commit 08728ff7db193b88e1a312f16cd0404357269f9e
parent fc85583bf898ab83bf2f390cfd910204d6aeaed8
Author: miksa234 <milutin@popovic.xyz>
Date: Tue, 16 Mar 2021 20:35:57 +0100
week5 almost done need to add worked exercises
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| A | week5.tex | | | 416 | +++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ |
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+\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{tikz}
+\usetikzlibrary{patterns,decorations.pathmorphing,positioning}
+
+\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{exercise}{Exercise}
+
+\theoremstyle{definition}
+\newtheorem{solution}{Solution}
+
+\newtheorem*{idea}{Proof Idea}
+
+
+\title{Notes on \\ Noncommutative Geometry and Particle Physics}
+\author{Popovic Milutin}
+\date{Week 5: 12.03 - 19.03}
+
+\begin{document}
+
+\maketitle
+\tableofcontents
+
+\section{Noncommutative Geometric Spaces }
+\subsection{Exercises}
+\begin{exercise}
+ Make the proof of the last theorem (see week4.pdf) explicit for $N=3$
+\end{exercise}
+\begin{solution}
+ For the C* algebra we have $A=\mathbb{C}^3$
+ For $H$ we have $H = (\mathbb{C}^2)^{\oplus 3} = H_2 \oplus H_2^1 \oplus H_2^2$.
+ The symmetric operator $D$ acting on $H$ and the representation $\pi (a)$:
+ \begin{align}
+ \pi((a(1), a(2), a(3)) &=
+ \begin{pmatrix}
+ a(1) & 0 \\ 0 & a(2)
+ \end{pmatrix} \oplus
+ \begin{pmatrix}
+ a(1) & 0 \\ 0 & a(3)
+ \end{pmatrix} \oplus
+ \begin{pmatrix}
+ a(2) & 0 \\ 0 & a(2)
+ \end{pmatrix} \nonumber \\
+ & =
+ \begin{pmatrix}
+ a(1) & 0 & 0 & 0 & 0 & 0 \\
+ 0 & a(2) & 0 & 0 & 0 & 0 \\
+ 0 & 0 & a(1) & 0 & 0 & 0 \\
+ 0 & 0 & 0 & a(3) & 0 & 0 \\
+ 0 & 0 & 0 & 0 & a(2) & 0 \\
+ 0 & 0 & 0 & 0 & 0 & a(3)
+ \end{pmatrix} \\
+ D &=
+ \begin{pmatrix}
+ 0 & x_1 \\ x_1 & 0
+ \end{pmatrix} \oplus
+ \begin{pmatrix}
+ 0 & x_1 \\ x_1 & 0
+ \end{pmatrix} \oplus
+ \begin{pmatrix}
+ 0 & x_1 \\ x_1 & 0
+ \end{pmatrix} \nonumber \\
+ &=
+ \begin{pmatrix}
+ 0 & x_1 & 0 & 0 & 0 & 0 \\
+ x_1 & 0 & 0 & 0 & 0 & 0 \\
+ 0 & 0 & 0 & x_2 & 0 & 0 \\
+ 0 & 0 & x_2 & 0 & 0 & 0 \\
+ 0 & 0 & 0 & 0 & 0 & x_3 \\
+ 0 & 0 & 0 & 0 & x_3 & 0 \\
+ \end{pmatrix} \\
+ \end{align}
+ Then the norm of the commutator would be the largest eigenvalue
+ \begin{align}
+ &||[D, \pi(a)]|| = ||D\pi(a) - \pi(a)D||\nonumber \\
+ &=
+ \left|\left|
+ \setlength{\arraycolsep}{0.1cm}
+ \renewcommand{\arraystretch}{0.1}
+ \begin{pmatrix}
+ 0 & x_1(a(2)-a(1)) & 0 & 0 & 0 & 0 \\
+ -x_1(a(2)-a(1)) & 0 & 0 & 0 & 0 & 0 \\
+ 0 & 0 & 0 & x_2(a(3)-a(1)) & 0 & 0 \\
+ 0 & 0 & -x_2(a(3)-a(1)) & 0 & 0 & 0 \\
+ 0 & 0 & 0 & 0 & 0 & x_3(a(3)-a(2)) \\
+ 0 & 0 & 0 & 0 & -x_3(a(2)-a(3)) & 0 \\
+ \end{pmatrix}\right|\right| \label{skew matrix}
+ \end{align}
+The matrix in Equation \ref{shew matrix} is a skew symmetric matrix its eigenvalues
+are $i\lambda_1, i\lambda_2, i\lambda_3, i\lambda_4$, where the $\lambda$'s are on the
+upper and lower diagonal check \url{https://en.wikipedia.org/wiki/Skew-symmetric_
+matrix#Skew-symmetrizable_matrix}. The matrix norm of would be the maximum of the norm of
+the larges eigenvalues:
+\begin{align}
+ ||[D, \pi(a)]|| = \max_{a\in A}\{x_i|a(j)-a(k)|\}
+\end{align}
+\end{solution}
+
+\begin{exercise}
+ Compute the metric on the space of three points given by $d_{ij} =
+ \sup_{a\in A}\{|a(i) - a(j)|: ||[D, \pi(a)]|| \leq 1\}$ for the set of data
+ $A = \mathbb{C}^3$ acting in the defining representation $H = \mathbb{C}^3$, and
+ \begin{align*}
+ D =
+ \begin{pmatrix}
+ 0 & d^{-1} & 0 \\
+ d^{-1} & 0 & 0 \\
+ 0 & 0 & 0
+ \end{pmatrix}
+ \end{align*}
+ for some $d \in \mathbb{R}$
+\end{exercise}
+\begin{solution}
+ We have $A=\mathbb{C}^3$, $H=\mathbb{C}^3$ and $D$ from above, then
+
+ \begin{align}
+ ||[D, \pi(a)]|| &= d^{-1}\left|\left|
+ \begin{pmatrix}
+ 0 & a(2)-a(1) & 0 \\
+ -(a(2)-a(1)) & 0 & 0 \\
+ 0 & 0 & 0
+ \end{pmatrix} \right|\right| \\
+ &= d^{-1} |a(2) - a(1)|
+ \end{align}
+\end{solution}
+\begin{exercise}
+ Show that $d_{ij}$ from Equation \ref{ext metric} is a metric on $\hat{A}$ by
+ establishing that:
+ \begin{align}
+ d_{ij} &= 0\;\;\; \Leftrightarrow \;\;\; i=j \label{metric 1} \\
+ d_{ij} &= d_{ji} \label{metric 2}\\
+ d_{ij} &\leq d_{ik} + d_{kj} \label{metric 3}
+ \end{align}
+ \begin{equation} \label{ext metric}
+ d_{ij} = \sup_{a\in A}\big\{|\text{Tr}(a(i)) - \text{Tr}((a(j))|: ||[D, a]|| \leq 1\big\}
+ \end{equation}
+\end{exercise}
+\begin{solution}
+For Equation \ref{metric 1} set $i=j$ in \ref{ext metric}.
+\begin{align*}
+ d_{ii} &= \sup_{a \in A}\{|\text{Tr}(a(i)) - \text{Tr}((a(i))|: ||[D, a]|| \leq
+ 1\big\} \\
+ &= \sup_{a \in A}\{0: ||[D, a]|| \leq 1\big\} = 0
+\end{align*}
+For Equation \ref{metric 2} obviously we have the commuting property of
+addition.
+\newline
+For Equation \ref{metric 3}, for $k=j$ then $d_{kj} = 0$ and the equality
+holds. For $i = k$ then $d_{ik} = 0$ and equality holds. Else set $d_{ik} =
+1$ and $d_{kj} = 1$ then $d_{ij} = 1 \leq d_{ik} + d_{kj} = 2$
+\end{solution}
+
+\subsection{Properties of Matrix Algebras}
+\begin{lemma}
+ If $A$ is a unital C* algebra that acts faithfully on a finite
+ dimensional Hilbert space, then $A$ is a matrix algebra of the Form:
+ \begin{equation}
+ A \simeq \bigoplus _{i=1}^N M_{n_i}(\mathbb{C})
+ \end{equation}
+\end{lemma}
+\begin{proof}
+ Since $A$ acts faithfully on a Hilbert space, then $A$ is a C*
+ subalgebra of a matrix algebra $L(H) = M_{\dim (H)}(\mathbb{C}
+ \Rightarrow A \simeq \text{Matrix algebra}$.
+\end{proof}
+
+\begin{question}
+ What does the author mean when he sais 'acts faithfully on a
+ Hilbertspace`? Then the representation is fully reducible, or that the
+ presentation is irreducible?
+\end{question}
+
+\begin{example}
+ $A = M_n(\mathbb{C})$ and $H=\mathbb{C}^n$, $A$ acts on $H$ with matrix
+ multiplication and standard inner product. $D$ on $H$ is a hermitian
+ matrix $n\times n$ matrix.
+\end{example}
+
+$D$ is referred to as a finite Dirac operator as in as its $\infty$
+dimensional on Riemannian Spin manifolds coming in Chapter 4. Now we
+introduce it as
+\begin{equation}
+ \frac{a(i)-a(j)}{d_{ij}}
+\end{equation}
+for each pair $i$, $j$ $\in X$ the finite dimensional discrete space.
+This appears in the entries in the commutator $[D, a]$ in the above
+exercises.
+\begin{definition}
+ Given an finite spectral triple $(A, H, D)$, the $A$-bimodule of
+ Connes' differential one form is:
+ \begin{equation}
+ \Omega _D ^1 (A) := \left\{ \sum _k a_k[D, b_k]: a_k, b_k \in A \right\}
+ \end{equation}
+\end{definition}
+
+\begin{question}
+ Is the Conne's differential one form the set of all '1st order
+ differential operators` given $A$, that act on $H$?
+\end{question}
+Then there is a map $d:A\rightarrow \Omega _D ^1 (A)$, $d = [D, \cdot]$.
+
+\begin{lemma}
+ Let $(A, H, D) = (M_n(\mathbb{C}, \mathbb{C}^n, D)$, with $D$ a hermitian
+ $n\times n$ matrix. If $D$ is not a multiple of the identity then:
+ \begin{equation}
+ \Omega _D ^1 (A) \simeq M_n(\mathbb{C}) = A
+ \end{equation}
+\end{lemma}
+
+\begin{proof}
+ Assume $D = \sum _i \lambda _i e_{ii}$ (diagonal), $\lambda _i \in \mathbb{R}$ and
+ $\{e_{ij}\}$ the basis of $M_n(\mathbb{C}$. For fixed $i$, $j$ choose $k$
+ such that $\lambda _k \neq \lambda _j$ then
+ \begin{align*}
+ \left(\frac{1}{\lambda _k - \lambda _j} e_{ik}\right) [D, e_{kj}] =
+ e_{ij}
+ \end{align*}
+ $e_{ij}\in \Omega _D ^1 (A)$ by the above definition. And $\Omega _D ^1
+ (A) \subset L(\mathbb{C}^n) = H \simeq M_n(\mathbb{C}) = A$
+\end{proof}
+
+\subsection{Morphisms Between Finite Spectral Triples}
+\begin{definition}
+ two finite spectral tripes $(A_1, H_1, D_1)$ and $(A_2, H_2, D_2)$ are
+ called unitarily equivalent if
+ \begin{itemize}
+ \item $A_1 = A_2$
+ \item $\exists \;\; U: H_1 \rightarrow H_2$, unitary with
+ \begin{enumerate}
+ \item $U\pi_1(a)U^* = \pi_2(a)$ with $a \in A_1$
+ \item $UD_1 U^* = D_2$
+ \end{enumerate}
+ \end{itemize}
+\end{definition}
+
+Some remarks
+\begin{itemize}
+ \item the above is an equivalence relation
+ \item spectral unitary equivalence is given by the unitaries of the
+ matrix algebra itself
+ \item for any such $U$ then $(A, H, D) \sim (A, H, UDU^*)$
+ \item $UDU^* = D + U[D, U^*]$ of the form of elements in
+ $\Omega _D^1 (A)$.
+\end{itemize}
+
+Extending the this relation we look again at the notion of equivalence from
+Morita equivalence of Matrix Algebras.
+\newline
+
+Given a Hilbert bimodule $E \in KK_f(B, A)$ and $(A, H, D)$ we construct
+a finite spectral triple on $B$, $(B, H', D')$
+\begin{equation}
+ H' = E \otimes _A H
+\end{equation}
+This extends the left action on $B$ with the right action and inherits the
+$\mathbb{C}$ valued inner product space.
+\begin{equation}
+ D'(e\otimes \xi) = e \otimes D \xi + \nabla (e) \xi \;\;\;\; e\in
+ E, a\in A
+\end{equation}
+Where $\nabla$ is called the \textit{connection on the right A-module E}
+associated with the derivation $d=[D, \cdot]$ and satisfying the
+\textit{Leibnitz Rule} which is
+\begin{equation}
+ \nabla(ae) = \nabla(e)a + e \otimes [D, a] \;\;\;\;\; e\in E,\; a\in A
+\end{equation}
+Then the linearity of the balanced tensor product $E \otimes _A H$ is
+satisfied
+\begin{align*}
+ D'(ea \otimes \xi - e \otimes a \xi) &= D'(ea \otimes \xi) - D'(e
+ \otimes \xi) \\
+ &= ea\otimes D\xi + \nabla(ae) \xi - e \otimes D(a\xi ) - \nabla (e)a
+ \xi \\
+ &= 0
+\end{align*}
+With the information thus far we can prove the following theorem
+\begin{theorem}
+ If $(A, H, D)$ a finite spectral triple, $E \in KK_f(B, A)$.
+ Then $(V, E\otimes _A H, D')$ is a finite spectral triple, provided that
+ $\nabla$ satisfies the compatibility condition
+ \begin{equation}
+ \langle e_1, \nabla e_2 \rangle _E - \langle \nabla e_1, e_2
+ \rangle _E = d\langle e_1, e_2 \rangle _E \;\;\;\; e_1, e_2 \in E
+ \end{equation}
+\end{theorem}
+\begin{proof}
+ $E\otimes _A H$ was shown in the previous section (text before the
+ theorem). The only thing left is to show that $D'$ is a symmetric
+ operator, this we can just compute. Let $e_1, e_2 \in E$ and $\xi _1,
+ \xi _2 \in H$ then
+ \begin{align*}
+ \langle e_1 \otimes \xi _1, D'(e_2 \otimes \xi_2)\rangle _{E\otimes _A H} &=
+ \langle \xi _1, \langle e_1, \nabla e_2\rangle _E \xi _2\rangle + \langle \xi _1 , \langle e_1, e_2\rangle _E D\xi
+ _2\rangle _H \\
+ &= \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
+ \xi _2\rangle _H \\
+ &+ \langle D\xi _1,\langle e_1, e_2\rangle _E \xi _2\rangle _H - \langle \xi _1, [D, \langle e_1, e_2\rangle _E] \xi
+ _2 \rangle _H \\
+ &= \langle D'(e_1 \otimes \xi _1), e_2 \otimes \xi _2\rangle _{E \otimes _A H}
+ \end{align*}
+\end{proof}
+
+\subsection{Graphing Finite Spectral Triples}
+\begin{definition}
+ A \textit{graph} is a ordered pair $(\Gamma ^{(0)}, \Gamma ^{(1)})$.
+ Where $\Gamma ^{(0)}$ is the set of vertices (nodes) and $\Gamma ^{(1)}$
+ a set of pairs of vertices (edges)
+\end{definition}
+\begin{figure}[h!]
+ \centering
+\begin{tikzpicture}[
+ mass/.style = {draw,circle, minimum size=0.2cm, inner sep=0pt, thick},
+ spring/.style = {decorate,decoration={zigzag, pre length=1cm,post length=1cm,segment length=5pt}},]
+ \node[mass] (m1) at (1,1.5) {};
+ \node[mass] (m2) at (-1,1.5) {};
+ \node[mass] (m3) at (0,0) {};
+
+ \draw (m1) -- (m2);
+ \draw (m1) -- (m3);
+ \draw (m2) -- (m3);
+ \end{tikzpicture}
+ \caption{A simple graph with three vertices and three edges}
+\end{figure}
+\begin{definition}
+ A $\Lambda$-decorated graph is given by an ordered pair $(\Gamma,
+ \Lambda)$ of a finite graph $\Gamma$ and a set of positive integers
+ $\Lambda$ with the labeling
+ \begin{itemize}
+ \item of the vetices $v\in \Gamma ^{(0)}$ given by $n(\nu) \in
+ \Lambda$
+ \item of the edges $e = (\nu _1, \nu _2) \in \Gamma ^{(1)}$ by
+ operators
+ \begin{itemize}
+ \item $D_e: \mathbb{C}^{n(\nu _1)} \rightarrow
+ \mathbb{C}^{n(\nu _2)}$
+ \item and $D_e^*: \mathbb{C}^{n(\nu _2)} \rightarrow
+ \mathbb{C}^{n(\nu _1)}$ its conjugate traspose
+ (pullback?)
+ \end{itemize}
+ \end{itemize}
+ such that
+ \begin{equation}
+ n(\Gamma ^{(0)}) = \Lambda
+ \end{equation}
+\end{definition}
+\begin{question}
+ Would then $D_e$ be the pullback?
+\end{question}
+\begin{question}
+ These graphs are important in the next chapter I should look
+ into it more, I don't understand much here, specific
+ how to construct them with the abstraction of a spectral triple...
+\end{question}
+
+The operator $D_e$ between $\textbf{n}_i$ and $\textbf{n}_j$ add up to
+$D_{ij}$
+\begin{align*}
+ D_{ij} = \sum\limits_{\substack{e = (\nu _1, \nu _2) \\ n(\nu _1) =
+ \textbf{n}_i \\ n(\nu _2) = \textbf{n}_j}} D_e
+\end{align*}
+
+\begin{theorem}
+ There is a on to one correspondence between finite spectral triples
+ modulo unitary equivalence and $\Lambda$-decorated graphs, given by
+ associating a finite spectral triples $(A, H, D)$ to a $\Lambda$ decorated
+ graph $(\Gamma, \Lambda)$ in the following way:
+ \begin{equation}
+ A = \bigoplus _{n\in \Lambda} M_n(\mathbb{C}); \;\;\;
+ H = \bigoplus _{\nu \in \Gamma ^{(0)}} \mathbb{C}^{n(\nu)}; \;\;\;
+ D = \sum _{e \in \Gamma ^{(1)}} D_e + D_e^*
+ \end{equation}
+\end{theorem}
+\begin{example}
+\begin{figure}[h!]
+ \centering
+ \begin{tikzpicture}[
+ mass/.style = {draw,circle, minimum size=0.3cm, inner sep=0pt, thick},
+ ]
+
+ \node[mass, label={\textbf{n}}] (m1) at (1,0) {};
+ \draw (m1) to [out=330, in=210, looseness=25] node[above] {$D_e$} (m1);
+ \end{tikzpicture}
+ \caption{A $\Lambda$-decorated Graph of $(M_n(\mathbb{C}), \mathbb{C}^n,
+ D = D_e + D_e^*)$}
+\end{figure}
+\end{example}
+\end{document}
