week6.tex (14607B)
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 \date{Week 6: 19.03 - 26.03} 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 6: 19.03 - 26.03} 83 84 85 \begin{document} 86 87 \maketitle 88 \tableofcontents 89 \newpage 90 \section{Finite Real Noncommutative Spaces} 91 \subsection{Finite Real Spectral Triples} 92 Add on to finite real spectral triples a \textit{real structure}. The 93 requirement is that $H$ is a $A$-$A$-bimodule (before only a $A$-left 94 module). 95 \newline 96 97 For this we introduce a $\mathbb{Z}_2$-grading $\gamma$ with 98 \begin{align} 99 &\gamma ^* = \gamma \\ 100 &\gamma ^2 = 1 \\ 101 &\gamma D = - D \gamma\\ 102 &\gamma a = a \gamma \;\;\;\; a\in A 103 \end{align} 104 105 \begin{definition} 106 A \textit{finite real spectral triple} is given by a finite spectral 107 triple $(A, H, D)$ and a anti-unitary operator $J:H\rightarrow H$ called 108 the \textit{real structure}, such that 109 \begin{align} 110 a^\circ := J a^* J^{-1} 111 \end{align} 112 is a right representation of $A$ on $H$, that is $(ab)^\circ = b^\circ 113 a^\circ$. With two requirements 114 \begin{align} 115 &[a, b^\circ] = 0\\ 116 &[[D, a],b^\circ] = 0. 117 \end{align} 118 They are called the \textit{commutant property}, and mean that the left 119 action of an element in $A$ and $\Omega _D^1(A)$ commutes with the right 120 action on $A$. 121 \end{definition} 122 \begin{definition} 123 The $KO$-dimension of a real spectral triple is determined by the sings 124 $\epsilon, \epsilon ' ,\epsilon '' \in \{-1, 1\}$ appearing in 125 \begin{align} 126 &J^2 = \epsilon \\ 127 &JD = \epsilon \ DJ\\ 128 &J\gamma = \epsilon '' \gamma J. 129 \end{align} 130 \end{definition} 131 \begin{table}[h!] 132 \centering 133 \caption{$KO$-dimension $k$ modulo $8$ of a real spectral triple} 134 \begin{tabular}{ c | c c c c c c c c} 135 \hline 136 $k$ & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ 137 \hline 138 $\epsilon$ & 1 & 1 & -1 & -1 & -1 & -1 & 1 & 1 \\ 139 $\epsilon '$ & 1 & -1 & 1 & 1 & 1 & -1 & 1 & 1 \\ 140 $\epsilon ''$ & 1 & & -1 & & 1 & & -1 & \\ 141 \hline 142 \end{tabular} 143 \end{table} 144 145 146 \begin{definition} 147 An opposite-algebra $A^\circ$ of a $A$ is defined to be equal to $A$ as a 148 vector space with the opposite product 149 \begin{align} 150 &a\circ b := ba\\ 151 &\Rightarrow a^\circ = Ja^* J^{-1} \;\;\; \text{defines the left 152 representation of $A^\circ$ on $H$} 153 \end{align} 154 \end{definition} 155 156 157 \begin{example} 158 Matrix algebra $M_N(\mathbb{C})$ acting on $H=M_N(\mathbb{C})$ by left 159 matrix multiplication with the Hilbert Schmidt inner product. 160 \begin{align} 161 \langle a , b \rangle = \text{Tr}(a^* b) 162 \end{align} 163 Then we define $\gamma (a) = a$ and $J(a) = a^*$ with $a\in H$. 164 Since $D$ mus be odd with respect to $\gamma$ it vanishes identically. 165 \end{example} 166 167 \begin{definition} 168 We call $\xi \in H$ \textbf{cyclic vector} in $A$ if: 169 \begin{align} 170 A\xi := { a\xi:\;\; a\in A} = H 171 \end{align} 172 173 We call $\xi \in H$ \textbf{separating vector} in $A$ if: 174 \begin{align} 175 a\xi = 0\;\; \Rightarrow \;\; a=0;\;\;\; a\in A 176 \end{align} 177 \end{definition} 178 179 \begin{MyExercise} 180 \textbf{ 181 In the previous example, show that the right action on $M_N(\mathbb{C})$ 182 on $H = M_N(\mathbb{C})$ as defined by $a \mapsto a^\circ$ 183 is given by right matrix multiplication. 184 }\newline 185 186 \begin{align} 187 a^\circ \xi = J a^* J^{-1}\xi = Ja^* \xi^* = J\xi a=\xi^* a 188 \end{align} 189 \end{MyExercise} 190 \begin{MyExercise} 191 \textbf{ 192 Let $A= \bigoplus _i M_{n_i}(\mathbb{C})$, represented on $H = \bigoplus_i \mathbb{C}^{n_i} 193 \otimes \mathbb{C}^{m_i}$, meaning that the irreducible representation $\textbf{n}_i$ has 194 multiplicity $m_i$. 195 \begin{enumerate} 196 \item Show that the commutant $A'$ of $A$ is $A'\simeq \bigoplus_i M_{m_i} (\mathbb{C})$. As a consequence show $A'' \simeq A$. 197 \item Show that if $\xi$ is a separating vector for $A$ than it is cyclic for $A'$. 198 \end{enumerate} 199 } 200 201 202 \begin{enumerate} 203 \item We know the multiplicity space is $V_i = \mathbb{C}^{m_i}$. We know that 204 for $T\in H$ and 205 $a\in A'$ to work we need $aT=Ta$ by laws of matrix multiplication we need 206 $A' \simeq \oplus _i M_{m_i}(\mathbb{C})$ for this to work since $H = \bigoplus_i 207 \mathbb{C}^{n_i} 208 \otimes \mathbb{C}^{m_i}$ 209 210 \item Suppose $\xi$ is cyclic for $A$ then $A'\xi = \{0\}$. Under the action of $A$ we 211 then have $A'A\xi = AA' \xi = 0 \Rightarrow A' = 0$.\\ 212 Suppose now $\xi$ is separating for $A'$, we have $A'\xi = \{0\}$. We can define a 213 projection in $A'$, $A\xi = P'$. With this projection we have $(1-P')\xi = 0 214 \Rightarrow 1-P' = 0 \Rightarrow A\xi = H$. 215 \end{enumerate} 216 \end{MyExercise} 217 \begin{MyExercise} 218 \textbf{ Suppose $(A, H, D = 0)$ is a finite spectral triple such that $H$ possesses a 219 cyclic and separating vector for $A$. 220 \begin{enumerate} 221 \item Show that the formula $S(a \xi) = a* \xi$ defines a anti-linear operator\\ 222 $S: H \rightarrow H$. 223 \item Show that $S$ is invertible 224 \item Let $J: H \rightarrow H$ be the operator in $S = J \Delta ^{1/2}$ with 225 $\Delta = S^*S$. Show that $J$ is anti-unitary 226 \end{enumerate} 227 } 228 229 230 \begin{enumerate} 231 \item By composition $S(a\xi) = a*\xi$ this is literally anti-linearity. Does this mean 232 $S\xi = \xi$? 233 \item Let $\xi \in H$ be cyclic then: $S(A\xi) = A^*\xi = A\xi = H$. The same has to work 234 for $S^{-1}$ if not then $\xi$ wouldn't exist. $S^{-1}(A^*\xi) = S^{-1}(H) = H$. 235 \item Since $S$ is bijective then $\Delta ^{1/2}$ and $J$ need to be bijective. 236 We also have $J = S \Delta^{-1/2}$ and $\Delta^* = \Delta$\\ 237 Now let $\xi _1 , \xi _2 \in H$ \begin{align} 238 <J \xi _1 , J \xi _2 > &= < J^*J\xi_1 , \xi_2>^* =\\ 239 &= <(\Delta ^{-1/2})^* S^* S \Delta ^{-1/2} \xi_1, \xi_2>^* = \\ 240 &= <(\Delta^{-1/2})^* \Delta \Delta^{-1/2} \xi_1, \xi_2>^* =\\ 241 &= <\Delta^{-1/2} \Delta^{1/2}\Delta^{1/2} \Delta^{-1/2} \xi_1, \xi_2>^* =\\ 242 &= <\xi _1, \xi_2>^* = <\xi_2 , \xi_1>. 243 \end{align} 244 \end{enumerate} 245 \end{MyExercise} 246 \subsection{Morphisms Between Finite Real Spectral Triples} 247 Extend unitary equivalence of finite spectral triples to real ones (with $J$ 248 and $\gamma$) 249 250 \begin{definition} 251 We call two finite real spectral triples $(A_1, H_1 ,D_1 ; J_1 , \gamma 252 _1)$ and $(A_2, H_2, D_2; J_2, \gamma _2)$ unitarily equivalent if $A_1 = 253 A_2$ and if there exists a unitary operator $U: H_1 \rightarrow H_2$ such 254 that 255 \begin{align} 256 &U\pi_1(a) U^* = \pi _2(a)\\ 257 &UD_1U^*=D_2\\ 258 &U\gamma _1 U^* = \gamma _2\\ 259 &UJ_1 U^* = J_2 260 \end{align} 261 \end{definition} 262 \begin{definition} 263 Let $E$ be a $B$-$A$ bimodule. The \textit{conjugate Module} $E^\circ$ is 264 given by the $A$-$B$-bimodule. 265 \begin{align} 266 E^\circ = \{\bar{e} : e\in E\} 267 \end{align} 268 with 269 \begin{align} 270 a \cdot \bar{e} \cdot b = b^* \bar{e} a^* \;\;\;\; \forall a\in A, b \in 271 B 272 \end{align} 273 \end{definition} 274 $E^\circ$ is not a Hilbert bimodule for $(A, B)$ because it doesn't have a 275 natural $B$-valued inner product. But there is a $A$-valued inner product on 276 the left $A$-module $E^\circ$ with 277 \begin{align} 278 \langle \bar{e}_1, \bar{e}_2 \rangle = \langle e_2 , e_1 \rangle 279 \;\;\;\; e_1, e_2 \in E 280 \end{align} 281 and linearity in $A$: 282 \begin{align} 283 \langle a \bar{e}_1, \bar{e}_2 \rangle = a \langle \bar{e}_1, \bar{e}_2 284 \rangle \;\;\;\; \forall a \in A. 285 \end{align} 286 287 \begin{MyExercise} 288 \textbf{Show that $E^\circ$ is a Hilbert bimodule $(B^{\circ}, A^{\circ})$ 289 }\newline 290 291 292 Straightforward show properties of the Hilbert bimodule and its $B^{\circ}$ 293 valued inner product. Let $\bar{e}_1, \bar{e}_2 \in E^{\circ}$ and $a^\circ \in A, 294 b^\circ \in B$. \\ 295 \begin{align} 296 <\bar{e}_1, a^\circ \bar{e}_2> &= <\bar{e}_1, Ja^*J^{-1} \bar{e}_2>=\\ 297 &= <\bar{e}_1 , J a^* e_2> = \\ 298 &= <J^{-1} e_1, a^* e_2> =\\ 299 & = <a^* e_1, e_2>= <J^{-1}(a^\circ)^* J e_1, e_2> = \\ 300 & = <J^{-1} (a^\circ)^* \bar{e}_1, e_2> =\\ 301 & = <(a^\circ)^* \bar{e}_1 , \bar{e}_2>. 302 \end{align} 303 304 Next $<\bar{e}_1, \bar{e}_2 b^\circ> = <\bar{e}_1, \bar{e_2}> b^\circ$. 305 \begin{align} 306 <\bar{e}_1, \bar{e}_2 b^\circ> &= <\bar{e}_1, \bar{e}_2 Jb^*J^{-1}> =\\ 307 &= <\bar{e}_1, \bar{e_2}> Jb^*J^{-1} = \\ 308 &= <\bar{e}_1, \bar{e}_2> b^\circ. 309 \end{align} 310 Then: 311 \begin{align} 312 (<\bar{e}_1, \bar{e}_2)>_{E^\circ})^* &= (<e_2, e_1>_E)^* =\\ 313 &= <e_1, e_2>_E^* = <\bar{e}_2, \bar{e}_2>_{E^\circ} 314 \end{align} 315 And of course $<\bar{e}, \bar{e}> = <e, e> \geq 0$ 316 \end{MyExercise} 317 318 \subsubsection{Construction of a Finite Real Spectral Triple from a Finite 319 Real Spectral Triple} 320 Given a Hilbert bimodule $E$ for $(B, A)$ we construct a spectral triple 321 $(B, H', D'; J', \gamma ')$ from $(A, H, D; J, \gamma)$ 322 323 For the $H'$ we make a $\mathbb{C}$-valued inner product on $H'$ by combining 324 the $A$ valued inner product on $E$ and $E^\circ$ with the 325 $\mathbb{C}$-valued inner product on $H$. 326 \begin{align} 327 H' := E\otimes _A H \otimes _A E^\circ 328 \end{align} 329 330 Then the action of $B$ on $H'$ is: 331 \begin{align} 332 b(e_2 \otimes \xi \otimes \bar{e}_2 ) = (be_1) \otimes \xi \otimes 333 \bar{e}_2 334 \end{align} 335 The right action of $B$ on $H'$ defined by action on the right component 336 $E^\circ$ 337 \begin{align} 338 J'(e_1 \otimes \xi \otimes \bar{e}_2) = e_2 \otimes J \xi \otimes 339 \bar{e}_1 340 \end{align} 341 with $b^\circ = J' b^* (J')^{-1}$, $b^* \in B$ action on $H'$. 342 \newline 343 344 345 \newpage 346 \begin{MyExercise} 347 \textbf{ Let $\nabla : E \Rightarrow E \otimes _A \Omega _d^1 (A)$ be a right connection on $E$ 348 consider the following anti-linear map: 349 \begin{align} 350 \tau : E \otimes_A \Omega _D^1 (A) &\rightarrow \Omega _D^1 (A) \otimes_A E^\circ\\ 351 e \otimes \omega &\mapsto -\omega ^* \otimes \bar{e} 352 \end{align} 353 Show that the map $\bar{\nabla} : E^\circ \rightarrow \Omega _D^1(A) \otimes E^\circ$ 354 with $\bar{\nabla}(\bar{e}) = \tau \circ \nabla(e)$ is a left connection, that means 355 show that it satisfied the left Leibniz rule: 356 \begin{equation} 357 \bar{\nabla}(a\bar{e}) = [D, a] \otimes \bar{e} + a \bar{\nabla}(\bar{e}) 358 \end{equation} 359 } 360 361 Hagime: 362 \begin{align} 363 &\text{For one:}\\ 364 &\tau \circ \nabla(ae) = \bar{\nabla}(a\bar{e}) = \bar{\nabla}(a^* \bar{e})\\ 365 &\text{For two:}\\ 366 &\tau \circ \nabla(ae) = \tau(\nabla(e)a) + \tau \circ(e \otimes d(a))=\\ 367 &=a^*\bar{\nabla}(\bar{e}) - d(a)^* \otimes \bar{e}. \\ 368 &= a^*\bar{\nabla}(\bar{e}) + d(a^*) \otimes \bar{e}. 369 \end{align} 370 \end{MyExercise} 371 Then the connections 372 \begin{align} 373 &\nabla: E \rightarrow E\otimes _A \Omega _D ^1(A) \\ 374 &\bar{\nabla}:E^\circ \rightarrow \Omega _D^1(A) \otimes _A E^\circ 375 \end{align} 376 give us the Dirac operator on $H' = E \otimes _A H \otimes _A E^\circ$ 377 \begin{align} 378 D'(e_1 \otimes \xi \otimes \bar{e}_2) = (\nabla e_1) \xi \otimes 379 \bar{e_2}+ e_1 \otimes D\xi \otimes \bar{e}_2 + e_1 \otimes 380 \xi(\bar{\nabla}\bar{e}_2) 381 \end{align} 382 383 And the right action of $\omega \in \Omega _D ^1(A)$ on $\xi \in H$ is 384 defined by 385 \begin{align} 386 \xi \mapsto \epsilon' J \omega ^* J^{-1}\xi 387 \end{align} 388 389 Finally for the grading 390 \begin{align} 391 \gamma ' = 1 \otimes \gamma \otimes 1 392 \end{align} 393 394 \begin{theorem} 395 Suppose $(A, H, D; J, \gamma)$ is a finite spectral triple of 396 $KO$-dimension $k$, let $\nabla$ be like above satisfying the 397 compatibility condition (like with finite spectral triples). 398 399 Then $(B, H',D'; J', \gamma')$ is a finite spectral triple of 400 $KO$-Dimension $k$. ($H', D', J', \gamma'$ like above) 401 \end{theorem} 402 403 \begin{proof} 404 The only thing left is to check if the $KO$-dimension is preserved, 405 for this we check if the $\epsilon$'s are the same. 406 \begin{align} 407 &(J')^2 = 1 \otimes J^2 \otimes 1 = \epsilon\\ 408 &J' \gamma '= \epsilon ''\gamma'J' 409 \end{align} 410 and for $\epsilon '$ 411 \begin{align} 412 J'D'(e_1 \otimes \xi \otimes \bar{e}_2)&=J'((\nabla e_1) \xi \otimes 413 \bar{e_2} + e_1 \otimes D\xi \otimes \bar{e}_2 + e_1 \otimes \xi (\tau 414 \nabla e_2))\\ 415 &= \epsilon' D'(e_2 \otimes J\xi \otimes \bar{e}_2)\\ 416 &= \epsilon' D'J'(e_1 \otimes \xi \bar{e}_2) 417 \end{align} 418 \end{proof} 419 420 \end{document}