commit 60bf195baed413f5060ced8b12e31c38f7608e56 parent 6c1434b922b99f7b363042a50c7da249f4e1ce52 Author: miksa234 <milutin@popovic.xyz> Date: Sat, 7 Aug 2021 18:59:31 +0200 checkpoint Diffstat:
19 files changed, 1757 insertions(+), 201 deletions(-)
diff --git a/src/thesis/back/packages.tex b/src/thesis/back/packages.tex @@ -22,74 +22,166 @@ \usetikzlibrary{calc,decorations.markings} \usepackage[colorlinks=true,naturalnames=true,plainpages=false,pdfpagelabels=true]{hyperref} %\usepackage[parfill]{parskip} - -\usepackage[backend=biber, sorting=none]{biblatex} -\addbibresource{thesis.bib} - -\numberwithin{equation}{section} - \usepackage{lipsum} +\newcounter{definition} +\newcounter{theorem} +\newcounter{lemma} +\newcounter{proposition} +\numberwithin{equation}{section} -% new commands just untill done rewriting stuff -\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}} +\usepackage[backend=biber, sorting=none]{biblatex} +\addbibresource{thesis.bib} -\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}~};}, +\usepackage{tikz} +\usepackage{tcolorbox} +\tcbuselibrary{skins,breakable} + +\colorlet{colexam}{black} +\newtcolorbox[use counter=definition]{mydefinition}{ + empty, + title={Definition~\thetcbcounter}, + attach boxed title to top left, + fontupper=\sl, + boxed title style={ + empty, + size=minimal, + bottomrule=1pt, + top=1pt, + left skip=0cm, + overlay= + {\draw[colexam,line width=1pt]([yshift=-0.4cm]frame.north + west)--([yshift=-0.4cm]frame.north east);}}, + coltitle=colexam, + fonttitle=\bfseries, + before=\par\medskip\noindent, + parbox=false, + boxsep=-1pt, + left=0.75cm, + right=3mm, + top=4pt, + breakable, + pad at break*=0mm, + vfill before first, + overlay unbroken={ + \draw[colexam,line width=1pt] + ([xshift=0.6cm, yshift=-0.5pt]frame.south + west)--([xshift=0.6cm,yshift=-1pt]frame.north west) + --([xshift=0.6cm]frame.south west)--([xshift=-13cm]frame.south east); }, + overlay last={ + \draw[colexam,line width=1pt] + ([xshift=0.6cm, yshift=-0.5pt]frame.south + west)--([xshift=0.6cm,yshift=-1pt]frame.north west) + --([xshift=0.6cm]frame.south west)--([xshift=-13cm]frame.south east); } } -\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} +\newtcolorbox[use counter=theorem]{mytheorem}{ + empty, + title={Theorem~\thetcbcounter}, + attach boxed title to top left, + fontupper=\sl, + boxed title style={ + empty, + size=minimal, + bottomrule=1pt, + top=1pt, + left skip=0cm, + overlay= + {\draw[colexam,line width=1pt]([yshift=-0.4cm]frame.north + west)--([yshift=-0.4cm]frame.north east);}}, + coltitle=colexam, + fonttitle=\bfseries, + before=\par\medskip\noindent, + parbox=false, + boxsep=0pt, + left=0.75cm, + right=3mm, + top=4pt, + breakable, + pad at break*=0mm, + vfill before first, + overlay unbroken={ + \draw[colexam,line width=1pt] + ([xshift=0.6cm, yshift=-0.5pt]frame.south + west)--([xshift=0.6cm,yshift=-1pt]frame.north west) + --([xshift=0.6cm]frame.south west)--([xshift=-13cm]frame.south east); }, + overlay last={ + \draw[colexam,line width=1pt] + ([xshift=0.6cm, yshift=-0.5pt]frame.south + west)--([xshift=0.6cm,yshift=-1pt]frame.north west) + --([xshift=0.6cm]frame.south west)--([xshift=-13cm]frame.south east); }, +} -\theoremstyle{theorem} -\newtheorem{proposition}{Proposition} +\newtcolorbox[use counter=lemma]{mylemma}{ + empty, + title={Lemma~\thetcbcounter}, + attach boxed title to top left, + fontupper=\sl, + boxed title style={ + empty, + size=minimal, + bottomrule=1pt, + top=1pt, + left skip=0cm, + overlay= + {\draw[colexam,line width=1pt]([yshift=-0.4cm]frame.north + west)--([yshift=-0.4cm]frame.north east);}}, + coltitle=colexam, + fonttitle=\bfseries, + before=\par\medskip\noindent, + parbox=false, + boxsep=0pt, + left=0.75cm, + right=3mm, + top=4pt, + breakable, + pad at break*=0mm, + vfill before first, + overlay unbroken={ + \draw[colexam,line width=1pt] + ([xshift=0.6cm, yshift=-0.5pt]frame.south + west)--([xshift=0.6cm,yshift=-1pt]frame.north west) + --([xshift=0.6cm]frame.south west)--([xshift=-13cm]frame.south east); }, + overlay last={ + \draw[colexam,line width=1pt] + ([xshift=0.6cm, yshift=-0.5pt]frame.south + west)--([xshift=0.6cm,yshift=-1pt]frame.north west) + --([xshift=0.6cm]frame.south west)--([xshift=-13cm]frame.south east); }, +} -\newtheorem*{idea}{Proof Idea} +\newtcolorbox[use counter=proposition]{myproposition}{ + empty, + title={Proposition~\thetcbcounter}, + attach boxed title to top left, + fontupper=\sl, + boxed title style={ + empty, + size=minimal, + bottomrule=-1pt, + top=1pt, + left skip=0cm, + overlay= + {\draw[colexam,line width=1pt]([yshift=-0.4cm]frame.north + west)--([yshift=-0.4cm]frame.north east);}}, + coltitle=colexam, + fonttitle=\bfseries, + before=\par\medskip\noindent, + parbox=false, + boxsep=0pt, + left=0.75cm, + right=3mm, + top=4pt, + breakable, + pad at break*=0mm, + vfill before first, + overlay unbroken={ + \draw[colexam,line width=1pt] + ([xshift=0.6cm, yshift=-0.5pt]frame.south + west)--([xshift=0.6cm,yshift=-1pt]frame.north west) + --([xshift=0.6cm]frame.south west)--([xshift=-13cm]frame.south east); }, + overlay last={ + \draw[colexam,line width=1pt] + ([xshift=0.6cm, yshift=-0.5pt]frame.south + west)--([xshift=0.6cm,yshift=-1pt]frame.north west) + --([xshift=0.6cm]frame.south west)--([xshift=-13cm]frame.south east); }, +} diff --git a/src/thesis/chapters/backup/acknowledgment.tex b/src/thesis/chapters/backup/acknowledgment.tex @@ -0,0 +1,3 @@ + +\section{Acknowledgment} +\lipsum[1] diff --git a/src/thesis/chapters/electroncg.tex b/src/thesis/chapters/backup/backelectroncg.tex diff --git a/src/thesis/chapters/backup/basics.tex b/src/thesis/chapters/backup/basics.tex @@ -0,0 +1,553 @@ +\subsection{Noncommutative Geometric Spaces} +\subsubsection{$*$-Algebra} +To grasp the idea of encoding geometrical data into a spectral triple we +introduce the first ingredient of a spectral triple, an unital $*$ algebra. +\begin{mydefinition} + A \textit{vector space} $A$ over $\mathbb{C}$ is called a \textit{complex, unital Algebra} if, \\ + $\forall a,b \in A$ : + \begin{align} + A \times A \rightarrow A\\ + (a,\ b)\ &\mapsto \ a\cdot b, + \end{align} + with an identity element: + \begin{align} + 1a = a1 =a. + \end{align} + Extending the definition, a $*$-algebra is an algebra $A$ with a \textit{conjugate linear map (involution)} $*:A\ \rightarrow A$, + $\forall a, b \in A$ satisfying + \begin{align} + (a\ b)^* &= b^*a^*, + (a^*)^* &= a. + \end{align} +\end{mydefinition} +In the following all unital algebras are referred to as algebras. + +\subsubsection{Finite Discrete Space} +Let us consider an example of an $*$-algebra of continuous functions $C(X)$ +on a discrete topological space $X$ with $N$ points. Functions of a +continuous $*$-algebra $C(X)$ assign values to $\mathbb{C}$, thus $f,\ g \in +C(X)$, $\lambda \in \mathbb{C}$ and $x \in X$ they provide the following structure: +\begin{itemize} + \item \textit{pointwise linear} \\ + $(f + g)(x) = f(x) + g(x)$,\\ + $(\lambda\ f)(x) = \lambda (f(x)),$ + \item \textit{pointwise multiplication} \\ + $f\ g\ (x) = f(x)g(x)$, + \item \textit{pointwise involution} \\ + $f^*(x) = \overline{f(x)}.$ +\end{itemize} +The $*$-algebra $C(X)$ is \textit{isomorphic} to a $*$-algebra $\mathbb{C}^N$ +with involution ($N$ number of points in $X$), we write $C(X) \simeq +\mathbb{C}^N$. Isomorphisms are bijective maps that preserve structure and +don't lose physical information. A function $f:X\ \rightarrow\ \mathbb{C}$ +can be represented with $N \times N$ diagonal matrices, where each diagonal +value represents the function value at the corresponding $i$-th point for $i += 1,...,N$. Because of matrix multiplication and hermitian conjugate of +matrices we have a preserving structure. + +Moreover we can \textit{map} between finite discrete spaces $X_1$ and $X_2$ with a +function +\begin{align} + \phi:\ X_1 \rightarrow\ X_2. +\end{align} +For every such map there exists a corresponding map +\begin{align} + \phi ^*:C(X_2)\ \rightarrow C(X_1), +\end{align} +which `pulls back' values even if $\phi$ is not bijective. +Note that the pullback doesn't map points back, but maps functions on an $*$-algebra $C(X)$. +The pullback, in literature often called a $*$-homomorphism or a $*$-algebra map under +pointwise product has the following properties +\begin{itemize} + \item $\phi ^*(f\ g) = \phi ^*(f)\ \phi ^*(g)$, + \item $\phi ^*(\overline{f}) = \overline{\phi ^*(f)}$, + \item $\phi ^*(\lambda\ f + g) = \lambda\ \phi ^*(f) + \phi ^*(g)$. +\end{itemize} +%------------ Exercise + The map $\phi :X_1\ \rightarrow \ X_2$ is an injective (surjective) map, + if only if the corresponding pullback $\phi ^* :C(X_2)\ \rightarrow \ + C(X_1)$ is surjective (injective). Let us say, that $X_1$ has $n$ points and + $X_2$ with $m$ points. Then there are three different cases, first $n=m$ and + obviously $\phi$ is bijective and $\phi ^*$ too. Then $n > m$, in this case + $\phi$ assigns $n$ points to $m$ points when $n > m$, which is by definition + surjective. On the other hand $\phi ^*$ assigns $m$ points to $n$ points when + $n > m$, which is by definition injective. Lastly $n < m $, which is + completely analogous to the case $n > m$. +%------------ Exercise + +\subsubsection{Matrix Algebras} +\begin{mydefinition} + A \textit{(complex) matrix algebra} A is a direct sum, for $n_i, N \in + \mathbb{N}$ + \begin{align} + A = \bigoplus _{i=1}^{N} M_{n_i}(\mathbb{C}). + \end{align} + The involution is the hermitian conjugate, a $*$ algebra with involution is referred to as + a matrix algebra +\end{mydefinition} +From a topological discrete space $X$, we can construct a $*$-algebra +$C(X)$ which is isomorphic to a matrix algebra $A$. Then the question instantly +arises, if we can construct $X$ given $A$? For a matrix algebra $A$, +which in most cases is not commutative, the answer is generally no. + +Thus there are two options. We can restrict ourselves to commutative matrix algebras, +which are the vast minority and not physically interesting. +Or we can allow more morphisms (isomorphisms) between matrix algebras. + +\subsubsection{Finite Inner Product Spaces and Representations} +Until now we looked at finite topological discrete spaces, moreover we can consider a +finite dimensional inner product space $H$ (finite Hilbert-spaces), with inner product +$(\cdot,\cdot)\rightarrow \mathbb{C}$. We denote $L(H)$ as the $*$-algebra of operators on $H$ +equipped with a product given by composition and involution of the adjoint, $T \mapsto T^*$. +Then $L(H)$ is a \textit{normed vector space} with +\begin{align} + \|T\|^2 &= \sup_{h \in H}\big\{(T\ h,\ T\ h): (h,\ h) \leq 1\big|\ T + \in L(H)\big \},\\ + \|T\| &= \sup\big\{\sqrt{\lambda}:\; \lambda \text{ eigenvalue of } T\big\}. +\end{align} +This allows us to define representations of $*$-algebras. +\begin{mydefinition} + The \textit{representation} of a finite dimensional $*$-algebra $A$ is a + pair $(H, \pi)$, where $H$ is a finite dimensional inner product space + and $\pi$ is a $*$-\textit{algebra map} + \begin{align} + \pi:A\ \rightarrow \ L(H). + \end{align} + We call the representation $(H, \pi)$ \textit{irreducible} if + \begin{itemize} + \item $H \neq \emptyset$, + \item only $\emptyset$ or $H$ is invariant under the action of $A$ on + $H$. + \end{itemize} +\end{mydefinition} +Here are some examples of reducible and irreducible representations +\begin{itemize} + \item For $A = M_n(\mathbb{C})$ the representation $H=\mathbb{C}^n$, $A$ acts as matrix multiplication\\ + $H$ is irreducible. + \item For $A = M_n(\mathbb{C})$ the representation $H=\mathbb{C}^n\oplus \mathbb{C}^n$, with $a \in A$ acting + in block form \\ $\pi: a \mapsto \big(\begin{smallmatrix} a & 0\\ 0 & a \end{smallmatrix}\big)$ is + reducible. +\end{itemize} +Naturally there are also certain equivalences between different +representations. +\begin{mydefinition} +Two representations of a $*$-algebra $A$, $(H_1, \pi _1)$ and +$(H_2, \pi _2)$ are called \textit{unitary equivalent} if there exists a map +$U: H_1 \rightarrow H_2$ such that. + \begin{align} + \pi _1(a) = U^* \pi _2(a) U + \end{align} +\end{mydefinition} + +Furthermore we define a mathematical structure called the structure space, +which will later become important, when speaking of the duality between a spectral +triple and a space. +\begin{mydefinition} + Let $A$ a $*$-algebra then, $\hat{A}$ is called the structure space of all \textit{unitary equivalence classes + of irreducible representations of A} +\end{mydefinition} +%------------- EXERCISE + Given a representation $(H, \pi)$ of a $*$-algebra $A$, the \textbf{commutant} $\pi (A)'$ of $\pi (A)$ is defined as a set + of operators in $L(H)$ that commute with all $\pi (a)$ + \begin{align} + \pi (A)' = \big\{T \in L(H):\ \pi(a)\ T = T\ \pi(a) \;\; \forall a\in + A\big\} + \end{align} + The commutant $\pi (A)'$ is also a $*$-algebra, because it has unital, + associative and involutive properties. + We note that $\pi (a) \in L(H)\ \forall a \in A$, unitary property is given + by the unital operator of the $*$-algebra of operators $L(H)$, which exists + by definition because H is a inner product space. Associativity is given by + the $*$-algebra of $L(H)$, where $L(H) \times L(H) \mapsto L(H)$, which is + associative by definition. The involutive property is also given by the $*$-algebra $L(H)$ + with a map $*: L(H) \mapsto L(H)$ only for a $T$ that commutes with $\pi (a)$. +%------------- EXERCISE + +%------------- EXERCISE + For a unital algebra $*$-algebra $A$, the matrices $M_n(A)$ with entries + in $A$ form a unital $*$-algebra, because unitary operation in $M_n(A)$ is given by the identity Matrix, which + has to exists in every entry in $M_n(A)$, and behaves like in $A$. Associativity is given by + matrix multiplication. Lastly involution is given by the conjugate transpose. + + A representation $\pi :A\ \rightarrow \ L(H)$ of a $*$-algebra $A$, for + $H^n = H \oplus ... \oplus H$, $n$ times. Then we have the following + representation $\tilde{\pi}:M_n(A) \rightarrow L(H^n)$ for the Matrix + Algebra with $\tilde{\pi}((a_{ij})) = (\tilde{\pi}(a_{ij})) \in M_n(A)$. + We have direct isomorphisms of $A \simeq M_n(A)$ and $H \simeq H^n$ + meaning $\tilde{\pi}$ is a valid reducible representation. + + Let $\tilde{\pi}:M_n(A) \rightarrow L(H^n)$ be a $*$ algebra + representation of $M_n(A)$, then $\pi: A \rightarrow L(H^n)$ is a representation of $A$. + The fact that $\tilde{\pi}$ and $\pi$ are unitary equivalent, there is + a map $U: H^n \rightarrow H^n$ given by $U=\mathbbm{1}_n$, thus + \begin{align} + \pi (a) &= \mathbbm{1}_n^*\ \tilde{\pi}((a_{ij})), \\ + \mathbbm{1}_n &= \tilde{\pi}((a_{ij})) = \pi (a_{ij}) + \Rightarrow a_{ij} = a\ \mathbbm{1}_n. + \end{align} +%------------- EXERCISE + + +A commutative matrix algebra can be used to reconstruct a discrete space. +The structure space $\hat{A}$ is used for this. Because $A \simeq +\mathbb{C}^N$ all +irreducible representation are of the form +\begin{align} + \pi _i:(\lambda_1,...,\lambda_N)\in \mathbb{C}^N \mapsto \lambda_i \in + \mathbb{C} +\end{align} +for $i = 1,...,N$ and thus $\hat{A} \simeq \{1,...,N\}$. +The conclusion is that, there is a duality between discrete spaces and +commutative matrix algebra this duality is called the \textit{finite +dimensional Gelfand duality} + +Our aim is to construct a duality between finite dimensional spaces and +\textit{equivalence classes} of matrix algebras, to preserve general +non-commutativity of matrices. Equivalence classes are described by a +generalized notion of isomorphisms between matrix algebras (\textit{Morita +Equivalence}) + +\subsubsection{Algebraic Modules} +An important notion for Morita Equivalence are algebraic modules, later +extended with Hilbert bimodules. +\begin{mydefinition} + Let $A$, $B$ be algebras (need not be matrix algebras) + \begin{enumerate} + \item \textit{left} A-module is a vector space $E$, that carries a left + representation of $A$, that is $\exists$ a bilinear map $\gamma: A + \times E \rightarrow E$ with + \begin{align} + (a_1\ a_2)\cdot e = a_1 \cdot (a_2 \cdot e);\;\;\; a_1, a_2 \in + A, e \in E. + \end{align} + \item \textit{right} B-module is a vector space $F$, that carries a + right representation of $A$, that is there exists a bilinear map + $\gamma: F \times B \rightarrow F$ with + \begin{align} + f \cdot (b_1\ b_2)= (f \cdot b_1) \cdot b_2;\;\;\; b_1, b_2 \in B, f \in F + \end{align} + \item \textit{left} A-module and \textit{right} B-module is a + \textit{bimodule}, a vector space $E$ satisfying + \begin{align} + a \cdot (e \cdot b)= (a \cdot e) \cdot b;\;\;\; a \in A, b \in B, e \in E + \end{align} + \end{enumerate} +\end{mydefinition} +An $A$-\textbf{module homomorphism} as linear map $\phi: E\rightarrow F$ which respects the +representation of A, e.g.\ for left module. +\begin{align} + \phi (a\ e) = a \phi (e); \;\;\; a \in A, e \in E. +\end{align} +We will use the notation +\begin{itemize} + \item ${}_A E$, for left $A$-module $E$; + \item ${}_A E_B$, for right $B$-module $F$; + \item ${}_A E_B$, for $A$-$B$-bimodule $E$, simply bimodule. +\end{itemize} +%------------------- EXERCISE +From a simple observation, we see that an arbitrary representation $\pi : A +\rightarrow L(H)$ of a $*$-algebra A, turns H into a left module ${}_A H$. If +$_A H$ than $(a_1\ a_2) h = a_1 (a_2\ h)$ for $a_1, a_2 \in A$ and $h \in H$. We +take the representation of an $a \in A$, $\pi (a)$, and write +\begin{align} + \big(\pi(a_1)\ \pi(a_2)\big)h = \pi(a_1)\big(\pi(a_2)\ h\big) = + \big(T_1\ T_2\big) h = T_1 \big(T_2\ h\big) +\end{align} +For $T_1, T_2 \in L(H)$, which operate naturally from the left on $h$. + +%------------------- EXERCISE +%------------------- EXERCISE + +Furthermore notice that that an $*$-algebra $A$ is a bimodule ${}_A A_A$ with +itself, given by the map +\begin{align} + \gamma: A\times A\times A \rightarrow A, +\end{align} +which is the inner product of a $*$-algebra. +%------------------- EXERCISE + +\subsubsection{Balanced Tensor Product and Hilbert Bimodules} + +\begin{mydefinition} + Let $A$ be an algebra, $E$ be a \textit{right} $A$-module and $F$ be a + \textit{left} $A$-module. The \textit{balanced tensor product} of $E$ and + $F$ forms a $A$-bimodule. + \begin{align} + E \otimes _A F := E \otimes F / \left\{\sum _i e_i a_i \otimes f_i - + e_i \otimes a_i f_i : \;\;\; a_i \in A,\ e_i \in E,\ f_i \in F + \right\}. + \end{align} +\end{mydefinition} +The $/$ denotes the quotient space. By that the operation $\otimes _A$ takes +two left/right modules and makes a bimodule with the help the tensor product of +the two modules and the quotient space that takes out all the elements from the +tensor product that don't preserver the left/right representation and that are +duplicates. +\begin{mydefinition} + Let $A$, $B$ be \textit{matrix algebras}. The \textit{Hilbert bimodule} for + $(A, B)$ is given by an $A$-$B$-bimodue $E$ and by an $B$-valued + \textit{inner product} $\langle \cdot,\cdot\rangle_E: E\times E \rightarrow + B$, which satisfies the following conditions for $e, e_1, e_2 \in + E,\ a \in A$ and $b \in B$ +\begin{align} + \langle e_1,\ a\cdot e_2\rangle_E &= \langle a^*\cdot e_1,\ e_2\rangle_E + \;\;\;\; & \text{sesquilinear in $A$},\\ + \langle e_1,\ e_2 \cdot b\rangle_E + &= \langle e_1,\ e_2\rangle_E b \;\;\;\; & \text{scalar in $B$},\\ + \langle e_1,\ e_2\rangle_E &= \langle e_2,\ e_1\rangle^*_E \;\;\;\; & + \text{hermitian}, \\ + \langle e,\ e\rangle_E &\ge 0 \;\;\;\; & \text{equality + holds iff $e=0$}. +\end{align} +We denote $KK_f(A,\ B)$ as the set of all \textit{Hilbert bimodules} of $(A,\ B)$. +\end{mydefinition} +%-------------- EXERCISE + +And indeed the Hilbert bimodule extension takes a representation $\pi:\ A \ +\rightarrow L(H)$ of a matrix algebra $A$ and turns $H$ into a Hilbert bimodule for +$(A, \mathbb{C})$, because the representation of $a \in A$, $\pi(a)=T \in L(H)$ fulfills +the conditions of the $\mathbb{C}$-valued inner product for $h_1, h_2 \in H$ +\begin{itemize} + \item $\langle h_1,\ \pi(a)\ h_2\rangle _\mathbb{C} = \langle h_1,\ T\ h_2\rangle _\mathbb{C} = + \langle T^* h_1, h_2\rangle _\mathbb{C}$, $T^*$ given by the adjoint, + \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, + \item $\langle h_1,\ h_2\rangle _\mathbb{C}^* = \langle h_2,\ h_1\rangle _\mathbb{C}$, hermitian because of the + $\mathbb{C}$-valued inner product + \item $\langle h_1,\ h_2\rangle \ge 0$, $\mathbb{C}$-valued inner product. +\end{itemize} +%-------------- EXERCISE + +%-------------- EXERCISE +Take again the $A-A$ bimodule given by an $*$-algebra $A$, it is in +$KK_f(A,\ A)$. This becomes clear by looking at the following inner product + $\langle \cdot,\cdot\rangle_A:A \times A \rightarrow A$: + \begin{align} + \langle a,\ a\rangle_A = a^*a' \;\;\;\; a,a'\in A. \label{eq:inner-product} + \end{align} + Simply checking the conditions in $\langle \cdot, \cdot\rangle _A$ for + $a, a_1, a_2 \in A$ + \begin{align} + \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, \\ + \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,\\ + \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. + \end{align} + +%-------------- EXERCISE + +%-------------- EXAMPLE +As an exemplar for overview consider a $*$ homomorphism between two matrix +algebras $\phi:A\rightarrow B$, we can construct a Hilbert bimodule +$E_{\phi} \in KK_f(A, B)$ in the following way. We let $E_{\phi}$ be $B$ in +as an vector space and an inner product from above in equation +\eqref{eq:inner-product}, with $A$ acting on the left with $\phi$. +\begin{align} + a\cdot b = \phi(a)\ b +\end{align} +for $a\in A, b\in E_{\phi}$. +%-------------- EXAMPLE + +\subsubsection{Kasparov Product and Morita Equivalence} +\begin{mydefinition} + Let $E \in KK_f(A, B)$ and $F \in KK_F(B, D)$ the \textit{Kasparov product} is defined as + with the balanced tensor product + \begin{align} + F \circ E := E \otimes _B F. + \end{align} + Then $F\circ E \in KK_f(A,D)$ is equipped with a $D$-valued inner product + \begin{align} + \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 + \end{align} +\end{mydefinition} + +%-------------- EXERCISE +The Kasparov product for $*$-algebra homomorphism $\phi: A \rightarrow B$ and +$\psi: B \rightarrow C$ are isomorphisms in the sense of +\begin{align} + E_{\psi} \circ E_{\phi}\ \equiv\ E_{\phi} \otimes _B E_{\psi}\ + \simeq\ + E_{\psi \circ \phi} \in KK_f(A,C). +\end{align} + +In the direct computation for elements $a \in A$, $b\in B$, and $c\in C$ which +is $\psi \circ \phi$ gives us +\begin{align} +a \cdot b \cdot c = \psi(\phi (a) \cdot b) \cdot c +\end{align} +An interesting case arises when looking at $E_{\text{id}_A} \simeq A \in KK_f(A,A)$ for +$\text{id}_A: A \rightarrow A$. This is obvious when we let $E_{\phi}$ be $A$ +with a natural right representation. It follows that $E_{\phi}\simeq A$, thus +an inner product, acting from the left on $A$ for $\phi$, $a', a\in A$ reads +\begin{align} + a'\ a = (\phi(a')\ a) \in A, +\end{align} +which is satisfied by $\phi = \text{id}_A$ + +\begin{mydefinition} + Let $A$, $B$ be \textit{matrix algebras}. They are called \textit{Morita equivalent} if there + exists an $E \in KK_f(A, B)$ and an $F \in KK_f(B, A)$ such that + \begin{align} + E \otimes _B F \simeq A \;\;\; \text{and} \;\;\; F \otimes _A E \simeq + B, + \end{align} + where $\simeq$ denotes the isomorphism between Hilbert bimodules and note + that $A$ or $B$ is a bimodule by itself. +\end{mydefinition} + +The modules $E$ and $F$ are each others inverse in regards to the Kasparov +Product, because we land in the same space as we started. To clarify, in +the definition we have $E \in KK_f(A, B)$. We start from $A$ and $E \otimes +_B F$, which lands in $A$. Oppositely we have $F \in KK_f(B, D)$ we start +from $B$ and $F \otimes _A E$, which lands in $B$. + + +%------------- EXERCISE +By definition $E \otimes _B F$ is a $A-D$ bimodule. Since +\begin{align} + E \otimes _B F = E \otimes F / \bigg\{\sum_i\ e_i\ b_i \otimes f_i - e_i + \otimes b_i\ f_i\ \big|\;\; e_i \in E_i,\ b_i \in B,\ f_i \in F\bigg\}, +\end{align} +the last part takes out all tensor product elements of $E$ and $F$ that don't +preserver the left/right representation and that are duplicates. + +Additionally $\langle \cdot,\cdot\rangle _{E\oplus _B F}$ defines a $D$ valued +inner product, as $\langle e_1,\ e_2\rangle _E \in B$ and $\langle f_1,\ f_2\rangle _F \in C$ by +definition. So for $\langle e_1,\ e_2\rangle _E =b$ we have +\begin{align} + \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 = \langle f_1,\ b\ f_2\rangle _F \in C +\end{align} +%------------- EXERCISE +%------------- EXAMPLE +Picking up the example of $(A, A)$, the Hilbert bimodule $A$, we can +consider an $E \in KK_f(A,B)$ for +\begin{align} + E \circ A = A\oplus _A E \simeq E. +\end{align} +We conclude, that $_A A_A$ is the identity element in the Kasparov product (up +to isomorphism). +%------------- EXAMPLE +%------------- EXAMPLE +Let us examine another example for $E = \mathbb{C}^n$, which is a +$(M_n(\mathbb{C}), \mathbb{C})$ Hilbert bimodule with the standard $\mathbb{C}$ +inner product. Further let $F = \mathbb{C}^n$, which is a $(\mathbb{C}, +M_n(\mathbb{C}))$ Hilbert bimodule by right matrix multiplication with +$M_n(\mathbb{C})$ valued inner product, we can write + \begin{align} + \langle v_1, v_2\rangle =\bar{v_1}v_2^t \;\; \in M_n(\mathbb{C}). + \end{align} +If we take the Kasparov product of $E$ and $F$ + \begin{align} + F\circ E\ &=\ E\otimes _{\mathbb{C}}F\ \;\;\;\;\;\; \simeq \ + M_n(\mathbb{C}),\\ + E\circ F\ &=\ F\otimes _{M_n(\mathbb{C})}E\ \simeq\ \mathbb{C}, + \end{align} +we see that $M_n(\mathbb{C})$ and $\mathbb{C}$ are Morita equivalent! +%------------- EXAMPLE + +\begin{mylemma} + Two matrix algebras are Morita Equivalent if, and only if their their structure spaces + are isomorphic as discreet spaces (have the same cardinality / same number + of elements). +\end{mylemma} +\begin{proof} + Let $A$, $B$ be \textit{Morita equivalent}. Then there exist $_A E_B$ and $_B F_A$ with + \begin{align} + E \otimes _B F \simeq A \;\;\; \text{and} \;\;\; F \otimes _A E \simeq + B. + \end{align} + Also consider $[(\pi _B, H)] \in \hat{B}$. We can construct a + representation of $A$, which reads + \begin{align} + \pi _A \rightarrow L(E \otimes _B H)\;\;\; \text{with} \;\;\; \pi _A(a) + (e \otimes v) = a e \otimes w + \end{align} + Vice versa, we have $[(\pi _A, W)] \in \hat{A}$ we can construct $\pi _B$ + as + \begin{align} + \pi _B: B \rightarrow L(F \otimes _A W) \;\;\; \text{and}\;\;\; \pi + _B(b) (f\otimes w) = bf\otimes w. + \end{align} + Now we need to show that the representation $\pi _A$ is irreducible if and + only if $\pi _B$ is irreducible. For $(\pi _B, H)$ to be irreducible, we + need $H \neq \emptyset$ and only $\emptyset$ or $H$ to be invariant under + the Action of $B$ on $H$. Than $E\otimes _B H$ and $E\otimes _B H \simeq A$ + cannot be empty, because $E$ preserves left representation of $A$. + + Lastly we need to check if the association of the class $[\pi _A]$ to $[\pi + _B]$ is independent of the choice of representatives $\pi _A$ and $\pi _B$. + The important thing is that $[\pi _A] \in \hat{A}$ respectively $[\pi _B] \in + \hat{B}$, hence any choice of representation is irreducible, because the + structure space denotes all unitary equivalence classes of irreducible + representations. + + Note that the statements $E \simeq H$ and $F \simeq W$ are not particularly + true, since all infinite dimensional Hilbert spaces are isomorphic. Here + we are looking at finite dimensional Hilbert spaces. Another thing to keep + in mind, is that for $[\pi _B, H] \in \hat{B}$ and looking at algebraic + bimodules, we know that $H$ is a bimodule of $B$, hence $E \otimes _B + H\simeq A$, and for $[\pi _A, W]$, which is the same. + Finally we can conclude, that these maps are each others inverses, thus + $\hat{A} \simeq \hat{B}$. +\end{proof} + +\begin{mylemma} + The matrix algebra $M_n(\mathbb{C})$ has a unique irreducible + representation (up to isomorphism) given by the defining representation on + $\mathbb{C}^n$. +\end{mylemma} +\begin{proof} + We know $\mathbb{C}^n$ is a irreducible representation of $A= + M_n(\mathbb{C})$. Let $H$ be irreducible and of dimension $k$, then we + define a map + \begin{align} + \phi : A\oplus...\oplus A &\rightarrow H^* \\ + (a_1,...,a_k)&\mapsto e^1\circ a_1^t+...+e^k\circ a_k^t, + \end{align} +where $\{e^1,...,e^k\}$ is the basis of the dual space $H^*$ and +$(\circ)$ being the pre-composition of elements in $H^*$ and $A$ acting on $H$. +This forms a morphism of $M_n(\mathbb{C})$ modules, provided a matrix $a \in A$ +acts on $H^*$ with $v\mapsto v\circ a^t$ ($v\in H^*$). Furthermore this +morphism is surjective, thus making the pullback $\phi ^*:H\mapsto (A^k)^*$ +injective. Now identify $(A^k)^*$ with $A^k$ as a $A$-module and note that +$A=M_n(\mathbb{C}) \simeq \oplus ^n \mathbb{C}^n$ as a n A module. It follows +that $H$ is a submodule of $A^k \simeq \oplus ^{nk}\mathbb{C}$. By +irreducibility $H \simeq \mathbb{C}$. +\end{proof} + +%---------------- EXAMPLE +Let us look at an examples for two matrix algebras $A$, and $B$. +\begin{align} + A = \bigoplus ^N_{i=1} M_{n_i}(\mathbb{C}), \;\;\; + B = \bigoplus ^M_{j=1} M_{m_j}(\mathbb{C}). +\end{align} +Let $\hat{A} \simeq \hat{B}$, this implies $N=M$. Further define $E$ with $A$ +acting by block-diagonal matrices on the first tensor and B acting in the same +manner on the second tensor. Define $F$ vice versa, ultimately reading +\begin{align} + E:= \bigoplus _{i=1}^N \mathbb{C}^{n_i} \otimes \mathbb{C}^{m_i}, \;\;\; + F:= \bigoplus _{i=1}^N \mathbb{C}^{m_i} \otimes \mathbb{C}^{n_i}. +\end{align} +When we calculate the Kasparov product we get the following +\begin{align} + E \otimes _B F &\simeq \bigoplus _{i=1}^N (\mathbb{C}^{n_i}\otimes\mathbb{C}^{m_i}) + \otimes _{M_{m_i}(\mathbb{C})} (\mathbb{C}^{m_i}\otimes\mathbb{C}^{n_i}) \\ + &\simeq \bigoplus _{i=1}^N \mathbb{C}^{n_i}\otimes + \left(\mathbb{C}^{m_i}\otimes _{M_{m_i}(\mathbb{C})}\mathbb{C}^{m_i}\right) + \oplus \mathbb{C}^{n_i} \\ + &\simeq \bigoplus _{i=1}^N + \mathbb{C}^{m_i}\otimes\mathbb{C}^{n_i} \simeq A. +\end{align} +On the other hand we get +\begin{align} + F \otimes _A E \simeq B. +\end{align} +%---------------- EXAMPLE + +To summarize, there is a duality between finite spaces and Morita equivalence classes of matrix algebras. By replacing $*$-homomorphism $A\rightarrow B$ with Hilbert bimodules $(A,B)$ we introduce +a richer structure of morphism between matrix algebras. diff --git a/src/thesis/chapters/backup/conclusion.tex b/src/thesis/chapters/backup/conclusion.tex @@ -0,0 +1,2 @@ +\section{Conclusion} +\lipsum[1] diff --git a/src/thesis/chapters/backup/diffgeo.tex b/src/thesis/chapters/backup/diffgeo.tex @@ -0,0 +1,128 @@ +\subsection{Excurse} +\textbf{Manifold:} A topological space that is locally Euclidean. +\newline +\textbf{Riemannian Manifold:}A Manifold equipped with a Riemannian +Metric, a +symmetric bilinear form on Vector Fields $\Gamma(TM)$ +\begin{align} + &g: \Gamma(TM) \times \Gamma(TM) \rightarrow C(M) \\ + \text{with}& \nonumber\\ + &g(X, Y) \in \mathbb{R} \;\;\; \text{if $X, Y \in \mathbb{R}$}\\ + &\text{$g$ is $C(M)$-bilinear } \forall f\in C(M):\;\; g(fX, Y) = + g(X, + fY) = fg(X,Y)\\ + &g(X,X) \begin{cases}\geq 0 \;\;\; \forall X \\ = 0 \;\;\; \forall X + =0 + \end{cases} +\end{align} +$g$ on $M$ gives rise to a distance function on $M$ +\begin{align} + d_g(x, y) = \inf_\gamma \left\{\int_0^1(\dot{\gamma}(t), + \dot{\gamma}(t))dt;\;\; \gamma(0) = x, \gamma(1) = y \right\} +\end{align} +Riemannian Manifold is called spin$^c$ if there exists a vector bundle $S +\rightarrow M$ with an algebra bundle isomorphism +\begin{align} + \mathbb{C}\text{I}(TM) &\simeq \text{End}(S)\;\;\; &\text{($dim(M)$ + even)}\\ + \mathbb{C}\text{I}(TM)^\circ &\simeq \text{End}(S)\;\;\; + &\text{($dim(M)$ odd)}\\ +\end{align} +$(M,S)$ is called the \textbf{spin$^c$ structure on $M$}. +\newline +$S$ is called the \textbf{spinor Bundle}. +\newline +$\Gamma(S)$ are the \textbf{spinors}. + +Riemannian spin$^c$ Manifold is called spin if there exists an +anti-unitary +operator $J_M:\Gamma(S) \rightarrow \Gamma(S)$ such that: +\begin{enumerate} + \item $J_M$ commutes with the action of real-valued continuous + functions + on $\Gamma(S)$. + \item $J_M$ commutes with $\text{Cliff}^-(M)$ (even case)\\ + $J_M$ commutes with $\text{Cliff}^-(M)^\circ$ (odd case) +\end{enumerate} +$(S, J_M)$ is called the \textbf{spin Structure on $M$} +\newline +$J_M$ is called the \textbf{charge conjugation}. + +\subsection{Operators of Laplace Type} +Let $M$ be a $n$ dimensional compact Riemannian manifold with $\partial M = 0$. +Then consider a vector bundle $V$ over $M$ (i.e. there is a vector space to +each point on $M$), so we can define smooth functions. We want to look at +arbitrary differential operators $D$ of Laplace type on $V$, they have the general +from +\begin{align} + D = -(g^{\mu\nu} \partial_\mu\partial_\nu + a^\sigma\partial_\sigma +b) +\end{align} +where $a^\sigma, b$ are matrix valued functions on $M$ and $g^{\mu\nu}$ is the +inverse metric on $M$. There is a unique connection on $V$ and a unique +endomorphism (matrix valued function) $E$ on $V$, then we can rewrite $D$ in +terms of $E$ and covariant derivatives +\begin{align} + D = -(g^{\mu\nu} \nabla_\mu \nabla_\nu +E) +\end{align} +Where the covariant derivative consists of $\nabla = \nabla^{[R]} +\omega$ the +standard Riemannian covariant derivative $\nabla^{[R]}$ and a "gauge" bundle +$\omega$ (fluctuations). WE can write $E$ and $\omega$ in terms of geometrical +identities +\begin{align} + \omega_\delta &= \frac{1}{2}g_{\nu\delta}(a^\nu + +g^{\mu\sigma}\Gamma^\nu_{\mu\sigma}I_V)\\ + E &= b - g^{\nu\mu}(\partial_\mu \omega_\nu + \omega_\nu \omega_\mu - + \omega_\sigma \Gamma^\sigma_{\nu\mu}) +\end{align} +where $I_V$ is the identity in $V$ and the Christoffel symbol +\begin{align} + \Gamma^\sigma_{\mu\nu} = g^{\sigma\varrho} \frac{1}{2} (\partial_\mu + g_{\nu\varrho} + \partial_\nu g_{\mu\varrho} - \partial_\varrho g_{\mu\nu}) +\end{align} +Furthermore we remind ourselves of the Riemmanian curvature tensor, Ricci +Tensor and the Scalar curavture. +\begin{align} + R^\mu_{\nu\varrho\sigma} &= \partial_\sigma \Gamma^{\mu}_{\nu\varrho} + -\partial_\varrho \Gamma^\mu_{\nu\sigma} + \Gamma^{\lambda}_{\nu\varrho}\Gamma^{\mu}_{\lambda\sigma} + \Gamma^{\lambda}_{\nu\sigma}\Gamma^{\mu}_{\lambda\varrho}\\ + R_{\mu\nu} &:= R^{\sigma}_{\mu\nu\sigma}\\ + R &:= R^\mu_{\ \mu} +\end{align} + +The we let $\{e_1, \dots, e_n\}$ be the local orthonormal frame of +$TM$(tangent bundle $M$), which will be noted with flat indices $i,j,k,l +\in\{1,\dots, n\}$, we use $e^k_\mu, e^\nu_j$ to transform between flat indices +and curved indices $\mu, \nu, \varrho$. +\begin{align} + e^\mu_j e^\nu_k g_{\mu\nu} &= \delta_{jk}\\ + e^\mu_j e^\nu_k \delta^{jk} &= g^{\mu\nu} \\ + e^j_\mu e^\mu_k &= \delta^j_k +\end{align} + +The Riemannian part of the covariant derivative contains the standard +Levi-Civita connection, so that for a $v_\nu$ we write +\begin{align} + \nabla_\mu^{[R]} v_\nu = \partial_\mu v_\nu - + \Gamma^{\varrho}_{\mu\nu}v_\varrho. +\end{align} +The extended covariant derivative reads then +\begin{align} + \nabla_\mu v^j = \partial_\mu v^j + \sigma^{jk}_\mu v_k. +\end{align} +the condition $\nabla_\mu e^k_\nu = 0$ gives us the general connection +\begin{align} + \sigma^{kl}_\mu = e^\nu_l\Gamma^{\varrho}_{\mu\nu}e^k_\varrho - e^\nu_l + \partial_\mu e^k_\nu +\end{align} +The we may define the field strength $\Omega_{\mu\nu}$ of the connection $\omega$ +\begin{align} + \Omega_{\mu\nu} = \partial_\mu \omega_\nu -\partial_\nu \omega_\mu + +\omega_\mu \omega_\nu -\omega_\nu\omega_\mu. +\end{align} +If we apply the covariant derivative on $\Omega$ we get +\begin{align} + \nabla_\varrho\Omega_{\mu\nu} = \partial_\varrho \Omega_{\mu\nu} - + \Gamma^{\sigma}_{\varrho \mu} \Omega_{\sigma\mu} + [\omega_\varrho, + \Omega_{\mu\nu}] +\end{align} diff --git a/src/thesis/chapters/backup/electroncg.tex b/src/thesis/chapters/backup/electroncg.tex @@ -0,0 +1,455 @@ +\subsection{Noncommutative Geometry of Electrodynamics} +In this chapter we describe Electrodynamics with the almost commutative +manifold $M\times F_X$ and the abelian gauge group $U(1)$. +We arrive at a unified description of gravity and electrodynamics although in the classical level. + +The almost commutative Manifold $M\times F_X$ describes a local gauge group +$U(1)$. The inner fluctuations of the Dirac operator relate to $Y_\mu$ the +gauge field of $U(1)$. According to the setup we ultimately arrive at two +serious problems. + +First of all in the Two-Point space $F_X$, the operator $D_F$ must vanish for +us to have a real structure. However this implies that the electrons +are massless, which would be absurd. + +The second problem arises when looking at the Euclidean action for a free +Dirac field +\begin{align} + S = - \int i \bar{\psi}(\gamma ^\mu\partial _\mu - m) \psi d^4x, +\end{align} +where $\psi,\ \bar{\psi}$ must be considered as independent variables, which +means that the fermionic action $S_f$ needs two independent Dirac spinors. +Let us try and construct two independent Dirac spinors with our data. To do +this we take a look at the decomposition of the basis and of the total +Hilbertspace $H = L^2(S) \otimes H_F$. For the orthonormal basis of $H_F$ we +can write $\{e, \bar{e}\}$ , where $\{e\}$ is the orthonormal basis of +$H_F^+$ and $\{\bar{e}\}$ the orthonormal basis of $H_F^-$. Accompanied with +the real structure we arrive at the following relations +\begin{align} + J_F e &= \bar{e} \;\;\;\;\;\; J_F \bar{e} = e, \\ + \gamma_F e &= e \;\;\;\;\;\; \gamma_F \bar{e} = \bar{e}. +\end{align} +Along with the decomposition of $L^2(S) = L^2(S)^+ \oplus L^2(S)^-$ and $\gamma = \gamma _M +\otimes \gamma _F$ we can obtain the positive eigenspace +\begin{align} + H^+ = L^2(S)^+ \otimes H_F^+ \oplus L^(S)^- \otimes H_F^-. +\end{align} +So, for a $\xi \in H^+$ we can write +\begin{align} + \xi = \psi _L \otimes e + \psi _R \otimes \bar{e} +\end{align} +where $\psi_L \in L^2(S)^+$ and $\psi _R \in L^2(S)^-$ are the two Wheyl +spinors. We denote that $\xi$ is only determined by one Dirac spinor $\psi := +\psi_L + \psi _R$, \textbf{but we require two independent spinors}. Our +conclusion is that the definition of the fermionic action gives too much +restrictions to the Two-Point space $F_X$. +\subsubsection{The Finite Space} +To solve the two problems we simply enlarge (double) the Hilbertspace. This +is visualized by introducing multiplicities in Krajewski Diagrams which will also +allow us to choose a nonzero Dirac operator that will connect the two +vertices and preserve real structure making our particles massive and +bringing anti-particles into the mix. + +We start of with the same algebra $C^\infty(M, \mathbb{C}^2)$, corresponding +to space $N= M\times X$. The Hilbertspace describes four particles, meaning +it has four orthonormal basis elements. It describes \textbf{left handed +electrons} and \textbf{right handed positrons}. Pointing this out, we have +$\{ \underbrace{e_R, e_L}_{\text{left-handed}}, \underbrace{\bar{e}_R, +\bar{e}_L}_{\text{right-handed}}\}$ the orthonormal basis for $H_F = +\mathbb{C}^4$. Accompanied with the real structure $J_F$, which allows us to +interchange particles with antiparticles by the following equations +\begin{align} + &J_F e_R = \bar{e}_R, \\ + &J_F e_L = \bar{e_L}, \\ + \nonumber \\ + &\gamma _F e_R = -e_R,\\ + &\gamma_F e_L = e_L, +\end{align} +where $J_F$ and $\gamma_F$ have to following properties +\begin{align} + &J_F^2 = 1,\\ + & J_F \gamma_F = - \gamma_F J_F. +\end{align} +By means of $\gamma_F$ we have two options to decompose the total +Hilbertspace $H$, firstly into +\begin{align} + H_F = \underbrace{H_F^+}_{\text{ONB } \{e_L, \bar{e}_L\}} + \oplus \underbrace{H_F^-}_{\text{ONB } \{e_R, \bar{e}_R\}}, +\end{align} +or alternatively into the eigenspace of particles and their +antiparticles (electrons and positrons) which is preferred in literature and +which we will use going further +\begin{align} + H_F = \underbrace{H_{e}}_{\text{ONB } \{e_L, e_R\}} \oplus + \underbrace{H_{\bar{e}}}_{\text{ONB } \{\bar{e}_L, \bar{e}_R\}}. +\end{align} +Here ONB means orthonormal basis. + +The action of $a \in A = \mathbb{C}^2$ on $H$ with respect to the ONB +$\{e_L, e_R, \bar{e}_L, \bar{e}_R\}$ is represented by +\begin{align}\label{eq:leftrightrepr} + a = + (a_1 , a_2 ) \mapsto + \begin{pmatrix} + a_1 &0 &0 &0\\ + 0&a_1 &0 &0\\ + 0 &0 &a_2 &0\\ + 0 &0 &0 &a_2\\ + \end{pmatrix} +\end{align} +Do note that this action commutes wit the grading and that +$[a, b^\circ] = 0$ with $b:= J_F b^*J_F$ because both the left and the right +action is given by diagonal matrices by equation \eqref{eq:leftrightrepr}. Note +that we are still left with $D_F = 0$ and the following spectral +triple +\begin{align}\label{eq:fedfail} + \left( \mathbb{C}^2, \mathbb{C}^2, D_F=0; J_F = + \begin{pmatrix} + 0 & C \\ C &0 + \end{pmatrix}, + \gamma _F = + \begin{pmatrix} + 1 & 0 \\ 0 &-1 + \end{pmatrix} + \right). + \end{align} +It can be represented in the following Krajewski diagram, +with two nodes of multiplicity two + \begin{figure}[H] \centering + \begin{tikzpicture}[ + dot/.style = {draw, circle, inner sep=0.06cm}, + bigdot/.style = {draw, circle, inner sep=0.09cm}, + no/.style = {}, + ] + \node[no](a) at (0,0) [label=left:$\textbf{1}^\circ$] {}; + \node[no](b) at (0, -1) [label=left:$\textbf{1}^\circ$] {}; + \node[no](c) at (0.5, 0.5) [label=above:$\textbf{1}$] {}; + \node[no](d) at (1.5, 0.5) [label=above:$\textbf{1}$] {}; + \node[dot](d0) at (1.5,0) [] {}; + \node[dot](d0) at (0.5,-1) [] {}; + \node[bigdot](d0) at (1.5,0) [] {}; + \node[bigdot](d0) at (0.5,-1) [] {}; + \end{tikzpicture} + \end{figure} +\subsubsection{A noncommutative Finite Dirac Operator} +To extend our spectral triple with a non-zero Operator, we need to take a +closer look at the Krajewski diagram above. Notice that edges only exist +between multiple vertices, meaning we can construct a Dirac operator mapping +between the two vertices. The operator can be represented by the following matrix +\begin{align}\label{eq:feddirac} + D_F = + \begin{pmatrix} + 0 & d & 0 & 0 \\ + \bar{d} & 0 & 0 & 0 \\ + 0 & 0 & 0 & \bar{d} \\ + 0 & 0 & d & 0 + \end{pmatrix} +\end{align} +We can now define the finite space $F_{ED}$. +\begin{align} + F_{ED} := (\mathbb{C}^2, \mathbb{C}^4, D_F; J_F, \gamma_F) +\end{align} +where $J_F$ and $\gamma_F$ are like in equation \eqref{eq:fedfail} and $D_F$ +from equation \eqref{eq:feddirac}. + +\subsubsection{Almost commutative Manifold of Electrodynamics} +The almost commutative manifold $M\times F_{ED}$ has KO-dimension 2, and is +represented by the following spectral triple +\begin{align}\label{eq:almost commutative manifold} + M\times F_{ED} := \big(C^\infty(M,\mathbb{C}^2),\ L^2(S)\otimes + \mathbb{C}^4,\ + D_M\otimes 1 +\gamma _M \otimes D_F;\; J_M\otimes J_F,\ \gamma_M\otimes + \gamma _F\big) +\end{align} +The algebra didn't change, thus we can decompose it like before +\begin{align} + C^\infty(M, \mathbb{C}^2) = C^\infty (M) \oplus C^\infty (M) +\end{align} +As for the Hilbertspace, we can decomposition it in the following way +\begin{align} + H = (L^2(S) \otimes H_e ) \oplus (L^2(S) \otimes H_{\bar{e}}). +\end{align} +Note that the one component of the algebra is acting on $L^2(S) \otimes H_e$, +and the other one acting on $L^2(S) \otimes H_{\bar{e}}$. In other words the components of +the decomposition of both the algebra and the Hilbertspace match by the action of +the algebra. + +The derivation of the gauge theory is the same for $F_{ED}$ as for the +Two-Point space $F_X$. We have $\mathfrak{B}(F) \simeq U(1)$ and for an +arbitrary gauge field $B_\mu = A_\mu - J_F A_\mu J_F^{-1}$ we can write +\begin{align} \label{field} + B_\mu = + \begin{pmatrix} + Y_\mu & 0 & 0 & 0 \\ + 0 & Y_\mu& 0 & 0 \\ + 0 & 0 & Y_\mu& 0 \\ + 0 & 0 & 0 & Y_\mu + \end{pmatrix} \;\;\;\;\;\ \text{for} \;\;\ Y_\mu (x) \in \mathbb{R}. +\end{align} +We have one single $U(1)$ gauge field $Y_\mu$, carrying the action of the +gauge group +\begin{align} + \text{$\mathfrak{B}$}(M\times F_{ED}) \simeq C^\infty(M, U(1)) +\end{align} + +The space $N = M\times X$ consists of two copies of $M$. +If $D_F = 0$ we have infinite distance between the two copies. Now have +hacked the spectral triple to have nonzero Dirac operator $D_F$. The new +Dirac operator still has a commuting relation with the algebra $[D_F, a] = 0$ +$\forall a \in A$, and we should note that the distance between the two +copies of $M$ is still infinite. This is purely an mathematically abstract +observation and doesn't affect physical results. + +\subsubsection{Spectral Action} +In this chapter we bring all our results together to establish an +Action functional to describe a physical system. It turns out that +the Lagrangian of the almost commutative manifold $M\times F_{ED}$ +corresponds to the Lagrangian of Electrodynamics on a curved +background manifold (+ gravitational Lagrangian), consisting of the spectral +action $S_b$ (bosonic) and of the fermionic action $S_f$. + +The simplest spectral action of a spectral triple $(A, H, D)$ is given by the +trace of a function of $D$. We also consider inner fluctuations of the Dirac +operator $D_\omega = D + \omega + \varepsilon' J\omega J^{-1}$ where $\omega = +\omega ^* \in \Omega_D^1(A)$. +\begin{mydefinition} + Let $f:\mathbb{R} \rightarrow \mathbb{R}$ be a suitable function + \textbf{positive and even}. The spectral action is then + \begin{align} + S_b [\omega] := \text{Tr}\big(f(\frac{D_\omega}{\Lambda})\big) + \end{align} + where $\Lambda$ is a real cutoff parameter. The minimal condition on $f$ + is that $f(\frac{D_\omega}{\Lambda})$ is a trace class operator. A trace + class operator is a compact operator with a well defined finite trace + independent of the basis. The subscript $b$ in $S_b$ stands for bosonic, + because in physical applications $\omega$ will describe bosonic fields. + + In addition to the bosonic action $S_b$ we can define a topological spectral + action $S_{top}$. Leaning on the grading $\gamma$ the topological spectral action is + \begin{align} + S_{\text{top}}[\omega] := \text{Tr}(\gamma\ + f(\frac{D_\omega}{\Lambda})). + \end{align} +\end{mydefinition} +\begin{mydefinition}\label{def:fermionic action} + The fermionic action is defined by + \begin{align} + S_f[\omega, \psi] = (J\tilde{\psi}, D_\omega \tilde{\psi}) + \end{align} + with $\tilde{\psi} \in H_{cl}^+ := \{\tilde{\psi}: \psi \in H^+\}$, where + $H_{cl}^+$ is a set of Grassmann variables in $H$ in the +1-eigenspace + of the grading $\gamma$. +\end{mydefinition} + +%---------------------- APPENDIX ?????????????-------------------- +\textbf{APPENDIX??} +Grassmann variables are a set of Basis vectors of a vector space, they +form a unital algebra over a vector field $V$, where the generators are +anti commuting, that is for Grassmann variables $\theta _i, \theta _j$ we have +\begin{align} + &\theta _i \theta _j = -\theta _j \theta _i \\ + &\theta _i x = x\theta _j \;\;\;\; x\in V \\ + &(\theta_i)^2 = 0 \;\;\; (\theta _i \theta _i = -\theta _i \theta _i) +\end{align} +%---------------------- APPENDIX ?????????????-------------------- +\begin{proposition} + The spectral action of the almost commutative manifold $M$ with $\dim(M) + =4$ with a fluctuated Dirac operator is + \begin{align} + \text{Tr}(f\frac{D_\omega}{\Lambda}) \sim \int_M \mathcal{L}(g_{\mu\nu}, + B_\mu, \Phi) \sqrt{g}\ d^4x + O(\Lambda^{-1}), + \end{align} + where + \begin{align} + \mathcal{L}(g_{\mu\nu}, B_\mu, \Phi) = + N\mathcal{L}_M(g_{\mu\nu}) + \mathcal{L}_B(B_\mu)+ + \mathcal{L}_\phi(g_{\mu\nu}, B_\mu, \Phi). + \end{align} + The Lagrangian $\mathcal{L}_M$ is of the spectral triple, represented by + the following term + $(C^\infty(M) , L^2(S), D_M)$ + \begin{align}\label{lagr} + \mathcal{L}_M(g_{\mu\nu}) := \frac{f_4 \Lambda ^4}{2\pi^2} - + \frac{f_2 \Lambda^2}{24\pi ^2}s - \frac{f(0)}{320\pi^2} C_{\mu\nu + \varrho \sigma}C^{\mu\nu \varrho \sigma}, + \end{align} + here $C^{\mu\nu \varrho \sigma}$ is defined in terms of the Riemannian + curvature tensor $R_{\mu\nu \varrho \sigma}$ and the Ricci tensor + $R_{\nu\sigma} = g^{\mu\varrho} R_{\mu\nu \varrho\sigma}$. + The kinetic term of the gauge field is described by the Lagrangian + $\mathcal{L}_B$, which takes the following shape + \begin{align} + \mathcal{L}_B(B_\mu) := \frac{f(0)}{24\pi^2} + \text{Tr}(F_{\mu\nu}F^{\mu\nu}). + \end{align} + Lastly $\mathcal{L}_\phi$ is the scalar-field Lagrangian with a boundary + term, given by + \begin{align} + \mathcal{L}_\phi(g_{\mu\nu}, B_\mu, \Phi) := + &-\frac{2f_2\Lambda^2}{4\pi^2}\text{Tr}(\Phi^2) + \frac{f(0)}{8\pi^2} + \text{Tr}(\Phi^4) + \frac{f(0)}{24\pi^2} + \Delta(\text{Tr}(\Phi^2))\nonumber\\ + &+ \frac{f(0)}{48\pi^2}s\text{Tr}(\Phi^2) + \frac{f(0)}{8\pi^2}\text{Tr}((D_\mu \Phi)(D^\mu \Phi)). + \end{align} +\end{proposition} +\begin{proof} + The dimension of the manifold $M$ is $\dim(M) = \text{Tr}(id) =4$. For + an $x \in M$, we have an asymptotic expansion of the term + $\text{Tr}(f(\frac{D_\omega}{\Lambda}))$ as $\Lambda$ goes to infinity, + which can be written as + \begin{align} + \text{Tr}(f(\frac{D_\omega}{\Lambda})) \simeq& \ 2f_4 \Lambda ^4 + a_0(D_\omega ^2)+ 2f_2\Lambda^2 a_2(D_\omega^2)\nonumber \\&+ f(0) a_4(D_\omega^4) + +O(\Lambda^{-1}).\label{eq:trheatkernel} + \end{align} + We have to note here that the heat kernel coefficients are zero for uneven $k$, + and they are dependent on the fluctuated Dirac operator + $D_\omega$. We can rewrite the heat kernel coefficients in terms of $D_M$, + for the first two terms $a_0$ and $a_2$ we use $N:= + \text{Tr}\mathbbm{1_{H_F}})$ and write + \begin{align} + a_0(D_\omega^2) &= Na_0(D_M^2),\\ + a_2(D_\omega^2 &= Na_2(D_M^2) - \frac{1}{4\pi^2}\int_M + \text{Tr}(\Phi^2)\sqrt{g}d^4x. + \end{align} + For $a_4$ we extend in terms of coefficients of $F$, \textbf{REWRITE: look week9.pdf + for the standard version} + \begin{align} + &\frac{1}{360}\text{Tr}(60sF)= -\frac{1}{6}S(Ns + 4 + \text{Tr}(\Phi^2))\\ + \nonumber\\ + &F^2 = \frac{1}{16}s^2\otimes 1 + 1\otimes \Phi^4 - \frac{1}{4} + \gamma^\mu\gamma^\nu \gamma^\varrho\gamma^\sigma F_{\mu\nu}F^{\mu\nu}+\\ + &\;\;\;\;\;\;\;+\gamma^\mu\gamma^\nu\otimes(D_\mu\Phi)(D_\nu + \Phi)+\frac{1}{2}s\otimes \Phi^2 + \ \text{traceless terms}\\ + \nonumber\\ + &\frac{1}{360}\text{Tr}(180F^2) = \frac{1}{8}s^2N + 2\text{Tr}(\Phi^4) + + \text{Tr}(F_{\mu\nu}F^{\mu\nu}) +\\ + &\;\;\;\;\;\;\;+2\text{Tr}((D_\mu\Phi)(D^\mu\Phi)) + + s\text{Tr}(\Phi^2)\\ + \nonumber\\ + &\frac{1}{360}\text{Tr}(-60\Delta F)= + \frac{1}{6}\Delta(Ns+4\text{Tr}(\Phi^2)). + \end{align} + The cross terms of the trace in $\Omega_{\mu\nu}^E\Omega^{E\mu\nu}$ + vanishes because of the antisymmetric property of the Riemannian + curvature tensor, thus we can write + \begin{align} + \Omega_{\mu\nu}^E\Omega^{E\mu\nu} = \Omega_{\mu\nu}^S\Omega^{S\mu\nu} + \otimes 1 - 1\otimes F_{\mu\nu}F^{\mu\nu} + 2i\Omega_{\mu\nu}^S + \otimes F^{\mu\nu}. + \end{align} + The trace of the cross term $\Omega^{S}_{\mu\nu}$ vanishes because + \begin{align} + \text{Tr}(\Omega^{S}_{\mu\nu}) = \frac{1}{4} + R_{\mu\nu\varrho\sigma}\text{Tr}(\gamma^\mu\gamma^\nu) = \frac{1}{4} + R_{\mu\nu\varrho\sigma}g^{\mu\nu} =0, + \end{align} + then the trace of the whole term is given by + \begin{align} + \frac{1}{360}\text{Tr}(30\Omega^E_{\mu\nu}\Omega^{E\mu\nu}) = + \frac{N}{24}R_{\mu\nu\varrho\sigma}R^{\mu\nu\varrho\sigma} + -\frac{1}{3}\text{Tr}(F_{\mu\nu}F^{\mu\nu}). + \end{align} + Finally plugging the results into the coefficient $a_4$ and simplifying we get + \begin{align} + a_4(x, D_\omega^4) &= Na_4(x, D_M^2) + \frac{1}{4\pi^2}\bigg(\frac{1}{12} s + \text{Tr}(\Phi^2) + \frac{1}{2}\text{Tr}(\Phi^4) \nonumber \\ + &+ \frac{1}{4} + \text{Tr}((D_\mu\Phi)(D^\mu \Phi)) + \frac{1}{6} + \Delta\text{Tr}(\Phi^2) + \frac{1}{6} + \text{Tr}(F_{\mu\nu}F^{\mu\nu})\bigg). + \end{align} + The only thing left is to substitute the heat kernel coefficients into the + heat kernel expansion in equation \eqref{eq:trheatkernel}. +\end{proof} + +\subsubsection{Fermionic Action} +We remind ourselves the definition of the fermionic action in definition +\ref{def:fermionic action} and the manifold we are dealing with in equation +\eqref{eq:almost commutative manifold}. The Hilbertspace $H_F$ is separated +into the particle-antiparticle states with ONB $\{e_R, e_L, \bar{e}_R, +\bar{e}_L\}$. The orthonormal basis of $H_F^+$ is $\{e_L, \bar{e}_R\}$ and +consequently for $H_F^-$, $\{e_R, \bar{e}_L\}$. We can decompose a spinor +$\psi \in L^2(S)$ in each of the eigenspaces $H_F^\pm$, $\psi = \psi_R+ +\psi_L$. That means for an arbitrary $\psi \in H^+$ we can write +\begin{align} + \psi = \chi_R \otimes e_R + \chi_L \otimes e_L + \psi_L \otimes + \bar{e}_R+ + \psi_R \otimes \bar{e}_L, +\end{align} +where $\chi_L, \psi_L \in L^2(S)^+$ and $\chi_R, \psi_R \in L^2(S)^-$. + +Since the fermionic action yields too much restriction on $F_{ED}$ (modified +Two-Point space $F_X$) we redefine it by taking account the fluctuated Dirac +operator +\begin{align} + D_\omega = D_M \otimes i + \gamma^\mu \otimes B_\mu + \gamma_M \otimes + D_F. +\end{align} +The Fermionic Action is +\begin{align} +S_F = (J\tilde{\xi}, D_\omega\tilde{\xi}) +\end{align} +for a $\xi \in H^+$. Then the straight forward calculation gives \begin{align} + \frac{1}{2}(J\tilde{\xi}, D_\omega\tilde{\xi}) + &=\frac{1}{2}(J\tilde{\xi}, (D_M \otimes + i)\tilde{\xi})\label{eq:fermionic1}\\ + &+\frac{1}{2}(J\tilde{\xi}, (\gamma^\mu \otimes B_\mu) + \tilde{\xi})\label{eq:fermionic2}\\ + &+\frac{1}{2}(J\tilde{\xi}, (\gamma_M\otimes + D_F)\tilde{\xi})\label{eq:fermionic3}, +\end{align} +(note that we add the constant $\frac{1}{2}$ to the action). +For the term in \eqref{eq:fermionic1} we calculate +\begin{align} + \frac{1}{2}(J\tilde{\xi}, (D_M\otimes 1)\tilde{\xi}) &= + \frac{1}{2}(J_M\tilde{\chi}_R,D_M\tilde{\psi}_L)+\nonumber + \frac{1}{2}(J_M\tilde{\chi}_L,D_M\tilde{\psi}_R)+ + \\&+\frac{1}{2}(J_M\tilde{\psi}_L,D_M\tilde{\psi}_R)+\nonumber + \frac{1}{2}(J_M\tilde{\chi}_R,D_M\tilde{\chi}_L)\\ + &= (J_M\tilde{\chi},D_M\tilde{\chi}). +\end{align} +For the term in \eqref{eq:fermionic2} we have +\begin{align} + \frac{1}{2}(J\tilde{\xi}, (\gamma^\mu \otimes B_\mu)\tilde{\xi})&= + -\frac{1}{2}(J_M\tilde{\chi}_R, \gamma^\mu Y_\mu\tilde{\psi}_R) + -\frac{1}{2}(J_M\tilde{\chi}_L, \gamma^\mu Y_\mu\tilde{\psi}_R)+\nonumber\\ + &+\frac{1}{2}(J_M\tilde{\psi}_L, \gamma^\mu Y_\mu\tilde{\chi}_R)+ + \frac{1}{2}(J_M\tilde{\psi}_R, \gamma^\mu Y_\mu\tilde{\chi}_L)=\nonumber\\ + &= -(J_M\tilde{\chi}, \gamma^\mu Y_\mu\tilde{\psi}). +\end{align} +And for \eqref{eq:fermionic3} we can write +\begin{align} + \frac{1}{2}(J\tilde{\xi}, (\gamma_M\otimes D_F)\tilde{\xi})&= + +\frac{1}{2}(J_M\tilde{\chi}_R, d\gamma_M\tilde{\chi}_R) + +\frac{1}{2}(J_M\tilde{\chi}_L, \bar{d}\gamma_M\tilde{\chi}_L)+\nonumber\\ + &+\frac{1}{2}(J_M\tilde{\chi}_L, \bar{d}\gamma_M\tilde{\chi}_L) + +\frac{1}{2}(J_M\tilde{\chi}_R, d\gamma_M\tilde{\chi}_R)=\nonumber\\ + &= i(J_M\tilde{\chi}, m\tilde{\psi}). +\end{align} +A small problem arises, we obtain a complex mass parameter $d$, but we can +write $d:=im$ for $m\in \mathbb{R}$, which stands for the real mass. + +Finally the fermionic action of $M\times F_{ED}$ takes the form + \begin{align} + S_f = -i\big(J_M\tilde{\chi}, \gamma(\nabla^S_\mu - i\Gamma_\mu) + \tilde{\Psi}\big) + \big(S_M\tilde{\chi}_L, \bar{d}\tilde{\psi}_L\big) - + \big(J_M\tilde{\chi}_R, d \tilde{\psi}_R\big). + \end{align} +Ultimately we arrive at the full Lagrangian of $M\times F_{ED}$, which is the +sum of purely gravitational Lagrangian + \begin{align} + \mathcal{L}_{grav}(g_{\mu\nu})=4\mathcal{L}_M(g_{\mu\nu})+ + \mathcal{L}_\phi (g_{\mu\nu}), + \end{align} +and the Lagrangian of electrodynamics + \begin{align} + \mathcal{L}_{ED} = -i\bigg\langle + J_M\tilde{\chi},\big(\gamma^\mu(\nabla^S_\mu - iY_\mu) -m\big)\tilde{\psi}) + \bigg\rangle + +\frac{f(0)}{6\pi^2} Y_{\mu\nu}Y^{\mu\nu}. + \end{align} + diff --git a/src/thesis/chapters/finitencg.tex b/src/thesis/chapters/backup/finitencg.tex diff --git a/src/thesis/chapters/backup/heatkernel.tex b/src/thesis/chapters/backup/heatkernel.tex @@ -0,0 +1,314 @@ +\subsection{Heat Kernel Expansion} +\subsubsection{The Heat Kernel} +The heat kernel $K(t; x, y; D)$ is the fundamental solution of the heat +equation +\begin{align} + (\partial _t + D_x)K(t;x, y;D) =0, +\end{align} +which depends on the operator $D$ of Laplacian type. + +For a flat manifold $M = \mathbb{R}^n$ and $D = D_0 := -\Delta_\mu\Delta^\mu +m^2$ the +Laplacian with a mass term and the initial condition +\begin{align} + K(0;x,y;D) = \delta(x,y), +\end{align} +takes the form of the standard fundamental solution +\begin{align}\label{eq:standard} + K(t;x,y;D_0) = (4\pi t)^{-n/2}\exp\left(-\frac{(x-y)^2}{4t}-tm^2\right). +\end{align} +Let us consider now a more general operator $D$ with a potential term or a +gauge field, the heat kernel reads then +\begin{align} + K(t;x,y;D) = \langle x|e^{-tD}|y\rangle. +\end{align} +We can expand the heat kernel in $t$, still having a +singularity from the equation \eqref{eq:standard} as $t \rightarrow 0$ thus the +expansion reads +\begin{align} + K(t;x,y;D) = K(t;x,y;D_0)\left(1 + tb_2(x,y) + t^2b_4(x,y) + \dots + \right), +\end{align} +where $b_k(x,y)$ become regular as $y \rightarrow x$. These coefficients are called the heat +kernel coefficients. +%%----------------------- KANN WEGGELASSEN WERDEN +%\newline +%\textbf{KANN WEGELASSEN WERDEN BIS ZUM NÄCHSTEN KAPITEL} +%Let's turn our attention to a propagator $D^{-1}(x,y)$ defined through the +%heat kernel, with an integral representation +%\begin{align} +% D^{-1} (x,y) = \int_0^\infty dt K(t;x,y;D). +%\end{align} +%If we assume the heat kernel vanishes for $t\rightarrow \infty$, we can +%integrate formally to get +%\begin{align} +% D^{-1}(x,y) \simeq +% 2(4\pi)^{-n/2}\sum_{j=0}\left(\frac{|x-y|}{2m}\right)^{-\frac{n}{2}+j+1} +% K_{-\frac{n}{2}+j+1}(|x-y|m)b_{2j}(x,y), +%\end{align} +%where $b_0 = 1$ and $K_\nu (z)$ is the Bessel function +%\begin{align} +% K_\nu(z) = \frac{1}{\pi} \int_0^\pi \cos(\nu\tau-z\sin(\tau))d\tau. +%\end{align} +%The Bessel function solves the following differential equation +%\begin{align} +% z^2 \frac{d^2K}{dz^2} + z \frac{dK}{dz} + (z^2 - \nu^2)=0. +%\end{align} +%By looking at an integral approximation for the propagator we conclude that +%the singularities of $D^{-1}$ coincide with the singularities of the heat +%kernel coefficients. Thus we can say, that a generating functional in terms of +%$\det(D)$ is called the one-loop effective action (quantum field theory) +%\begin{align} +% W = \frac{1}{2}\ln(\det D). +%\end{align} +%We have a direct relation with one-loop effective action $W$ and the +%heat kernel. Furthermore notice that for each eigenvalue $\lambda >0$ of $D$ +%we can write the identity. +%\begin{align} +% \ln \lambda = -\int_0^\infty \frac{e^{-t\lambda}}{t}dt +%\end{align} +%This expression is correct up to an infinite constant which does not depend +%on the eigenvalue $\lambda$, thus we can ignore it. By substituting +%$\ln(\det D) = \text{Tr}(\ln D)$ we can rewrite the one-loop effective action +%$W$ into +%\begin{align} +% W = -\frac{1}{2} \int_0^\infty dt \frac{K(t, D)}{t}, +%\end{align} +%where +%\begin{align} +% K(t, D) = \text{Tr}(e^{-tD}) = \int d^n x \sqrt{g}K(t;x,x;D). +%\end{align} +%The problem now is that the integral of $W$ is divergent at both limits. Yet +%the divergences at $t\rightarrow \infty$ are caused by $\lambda \leq 0$ of $D$ +%(infrared divergences) and can be ignored. The divergences at $t\rightarrow 0$ +%are cutoff at $t=\Lambda^{-2}$, simply written as +%\begin{align} +% W_\Lambda = -\frac{1}{2} \int_{\Lambda^{-2}}^\infty dt \frac{K(t, D)}{t}. +%\end{align} +%We can calculate $W_\Lambda$ up to an order of $\lambda ^0$ +%\begin{align} +% W_\Lambda &= -(4\pi)^{-n/2} \int d^n x\sqrt{g}\bigg( +% \sum_{2(j+l)<n}\Lambda^{n-2j-2l}b_{2j}(x,x) \frac{(-m^2)^l l!}{n-2j-2l} +\\ +% &+ \sum_{2(j+l) =n }\ln(\Lambda) (-m^2)^l l! b_{2j}(x,x) +% \mathcal{O}(\lambda^0) \bigg) +%\end{align} +%There is an divergence at $b_2(x,x)$ for $k\leq n$. Computing the limit +%$\Lambda \rightarrow \infty$ we get +%\begin{align} +% -\frac{1}{2}(4\pi)^{n/2}m^n\int d^n x\sqrt{g} \sum_{2j>n} +% \frac{b_{2j}(x,x)}{m^{2j}}\Gamma(2j-n), +%\end{align} +%where $\Gamma$ stands for the gamma function. +%%----------------------- KANN WEGGELASSEN WERDEN + + +\subsubsection{Spectral Functions} +Manifolds $M$ with a disappearing boundary condition for the operator $e^{-tD}$ for $t>0$ is a +trace class operator on $L^2(V)$. Meaning for any smooth function $f$ on $M$ +we can define +\begin{align} + K(t,f,D) := \text{Tr}_{L^2}(fe^{-tD}), +\end{align} +or alternately write an integral representation +\begin{align} + K(t, f, D) = \int_M d^n x \sqrt{g} \text{Tr}_V(K(t;x,x;D)f(x)), +\end{align} +in the regular limit $y\rightarrow y$. We can write the Heat Kernel in terms +of the spectrum of $D$. So for an orthonormal basis $\{\phi_\lambda\}$ of +eigenfunctions for $D$, which corresponds to the eigenvalue $\lambda$, we +can rewrite the heat kernel into +\begin{align} + K(t;x,y;D) = \sum_\lambda \phi^\dagger_\lambda(x) + \phi_\lambda(y)e^{-t\lambda}. +\end{align} +An asymptotic expansion as $t \rightarrow 0$ for the trace is then +\begin{align} + \text{Tr}_{L^2}(fe^{-tD}) \simeq \sum_{k\geq 0}t^{(k-n)/2}a_k(f,D), +\end{align} +where +\begin{align} + a_k(f,D) = (4\pi)^{-n/2} \int_M d^4x \sqrt{g} b_k(x,x) f(x). +\end{align} +\subsubsection{General Formulae} +Let us summarize what we have obtained in the last chapter, we considered a +compact Riemannian manifold $M$ without boundary condition, a vector bundle +$V$ over $M$ to define functions which carry discrete (spin or gauge) +indices, an operator $D$ of Laplace type over $V$ and smooth function $f$ on +$M$. + +There is an asymptotic expansion where the heat kernel coefficients with an +odd index $k=2j+1$ vanish $a_{2j+1}(f,D) = 0$. On the other hand coefficients +with an even index are locally computable in terms of geometric invariants +\begin{align} + a_k(f,D) &= \text{Tr}_V\left(\int_M d^n x\sqrt{g}(f(x)a_k(x;D)\right) + =\nonumber\\ + &=\sum_I \text{Tr}_V\left(\int_M d^nx \sqrt{g}(fu^I + \mathcal{A}^I_k(D))\right). +\end{align} +We denote $\mathcal{A}^I_k$ as all possible independent invariants of +dimension $k$, and $u^I$ are constants. The invariants are constructed from +$E, \Omega, R_{\mu\nu\varrho\sigma}$ and their derivatives If $E$ has +dimension two, then the derivative has dimension one. So if $k=2$ there are +only two independent invariants, $E$ and $R$. This corresponds to the +statement $a_{2j+1}=0$. + +If we consider $M = M_1 \times M_2$ with coordinates $x_1$ and $x_2$ and a +decomposed Laplace style operator $D = D_1 \otimes 1 + 1 \otimes D_2$ we can +separate functions acting on operators and on coordinates linearly by the +following +\begin{align} + e^{-tD} &= e^{-tD_1} \otimes e^{-tD_2},\\ + f(x_1, x_2) &= f_1(x_1)f_2(x_2), +\end{align} +thus the heat kernel coefficients are separated by +\begin{align} + a_k(x;D) &= \sum_{p+q=k} a_p(x_1; D_1)a_q(x_2;D_2) +\end{align} +If we know the eigenvalues of $D_1$ are known, $l^2, l\in \mathbb{Z}$, we +can obtain the heat kernel asymmetries with the Poisson summation formula +giving us an approximation in the order of $e^{-1/t}$ +\begin{align} + K(t, D_1) &= \sum_{l\in\mathbb{Z}} e^{-tl^2} = \sqrt{\frac{\pi}{t}} + \sum_{l\in\mathbb{Z}} e^{-\frac{\pi^2l^2}{t}} = \nonumber \\ + &\simeq \sqrt{\frac{\pi}{t}} + \mathcal{O}(e^{-1/t}). +\end{align} +The exponentially small terms have no effect on the heat kernel +coefficients and that the only nonzero coefficient is $a_0(1, D_1) = +\sqrt{\pi}$, therefore the heat coefficients can be written as +\begin{align} + a_k(f(x^2), D) = \sqrt{\pi}\int_{M_2} + d^{n-1}x\sqrt{g}\sum_I\text{Tr}_V\left(f(x^2)u^I_{(n-1)} + \mathcal{A}^I_n(D_2)\right). +\end{align} + +Because all of the geometric invariants associated with $D$ are in the $D_2$ +part, they are independent of $x_1$. Ultimately meaning we are free to choose +$M_1$. For $M_1 = S^1$ with $x\in (0, 2\pi)$ and $D_1=-\partial_{x_1}^2$ +we can rewrite the heat kernel coefficients into +\begin{align} + a_k(f(x_2), D) &= \int_{S^1\times M_2}d^nx \sqrt{g} \sum_I + \text{Tr}_V(f(x_2) u_{(n)}^I \mathcal{A}^I_k(D_2))= \nonumber\\ + &= 2\pi \int_{M_2} d^nx\sqrt{g} \sum_I\text{Tr}_V(f(x_2) u_{(n)}^I + \mathcal{A}^I_k(D_2)). +\end{align} +Computing the two equations above we see that +\begin{align} + u_{(n)}^I = \sqrt{4\pi} u^I_{(n+1)} +\end{align} + +\subsubsection{Heat Kernel Coefficients} +To calculate the heat kernel coefficients we need the following variational +equations +\begin{align} + &\frac{d}{d\varepsilon}\bigg|_{\varepsilon=0}a_k(1, e^{-2\varepsilon f}D) = + (n-k) a_k(f, D),\label{eq:var1}\\ + &\frac{d}{d\varepsilon}\bigg|_{\varepsilon=0}a_k(1, D-\varepsilon F) = + a_{k-2}(F,D),\label{eq:var2}\\ + &\frac{d}{d\varepsilon}\bigg|_{\varepsilon=0}a_k(e^{-2\varepsilon f}F, + e^{-2\varepsilon f}D) = + 0\label{eq:var3}. +\end{align} +Let us explain the equations above. To get the first equation \eqref{eq:var1} +we differentiate \begin{align} + \frac{d}{d\varepsilon}\bigg|_{\varepsilon=0} \text{Tr}(\exp(-e^{-2\varepsilon + f}tD) = \text{Tr}(2ftDe^{-tD}) = -2t\frac{d}{dt}\text{Tr}(fe^{-tD})) +\end{align} +then we expand both sides in $t$ and get \eqref{eq:var1}. Equation \eqref{eq:var2} is derived similarly. + +For equation \eqref{eq:var3} we consider the following operator +\begin{align} + D(\varepsilon,\delta) = e^{-2\varepsilon f}(D-\delta F) +\end{align} +for $k=n$ we use equation \eqref{eq:var1} and we get +\begin{align} + \frac{d}{d\varepsilon}\bigg|_{\varepsilon=0}a_n(1,D(\varepsilon,\delta)) + =0, +\end{align} +then we take the variation in terms of $\delta$, evaluated at $\delta =0$ and +swap the differentiation, allowed by theorem of Schwarz +\begin{align} + 0 &= + \frac{d}{d\delta}\bigg|_{\delta=0}\frac{d}{d\varepsilon}\bigg|_{\varepsilon=0}a_n(1, + D(\varepsilon,\delta)) =\nonumber\\ + &=\frac{d}{d\varepsilon}\bigg|_{\varepsilon=0}\frac{d}{d\delta}\bigg|_{\delta=0}a_n(1, + D(\varepsilon,\delta)) =\nonumber\\ + &=a_{n-2} ( e^{-2\varepsilon f}F, e^{-2\varepsilon f}D), +\end{align} +which gives us equation \eqref{eq:var3}. + +Now that we have established the ground basis, we can calculate the constants +$u^I$, and by that the first three heat kernel coefficients read +\begin{align} + a_0(f, D) &= (4\pi)^{-n/2}\int_Md^n x\sqrt{g} \text{Tr}_V(a_0 f),\\ + a_2(f, D) &= (4\pi)^{-n/2}\frac{1}{6}\int_Md^n + x\sqrt{g}\text{Tr}_V)(f\alpha _1 E+\alpha _2 R),\\ + a_4(f, D) &= (4\pi)^{-n/2}\frac{1}{360}\int_Md^n + x\sqrt{g}\text{Tr}_V(f(\alpha_3 E_{,kk} + \alpha_4\ R\ E + \alpha_5 E^2 + \alpha_6 R_{,kk} + \nonumber\\ + &+\alpha_7 R^2 + \alpha_8 R_{ij}R_{ij} + \alpha_9 + R_{ijkl}R_{ijkl} +\alpha_{10} \Omega_{ij}\Omega{ij})), +\end{align} +where the comma subscript $,$ denotes the derivative and constants $\alpha_I$ +do not depend on the dimension of the Manifold and we can compute them with +our variational identities. + +The first coefficient $\alpha_0$ can be read from the heat kernel expansion of +the Laplacian on $S^1$ (above), $\alpha_0 = 1$. For $\alpha_1$ we use +\eqref{eq:var2}, the coefficient $k = 2$ is +\begin{align} + \frac{1}{6} \int_M d^n x\sqrt{g} \text{Tr}_V(\alpha_1F) = \int_M d^n + x\sqrt{g} \text{Tr}_V(F), +\end{align} +which means $\alpha_1 = 6$. Looking at the coefficient $k=4$ we have +\begin{align} + \frac{1}{360}\int_Md^n x \sqrt{g}\text{Tr}_V(\alpha_4\ F\ R + 2\alpha_5\ F\ E) + = \frac{1}{6} \int_Md^n x\sqrt{g}\text{Tr}_V(\alpha_1\ F\ E + \alpha_2\ F\ R), +\end{align} +thus $\alpha_4 = 60\alpha_2$ and $\alpha_5 = 180$. + +By applying \eqref{eq:var3} to $n=4$ we get +\begin{align} + \frac{d}{d\varepsilon}|_{\varepsilon=0} a_2(e^{-2\varepsilon f}F, + e^{-2\varepsilon f}D) = 0. +\end{align} +Collecting the terms with $\text{Tr}_V(\int_Md^nx\sqrt{g}(Ff_{,jj}))$ we +obtain $\alpha_1 = 6\alpha_2$, that is $\alpha_2 = 1$, so $\alpha_4 = 60$. + +Now we let $M=M_1\times M_2$ and split $D = -\Delta_1 -\Delta_2$, where +$\Delta_{1/2}$ are Laplacians for $M_1, M_2$. This allows us to decompose the heat +kernel coefficient for $k=4$ into +\begin{align} + a_4(1,-\Delta_1-\Delta_2) =& a_4(1, -\Delta_1) a_0(1, + -\Delta_2)\nonumber+ \\ + &+a_2(1,-\Delta_1) a_2(1,-\Delta_2)\nonumber \\ + &+ a_0(1,-\Delta_1)a_4(1,-\Delta_2), +\end{align} +with $E=0$ and $\Omega =0$ and by calculating the terms with $R_1R_2$ (scalar +curvature of $M_{1/2}$) we obtain $\frac{2}{360}\alpha_7 = +(\frac{\alpha_2}{6})^2$, thus $\alpha_7 = 5$. + +For $n=6$ we get +\begin{align} + 0 &= \text{Tr}_V(\int_Md^nx\sqrt{g} + (F(-2\alpha_3-10\alpha_4+4\alpha_5)f_{,kk}E +\nonumber\\ + &+(2\alpha_3 + 10\alpha_6)f_{,iijj}+\nonumber\\ + &+(2\alpha_4 -2\alpha_6 - 20\alpha_7 -2\alpha_8)f_{,ii}R\nonumber\\ + &+(-8\alpha_8 -8\alpha_6)f_{,ij}R_{ij})) +\end{align} +we obtain $\alpha_3 = 60$, $\alpha_6=12$, $\alpha_8 = -2$ and $\alpha_9 = 2$ + +To get $\alpha_{10}$ we use the Gauss-Bonnet theorem, ultimately giving us +$\alpha_{10}=30$. We leave out this lengthy calculation and refer to +\cite{heatkernel} for further reading. + +Let us summarize our calculations which ultimately give us the following heat kernel +coefficients +\begin{align} + \alpha_0(f, D) &= (4\pi)^{-n/2}\int_M d^n x \sqrt{g} \text{Tr}_V(f),\\ + \alpha_2(f, D) &= (4\pi)^{-n/2}\frac{1}{6}\int_M d^n x \sqrt{g} + \text{Tr}_V(f(6E+R)),\\ + \alpha_4(f, D) &= (4\pi)^{-n/2}\frac{1}{360}\int_M d^n x \sqrt{g} + \text{Tr}_V(f(60E_{,kk}+60RE+ 180E^2 +\\ + &+12R_{,kk} + 5R^2 - 2 R_{ij}R_{ij} + 2R_{ijkl}R_{ijkl} +30\Omega_{ij}\Omega_{ij})). +\end{align} + diff --git a/src/thesis/chapters/backup/intro.tex b/src/thesis/chapters/backup/intro.tex @@ -0,0 +1,2 @@ +\section{Introduction} +\lipsum[2] diff --git a/src/thesis/chapters/backup/main_sec.tex b/src/thesis/chapters/backup/main_sec.tex @@ -0,0 +1 @@ +\section{Main Section} diff --git a/src/thesis/chapters/realncg.tex b/src/thesis/chapters/backup/realncg.tex diff --git a/src/thesis/chapters/twopointspace.tex b/src/thesis/chapters/backup/twopointspace.tex diff --git a/src/thesis/chapters/basics.tex b/src/thesis/chapters/basics.tex @@ -1,41 +1,46 @@ \subsection{Noncommutative Geometric Spaces} \subsubsection{$*$-Algebra} To grasp the idea of encoding geometrical data into a spectral triple we -introduce the first ingredient of a spectral triple, an unital $C^*$ algebra. -\begin{definition} +introduce the first ingredient of a spectral triple, an unital $*$ algebra. +\begin{mydefinition} A \textit{vector space} $A$ over $\mathbb{C}$ is called a \textit{complex, unital Algebra} if, \\ $\forall a,b \in A$ : - \begin{enumerate} - \item - $A \times A \rightarrow A$, - $(a,\ b)\ \mapsto \ a\cdot b$, - \item with an identity element: - $1a = a1 =a$. - \end{enumerate} + \begin{align} + A \times A \rightarrow A\\ + (a,\ b)\ &\mapsto \ a\cdot b, + \end{align} + with an identity element: + \begin{align} + 1a = a1 =a. + \end{align} Extending the definition, a $*$-algebra is an algebra $A$ with a \textit{conjugate linear map (involution)} $*:A\ \rightarrow A$, - $\forall a, b \in A$ satisfying: - \begin{enumerate} - \item - $(a\ b)^* = b^*a^*$, - \item - $(a^*)^* = a$. - \end{enumerate} -\end{definition} + $\forall a, b \in A$ satisfying + \begin{align} + (a\ b)^* &= b^*a^*,\\ + (a^*)^* &= a. + \end{align} +\end{mydefinition} In the following all unital algebras are referred to as algebras. \subsubsection{Finite Discrete Space} -Let us consider an example of an $*$-algebra of continuous functions $C(X)$ +Let us consider an example, a $*$-algebra of continuous functions $C(X)$ on a discrete topological space $X$ with $N$ points. Functions of a -continuous $*$-algebra $C(X)$ assign values to $\mathbb{C}$, thus $f,\ g \in +continuous $*$-algebra $C(X)$ assign values to $\mathbb{C}$ and for $f,\ g \in C(X)$, $\lambda \in \mathbb{C}$ and $x \in X$ they provide the following structure: \begin{itemize} - \item \textit{pointwise linear} \\ - $(f + g)(x) = f(x) + g(x)$,\\ - $(\lambda\ f)(x) = \lambda (f(x)),$ - \item \textit{pointwise multiplication} \\ - $f\ g\ (x) = f(x)g(x)$, - \item \textit{pointwise involution} \\ - $f^*(x) = \overline{f(x)}.$ + \item \textit{pointwise linear} + \begin{align} + (f + g)(x) &= f(x) + g(x),\\ + (\lambda\ f)(x) &= \lambda (f(x)), + \end{align} + \item \textit{pointwise multiplication} + \begin{align} + f\ g\ (x) = f(x)g(x), + \end{align} + \item \textit{pointwise involution} + \begin{align} + $f^*(x) = \overline{f(x)}. + \end{align} \end{itemize} The $*$-algebra $C(X)$ is \textit{isomorphic} to a $*$-algebra $\mathbb{C}^N$ with involution ($N$ number of points in $X$), we write $C(X) \simeq @@ -43,7 +48,7 @@ with involution ($N$ number of points in $X$), we write $C(X) \simeq don't lose physical information. A function $f:X\ \rightarrow\ \mathbb{C}$ can be represented with $N \times N$ diagonal matrices, where each diagonal value represents the function value at the corresponding $i$-th point for $i -= 1,...,N$. Because of matrix multiplication and hermitian conjugate of += 1,...,N$. Matrix multiplication and hermitian conjugation of matrices we have a preserving structure. Moreover we can \textit{map} between finite discrete spaces $X_1$ and $X_2$ with a @@ -56,7 +61,7 @@ For every such map there exists a corresponding map \phi ^*:C(X_2)\ \rightarrow C(X_1), \end{align} which `pulls back' values even if $\phi$ is not bijective. -Note that the pullback doesn't map points back, but maps functions on an $*$-algebra $C(X)$. +Note that the pullback does not map points back, but maps functions on an $*$-algebra $C(X)$. The pullback, in literature often called a $*$-homomorphism or a $*$-algebra map under pointwise product has the following properties \begin{itemize} @@ -66,8 +71,8 @@ pointwise product has the following properties \end{itemize} %------------ Exercise The map $\phi :X_1\ \rightarrow \ X_2$ is an injective (surjective) map, - if only if the corresponding pullback $\phi ^* :C(X_2)\ \rightarrow \ - C(X_1)$ is surjective (injective). Let us say, that $X_1$ has $n$ points and + if only and if the corresponding pullback $\phi ^* :C(X_2)\ \rightarrow \ + C(X_1)$ is surjective (injective). To clarify let us say that $X_1$ has $n$ points and $X_2$ with $m$ points. Then there are three different cases, first $n=m$ and obviously $\phi$ is bijective and $\phi ^*$ too. Then $n > m$, in this case $\phi$ assigns $n$ points to $m$ points when $n > m$, which is by definition @@ -76,29 +81,28 @@ pointwise product has the following properties completely analogous to the case $n > m$. %------------ Exercise -\subsubsection{Matrix Algebras} -\begin{definition} +\begin{mydefinition} A \textit{(complex) matrix algebra} A is a direct sum, for $n_i, N \in \mathbb{N}$ \begin{align} A = \bigoplus _{i=1}^{N} M_{n_i}(\mathbb{C}). \end{align} - The involution is the hermitian conjugate, a $*$ algebra with involution is referred to as + The involution is the hermitian conjugate. A $*$ algebra with involution is referred to as a matrix algebra -\end{definition} -From a topological discrete space $X$, we can construct a $*$-algebra -$C(X)$ which is isomorphic to a matrix algebra $A$. Then the question instantly -arises, if we can construct $X$ given $A$? For a matrix algebra $A$, -which in most cases is not commutative, the answer is generally no. - -Thus there are two options. We can restrict ourselves to commutative matrix algebras, -which are the vast minority and not physically interesting. +\end{mydefinition} + +To summarize, from a topological discrete space $X$, we can construct a +$*$-algebra $C(X)$ which is isomorphic to a matrix algebra $A$. Then the +question instantly arises, if we can construct $X$ given $A$? For a matrix +algebra $A$, which in most cases is not commutative, the answer is generally +no. Hence there are two options. We can restrict ourselves to commutative +matrix algebras, which are the vast minority and not physically interesting. Or we can allow more morphisms (isomorphisms) between matrix algebras. \subsubsection{Finite Inner Product Spaces and Representations} -Until now we looked at finite topological discrete spaces, moreover we can consider a -finite dimensional inner product space $H$ (finite Hilbert-spaces), with inner product -$(\cdot,\cdot)\rightarrow \mathbb{C}$. We denote $L(H)$ as the $*$-algebra of operators on $H$ +Until now we have looked at finite topological discrete spaces, moreover we can consider a +finite dimensional inner product space $H$ (finite Hilbertspaces), with inner product +$(\cdot,\cdot)\rightarrow \mathbb{C}$. We denote $L(H)$ as the $*$-algebra of operators on $H$ equipped with a product given by composition and involution of the adjoint, $T \mapsto T^*$. Then $L(H)$ is a \textit{normed vector space} with \begin{align} @@ -106,8 +110,8 @@ Then $L(H)$ is a \textit{normed vector space} with \in L(H)\big \},\\ \|T\| &= \sup\big\{\sqrt{\lambda}:\; \lambda \text{ eigenvalue of } T\big\}. \end{align} -This allows us to define representations of $*$-algebras. -\begin{definition} +The Hilbert space allows us to define representations of $*$-algebras. +\begin{mydefinition} The \textit{representation} of a finite dimensional $*$-algebra $A$ is a pair $(H, \pi)$, where $H$ is a finite dimensional inner product space and $\pi$ is a $*$-\textit{algebra map} @@ -120,7 +124,7 @@ This allows us to define representations of $*$-algebras. \item only $\emptyset$ or $H$ is invariant under the action of $A$ on $H$. \end{itemize} -\end{definition} +\end{mydefinition} Here are some examples of reducible and irreducible representations \begin{itemize} \item For $A = M_n(\mathbb{C})$ the representation $H=\mathbb{C}^n$, $A$ acts as matrix multiplication\\ @@ -131,22 +135,22 @@ Here are some examples of reducible and irreducible representations \end{itemize} Naturally there are also certain equivalences between different representations. -\begin{definition} +\begin{mydefinition} Two representations of a $*$-algebra $A$, $(H_1, \pi _1)$ and $(H_2, \pi _2)$ are called \textit{unitary equivalent} if there exists a map $U: H_1 \rightarrow H_2$ such that. \begin{align} \pi _1(a) = U^* \pi _2(a) U \end{align} -\end{definition} +\end{mydefinition} Furthermore we define a mathematical structure called the structure space, -which will later become important, when speaking of the duality between a spectral -triple and a space. -\begin{definition} +which will become important later when speaking of the duality between a +spectral triple and a space. +\begin{mydefinition} Let $A$ a $*$-algebra then, $\hat{A}$ is called the structure space of all \textit{unitary equivalence classes of irreducible representations of A} -\end{definition} +\end{mydefinition} %------------- EXERCISE Given a representation $(H, \pi)$ of a $*$-algebra $A$, the \textbf{commutant} $\pi (A)'$ of $\pi (A)$ is defined as a set of operators in $L(H)$ that commute with all $\pi (a)$ @@ -154,9 +158,9 @@ triple and a space. \pi (A)' = \big\{T \in L(H):\ \pi(a)\ T = T\ \pi(a) \;\; \forall a\in A\big\} \end{align} - The commutant $\pi (A)'$ is also a $*$-algebra, because it has unital, + The commutant $\pi (A)'$ is also a $*$-algebra, since it has unital, associative and involutive properties. - We note that $\pi (a) \in L(H)\ \forall a \in A$, unitary property is given + That is $\pi (a) \in L(H)\ \forall a \in A$, unitary property is given by the unital operator of the $*$-algebra of operators $L(H)$, which exists by definition because H is a inner product space. Associativity is given by the $*$-algebra of $L(H)$, where $L(H) \times L(H) \mapsto L(H)$, which is @@ -166,19 +170,21 @@ triple and a space. %------------- EXERCISE For a unital algebra $*$-algebra $A$, the matrices $M_n(A)$ with entries - in $A$ form a unital $*$-algebra, because unitary operation in $M_n(A)$ is given by the identity Matrix, which - has to exists in every entry in $M_n(A)$, and behaves like in $A$. Associativity is given by - matrix multiplication. Lastly involution is given by the conjugate transpose. - - A representation $\pi :A\ \rightarrow \ L(H)$ of a $*$-algebra $A$, for - $H^n = H \oplus ... \oplus H$, $n$ times. Then we have the following + in $A$ form a unital $*$-algebra, because the unitary operation in + $M_n(A)$ is given by the identity Matrix, which has to exists in every + entry in $M_n(A)$, and behaves like in $A$. Associativity is given by + matrix multiplication. Lastly involution is given by the conjugate + transpose. + + Consider a representation $\pi :A\ \rightarrow \ L(H)$ of a $*$-algebra + $A$ and set $H^n = H \oplus ... \oplus H$, $n$ times. Then we have the following representation $\tilde{\pi}:M_n(A) \rightarrow L(H^n)$ for the Matrix - Algebra with $\tilde{\pi}((a_{ij})) = (\tilde{\pi}(a_{ij})) \in M_n(A)$. - We have direct isomorphisms of $A \simeq M_n(A)$ and $H \simeq H^n$ - meaning $\tilde{\pi}$ is a valid reducible representation. + Algebra with $\tilde{\pi}((a_{ij})) = (\tilde{\pi}(a_{ij})) \in M_n(A)$, + since a direct isomorphisms of $A \simeq M_n(A)$ and $H \simeq H^n$ + exists. Meaning $\tilde{\pi}$ is a valid reducible representation. - Let $\tilde{\pi}:M_n(A) \rightarrow L(H^n)$ be a $*$ algebra - representation of $M_n(A)$, then $\pi: A \rightarrow L(H^n)$ is a representation of $A$. + By looking at $\tilde{\pi}:M_n(A) \rightarrow L(H^n)$ a $*$ algebra + representation of $M_n(A)$. We see that $\pi: A \rightarrow L(H^n)$ is a representation of $A$. The fact that $\tilde{\pi}$ and $\pi$ are unitary equivalent, there is a map $U: H^n \rightarrow H^n$ given by $U=\mathbbm{1}_n$, thus \begin{align} @@ -189,10 +195,8 @@ triple and a space. %------------- EXERCISE -A commutative matrix algebra can be used to reconstruct a discrete space. -The structure space $\hat{A}$ is used for this. Because $A \simeq -\mathbb{C}^N$ all -irreducible representation are of the form +With help of the structure space $\hat{A}$, a commutative matrix algebra can be used to reconstruct a discrete space. +Since $A \simeq \mathbb{C}^N$ all irreducible representation are of the form \begin{align} \pi _i:(\lambda_1,...,\lambda_N)\in \mathbb{C}^N \mapsto \lambda_i \in \mathbb{C} @@ -210,8 +214,8 @@ Equivalence}) \subsubsection{Algebraic Modules} An important notion for Morita Equivalence are algebraic modules, later -extended with Hilbert bimodules. -\begin{definition} +extended by Hilbert bimodules. +\begin{mydefinition} Let $A$, $B$ be algebras (need not be matrix algebras) \begin{enumerate} \item \textit{left} A-module is a vector space $E$, that carries a left @@ -233,7 +237,7 @@ extended with Hilbert bimodules. a \cdot (e \cdot b)= (a \cdot e) \cdot b;\;\;\; a \in A, b \in B, e \in E \end{align} \end{enumerate} -\end{definition} +\end{mydefinition} An $A$-\textbf{module homomorphism} as linear map $\phi: E\rightarrow F$ which respects the representation of A, e.g.\ for left module. \begin{align} @@ -269,7 +273,7 @@ which is the inner product of a $*$-algebra. \subsubsection{Balanced Tensor Product and Hilbert Bimodules} -\begin{definition} +\begin{mydefinition} Let $A$ be an algebra, $E$ be a \textit{right} $A$-module and $F$ be a \textit{left} $A$-module. The \textit{balanced tensor product} of $E$ and $F$ forms a $A$-bimodule. @@ -278,13 +282,13 @@ which is the inner product of a $*$-algebra. e_i \otimes a_i f_i : \;\;\; a_i \in A,\ e_i \in E,\ f_i \in F \right\}. \end{align} -\end{definition} -The $/$ denotes the quotient space. By that the operation $\otimes _A$ takes +\end{mydefinition} +The symbol $/$ denotes the quotient space. By that the operation $\otimes _A$ takes two left/right modules and makes a bimodule with the help the tensor product of the two modules and the quotient space that takes out all the elements from the tensor product that don't preserver the left/right representation and that are duplicates. -\begin{definition} +\begin{mydefinition} Let $A$, $B$ be \textit{matrix algebras}. The \textit{Hilbert bimodule} for $(A, B)$ is given by an $A$-$B$-bimodue $E$ and by an $B$-valued \textit{inner product} $\langle \cdot,\cdot\rangle_E: E\times E \rightarrow @@ -301,7 +305,7 @@ duplicates. holds iff $e=0$}. \end{align} We denote $KK_f(A,\ B)$ as the set of all \textit{Hilbert bimodules} of $(A,\ B)$. -\end{definition} +\end{mydefinition} %-------------- EXERCISE And indeed the Hilbert bimodule extension takes a representation $\pi:\ A \ @@ -353,7 +357,7 @@ for $a\in A, b\in E_{\phi}$. %-------------- EXAMPLE \subsubsection{Kasparov Product and Morita Equivalence} -\begin{definition} +\begin{mydefinition} Let $E \in KK_f(A, B)$ and $F \in KK_F(B, D)$ the \textit{Kasparov product} is defined as with the balanced tensor product \begin{align} @@ -364,7 +368,7 @@ for $a\in A, b\in E_{\phi}$. \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 \end{align} -\end{definition} +\end{mydefinition} %-------------- EXERCISE The Kasparov product for $*$-algebra homomorphism $\phi: A \rightarrow B$ and @@ -389,7 +393,7 @@ an inner product, acting from the left on $A$ for $\phi$, $a', a\in A$ reads \end{align} which is satisfied by $\phi = \text{id}_A$ -\begin{definition} +\begin{mydefinition} Let $A$, $B$ be \textit{matrix algebras}. They are called \textit{Morita equivalent} if there exists an $E \in KK_f(A, B)$ and an $F \in KK_f(B, A)$ such that \begin{align} @@ -398,7 +402,7 @@ which is satisfied by $\phi = \text{id}_A$ \end{align} where $\simeq$ denotes the isomorphism between Hilbert bimodules and note that $A$ or $B$ is a bimodule by itself. -\end{definition} +\end{mydefinition} The modules $E$ and $F$ are each others inverse in regards to the Kasparov Product, because we land in the same space as we started. To clarify, in @@ -451,11 +455,11 @@ If we take the Kasparov product of $E$ and $F$ we see that $M_n(\mathbb{C})$ and $\mathbb{C}$ are Morita equivalent! %------------- EXAMPLE -\begin{theorem} +\begin{mylemma} Two matrix algebras are Morita Equivalent if, and only if their their structure spaces are isomorphic as discreet spaces (have the same cardinality / same number of elements). -\end{theorem} +\end{mylemma} \begin{proof} Let $A$, $B$ be \textit{Morita equivalent}. Then there exist $_A E_B$ and $_B F_A$ with \begin{align} @@ -497,11 +501,11 @@ we see that $M_n(\mathbb{C})$ and $\mathbb{C}$ are Morita equivalent! $\hat{A} \simeq \hat{B}$. \end{proof} -\begin{lemma} +\begin{mylemma} The matrix algebra $M_n(\mathbb{C})$ has a unique irreducible representation (up to isomorphism) given by the defining representation on $\mathbb{C}^n$. -\end{lemma} +\end{mylemma} \begin{proof} We know $\mathbb{C}^n$ is a irreducible representation of $A= M_n(\mathbb{C})$. Let $H$ be irreducible and of dimension $k$, then we @@ -550,5 +554,7 @@ On the other hand we get \end{align} %---------------- EXAMPLE -To summarize, there is a duality between finite spaces and Morita equivalence classes of matrix algebras. By replacing $*$-homomorphism $A\rightarrow B$ with Hilbert bimodules $(A,B)$ we introduce -a richer structure of morphism between matrix algebras. +To summarize, there is a duality between finite spaces and Morita equivalence +classes of matrix algebras. By replacing $*$-homomorphism $A\rightarrow B$ +with Hilbert bimodules $(A,B)$ we introduce a richer structure of morphism +between matrix algebras. diff --git a/src/thesis/chapters/electroncg.tex b/src/thesis/chapters/electroncg.tex @@ -212,7 +212,7 @@ The simplest spectral action of a spectral triple $(A, H, D)$ is given by the trace of a function of $D$. We also consider inner fluctuations of the Dirac operator $D_\omega = D + \omega + \varepsilon' J\omega J^{-1}$ where $\omega = \omega ^* \in \Omega_D^1(A)$. -\begin{definition} +\begin{mydefinition} Let $f:\mathbb{R} \rightarrow \mathbb{R}$ be a suitable function \textbf{positive and even}. The spectral action is then \begin{align} @@ -230,8 +230,8 @@ operator $D_\omega = D + \omega + \varepsilon' J\omega J^{-1}$ where $\omega = S_{\text{top}}[\omega] := \text{Tr}(\gamma\ f(\frac{D_\omega}{\Lambda})). \end{align} -\end{definition} -\begin{definition}\label{def:fermionic action} +\end{mydefinition} +\begin{mydefinition}\label{def:fermionic action} The fermionic action is defined by \begin{align} S_f[\omega, \psi] = (J\tilde{\psi}, D_\omega \tilde{\psi}) @@ -239,7 +239,7 @@ operator $D_\omega = D + \omega + \varepsilon' J\omega J^{-1}$ where $\omega = with $\tilde{\psi} \in H_{cl}^+ := \{\tilde{\psi}: \psi \in H^+\}$, where $H_{cl}^+$ is a set of Grassmann variables in $H$ in the +1-eigenspace of the grading $\gamma$. -\end{definition} +\end{mydefinition} %---------------------- APPENDIX ?????????????-------------------- \textbf{APPENDIX??} @@ -252,7 +252,7 @@ anti commuting, that is for Grassmann variables $\theta _i, \theta _j$ we have &(\theta_i)^2 = 0 \;\;\; (\theta _i \theta _i = -\theta _i \theta _i) \end{align} %---------------------- APPENDIX ?????????????-------------------- -\begin{proposition} +\begin{myproposition} The spectral action of the almost commutative manifold $M$ with $\dim(M) =4$ with a fluctuated Dirac operator is \begin{align} @@ -293,7 +293,7 @@ anti commuting, that is for Grassmann variables $\theta _i, \theta _j$ we have &+ \frac{f(0)}{48\pi^2}s\text{Tr}(\Phi^2) \frac{f(0)}{8\pi^2}\text{Tr}((D_\mu \Phi)(D^\mu \Phi)). \end{align} -\end{proposition} +\end{myproposition} \begin{proof} The dimension of the manifold $M$ is $\dim(M) = \text{Tr}(id) =4$. For an $x \in M$, we have an asymptotic expansion of the term diff --git a/src/thesis/chapters/finitencg.tex b/src/thesis/chapters/finitencg.tex @@ -12,7 +12,7 @@ j \in X}$ of \textit{real non-negative} entries in $X$ such that In the commutative case, the algebra $A$ is commutative and can describe the metric on $X$ in terms of algebraic data. -\begin{theorem} +\begin{mytheorem} Let $d_{ij}$ be a metric on $X$ a finite discrete space with $N$ points, $A = \mathbb{C}^N$ with elements $a = (a(i))_{i=1}^N$ such that $\hat{A} \simeq X$. Then there exists a representation $\pi$ of $A$ on a finite-dimensional inner product space $H$ and a symmetric @@ -21,7 +21,7 @@ metric on $X$ in terms of algebraic data. d_{ij} = \sup_{a\in A}\bigg\{\big|a(i)-a(j)\big| : |\big|\big[D, \pi(a)]\big|\big| \leq 1\bigg\} \end{equation} -\end{theorem} +\end{mytheorem} \begin{proof} We claim that this would follow from the equality: @@ -197,20 +197,20 @@ formula on a structure space of $A$. Finally we have all three ingredients to define a finite spectral triple, an mathematical structure which encodes finite discrete geometry into algebraic data. -\begin{definition} +\begin{mydefinition} A \textit{finite spectral triple} is a tripe $(A, H, D)$, where $A$ is a unital $*$-algebra, faithfully represented on a finite-dimensional Hilbert space $H$, with a symmetric operator $D: H \rightarrow H$. (Note that $A$ is automatically a matrix algebra.) -\end{definition} +\end{mydefinition} \subsubsection{Properties of Matrix Algebras} -\begin{lemma} +\begin{mylemma} If $A$ is a unital C* algebra acting faithfully on a finite dimensional Hilbert space, then $A$ is a matrix algebra of the Form: \begin{align} A \simeq \bigoplus _{i=1}^N M_{n_i}(\mathbb{C}) \end{align} -\end{lemma} +\end{mylemma} \begin{proof} The wording 'acting faithfully on a Hilbertspace' means that the $*$-representation is injective, or for a $*$-homomorphism that means @@ -224,14 +224,14 @@ A simple illustration would be for an algebra $A = M_n(\mathbb{C})$ and $H=\mathbb{C}^n$. Since $A$ acts on $H$ with matrix multiplication and standard inner product and $D$ on $H$ is a hermitian matrix $n\times n$ matrix. -\begin{definition} +\begin{mydefinition} Given an finite spectral triple $(A, H, D)$, the $A$-bimodule of Connes' differential one-forms is \begin{align}\label{eq:connesoneforms} \Omega _D ^1 (A) := \left\{ \sum _k a_k[D, b_k]: a_k, b_k \in A \right\}. \end{align} -\end{definition} +\end{mydefinition} Then there is a map $d:A\rightarrow \Omega _D ^1 (A)$, $d = [D, \cdot]$. Where $d$ is a derivation of the $*$-algebra in the sense that \begin{align} @@ -262,14 +262,14 @@ rewriting the defining equation \eqref{eq:connesoneforms} into &= \sum _k\ a_k'\ [D, b_k'] \end{align} -\begin{lemma} +\begin{mylemma} Let $\big(A, H, D\big) = \big(M_n(\mathbb{C}), \mathbb{C}^n, D\big)$, where $D$ is a hermitian $n\times n$ matrix. If $D$ is not a multiple of the identity then \begin{align} \Omega _D ^1 (A) \simeq M_n(\mathbb{C}) = A \end{align} -\end{lemma} +\end{mylemma} \begin{proof} Assume $D = \sum _i \lambda _i e_{ii}$ is diagonal, $\lambda _i \in \mathbb{R}$ and $\{e_{ij}\}$ is the basis of $M_n(\mathbb{C})$. Then for fixed $i$, $j$ choose $k$ @@ -304,7 +304,7 @@ Next we will define an equivalence relation between finite spectral triples, cal spectral unitary equivalence, which is given by the unitarity of the two matrix algebras themselves, and an additional map $U$ which allows us to associate a one operator to another second operator. -\begin{definition} +\begin{mydefinition} Two finite spectral tripes $(A_1, H_1, D_1)$ and $(A_2, H_2, D_2)$ are called unitary equivalent if $A_1 = A_2$ and there exists a map $U:\ H_1 \rightarrow H_2$ that satisfies @@ -312,7 +312,7 @@ one operator to another second operator. U\ \pi_1(a)\ U^* &= \pi_2(a)\;\;\;\; \text{with} \;\;\; a \in A_1,\\ U\ D_1\ U^* &= D_2. \end{align} -\end{definition} +\end{mydefinition} Notice that for any such $U$ we have the relation $(A, H, D) \sim (A, H, UDU^*)$. And hence $U\ D\ U^* = D + U[D, U^*]$ are of the form of elements in $\Omega _D^1 (A)$. @@ -353,11 +353,11 @@ $U_{23}: H_2 \rightarrow H_3$ are %-------------- EXERCISE In order to extend this relation we take a look at Morita equivalence of Matrix Algebras. -\begin{definition} +\begin{mydefinition} Let $A$ be an algebra. We say that $I \subset A$, as a vector space, is a right(left) ideal if $a\ b \in I$ for $a \in A$ and $b\in I$ (or $b\ a \in I$, $b\in I$, $a\in A$). We call a left-right ideal simply an ideal. -\end{definition} +\end{mydefinition} Given a Hilbert bimodule $E \in KK_f(B, A)$ and $(A, H, D)$ we construct a finite spectral triple on $B$, $(B, H', D')$ @@ -391,7 +391,7 @@ Hence $D'$ is well defined on $E \otimes _A H$ &= 0. \end{align} With the information thus far we can prove the following theorem -\begin{theorem} +\begin{mytheorem} 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 @@ -399,7 +399,7 @@ With the information thus far we can prove the following theorem \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} +\end{mytheorem} \begin{proof} $E\otimes _A H$ was previously. 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, @@ -465,11 +465,11 @@ So any such connection is of the form \end{align} %\subsubsection{Graphing Finite Spectral Triples} -%\begin{definition} +%\begin{mydefinition} % 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} +%\end{mydefinition} %\begin{figure}[h!] % \centering %\begin{tikzpicture}[ @@ -499,7 +499,7 @@ So any such connection is of the form %% $A$-submodule, because elements in $W^{\perp}$ need to satisfy the %% bimodularity. %%\end{MyExercise} -%\begin{definition} +%\begin{mydefinition} % 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 @@ -520,7 +520,7 @@ So any such connection is of the form % \begin{equation} % n(\Gamma ^{(0)}) = \Lambda % \end{equation} -%\end{definition} +%\end{mydefinition} %\begin{question} % Would then $D_e$ be the pullback? %\end{question} @@ -537,7 +537,7 @@ So any such connection is of the form % \textbf{n}_i \\ n(\nu _2) = \textbf{n}_j}} D_e %\end{align} % -%\begin{theorem} +%\begin{mytheorem} % 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 @@ -547,7 +547,7 @@ So any such connection is of the form % H = \bigoplus _{\nu \in \Gamma ^{(0)}} \mathbb{C}^{n(\nu)}; \;\;\; % D = \sum _{e \in \Gamma ^{(1)}} D_e + D_e^* % \end{equation} -%\end{theorem} +%\end{mytheorem} % \begin{figure}[h!] % \centering % \begin{tikzpicture}[ diff --git a/src/thesis/chapters/realncg.tex b/src/thesis/chapters/realncg.tex @@ -15,7 +15,7 @@ properties \gamma a &= a \gamma, \;\;\;\; a\in A. \end{align} Then we can define a finite real spectral triple. -\begin{definition} +\begin{mydefinition} A \textit{finite real spectral triple} is given by a finite spectral triple $(A, H, D)$ and a anti-unitary operator $J:H\rightarrow H$ called the \textit{real structure}, such that @@ -31,8 +31,8 @@ Then we can define a finite real spectral triple. The two properties are called the \textit{commutant property}, they require that the left action of an element in $A$ and $\Omega _D^1(A)$ commutes with the right action on $A$. -\end{definition} -\begin{definition} +\end{mydefinition} +\begin{mydefinition} The $KO$-dimension of a real spectral triple is determined by the sings $\epsilon, \epsilon ' ,\epsilon '' \in \{-1, 1\}$ appearing in \begin{align} @@ -40,7 +40,7 @@ Then we can define a finite real spectral triple. J\ D &= \epsilon \ D\ J,\\ J\ \gamma &= \epsilon''\ \gamma\ J. \end{align} -\end{definition} +\end{mydefinition} \begin{table}[h!] \centering \caption{$KO$-dimension $k$ modulo $8$ of a real spectral triple} @@ -59,7 +59,7 @@ Even thought the KO-dimension of a real spectral triple is important, we will not be doing in-depth introduction of the KO-dimension, for this we reference to \cite{ncgwalter}. -\begin{definition} +\begin{mydefinition} An opposite-algebra $A^\circ$ of a $A$ is defined to be equal to $A$ as a vector space with the opposite product \begin{align} @@ -67,7 +67,7 @@ vector space with the opposite product &\Rightarrow a^\circ = Ja^* J^{-1}, \end{align} which defines the left representation of $A^\circ$ on $H$ -\end{definition} +\end{mydefinition} %------------EXAMPLE EXERCISE @@ -92,7 +92,7 @@ multiplication %------------EXAMPLE EXERCISE -\begin{definition} +\begin{mydefinition} We call $\xi \in H$ \textbf{cyclic vector} in $A$ if: \begin{align} A\xi := { a\xi:\;\; a\in A} = H @@ -101,7 +101,7 @@ multiplication \begin{align} a\xi = 0\;\; \Rightarrow \;\; a=0;\;\;\; a\in A \end{align} -\end{definition} +\end{mydefinition} Suppose $(A, H, D = 0)$ is a finite spectral triple such that $H$ possesses a cyclic and separating vector for $A$ and let $J: H \rightarrow H$ be the operator in $S = J \Delta ^{1/2}$ with $\Delta = S^*S$ . By composition @@ -125,7 +125,7 @@ which concludes the anti-unitarity by definition. \subsubsection{Morphisms Between Finite Real Spectral Triples} Like the unitary equivalence relation for finite spectral triples, we can it to finite real spectral triples. -\begin{definition} +\begin{mydefinition} We call two finite real spectral triples $(A_1, H_1 ,D_1 ; J_1 , \gamma _1)$ and $(A_2, H_2, D_2; J_2, \gamma _2)$ unitarily equivalent if $A_1 = A_2$ and if there exists a unitary operator $U: H_1 \rightarrow H_2$ such @@ -136,8 +136,8 @@ to finite real spectral triples. U \gamma _1\ U^* &= \gamma _2,\\ U\ J_1\ U^* &= J_2. \end{align} -\end{definition} -\begin{definition} +\end{mydefinition} +\begin{mydefinition} Let $E$ be a $B$-$A$ bimodule. The \textit{conjugate Module} $E^\circ$ is given by the $A$-$B$-bimodule. \begin{align} @@ -148,7 +148,7 @@ to finite real spectral triples. a \cdot \bar{e} \cdot b = b^*\ \bar{e}\ a^*, \;\;\;\; \forall a\in A, b \in B. \end{align} -\end{definition} +\end{mydefinition} We bear in mind that $E^\circ$ is not a Hilbert bimodule for $(A, B)$ because it doesn't have a natural $B$-valued inner product. But there is a $A$-valued inner product on the left $A$-module $E^\circ$ with @@ -239,13 +239,13 @@ Finally for the grading we have \end{align} Summarizing we can write down the following theorem -\begin{theorem} +\begin{mytheorem} Suppose $(A, H, D; J, \gamma)$ is a finite spectral triple of $KO$-dimension $k$, let $\nabla$ be a connection satisfying the compatibility condition (same as with finite spectral triples). Then $(B, H',D'; J', \gamma')$ is a finite spectral triple of $KO$-Dimension $k$. ($H', D', J', \gamma'$) -\end{theorem} +\end{mytheorem} \begin{proof} The only thing left is to check if the $KO$-dimension is preserved, diff --git a/src/thesis/chapters/twopointspace.tex b/src/thesis/chapters/twopointspace.tex @@ -144,13 +144,13 @@ help us break down and search for the gauge group of the Two-Point $F_X$ space which we then extend to $M\times F_X$. We will only be diving superficially into this chapter, for further reading we refer to \cite{ncgwalter}. -\begin{definition} +\begin{mydefinition} Gauge Group of a real spectral triple is given by \begin{align} \mathfrak{B}(A, H; J) := \{ U = uJuJ^{-1} | u\in U(A)\} \end{align} -\end{definition} -\begin{definition} +\end{mydefinition} +\begin{mydefinition} A *-automorphism of a *-algebra $A$ is a linear invertible map \begin{align} @@ -170,7 +170,7 @@ Gauge Group of a real spectral triple is given by U(A) = \{ u\in A|\;\; uu^* = u^*u=1\} \;\;\; \text{(unitary)} \end{align} -\end{definition} +\end{mydefinition} The Gauge group of $F_X$ is given by the quotient $U(A)/U(A_J)$. We want a nontrivial Gauge group so we need to choose a $U(A_J) \neq U(A)$ and $U((A_F)_{J_F}) \neq U(A_F)$. @@ -185,9 +185,9 @@ which means the operator vanishes, and the spectral triple representation is Here $C$ is the complex conjugation, and $F_X$ is a real even finite spectral triple (space) of KO-dimension 6. -\begin{proposition} +\begin{myproposition} The Gauge group of the Two-Point space $\mathfrak{B}(F_X)$ is $U(1)$. -\end{proposition} +\end{myproposition} \begin{proof} Note that $U(A_F) = U(1) \times U(1)$. We need to show that $U(A_F) \cap U(A_F)_{J_F}) \simeq U(1)$, such that $\mathfrak{B}(F) \simeq U(1)$. So @@ -225,7 +225,7 @@ where $Y_\mu$ the $U(1)$ Gauge field is defined as i\ u(1)). \end{align} -\begin{proposition} +\begin{myproposition} The inner fluctuations of the almost-commutative manifold $M\times F_X$ are parameterized by a $U(1)$-gauge field $Y_\mu$ as \begin{align} @@ -237,5 +237,5 @@ where $Y_\mu$ the $U(1)$ Gauge field is defined as Y_\mu \mapsto Y_\mu - i\ u\partial_\mu u^*; \;\;\;\;\; (u\in \mathfrak{B}(M\times F_X)). \end{align} -\end{proposition} +\end{myproposition} diff --git a/src/thesis/main.pdf b/src/thesis/main.pdf Binary files differ.