2_finitencg.tex (27082B)
1 \subsection{Finite Spectral Triples\label{sec:2}} 2 \subsubsection{Metric on Finite Discrete Spaces} 3 We can describe our finite discrete space $X$ by a structure space $\hat{A}$ 4 of a matrix algebra $A$. To establish a distance between two points in $X$ (as 5 we would in a metric space) we use an array $\{d_{ij}\}_{i, j \in X}$ of 6 \textit{real non-negative} entries in $X$ such that 7 \begin{itemize} 8 \item $d_{ij} = d_{ji}$ Symmetric 9 \item $d_{ij} \leq d_{ik} d_{kj}$ Triangle Inequality 10 \item $d_{ij} = 0$ for $i=j$ 11 \end{itemize} 12 13 In the commutative case, the algebra $A$ is commutative and can describe the 14 metric on $X$ in terms of algebraic data. 15 \begin{mytheorem} 16 Let $d_{ij}$ be a metric on $X$ a finite discrete space with $N$ points, $A = \mathbb{C}^N$ 17 with elements $a = (a(i))_{i=1}^N$ such that $\hat{A} \simeq X$. Then there exists a 18 representation $\pi$ of $A$ on a finite-dimensional inner product space $H$ and a symmetric 19 operator $D$ on $H$ such that 20 \begin{equation} 21 d_{ij} = \sup_{a\in A}\bigg\{\big|a(i)-a(j)\big| : |\big|\big[D, 22 \pi(a)]\big|\big| \leq 1\bigg\} 23 \end{equation} 24 \end{mytheorem} 25 26 \begin{proof} 27 We claim that this would follow from the equality: 28 \begin{equation} 29 \big|\big|[D, \pi(a)]\big|\big| = \max_{k\neq l} 30 \bigg\{\frac{1}{d_{kl}}\big|a(k) - a(l)\big|\bigg\} 31 \label{induction} 32 \end{equation} 33 This can be proven with induction. Let us set $N=2$, 34 $H=\mathbb{C}^2$, $\pi:A\rightarrow L(H)$ and a hermitian matrix $D$. 35 \begin{align} 36 \pi(a) = 37 \begin{pmatrix} 38 a(1) & 0 \\ 39 0 & a(2) 40 \end{pmatrix} 41 \;\;\;\; 42 D = 43 \begin{pmatrix} 44 0 & (d_{12})^{-1} \\ 45 (d_{21})^{-1} & 0 46 \end{pmatrix} 47 \end{align} 48 Then we compute the commutator 49 \begin{align} 50 \big|\big|[D, \pi(a)]\big|\big| = (d_{12})^{-1} \big| a(1) - a(2)\big| 51 \end{align} 52 53 For the case $A=\mathbb{C}^3$, we have $H = (\mathbb{C}^2)^{\oplus 3} = H_2 54 \oplus H_2^1 \oplus H_2^2$. The representation $\pi (a)$ reads 55 \begin{align} 56 \pi((a(1), a(2), a(3)) &= 57 \begin{pmatrix} 58 a(1) & 0 \\ 0 & a(2) 59 \end{pmatrix} \oplus 60 \begin{pmatrix} 61 a(1) & 0 \\ 0 & a(3) 62 \end{pmatrix} \oplus 63 \begin{pmatrix} 64 a(2) & 0 \\ 0 & a(2) 65 \end{pmatrix} \nonumber \\ 66 & = \text{diag}\big(a(1), a(2), a(1), a(3), a(2), 67 a(3)\big) 68 \end{align} 69 And the operator $D$ takes the form 70 \begin{align} 71 D &= 72 \begin{pmatrix} 73 0 & x_1 \\ x_1 & 0 74 \end{pmatrix} \oplus 75 \begin{pmatrix} 76 0 & x_2 \\ x_2 & 0 77 \end{pmatrix} \oplus 78 \begin{pmatrix} 79 0 & x_3 \\ x_3 & 0 80 \end{pmatrix} \nonumber \\ 81 &= 82 \begin{pmatrix} 83 0 & x_1 & 0 & 0 & 0 & 0 \\ 84 x_1 & 0 & 0 & 0 & 0 & 0 \\ 85 0 & 0 & 0 & x_2 & 0 & 0 \\ 86 0 & 0 & x_2 & 0 & 0 & 0 \\ 87 0 & 0 & 0 & 0 & 0 & x_3 \\ 88 0 & 0 & 0 & 0 & x_3 & 0 \\ 89 \end{pmatrix}. 90 \end{align} 91 Then the norm of the commutator would be the largest eigenvalue 92 \begin{align}\label{eq:skew matrix} 93 &\big|\big|[D, \pi(a)]\big|\big| = \big|\big|D\pi(a) - \pi(a)D\big|\big|, 94 \end{align} 95 where the matrix in the norm from equation \eqref{eq:skew matrix} is a 96 skew symmetric matrix. Its eigenvalues are $i\lambda_1, i\lambda_2, 97 i\lambda_3, i\lambda_4$. The $\lambda$'s are on the upper and lower 98 diagonal. The matrix norm would be the maximum of the norm with the 99 larges eigenvalues: 100 \begin{align} 101 \big|\big|[D, \pi(a)]\big|\big| = \max_{a\in A}\bigg\{&x_1\big|a(2)-a(1)\big|,\nonumber\\ &x_2\big|(a(3)-a(1))\big|,\nonumber\\ 102 &x_3\big|(a(3)-a(2))\big|\bigg\}. 103 \end{align} 104 Hence the metric turns out to be 105 \begin{align} 106 d = 107 \begin{pmatrix} 108 0 & a(1)-a(2) & a(1)-a(3)\\ 109 a(2)-a(1) & 0 & a(2)-a(3)\\ 110 a(3)-a(1) & a(3)-a(2) & 0 111 \end{pmatrix}. 112 \end{align} 113 114 Suppose this holds for $N$ with $\pi_N$, $H_N = \mathbb{C}^N$ and $D_N$. 115 Then it has to hold for $N+1$ with $H_{N+1} = H_{N} \oplus \bigoplus_{i=1}^N 116 H_N^i$, since the representation reads 117 \begin{align} 118 \pi_{N+1}(a(1),\dots,a(N+1)) &= \pi_N(a(1),\dots,a(N)) 119 \oplus 120 \begin{pmatrix} 121 a(1) & 0 \\ 122 0 & a(N+1) 123 \end{pmatrix} \oplus \nonumber\\ 124 &\oplus \cdots \oplus 125 \begin{pmatrix} 126 a(N) & 0 \\ 127 0 1 & a(N+1) 128 \end{pmatrix}. 129 \end{align} 130 And the operator $D_{N+1}$ is 131 \begin{align} 132 D_{N+1} &= D_N 133 \oplus 134 \begin{pmatrix} 135 0 & (d_{1(N+1)})^{-1} \\ 136 (d_{1(N+1)})^{-1} & 0 137 \end{pmatrix}\oplus \nonumber \\ 138 &\oplus \cdots \oplus 139 \begin{pmatrix} 140 0 & (d_{N(N+1)})^{-1} \\ 141 (d_{N(N+1)})^{-1} & 0 142 \end{pmatrix}. 143 \end{align} 144 From this follows equation \eqref{induction}. 145 Hence we can continue the proof by setting for fixed $i, j$, $a(k) = 146 d_{ik}$, which then gives $|a(i) - a(j)| = d_{ij}$ and thereby it follows 147 that 148 \begin{align} 149 \frac{1}{d_{kl}} \big| a(k) - a(l) \big| = \frac{1}{d_{kl}} \big| 150 d_{ik} - d_{il} \big| \leq 1. 151 \end{align} 152 \end{proof} 153 154 %---------------- EXERCISE 155 To get a better understanding of the results of the theorem let us compute a 156 metric on the space of three points given by $d_{ij} = \sup_{a\in A}\{|a(i) - 157 a(j)|: ||[D, \pi(a)]|| \leq 1\}$ for the set of data $A = \mathbb{C}^3$ acting 158 in the defining representation $H = \mathbb{C}^3$, and 159 \begin{align} 160 D = 161 \begin{pmatrix} 162 0 & d^{-1} & 0 \\ d^{-1} & 0 & 0 \\ 0 & 0 & 0 163 \end{pmatrix}, 164 \end{align} 165 for some $d \in \mathbb{R}$. 166 From the data $A=\mathbb{C}^3$, $H=\mathbb{C}^3$ and $D$ we compute the 167 commutator 168 \begin{align} 169 \big|\big|[D, \pi(a)]\big|\big| &= d^{-1}\left|\left| 170 \begin{pmatrix} 171 0 & a(2)-a(1) & 0 \\ 172 -(a(2)-a(1)) & 0 & 0 \\ 173 0 & 0 & 0 174 \end{pmatrix} \right|\right|. 175 \end{align} 176 Hence the metric is 177 \begin{align} 178 d = 179 \begin{pmatrix} 180 0 & a(1)-a(2) & a(1) \\ 181 a(2)-a(1) & 0 & a(2) \\ 182 -a(1) & -a(2) & 0 183 \end{pmatrix}. 184 \end{align} 185 %---------------- EXERCISE 186 187 The translation of the metric on $X$ into algebraic data assumes commutativity 188 in $A$, this can be extended to a noncommutative matrix algebra, by the 189 following metric on a structure space $\hat{A}$ of a matrix algebra 190 $M_{n_i}(\mathbb{C}$ 191 \begin{equation} 192 d_{ij} = \sup_{a\in A}\big\{|\text{Tr}(a(i)) - \text{Tr}((a(j))|: ||[D, 193 a]|| \leq 1\big\}.\label{eq:discretemetric} 194 \end{equation} 195 Equation \eqref{eq:discretemetric} is special case of the Connes' distance 196 formula on a structure space of $A$. 197 198 Finally we have all three ingredients to define a finite spectral triple, an 199 mathematical structure which encodes finite discrete geometry into algebraic data. 200 \begin{mydefinition} 201 A \textit{finite spectral triple} is a tripe $(A, H, D)$, where $A$ is a unital $*$-algebra, 202 faithfully represented on a finite-dimensional Hilbert space $H$, with a symmetric operator 203 $D: H \rightarrow H$. (Note that $A$ is automatically a matrix algebra.) 204 \end{mydefinition} 205 206 \subsubsection{Properties of Matrix Algebras} 207 \begin{mylemma} 208 If $A$ is a unital C* algebra acting faithfully on a finite 209 dimensional Hilbert space, then $A$ is a matrix algebra of the Form: 210 \begin{align} 211 A \simeq \bigoplus _{i=1}^N M_{n_i}(\mathbb{C}). 212 \end{align} 213 The wording 'acting faithfully on a Hilbertspace' means that the 214 $*$-representation is injective, or for a $*$-homomorphism that means 215 one-to-one correspondence 216 \end{mylemma} 217 \begin{proof} 218 Since $A$ acts faithfully on a Hilbert 219 space, this means that $A$ is a $*$ subalgebra of a matrix algebra $L(H) = M_{\dim 220 (H)}(\mathbb{C})$. Hence it follows, that $A$ is isomorphic to a matrix 221 algebra. 222 \end{proof} 223 224 A simple illustration would be $A = M_n(\mathbb{C})$ for the algebra and 225 $H=\mathbb{C}^n$ for the Hilbertspace. Since $A$ acts on $H$ with matrix 226 multiplication and standard inner product and the operator $D$ on $H$ is a 227 hermitian $n\times n$ matrix. 228 229 \begin{mydefinition} 230 Given an finite spectral triple $(A, H, D)$, the $A$-bimodule of 231 Connes' differential one-forms is 232 \begin{align}\label{eq:connesoneforms} 233 \Omega _D ^1 (A) := \left\{ \sum _k a_k[D, b_k]: a_k, b_k \in A 234 \right\}. 235 \end{align} 236 \end{mydefinition} 237 Additionally there is a map $d:A\rightarrow \Omega _D ^1 (A)$, $d = [D, 238 \cdot]$, where $d$ is a derivation of the $*$-algebra in the sense that 239 \begin{align} 240 d(a\ b) = d(a)b + ad(b), \\ 241 d(a^*) = -d(a)^*. 242 \end{align} 243 Since we have $d(\cdot) = [D, \cdot]$, we can easily check the above equations 244 \begin{align} 245 d(a\ b) &= [D, a\ b] = [D, a]b + a[D,b]\nonumber\\ 246 &= d(a)\ b + a\ d(b). 247 \end{align} 248 And 249 \begin{align} 250 d(a^*) &= [D, a^*] = Da^* - a^*D \nonumber\\ 251 &=-(D^*\ a - a\ D^*) = -[D^*, a] \nonumber\\ 252 &= -d(a)^*. 253 \end{align} 254 Furthermore $\Omega _D^1 (A)$ is an $A$-bimodule, which can be seen by 255 rewriting the defining equation \eqref{eq:connesoneforms} into 256 \begin{align} 257 a\ (a_k[D, b_k])\ b &= a\ a_k(D\ b_k - b_k\ D)\ b = \nonumber\\ 258 &= a\ a_k(D\ b_k\ b - b_k\ D\ b)=\nonumber\\ 259 &= a\ a_k(D\ b_k\ b - b_k\ D\ b - b_k\ 260 b\ D +b_k\ b\ D)= 261 \nonumber\\ 262 &= a\ a_k(D\ b_k\ b-b_k\ b\ D + b_k\ b\ D - b_k\ D\ b) = \nonumber \\ 263 &= a\ a_k [D, b_k\ b] + a\ a_k\ b [D, b]=\nonumber\\ 264 &= \sum _k\ a_k'\ [D, b_k'] 265 \end{align} 266 267 \begin{mylemma} 268 Let $\big(A, H, D\big) = \big(M_n(\mathbb{C}), \mathbb{C}^n, D\big)$, where 269 $D$ is a hermitian $n\times n$ matrix. If $D$ is not a multiple of the 270 identity then 271 \begin{align} 272 \Omega _D ^1 (A) \simeq M_n(\mathbb{C}) = A 273 \end{align} 274 \end{mylemma} 275 \begin{proof} 276 Assume $D = \sum _i \lambda _i e_{ii}$ is diagonal, $\lambda _i \in \mathbb{R}$ and 277 $\{e_{ij}\}$ is the basis of $M_n(\mathbb{C})$. Then for fixed $i$, $j$ choose $k$ 278 such that $\lambda _k \neq \lambda _j$, hence we have 279 \begin{align} \label{eq:basis} 280 \left(\frac{1}{\lambda _k - \lambda _j} e_{ik}\right) [D, e_{kj}] = 281 e_{ij}, 282 \end{align} 283 for $e_{ij}\in \Omega _D ^1 (A)$ by the above definition 284 \eqref{eq:connesoneforms}. Ultimately we have 285 \begin{align} 286 \Omega _D ^1 287 (A) \subset L(\mathbb{C}^n) = H \simeq M_n(\mathbb{C}) = A 288 \end{align} 289 \end{proof} 290 291 Consider an example 292 \begin{align} 293 \left(A=\mathbb{C}^2, H=\mathbb{C}^2, 294 D = \begin{pmatrix} 0 & \lambda \\ \bar{\lambda} & 0 295 \end{pmatrix}\right) 296 \end{align} 297 with $\lambda \neq 0$. We can show that $\Omega _D^1(A) 298 \simeq M_2(\mathbb{C})$. The Hilbert Basis $D$ can be extended in terms of 299 the basis of $M_2(\mathbb{C})$, plugging this into equation 300 \eqref{eq:basis} will get us the same cyclic result and thus 301 $\Omega _D^1(A) \simeq M_2(\mathbb{C})$. 302 303 \subsubsection{Morphisms Between Finite Spectral Triples} 304 Next we will define an equivalence relation between finite spectral triples, called 305 spectral unitary equivalence. This equivalence relation is given by the unitarity of the 306 two matrix algebras themselves, and an additional map $U$ which allows us to associate 307 one operator to a second operator. 308 \begin{mydefinition} 309 Two finite spectral tripes $(A_1, H_1, D_1)$ and $(A_2, H_2, D_2)$ are 310 called unitary equivalent if $A_1 = A_2$ and there exists a map $U:\ H_1 311 \rightarrow H_2$ that satisfies 312 \begin{align} 313 U\ \pi_1(a)\ U^* &= \pi_2(a)\;\;\;\; \text{with} \;\;\; a \in A_1,\\ 314 U\ D_1\ U^* &= D_2. 315 \end{align} 316 \end{mydefinition} 317 Notice that for any such $U$ we have the relation $(A, H, D) \sim (A, H,\ UDU^*)$. 318 And hence 319 \begin{align} 320 U\ D\ U^* = D + U[D,\ U^*], 321 \end{align} 322 are of the form of elements in $\Omega _D^1 (A)$. 323 324 %-------------- EXERCISE 325 To make it clear that the above definition is an equivalence relation between 326 finite spectral triples, we need to see if the relation satisfies 327 reflexivity, symmetry and transitivity. Let us look then at three spectral 328 triples $(A_1, H_1, D_1)$, $(A_2, H_2, D_2)$ and $(A_3, H_3, D_3)$. 329 For reflexivity $(A_1, H_1, D_1) \sim (A_1, H_1, D_1)$. So there 330 exists the unitary map $U: H_1 \rightarrow H_1$, which is the identity 331 and always exists. On the other hand the symmetry condition requires 332 \begin{align} 333 (A_1, H_1, D_1) \sim (A_2, H_2, D_2) \Leftrightarrow 334 (A_2, H_2, D_2) \sim (A_1, H_1, D_1). 335 \end{align} 336 Because $U$ is unitary we can rewrite for the representation for $A_1$ 337 \begin{align} 338 &U\pi_1(a)U^* = \pi_2(a) \;\;\; | \cdot U^*\boxdot U\nonumber \\ 339 &U^*U\pi_1(a)U^*U = \pi_1(a) = U^*\pi_2(a)U. 340 \end{align} 341 The same relation applies for the symmetric operator $D$. 342 Lastly for transitivity the condition is 343 \begin{align} 344 (A_1, H_1, D_1) &\sim (A_2, H_2, D_2) \;\;\; \text{and} \;\;\; 345 (A_2, H_2, D_2) \sim (A_3, H_3, D_3) \nonumber\\ 346 &\Rightarrow (A_1, H_1, D_1) \sim (A_3, H_3, D_3). 347 \end{align} 348 Therefore the two unitary maps $U_{12}:H_1 \rightarrow H_2$ and 349 $U_{23}: H_2 \rightarrow H_3$ are 350 \begin{align} 351 U_{23}\ U_{12}\ \pi_1(a)\ U^*_{12}\ U^*_{23} &= U_{23}\ 352 \pi_2(a)\ U_{23}^*\nonumber \\ 353 &= \pi_3(a), \\ 354 U_{23}\ U_{12}\ D_1\ U^*_{12}\ U^*_{23} &= U_{23}\ 355 D_2 U_{23}^* \nonumber\\ 356 &= D_3. 357 \end{align} 358 %-------------- EXERCISE 359 360 In order to extend the equivalence relation we take a look at Morita 361 equivalence of Matrix Algebras. 362 \begin{mydefinition} 363 Let $A$ be an algebra. We say that $I \subset A$, as a vector space, is a 364 right(left) ideal if $a\ b \in I$ for $a \in A$ and $b\in I$ (or $b\ a \in 365 I$, $b\in I$, $a\in A$). We call a left-right ideal simply an ideal. 366 \end{mydefinition} 367 368 Given a Hilbert bimodule $E \in KK_f(B, A)$ and $(A, H, D)$ we construct 369 a finite spectral triple on $B$, $(B, H', D')$ 370 \begin{equation} 371 H' = E \otimes _A H. 372 \end{equation} 373 We might define $D'$ with 374 \begin{align} 375 D'(e \otimes \xi) = e\otimes D\xi 376 \end{align} 377 Although this would not satisfy the ideal defining the balanced tensor 378 product over $A$, which is generated by elements of the form 379 \begin{align} 380 e\ a \otimes \xi - e\otimes a\ \xi, \;\;\;\;\; e\in E, a\in A, \xi \in 381 H. 382 \end{align} 383 This inherits the left action on $B$ from $E$ and has a $\mathbb{C}$ 384 valued inner product space. $B$ also satisfies the ideal 385 \begin{equation} 386 D'(e\otimes \xi) = e \otimes D\ \xi + \nabla (e)\ \xi, \;\;\;\; e\in 387 E, a\in A, 388 \end{equation} 389 where $\nabla$ is called the \textit{connection on the right A-module E} 390 associated with the derivation $d=[D, \cdot]$. The connection needs to 391 satisfy the \textit{Leibniz Rule} 392 \begin{equation} 393 \nabla(ae) = \nabla(e)a + e \otimes [D, a], \;\;\;\;\; e\in E,\; a\in A. 394 \end{equation} 395 Hence $D'$ is well defined on $E \otimes _A H$ 396 \begin{align} 397 D'(e\ a \otimes \xi - e \otimes a\ \xi) &= D'(e\ a \otimes \xi) - D'(e 398 \otimes \xi) \nonumber\\ 399 &= e\ a\otimes D\ \xi + \nabla(a\ e)\ \xi - e \otimes D(a\ \xi) - \nabla 400 (e)\ a\ \xi \nonumber\\ 401 &= 0. 402 \end{align} 403 With the information thus far we can prove the following theorem 404 \begin{mytheorem} 405 If $(A, H, D)$ is a finite spectral triple and $E \in KK_f(B, A)$, 406 then $(V, E\otimes _A H, D')$ is a finite spectral triple, provided that 407 $\nabla$ satisfies the compatibility condition 408 \begin{equation} 409 \langle e_1, \nabla e_2 \rangle _E - \langle \nabla e_1, e_2 410 \rangle _E = d\langle e_1, e_2 \rangle _E \;\;\;\; e_1, e_2 \in E 411 \end{equation} 412 \end{mytheorem} 413 \begin{proof} 414 The computation for $E\otimes _A H$ is above . The only thing left is to 415 show is, that $D'$ is a symmetric operator. We can prove this by 416 computing for $e_1, e_2 \in E$ and $\xi _1, \xi _2 \in H$ then 417 \begin{align} 418 \langle e_1 \otimes \xi _1, D'(e_2 \otimes \xi_2)\rangle _{E\otimes _A H} &= 419 \langle \xi _1, \langle e_1, \nabla e_2\rangle _E\ \xi _2\rangle 420 \langle \xi _1 , \langle e_1, e_2\rangle _E\ D\ \xi_2\rangle_H \nonumber\\ 421 &= \langle \xi _1, \langle \nabla e_1, e_2\rangle _E\ \xi _2\rangle _H + \langle \xi _1, d\langle e_1, e_2\rangle _E 422 \ \xi _2\rangle _H \nonumber\\ 423 &+ \langle D\ \xi _1,\langle e_1, e_2\rangle _E\ \xi _2\rangle _H - 424 \langle \xi _1, [D, \langle e_1, e_2\rangle _E]\ \xi 425 _2 \rangle _H \nonumber\\ 426 &= \langle D'(e_1 \otimes \xi _1), e_2 \otimes \xi _2\rangle _{E \otimes _A H} 427 \end{align} 428 \end{proof} 429 430 Let us examine the scenario where we consider the difference of connections $\nabla$ and 431 $\nabla'$ on a right $A$-module $E$. Since both connections need to satisfy 432 the Leibniz rule, the difference also should 433 \begin{align} 434 \nabla(ea)-\nabla'(ea)&=\nabla(e) + e\otimes[D, a]\nonumber\\ 435 &-(\nabla'(e)a + e\otimes[D',a])\nonumber\\ 436 &=\bar{\nabla}a + e\otimes(Da-aD-D'a+aD')\nonumber\\ 437 &=\bar{\nabla}a + e\otimes((D-D')a-a(D-D'))\nonumber\\ 438 &=\bar{\nabla}a + e\otimes[D', a]\nonumber\\ 439 &=\bar{\nabla}(ea). 440 \end{align} 441 Therefore $\nabla-\nabla'$ is a right $A$-linear map 442 $E \rightarrow E\otimes _A \Omega _D^1(A)$. 443 444 To get a better grasp of the results let us construct a finite spectral 445 triple $(A, H', D')$ from $(A, H, D)$. The derivation $d(\cdot):A \rightarrow 446 A\otimes _A \Omega_D^1(A)=\Omega_D^1(A)$ is a connection on $A$, i.e.\ considered a 447 right $A$-module 448 \begin{align} 449 \nabla(e \cdot a) = d(a), 450 \end{align} 451 hence $A\otimes_A H\simeq H$. Next we can construct the operator $D'$ 452 for the connection $d(\cdot)$ 453 \begin{align} 454 D'(a\xi) = a(D\xi) + (\nabla a) \xi = D(a\xi). 455 \end{align} 456 By using the identity element in the connection relation we conclude 457 \begin{align} 458 \nabla (e\cdot a) = \nabla(e) a + 1 \otimes d(a)=d(a) \nabla(e) a, 459 \end{align} 460 thus any connection $\nabla: A\rightarrow A\otimes_A \Omega_D^1(A)$ is 461 given by 462 \begin{align}\label{eq: unique connection} 463 \nabla = d + \omega, 464 \end{align} 465 where $\omega \in \Omega_D^1(A)$. This becomes clear when looking at the 466 difference operator $D'$ with the connection on $A$, which is given by 467 \begin{align} 468 D'(a\otimes \xi) &= D'(a \xi) = a(D\xi) + (\nabla a)\xi \nonumber \\ 469 &=a(D\xi) + \nabla(e \cdot a) \xi \nonumber\\ 470 &= D(a\xi) + \nabla(e) (a\xi), 471 \end{align} 472 hence any such connection is of the form as in equation \eqref{eq: unique connection} 473 474 %\subsubsection{Graphing Finite Spectral Triples} 475 %\begin{mydefinition} 476 % A \textit{graph} is a ordered pair $(\Gamma ^{(0)}, \Gamma ^{(1)})$. 477 % Where $\Gamma ^{(0)}$ is the set of vertices (nodes) and $\Gamma ^{(1)}$ 478 % a set of pairs of vertices (edges) 479 %\end{mydefinition} 480 %\begin{figure}[h!] 481 % \centering 482 %\begin{tikzpicture}[ 483 % mass/.style = {draw,circle, minimum size=0.2cm, inner sep=0pt, thick}, 484 % spring/.style = {decorate,decoration={zigzag, pre length=1cm,post length=1cm,segment length=5pt}},] 485 % \node[mass] (m1) at (1,1.5) {}; 486 % \node[mass] (m2) at (-1,1.5) {}; 487 % \node[mass] (m3) at (0,0) {}; 488 % 489 % \draw (m1) -- (m2); 490 % \draw (m1) -- (m3); 491 % \draw (m2) -- (m3); 492 % \end{tikzpicture} 493 % \caption{A simple graph with three vertices and three edges} 494 %\end{figure} 495 %%\begin{MyExercise} 496 %% \textbf{ 497 %% Show that any finite-dimensional faithful representation $H$ of a matrix 498 %% algebra $A$ is completely reducible. To do that show that the complement 499 %% $W^{\perp}$ of an $A$-submodule $W\subset H$ is also an $A$-submodule 500 %% of $H$. 501 %%}\newline 502 %% 503 %% $A\simeq \bigoplus_{i=1}^N M_{n_i}(\mathbb{C})$ is the matrix algebra 504 %% then $H$ is a Hilbert $A$-bimodule and $W$ a submodule of $A$. 505 %% Because we have $H = W \cup W^{\perp}$, then $W^{\perp}$ is naturally a 506 %% $A$-submodule, because elements in $W^{\perp}$ need to satisfy the 507 %% bimodularity. 508 %%\end{MyExercise} 509 %\begin{mydefinition} 510 % A $\Lambda$-decorated graph is given by an ordered pair $(\Gamma, 511 % \Lambda)$ of a finite graph $\Gamma$ and a set of positive integers 512 % $\Lambda$ with the labeling 513 % \begin{itemize} 514 % \item of the vetices $v\in \Gamma ^{(0)}$ given by $n(\nu) \in 515 % \Lambda$ 516 % \item of the edges $e = (\nu _1, \nu _2) \in \Gamma ^{(1)}$ by 517 % operators 518 % \begin{itemize} 519 % \item $D_e: \mathbb{C}^{n(\nu _1)} \rightarrow 520 % \mathbb{C}^{n(\nu _2)}$ 521 % \item and $D_e^*: \mathbb{C}^{n(\nu _2)} \rightarrow 522 % \mathbb{C}^{n(\nu _1)}$ its conjugate traspose 523 % (pullback?) 524 % \end{itemize} 525 % \end{itemize} 526 % such that 527 % \begin{equation} 528 % n(\Gamma ^{(0)}) = \Lambda 529 % \end{equation} 530 %\end{mydefinition} 531 %\begin{question} 532 % Would then $D_e$ be the pullback? 533 %\end{question} 534 %\begin{question} 535 % These graphs are important in the next chapter I should look 536 % into it more, I don't understand much here, specific 537 % how to construct them with the abstraction of a spectral triple... 538 %\end{question} 539 % 540 %The operator $D_e$ between $\textbf{n}_i$ and $\textbf{n}_j$ add up to 541 %$D_{ij}$ 542 %\begin{align} 543 % D_{ij} = \sum\limits_{\substack{e = (\nu _1, \nu _2) \\ n(\nu _1) = 544 % \textbf{n}_i \\ n(\nu _2) = \textbf{n}_j}} D_e 545 %\end{align} 546 % 547 %\begin{mytheorem} 548 % There is a on to one correspondence between finite spectral triples 549 % modulo unitary equivalence and $\Lambda$-decorated graphs, given by 550 % associating a finite spectral triples $(A, H, D)$ to a $\Lambda$ decorated 551 % graph $(\Gamma, \Lambda)$ in the following way: 552 % \begin{equation} 553 % A = \bigoplus _{n\in \Lambda} M_n(\mathbb{C}); \;\;\; 554 % H = \bigoplus _{\nu \in \Gamma ^{(0)}} \mathbb{C}^{n(\nu)}; \;\;\; 555 % D = \sum _{e \in \Gamma ^{(1)}} D_e + D_e^* 556 % \end{equation} 557 %\end{mytheorem} 558 % \begin{figure}[h!] 559 % \centering 560 % \begin{tikzpicture}[ 561 % mass/.style = {draw,circle, minimum size=0.3cm, inner sep=0pt, thick}, 562 % ] 563 % 564 % \node[mass, label={\textbf{n}}] (m1) at (1,0) {}; 565 % \draw (m1) to [out=330, in=210, looseness=25] node[above] {$D_e$} (m1); 566 % \end{tikzpicture} 567 % \caption{A $\Lambda$-decorated Graph of $(M_n(\mathbb{C}), \mathbb{C}^n, 568 % D = D_e + D_e^*)$} 569 %\end{figure} 570 % 571 %%\begin{MyExercise} 572 %% \textbf{ 573 %% Draw a $\Lambda$ decorated graph corresponding to the spectral triple 574 %% $(A=\mathbb{C}^3, H=\mathbb{C}^3, D=\begin{pmatrix}0 & \lambda & 0\\ 575 %% \bar{\lambda} &0 &0 \\ 0&0&0\end{pmatrix})$ 576 %%}\newline 577 %% 578 %%\centering 579 %%\begin{tikzpicture}[ 580 %% mass/.style = {draw,circle, minimum size=0.4cm, inner sep=0pt, thick}, 581 %% spring/.style = {decorate,decoration={zigzag, pre length=1cm,post length=1cm,segment length=5pt}},] 582 %% \node[mass] (m1) at (-1,1.5) {\textbf{1}}; 583 %% \node[mass] (m2) at (1,1.5) {\textbf{2}}; 584 %% \node[mass] (m3) at (3,1.5) {\textbf{3}}; 585 %% 586 %% \draw[style=thick, -] (1.1,1.7) -- (-1.1,1.7); 587 %% \draw[style=thick, -] (1.1,1.3) -- (-1.1,1.3); 588 %% \end{tikzpicture} 589 %% % \captionof{figure}{Solution} 590 %%\end{MyExercise} 591 %%\begin{MyExercise} 592 %% \textbf{ 593 %% Use $\Lambda$-decorated graphs to classify all finite spectral triples 594 %% (modulo unitary equivalence) on the matrix algebra 595 %% $A=\mathbb{C}\oplus M_2(\mathbb{C})$ 596 %%}\newline 597 %% 598 %% \centering 599 %%\begin{tikzpicture}[ 600 %% mass/.style = {draw,circle, minimum size=0.4cm, inner sep=0pt, thick}, 601 %% spring/.style = {decorate,decoration={zigzag, pre length=1cm,post length=1cm,segment length=5pt}},] 602 %% \node[mass] (m1) at (-1,1) {\textbf{1}}; 603 %% \node[mass] (m2) at (1,1) {\textbf{2}}; 604 %% \node[mass] (m3) at (3,1) {\textbf{3}}; 605 %% 606 %% \node[mass] (m4) at (-1,0) {\textbf{1}}; 607 %% \node[mass] (m5) at (1,0) {\textbf{2}}; 608 %% \node[mass] (m6) at (3,0) {\textbf{3}}; 609 %% 610 %% \node[mass] (m7) at (-1,-1) {\textbf{1}}; 611 %% \node[mass] (m8) at (1,-1) {\textbf{2}}; 612 %% \node[mass] (m9) at (3,-1) {\textbf{3}}; 613 %% 614 %% \node[mass] (m10) at (-1,-2) {\textbf{1}}; 615 %% \node[mass] (m11) at (1,-2) {\textbf{2}}; 616 %% \node[mass] (m12) at (3,-2) {\textbf{3}}; 617 %% 618 %% \draw[style=thick, -] (1.1,0.2) -- (-1.1,0.2); 619 %% \draw[style=thick, -] (1.1,-0.2) -- (-1.1,-0.2); 620 %% \draw[style=thick, -] (m7) to [out=330, in=210, looseness=10] node[above] {} (m7); 621 %% \draw[style=thick, -] (m10) -- (m11) ; 622 %% 623 %%\end{tikzpicture} 624 %%% \captionof{figure}{Solution $A=M_3(\mathbb{C})$} 625 %%\end{MyExercise} 626 %\subsubsection{Graph Construction of Finite Spectral Triples} 627 %\textbf{Algebra:}We know if a acts on a finite dimensional Hilbert space then 628 %this C* algebra is isomorphic to a matrix algebra so $A \simeq 629 %\bigoplus_{i=1}^{N}M_{n_i}(\mathbb{C})$. Where $i\in 630 %\hat{A}$ represents an equivalence class and runs from $1$ to $N$, 631 %thus $\hat{A}\simeq\{1,\dots, N\}$. We label equivalence classes by 632 %$\textbf{n}_i$, then $\hat{A}\simeq\{\textbf{n}_1,\dots,\textbf{n}_N\}$. 633 %\newline 634 % 635 %\textbf{Hilbert Space:} Since every Hilbert space that acts faithfully on a 636 %C* algebra is completely reducible, it is isomorphic to the composition 637 %of irreducible representations. $H \simeq \bigoplus_{i=1}^N\mathbb{C}^{n_i} 638 %\otimes V_i$. Where all $V_i$'s are Vector spaces, their dimension is the 639 %multiplicity of the representation landed by $\textbf{n}_i$ to $V_i$ itself 640 %by the multiplicity space. 641 %\newline 642 % 643 %\textbf{Finite Dirac Operator:} $D_{ij}$ is connecting nodes $\textbf{n}_i$ 644 %and $\textbf{n}_j$, with a symmetric map $D_{ij}:\mathbb{C}^{n_i}\otimes V_i 645 %\rightarrow \mathbb{C}^{n_j}\otimes V_j$ 646 %\newline 647 % 648 %To draw a graph, draw nodes in position $\textbf{n}_i\in \hat{A}$. 649 %Multiple nodes at the same position represent multiplicities in $H$. 650 %Draw lines between nodes to represent $D_{ij}$. 651 % 652 %\begin{figure}[h!] 653 % \centering 654 %\begin{tikzpicture} 655 % \node[draw, label=above:{$\textbf{n}_1$},circle, thick] at (-3,0) {}; 656 % \node[label=above:{$\dots$}] at (-2,0) {}; 657 % \node[draw, label=above:{$\textbf{n}_i$},circle, thick] at (-1,0) {}; 658 % \node[label=above:{$\dots$}] at (0,0) {}; 659 % \node[draw, label=above:{$\textbf{n}_j$},circle, thick] at (1,0) {}; 660 % \node[draw, label=above:{},circle, thick, inner sep=0cm, minimum 661 % size=0.2cm] at (1,0) {}; 662 % \node[label=above:{$\dots$}] at (2,0) {}; 663 % \node[draw, label=above:{$\textbf{n}_N$},circle, thick] at (3,0) {}; 664 % 665 % \draw[style=thick, -] (-1,-0.2) -- (1,-0.2); 666 % \draw[style=thick, -] (-1,0.2) -- (1,0.2); 667 % \path[style=thick, -] (-1,-0.2) edge[bend right=15] 668 % node[pos=0.5,below] {} (3,-0.2); 669 % \end{tikzpicture} 670 % \caption{Example} 671 %\end{figure}