3_realncg.tex (12338B)
1 \subsection{Finite Real Noncommutative Spaces\label{sec:3}} 2 \subsubsection{Finite Real Spectral Triples} 3 In this chapter we supplement the finite spectral triples with a \textit{real 4 structure}. We additionally require a symmetry condition that that $H$ is an 5 $A$-$A$-bimodule rather than only a $A$-left module. This ansatz has tight 6 bounds with physical properties such as charge conjugation, into which we will 7 dive in deeper in later chapters. In regards to this we will need to set a basis 8 of definitions to get an overview. 9 First we introduce a $\mathbb{Z}_2$-grading $\gamma$ with the following 10 properties 11 \begin{align} 12 \gamma ^* &= \gamma, \\ 13 \gamma ^2 &= 1, \\ 14 \gamma D &= - D \gamma,\\ 15 \gamma a &= a \gamma, \;\;\;\; a\in A. 16 \end{align} 17 Then we can define a finite real spectral triple. 18 \begin{mydefinition} 19 A \textit{finite real spectral triple} is given by a finite spectral 20 triple $(A, H, D)$ and a anti-unitary operator $J:H\rightarrow H$ called 21 the \textit{real structure}, such that 22 \begin{align} 23 a^\circ := J\ a^*\ J^{-1}, 24 \end{align} 25 is a right representation of $A$ on $H$, that is $(ab)^\circ = b^\circ 26 a^\circ$. With two requirements 27 \begin{align} 28 &[a, b^\circ] = 0,\\ 29 &[[D, a],\ b^\circ] = 0. 30 \end{align} 31 The two properties are called the \textit{commutant property}, they 32 require that the left action of an element in $A$ and $\Omega _D^1(A)$ commutes with the right 33 action on $A$. 34 \end{mydefinition} 35 \begin{mydefinition} 36 The $KO$-dimension of a real spectral triple is determined by the sings 37 $\epsilon, \epsilon ' ,\epsilon '' \in \{-1, 1\}$ appearing in 38 \begin{align} 39 J^2 &= \epsilon, \\ 40 J\ D &= \epsilon \ D\ J,\\ 41 J\ \gamma &= \epsilon''\ \gamma\ J. 42 \end{align} 43 \end{mydefinition} 44 \begin{table}[h!] 45 \centering 46 \caption{$KO$-dimension $k$ modulo $8$ of a real spectral triple} 47 \begin{tabular}{ c | c c c c c c c c} 48 \hline 49 $k$ & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ 50 \hline 51 $\epsilon$ & 1 & 1 & -1 & -1 & -1 & -1 & 1 & 1 \\ 52 $\epsilon '$ & 1 & -1 & 1 & 1 & 1 & -1 & 1 & 1 \\ 53 $\epsilon ''$ & 1 & & -1 & & 1 & & -1 & \\ 54 \hline 55 \end{tabular} 56 \end{table} 57 \noindent 58 Even thought the KO-dimension of a real spectral triple is important, we will 59 not be doing in-depth introduction of the KO-dimension, for this we reference 60 again to \cite{ncgwalter}. 61 62 \begin{mydefinition} 63 An opposite-algebra $A^\circ$ of a $A$ is defined to be equal to $A$ as a 64 vector space with the opposite product 65 \begin{align} 66 &a\circ b := ba\\ 67 &\Rightarrow a^\circ = Ja^* J^{-1}, 68 \end{align} 69 which defines the left representation of $A^\circ$ on $H$ 70 \end{mydefinition} 71 72 73 %%------------EXAMPLE EXERCISE 74 %Let us examine an example of a matrix algebra $M_N(\mathbb{C})$ acting on 75 %$H=M_N(\mathbb{C})$ by left matrix multiplication with the Hilbert Schmidt 76 %inner product. 77 %\begin{align} 78 % \langle a , b \rangle = \text{Tr}(a^* b). 79 %\end{align} 80 %We can define $\gamma (a) = a$ and $J(a) = a^*$ with $a\in H$. Since $D$ 81 %must be odd with respect to $\gamma$ it vanishes identically. Furthermore we 82 %know the multiplicity space is $V_i = \mathbb{C}^{m_i}$, and also we know 83 %that for $T\in H$ and $a\in A'$ to work we need $a\ T=T\ a$. Thus by laws of 84 %matrix multiplication we need $A' \simeq \bigoplus _i M_{m_i}(\mathbb{C})$. For 85 %this to work we naturally need $H = \bigoplus_i \mathbb{C}^{n_i} \otimes 86 %\mathbb{C}^{m_i}$. Hence the right action of $M_N(\mathbb{C})$ on $H = 87 %M_N(\mathbb{C})$ as defined by $a \mapsto a^\circ$ is given by right matrix 88 %multiplication 89 %\begin{align} 90 % a^\circ \xi = J a^* J^{-1}\xi = Ja^* \xi^* = J\xi a=\xi^* a 91 %\end{align} 92 % 93 %%------------EXAMPLE EXERCISE 94 95 \begin{mydefinition} 96 We call $\xi \in H$ \textbf{cyclic vector} in $A$ if: 97 \begin{align} 98 A\xi := { a\xi:\;\; a\in A} = H 99 \end{align} 100 We call $\xi \in H$ \textbf{separating vector} in $A$ if: 101 \begin{align} 102 a\xi = 0\;\; \Rightarrow \;\; a=0;\;\;\; a\in A 103 \end{align} 104 \end{mydefinition} 105 %------------------- EXERCISE 106 Suppose $(A, H, D = 0)$ is a finite spectral triple such that $H$ possesses a 107 cyclic and separating vector for $A$ and let 108 \begin{align} 109 J: H \rightarrow H 110 \end{align} 111 be the operator in $S = J \Delta ^{1/2}$ with $\Delta = S^*S$. By composition 112 $S(a\xi) = a*\xi$ this is literally anti-linearity, then $S(a \xi) = a* \xi$ 113 defines a anti-linear operator. Furthermore the operator $S$ is invertible 114 because, if a $\xi \in H$ is cyclic then we have 115 \begin{align} 116 S(A\xi) = A^*\xi = A\xi = H. 117 \end{align} 118 Vice versa the same has to work for $S^{-1}$, otherwise $\xi$ wouldn't exist, 119 hence 120 \begin{align} 121 S^{-1}(A^*\xi) = S^{-1}(H) = H. 122 \end{align} 123 Additionally $J$ is anti-unitary because firstly, $S$ is bijective thus 124 $\Delta ^{1/2}$ and $J$ need to be bijective. Also have $J = S 125 \Delta^{-1/2}$ and $\Delta^* = \Delta$, so for a $\xi _1 , \xi _2 \in H$ we 126 can write 127 \begin{align} 128 \langle J \xi _1 , J \xi _2 \rangle &= \langle J^*J\xi_1 , \xi_2\rangle ^* =\nonumber\\ 129 &= \langle (\Delta ^{-1/2})^* S^* S \Delta ^{-1/2} \xi_1, \xi_2\rangle ^* =\nonumber \\ 130 &= \langle (\Delta^{-1/2})^* \Delta \Delta^{-1/2} \xi_1, \xi_2\rangle ^* =\nonumber\\ 131 &= \langle \Delta^{-1/2} \Delta^{1/2}\Delta^{1/2} \Delta^{-1/2} \xi_1, \xi_2\rangle ^* 132 =\nonumber\\ 133 &= \langle \xi _1, \xi_2\rangle ^* = \langle \xi_2 , \xi_1\rangle , 134 \end{align} 135 which concludes the anti-unitarity by definition. 136 %------------------- EXERCISE 137 \subsubsection{Morphisms Between Finite Real Spectral Triples} 138 Like the unitary equivalence relation for finite spectral triples, we can it 139 extend it to finite real spectral triples. 140 \begin{mydefinition} 141 We call two finite real spectral triples $(A_1, H_1 ,D_1 ; J_1 , \gamma_1)$ 142 and $(A_2, H_2, D_2; J_2, \gamma _2)$ unitarily equivalent if $A_1 = 143 A_2$ and if there exists a unitary operator $U: H_1 \rightarrow H_2$ such 144 that 145 \begin{align} 146 U\ \pi_1(a)\ U^* &= \pi _2(a),\\ 147 U\ D_1\ U^* &= D_2,\\ 148 U \gamma _1\ U^* &= \gamma _2,\\ 149 U\ J_1\ U^* &= J_2. 150 \end{align} 151 \end{mydefinition} 152 \begin{mydefinition} 153 Let $E$ be a $B$-$A$ bimodule. The \textit{conjugate Module} $E^\circ$ is 154 given by the $A$-$B$-bimodule. 155 \begin{align} 156 E^\circ = \{\bar{e} : e\in E\}, 157 \end{align} 158 with 159 \begin{align} 160 a \cdot \bar{e} \cdot b = b^*\ \bar{e}\ a^*, \;\;\;\; \forall a\in A, b \in 161 B. 162 \end{align} 163 \end{mydefinition} 164 We bear in mind that $E^\circ$ is not a Hilbert bimodule for $(A, B)$ because 165 it doesn't have a natural $B$-valued inner product. But there is a $A$-valued 166 inner product on the left $A$-module $E^\circ$ with 167 \begin{align} 168 \langle \bar{e}_1, \bar{e}_2 \rangle = \langle e_2 , e_1 \rangle, 169 \;\;\;\; e_1, e_2 \in E. 170 \end{align} 171 And linearity in $A$ by the terms 172 \begin{align} 173 \langle a\ \bar{e}_1, \bar{e}_2 \rangle = a \langle \bar{e}_1, \bar{e}_2 174 \rangle, \;\;\;\; \forall a \in A. 175 \end{align} 176 177 %------------- EXERCISE 178 It turns out that $E^\circ$ is a Hilbert bimodule 179 of $(B^{\circ}, A^{\circ})$. A straightforward calculation of the properties of the Hilbert bimodule and its $B^{\circ}$ 180 valued inner product gives the results. For $\bar{e}_1, \bar{e}_2 \in E^{\circ}$ and $a^\circ \in A, 181 b^\circ \in B$ we write 182 \begin{align} 183 \langle\bar{e}_1, a^\circ \bar{e}_2\rangle &= \langle\bar{e}_1, Ja^*J^{-1} 184 \bar{e}_2\rangle=\nonumber\\ 185 &= \langle\bar{e}_1 , J a^* e_2\rangle \nonumber \\ 186 &= \langle J^{-1} e_1, a^* e_2\rangle \nonumber\\ 187 & = \langle a^* e_1, e_2\rangle= \langle J^{-1}(a^\circ)^* J e_1, e_2\rangle \nonumber\\ 188 & = \langle J^{-1} (a^\circ)^* \bar{e}_1, e_2\rangle \nonumber\\ 189 & = \langle (a^\circ)^* \bar{e}_1 , \bar{e}_2\rangle. 190 \end{align} 191 Next for $\langle\bar{e}_1, \bar{e}_2 b^\circ\rangle = \langle\bar{e}_1, 192 \bar{e_2}\rangle b^\circ$ we obtain 193 \begin{align} 194 \langle\bar{e}_1, \bar{e}_2 b^\circ\rangle &= \langle\bar{e}_1, \bar{e}_2 Jb^*J^{-1}\rangle 195 \nonumber\\ 196 &= \langle\bar{e}_1, \bar{e_2}\rangle Jb^*J^{-1} \nonumber \\ 197 &= \langle\bar{e}_1, \bar{e}_2\rangle b^\circ. 198 \end{align} 199 Additionally we get 200 \begin{align} 201 (\langle\bar{e}_1, \bar{e}_2)\rangle_{E^\circ})^* &= (\langle e_2, e_1\rangle_E)^*\nonumber\\ 202 &= \langle e_1, e_2\rangle_E^* \nonumber\\ 203 &= \langle\bar{e}_2, \bar{e}_2\rangle_{E^\circ}. 204 \end{align} 205 And finally we have 206 \begin{align} 207 \langle\bar{e}, \bar{e}\rangle = \langle e, e\rangle \geq 0 208 \end{align} 209 %------------- EXERCISE 210 211 Given the results thus far, given a Hilbert bimodule $E$ for $(B, A)$ one can 212 construct a spectral triple $(B, H', D'; J', \gamma ')$ from $(A, H, D; J, 213 \gamma)$. For $H'$ we make a $\mathbb{C}$-valued inner product on $H'$ by combining 214 the $A$ valued inner product on $E$ and $E^\circ$ with the 215 $\mathbb{C}$-valued inner product on $H$ by defining 216 \begin{align} 217 H' := E\otimes _A H \otimes _A E^\circ. 218 \end{align} 219 Then the action of $B$ on $H'$ takes the following form 220 \begin{align} 221 b(e_2 \otimes \xi \otimes \bar{e}_2 ) = (be_1) \otimes \xi \otimes 222 \bar{e}_2. 223 \end{align} 224 The right action of $B$ on $H'$ defined by action on the right components of 225 $E^\circ$ is 226 \begin{align} 227 J'(e_1 \otimes \xi \otimes \bar{e}_2) = e_2 \otimes J \xi \otimes 228 \bar{e}_1, 229 \end{align} 230 where $b^\circ = J' b^* (J')^{-1}$ and $b^* \in B$ is the action on $H'$. 231 Hence the connection reads 232 \begin{align} 233 &\nabla: E \rightarrow E\otimes _A \Omega _D ^1(A) \\ 234 &\bar{\nabla}:E^\circ \rightarrow \Omega _D^1(A) \otimes _A E^\circ, 235 \end{align} 236 which gives the Dirac operator on $H' = E \otimes _A H \otimes _A 237 E^\circ$ as 238 \begin{align} 239 D'(e_1 \otimes \xi \otimes \bar{e}_2) = (\nabla e_1) \xi \otimes 240 \bar{e_2}+ e_1 \otimes D\xi \otimes \bar{e}_2 + e_1 \otimes 241 \xi(\bar{\nabla}\bar{e}_2). 242 \end{align} 243 And the right action of $\omega \in \Omega _D ^1(A)$ on $\xi \in H$ is 244 defined by 245 \begin{align} 246 \xi \mapsto \epsilon' J \omega ^* J^{-1}\xi. 247 \end{align} 248 Finally for the grading one obtains 249 \begin{align} 250 \gamma ' = 1 \otimes \gamma \otimes 1. 251 \end{align} 252 253 Summarizing we can write down the following theorem 254 \begin{mytheorem} 255 Suppose $(A, H, D; J, \gamma)$ is a finite spectral triple of 256 $KO$-dimension $k$, let $\nabla$ be a connection satisfying the 257 compatibility condition (same as with finite spectral triples). 258 Then $(B, H',D'; J', \gamma')$ is a finite spectral triple of 259 $KO$-Dimension $k$. ($H', D', J', \gamma'$) 260 \end{mytheorem} 261 262 \begin{proof} 263 The only thing left is to check is, if the $KO$-dimension is preserved. 264 That is one needs to check if if the $\epsilon$'s are the same. 265 \begin{align} 266 &(J')^2 = 1 \otimes J^2 \otimes 1 = \epsilon,\\ 267 &J' \gamma '= \epsilon ''\gamma'J'. 268 \end{align} 269 Lastly for $\epsilon '$ one obtains 270 \begin{align} 271 J'D'(e_1 \otimes \xi \otimes \bar{e}_2)&=J'\big((\nabla e_1) \xi \otimes 272 \bar{e_2} + e_1 \otimes D\xi \otimes \bar{e}_2 + e_1 \otimes \xi (\tau 273 \nabla e_2)\big)\nonumber \\ 274 &= \epsilon' D'\left(e_2 \otimes J\xi \otimes \bar{e}_2\right)\nonumber\\ 275 &= \epsilon' D'J'\left(e_1 \otimes \xi \bar{e}_2\right) 276 \end{align} 277 \end{proof} 278 279 Let us take a look at $\nabla : E \Rightarrow E \otimes _A \Omega _d^1 (A)$, 280 the right connection on $E$ and consider the following anti-linear map 281 \begin{align} 282 \tau : E \otimes_A \Omega _D^1 (A) &\rightarrow \Omega _D^1 (A) \otimes_A E^\circ\\ 283 e \otimes \omega &\mapsto -\omega ^* \otimes \bar{e}. 284 \end{align} 285 Interestingly the map $\bar{\nabla} : E^\circ \rightarrow \Omega _D^1(A) \otimes E^\circ$ 286 with $\bar{\nabla}(\bar{e}) = \tau \circ \nabla(e)$ is a left connection, that means 287 show that it satisfied the left Leibniz rule, for one 288 \begin{align} 289 \tau \circ \nabla(ae) = \bar{\nabla}(a\bar{e}) = \bar{\nabla}(a^* 290 \bar{e}). 291 \end{align} 292 And for two 293 \begin{align} 294 \tau \circ \nabla(ae) &= \tau(\nabla(e)a) + \tau \circ(e \otimes 295 d(a))\nonumber \\ 296 &=a^*\bar{\nabla}(\bar{e}) - d(a)^* \otimes \bar{e}. \nonumber\\ 297 &= a^*\bar{\nabla}(\bar{e}) + d(a^*) \otimes \bar{e}. 298 \end{align} 299