commit a4d2f621591e31344cda27e813e358f87b908ab8
parent c708e7589668f5f78a3704d70d0776d8b65a494a
Author: miksa <milutin@popovic.xyz>
Date: Wed, 28 Apr 2021 13:19:05 +0200
almost done week7
Diffstat:
4 files changed, 582 insertions(+), 4 deletions(-)
diff --git a/pdfs/week5.pdf b/pdfs/week5.pdf
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diff --git a/pdfs/week7.pdf b/pdfs/week7.pdf
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diff --git a/src/week5.tex b/src/week5.tex
@@ -327,10 +327,11 @@ Then there is a map $d:A\rightarrow \Omega _D ^1 (A)$, $d = [D, \cdot]$.
Begin
\begin{align*}
- a(a_k[D, b_k])b &= a_k'(Db_k - b_k D) b = \\
- &= a_k'(Db_k b - b_k D b) = a_k(Db_k b - b_k Db -Db_k +Db_kb)=\\
- &= a_k'([D, b_kb] - b_k D b + D b_k + \cdots - \cdots) = \\
- &=\sum_k a_k' [D, b_k']
+ a(a_k[D, b_k])b &= aa_k(Db_k - b_k D) b = \\
+ &= aa_k(Db_k b - b_k D b) = aa_k(Db_k b - b_k Db - b_kbD +b_kbD)=\\
+ &= aa_k(Db_kb-b_kbD + b_k b D - b_k D b) = \\
+ &= aa_k [D, b_kb] + aa_k b [D, b]=\\
+ &= \sum _k a_k' [D, b_k']
\end{align*}
\end{MyExercise}
diff --git a/src/week7.tex b/src/week7.tex
@@ -0,0 +1,577 @@
+\documentclass[a4paper]{article}
+
+\usepackage[T1]{fontenc}
+\usepackage[utf8]{inputenc}
+
+\usepackage{mathptmx}
+
+\usepackage{subcaption}
+\usepackage[shortlabels]{enumitem}
+\usepackage{amsmath,amssymb}
+\usepackage{amsthm}
+\usepackage{bbm}
+\usepackage{graphicx}
+\usepackage[colorlinks=true,naturalnames=true,plainpages=false,pdfpagelabels=true]{hyperref}
+\usepackage[parfill]{parskip}
+
+\usepackage{pst-node}
+\usepackage{tikz-cd}
+
+\usepackage{tikz}
+\usetikzlibrary{patterns,decorations.pathmorphing,positioning}
+
+\usepackage[framemethod=TikZ]{mdframed}
+
+\tikzstyle{titlered} =
+ [draw=black, thick, fill=white,%
+ text=black, rectangle,
+ right, minimum height=.7cm]
+
+\newcounter{exercise}
+
+\renewcommand*\theexercise{Exercise~\arabic{exercise}}
+
+\makeatletter
+\mdfdefinestyle{exercisestyle}{%
+ outerlinewidth=1em,%
+ outerlinecolor=white,%
+ leftmargin=-1em,%
+ rightmargin=-1em,%
+ middlelinewidth=1.2pt,%
+ roundcorner=5pt,%
+ linecolor=black,%
+ backgroundcolor=blue!5,
+ innertopmargin=1.2\baselineskip,
+ skipabove={\dimexpr0.5\baselineskip+\topskip\relax},
+ skipbelow={-1em},
+ needspace=3\baselineskip,
+ frametitlefont=\sffamily\bfseries,
+ settings={\global\stepcounter{exercise}},
+ singleextra={%
+ \node[titlered,xshift=1cm] at (P-|O) %
+ {~\mdf@frametitlefont{\theexercise}~};},%
+ firstextra={%
+ \node[titlered,xshift=1cm] at (P-|O) %
+ {~\mdf@frametitlefont{\theexercise}~};},
+}
+\makeatother
+
+\newenvironment{MyExercise}%
+{\begin{mdframed}[style=exercisestyle]}{\end{mdframed}}
+
+\theoremstyle{definition}
+\newtheorem{definition}{Definition}
+
+\theoremstyle{definition}
+\newtheorem{question}{Question}
+
+\theoremstyle{definition}
+\newtheorem{example}{Example}
+
+\theoremstyle{theorem}
+\newtheorem{theorem}{Theorem}
+
+\theoremstyle{theorem}
+\newtheorem{lemma}{Lemma}
+
+\theoremstyle{theorem}
+\newtheorem{proposition}{Proposition}
+
+\newtheorem*{idea}{Proof Idea}
+
+
+\title{Notes on \\ Noncommutative Geometry and Particle Physics}
+\author{Popovic Milutin}
+\date{Week 7: 23.04 - 27.04}
+
+\begin{document}
+
+\maketitle
+\tableofcontents
+
+\section{Classification of Finite Real Spectral Triples}
+
+Here we classify finite real spectral triples modulo unitary equivalence with
+\textit{Krajewski Diagrams}. We extend $\Lambda$-decorated graphs to the case of
+real spectral triples (grading and real structure).
+
+\textbf{The Algebra:}Like before:
+\begin{align}
+ A\simeq \bigoplus_{i=1}^N M_{n_i}(\mathbb{C}) \;\;\;\;\;\;\; \text{with} \;\;\; \hat{A} = \{\textbf{n}_1, \dots, \textbf{n}_N\}
+\end{align}
+Where $\textbf{n}_i$ are irreducible representation of $A$ on
+$\mathbb{C}^{n_i}$
+
+\textbf{The Hilbertspace:}Faithful irreducible representation on $A$ are the
+direct sum of $\mathbb{C}^{n_i}$'s, which act on $A$ by left block-diagonal
+matrix multiplication.
+\begin{align}
+ \bigoplus_{i=1}^N \mathbb{C}^{n_i}
+\end{align}
+Furthermore we need a representation of $A^\circ$ on $H$ that commutes with
+$A$. That is
+\begin{align}
+ A^\circ \simeq &\bigoplus_{i=1}^N M_{n_i}(\mathbb{C})^\circ \\
+ \text{with} \;\;\; &\hat{A}^\circ = \{\textbf{n}_1^\circ, \dots,
+ \textbf{n}_N^\circ\} \\
+ \text{and} \;\;\; &\bigoplus_{i=1}^N \mathbb{C}^{n_i\circ}
+\end{align}
+And we need the multiplicity space $V_{ij}$ of $\mathbb{C}^{n_i} \otimes
+\mathbb{C}^{n_j\circ}$.
+Thus making the Hilbertspace:
+\begin{align}
+ H=\bigoplus_{i,j=1}^N \mathbb{C}^{n_i} \otimes \mathbb{C}^{n_j\circ}
+ \otimes V_{ij}
+\end{align}
+\begin{itemize}
+ \item $\textbf{n}_i$, $\textbf{n}_j^\circ$ form a grid
+ \item if there is a node at $(\textbf{n}_i$, $\textbf{n}_j^\circ)$ then
+ $\mathbb{C}^{n_i} \otimes \mathbb{C}^{n_j\circ}$ is nonzero in $H$.
+ \item multiplicity implies multiple nodes
+\end{itemize}
+
+\begin{example}
+ $A = \mathbb{C} \oplus M_2 (\mathbb{C})$, two options of the Hilbertspace.
+ \begin{figure}[h!] \centering
+ \begin{tikzpicture}[
+ dot/.style = {draw, circle, inner sep=0.06cm},
+ no/.style = {},
+ ]
+ \node[no](a) at (0,0) [label=left:$\textbf{1}^\circ$] {};
+ \node[no](b) at (0, -1) [label=left:$\textbf{2}^\circ$] {};
+ \node[no](c) at (1, 0.5) [label=above:$\textbf{1}$] {};
+ \node[no](d) at (2, 0.5) [label=above:$\textbf{2}$] {};
+ \node[dot](d0) at (1,0) [] {};
+ \node[dot](d0) at (2,-1) [] {};
+
+ \node[no](a1) at (6,0) [label=left:$\textbf{1}^\circ$] {};
+ \node[no](b2) at (6, -1) [label=left:$\textbf{2}^\circ$] {};
+ \node[no](c2) at (7, 0.5) [label=above:$\textbf{1}$] {};
+ \node[no](d2) at (8, 0.5) [label=above:$\textbf{2}$] {};
+ \node[dot](d0) at (7,0) [] {};
+ \node[dot](d0) at (8,0) [] {};
+
+
+ \end{tikzpicture}
+ \end{figure}
+
+The first diagram corresponds to $H_1 = \mathbb{C} \oplus M_2(\mathbb{C})$,
+to the second $H_2 = \mathbb{C} \oplus \mathbb{C}^2$.
+\end{example}
+
+\begin{MyExercise}
+ \textbf{Let $J$ be an anti-unitary operator on a finite-dimensional Hilbert space.
+ Show that $J^2$ is an unitary operator
+ } \newline
+
+ Straight forward, say $J:\; H \rightarrow H$, then let $\xi_1, \xi_2 \in
+ H$:
+
+ \centering
+ \begin{align}
+ <J^2 \xi_1, J^2 \xi_2> &= <J(J\xi_1), J(J\xi_2)> =\\
+ &= <J\xi_2, J\xi_1> = <\xi_1, \xi_2>
+ \end{align}
+\end{MyExercise}
+
+\textbf{The real Structure:} $J:\; H \rightarrow H$.
+\begin{lemma}
+ \label{lemma}
+ Let $J$ be an anti-unitary operator on a finite-dimensional Hilbertspace
+ $H$ with $J^2 = \pm 1 $
+ \begin{enumerate}
+ \item If $J^2 = 1 \;\; \Rightarrow \;\; \exists$ an ONB $\{e_k\}$ of $H$\\
+ with $Je_k = e_k$.
+ \item If $J^2 = -1 \;\; \Rightarrow \;\; \exists$ an ONB $\{e_k, f_k\}$ of $H$\\
+ with $Je_k = f_k$ and consequently $Jf_k = -e_k$.
+ \end{enumerate}
+\end{lemma}
+\begin{proof}
+ \textbf{1.} $J^2 = 1$\newline
+
+ $v\in H$ and set:
+ \begin{align}
+ e_1 :=
+ \begin{cases}
+ c (v + Jv)\;\;\; \text{if}\;\;\; Jv \neq -v \\
+ iv\;\;\; \text{if}\;\;\; Jv = -v
+ \end{cases}
+ \end{align}
+ Where $c$ is a normalization constant, then take $Je_1$
+ \begin{align}
+ &J(v + Jv) = Jv + J^2v= v + Jv \;\;\;\; \text{and} \\
+ &J(iv) = -iJv = iv\\
+ &\Rightarrow Je_1 = e_1
+ \end{align}
+ Take $v'\perp e_1$ making:
+ \begin{align}
+ <e_1 , Jv'> = <J^2 v', Je_1> = <v' , Je_1>= <v', e_1> =0
+ \end{align}
+ Construct $e_2 \perp e_1$ with $v'$:
+ \begin{align}
+ e_2 :=
+ \begin{cases}
+ c (v' + Jv')\;\;\; \text{if}\;\;\; Jv' \neq -v' \\
+ iv'\;\;\; \text{if}\;\;\; Jv' = -v'
+ \end{cases}
+ \end{align}
+ Do this $k$ times and get $\{e_k\}$ ONB of $H$ for $J^2 = 1$.
+ \newline
+
+ \textbf{2.} $J^2 = -1$\newline
+ $v \in H$ and set $e_1 = cv$, $c$ normalization constant.
+ Then we set $f_1 = Je_1$ with $f_1 \perp e_1$, this is automatically the
+ case because:
+ \begin{align}
+ <f_1, e_1> &= <Je_1, e_1> = -<Je_1 , J^2e_1> =\\
+ &= -<Je_1, e_1> = -<f_1, e_1>
+ \end{align}
+ this only holds for 0. Then take some $v' \perp e_1, f_1$ and set\\
+ $e_2 =c 'v'$ and $f_2 = Je_2 \perp e_2, f_1, e_1$.
+ \begin{align}
+ &<e_1, f_2> = <e_1, Je_2> = -<J^2e_1, Je_2> = -<e_2, Je_1> = -<e_2,
+ f_1>=0\\
+ &<f_1, f_2> = <Je_1, Je_2> = <e_2, e_1> = 0.
+ \end{align}
+ Do this $k$ times and get $\{e_k, f_k\}$ ONB of $H$ for $J^2 = -1$
+
+\end{proof}
+
+Apply Lemma \ref{lemma} to the real structure $J$ on a spectral triple. $J$
+implements right action of $A$ on $H$ with
+\begin{align}
+ a^\circ = Ja^* J^{-1}
+\end{align}
+and satisfying $[a, b^\circ]=0$. With the block form of $A$, this implies
+\begin{align}
+ J(a^*_1 \oplus \cdots \oplus a_N^*) = (a^\circ_1 \oplus \cdots \oplus
+ a_N^\circ)J.
+\end{align}
+With this we can conclude that the Krajewski diagram for a real finite spectral
+triple is symmetric along the diagonal.$J$ hast then the following bilinear
+mapping:
+\begin{align}
+ J:\;\; \mathbb{C}^{n_i} \otimes \mathbb{C}^{n_j\circ} \otimes V_{ij}
+ \rightarrow \mathbb{C}^{n_j} \otimes \mathbb{C}^{n_i\circ} \otimes V_{ji}.
+\end{align}
+
+\begin{proposition}
+ \label{proposition}
+ Let $J$ be a real structure on a finite real spectral triple $(A, H , D;
+ J)$.
+ \begin{enumerate}
+ \item If $J^2 = 1$ (K0-dimension 0, 1, 6, 7) $Rightarrow \;\; \exists$
+ an ONB $\{e_k^{(ij)}\}$\\ with $e_k^{(ij)} \in \mathbb{C}^{n_i} \otimes
+ \mathbb{C}^{n_j\circ} \otimes V_{ij}$ such that
+ \begin{align}
+ Je_k^{(ij)} = e_k^{(ij)} \;\;\; (i, j = 1,\dots,N;\; k=1,\dots
+ dim(V_{ij}))
+ \end{align}
+ \item If $J^2 = -1$ (KO-dimension 2, 3, 4, 5) $\Rightarrow \;\; \exists$
+ ONB $\{e_k^{(ij)}, f_k^{(ji)}\}$ \\
+ with $e_k^{(ij)} \in \mathbb{C}^{n_i} \otimes \mathbb{C}^{n_j\circ}
+ \otimes V_{ij}$ and $f_k^{(ji)} \in \mathbb{C}^{n_j} \otimes
+ \mathbb{C}^{n_i\circ} \otimes V_{ji}$ such that
+ \begin{align}
+ Je_k^{(ij)} = f_k^{(ji)} \;\;\; (i\leq j=1,\dots, N;\;
+ k=1,\dots,dim(V_{ji})).
+ \end{align}
+ \end{enumerate}
+\end{proposition}
+\begin{proof}
+ Similar to Lemma \ref{lemma}.
+\end{proof}
+
+For whatever unknown reasons this implies that in the case of KO-dimension 2,
+3, 4, 5, diagonals $H_ii$ need to have even multiplicity.
+
+\textbf{The finite Dirac Operator:} Is a mapping between $H_{ij}$ to $H_{kl}$
+
+\begin{align}
+ D_{ij,kl}: \; \mathbb{C}^{n_i} \otimes
+ \mathbb{C}^{n_j\circ}\otimes V_{ij} \rightarrow \mathbb{C}^{n_k} \otimes
+ \mathbb{C}^{n_l\circ}\otimes V_{kl}
+\end{align}
+We have $D_{kl,ij} = D^*_{ij, kl}$. And in the diagram we have a line between
+the nodes $(\textbf{n}_i, \textbf{n}_j^\circ)$ and $(\textbf{n}_l,
+\textbf{n}_k^\circ)$. But instead of drawing directional lines draw a single
+undirected line that represents both $D_{ij, kl}$ and the adjoint $D_{kl, ij}$.
+
+\begin{lemma}
+ The conditions $JD = \pm DJ$ and $[[D,a], b^\circ] = 0$ imply that the
+ connections in the diagram run only vertically or horizontally and thereby
+ the diagonal symmetry between the nodes is preserved.
+\end{lemma}
+\begin{proof}
+ The condition $JD = \pm DJ$ has the following commutative diagram.
+
+\[
+\begin{tikzcd}
+ \mathbb{C}^{n_i\circ}\otimes \mathbb{C}^{n_j\circ}\otimes V_{ij}
+ \arrow[r,"D"] \arrow[d,swap,"J"] &
+ \mathbb{C}^{n_k\circ}\otimes \mathbb{C}^{n_l\circ}\otimes V_{kl} \arrow[d,"J"] \\
+\mathbb{C}^{n_j\circ}\otimes \mathbb{C}^{n_i\circ}\otimes V_{ji} \arrow[r,"\pm D"] &
+\mathbb{C}^{n_l\circ}\otimes \mathbb{C}^{n_k\circ}\otimes V_{lk}
+\end{tikzcd}
+\]
+ Relating $D_{ij, kl}$ to $D_{ji, lk}$ and maintaining diagonal symmetry.
+ Wit the condition $[[D, a], b^\circ]=0$ for the diagonal elements $a =
+ \lambda_1\mathbb{I}_{n_1}\oplus \cdots \oplus \lambda_N \mathbb{I}_{n_N}
+ \in A$ and $b = \mu_1\mathbb{I}_{n_1}\oplus \cdots \oplus \mu_N
+ \mathbb{I}_{n_N} \in A$, with some $\lambda _i , \mu _i \in \mathbb{C}$, we
+ can commute:
+ \begin{align}
+ D_{ij, kl} (\lambda _i - \lambda _k)(\bar{\mu}_j - \bar{\mu}_l)= 0
+ \end{align}
+ $\forall \lambda _i , \mu _j \in \mathbb{C}$, thus $D_ij, kl = 0$ for
+ $i\neq j$ or $j\neq i $.
+\end{proof}
+\textbf{The Grading:} $\gamma : \; H \rightarrow H$ each node gets labeled by a $+$
+or a $-$ sign.
+
+\begin{itemize}
+ \item D only connects nodes with different signs
+ \item If $(\textbf{n}_i, \textbf{n}_j^\circ)$ has a $\pm$ sing then
+ $(\textbf{n}_j, \textbf{n}_i^\circ)$ has a $\mp$, $\varepsilon''$ sign\\
+ according to $J\gamma = \varepsilon'' \gamma J$
+\end{itemize}
+
+\begin{definition}
+ A Krajewski Diagram of KO-dimension $k$ is an ordered pair $(\Gamma,
+ \Lambda)$ where $\Gamma$ is a finite graph and $\Lambda$ is a set of
+ positive integers with a labeling:
+
+ \begin{itemize}
+ \item of $v \in \Gamma^{(0)}$ of vertices by elements $\iota (v) =
+ (n(v), m(v))\; \in \; \Lambda \times \Lambda$, an edge from $v$ to
+ $v'$ implies that either $n(v) = n(v')$ or $m(v) = m('v)$ or both
+ \item of $e = (v_1, v_2) \in \Gamma^{(1)}$ edges with non-zero
+ operators $D_e$ and their adjoints $D_e^*$:
+ \begin{align*}
+ &D_e:\mathbb{C}^{n(v_1)} \rightarrow
+ \mathbb{C}^{n(v_2)}\;\;\;\;\; &\text{if} \;\;\;\; m(v_1) = m(v_2)\\
+ &D_e:\mathbb{C}^{m(v_1)} \rightarrow
+ \mathbb{C}^{m(v_2)}\;\;\;\;\; &\text{if} \;\;\;\; n(v_1) = n(v_2)
+ \end{align*}
+ \end{itemize}
+ Together with an involutive graph automorphism $j:\Gamma \Rightarrow
+ \Gamma$ such that the following conditions hold:
+ \begin{enumerate}
+ \item every row or column in $\Gamma \times \Gamma$ has non-empty
+ intersection with $\iota(\Gamma)$
+ \item for each vertex $v$ we have $n(j(v)) = m(v)$
+ \item for each edge $e$ we have $D_e = \epsilon' D_{j(e)}$
+ \item if KO dimension $k$ is even, then the vertices are labeled by
+ $\pm 1$ and the edges only connect opposite signs. The signs at $v$
+ and $j(v)$ differ by a factor of $\epsilon$
+ \item if the K0-dimension is 2, 3, 4, 5 then the inverse image of
+ $\iota$ of the diagonal elements in $\Lambda \times \Lambda$
+ contains an even number of vertices of $\Gamma$
+ \end{enumerate}
+\end{definition}
+With this definition we can label different vertices by the same element in
+$\Lambda \times \Lambda$ (accounting for the multiplicities in $V_{ij}$)
+\newline
+
+\textbf{Diagram:} To sum it up we have the following diagram
+\begin{itemize}
+ \item Node at $(\textbf{n}_i, \textbf{n}_j^\circ)$ for each vertex with that label
+ \item Operators $D_e$ add up to $D_{ij,kl}$ connecting nodes $(\textbf{n}_i,
+ \textbf{n}_j^\circ)$ with $(\textbf{n}_k, \textbf{n}_l^\circ)$
+ \begin{align}
+ D_{ij, kl} = \sum\limits_{\substack{e=(v_1, v_2) \in \Gamma^{(1)}
+ \\ \iota(v_1) = (\textbf{n}_i, \textbf{n}_j)\\
+ \iota(v_2)=(\textbf{n}_k, \textbf{n}_l)}} D_e
+ \end{align}
+ \item only vertical or horizontal connections
+\end{itemize}
+
+\begin{theorem}
+ There is a one-to-one correspondence between finite real spectral triples
+ $(A, H, D; J, \gamma)$
+ of K0-dimension $k$ modulo unitary equivalence and Krajewski diagrams of
+ KO-dimension $k$ in the following way:
+
+ \begin{align}
+ & A = \bigoplus_{n \in \Lambda} M_n(\mathbb{C})\\
+ & H = \bigoplus_{v \in \Gamma^{(0)}} \mathbb{C}^{n(v)} \otimes
+ \mathbb{C}^{m(v)\circ}\\
+ & D = \sum_{e\in \Gamma^{(1)}} D_e + D_e^*
+ \end{align}
+ The real structure $J:H\rightarrow H$ is given as as in Proposition
+ \ref{proposition} with a basis dictated by a graph automorphism $j: \Gamma
+ \rightarrow \Gamma$. The grading $\gamma$ is difened by setting $\gamma =
+ \pm 1$ on $\mathbb{C}^{n(v)} \otimes \mathbb{C}^{m(v)\circ} \subset H$
+ according to the labeling $\pm$ of the vertex $v$.
+\end{theorem}
+
+\begin{example}
+ $A = M_n(\mathbb{C})$ with $\hat{A} = {\textbf{n}}$. We have the following
+ Krajewski diagram.
+ \begin{figure}[h!] \centering
+ \begin{tikzpicture}[
+ dot/.style = {draw, circle, inner sep=0.06cm},
+ no/.style = {},
+ ]
+ \node[no](a) at (0,0) [label=left:$\textbf{n}^\circ$] {};
+ \node[no](c) at (0.25, 0.25) [label=above:$\textbf{n}$] {};
+ \node[dot](d0) at (0.25,0) [] {};
+ \end{tikzpicture}
+ \end{figure}
+ \begin{itemize}
+ \item We can label the node either with a $+$ or a $-$ sign, the choice being
+ irrelevant
+ \item $H = \mathbb{C}^n \otimes \mathbb{C}^{n\circ} \simeq
+ M_n(\mathbb{C})$
+ \item $\gamma$ trivial grading ($+1$)
+ \item $J$ is a combination of complex conjugation and the flip
+ $n\otimes n^\circ$ ($\Rightarrow M_n(\mathbb{C})$ as matrix adjoint)
+ \item Because node label is $\pm$ there is no non-zero Dirac operator
+ \item $\Rightarrow (A = M_n(\mathbb{C}), H=M_n(\mathbb{C}) , D=0; J=(\cdot)^*,
+ \gamma = 1)$
+ \end{itemize}
+\end{example}
+\section{Real Algebras and Krajewski Diagrams}
+
+\begin{definition}
+ A real Algebra is a Vector space $A$ over $\mathbb{R}$ with $A\times A
+ \rightarrow A$, $(a, b) \mapsto ab$ and $1a = a1 = a \;\; \forall a\in A$
+\end{definition}
+
+A real *-algebra is a real algebra with a bilinear map $*:A \rightarrow A$
+such that $(ab)^* = b^*a^*$ and $(a^*)^* \;\;\; \forall a,b\in A$
+\begin{example}
+ Real *-algebra of quaternions $\mathbb{H}$ subalgebra of $M_2(\mathbb{C})$.
+ \begin{align}
+ \mathbb{H} = \{ \begin{pmatrix}\alpha & \beta \\ -\bar{\beta} &
+ \bar{\alpha}\end{pmatrix} : \alpha, \beta \in
+ \mathbb{C}\}
+ \end{align}
+ $\mathbb{H}$ consists of matricies that commute in $M_2(\mathbb{C})$ with
+ the operator $I$ defined by:
+ \begin{align}
+ I\begin{pmatrix}v_1 \\ v_2\end{pmatrix} = \begin{pmatrix}-\bar{v}_2 \\
+ \bar{v}_1\end{pmatrix}
+ \end{align}
+ The involution is the hermitian conjugation of $M_2(\mathbb{C})$.
+\end{example}
+\begin{MyExercise}
+ \textbf{
+ \begin{enumerate}
+ \item Show that $\mathbb{H}$ is a real *-algebra which contains a
+ real subalgebra isomorphic to $\mathbb{C}$.
+ \item Show that $\mathbb{H} \otimes_\mathbb{R} \mathbb{C} \simeq
+ M_2(\mathbb{C})$ as complex *-algebras.
+ \item Show that $M_k(\mathbb{H})$ is areal *-algebra for any $k$
+ \item Show that $M_k(\mathbb{H} \otimes_{\mathbb{R}} \mathbb{C}
+ \simeq M_{2k}(\mathbb{C})$ as complex *algebras.
+ \end{enumerate}
+ }
+\end{MyExercise}
+\begin{definition}
+ A representation of a finite-dimensional real * algebra $A$ is a pair $(\pi
+ , H$), $H$- Hilbertspace, $\pi : A \rightarrow L(H)$
+\end{definition}
+\begin{MyExercise}
+ Show that there is a one-to-one correspondence between Hilbertspace
+ representations of real *-algebras $A$ and complex representations of its
+ complexification $A\otimes _\mathbb{R} \mathbb{C}$. Conclude that the
+ unique irreducible Hilbertspace representation of $M_k(\mathbb{H})$ is
+ $\mathbb{C}^{2k}$
+\end{MyExercise}
+\begin{lemma}
+ Real *-algebra $A$ represented faithfully on a finite dimensional
+ Hilbertspace $H$ through a real linear *-algebra map $\pi: A \rightarrow
+ L(H)$ hen $A$ is a matrix algebra.
+ \begin{align}
+ A \simeq \bigoplus _{i=1}^N M_{n_i} (\mathbb{F}_i)
+ \end{align}
+ Where $\mathbb{F}_i = \mathbb{R}, \mathbb{C}, \mathbb{H}$ depending on $i$.
+\end{lemma}
+\begin{proof}
+ $\pi$ allows $A$ to be considered as a real *-subalgebra of
+ $M_{dim(H)}(\mathbb{C}) \Rightarrow A+iA$ complex *-subalgebra of
+ $M_{dim(H)}(\mathbb{C})$. Then $A+iA$ is a matrix algebra and $A+iA =
+ M_k(\mathbb{C})$ for $k \geq 1$. Thus we have
+ \begin{align}
+ A \cap iA =
+ \begin{cases}
+ \{0\} \;\;\;\; \text{if $A = M_k(\mathbb{C})$}\\
+ A+iA = M_k(\mathbb{C})
+ \end{cases}
+ \end{align}
+ Furthermore $A$ is a fixed point algebra of an anti-linear automorphism
+ $\alpha$ of $M_k(\mathbb{C})$ with $\alpha(a+ib) = a-ib$ for $a, b \in A$.
+ Implement $\alpha$ by an anti-linear isometry $I$ on $\mathbb{C}^n$ such
+ that $\alpha (x) = I\times I^{-1}\;\;\;\ \forall x\in M_k(\mathbb{C})$.
+ Now since $\alpha^2 = 1$, $I^2$ commutes with $M_k(\mathbb{C})$ and is
+ proportional to a complex scalar $I^2 = \pm 1 $ and A is the commutant of
+ $I$
+ \begin{itemize}
+ \item if $I^2 = 1 \;\;\ \Rightarrow \;\; \exists \;\;\ \{e_i\}$ ONB of
+ $\mathbb{C}^k$ with $Ie_i = e_i$, then $A=M_k(\mathbb{R})$
+ \item if $I^2 = -1 \;\;\ \Rightarrow \;\; \exists \;\;\ \{e_i,f_i\}$ ONB of
+ $\mathbb{C}^k$ with $Ie_i = f_i$, ($k$ even)\\
+ Therefor $I$ must be a $k/2 \times k/2$ matrix because of commutation with
+ $M_k(\mathbb{C})$, then $A = M_{k/2} (\mathbb{H})$
+ \end{itemize}
+\end{proof}
+The Krajewski diagrams can also classify real algebras, as long as we take
+$\mathbb{F}_i$ for each $i$ into account. That is we enhance the set $\Lambda$
+to be
+\begin{align}
+ \Lambda = \{ \textbf{n}_1 \mathbb{F}_1,\dots, \textbf{n}_N \mathbb{F}_N\}
+\end{align}
+Reducing in to the previous $\Lambda$ if all $\mathbb{F}_i = \mathbb{C}$.
+\section{Classification of Irreducible Geometries}
+Classify irreducible real spectral triples based on $M_N(\mathbb{C} \oplus
+M_N(\mathbb{C})$ for some $N$
+\begin{definition}
+ A finite real spectral triple $(A, H, D; J, \gamma)$ is called irreducible
+ if the triple $(A, H, J)$ is irreducible, that is when
+ \begin{enumerate}
+ \item The representation of $A$ and $J$ on $H$ are irreducible
+ \item The action of $A$ on $H$ has a separating vector
+ \end{enumerate}
+\end{definition}
+
+\begin{theorem}
+ Let $(A, H, D; J, \gamma)$ be an irreducible finite real spectral triple of
+ KO-dimension 6. Then exists a positive integer $N$ such that $A \simeq
+ M_N(\mathbb{C}) \oplus M_N(\mathbb{C})$.
+\end{theorem}
+\begin{proof}
+ Let $(A, H, D; J, \gamma)$ be an arbitrary finite real spectral triple,
+ corresponding to
+ \begin{align}
+ &A = \bigoplus_i^{N} M_{n_i}(\mathbb{C})\\
+ &H = \bigoplus_{i,j=1}^N \mathbb{C}^{n_i} \otimes
+ \mathbb{C}^{n_j\circ} \otimes V_{ij}
+ \end{align}
+ Remember that each $\mathbb{C}^{n_i} \otimes \mathbb{C}^{n_j}$ is a
+ irreducible representation of $A$. In order for $H$ to support the real
+ structure $J$ we need both $\mathbb{C}^{n_i} \otimes \mathbb{C}^{n_j}$
+ and $\mathbb{C}^{n_j} \otimes \mathbb{C}^{n_i}$. With Lemma \ref{lemma}
+ with $J^2 = 1$ with multiplicity $dim(V_{ij}) = 1$ we have such a
+ structure. Hence
+ \begin{align}
+ H = \mathbb{C}^{n_i} \otimes \mathbb{C}^{n_j} \oplus \mathbb{C}^{n_j}
+ \otimes \mathbb{C}^{n_i}
+ \end{align}
+ For $i,j \in \{1, \dots, N\}$
+ \newline
+
+ For the second condition (existence of the separating vector). The
+ representations of $A$ in $H$ are only faithful if $A = M_{n_i}(\mathbb{C})
+ \oplus M_{n_j}(\mathbb{C})$. The stronger condition applies $n_i = n_j$
+ then we have $A' \xi = H$ with the commutant of $A$ and $\xi \in H$ the
+ separating vector. Normally since $A' = M_{n_j}(\mathbb{C}) \oplus
+ M_{n_i}(\mathbb{C})$ with $dim(A') = n_i^2 + n_j^2$ and $dim(H) = 2n_i n_j$
+ we have a equality $n_i = n_j$.
+\end{proof}
+\end{document}
+
+
+
+
+
