week7.tex (25325B)
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{pst-node} 18 \usepackage{tikz-cd} 19 20 \usepackage{tikz} 21 \usetikzlibrary{patterns,decorations.pathmorphing,positioning} 22 23 \usepackage[framemethod=TikZ]{mdframed} 24 25 \tikzstyle{titlered} = 26 [draw=black, thick, fill=white,% 27 text=black, rectangle, 28 right, minimum height=.7cm] 29 30 \newcounter{exercise} 31 32 \renewcommand*\theexercise{Exercise~\arabic{exercise}} 33 34 \makeatletter 35 \mdfdefinestyle{exercisestyle}{% 36 outerlinewidth=1em,% 37 outerlinecolor=white,% 38 leftmargin=-1em,% 39 rightmargin=-1em,% 40 middlelinewidth=1.2pt,% 41 roundcorner=5pt,% 42 linecolor=black,% 43 backgroundcolor=blue!5, 44 innertopmargin=1.2\baselineskip, 45 skipabove={\dimexpr0.5\baselineskip+\topskip\relax}, 46 skipbelow={-1em}, 47 needspace=3\baselineskip, 48 frametitlefont=\sffamily\bfseries, 49 settings={\global\stepcounter{exercise}}, 50 singleextra={% 51 \node[titlered,xshift=1cm] at (P-|O) % 52 {~\mdf@frametitlefont{\theexercise}~};},% 53 firstextra={% 54 \node[titlered,xshift=1cm] at (P-|O) % 55 {~\mdf@frametitlefont{\theexercise}~};}, 56 } 57 \makeatother 58 59 \newenvironment{MyExercise}% 60 {\begin{mdframed}[style=exercisestyle]}{\end{mdframed}} 61 62 \theoremstyle{definition} 63 \newtheorem{definition}{Definition} 64 65 \theoremstyle{definition} 66 \newtheorem{question}{Question} 67 68 \theoremstyle{definition} 69 \newtheorem{example}{Example} 70 71 \theoremstyle{theorem} 72 \newtheorem{theorem}{Theorem} 73 74 \theoremstyle{theorem} 75 \newtheorem{lemma}{Lemma} 76 77 \theoremstyle{theorem} 78 \newtheorem{proposition}{Proposition} 79 80 \newtheorem*{idea}{Proof Idea} 81 82 \title{University of Vienna\\ Faculty of Physics\\ \vspace{1.25cm} 83 Notes on\\ Noncommutative Geometry and Particle Phyiscs} 84 \author{Milutin Popovic \\ Supervisor: Dr. Lisa 85 Glaser} 86 \date{Week 7: 23.04 - 27.04} 87 88 \begin{document} 89 90 \maketitle 91 \tableofcontents 92 \newpage 93 \section{Classification of Finite Real Spectral Triples} 94 95 Here we classify finite real spectral triples modulo unitary equivalence with 96 \textit{Krajewski Diagrams}. We extend $\Lambda$-decorated graphs to the case of 97 real spectral triples (grading and real structure). 98 99 \textbf{The Algebra:}Like before: 100 \begin{align} 101 A\simeq \bigoplus_{i=1}^N M_{n_i}(\mathbb{C}) \;\;\;\;\;\;\; \text{with} \;\;\; \hat{A} = \{\textbf{n}_1, \dots, \textbf{n}_N\} 102 \end{align} 103 Where $\textbf{n}_i$ are irreducible representation of $A$ on 104 $\mathbb{C}^{n_i}$ 105 106 \textbf{The Hilbertspace:}Faithful irreducible representation on $A$ are the 107 direct sum of $\mathbb{C}^{n_i}$'s, which act on $A$ by left block-diagonal 108 matrix multiplication. 109 \begin{align} 110 \bigoplus_{i=1}^N \mathbb{C}^{n_i} 111 \end{align} 112 Furthermore we need a representation of $A^\circ$ on $H$ that commutes with 113 $A$. That is 114 \begin{align} 115 A^\circ \simeq &\bigoplus_{i=1}^N M_{n_i}(\mathbb{C})^\circ \\ 116 \text{with} \;\;\; &\hat{A}^\circ = \{\textbf{n}_1^\circ, \dots, 117 \textbf{n}_N^\circ\} \\ 118 \text{and} \;\;\; &\bigoplus_{i=1}^N \mathbb{C}^{n_i\circ} 119 \end{align} 120 And we need the multiplicity space $V_{ij}$ of $\mathbb{C}^{n_i} \otimes 121 \mathbb{C}^{n_j\circ}$. 122 Thus making the Hilbertspace: 123 \begin{align} 124 H=\bigoplus_{i,j=1}^N \mathbb{C}^{n_i} \otimes \mathbb{C}^{n_j\circ} 125 \otimes V_{ij} 126 \end{align} 127 \begin{itemize} 128 \item $\textbf{n}_i$, $\textbf{n}_j^\circ$ form a grid 129 \item if there is a node at $(\textbf{n}_i$, $\textbf{n}_j^\circ)$ then 130 $\mathbb{C}^{n_i} \otimes \mathbb{C}^{n_j\circ}$ is nonzero in $H$. 131 \item multiplicity implies multiple nodes 132 \end{itemize} 133 134 \begin{example} 135 $A = \mathbb{C} \oplus M_2 (\mathbb{C})$, two options of the Hilbertspace. 136 \begin{figure}[h!] \centering 137 \begin{tikzpicture}[ 138 dot/.style = {draw, circle, inner sep=0.06cm}, 139 no/.style = {}, 140 ] 141 \node[no](a) at (0,0) [label=left:$\textbf{1}^\circ$] {}; 142 \node[no](b) at (0, -1) [label=left:$\textbf{2}^\circ$] {}; 143 \node[no](c) at (1, 0.5) [label=above:$\textbf{1}$] {}; 144 \node[no](d) at (2, 0.5) [label=above:$\textbf{2}$] {}; 145 \node[dot](d0) at (1,0) [] {}; 146 \node[dot](d0) at (2,-1) [] {}; 147 148 \node[no](a1) at (6,0) [label=left:$\textbf{1}^\circ$] {}; 149 \node[no](b2) at (6, -1) [label=left:$\textbf{2}^\circ$] {}; 150 \node[no](c2) at (7, 0.5) [label=above:$\textbf{1}$] {}; 151 \node[no](d2) at (8, 0.5) [label=above:$\textbf{2}$] {}; 152 \node[dot](d0) at (7,0) [] {}; 153 \node[dot](d0) at (8,0) [] {}; 154 155 156 \end{tikzpicture} 157 \end{figure} 158 159 The first diagram corresponds to $H_1 = \mathbb{C} \oplus M_2(\mathbb{C})$, 160 to the second $H_2 = \mathbb{C} \oplus \mathbb{C}^2$. 161 \end{example} 162 163 \begin{MyExercise} 164 \textbf{Let $J$ be an anti-unitary operator on a finite-dimensional Hilbert space. 165 Show that $J^2$ is an unitary operator 166 } \newline 167 168 Straight forward, say $J:\; H \rightarrow H$, then let $\xi_1, \xi_2 \in 169 H$: 170 171 \centering 172 \begin{align} 173 <J^2 \xi_1, J^2 \xi_2> &= <J(J\xi_1), J(J\xi_2)> =\\ 174 &= <J\xi_2, J\xi_1> = <\xi_1, \xi_2> 175 \end{align} 176 \end{MyExercise} 177 178 \textbf{The real Structure:} $J:\; H \rightarrow H$. 179 \begin{lemma} 180 \label{lemma} 181 Let $J$ be an anti-unitary operator on a finite-dimensional Hilbertspace 182 $H$ with $J^2 = \pm 1 $ 183 \begin{enumerate} 184 \item If $J^2 = 1 \;\; \Rightarrow \;\; \exists$ an ONB $\{e_k\}$ of $H$\\ 185 with $Je_k = e_k$. 186 \item If $J^2 = -1 \;\; \Rightarrow \;\; \exists$ an ONB $\{e_k, f_k\}$ of $H$\\ 187 with $Je_k = f_k$ and consequently $Jf_k = -e_k$. 188 \end{enumerate} 189 \end{lemma} 190 \begin{proof} 191 \textbf{1.} $J^2 = 1$\newline 192 193 $v\in H$ and set: 194 \begin{align} 195 e_1 := 196 \begin{cases} 197 c (v + Jv)\;\;\; \text{if}\;\;\; Jv \neq -v \\ 198 iv\;\;\; \text{if}\;\;\; Jv = -v 199 \end{cases} 200 \end{align} 201 Where $c$ is a normalization constant, then take $Je_1$ 202 \begin{align} 203 &J(v + Jv) = Jv + J^2v= v + Jv \;\;\;\; \text{and} \\ 204 &J(iv) = -iJv = iv\\ 205 &\Rightarrow Je_1 = e_1 206 \end{align} 207 Take $v'\perp e_1$ making: 208 \begin{align} 209 <e_1 , Jv'> = <J^2 v', Je_1> = <v' , Je_1>= <v', e_1> =0 210 \end{align} 211 Construct $e_2 \perp e_1$ with $v'$: 212 \begin{align} 213 e_2 := 214 \begin{cases} 215 c (v' + Jv')\;\;\; \text{if}\;\;\; Jv' \neq -v' \\ 216 iv'\;\;\; \text{if}\;\;\; Jv' = -v' 217 \end{cases} 218 \end{align} 219 Do this $k$ times and get $\{e_k\}$ ONB of $H$ for $J^2 = 1$. 220 \newline 221 222 \textbf{2.} $J^2 = -1$\newline 223 $v \in H$ and set $e_1 = cv$, $c$ normalization constant. 224 Then we set $f_1 = Je_1$ with $f_1 \perp e_1$, this is automatically the 225 case because: 226 \begin{align} 227 <f_1, e_1> &= <Je_1, e_1> = -<Je_1 , J^2e_1> =\\ 228 &= -<Je_1, e_1> = -<f_1, e_1> 229 \end{align} 230 this only holds for 0. Then take some $v' \perp e_1, f_1$ and set\\ 231 $e_2 =c 'v'$ and $f_2 = Je_2 \perp e_2, f_1, e_1$. 232 \begin{align} 233 &<e_1, f_2> = <e_1, Je_2> = -<J^2e_1, Je_2> = -<e_2, Je_1> = -<e_2, 234 f_1>=0\\ 235 &<f_1, f_2> = <Je_1, Je_2> = <e_2, e_1> = 0. 236 \end{align} 237 Do this $k$ times and get $\{e_k, f_k\}$ ONB of $H$ for $J^2 = -1$ 238 239 \end{proof} 240 241 Apply Lemma \ref{lemma} to the real structure $J$ on a spectral triple. $J$ 242 implements right action of $A$ on $H$ with 243 \begin{align} 244 a^\circ = Ja^* J^{-1} 245 \end{align} 246 and satisfying $[a, b^\circ]=0$. With the block form of $A$, this implies 247 \begin{align} 248 J(a^*_1 \oplus \cdots \oplus a_N^*) = (a^\circ_1 \oplus \cdots \oplus 249 a_N^\circ)J. 250 \end{align} 251 With this we can conclude that the Krajewski diagram for a real finite spectral 252 triple is symmetric along the diagonal.$J$ hast then the following bilinear 253 mapping: 254 \begin{align} 255 J:\;\; \mathbb{C}^{n_i} \otimes \mathbb{C}^{n_j\circ} \otimes V_{ij} 256 \rightarrow \mathbb{C}^{n_j} \otimes \mathbb{C}^{n_i\circ} \otimes V_{ji}. 257 \end{align} 258 259 \begin{proposition} 260 \label{proposition} 261 Let $J$ be a real structure on a finite real spectral triple $(A, H , D; 262 J)$. 263 \begin{enumerate} 264 \item If $J^2 = 1$ (K0-dimension 0, 1, 6, 7) $Rightarrow \;\; \exists$ 265 an ONB $\{e_k^{(ij)}\}$\\ with $e_k^{(ij)} \in \mathbb{C}^{n_i} \otimes 266 \mathbb{C}^{n_j\circ} \otimes V_{ij}$ such that 267 \begin{align} 268 Je_k^{(ij)} = e_k^{(ij)} \;\;\; (i, j = 1,\dots,N;\; k=1,\dots 269 dim(V_{ij})) 270 \end{align} 271 \item If $J^2 = -1$ (KO-dimension 2, 3, 4, 5) $\Rightarrow \;\; \exists$ 272 ONB $\{e_k^{(ij)}, f_k^{(ji)}\}$ \\ 273 with $e_k^{(ij)} \in \mathbb{C}^{n_i} \otimes \mathbb{C}^{n_j\circ} 274 \otimes V_{ij}$ and $f_k^{(ji)} \in \mathbb{C}^{n_j} \otimes 275 \mathbb{C}^{n_i\circ} \otimes V_{ji}$ such that 276 \begin{align} 277 Je_k^{(ij)} = f_k^{(ji)} \;\;\; (i\leq j=1,\dots, N;\; 278 k=1,\dots,dim(V_{ji})). 279 \end{align} 280 \end{enumerate} 281 \end{proposition} 282 \begin{proof} 283 Similar to Lemma \ref{lemma}. 284 \end{proof} 285 286 For whatever unknown reasons this implies that in the case of KO-dimension 2, 287 3, 4, 5, diagonals $H_ii$ need to have even multiplicity. 288 289 \textbf{The finite Dirac Operator:} Is a mapping between $H_{ij}$ to $H_{kl}$ 290 291 \begin{align} 292 D_{ij,kl}: \; \mathbb{C}^{n_i} \otimes 293 \mathbb{C}^{n_j\circ}\otimes V_{ij} \rightarrow \mathbb{C}^{n_k} \otimes 294 \mathbb{C}^{n_l\circ}\otimes V_{kl} 295 \end{align} 296 We have $D_{kl,ij} = D^*_{ij, kl}$. And in the diagram we have a line between 297 the nodes $(\textbf{n}_i, \textbf{n}_j^\circ)$ and $(\textbf{n}_l, 298 \textbf{n}_k^\circ)$. But instead of drawing directional lines draw a single 299 undirected line that represents both $D_{ij, kl}$ and the adjoint $D_{kl, ij}$. 300 301 \begin{lemma} 302 The conditions $JD = \pm DJ$ and $[[D,a], b^\circ] = 0$ imply that the 303 connections in the diagram run only vertically or horizontally and thereby 304 the diagonal symmetry between the nodes is preserved. 305 \end{lemma} 306 \begin{proof} 307 The condition $JD = \pm DJ$ has the following commutative diagram. 308 309 \[ 310 \begin{tikzcd} 311 \mathbb{C}^{n_i\circ}\otimes \mathbb{C}^{n_j\circ}\otimes V_{ij} 312 \arrow[r,"D"] \arrow[d,swap,"J"] & 313 \mathbb{C}^{n_k\circ}\otimes \mathbb{C}^{n_l\circ}\otimes V_{kl} \arrow[d,"J"] \\ 314 \mathbb{C}^{n_j\circ}\otimes \mathbb{C}^{n_i\circ}\otimes V_{ji} \arrow[r,"\pm D"] & 315 \mathbb{C}^{n_l\circ}\otimes \mathbb{C}^{n_k\circ}\otimes V_{lk} 316 \end{tikzcd} 317 \] 318 Relating $D_{ij, kl}$ to $D_{ji, lk}$ and maintaining diagonal symmetry. 319 Wit the condition $[[D, a], b^\circ]=0$ for the diagonal elements $a = 320 \lambda_1\mathbb{I}_{n_1}\oplus \cdots \oplus \lambda_N \mathbb{I}_{n_N} 321 \in A$ and $b = \mu_1\mathbb{I}_{n_1}\oplus \cdots \oplus \mu_N 322 \mathbb{I}_{n_N} \in A$, with some $\lambda _i , \mu _i \in \mathbb{C}$, we 323 can commute: 324 \begin{align} 325 D_{ij, kl} (\lambda _i - \lambda _k)(\bar{\mu}_j - \bar{\mu}_l)= 0 326 \end{align} 327 $\forall \lambda _i , \mu _j \in \mathbb{C}$, thus $D_ij, kl = 0$ for 328 $i\neq j$ or $j\neq i $. 329 \end{proof} 330 \textbf{The Grading:} $\gamma : \; H \rightarrow H$ each node gets labeled by a $+$ 331 or a $-$ sign. 332 333 \begin{itemize} 334 \item D only connects nodes with different signs 335 \item If $(\textbf{n}_i, \textbf{n}_j^\circ)$ has a $\pm$ sing then 336 $(\textbf{n}_j, \textbf{n}_i^\circ)$ has a $\mp$, $\varepsilon''$ sign\\ 337 according to $J\gamma = \varepsilon'' \gamma J$ 338 \end{itemize} 339 340 \begin{definition} 341 A Krajewski Diagram of KO-dimension $k$ is an ordered pair $(\Gamma, 342 \Lambda)$ where $\Gamma$ is a finite graph and $\Lambda$ is a set of 343 positive integers with a labeling: 344 345 \begin{itemize} 346 \item of $v \in \Gamma^{(0)}$ of vertices by elements $\iota (v) = 347 (n(v), m(v))\; \in \; \Lambda \times \Lambda$, an edge from $v$ to 348 $v'$ implies that either $n(v) = n(v')$ or $m(v) = m('v)$ or both 349 \item of $e = (v_1, v_2) \in \Gamma^{(1)}$ edges with non-zero 350 operators $D_e$ and their adjoints $D_e^*$: 351 \begin{align} 352 &D_e:\mathbb{C}^{n(v_1)} \rightarrow 353 \mathbb{C}^{n(v_2)}\;\;\;\;\; &\text{if} \;\;\;\; m(v_1) = m(v_2)\\ 354 &D_e:\mathbb{C}^{m(v_1)} \rightarrow 355 \mathbb{C}^{m(v_2)}\;\;\;\;\; &\text{if} \;\;\;\; n(v_1) = n(v_2) 356 \end{align} 357 \end{itemize} 358 Together with an involutive graph automorphism $j:\Gamma \Rightarrow 359 \Gamma$ such that the following conditions hold: 360 \begin{enumerate} 361 \item every row or column in $\Gamma \times \Gamma$ has non-empty 362 intersection with $\iota(\Gamma)$ 363 \item for each vertex $v$ we have $n(j(v)) = m(v)$ 364 \item for each edge $e$ we have $D_e = \epsilon' D_{j(e)}$ 365 \item if KO dimension $k$ is even, then the vertices are labeled by 366 $\pm 1$ and the edges only connect opposite signs. The signs at $v$ 367 and $j(v)$ differ by a factor of $\epsilon$ 368 \item if the K0-dimension is 2, 3, 4, 5 then the inverse image of 369 $\iota$ of the diagonal elements in $\Lambda \times \Lambda$ 370 contains an even number of vertices of $\Gamma$ 371 \end{enumerate} 372 \end{definition} 373 With this definition we can label different vertices by the same element in 374 $\Lambda \times \Lambda$ (accounting for the multiplicities in $V_{ij}$) 375 \newline 376 377 \textbf{Diagram:} To sum it up we have the following diagram 378 \begin{itemize} 379 \item Node at $(\textbf{n}_i, \textbf{n}_j^\circ)$ for each vertex with that label 380 \item Operators $D_e$ add up to $D_{ij,kl}$ connecting nodes $(\textbf{n}_i, 381 \textbf{n}_j^\circ)$ with $(\textbf{n}_k, \textbf{n}_l^\circ)$ 382 \begin{align} 383 D_{ij, kl} = \sum\limits_{\substack{e=(v_1, v_2) \in \Gamma^{(1)} 384 \\ \iota(v_1) = (\textbf{n}_i, \textbf{n}_j)\\ 385 \iota(v_2)=(\textbf{n}_k, \textbf{n}_l)}} D_e 386 \end{align} 387 \item only vertical or horizontal connections 388 \end{itemize} 389 390 \begin{theorem} 391 There is a one-to-one correspondence between finite real spectral triples 392 $(A, H, D; J, \gamma)$ 393 of K0-dimension $k$ modulo unitary equivalence and Krajewski diagrams of 394 KO-dimension $k$ in the following way: 395 396 \begin{align} 397 & A = \bigoplus_{n \in \Lambda} M_n(\mathbb{C})\\ 398 & H = \bigoplus_{v \in \Gamma^{(0)}} \mathbb{C}^{n(v)} \otimes 399 \mathbb{C}^{m(v)\circ}\\ 400 & D = \sum_{e\in \Gamma^{(1)}} D_e + D_e^* 401 \end{align} 402 The real structure $J:H\rightarrow H$ is given as as in Proposition 403 \ref{proposition} with a basis dictated by a graph automorphism $j: \Gamma 404 \rightarrow \Gamma$. The grading $\gamma$ is difened by setting $\gamma = 405 \pm 1$ on $\mathbb{C}^{n(v)} \otimes \mathbb{C}^{m(v)\circ} \subset H$ 406 according to the labeling $\pm$ of the vertex $v$. 407 \end{theorem} 408 409 \begin{example} 410 $A = M_n(\mathbb{C})$ with $\hat{A} = {\textbf{n}}$. We have the following 411 Krajewski diagram. 412 \begin{figure}[h!] \centering 413 \begin{tikzpicture}[ 414 dot/.style = {draw, circle, inner sep=0.06cm}, 415 no/.style = {}, 416 ] 417 \node[no](a) at (0,0) [label=left:$\textbf{n}^\circ$] {}; 418 \node[no](c) at (0.25, 0.25) [label=above:$\textbf{n}$] {}; 419 \node[dot](d0) at (0.25,0) [] {}; 420 \end{tikzpicture} 421 \end{figure} 422 \begin{itemize} 423 \item We can label the node either with a $+$ or a $-$ sign, the choice being 424 irrelevant 425 \item $H = \mathbb{C}^n \otimes \mathbb{C}^{n\circ} \simeq 426 M_n(\mathbb{C})$ 427 \item $\gamma$ trivial grading ($+1$) 428 \item $J$ is a combination of complex conjugation and the flip 429 $n\otimes n^\circ$ ($\Rightarrow M_n(\mathbb{C})$ as matrix adjoint) 430 \item Because node label is $\pm$ there is no non-zero Dirac operator 431 \item $\Rightarrow (A = M_n(\mathbb{C}), H=M_n(\mathbb{C}) , D=0; J=(\cdot)^*, 432 \gamma = 1)$ 433 \end{itemize} 434 \end{example} 435 \section{Real Algebras and Krajewski Diagrams} 436 437 \begin{definition} 438 A real Algebra is a Vector space $A$ over $\mathbb{R}$ with $A\times A 439 \rightarrow A$, $(a, b) \mapsto ab$ and $1a = a1 = a \;\; \forall a\in A$ 440 \end{definition} 441 442 A real *-algebra is a real algebra with a bilinear map $*:A \rightarrow A$ 443 such that $(ab)^* = b^*a^*$ and $(a^*)^* \;\;\; \forall a,b\in A$ 444 \begin{example} 445 Real *-algebra of quaternions $\mathbb{H}$ subalgebra of $M_2(\mathbb{C})$. 446 \begin{align} 447 \mathbb{H} = \{ \begin{pmatrix}\alpha & \beta \\ -\bar{\beta} & 448 \bar{\alpha}\end{pmatrix} : \alpha, \beta \in 449 \mathbb{C}\} 450 \end{align} 451 $\mathbb{H}$ consists of matrices that commute in $M_2(\mathbb{C})$ with 452 the operator $I$ defined by: 453 \begin{align} 454 I\begin{pmatrix}v_1 \\ v_2\end{pmatrix} = \begin{pmatrix}-\bar{v}_2 \\ 455 \bar{v}_1\end{pmatrix} 456 \end{align} 457 The involution is the hermitian conjugation of $M_2(\mathbb{C})$. 458 \end{example} 459 \begin{MyExercise} 460 \textbf{ 461 \begin{enumerate} 462 \item Show that $\mathbb{H}$ is a real *-algebra which contains a 463 real subalgebra isomorphic to $\mathbb{C}$. 464 \item Show that $\mathbb{H} \otimes_\mathbb{R} \mathbb{C} \simeq 465 M_2(\mathbb{C})$ as complex *-algebras. 466 \item Show that $M_k(\mathbb{H})$ is areal *-algebra for any $k$ 467 \item Show that $M_k(\mathbb{H} \otimes_{\mathbb{R}} \mathbb{C} 468 \simeq M_{2k}(\mathbb{C})$ as complex *algebras. 469 \end{enumerate} 470 } 471 1). Let us take some $a, b \in \mathbb{H}$ with 472 \begin{align} 473 a = \begin{pmatrix} 474 \alpha & \beta \\ 475 -\bar{\beta} & \bar{\alpha} 476 \end{pmatrix} \;\;\;\; 477 b = \begin{pmatrix} 478 \gamma & \delta \\ 479 -\bar{\delta} & \bar{\gamma} 480 \end{pmatrix} 481 \end{align} 482 where $\alpha, \beta, \gamma, \delta \in \mathbb{C}$. Since 483 $\mathbb{H}$ is represented in standard $2x2$ matrices, the involution 484 is just subsequent from there, the only thing left to show is the 485 closure $ab \in \mathbb{H}$. 486 \begin{align} 487 ab &= 488 \begin{pmatrix} 489 \alpha & \beta \\ 490 -\bar{\beta} & \bar{\alpha} 491 \end{pmatrix} 492 \begin{pmatrix} 493 \gamma & \delta \\ 494 -\bar{\delta} & \bar{\gamma} 495 \end{pmatrix} =\\ 496 &= 497 \begin{pmatrix} 498 \alpha\beta - \beta\bar{\delta}& \alpha\delta + \beta \bar{\gamma}\\ 499 -(\bar{\alpha}\bar{\delta} + \bar{\beta}\gamma) & 500 \bar{\alpha}\gamma-\bar{\beta}\delta 501 \end{pmatrix} = 502 \begin{pmatrix} 503 \xi& \psi\\ 504 -\bar{\psi} & \bar{\xi} 505 \end{pmatrix} \in \mathbb{H} 506 \end{align} 507 where $\xi, \psi \in \mathbb{C}$ because of closure of complex numbers 508 in regards to multiplication and addition, which is $\mathbb{R} 509 \otimes_{\mathbb{R}}\mathbb{C} \simeq \mathbb{C}$, e.g. $\beta \cdot c 510 \in \mathbb{C}$ with $c \in \mathbb{C}$. 511 \newline 512 2)For $\mathbb{H}\otimes_\mathbb{R} \mathbb{C}$ we have for some $h \in 513 \mathbb{H}$ and $c \in mathbb{C}$ 514 \begin{align} 515 h\otimes c &= 516 \begin{pmatrix} 517 \alpha & \beta \\ 518 -\bar{\beta} & \bar{\alpha} 519 \end{pmatrix}\otimes c = \\ 520 &= 521 \begin{pmatrix} 522 \alpha c & \beta c \\ 523 -\bar{\beta} c & \bar{\alpha} c 524 \end{pmatrix} \simeq M_2(\mathbb{C}) 525 \end{align} 526 because again of $\mathbb{R} \otimes_\mathbb{R} \mathbb{C} \simeq 527 \mathbb{C}$. 528 \newline 529 3)We know that $\mathbb{H}$ is a real subalgebra of $M_2(\mathbb{C})$, 530 so $M_k(\mathbb{H})$ is just an extension and an real subalgebra of 531 $M_{2k}(\mathbb{C})$. 532 \newline 533 4) Here we use what we have learned 534 \begin{align} 535 M_k(\mathbb{H})\otimes_\mathbb{R} \mathbb{C} \simeq 536 M_k(M_2(\mathbb{C})) = M_{2k}(\mathbb{C}) 537 \end{align} 538 \end{MyExercise} 539 \begin{definition} 540 A representation of a finite-dimensional real * algebra $A$ is a pair $(\pi 541 , H$), $H$- Hilbertspace, $\pi : A \rightarrow L(H)$ 542 \end{definition} 543 \begin{MyExercise} 544 Show that there is a one-to-one correspondence between Hilbertspace 545 representations of real *-algebras $A$ and complex representations of its 546 complexification $A\otimes _\mathbb{R} \mathbb{C}$. Conclude that the 547 unique irreducible Hilbertspace representation of $M_k(\mathbb{H})$ is 548 $\mathbb{C}^{2k}$ 549 \end{MyExercise} 550 \begin{lemma} 551 Real *-algebra $A$ represented faithfully on a finite dimensional 552 Hilbertspace $H$ through a real linear *-algebra map $\pi: A \rightarrow 553 L(H)$ hen $A$ is a matrix algebra. 554 \begin{align} 555 A \simeq \bigoplus _{i=1}^N M_{n_i} (\mathbb{F}_i) 556 \end{align} 557 Where $\mathbb{F}_i = \mathbb{R}, \mathbb{C}, \mathbb{H}$ depending on $i$. 558 \end{lemma} 559 \begin{proof} 560 $\pi$ allows $A$ to be considered as a real *-subalgebra of 561 $M_{dim(H)}(\mathbb{C}) \Rightarrow A+iA$ complex *-subalgebra of 562 $M_{dim(H)}(\mathbb{C})$. Then $A+iA$ is a matrix algebra and $A+iA = 563 M_k(\mathbb{C})$ for $k \geq 1$. Thus we have 564 \begin{align} 565 A \cap iA = 566 \begin{cases} 567 \{0\} \;\;\;\; \text{if $A = M_k(\mathbb{C})$}\\ 568 A+iA = M_k(\mathbb{C}) 569 \end{cases} 570 \end{align} 571 Furthermore $A$ is a fixed point algebra of an anti-linear automorphism 572 $\alpha$ of $M_k(\mathbb{C})$ with $\alpha(a+ib) = a-ib$ for $a, b \in A$. 573 Implement $\alpha$ by an anti-linear isometry $I$ on $\mathbb{C}^n$ such 574 that $\alpha (x) = I\times I^{-1}\;\;\;\ \forall x\in M_k(\mathbb{C})$. 575 Now since $\alpha^2 = 1$, $I^2$ commutes with $M_k(\mathbb{C})$ and is 576 proportional to a complex scalar $I^2 = \pm 1 $ and A is the commutant of 577 $I$ 578 \begin{itemize} 579 \item if $I^2 = 1 \;\;\ \Rightarrow \;\; \exists \;\;\ \{e_i\}$ ONB of 580 $\mathbb{C}^k$ with $Ie_i = e_i$, then $A=M_k(\mathbb{R})$ 581 \item if $I^2 = -1 \;\;\ \Rightarrow \;\; \exists \;\;\ \{e_i,f_i\}$ ONB of 582 $\mathbb{C}^k$ with $Ie_i = f_i$, ($k$ even)\\ 583 Therefor $I$ must be a $k/2 \times k/2$ matrix because of commutation with 584 $M_k(\mathbb{C})$, then $A = M_{k/2} (\mathbb{H})$ 585 \end{itemize} 586 \end{proof} 587 The Krajewski diagrams can also classify real algebras, as long as we take 588 $\mathbb{F}_i$ for each $i$ into account. That is we enhance the set $\Lambda$ 589 to be 590 \begin{align} 591 \Lambda = \{ \textbf{n}_1 \mathbb{F}_1,\dots, \textbf{n}_N \mathbb{F}_N\} 592 \end{align} 593 Reducing in to the previous $\Lambda$ if all $\mathbb{F}_i = \mathbb{C}$. 594 \section{Classification of Irreducible Geometries} 595 Classify irreducible real spectral triples based on $M_N(\mathbb{C} \oplus 596 M_N(\mathbb{C})$ for some $N$ 597 \begin{definition} 598 A finite real spectral triple $(A, H, D; J, \gamma)$ is called irreducible 599 if the triple $(A, H, J)$ is irreducible, that is when 600 \begin{enumerate} 601 \item The representation of $A$ and $J$ on $H$ are irreducible 602 \item The action of $A$ on $H$ has a separating vector 603 \end{enumerate} 604 \end{definition} 605 606 \begin{theorem} 607 Let $(A, H, D; J, \gamma)$ be an irreducible finite real spectral triple of 608 KO-dimension 6. Then exists a positive integer $N$ such that $A \simeq 609 M_N(\mathbb{C}) \oplus M_N(\mathbb{C})$. 610 \end{theorem} 611 \begin{proof} 612 Let $(A, H, D; J, \gamma)$ be an arbitrary finite real spectral triple, 613 corresponding to 614 \begin{align} 615 &A = \bigoplus_i^{N} M_{n_i}(\mathbb{C})\\ 616 &H = \bigoplus_{i,j=1}^N \mathbb{C}^{n_i} \otimes 617 \mathbb{C}^{n_j\circ} \otimes V_{ij} 618 \end{align} 619 Remember that each $\mathbb{C}^{n_i} \otimes \mathbb{C}^{n_j}$ is a 620 irreducible representation of $A$. In order for $H$ to support the real 621 structure $J$ we need both $\mathbb{C}^{n_i} \otimes \mathbb{C}^{n_j}$ 622 and $\mathbb{C}^{n_j} \otimes \mathbb{C}^{n_i}$. With Lemma \ref{lemma} 623 with $J^2 = 1$ with multiplicity $dim(V_{ij}) = 1$ we have such a 624 structure. Hence 625 \begin{align} 626 H = \mathbb{C}^{n_i} \otimes \mathbb{C}^{n_j} \oplus \mathbb{C}^{n_j} 627 \otimes \mathbb{C}^{n_i} 628 \end{align} 629 For $i,j \in \{1, \dots, N\}$ 630 \newline 631 632 For the second condition (existence of the separating vector). The 633 representations of $A$ in $H$ are only faithful if $A = M_{n_i}(\mathbb{C}) 634 \oplus M_{n_j}(\mathbb{C})$. The stronger condition applies $n_i = n_j$ 635 then we have $A' \xi = H$ with the commutant of $A$ and $\xi \in H$ the 636 separating vector. Normally since $A' = M_{n_j}(\mathbb{C}) \oplus 637 M_{n_i}(\mathbb{C})$ with $dim(A') = n_i^2 + n_j^2$ and $dim(H) = 2n_i n_j$ 638 we have a equality $n_i = n_j$. 639 \end{proof} 640 \end{document} 641 642 643 644 645