The result is the following theorem can be proved. 469 \begin{theorem} 470 Let $d_{ij}$ be a metric on $X$ a finite discrete space with $N$ points, $A = \mathbb{C}^N$ 471 with elements $a = (a(i))_{i=1}^N$ such that $\hat{A} \simeq X$. Then there exists a 472 representation $\pi$ of $A$ on a finite-dimensional inner product space $H$ and a symmetric 473 operator $D$ on $H$ such that 474 \begin{equation} 475 d_{ij} = \sup_{a\in A}\{|a(i)-a(j)| : ||[D, \pi(a)]|| \leq 1\} 476 \end{equation} 477 \end{theorem} 478 479 \begin{proof} 480 Claim that this would follow from the equality: 481 \begin{equation} 482 ||[D, \pi(a)]|| = \max_{k\neq l} \big\{\frac{1}{d_{kl}}|a(k) - a(l)|\big\} 483 \label{induction} 484 \end{equation} 485 This can be proved with induction. Set $N=2$ then $H=\mathbb{C}^2$, $\pi:A\rightarrow L(H)$ and 486 a hermitian matrix $D$. 487 \begin{align} 488 \pi(a) = 489 \begin{pmatrix} 490 a(1) & 0 \\ 491 0 & a(2) 492 \end{pmatrix} 493 \;\;\;\; 494 D = 495 \begin{pmatrix} 496 0 & (d_{12})^{-1} \\ 497 (d_{21})^{-1} & 0 498 \end{pmatrix} 499 \end{align} 500 Then: 501 \begin{align} 502 ||[D, \pi(a)]|| = (d_{12})^{-1} | a(1) - a(2)| 503 \end{align} 504 Suppose this holds for $N$ with $\pi_N$, $H_N = \mathbb{C}^N$ and $D_N$. 505 Then it holds for $N+1$ with $H_{N+1} = H_{N} \oplus \bigoplus_{i=1}^N H_N^i$ and 506 \begin{align} 507 \pi_{N+1}(a(1),\dots,a(N+1)) = \pi_N(a(1),\dots,a(N)) 508 \oplus 509 \begin{pmatrix} 510 a(1) & 0 \\ 511 0 & a(N+1) 512 \end{pmatrix} 513 \oplus \cdots \oplus 514 \begin{pmatrix} 515 a(N) & 0 \\ 516 0 1 & a(N+1) 517 \end{pmatrix} 518 \end{align} 519 And $D_{N+1}$: 520 \begin{align} 521 D_{N+1} = D_N 522 \oplus 523 \begin{pmatrix} 524 0 & (d_{1(N+1)})^{-1} \\ 525 (d_{1(N+1)})^{-1} & 0 526 \end{pmatrix} 527 \oplus \cdots \oplus 528 \begin{pmatrix} 529 0 & (d_{N(N+1)})^{-1} \\ 530 (d_{N(N+1)})^{-1} & 0 531 \end{pmatrix} 532 \end{align} 533 From this follows \ref{induction}. 534 Then we can continue the proof, we set for fixed $i, j$, $a(k) = d_{ik}$, which gives 535 $|a(i) - a(j)| = d_{ij}$ 536 \begin{align} 537 \Rightarrow \frac{1}{d_{kl}} | a(k) - a(l) | = \frac{1}{d_{kl}} | d_{ik} - d_{il} | \leq 1 538 \end{align} 539 \end{proof} 540 541 The translation of the metric on $X$ into algebraic data assumes commutativity in $A$, but this can be 542 extended to noncommutative matrix algebra, with the following metric on a structure space $\hat{A}$ 543 of a matrix algebra $M_{n_i}(\mathbb{C}$ 544 \begin{equation} 545 d_{ij} = \sup_{a\in A}\big\{|\text{Tr}(a(i)) - \text{Tr}((a(j))|: ||[D, a]|| \leq 1\big\} 546 \end{equation} 547 This is special case of the Connes' distance formula on a structure space of $A$. 548 549 \begin{definition} 550 A \textit{finite spectral triple} is a tripe $(A, H, D)$, where $A$ is a unital $*$-algebra, 551 faithfully represented on a finite-dimensional Hilbert space $H$, with a symmetric operator 552 $D: H \rightarrow H$. 553 \end{definition} 554 $A$ is automatically a matrix algebra. 555 556 \end{document}