week2.tex (17201B)
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 \title{University of Vienna\\ Faculty of Physics\\ \vspace{1.25cm} 78 Notes on\\ Noncommutative Geometry and Particle Phyiscs} 79 \author{Milutin Popovic \\ Supervisor: Dr. Lisa 80 Glaser} 81 \date{Week 2: 12.02 - 19.02} 82 83 \begin{document} 84 85 \maketitle 86 \tableofcontents 87 \newpage 88 89 \section{Noncommutative Geometric Spaces} 90 \subsection{Noncommutative Matrix Algebras} 91 \subsubsection{Balanced Tensor Product and Hilbert Bimodules} 92 93 \begin{definition} 94 Let $A$ be an algebra, $E$ be a \textit{right} $A$-module and $F$ be a \textit{left} $A$-module. 95 The \textit{balanced tensor product} of $E$ and $F$ forms a $A$-bimodule. 96 \begin{align} 97 E \otimes _A F := E \otimes F / \left\{\sum _i e_i a_i \otimes f_i - e_i \otimes a_i f_i : \;\;\; 98 a_i \in A,\ e_i \in E,\ f_i \in F \right\} 99 \end{align} 100 \end{definition} 101 Note $/$ denotes the quotient space. So $\otimes _A$ takes two left/right modules and makes a 102 bimodule with the help the tensor product of the two modules and the quotient space that takes 103 out all the elements from the tensor product that dont preserver the left/right representation and that 104 are duplicates. 105 \begin{definition} 106 Let $A$, $B$ be \textit{matrix algebras}. The \textit{Hilbert bimodule} for $(A, B)$ is given by 107 \begin{itemize} 108 \item $E$, an $A$-$B$-bimodue $E$ and by 109 \item an $B$-valued \textit{inner product} $\langle \cdot,\cdot\rangle_E: E\times E \rightarrow B$ 110 \end{itemize} 111 $\langle \cdot,\cdot\rangle_E$ needs to satisfy the following for $e, e_1, e_2 \in E,\ a \in A$ and $b \in B$. 112 \begin{align} 113 \langle e_1, a\cdot e_2\rangle_E &= \langle a^*\cdot e_1, e_2\rangle_E \;\;\;\; & \text{sesquilinear in $A$}\\ 114 \langle e_1, e_2 \cdot b\rangle_E &= \langle e_1, e_2\rangle_E b \;\;\;\; & \text{scalar in $B$} \\ 115 \langle e_1, e_2\rangle_E &= \langle e_2,e_1\rangle^*_E \;\;\;\; & \text{hermitian} \\ 116 \langle e, e\rangle_E &\ge 0 \;\;\;\; & \text{equality holds iff $e=0$} 117 \end{align} 118 119 \end{definition} 120 121 We denote $KK_f(A,B)$ the set of all \textit{Hilbert bimodules} of $(A,B)$. 122 123 \begin{MyExercise} 124 \textbf{ 125 Check that a representation $\pi:\ A \ \rightarrow L(H)$ of a matrix algebra $A$ turns $H$ into 126 a Hilbert bimodule for $(A, \mathbb{C})$. 127 \label{ex: bimodule} 128 }\newline 129 130 131 We check if the representation of $a \in A$, $\pi(a)=T \in L(H)$ fulfills 132 the conditions on the $\mathbb{C}$-valued inner product for $h_1, h_2 \in H$: 133 \begin{itemize} 134 \item $\langle h_1, \pi(a) h)2\rangle _\mathbb{C} = \langle h_1, T h_2\rangle _\mathbb{C} = 135 \langle T^* h_1, h_2\rangle _\mathbb{C}$, $T^*$ given by the adjoint 136 \item $\langle h_1, h_2 \pi(a)\rangle _\mathbb{C} = \langle h_1, h_2 T\rangle _\mathbb{C} = \langle h_1, h_2\rangle _\mathbb{C}$, $T$ acts from the left 137 \item $\langle h_1, h_2\rangle _\mathbb{C}^* = \langle h_2,h_1\rangle _\mathbb{C}$, hermitian because of the 138 $\mathbb{C}$-valued inner product 139 \item $\langle h_1, h_2\rangle \ge 0$, $\mathbb{C}$-valued inner product. 140 \end{itemize} 141 \end{MyExercise} 142 143 \begin{MyExercise} 144 \textbf{ 145 Show that the $A-A$ bimodule given by $A$ is in $KK_f(A,A)$ by taking the following inner product 146 $\langle \cdot,\cdot\rangle_A:A \times A \rightarrow A$: 147 \begin{align} 148 \langle a, a\rangle_A = a^*a' \;\;\;\; a,a'\in A 149 \end{align} 150 \label{exercise: inner-product} 151 }\newline 152 153 154 We check again the conditions on $\langle \cdot, \cdot\rangle _A$, let $a, a_1, a_2 \in A$: 155 \begin{itemize} 156 \item $\langle a_1, a\cdot a_2\rangle _A = a^*\ a\cdot a_2 = (a^*a_1)^* a_2 = \langle a^*a_1, a_2\rangle $ 157 \item $\langle a_1, a_2 \cdot a\rangle _A = a^*_1 (a_2\cdot a) = (a^*a_2)\cdot a = \langle a_1, a_2\rangle _A a$ 158 \item $\langle a_1, a_2\rangle _A^* = (a_1^* a_2)^* = a_2^*(a_1^*)^* = a_2^* a_1 = \langle a_2, a_1\rangle $ 159 \end{itemize} 160 \end{MyExercise} 161 162 \begin{example} 163 Consider a $*$ homomorphism between two matrix algebras $\phi:A\rightarrow B$. 164 From it we can construct a Hilbert bimodule $E_{\phi} \in KK_f(A, B)$ in the following way. 165 We let $E_{\phi}$ be $B$ in the vector space sense and an inner product from the above 166 Exercise \ref{exercise: inner-product}, with $A$ acting on the left with $\phi$. 167 \begin{align} 168 a\cdot b = \phi(a)b \;\;\;\; a\in A, b\in E_{\phi} 169 \end{align} 170 \end{example} 171 172 173 174 \subsubsection{Kasparov Product and Morita Equivalence} 175 \begin{definition} 176 Let $E \in KK_f(A, B)$ and $F \in KK_F(B, D)$ the \textit{Kasparov product} is defined as 177 with the balanced tensor product 178 \begin{align} 179 F \circ E := E \otimes _B F 180 \end{align} 181 Such that $F\circ E \in KK_f(A,D)$ with a $D$-valued inner product. 182 \begin{align} 183 \langle e_1 \otimes f_1, e_2 \otimes f_2\rangle _{E\otimes _B F} = \langle f_1,\langle e_1, e_2\rangle _E f_2\rangle _F 184 \end{align} 185 \end{definition} 186 187 \begin{question} 188 How do we go from $\langle e_1 \otimes f_1, e_2 \otimes f_2\rangle _{E\otimes _B F}$ to $ 189 \langle f_1,\langle e_1, e_2\rangle _E f_2\rangle _F$ \label{q: tensorproduct}\\ 190 This statement is still in the definition. 191 \end{question} 192 193 %\begin{question} 194 %What is the meaning of `associative up to isomorphism'? Isomorphism of $F \circ E$ or of $A, B$ or $D$? 195 %\end{question} 196 197 \begin{MyExercise} 198 \textbf{ 199 Show that the association $\phi \leadsto E_\phi$ (from the previous Example) is natural 200 in the sense 201 \begin{enumerate} 202 \item $E_{\text{id}_A} \simeq A \in KK_f(A,A)$ 203 \item for $*$-algebra homomorphism $\phi: A \rightarrow B$ and $\psi: B \rightarrow C$ we have 204 an isomorphism 205 \begin{align} 206 E_{\psi} \circ E_{\phi}\ \equiv\ E_{\phi} \otimes _B E_{\psi}\ \simeq\ 207 E_{\psi \circ \phi} \in KK_f(A,C) 208 \end{align} 209 \end{enumerate} 210 } 211 \begin{enumerate} 212 \item $\text{id}_A: A \rightarrow A$.\\ 213 To construct $E_{\phi}\in KK_f(A,A)$, we let $E_{\phi}$ be $A$ with a natural right 214 representation, so $\Rightarrow E_{\phi}\simeq A$.\\ 215 With an inner product, acting on $A$ from the left with $\phi$, $a', a\in A$\\ 216 $a'a = (\phi(a') a) \in A $, which is satisfied by $\text{id}_A$, so $\phi = \text{id}_A$. 217 \item $a \cdot b \cdot c = \psi(\phi (a) \cdot b) \cdot c$ for $a \in A$, $b\in B$, and $c\in C$ 218 which is $\psi \circ \phi$ 219 \end{enumerate} 220 \end{MyExercise} 221 222 \begin{MyExercise} 223 \textbf{ 224 In the definition of Morita equivalence: 225 \begin{enumerate} 226 \item Check that $E \otimes _B F$ is a $A-D$ bimodule 227 \item Check that $\langle \cdot,\cdot\rangle _{E\oplus _B F}$ defines a $D$ valued inner product 228 \item Check that $\langle a^*(e_1 \otimes f_1), e_2 \otimes f_2\rangle _{E \otimes _B F} = \langle e_1 \otimes f_1, a(e_2 \otimes f_2)\rangle _{E \otimes _B F}$. 229 \end{enumerate} 230 } 231 \begin{enumerate} 232 \item $E \otimes _B F = E \otimes F / \{\sum_i e_i b_i \otimes f_i - e_i \otimes b_i f_i; 233 e_i \in E_i, b_i \in B, f_i \in F\}$ the last part takes out all tensor product elements of 234 $E$ and $F$ that don't preserver the left/right representation and that are duplicates. 235 \item $\langle e_1, e_2\rangle _E \in B$ and $\langle f_1, f_2\rangle _F \in C$ by definition. So let $\langle e_1, e_2\rangle _E =b$. \\ 236 Then $\langle e_1 \otimes f_1, e_2 \otimes f_2\rangle _{E\otimes _B F} = \langle f_1, \langle e_1, e_2\rangle _E f_2\rangle _F = 237 \langle f_1, b f_2\rangle _F \in C$ 238 \item Check Question \ref{q: tensorproduct}.\\ 239 But let $G := E\otimes _B F \in KK_f(A,C)$ then $\forall g_1, g_2 \in G$ and $a \in A$ we need 240 by definition $\langle g_1, ag_2\rangle _G = \langle a^*g_1, g_2\rangle _G$ and we set $g_1 = e_1 \otimes f_1$ and 241 $g_2 = e_2 \otimes f_2$ for some $e_1, e_2 \in E$ and $f_1, f_2 \in F$, or else 242 $G \notin KK_f(A,C)$ which would violate the Kasparov product 243 \end{enumerate} 244 \end{MyExercise} 245 246 \begin{definition} 247 Let $A$, $B$ be \textit{matrix algebras}. They are called \textit{Morita equivalent} if there 248 exists an $E \in KK_f(A, B)$ and an $F \in KK_f(B, A)$ such that: 249 \begin{align} 250 E \otimes _B F \simeq A \;\;\; \text{and} \;\;\; F \otimes _A E \simeq B 251 \end{align} 252 Where $\simeq$ denotes the isomorphism between Hilbert bimodules, note that $A$ or $B$ is a bimodule by 253 itself. 254 \end{definition} 255 256 \begin{question} 257 Why are $E$ and $F$ each others inverse in the Kasparov Product? \\ 258 They are each others inverse with respect to the Kasparov Product because we land in the same space as we started. 259 In the definition we have $E \in KK_f(A, B)$ we start from $A$ and $E \otimes _B F$ lands in $A$.\\ 260 On the other hand we have $F \in KK_f(B, D)$ we start from $B$ and $F \otimes _A E$ lands in $B$. 261 \end{question} 262 263 \begin{example} 264 \ 265 \begin{itemize} 266 \item Hilber bimodule of $(A,A)$ is $A$ 267 \item Let $E \in KK_f(A,B)$, we take $E \circ A = A\oplus _A E \simeq E$ 268 \item we conclude, that $_A A_A$ is the identity in the Kasparov product (up to isomorphism) 269 \end{itemize} 270 \end{example} 271 272 \begin{example} 273 Let $E = \mathbb{C}^n$, which is a $(M_n(\mathbb{C}), \mathbb{C})$ Hilbert bimodule with the 274 standard $\mathbb{C}$ inner product.\\ 275 On the other hand let $F = \mathbb{C}^n$, which is a $(\mathbb{C}, M_n(\mathbb{C}))$ Hilbert 276 bimodule by right matrix multiplication with $M_n(\mathbb{C})$ valued inner product: 277 \begin{align} 278 \langle v_1, v_2\rangle =\bar{v_1}v_2^t \;\; \in M_n(\mathbb{C}) 279 \end{align} 280 Now we take the Kasparov product of $E$ and $F$: 281 \begin{itemize} 282 \item $F\circ E\ =\ E\otimes _{\mathbb{C}}F\ \;\;\;\;\;\; \simeq \ M_n(\mathbb{C})$ 283 \item $E\circ F\ =\ F\otimes _{M_n(\mathbb{C})}E\ \simeq\ \mathbb{C}$ 284 \end{itemize} 285 $M_n(\mathbb{C})$ and $\mathbb{C}$ are Morita equivalent 286 \end{example} 287 288 \begin{theorem} 289 Two matrix algebras are Morita Equivalent iff their their Structure spaces 290 are isomorphic as discreet spaces (have the same cardinality / same number of elements) 291 \end{theorem} 292 \begin{proof} 293 Let $A$, $B$ be \textit{Morita equivalent}. So there exists $_A E_B$ and $_B F_A$ with 294 \begin{align} 295 E \otimes _B F \simeq A \;\;\; \text{and} \;\;\; F \otimes _A E \simeq B 296 \end{align} 297 Consider $[(\pi _B, H)] \in \hat{B}$ than we construct a representation of $A$, 298 \begin{align} 299 \pi _A \rightarrow L(E \otimes _B H)\;\;\; \text{with} \;\;\; \pi _A(a) (e \otimes v) = a e \otimes w 300 \end{align} 301 \begin{question} 302 Is $E \simeq H$ and $F \simeq W$? \\ 303 Not in particular, there is a theorem that all infinite dimensional Hilbert spaces are isomorphic. 304 Here we are looking at finite dimensional Hilbert spaces.\\ 305 Another thing to is that $[\pi _B, H] \in \hat{B}$ and looking at Exercise \ref{ex: bimodule} 306 we know that $H$ is a bimodule of $B$, hence $E \otimes _B H\simeq A$, and for $[\pi _A, W]$ 307 the same. 308 \end{question} 309 \textit{vice versa}, consider $[(\pi _A, W)] \in \hat{A}$ we can construct $\pi _B$ 310 \begin{align} 311 \pi _B: B \rightarrow L(F \otimes _A W) \;\;\; \text{and}\;\;\; \pi _B(b) (f\otimes w) = bf\otimes w 312 \end{align} 313 These maps are each others inverses, thus $\hat{A} \simeq \hat{B}$ 314 \end{proof} 315 316 \begin{MyExercise} 317 \textbf{ 318 Fill in the gaps in the above proof: 319 \begin{enumerate} 320 \item show that the representation of $\pi _A$ defined is irreducible iff $\pi _B$ is. 321 \item Show that the association of the class $[\pi _A]$ to $[\pi _B]$ is independent 322 of the choice of representatives $\pi _A$ and $\pi _B$ 323 \end{enumerate} 324 } 325 326 \begin{enumerate} 327 \item $(\pi _B, H)$ is irreducible means $H \neq \emptyset$ and only $\emptyset$ or $H$ 328 is invariant under the Action of $B$ on $H$. 329 Than $E\otimes _B H$ cannot be empty, because also $E$ preserves left representation of $A$ 330 and also $E\otimes _B H \simeq A$. 331 \item The important thing is that $[\pi _A] \in \hat{A}$ respectively $[\pi _B] \in \hat{B}$, 332 hence any choice of representation is irreducible, because the structure space denotes all unitary 333 equivalence classes of irreducible representations. 334 \end{enumerate} 335 \end{MyExercise} 336 337 \begin{lemma} 338 The matrix algebra $M_n(\mathbb{C})$ has a unique irreducible representation (up to isomorphism) 339 given by the defining representation on $\mathbb{C}^n$. 340 \end{lemma} 341 \begin{proof} 342 We know $\mathbb{C}^n$ is a irreducible representation of $A= M_n(\mathbb{C})$. Let $H$ be irreducible 343 and of dimension $k$, then we define a map 344 \begin{align} 345 \phi : A\oplus...\oplus A &\rightarrow H^* \\ 346 (a_1,...,a_k) &\mapsto e^1\circ a_1^t+...+e^k\circ a_k^t 347 \end{align} 348 With $\{e^1,...,e^k\}$ being the basis of the dual space $H^*$ and $(\circ)$ being the pre-composition 349 of elements in $H^*$ and $A$ acting on $H$. This forms a morphism of $M_n(\mathbb{C})$ modules, 350 provided a matrix $a \in A$ acts on $H^*$ with $v\mapsto v\circ a^t$ ($v\in H^*$). 351 Furthermore this morphism is surjective, thus making the pullback $\phi ^*:H\mapsto (A^k)^*$ injective. 352 Now identify $(A^k)^*$ with $A^k$ as a $A$-module and note that 353 $A=M_n(\mathbb{C}) \simeq \oplus ^n \mathbb{C}^n$ as a n A module. 354 It follows that $H$ is a submodule of $A^k \simeq \oplus ^{nk}\mathbb{C}$. By irreducibility 355 $H \simeq \mathbb{C}$. 356 \end{proof} 357 358 \begin{example} 359 Consider two matrix algebras $A$, and $B$. 360 \begin{align} 361 A = \bigoplus ^N_{i=1} M_{n_i}(\mathbb{C}) \;\;\; B = \bigoplus ^M_{j=1} M_{m_j}(\mathbb{C}) 362 \end{align} 363 Let $\hat{A} \simeq \hat{B}$ that implies $N=M$ and define $E$ with $A$ acting by block-diagonal 364 matrices on the first tensor and B acting in the same way on the second tensor. Define $F$ vice versa. 365 \begin{align} 366 E:= \bigoplus _{i=1}^N \mathbb{C}^{n_i} \otimes \mathbb{C}^{m_i} \;\;\; 367 F:= \bigoplus _{i=1}^N \mathbb{C}^{m_i} \otimes \mathbb{C}^{n_i} 368 \end{align} 369 Then we calculate the Kasparov product. 370 \begin{align} 371 E \otimes _B F &\simeq \bigoplus _{i=1}^N (\mathbb{C}^{n_i}\otimes\mathbb{C}^{m_i}) 372 \otimes _{M_{m_i}(\mathbb{C})} (\mathbb{C}^{m_i}\otimes\mathbb{C}^{n_i}) \\ 373 &\simeq \bigoplus _{i=1}^N \mathbb{C}^{n_i}\otimes 374 \left(\mathbb{C}^{m_i}\otimes _{M_{m_i}(\mathbb{C})}\mathbb{C}^{m_i}\right) 375 \oplus \mathbb{C}^{n_i} \\ 376 &\simeq \bigoplus _{i=1}^N \mathbb{C}^{m_i}\otimes\mathbb{C}^{n_i} \simeq A 377 \end{align} 378 and from $F \otimes _A E \simeq B$. 379 \end{example} 380 381 We conclude that. 382 \begin{itemize} 383 \item There is a duality between finite spaces and Morita equivalence classes of matrix algebras. 384 \item By replacing $*$-homomorphism $A\rightarrow B$ with Hilbert bimodules $(A,B)$ we introduce 385 a richer structure of morphism between matrix algebras. 386 \end{itemize} 387 388 \end{document}