week5.tex (27284B)
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 \usepackage[framemethod=TikZ]{mdframed} 21 22 \tikzstyle{titlered} = 23 [draw=black, thick, fill=white,% 24 text=black, rectangle, 25 right, minimum height=.7cm] 26 27 \newcounter{exercise} 28 29 \renewcommand*\theexercise{Exercise~\arabic{exercise}} 30 31 \makeatletter 32 \mdfdefinestyle{exercisestyle}{% 33 outerlinewidth=1em,% 34 outerlinecolor=white,% 35 leftmargin=-1em,% 36 rightmargin=-1em,% 37 middlelinewidth=1.2pt,% 38 roundcorner=5pt,% 39 linecolor=black,% 40 backgroundcolor=blue!5, 41 innertopmargin=1.2\baselineskip, 42 skipabove={\dimexpr0.5\baselineskip+\topskip\relax}, 43 skipbelow={-1em}, 44 needspace=3\baselineskip, 45 frametitlefont=\sffamily\bfseries, 46 settings={\global\stepcounter{exercise}}, 47 singleextra={% 48 \node[titlered,xshift=1cm] at (P-|O) % 49 {~\mdf@frametitlefont{\theexercise}~};},% 50 firstextra={% 51 \node[titlered,xshift=1cm] at (P-|O) % 52 {~\mdf@frametitlefont{\theexercise}~};}, 53 } 54 \makeatother 55 56 \newenvironment{MyExercise}% 57 {\begin{mdframed}[style=exercisestyle]}{\end{mdframed}} 58 59 \theoremstyle{definition} 60 \newtheorem{definition}{Definition} 61 62 \theoremstyle{definition} 63 \newtheorem{question}{Question} 64 65 \theoremstyle{definition} 66 \newtheorem{example}{Example} 67 68 \theoremstyle{theorem} 69 \newtheorem{theorem}{Theorem} 70 71 \theoremstyle{theorem} 72 \newtheorem{lemma}{Lemma} 73 74 \newtheorem*{idea}{Proof Idea} 75 76 77 78 \title{University of Vienna\\ Faculty of Physics\\ \vspace{1.25cm} 79 Notes on\\ Noncommutative Geometry and Particle Phyiscs} 80 \author{Milutin Popovic \\ Supervisor: Dr. Lisa 81 Glaser} 82 \date{Week 5: 12.03 - 19.03} 83 84 \begin{document} 85 86 \maketitle 87 \tableofcontents 88 \newpage 89 90 \section{Noncommutative Geometric Spaces } 91 \subsection{Exercises} 92 93 \begin{MyExercise} 94 \textbf{ 95 Make the proof of the last theorem (see week4.pdf) explicit for $N=3$. 96 }\newline 97 98 For the C* algebra we have $A=\mathbb{C}^3$ 99 For $H$ we have $H = (\mathbb{C}^2)^{\oplus 3} = H_2 \oplus H_2^1 \oplus H_2^2$. 100 The symmetric operator $D$ acting on $H$ and the representation $\pi (a)$: 101 \begin{align} 102 \pi((a(1), a(2), a(3)) &= 103 \begin{pmatrix} 104 a(1) & 0 \\ 0 & a(2) 105 \end{pmatrix} \oplus 106 \begin{pmatrix} 107 a(1) & 0 \\ 0 & a(3) 108 \end{pmatrix} \oplus 109 \begin{pmatrix} 110 a(2) & 0 \\ 0 & a(2) 111 \end{pmatrix} \nonumber \\ 112 & = 113 \begin{pmatrix} 114 a(1) & 0 & 0 & 0 & 0 & 0 \\ 115 0 & a(2) & 0 & 0 & 0 & 0 \\ 116 0 & 0 & a(1) & 0 & 0 & 0 \\ 117 0 & 0 & 0 & a(3) & 0 & 0 \\ 118 0 & 0 & 0 & 0 & a(2) & 0 \\ 119 0 & 0 & 0 & 0 & 0 & a(3) 120 \end{pmatrix} \\ 121 D &= 122 \begin{pmatrix} 123 0 & x_1 \\ x_1 & 0 124 \end{pmatrix} \oplus 125 \begin{pmatrix} 126 0 & x_2 \\ x_2 & 0 127 \end{pmatrix} \oplus 128 \begin{pmatrix} 129 0 & x_3 \\ x_3 & 0 130 \end{pmatrix} \nonumber \\ 131 &= 132 \begin{pmatrix} 133 0 & x_1 & 0 & 0 & 0 & 0 \\ 134 x_1 & 0 & 0 & 0 & 0 & 0 \\ 135 0 & 0 & 0 & x_2 & 0 & 0 \\ 136 0 & 0 & x_2 & 0 & 0 & 0 \\ 137 0 & 0 & 0 & 0 & 0 & x_3 \\ 138 0 & 0 & 0 & 0 & x_3 & 0 \\ 139 \end{pmatrix} \\ 140 \end{align} 141 Then the norm of the commutator would be the largest eigenvalue 142 \begin{align} 143 &||[D, \pi(a)]|| = ||D\pi(a) - \pi(a)D||\nonumber\\ 144 &= 145 \left|\left| 146 \setlength{\arraycolsep}{0.1cm} 147 \renewcommand{\arraystretch}{0.1} 148 \begin{pmatrix} 149 0 & x_1(a(2)-a(1)) & 0 & 0 & 0 & 0 \\ 150 -x_1(a(2)-a(1)) & 0 & 0 & 0 & 0 & 0 \\ 151 0 & 0 & 0 & x_2(a(3)-a(1)) & 0 & 0 \\ 152 0 & 0 & -x_2(a(3)-a(1)) & 0 & 0 & 0 \\ 153 0 & 0 & 0 & 0 & 0 & x_3(a(3)-a(2)) \\ 154 0 & 0 & 0 & 0 & -x_3(a(2)-a(3)) & 0 \\ 155 \end{pmatrix}\right|\right| \label{skew matrix} 156 \end{align} 157 The matrix in Equation \ref{shew matrix} is a skew symmetric matrix its eigenvalues 158 are $i\lambda_1, i\lambda_2, i\lambda_3, i\lambda_4$, where the $\lambda$'s are on the 159 upper and lower diagonal check \url{https://en.wikipedia.org/wiki/Skew-symmetric_ 160 matrix#Skew-symmetrizable_matrix}. The matrix norm of would be the maximum of the norm of 161 the larges eigenvalues: 162 \begin{align} 163 ||[D, \pi(a)]|| = \max_{a\in A}\{&x_1|a(2)-a(1)|,\\ 164 &x_2|(a(3)-a(1))|,\nonumber\\ 165 &x_3|(a(3)-a(2))|,\}\nonumber 166 \end{align} 167 The metric is then: 168 \begin{align} 169 d = 170 \begin{pmatrix} 171 0 & a(1)-a(2) & a(1)-a(3)\\ 172 a(2)-a(1) & 0 & a(2)-a(3)\\ 173 a(3)-a(1) & a(3)-a(2) & 0 174 \end{pmatrix} 175 \end{align} 176 \end{MyExercise} 177 178 \begin{MyExercise} 179 \textbf{ 180 Compute the metric on the space of three points given by $d_{ij} = 181 \sup_{a\in A}\{|a(i) - a(j)|: ||[D, \pi(a)]|| \leq 1\}$ for the set of data 182 $A = \mathbb{C}^3$ acting in the defining representation $H = \mathbb{C}^3$, and 183 \begin{align} 184 D = 185 \begin{pmatrix} 186 0 & d^{-1} & 0 \\ 187 d^{-1} & 0 & 0 \\ 188 0 & 0 & 0 189 \end{pmatrix} 190 \end{align} 191 for some $d \in \mathbb{R}$ 192 }\newline 193 194 We have $A=\mathbb{C}^3$, $H=\mathbb{C}^3$ and $D$ from above, then 195 196 \begin{align} 197 ||[D, \pi(a)]|| &= d^{-1}\left|\left| 198 \begin{pmatrix} 199 0 & a(2)-a(1) & 0 \\ 200 -(a(2)-a(1)) & 0 & 0 \\ 201 0 & 0 & 0 202 \end{pmatrix} \right|\right| 203 \end{align} 204 The metric is then 205 \begin{align} 206 d = 207 \begin{pmatrix} 208 0 & a(1)-a(2) & a(1) \\ 209 a(2)-a(1) & 0 & a(2) \\ 210 -a(1) & -a(2) & 0 211 \end{pmatrix} 212 \end{align} 213 \end{MyExercise} 214 215 \begin{MyExercise} 216 \textbf{ 217 Show that $d_{ij}$ from Equation \ref{ext metric} is a metric on $\hat{A}$ by 218 establishing that: 219 \begin{align} 220 d_{ij} &= 0\;\;\; \Leftrightarrow \;\;\; i=j \label{metric 1} \\ 221 d_{ij} &= d_{ji} \label{metric 2}\\ 222 d_{ij} &\leq d_{ik} + d_{kj} \label{metric 3} 223 \end{align} 224 \begin{equation} \label{ext metric} 225 d_{ij} = \sup_{a\in A}\big\{|\text{Tr}(a(i)) - \text{Tr}((a(j))|: ||[D, a]|| \leq 1\big\} 226 \end{equation} 227 }\newline 228 229 For Equation \ref{metric 1} set $i=j$ in \ref{ext metric}. 230 \begin{align} 231 d_{ii} &= \sup_{a \in A}\{|\text{Tr}(a(i)) - \text{Tr}((a(i))|: ||[D, a]|| \leq 232 1\big\} \\ 233 &= \sup_{a \in A}\{0: ||[D, a]|| \leq 1\big\} = 0 234 \end{align} 235 For Equation \ref{metric 2} obviously we have the commuting property of 236 addition. 237 \newline 238 For Equation \ref{metric 3}, for $k=j$ then $d_{kj} = 0$ and the equality 239 holds. For $i = k$ then $d_{ik} = 0$ and equality holds. Else set $d_{ik} = 240 1$ and $d_{kj} = 1$ then $d_{ij} = 1 \leq d_{ik} + d_{kj} = 2$ 241 \end{MyExercise} 242 243 \subsection{Properties of Matrix Algebras} 244 \begin{lemma} 245 If $A$ is a unital C* algebra that acts faithfully on a finite 246 dimensional Hilbert space, then $A$ is a matrix algebra of the Form: 247 \begin{equation} 248 A \simeq \bigoplus _{i=1}^N M_{n_i}(\mathbb{C}) 249 \end{equation} 250 \end{lemma} 251 \begin{proof} 252 Since $A$ acts faithfully on a Hilbert space, then $A$ is a C* 253 subalgebra of a matrix algebra $L(H) = M_{\dim (H)}(\mathbb{C} 254 \Rightarrow A \simeq \text{Matrix algebra}$. 255 \end{proof} 256 257 \begin{question} 258 What does the author mean when he sais 'acts faithfully on a 259 Hilbertspace`? Then the representation is fully reducible, or that the 260 presentation is irreducible? 261 \newline 262 263 For a *-representation 'faithful` if it is injective. For a 264 *-homomorphism 'faithful` means one-to-one correspondance 265 \end{question} 266 267 \begin{example} 268 $A = M_n(\mathbb{C})$ and $H=\mathbb{C}^n$, $A$ acts on $H$ with matrix 269 multiplication and standard inner product. $D$ on $H$ is a hermitian 270 matrix $n\times n$ matrix. 271 \end{example} 272 273 $D$ is referred to as a finite Dirac operator as in as its $\infty$ 274 dimensional on Riemannian Spin manifolds coming in Chapter 4. 275 \newline 276 277 Now can introduce a 'differential 'geometric structure` on the finite space X 278 with the \textbf{devided difference} 279 \begin{equation} 280 \frac{a(i)-a(j)}{d_{ij}} 281 \end{equation} 282 for each pair $i$, $j$ $\in X$ the finite dimensional discrete space $X$. 283 This appears in the entries in the commutator $[D, a]$ in the above 284 exercises. 285 286 \begin{definition} 287 Given an finite spectral triple $(A, H, D)$, the $A$-bimodule of 288 Connes' differential one-forms is: 289 \begin{equation} 290 \Omega _D ^1 (A) := \left\{ \sum _k a_k[D, b_k]: a_k, b_k \in A \right\} 291 \end{equation} 292 \end{definition} 293 294 \begin{question} 295 Is the Conne's differential one form the set of all '1st order 296 differential operators` given $A$, that act on $H$? 297 \end{question} 298 Then there is a map $d:A\rightarrow \Omega _D ^1 (A)$, $d = [D, \cdot]$. 299 \begin{MyExercise} 300 \textbf{ 301 Verify that 'd` is a derivation of the C* algebra 302 \begin{align} 303 d(ab) = d(a)b + ad(b) \\ 304 d(a^*) = -d(a)^* 305 \end{align} 306 }\newline 307 308 For the record $d(\cdot) = [D, \cdot]$, then we have 309 \begin{enumerate} 310 \item 311 \begin{align} 312 d(ab) &= [D, ab] = [D, a]b + a[D,b]\\ 313 &= d(a)b + ad(b) 314 \end{align} 315 \item 316 \begin{align} 317 d(a^*) &= [D, a^*] = Da^* - a^*D \\ 318 &=-(D^*a - aD^*) = -[D^*, a] \\ 319 &= -d(a)^* 320 \end{align} 321 \end{enumerate} 322 \end{MyExercise} 323 \begin{MyExercise} 324 \textbf{ 325 Verify that $\Omega _D^1 (A)$ is an $A$-bimodule by rewriting 326 } 327 \begin{align} 328 a(a_k[D, b_k])b = \sum_k a'_k[D, b'_k] \;\;\;\; a'_k, b'_k \in A 329 \end{align} 330 \newline 331 332 Begin 333 \begin{align} 334 a(a_k[D, b_k])b &= aa_k(Db_k - b_k D) b = \\ 335 &= aa_k(Db_k b - b_k D b) = aa_k(Db_k b - b_k Db - b_kbD +b_kbD)=\\ 336 &= aa_k(Db_kb-b_kbD + b_k b D - b_k D b) = \\ 337 &= aa_k [D, b_kb] + aa_k b [D, b]=\\ 338 &= \sum _k a_k' [D, b_k'] 339 \end{align} 340 341 \end{MyExercise} 342 343 \begin{lemma} 344 Let $(A, H, D) = (M_n(\mathbb{C}, \mathbb{C}^n, D)$, with $D$ a hermitian 345 $n\times n$ matrix. If $D$ is not a multiple of the identity then: 346 \begin{equation} 347 \Omega _D ^1 (A) \simeq M_n(\mathbb{C}) = A 348 \end{equation} 349 \end{lemma} 350 351 \begin{proof} 352 Assume $D = \sum _i \lambda _i e_{ii}$ (diagonal), $\lambda _i \in \mathbb{R}$ and 353 $\{e_{ij}\}$ the basis of $M_n(\mathbb{C}$. For fixed $i$, $j$ choose $k$ 354 such that $\lambda _k \neq \lambda _j$ then 355 \begin{align} \label{basis} 356 \left(\frac{1}{\lambda _k - \lambda _j} e_{ik}\right) [D, e_{kj}] = 357 e_{ij} 358 \end{align} 359 $e_{ij}\in \Omega _D ^1 (A)$ by the above definition. And $\Omega _D ^1 360 (A) \subset L(\mathbb{C}^n) = H \simeq M_n(\mathbb{C}) = A$ 361 \end{proof} 362 363 \begin{MyExercise} 364 \textbf{ 365 Consider $(A=\mathbb{C}^2, H=\mathbb{C}^2, 366 D = \begin{pmatrix} 0 & \lambda \\ \bar{\lambda} & 0 367 \end{pmatrix})$ with $\lambda \neq 0$. Show that $\Omega _D^1(A) 368 \simeq M_2(\mathbb{C})$ 369 } 370 \newline 371 372 Because of the Hilbert Basis $D$ can be extended in terms of 373 the basis of $M_2(\mathbb{C})$, plugging this into Equation 374 \ref{basis} will get us the same cyclic result, thus 375 $\Omega _D^1(A) \simeq M_2(\mathbb{C})$ 376 \ 377 \end{MyExercise} 378 379 \subsection{Morphisms Between Finite Spectral Triples} 380 \begin{definition} 381 two finite spectral tripes $(A_1, H_1, D_1)$ and $(A_2, H_2, D_2)$ are 382 called unitarily equivalent if 383 \begin{itemize} 384 \item $A_1 = A_2$ 385 \item $\exists \;\; U: H_1 \rightarrow H_2$, unitary with 386 \begin{enumerate} 387 \item $U\pi_1(a)U^* = \pi_2(a)$ with $a \in A_1$ 388 \item $UD_1 U^* = D_2$ 389 \end{enumerate} 390 \end{itemize} 391 \end{definition} 392 393 Some remarks 394 \begin{itemize} 395 \item the above is an equivalence relation 396 \item spectral unitary equivalence is given by the unitaries of the 397 matrix algebra itself 398 \item for any such $U$ then $(A, H, D) \sim (A, H, UDU^*)$ 399 \item $UDU^* = D + U[D, U^*]$ of the form of elements in 400 $\Omega _D^1 (A)$. 401 \end{itemize} 402 403 \begin{MyExercise} 404 \textbf{ 405 Show that the unitary equivalence between finite spectral 406 triples is a equivalence relation 407 }\newline 408 409 An equivalence relation needs to satisfy reflexivity, symmetry 410 transitivity. 411 Let $(A_1, H_1, D_1)$, $(A_2, H_2, D_2)$ and $(A_3, H_3, D_3)$ 412 be three finite spectral triples. 413 \newline 414 415 For reflexivity $(A_1, H_1, D_1) \sim (A_1, H_1, D_1)$. So there 416 exists a $U: H_1 \rightarrow H_1$ unitary, which is the identity 417 and always exists. 418 \newline 419 420 For symmetry we need 421 \begin{align} 422 (A_1, H_1, D_1) \sim (A_2, H_2, D_2) \Leftrightarrow 423 (A_2, H_2, D_2) \sim (A_1, H_1, D_1) 424 \end{align} 425 because $U$ is unitary: 426 \begin{align} 427 &U\pi_1(a)U^* = \pi_2(a) \;\;\; | \cdot U^*\boxdot U \\ 428 &U^*U\pi_1(a)U^*U = \pi_1(a) = U^*\pi_2(a)U \\ 429 \end{align} 430 The same with the symmetric operator $D$. 431 \newline 432 433 For transitivity we need 434 \begin{align} 435 (A_1, H_1, D_1) &\sim (A_2, H_2, D_2) \;\;\; \text{and} \;\;\; 436 (A_2, H_2, D_2) \sim (A_3, H_3, D_3) \\ 437 &\Rightarrow (A_1, H_1, D_1) \sim (A_3, H_3, D_3) 438 \end{align} 439 There are two unitary maps $U_{12}:H_1 \rightarrow H_2$ and 440 $U_{23}: H_2 \rightarrow H_3$ then 441 \begin{align} 442 U_{23}U_{12} \pi_1(a) U^*_{12}U^*_{23} &= U_{23} 443 \pi_2(a) U_23^* \\ 444 &= \pi_3(a) \\ 445 U_{23}U_{12} D_1U^*_{12}U^*_{23} &= U_{23} 446 D_2 U_23^* \\ 447 &= D_3 448 \end{align} 449 \end{MyExercise} 450 451 Extending the this relation we look again at the notion of equivalence from 452 Morita equivalence of Matrix Algebras. 453 \newline 454 455 \begin{definition} 456 Let $A$ be an algebra. We say that $I \subset A$, as a vector space, is a 457 right(left) ideal if $ab \in I$ for $a \in A$ and $b\in I$ (or $ba \in 458 I$, $b\in I$, $a\in A$). We call a left-right ideal simply an ideal. 459 \end{definition} 460 461 Given a Hilbert bimodule $E \in KK_f(B, A)$ and $(A, H, D)$ we construct 462 a finite spectral triple on $B$, $(B, H', D')$ 463 \begin{equation} 464 H' = E \otimes _A H 465 \end{equation} 466 We might define $D'$ with $D'(e \otimes \xi) = e\otimes D\xi$, thought this 467 would not satisfy the ideal defining the balanced tensor product over $A$, 468 which is generated by elements of the form 469 \begin{align} 470 e a \otimes \xi - e\otimes a \xi ;\;\;\;\; e\in E, a\in A, \xi \in H 471 \end{align} 472 This inherits the left action on $B$ from $E$ and has a $\mathbb{C}$ 473 valued inner product space. $B$ also satisfies the ideal. 474 \begin{equation} 475 D'(e\otimes \xi) = e \otimes D \xi + \nabla (e) \xi \;\;\;\; e\in 476 E, a\in A 477 \end{equation} 478 Where $\nabla$ is called the \textit{connection on the right A-module E} 479 associated with the derivation $d=[D, \cdot]$ and satisfying the 480 \textit{Leibnitz Rule} which is 481 \begin{equation} 482 \nabla(ae) = \nabla(e)a + e \otimes [D, a] \;\;\;\;\; e\in E,\; a\in A 483 \end{equation} 484 Then $D'$ is well defined on $E \otimes _A H$: 485 \begin{align} 486 D'(ea \otimes \xi - e \otimes a \xi) &= D'(ea \otimes \xi) - D'(e 487 \otimes \xi) \\ 488 &= ea\otimes D\xi + \nabla(ae) \xi - e \otimes D(a\xi ) - \nabla (e)a 489 \xi \\ 490 &= 0. 491 \end{align} 492 With the information thus far we can prove the following theorem 493 \begin{theorem} 494 If $(A, H, D)$ a finite spectral triple, $E \in KK_f(B, A)$. 495 Then $(V, E\otimes _A H, D')$ is a finite spectral triple, provided that 496 $\nabla$ satisfies the compatibility condition 497 \begin{equation} 498 \langle e_1, \nabla e_2 \rangle _E - \langle \nabla e_1, e_2 499 \rangle _E = d\langle e_1, e_2 \rangle _E \;\;\;\; e_1, e_2 \in E 500 \end{equation} 501 \end{theorem} 502 \begin{proof} 503 $E\otimes _A H$ was shown in the previous section (text before the 504 theorem). The only thing left is to show that $D'$ is a symmetric 505 operator, this we can just compute. Let $e_1, e_2 \in E$ and $\xi _1, 506 \xi _2 \in H$ then 507 \begin{align} 508 \langle e_1 \otimes \xi _1, D'(e_2 \otimes \xi_2)\rangle _{E\otimes _A H} &= 509 \langle \xi _1, \langle e_1, \nabla e_2\rangle _E \xi _2\rangle + \langle \xi _1 , \langle e_1, e_2\rangle _E D\xi 510 _2\rangle _H \\ 511 &= \langle \xi _1, \langle \nabla e_1, e_2\rangle _E \xi _2\rangle _H + \langle \xi _1, d\langle e_1, e_2\rangle _E 512 \xi _2\rangle _H \\ 513 &+ \langle D\xi _1,\langle e_1, e_2\rangle _E \xi _2\rangle _H - \langle \xi _1, [D, \langle e_1, e_2\rangle _E] \xi 514 _2 \rangle _H \\ 515 &= \langle D'(e_1 \otimes \xi _1), e_2 \otimes \xi _2\rangle _{E \otimes _A H} 516 \end{align} 517 \end{proof} 518 519 \begin{MyExercise} 520 \textbf{ 521 Let $\nabla$ and $\nabla'$ be two connections on a right $A$-module 522 $E$. Show that $\nabla - \nabla'$ is a right $A$-linear map 523 $E \rightarrow E\otimes _A \Omega _D^1(A)$ 524 }\newline 525 526 Both $\nabla$ and $\nabla'$ need to satisfy the Leiblitz rule, so 527 let's see if $\nabla - \nabla'$ does. 528 529 \begin{align} 530 \nabla(ea)-\nabla'(ea)&=\nabla(e) + e\otimes[D, a]\\ 531 &-(\nabla'(e)a + e\otimes[D',a])\\ 532 &=\bar{\nabla}a + e\otimes(Da-aD-D'a+aD')\\ 533 &=\bar{\nabla}a + e\otimes((D-D')a-a(D-D'))\\ 534 &=\bar{\nabla}a + e\otimes[D', a]\\ 535 &=\bar{\nabla}(ea) 536 \end{align} 537 Therefore $\nabla-\nabla'$ is a linear map. 538 \end{MyExercise} 539 540 \begin{MyExercise} 541 \textbf{ 542 Construct a finite spectral triple $(A, H', D')$ from $(A, H, D)$ 543 \begin{enumerate} 544 \item show that the derivation $d(\cdot):A \rightarrow A\otimes _A 545 \Omega_D^1(A)=\Omega_D^1(A)$ is a connection on $A$ 546 considered a right $A$-module 547 \item Upon identifying $A\otimes_A H\simeq H$, what is $D'$ 548 when the connection is $d(\cdot)$. 549 \item Use 1) and 2) to show that any connection $\nabla: 550 A\rightarrow A\otimes_A \Omega_D^1(A)$ is given by 551 \begin{align} 552 \nabla = d + \omega 553 \end{align} 554 where $\omega \in \Omega_D^1(A)$ 555 \item Upon identifying $A\otimes_A H \simeq H$, what is the 556 difference operator $D'$ with the connection on $A$ given by 557 $\nabla = d + \omega$ 558 \end{enumerate} 559 } 560 \begin{enumerate} 561 \item $\nabla(e \cdot a) = d(a)$ 562 \item 563 $D'(a\xi) = a(D\xi) + (\nabla a) \xi = D(a\xi)$ 564 \item Use the identity element $e \in A$\\ 565 $\nabla (e\cdot a) = \nabla(e) a + 1 \otimes d(a)=d(a) 566 \nabla(e) a$ 567 \item $D'(a\otimes \xi) = D'(a \xi) = a(D\xi) + (\nabla a)\xi = 568 a(D\xi) + \nabla(e \cdot a) \xi \\ 569 = D(a\xi) + \nabla(e) (a\xi)$ 570 \end{enumerate} 571 \end{MyExercise} 572 573 \subsection{Graphing Finite Spectral Triples} 574 \begin{definition} 575 A \textit{graph} is a ordered pair $(\Gamma ^{(0)}, \Gamma ^{(1)})$. 576 Where $\Gamma ^{(0)}$ is the set of vertices (nodes) and $\Gamma ^{(1)}$ 577 a set of pairs of vertices (edges) 578 \end{definition} 579 \begin{figure}[h!] 580 \centering 581 \begin{tikzpicture}[ 582 mass/.style = {draw,circle, minimum size=0.2cm, inner sep=0pt, thick}, 583 spring/.style = {decorate,decoration={zigzag, pre length=1cm,post length=1cm,segment length=5pt}},] 584 \node[mass] (m1) at (1,1.5) {}; 585 \node[mass] (m2) at (-1,1.5) {}; 586 \node[mass] (m3) at (0,0) {}; 587 588 \draw (m1) -- (m2); 589 \draw (m1) -- (m3); 590 \draw (m2) -- (m3); 591 \end{tikzpicture} 592 \caption{A simple graph with three vertices and three edges} 593 \end{figure} 594 \begin{MyExercise} 595 \textbf{ 596 Show that any finite-dimensional faithful representation $H$ of a matrix 597 algebra $A$ is completely reducible. To do that show that the complement 598 $W^{\perp}$ of an $A$-submodule $W\subset H$ is also an $A$-submodule 599 of $H$. 600 }\newline 601 602 $A\simeq \bigoplus_{i=1}^N M_{n_i}(\mathbb{C})$ is the matrix algebra 603 then $H$ is a Hilbert $A$-bimodule and $W$ a submodule of $A$. 604 Because we have $H = W \cup W^{\perp}$, then $W^{\perp}$ is naturally a 605 $A$-submodule, because elements in $W^{\perp}$ need to satisfy the 606 bimodularity. 607 \end{MyExercise} 608 \begin{definition} 609 A $\Lambda$-decorated graph is given by an ordered pair $(\Gamma, 610 \Lambda)$ of a finite graph $\Gamma$ and a set of positive integers 611 $\Lambda$ with the labeling 612 \begin{itemize} 613 \item of the vetices $v\in \Gamma ^{(0)}$ given by $n(\nu) \in 614 \Lambda$ 615 \item of the edges $e = (\nu _1, \nu _2) \in \Gamma ^{(1)}$ by 616 operators 617 \begin{itemize} 618 \item $D_e: \mathbb{C}^{n(\nu _1)} \rightarrow 619 \mathbb{C}^{n(\nu _2)}$ 620 \item and $D_e^*: \mathbb{C}^{n(\nu _2)} \rightarrow 621 \mathbb{C}^{n(\nu _1)}$ its conjugate traspose 622 (pullback?) 623 \end{itemize} 624 \end{itemize} 625 such that 626 \begin{equation} 627 n(\Gamma ^{(0)}) = \Lambda 628 \end{equation} 629 \end{definition} 630 \begin{question} 631 Would then $D_e$ be the pullback? 632 \end{question} 633 \begin{question} 634 These graphs are important in the next chapter I should look 635 into it more, I don't understand much here, specific 636 how to construct them with the abstraction of a spectral triple... 637 \end{question} 638 639 The operator $D_e$ between $\textbf{n}_i$ and $\textbf{n}_j$ add up to 640 $D_{ij}$ 641 \begin{align} 642 D_{ij} = \sum\limits_{\substack{e = (\nu _1, \nu _2) \\ n(\nu _1) = 643 \textbf{n}_i \\ n(\nu _2) = \textbf{n}_j}} D_e 644 \end{align} 645 646 \begin{theorem} 647 There is a on to one correspondence between finite spectral triples 648 modulo unitary equivalence and $\Lambda$-decorated graphs, given by 649 associating a finite spectral triples $(A, H, D)$ to a $\Lambda$ decorated 650 graph $(\Gamma, \Lambda)$ in the following way: 651 \begin{equation} 652 A = \bigoplus _{n\in \Lambda} M_n(\mathbb{C}); \;\;\; 653 H = \bigoplus _{\nu \in \Gamma ^{(0)}} \mathbb{C}^{n(\nu)}; \;\;\; 654 D = \sum _{e \in \Gamma ^{(1)}} D_e + D_e^* 655 \end{equation} 656 \end{theorem} 657 \begin{figure}[h!] 658 \centering 659 \begin{tikzpicture}[ 660 mass/.style = {draw,circle, minimum size=0.3cm, inner sep=0pt, thick}, 661 ] 662 663 \node[mass, label={\textbf{n}}] (m1) at (1,0) {}; 664 \draw (m1) to [out=330, in=210, looseness=25] node[above] {$D_e$} (m1); 665 \end{tikzpicture} 666 \caption{A $\Lambda$-decorated Graph of $(M_n(\mathbb{C}), \mathbb{C}^n, 667 D = D_e + D_e^*)$} 668 \end{figure} 669 670 \begin{MyExercise} 671 \textbf{ 672 Draw a $\Lambda$ decorated graph corresponding to the spectral triple 673 $(A=\mathbb{C}^3, H=\mathbb{C}^3, D=\begin{pmatrix}0 & \lambda & 0\\ 674 \bar{\lambda} &0 &0 \\ 0&0&0\end{pmatrix})$ 675 }\newline 676 677 \centering 678 \begin{tikzpicture}[ 679 mass/.style = {draw,circle, minimum size=0.4cm, inner sep=0pt, thick}, 680 spring/.style = {decorate,decoration={zigzag, pre length=1cm,post length=1cm,segment length=5pt}},] 681 \node[mass] (m1) at (-1,1.5) {\textbf{1}}; 682 \node[mass] (m2) at (1,1.5) {\textbf{2}}; 683 \node[mass] (m3) at (3,1.5) {\textbf{3}}; 684 685 \draw[style=thick, -] (1.1,1.7) -- (-1.1,1.7); 686 \draw[style=thick, -] (1.1,1.3) -- (-1.1,1.3); 687 \end{tikzpicture} 688 % \captionof{figure}{Solution} 689 \end{MyExercise} 690 \begin{MyExercise} 691 \textbf{ 692 Use $\Lambda$-decorated graphs to classify all finite spectral triples 693 (modulo unitary equivalence) on the matrix algebra 694 $A=\mathbb{C}\oplus M_2(\mathbb{C})$ 695 }\newline 696 697 \centering 698 \begin{tikzpicture}[ 699 mass/.style = {draw,circle, minimum size=0.4cm, inner sep=0pt, thick}, 700 spring/.style = {decorate,decoration={zigzag, pre length=1cm,post length=1cm,segment length=5pt}},] 701 \node[mass] (m1) at (-1,1) {\textbf{1}}; 702 \node[mass] (m2) at (1,1) {\textbf{2}}; 703 \node[mass] (m3) at (3,1) {\textbf{3}}; 704 705 \node[mass] (m4) at (-1,0) {\textbf{1}}; 706 \node[mass] (m5) at (1,0) {\textbf{2}}; 707 \node[mass] (m6) at (3,0) {\textbf{3}}; 708 709 \node[mass] (m7) at (-1,-1) {\textbf{1}}; 710 \node[mass] (m8) at (1,-1) {\textbf{2}}; 711 \node[mass] (m9) at (3,-1) {\textbf{3}}; 712 713 \node[mass] (m10) at (-1,-2) {\textbf{1}}; 714 \node[mass] (m11) at (1,-2) {\textbf{2}}; 715 \node[mass] (m12) at (3,-2) {\textbf{3}}; 716 717 \draw[style=thick, -] (1.1,0.2) -- (-1.1,0.2); 718 \draw[style=thick, -] (1.1,-0.2) -- (-1.1,-0.2); 719 \draw[style=thick, -] (m7) to [out=330, in=210, looseness=10] node[above] {} (m7); 720 \draw[style=thick, -] (m10) -- (m11) ; 721 722 \end{tikzpicture} 723 % \captionof{figure}{Solution $A=M_3(\mathbb{C})$} 724 \end{MyExercise} 725 \subsubsection{Graph Construction of Finite Spectral Triples} 726 \textbf{Algebra:}We know if a acts on a finite dimensional Hilbert space then 727 this C* algebra is isomorphic to a matrix algebra so $A \simeq 728 \bigoplus_{i=1}^{N}M_{n_i}(\mathbb{C})$. Where $i\in 729 \hat{A}$ represents an equivalence class and runs from $1$ to $N$, 730 thus $\hat{A}\simeq\{1,\dots, N\}$. We label equivalence classes by 731 $\textbf{n}_i$, then $\hat{A}\simeq\{\textbf{n}_1,\dots,\textbf{n}_N\}$. 732 \newline 733 734 \textbf{Hilbert Space:} Since every Hilbert space that acts faithfully on a 735 C* algebra is completely reducible, it is isomorphic to the composition 736 of irreducible representations. $H \simeq \bigoplus_{i=1}^N\mathbb{C}^{n_i} 737 \otimes V_i$. Where all $V_i$'s are Vector spaces, their dimension is the 738 multiplicity of the representation landed by $\textbf{n}_i$ to $V_i$ itself 739 by the multiplicity space. 740 \newline 741 742 \textbf{Finite Dirac Operator:} $D_{ij}$ is connecting nodes $\textbf{n}_i$ 743 and $\textbf{n}_j$, with a symmetric map $D_{ij}:\mathbb{C}^{n_i}\otimes V_i 744 \rightarrow \mathbb{C}^{n_j}\otimes V_j$ 745 \newline 746 747 To draw a graph, draw nodes in position $\textbf{n}_i\in \hat{A}$. 748 Multiple nodes at the same position represent multiplicities in $H$. 749 Draw lines between nodes to represent $D_{ij}$. 750 751 \begin{figure}[h!] 752 \centering 753 \begin{tikzpicture} 754 \node[draw, label=above:{$\textbf{n}_1$},circle, thick] at (-3,0) {}; 755 \node[label=above:{$\dots$}] at (-2,0) {}; 756 \node[draw, label=above:{$\textbf{n}_i$},circle, thick] at (-1,0) {}; 757 \node[label=above:{$\dots$}] at (0,0) {}; 758 \node[draw, label=above:{$\textbf{n}_j$},circle, thick] at (1,0) {}; 759 \node[draw, label=above:{},circle, thick, inner sep=0cm, minimum 760 size=0.2cm] at (1,0) {}; 761 \node[label=above:{$\dots$}] at (2,0) {}; 762 \node[draw, label=above:{$\textbf{n}_N$},circle, thick] at (3,0) {}; 763 764 \draw[style=thick, -] (-1,-0.2) -- (1,-0.2); 765 \draw[style=thick, -] (-1,0.2) -- (1,0.2); 766 \path[style=thick, -] (-1,-0.2) edge[bend right=15] 767 node[pos=0.5,below] {} (3,-0.2); 768 \end{tikzpicture} 769 \caption{Example} 770 \end{figure} 771 772 773 \end{document}