week1.tex (16785B)
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 \title{University of Vienna\\ Faculty of Physics\\ \vspace{1.25cm} 77 Notes on\\ Noncommutative Geometry and Particle Phyiscs} 78 \date{Week 1: 05.02 - 12.02} 79 \author{Milutin Popovic \\ Supervisor: Dr. Lisa 80 Glaser} 81 82 \begin{document} 83 84 \maketitle 85 \tableofcontents 86 \newpage 87 88 \section{Noncommutative Geometric Spaces} 89 \subsection{Matrix Algebras and Finite Spaces} 90 \subsubsection{$*$-Algebra} 91 \begin{definition} 92 A \textit{vector space} $A$ over $\mathbb{C}$ is called a \textit{complex, unital Algebra} if, \\ 93 $\forall a,b \in A$ : 94 \begin{enumerate} 95 \item 96 $A \times A \rightarrow A$ \hspace{0.1\textwidth} \textit{bilinear} \\ 97 $(a, b)\ \mapsto \ a\cdot b$ 98 \item 99 $1a = a1 =a$ \hspace{0.08\textwidth} \textit{unital} \\ 100 \end{enumerate} 101 \end{definition} 102 103 \begin{definition} 104 A $*$-algebra is an algebra $A$ with a \textit{conjugate linear map (involution)} $*:A\ \rightarrow A$, 105 $\forall a, b \in A$ satisfying: 106 \begin{enumerate} 107 \item 108 $(ab)^* = b^*a^*$ \hspace{0.05\textwidth} \textit{antidistributive} 109 \item 110 $(a^*)^* = a$ \hspace{0.1\textwidth} \textit{closure} 111 \end{enumerate} 112 \end{definition} 113 In the following all unital algebras are referred to as algebras. 114 115 \subsubsection{Functions on Discrete Spaces} 116 Let $X$ be a \textit{discretized topological} space with $N$ points. 117 Consider functions of a continuous $*$-algebra $C(X)$ assigning values to $\mathbb{C}$, for $f, g \in C(X)$, 118 $\lambda \in \mathbb{C}$ and $x \in X$ they provide the following structures: 119 120 \begin{itemize} 121 \item \textit{pointwise linear} \\ 122 $(f + g)(x) = f(x) + g(x)$\\ 123 $(\lambda f)(x) = \lambda (f(x))$ 124 \item \textit{pointwise multiplication} \\ 125 $fg(x) = f(x)g(x)$ \hspace{0.1\textwidth} same as $(fg)(x) = f(x)g(x))$? 126 \item \textit{pointwise involution} \\ 127 $f^*(x) = \overline{f(x)}$ 128 \end{itemize} 129 130 \begin{question} 131 Mathematical difference between Topological Discreet Spaces and just Discreet Spaces? 132 \end{question} 133 134 The author indicates that $\mathbb{C}$-valued functions on $X$ are automatically continuous. 135 \begin{idea} 136 CAN WE USE THE METRIC? NO! 137 We know that $X$ is a \textit{finite discrete space}, meaning in an $\epsilon$-$\delta$ approach 138 for each $x \in X$ the only $y \in X$, that is small enough is $x$ by itself, which implies 139 $\epsilon$ is always bigger than zero, thus every function $f:X\ \rightarrow\ \mathbb{C}$ is continuous. 140 \end{idea} 141 142 \subsubsection{Isomorphism Property} 143 Furthermore $C(X)$ $*$-algebra is \textit{isomorphic} to a $*$-algebra $\mathbb{C}^N$ with involution 144 ($N$ number of points in $X$), written as $C(X) \simeq \mathbb{C}^N$. 145 A function $f:X\ \rightarrow\ \mathbb{C}$ can be represented with $N \times N$ diagonal matrices, 146 where the value $(ii)$ is the value of the function at the corresponding 147 $i$-th point ($i = 1,...,N$). The structure is preserved because of the definitions of 148 matrix multiplication and the hermitian conjugate of matrices. 149 150 \begin{question} 151 Can isomorphisms between $C(X)$ and $\mathbb{C}^N$ be shown with matrix factorization? 152 \end{question} 153 Isomorphisms are bijective preserve structure and don't lose physical information/ 154 155 \subsubsection{Mapping Finite Discrete Spaces} 156 157 \begin{definition} 158 A \textit{map} between finite discrete spaces $X_1$ and $X_2$ is a function $\phi:\ X_1 \rightarrow\ X_2$ 159 \end{definition} 160 161 For every map between finite discrete spaces there exists a corresponding map \\ 162 $\phi ^*:C(X_2)\ \rightarrow C(X_1)$, which `pulls back' values even if $\phi$ is not bijective. 163 Note that the pullback doesn't map points back, but maps functions on an $*$-algebra $C(X)$. 164 165 166 This map is called a pullback (or a $*$-homomorphism or a $*$-algebra map under pointwise product). 167 Under the pointwise product: 168 \begin{itemize} 169 \item $\phi ^*(fg) = \phi ^*(f) \phi ^*(g)$ 170 \item $\phi ^*(\overline{f}) = \overline{\phi ^*(f)}$ 171 \item $\phi ^*(\lambda f + g) = \lambda \phi ^*(f) + \phi ^*(g)$ 172 \end{itemize} 173 174 \begin{question} 175 $\phi$ is in most cases not bijective, so how can we prove that there exists such a 176 pullback for every map between discrete spaces which preserves information? For bijective 177 it is given by its inverse, which by definition exists because $\phi$ is a map. 178 Or I didn't understand this correctly? 179 \end{question} 180 181 \begin{MyExercise} 182 \textbf{ 183 Show that $\phi :X_1\ \rightarrow \ X_2$ is injective (surjective) map of finite spaces iff 184 $\phi ^* :C(X_2)\ \rightarrow \ C(X_1)$ is surjective (injective). 185 }\newline 186 187 Consider $X_1$ with $n$ points and $X_2$ with $m$ points. Then there are three cases: 188 \begin{enumerate} 189 \item $n=m$ \\ 190 Obviously $\phi$ is bijective and $\phi ^*$ too. 191 \item $n \rangle m$ \\ 192 $\phi$ assigns $n$ points to $m$ points when $n \rangle m$, 193 which is by definition surjective. \\ 194 $\phi ^*$ assigns $m$ points to $n$ points when $n \rangle m$, 195 which is by definition injective. \\ 196 \item $n \langle m $ \\ 197 analogous 198 \end{enumerate} 199 \end{MyExercise} 200 201 \subsubsection{Matrix Algebras} 202 \begin{definition} 203 A \textit{(complex) matrix algebra} A is a direct sum, for $n_i, N \in \mathbb{N}$. 204 \begin{align} 205 A = \bigoplus _{i=1}^{N} M_{n_i}(\mathbb{C}) 206 \end{align} 207 The involution is the hermitian conjugate, a $*$ algebra with involution is referred to as 208 a matrix algebra 209 \end{definition} 210 211 So from a topological discrete space $X$, we can construct a $*$-algebra $C(X)$ which is isomorphic 212 to a matrix algebra $A$. The question is can we construct $X$ given $A$? $A$ is a matrix algebra, 213 which are in most cases is not commutative, so the answer is generally no. 214 215 There are two options. We can restrict ourselves to commutative matrix algebras, 216 which are the vast minority and not physically interesting. 217 Or we can allow more morphisms(isomorphisms) between matrix algebras. 218 219 \begin{question} 220 Why are non-commutative algebras not physically interesting? 221 Maybe too far fetched,but because physical observables (QM-Operators) are not commutative? 222 \end{question} 223 Exactly. 224 225 \subsubsection{Finite Inner Product Spaces and Representations} 226 Until now we looked at a finite topological discrete space, moreover we can consider a 227 finite dimensional inner product space $H$ (finite Hilbert-spaces), with inner product 228 $(\cdot,\cdot)\rightarrow \mathbb{C}$. $L(H)$ is the $*$-algebra of operators on $H$ 229 with product given by composition and involution given by the adjoint, $T \mapsto T^*$. 230 $L(H)$ is a \textit{normed vector space} with 231 \begin{align} 232 &\|T\|^2 = \text{sup}_{h \in H}\{(Th,Th): (h,h) \leq 1\} \hspace{0.1\textwidth} T \in L(H) \\ 233 &\|T\| = \text{sup}\{\sqrt{\lambda}: \lambda \text{ eigenvalue of } T\} 234 \end{align} 235 236 237 \begin{definition} 238 The \textit{representation} of a finite dimensional $*$-algebra A is a pair $(H, \pi)$. 239 $H$ is a finite, dimensional inner product space and $\pi$ is a $*$-\textit{algebra map} 240 \begin{align} 241 \pi:A\ \rightarrow \ L(H) 242 \end{align} 243 \end{definition} 244 \begin{definition} 245 $(H, \pi)$ is called \textit{irreducible} if: 246 \begin{itemize} 247 \item $H \neq \emptyset$ 248 \item only $\emptyset$ or $H$ is invariant under the action of $A$ on $H$ 249 \end{itemize} 250 \end{definition} 251 252 Examples for reducible and irreducible representations 253 \begin{itemize} 254 \item $A = M_n(\mathbb{C})$, representation $H=\mathbb{C}^n$, $A$ acts as matrix multiplication\\ 255 $H$ is irreducible. 256 \item $A = M_n(\mathbb{C})$, representation $H=\mathbb{C}^n\oplus \mathbb{C}^n$, with $a \in A$ acting 257 in block form \\ $\pi: a \mapsto \big(\begin{smallmatrix} a & 0\\ 0 & a \end{smallmatrix}\big)$ is 258 reducible. 259 \end{itemize} 260 261 \begin{definition} 262 Let $(H_1, \pi _1)$ and $(H_2, \pi _2)$ be representations of a $*$-algebra $A$. They are called 263 \textit{unitary equivalent} if there exists a map $U: H_1 \rightarrow H_2$ such that. 264 \begin{align} 265 \pi _1(a) = U^* \pi _2(a) U 266 \end{align} 267 \end{definition} 268 269 \begin{question} 270 In matrix representation this is diagonalization condition? (unitary diagonalization) 271 \end{question} 272 Yes 273 274 \begin{definition} 275 $A$ a $*$-algebra then, $\hat{A}$ is called the structure space of all \textit{unitary equivalence classes 276 of irreducible representations of A} 277 \end{definition} 278 279 \begin{question} 280 Gelfand duality and the spectrum of $\hat{A}$, examples Fourier-Transform and Laplace-Transform 281 for simple spaces. 282 \end{question} 283 More on that in later chapters. 284 285 \begin{MyExercise} 286 \textbf{ 287 Given $(H, \pi)$ of a $*$-algebra $A$, the \textbf{commutant} $\pi (A)'$ of $\pi (A)$ is defined as a set 288 of operators in $L(H)$ that commute with all $\pi (a)$ 289 \begin{align} 290 \pi (A)' = \{T \in L(H):\pi (a)T = T\pi (a) \;\;\; \forall a\in A\} 291 \end{align} 292 \begin{enumerate} 293 \item Show that $\pi (A)'$ is a $*$-algebra. 294 \item Show that a representation $(H, \pi)$ of $A$ is irreducible iff the commutant $\pi (A)'$ 295 consists of multiples of the identity 296 \end{enumerate} 297 } 298 299 1. To show that $\pi (A)'$ is a $*$-algebra we have to show that it is unital, associative and involute. 300 And note that $\pi (a) \in L(H)\ \forall a \in A$. 301 Unitarity is given by the unital operator of the $*$-algebra of operators $L(H)$, which exists by definition 302 because H is a inner product space. Associativity is given by $*$-algebra of $L(H)$, $L(H) \times L(H) \mapsto L(H)$, 303 which is associative by definition. Involutnes is also given by the $*$-algebra $L(H)$ 304 with a map $*: L(H) \mapsto L(H)$ only for $T$ that commute with $\pi (a)$. 305 \\ 306 2.? 307 \end{MyExercise} 308 309 \begin{MyExercise} 310 \textbf{ 311 \begin{enumerate} 312 \item If $A$ is a unital $*$-algebra, show that the $n \times n$ matrices $M_n(A)$ with entries 313 in $A$ form a unital $*$-algebra. 314 \item Let $\pi :A\ \rightarrow \ L(H)$ be a representation of a $*$-algebra $A$ and set 315 $H^n = H \oplus ... \oplus H$, $n$ times. Show that $\tilde{\pi}:M_n(A) \rightarrow L(H^n)$ 316 of $M_n(A)$ with\\ 317 $\tilde{\pi}((a_{ij})) = (\tilde{\pi}(a_{ij})) \in M_n(A)$. 318 \item Let $\tilde{\pi}:M_n(A) \rightarrow L(H^n)$ be a $*$ algebra representation of $M_n(A)$. 319 Show that $\pi: A \rightarrow L(H^n)$ is a representation of $A$. 320 \end{enumerate} 321 } 322 1. We know $A$ is a $*$ algebra. Unitary operaton in $M_n(A)$ is given by the identity Matrix, which 323 has to exists because every entry in $M_n(A)$ has to behave like in $A$. Associativity is given by 324 matrix multiplication. Involutnes is given by the conjugate transpose.\\ 325 2. $A \simeq M_n(A)$ and $H \simeq H^n$ meaning $\tilde{\pi}$ is a valid reducible representation.\\ 326 3. $\tilde{\pi}$ and $\pi$ are unitary equivalent, there is a map $U: H^n \rightarrow H^n$ given by 327 $U=\mathbbm{1}_n$:\\ 328 $\pi (a) = \mathbbm{1}_n^*\ \tilde{\pi}((a_{ij}))\ \mathbbm{1}_n = \tilde{\pi}((a_{ij})) = \pi (a_{ij}) 329 \Rightarrow a_{ij} = a\mathbbm{1}_n$. 330 \end{MyExercise} 331 332 \subsection{Commutative Matrix Algebras} 333 \begin{itemize} 334 \item Commutative matrix algebras can be used to reconstruct a discrete space given 335 a matrix \textit{commutative} matrix algebra. 336 \item The structure space $\hat{A}$ is used for this. Because $A \simeq \mathbb{C}^N$ we all any 337 irreducible representation are of the form 338 $\pi _i:(\lambda_1,...,\lambda_N)\in \mathbb{C}^N \mapsto \lambda_i \in \mathbb{C}$ \\ 339 for $i = 1,...,N \Rightarrow \hat{A} \simeq \{1,...,N\}.$ 340 \item Conclusion is that there is a duality between discrete spaces and commutative matrix algebra 341 this duality is called the \textit{finite dimensional Gelfand duality} 342 \end{itemize} 343 344 \subsection{Noncommutative Matrix Algebras} 345 Aim is to construct duality between finite dimensional spaces and \textit{equivalence classes} 346 of matrix algebras, to preserve general non-commutivity of matrices. 347 \begin{itemize} 348 \item Equivalence classes are described by a generalized notion of ispomorphisms between matrix 349 algebras (\textit{Morita Equivalence}) 350 \end{itemize} 351 352 \subsubsection{Algebraic Modules} 353 \begin{definition} 354 Let $A$, $B$ be algebras (need not be matrix algebras) 355 \begin{enumerate} 356 \item \textit{left} A-module is a vector space $E$, that carries a left representation of $A$, 357 that is $\exists$ a bilinear map $\gamma: A \times E \rightarrow E$ with 358 \begin{align} 359 (a_1a_2)\cdot e = a_1 \cdot (a_2 \cdot e);\;\;\; a_1, a_2 \in A, e \in E 360 \end{align} 361 \item \textit{right} B-module is a vector space $F$, that carries a right representation of $A$, 362 that is $\exists$ a bilinear map $\gamma: F \times B \rightarrow F$ with 363 \begin{align} 364 f \cdot (b_1b_2)= (f \cdot b_1) \cdot b_2;\;\;\; b_1, b_2 \in B, f \in F 365 \end{align} 366 \item \textit{left} A-module and \textit{right} B-module is a \textit{bimodule}, a vector space $E$ 367 satisfying 368 \begin{align} 369 a \cdot (e \cdot b)= (a \cdot e) \cdot b;\;\;\; a \in A, b \in B, e \in E 370 \end{align} 371 \end{enumerate} 372 \end{definition} 373 Notion of A-\textbf{module homomorphism} as linear map $\phi: E\rightarrow F$ which respects the 374 representation of A, e.g. for left module. 375 \begin{align} 376 \phi (ae) = a \phi (e); \;\;\; a \in A, e \in E. 377 \end{align} 378 Remark on the notation 379 \begin{itemize} 380 \item ${}_A E$ left $A$-module $E$; 381 \item ${}_A E_B$ right $B$-module $F$; 382 \item ${}_A E_B$ $A$-$B$-bimodule $E$; 383 \end{itemize} 384 385 \begin{MyExercise} 386 \textbf{ 387 Check that a representation of $\pi : A \rightarrow L(H)$ of a $*$-algebra A turns H into a 388 left module ${}_A H$. 389 }\newline 390 391 Not quite sure but \\ 392 $a \in A$, $h_1, h_2 \in H$, we know $\pi (a) = T \in L(H)$ than 393 \begin{align} 394 \langle \pi (a) h_1, \pi (a) h_2\rangle = \langle T h_1, T h_2\rangle = \langle T^*T h_1, h_2\rangle = \langle h_1, h_2\rangle 395 \end{align} 396 Or maybe this \\ 397 If $_A H$ than $(a_1a_2) h = a_1 (a_2 h)$ for $a_1, a_2 \in A$ and $h \in H$.\\ 398 Then we take the representation of an $a \in A$, $\pi (a)$: 399 \begin{align} 400 (\pi(a_1)\pi(a_2))h = \pi(a_1)(\pi(a_2) h) = (T_1T_2) h = T_1 (T_2 h) 401 \end{align} 402 For $T_1, T_2 \in L(H)$, which operate naturally from the left on $h$. 403 \end{MyExercise} 404 405 \begin{MyExercise} 406 \textbf{ 407 Show that $A$ is a bimodule ${}_A A_A$ with itself. 408 }\newline 409 410 $\gamma: A\times A\times A \rightarrow A$ which is given by the inner product of the $*$-algebra. 411 \end{MyExercise} 412 413 \end{document}