commit 9c62b5a750a93f4f219109435edd51149c7f239b
parent 91eb67f586e45f2e72abd9410eeb110be9a6d5e3
Author: miksa <milutin@popovic.xyz>
Date: Tue, 20 Jul 2021 18:26:10 +0200
working from the end to the start
Diffstat:
15 files changed, 2350 insertions(+), 13 deletions(-)
diff --git a/src/thesis/back/packages.tex b/src/thesis/back/packages.tex
@@ -1,9 +1,12 @@
-\usepackage[utf8]{inputenc} % für direkte Eingabe von ß, ö, ä
+\usepackage[utf8]{inputenc}
+\usepackage{mathptmx}
+
%\usepackage{ngerman} % Sprachanpassung Deutsch
-\usepackage{graphicx} % Graphikeinbindung
-\usepackage{geometry} % Seitenränder
-\geometry{a4paper,left=25mm,right=20mm, top=10mm, bottom=20mm}
+\usepackage{graphicx}
+\usepackage{geometry}
+%\geometry{a4paper,left=25mm,right=25mm, top=20mm, bottom=30mm}
+\geometry{a4paper, top=15mm}
\usepackage{subcaption}
\usepackage[shortlabels]{enumitem}
@@ -21,6 +24,72 @@
%\usepackage[parfill]{parskip}
\usepackage[backend=biber, sorting=none]{biblatex}
-\addbibresource{uni.bib}
+\addbibresource{thesis.bib}
+
+\numberwithin{equation}{section}
\usepackage{lipsum}
+
+
+% 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}}
+
+\makeatletter
+\mdfdefinestyle{exercisestyle}{%
+ outerlinewidth=1em,%
+ outerlinecolor=white,%
+ leftmargin=-1em,%
+ rightmargin=-1em,%
+ middlelinewidth=1.2pt,%
+ roundcorner=5pt,%
+ linecolor=black,%
+ backgroundcolor=blue!5,
+ innertopmargin=1.2\baselineskip,
+ skipabove={\dimexpr0.5\baselineskip+\topskip\relax},
+ skipbelow={-1em},
+ needspace=3\baselineskip,
+ frametitlefont=\sffamily\bfseries,
+ settings={\global\stepcounter{exercise}},
+ singleextra={%
+ \node[titlered,xshift=1cm] at (P-|O) %
+ {~\mdf@frametitlefont{\theexercise}~};},%
+ firstextra={%
+ \node[titlered,xshift=1cm] at (P-|O) %
+ {~\mdf@frametitlefont{\theexercise}~};},
+}
+\makeatother
+
+\newenvironment{MyExercise}%
+{\begin{mdframed}[style=exercisestyle]}{\end{mdframed}}
+
+\theoremstyle{definition}
+\newtheorem{definition}{Definition}
+
+\theoremstyle{definition}
+\newtheorem{question}{Question}
+
+\theoremstyle{definition}
+\newtheorem{example}{Example}
+
+\theoremstyle{theorem}
+\newtheorem{theorem}{Theorem}
+
+\theoremstyle{theorem}
+\newtheorem{lemma}{Lemma}
+
+
+\theoremstyle{theorem}
+\newtheorem{proposition}{Proposition}
+
+\newtheorem*{idea}{Proof Idea}
diff --git a/src/thesis/back/title.tex b/src/thesis/back/title.tex
@@ -2,7 +2,7 @@
\begin{center}
\begin{figure}
-\hspace{9.1cm}
+\hspace{8.8cm}
\includegraphics[width=8cm]{pics/uni_logo}
\end{figure}
\vspace*{2cm}
@@ -34,7 +34,7 @@
{ \fontsize{12}{0} \selectfont Vienna, July 2021}\\
-\vspace*{5.5cm}
+\vspace*{3.5cm}
\begin{tabular}{p{9cm}p{11.25cm}}
\fontsize{12}{0} \selectfont degree programme code as it appears on / &
diff --git a/src/thesis/chapters/1 b/src/thesis/chapters/1
@@ -0,0 +1,754 @@
+\subsection{Noncommutative Geometry of Electrodynamics}
+\subsubsection{The Two-Point Space}
+One of the basics forms of a noncommutative space is the the two point space $X
+:= \{x, y\}$, it can be represented by the following spectral triple
+\begin{align}
+ F_x := (C(X) = \mathbb{C}^2, H_F, D_F; J_F, \gamma _f).
+\end{align}
+There are three properties of $F_x$ that stand out, first of all the action of
+$C(X)$ on $H_F$ is faithful for $dim(H_F) \geq 2$. Thus we can make a simple
+choice for the Hilbertspace, $H_F = \mathbb{C}^2$. Furthermore $\gamma_F$ is
+the $\mathbb{Z}_2$ grading, which allows us to decompose $H_F$ into
+\begin{align}
+ H_F = H_F^+ \otimes H_F^- = \mathbb{C} \otimes \mathbb{C},
+\end{align}
+where
+\begin{align}
+ H_F^\pm = \{\psi \in H_F |\; \gamma_F\psi = \pm \psi\},
+\end{align}
+are two eigenspaces. And lastly the Dirac operator $D_F$ lets us
+interchange between $H_F^\pm$,
+\begin{align}
+ D_F =
+ \begin{pmatrix}0 & t \\ \bar{t} & 0\end{pmatrix}, \;\;\;\;\;
+ \text{with} \;\; t\in\mathbb{C}.
+\end{align}
+
+ The Two-Point space $F_x$ can only have a real structure if the Dirac
+ operator vanishes, i.e. $D_F = 0$, in that case we have KO-dimension of 0,
+ 2 or 6.
+
+ To elaborate on this, we know that there are two diagram representations of
+ $F_x$ at $\underbrace{\mathbb{C} \oplus \mathbb{C}}_{C(X)}$ on
+ $\underbrace{\mathbb{C} \oplus\mathbb{C}}_{H_F}$, which are:
+ \begin{figure}[h!] \centering
+ \begin{tikzpicture}[
+ dot/.style = {draw, circle, inner sep=0.06cm},
+ no/.style = {},
+ ]
+ \node[no](a) at (0,0) [label=left:$\textbf{1}^\circ$] {};
+ \node[no](b) at (0, -1) [label=left:$\textbf{1}^\circ$] {};
+ \node[no](c) at (1, 0.5) [label=above:$\textbf{1}$] {};
+ \node[no](d) at (2, 0.5) [label=above:$\textbf{1}$] {};
+ \node[dot](d0) at (2,0) [] {};
+ \node[dot](d0) at (1,-1) [] {};
+
+ \node[no](a1) at (6,0) [label=left:$\textbf{1}^\circ$] {};
+ \node[no](b2) at (6, -1) [label=left:$\textbf{1}^\circ$] {};
+ \node[no](c2) at (7, 0.5) [label=above:$\textbf{1}$] {};
+ \node[no](d2) at (8, 0.5) [label=above:$\textbf{1}$] {};
+ \node[dot](d0) at (7,0) [] {};
+ \node[dot](d0) at (8,-1) [] {};
+ \end{tikzpicture}
+ \end{figure}\newline
+If the Two-Point space $F_x$ would be a real spectral triple then $D_F$ can
+only go vertically or horizontally. This would mean that $D_F$ vanishes.
+
+The diagram on the left has KO-dimension 2 and 6, the diagram on the
+right has KO-dimension 0 and 4. Yet KO-dimension 4 is not ruled out because
+$dim(H_F^\pm) = 1$ (see Lemma 3.8 Book), which ultimately means $J_F^2 = -1$ is
+not allowed.
+\subsubsection{The product Space}
+By Extending the Two-Point space with a four dimensional Riemannian spin
+manifold, we get an almost commutative manifold $M\times F_x$, given by
+\begin{align}
+ M\times F_x = (C^\infty(M), \mathbb{C}^2), L^2(S)\otimes \mathbb{C}^2,
+ D_M\otimes 1 ; J_M\otimes J_F, \gamma_M \otimes \gamma_F),
+\end{align}
+where
+\begin{align}
+ C^\infty(M, \mathbb{C}^2) \simeq C^\infty(M) \oplus C^\infty(M).
+\end{align}
+According to Gelfand duality the algebra $C^\infty(M, \mathbb{C}^2)$ of the
+spectral triple corresponds to the space
+\begin{align}
+ N:= M\otimes X \simeq M\sqcup X
+\end{align}
+Keep in mind that we still need to find an appropriate real structure on the
+Riemannian spin manifold, $J_M$.
+\newline
+The total Hilbertspace can be decomposed into $H = L^2(S) \oplus L^2(S)$, such
+that for
+\newline
+$\underbrace{a,b\in C^\infty(M)}_{(a, b) \in C^\infty(N)}$
+and $\underbrace{\psi, \phi \in L^2(S)}_{(\psi, \phi) \in H}$ we have
+\begin{align}
+ (a, b)(\psi, \phi) = (a\psi, b\phi)
+\end{align}
+Along with the decomposition of the total Hilbertspace we can consider a
+distance formula on $M\times F_x$ with
+\begin{align}\label{eq:commutator inequality}
+ d_{D_F}(x,y) = \sup\left\{ |a(x) - a(y)|:a\in A_F, ||[D_F, a]|| \leq
+ 1 \right\}.
+\end{align}
+To calculate the distance between two points on the Two-Point space $X= \{x,
+y\}$, between $x$ and $y$, we consider an $a \in \mathbb{C}^2 = C(X)$, which is
+specified by two complex numbers $a(x)$ and $a(y)$. Then we simplify the
+commutator inequality in \ref{eq:commutator inequality}
+\begin{align}
+ &||[D_F , a]|| = ||(a(y) - a(x))\begin{pmatrix}0 &t\\\bar{t} &0
+ \end{pmatrix}|| \leq 1,\\
+ &\Rightarrow |a(y) - a(x)|\leq \frac{1}{|t|},
+\end{align}
+and the supremum gives us the distance
+\begin{align}
+ d_{D_F} (x,y) = \frac{1}{|t|}.
+\end{align}
+Note that if there exists $J_M$ (real structure) $\Rightarrow t=0$ then
+$d_{D_F}(x,y) \rightarrow \infty$!
+\newline
+
+Now let $p \in M$, then take two points on $N=M\times X$, $(p, x)$ and
+$(p,y)$ and $a \in C^\infty(N)$ is determined by $a_x(p):=a(p, x)$ and
+$a_y(p):=a(p, y)$. The distance between these two points is then
+\begin{align}
+ d_{D_F\otimes 1}(n_1, n_2) = \sup \left\{ |a(n_1) - a(n_2)|: a\in
+ A, ||[D\otimes 1, a]||\right\}
+\end{align}
+\textbf{Remark}: If $n_1 = (p,x)$ and $n_2 = (q, x)$ for $p,q \in M$ then
+\begin{align}
+ d_{D_M \otimes 1} (n_1, n_2) = |a_x(p) - a_x(q)| \;\;\; a_x\in
+ C^\infty(M) \;\; \text{with} \;\; ||[D\otimes 1, a_x]|| \leq 1
+\end{align}
+The distance turns to the geodestic distance formula
+\begin{align}
+ d_{D_M\otimes1}(n_1, n_2) = d_g(p, q)
+\end{align}
+
+However if $n_1 = (p, x)$ and $n_2 = (q, y)$ then the two conditions are
+$||[D_M, a_x]|| \leq 1$ and $||[D_M, a_y|| \leq 1$. They have no
+restriction which results in the distance being infinite! And $N =
+M\times X$ is given by two disjoint copies of M which are separated by
+infinite distance
+
+\textbf{Note}: distance is only finite if $[D_F, a] \neq 1$. The
+commutator
+generates a scalar field say $\phi$ and the finiteness of the distance is
+related to the existence of scalar fields.
+\subsubsection{$U(1)$ Gauge Group}
+Here we determine the Gauge theory corresponding to the almost
+commutative
+Manifold $M\times F_x$.
+
+\textbf{Gauge Group of a Spectral Triple}:
+\begin{align}
+ \mathfrak{B}(A, H; J) := \{ U = uJuJ^{-1} | u\in U(A)\}
+\end{align}
+\begin{definition}
+ A *-automorphism of a *-algebra $A$ is a linear invertible
+ map
+ \begin{align}
+ &\alpha:A \rightarrow A\;\;\; \text{with}\\
+ \nonumber\\
+ &\alpha(ab) = \alpha(a)\alpha(b)\\
+ &\alpha(a)^* = \alpha(a^*)
+ \end{align}
+ The \textbf{Group of automorphisms of the *-Algebra $A$} is
+ $(A)$.\newline
+ The automorphism $\alpha$ is called \textbf{inner} if
+ \begin{align}
+ \alpha(a) = u a u^* \;\;\; \text{for} \;\; U(A)
+ \end{align}
+ where $U(A)$ is
+ \begin{align}
+ U(A) = \{ u\in A|\;\; uu^* = u^*u=1\} \;\;\;
+ \text{(unitary)}
+ \end{align}
+\end{definition}
+The Gauge group is given by the quotient $U(A)/U(A_J)$.
+We want a nontrivial Gauge group so we need to choose $U(A_J) \neq
+U(A)$ which is the same as $U((A_F)_{J_F}) \neq
+U(A_F)$.
+We consider $F_x$ to be
+\begin{align}
+ F_x := \left(\mathbb{C}^2,\mathbb{C}^2, D_F =\begin{pmatrix}
+ 0&0\\0&0\end{pmatrix}; J_f =\begin{pmatrix}
+ 0&C\\C&0\end{pmatrix},
+ \gamma_F = \begin{pmatrix}1&0\\0&-1\end{pmatrix}\right).
+\end{align}
+Here $C$ is the complex conjugation, and $F_X$ is a real even finite
+ spectral triple (space) with $KO-dim=6$
+
+\begin{proposition}
+ The Gauge group $\mathfrak{B}(F)$ of the two point space is given by
+ $U(1)$.
+\end{proposition}
+\begin{proof}
+ Note that $U(A_F) = U(1) \times U(1)$. We need to show that
+ $U(\mathcal{A}_F)
+ \cap U(A_F)_{J_F}) \simeq U(1)$, such that $\mathfrak{B}(F)
+ \simeq U(1)$.\newline
+
+ So for $a \in \mathbb{C}^2$ to be in $(A_F)_{J_F}$ it has
+ to satisfy $J_F a^* J_F = a$.
+ \begin{align}
+ J_F a^* J^{-1} =
+ \begin{pmatrix}0&C\\C&0\end{pmatrix}
+ \begin{pmatrix}\bar{a}_1&0\\0&\bar{a}_2\end{pmatrix}
+ \begin{pmatrix}0&C\\C&0\end{pmatrix}
+ =
+ \begin{pmatrix}a_2&0\\0&a_1\end{pmatrix}
+ \end{align}
+ Which is only the case if $a_1 = a_2$. So we have
+ $(A_F)_{J_F} \simeq \mathbb{C}$, whose unitary elements
+ from $U(1)$ are contained in the diagonal subgroup of
+ $U(\mathcal{A}_F)$.
+\end{proof}
+
+Now we need to find the exact from of the field $B_\mu$ to calculate the
+spectral action of a spectral triple. Since $(A_F)_{J_F} \simeq
+\mathbb{C}$ we find that $\mathfrak{h}(F) = \mathfrak{u}((A_F)_{J_F})
+\simeq i\mathbb{R}$. Where $\mathfrak{h}(F)$ is the Lie Algebra on $F$
+and $\mathfrak{u}((A_F)_{J_F})$ is the Lie algebra of the unitary group
+$(A_F)_{J_F}$.\newline
+
+An arbitrary hermitian field $A_\mu = -ia\partial _\mu b$ is given by
+two
+$U(1)$ Gauge fields $X_\mu^1, X_\mu^2 \in C^\infty(M, \mathbb{R})$.
+However $A_\mu$ appears in combination $A_\mu - J_F A_\mu J_F^{-1}$:
+\begin{align}
+ B_\mu = A_\mu - J_F A_\mu J_F^{-1} =
+ \begin{pmatrix}X_\mu^1&0\\0&X_\mu^2 \end{pmatrix}
+ -
+ \begin{pmatrix}X_\mu^2&0\\0&X_\mu^1 \end{pmatrix}
+ =:
+ \begin{pmatrix}Y_\mu&0\\0&-Y_\mu \end{pmatrix}=
+ Y_\mu \otimes \gamma _F
+\end{align}
+where $Y_\mu$ the $U(1)$ Gauge field is defined as
+\begin{align}
+ Y_\mu := X_\mu^1 - X_\mu^2 \in C^\infty(M, \mathbb{R}) = C^\infty(M,
+ i\ u(1)).
+\end{align}
+
+\begin{proposition}
+ The inner fluctuations of the almost-commutative manifold $M\times
+ F_x$ described above are parametrized by a $U(1)$-gauge field $Y_\mu$
+ as
+ \begin{align}
+ D \mapsto D' = D + \gamma ^\mu Y_\mu \otimes \gamma_F
+ \end{align}
+ The action of the gauge group $\mathfrak{B}(M\times F_X) \simeq
+ C^\infty (M, U(1))$ on $D'$ is implemented by
+ \begin{align}
+ Y_\mu \mapsto Y_\mu - i\ u\partial_\mu u^*; \;\;\;\;\; (u\in
+ \mathfrak{B}(M\times F_X)).
+ \end{align}
+\end{proposition}
+
+\subsection{Electrodynamics}
+Now we use the almost commutative Manifold and the abelian gauge group
+$U(1)$ to describe Electrodynamics. We arrive at a unified description of
+gravity and electrodynamics although in the classical level.
+\newline
+
+The almost commutative Manifold $M\times F_X$ describes a local gauge group
+$U(1)$. The inner fluctuations of the Dirac operator describe $Y_\mu$ the
+gauge field of $U(1)$. There arise two Problems:
+\newline
+(1): With $F_X$, $D_F$ must vanish, however this implies that the electrons
+are massless (this we do not want)
+\newline
+
+(2): The Euclidean action for a free Dirac field is
+\begin{align}
+ S = - \int i \bar{\psi}(\gamma ^\mu\partial _\mu - m) \psi d^4x,
+\end{align}
+$\psi,\ \bar{\psi}$ must be considered as independent variables, which means
+$S_F$ need two independent Dirac Spinors. We write $\{e, \bar{e}\}$ for the
+ONB of $H_F$, where $\{e\}$ is the ONB of $H_F^+$ and $\{\bar{e}\}$ the ONB
+of $H_F^-$ with the real structure this gives us 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}
+The total Hilbertspace is $H = L^2(S) \otimes H_F$, with $\gamma _F$ we can
+decompose $L^2(S) = L^2(S)^+ \oplus L^2(S)^-$, so with $\gamma = \gamma _M
+\otimes \gamma _F$ we can obtain the positive eigenspace $H^+$
+\begin{align}
+ H^+ = L^2(S)^+ \otimes H_F^+ \oplus L^(S)^- \otimes H_F^-.
+\end{align}
+For a $\xi \i H^+$ 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}. This is
+too much restriction for $F_X$.
+\subsubsection{The Finite Space}
+Here we solve the two problems by enlarging(doubling) the Hilbertspace. This
+is done by introducing multiplicities in Krajewski Diagrams which will also
+allow us to choose a nonzero Dirac operator which will connect the two
+vertices (next chapter).
+\newline
+
+We start of with the same algebra $C^\infty(M, \mathbb{C}^2)$, corresponding
+to space $N= M\times X \simeq M\sqcup M$.
+\newline
+
+The Hilbertspace will describe four particles,
+\begin{itemize}
+ \item left handed electrons
+ \item right handed positrons
+\end{itemize}
+Thus we have $\{ \underbrace{e_R, e_L}_{\text{left-handed}},
+\underbrace{\bar{e}_R, \bar{e}_L}_{\text{right-handed}}\}$ the ONB for $H_F
+\mathbb{C}^4$.
+\newline
+Then with $J_F$ we interchange particles with antiparticles we have the
+following properties
+\begin{align}
+ &J_F e_R = \bar{e}_R \;\;\;\;\; &J_F e_L = \bar{e_L} \\
+ &\gamma _F e_R = -e_R \;\;\;\;\; &\gamma_F e_L = e_L \\
+ \text{and}& \nonumber \\
+ &J_F^2 = 1 \;\;\;\;\; & J_F \gamma_F = - \gamma_F J_F
+\end{align}
+This corresponds to KO-dim$= 6$. Then $\gamma_F$ allows us to can decompose
+$H$
+\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}
+Alternatively we can decompose $H$ into the eigenspace of particles and their
+antiparticles (electrons and positrons) 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}
+Now 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}
+ a =
+ \begin{pmatrix}a_1 & a_2 \end{pmatrix} \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.
+\begin{proposition}
+ The data
+ \begin{align}
+ \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}
+ defines a real even spectral triple of KO-dimension 6.
+\end{proposition}
+This spectral triple 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}
+Add a non-zero Dirac Operator to $F_{ED}$. From the Krajewski Diagram, we see
+that edges only exist between the multiple vertices. So we construct a Dirac
+operator mapping between the two vertices.
+\begin{align}\label{dirac}
+ 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 consider 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 $\gamma_F$ like before, $D_F$ like above.
+\subsubsection{The almost-commutative Manifold}
+The almost commutative manifold $M\times F_{ED}$ has KO-dim$=2$, it is the
+following spectral triple
+\begin{align}
+ M\times F_{ED} := \left(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\right)
+\end{align}
+
+The algebra decomposition is like before
+\begin{align}
+ C^\infty(M, \mathbb{C}^2) = C^\infty (M) \oplus C^\infty (M)
+\end{align}
+
+The Hilbertspace decomposition is
+\begin{align}
+ H = (L^2(S) \otimes H_e ) \oplus (L^2(S) \otimes H_{\bar{e}}).
+\end{align}
+Here we have the one component of the algebra acting on $L^2(S) \otimes H_e$,
+and the other one acting on $L^2(S) \otimes H_{\bar{e}}$
+\newline
+
+The derivation of the gauge theory is the same for $F_{ED}$ as for $F_X$, we
+have $\mathfrak{B}(F) \simeq U(1)$ and for $B_\mu = A_\mu - J_F A_\mu
+J_F^{-1}$
+\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}
+
+Our space $N = M\times X \simeq M\sqcup M$ consists of two compies of $M$.
+If $D_F = 0$ we have infinite distance between the two copies. Now we have $D_F$
+nonzero but $[D_F, a] = 0$ $\forall a \in A$ which still yields infinite
+distance.
+\begin{question}
+ What does this imply (physically, mathematically)? Why can we continue
+ even thought we have infinite distance between the same manifold? What do
+ we get if we fix this?
+\end{question}
+\subsubsection{The Spectral Action}
+Here we calculate the Lagrangian of the almost commutative Manifold $M\times
+F_{ED}$, which corresponds to the Lagrangian of Electrodynamics on a curved
+background Manifold (+ gravitational Lagrangian). It consists of the spectral
+action $S_b$ (bosonic) and of the fermionic action $S_f$.
+
+The simples spectral action of a spectral triple $(A, H, D)$ is given by the
+trace of some function of $D$, we also allow 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}
+ 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}f(\frac{D_\omega}{\Lambda})
+ \end{align}
+ where $\Lambda$ is a real cutoff parameter. The minimal condition on $f$
+ is that $f(\frac{D_\omega}{\Lambda})$ is a traclass operator, which mean
+ that it should be compact operator with well defined finite trace
+ independent of the basis. The subscript $b$ of $S_b$ refers to bosonic,
+ because in physical applications $\omega$ will describe bosonic fields.
+
+ Furthermore there is a topological spectral action, defined with the
+ grading $\gamma$
+ \begin{align}
+ S_{\text{top}}[\omega] := \text{Tr}(\gamma\
+ f(\frac{D_\omega}{\Lambda})).
+ \end{align}
+\end{definition}
+\begin{definition}
+ The fermionic action is defined by
+ \begin{align}
+ S_f[\omega, \psi] = (J\tilde{\psi}, D_\omega \tilde{\psi})
+ \end{align}
+ with $\tilde{\psi} \in H_{cl}^+ := \{\tilde{\psi}: \psi \in H^+\}$.
+ $H_{cl}^+$ is the set of Grassmann variables in $H$ in the +1-eigenspace
+ of the grading $\gamma$.
+\end{definition}
+The grasmann variables are a set of Basis vectors of a vector space, they
+form a unital algebra over a vector field say $V$ where the generators are anti commuting, that is for
+$\theta _i, \theta _j$ some Grassmann variables 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}
+\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}
+ with
+ \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}
+ where $N=4$ and $\mathcal{L}_M$ is the Lagrangian of the spectral triple
+ $(C^\infty(M) , L^2(S), D_M)$
+ \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
+ $R_{\nu\sigma} = g^{\mu\varrho} R_{\mu\nu \varrho\sigma}$.
+
+
+ Furthermore $\mathcal{L}_B$ describes the kinetic term of the gauge field
+ \begin{align}
+ \mathcal{L}_B(B_\mu) := \frac{f(0)}{24\pi^2}
+ \text{Tr}(F_{\mu\nu}F^{\mu\nu}).
+ \end{align}
+ Last $\mathcal{L}_\phi$ is the scalar-field Lagrangian with a boundary
+ term.
+ \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))\\
+ &+ \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 our manifold $m$ is $\dim(m) = \text{tr}(id) =4 $. let us
+ take a $x \in m$, we have an asymtotic expansion of
+ $\text{tr}(f(\frac{d_\omega}{\lambda}))$ as $\lambda \rightarrow \infty$
+ \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) \\&+ f(0) a_4(d_\omega^4)
+ +o(\lambda^{-1}).
+ \end{align}
+ note that the heat kernel coefficients are zero for uneven $k$,
+ furthermore 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 we note that $n:= \text{tr}\mathbbm{1_{h_f}})$
+ \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 need to extend in terms of coefficients of $f$, 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}
+ now for the cross terms of $\omega_{\mu\nu}^e\omega^{e\mu\nu}$ the trace
+ vanishes because of the anti-symmetric properties of the riemannian
+ curvature tensor
+ \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 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}
+ and the trace of the whole term is
+ \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}
+ plugging the results into $a_4$ and simplifying we can write
+ \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) \\
+ &+ \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 plug in the heat kernel coefficients into the
+ heat kernel expansion above.
+\end{proof}
+
+Here on we go and calculate the spectral action of $M\times F_{ED}$
+\begin{proposition}
+ The Spectral action of $M\times F_{ED}$ is
+ \begin{align}
+ \text{Tr}(f\frac{D_\omega}{\Lambda}) \sim \int_M \mathcal{L}(g_{\mu\nu},
+ Y_\mu) \sqrt{g}\ d^4x + O(\Lambda^{-1})
+ \end{align}
+ where the Lagrangian is
+ \begin{align}
+ \mathcal{L}(g_{\mu\nu}, Y_\mu) =
+ 4\mathcal{L}_M(g_{\mu\nu})+
+ \mathcal{L}_Y(Y_\mu)+
+ \mathcal{L}_\phi(g_{\mu\nu}, d)
+ \end{align}
+ here the $d$ in $\mathcal{L}_\phi$ is from $D_F$ in equation
+ \ref{dirac}. The Lagrangian $\mathcal{L}_M$ is like in equation
+ \ref{lagr}. The Lagrangian $\mathcal{L}_Y$ is the kinetic term of the
+ $U(1)$ gauge field $Y_\mu$
+ \begin{align}
+ \mathcal{L}_Y(Y_\mu):= \frac{f(0)}{6\pi^2}
+ Y_{\mu\nu}Y^{\mu\nu}\;\;\;\;\;\;\;\;\text{with}\;\;\; Y_{\mu\nu} =
+ \partial_\mu Y_\nu -
+ \partial_\nu Y_\mu.
+ \end{align}
+ Then there is $\mathcal{L}_\phi$, which has two constant terms
+ (disregarding the boundary term) that add up to the Cosmological Constant
+ and a term that for the Einstein-Hilbert action
+ \begin{align}
+ \mathcal{L}_\phi(g_{\mu\nu}, d) := \frac{2f_2 \Lambda ^2}{\pi^2}
+ |d|^2 + \frac{f(0)}{2\pi^2} |d|^4 + \frac{f(0)}{12\pi ^2} s |d|^2.
+ \end{align}
+\end{proposition}
+\begin{proof}
+ The Trace of $\mathbb{C}^4$ (the Hilbertspace) gives $N=4$. With $B_\mu$
+ like in equation \ref{field} we have $\text{Tr}(F_{\mu\nu}
+ F^{\mu\nu})=4Y_{\mu\nu}Y^{\mu\nu}$. This provides $\mathcal{L}_Y$.
+ Furthermore we have $\Phi^2 = D_F^2 = |d|^2$ and $\mathcal{L}_\phi$ only
+ give numerical contributions to the cosmological constant and the
+ Einstein-Hilbert action.
+
+ The proof is relying itself on just plugging the terms into the previous
+ proposition, for which I didn't write the proof for.
+\end{proof}
+
+
+\subsection{fermionic action}
+a quick reminder with what we are dealing with, the fermionic action is defined
+in the following way.
+\begin{definition}
+ the fermionic action is defined by
+ \begin{align}
+ s_f[\omega, \psi] = (j\tilde{\psi}, d_\omega \tilde{\psi})
+ \end{align}
+ with $\tilde{\psi} \in h_{cl}^+ := \{\tilde{\psi}: \psi \in h^+\}$.
+ $h_{cl}^+$ is the set of grassmann variables in $h$ in the +1-eigenspace
+ of the grading $\gamma$.
+\end{definition}
+
+the almostcommutative manifold we are dealing with is the following
+\begin{align}
+ &m\times f_{ed} := \left(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\right).\\
+ \nonumber\\
+ &\text{where:} \nonumber \\
+ &c^\infty(m,\mathbb{c}^2) = c^\infty(m) \otimes c^\infty(m)
+ &\mathcal{h} = \mathcal{h}^+ \otimes \mathcal{h}^-\\
+ &\mathcal{h} = l^2(s)^+ \otimes h_f^+ \oplus l^2(s)^- \otimes h_f^-.
+\end{align}
+where $h_f$ is separated into the particle-anitparticle states with onb $\{e_r,
+e_l, \bar{e}_r, \bar{e}_l\}$. the onb of $h_f^+$ is $\{e_l, \bar{e}_r\}$ and
+for $h_f^-$ we have $\{e_r, \bar{e}_l\}$. furthermore we can decompose a spinor
+$\psi \in l^2(s)$ for each of the eigenspaces $h_f^\pm$, $\psi = \psi_r
+\psi_l$. thus we can write for an arbitrary $\psi \in \mathcal{h}^+$
+\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}
+for $\chi_l, \psi_l \in l^2(s)^+$ and $\chi_r, \psi_r \in l^2(s)^-$.
+\begin{proposition}
+ we can define the action of the fermionic art of $m\times f_{ed}$ in the
+ following way
+ \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}
+\end{proposition}
+\begin{proof}
+ we take the fluctuated dirac operator
+ \begin{align}
+ d_\omega = d_m \otimes i + \gamma^\mu \otimes b_\mu + \gamma_m \otimes
+ d_f
+ \end{align}
+\end{proof}
+the fermionic action is $s_f = (j\tilde{\xi}, d_\omega\tilde{\xi})$ for a $\xi
+\in \mathcal{h}^+$, we can begin to calculate (note that we add the constant
+$\frac{1}{2}$ to the action)
+\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:1}\\
+ &+\frac{1}{2}(j\tilde{\xi}, (\gamma^\mu \otimes b_\mu)
+ \tilde{\xi})\label{eq:2}\\
+ &+\frac{1}{2}(j\tilde{\xi}, (\gamma_m\otimes
+ d_f)\tilde{\xi})\label{eq:3}.
+\end{align}
+for equation \ref{eq:1} 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)+
+ \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)+
+ \frac{1}{2}(j_m\tilde{\chi}_r,d_m\tilde{\chi}_l)\\
+ &= (j_m\tilde{\chi},d_m\tilde{\chi}).
+\end{align}
+for equation \ref{eq:2} 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)+\\
+ &+\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)=\\
+ &= -(j_m\tilde{\chi}, \gamma^\mu y_\mu\tilde{\psi}).
+\end{align}
+for equation \ref{eq:3} we have
+\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)+\\
+ &+\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)=\\
+ &= i(j_m\tilde{\chi}, m\tilde{\psi})
+\end{align}
+note that we obtain a complex mass parameter $d$, so we write $d:=im$ for $m\in \mathbb{r}$,
+which stands for the real mass and we obtain a nice result
+
+\begin{theorem}
+ the full lagrangian of $m\times f_{ed}$ 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}
+
+\end{theorem}
diff --git a/src/thesis/front/acknowledgment.tex b/src/thesis/chapters/acknowledgment.tex
diff --git a/src/thesis/chapters/basics.tex b/src/thesis/chapters/basics.tex
@@ -0,0 +1,325 @@
+
+\subsection{Noncommutative Geometric Spaces}
+\subsubsection{Matrix Algebras and Finite Spaces}
+\subsubsection{$*$-Algebra}
+\begin{definition}
+ A \textit{vector space} $A$ over $\mathbb{C}$ is called a \textit{complex, unital Algebra} if, \\
+ $\forall a,b \in A$ :
+ \begin{enumerate}
+ \item
+ $A \times A \rightarrow A$ \hspace{0.1\textwidth} \textit{bilinear} \\
+ $(a, b)\ \mapsto \ a\cdot b$
+ \item
+ $1a = a1 =a$ \hspace{0.08\textwidth} \textit{unital} \\
+ \end{enumerate}
+\end{definition}
+
+\begin{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
+ $(ab)^* = b^*a^*$ \hspace{0.05\textwidth} \textit{antidistributive}
+ \item
+ $(a^*)^* = a$ \hspace{0.1\textwidth} \textit{closure}
+ \end{enumerate}
+\end{definition}
+In the following all unital algebras are referred to as algebras.
+
+\subsubsection{Functions on Discrete Spaces}
+Let $X$ be a \textit{discretized topological} space with $N$ points.
+Consider functions of a continuous $*$-algebra $C(X)$ assigning values to $\mathbb{C}$, for $f, g \in C(X)$,
+$\lambda \in \mathbb{C}$ and $x \in X$ they provide the following structures:
+
+\begin{itemize}
+ \item \textit{pointwise linear} \\
+ $(f + g)(x) = f(x) + g(x)$\\
+ $(\lambda f)(x) = \lambda (f(x))$
+ \item \textit{pointwise multiplication} \\
+ $fg(x) = f(x)g(x)$ \hspace{0.1\textwidth} same as $(fg)(x) = f(x)g(x))$?
+ \item \textit{pointwise involution} \\
+ $f^*(x) = \overline{f(x)}$
+\end{itemize}
+
+\begin{question}
+ Mathematical difference between Topological Discreet Spaces and just Discreet Spaces?
+\end{question}
+
+The author indicates that $\mathbb{C}$-valued functions on $X$ are automatically continuous.
+\begin{idea}
+ CAN WE USE THE METRIC? NO!
+ We know that $X$ is a \textit{finite discrete space}, meaning in an $\epsilon$-$\delta$ approach
+ for each $x \in X$ the only $y \in X$, that is small enough is $x$ by itself, which implies
+ $\epsilon$ is always bigger than zero, thus every function $f:X\ \rightarrow\ \mathbb{C}$ is continuous.
+\end{idea}
+
+\subsubsection{Isomorphism Property}
+Furthermore $C(X)$ $*$-algebra is \textit{isomorphic} to a $*$-algebra $\mathbb{C}^N$ with involution
+($N$ number of points in $X$), written as $C(X) \simeq \mathbb{C}^N$.
+A function $f:X\ \rightarrow\ \mathbb{C}$ can be represented with $N \times N$ diagonal matrices,
+where the value $(ii)$ is the value of the function at the corresponding
+$i$-th point ($i = 1,...,N$). The structure is preserved because of the definitions of
+matrix multiplication and the hermitian conjugate of matrices.
+
+\begin{question}
+ Can isomorphisms between $C(X)$ and $\mathbb{C}^N$ be shown with matrix factorization?
+\end{question}
+ Isomorphisms are bijective preserve structure and don't lose physical information/
+
+\subsubsection{Mapping Finite Discrete Spaces}
+
+\begin{definition}
+ A \textit{map} between finite discrete spaces $X_1$ and $X_2$ is a function $\phi:\ X_1 \rightarrow\ X_2$
+\end{definition}
+
+For every map between finite discrete spaces there exists a corresponding map \\
+$\phi ^*:C(X_2)\ \rightarrow C(X_1)$, 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 $C(X)$.
+
+
+This map is called a pullback (or a $*$-homomorphism or a $*$-algebra map under pointwise product).
+Under the pointwise product:
+\begin{itemize}
+ \item $\phi ^*(fg) = \phi ^*(f) \phi ^*(g)$
+ \item $\phi ^*(\overline{f}) = \overline{\phi ^*(f)}$
+ \item $\phi ^*(\lambda f + g) = \lambda \phi ^*(f) + \phi ^*(g)$
+\end{itemize}
+
+\begin{question}
+ $\phi$ is in most cases not bijective, so how can we prove that there exists such a
+ pullback for every map between discrete spaces which preserves information? For bijective
+ it is given by its inverse, which by definition exists because $\phi$ is a map.
+ Or I didn't understand this correctly?
+\end{question}
+
+\begin{MyExercise}
+ \textbf{
+ Show that $\phi :X_1\ \rightarrow \ X_2$ is injective (surjective) map of finite spaces iff
+ $\phi ^* :C(X_2)\ \rightarrow \ C(X_1)$ is surjective (injective).
+}\newline
+
+ Consider $X_1$ with $n$ points and $X_2$ with $m$ points. Then there are three cases:
+ \begin{enumerate}
+ \item $n=m$ \\
+ Obviously $\phi$ is bijective and $\phi ^*$ too.
+ \item $n \rangle m$ \\
+ $\phi$ assigns $n$ points to $m$ points when $n \rangle m$,
+ which is by definition surjective. \\
+ $\phi ^*$ assigns $m$ points to $n$ points when $n \rangle m$,
+ which is by definition injective. \\
+ \item $n \langle m $ \\
+ analogous
+ \end{enumerate}
+\end{MyExercise}
+
+\subsubsection{Matrix Algebras}
+\begin{definition}
+ 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{definition}
+
+So from a topological discrete space $X$, we can construct a $*$-algebra $C(X)$ which is isomorphic
+to a matrix algebra $A$. The question is can we construct $X$ given $A$? $A$ is a matrix algebra,
+which are in most cases is not commutative, so the answer is generally no.
+
+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.
+
+\begin{question}
+ Why are non-commutative algebras not physically interesting?
+ Maybe too far fetched,but because physical observables (QM-Operators) are not commutative?
+\end{question}
+Exactly.
+
+\subsubsection{Finite Inner Product Spaces and Representations}
+Until now we looked at a finite topological discrete space, moreover we can consider a
+finite dimensional inner product space $H$ (finite Hilbert-spaces), with inner product
+$(\cdot,\cdot)\rightarrow \mathbb{C}$. $L(H)$ is the $*$-algebra of operators on $H$
+with product given by composition and involution given by the adjoint, $T \mapsto T^*$.
+$L(H)$ is a \textit{normed vector space} with
+\begin{align}
+ &\|T\|^2 = \text{sup}_{h \in H}\{(Th,Th): (h,h) \leq 1\} \hspace{0.1\textwidth} T \in L(H) \\
+ &\|T\| = \text{sup}\{\sqrt{\lambda}: \lambda \text{ eigenvalue of } T\}
+\end{align}
+
+
+\begin{definition}
+ The \textit{representation} of a finite dimensional $*$-algebra A is a pair $(H, \pi)$.
+ $H$ is a finite, dimensional inner product space and $\pi$ is a $*$-\textit{algebra map}
+ \begin{align}
+ \pi:A\ \rightarrow \ L(H)
+ \end{align}
+\end{definition}
+\begin{definition}
+ $(H, \pi)$ is called \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{definition}
+
+Examples for reducible and irreducible representations
+\begin{itemize}
+ \item $A = M_n(\mathbb{C})$, representation $H=\mathbb{C}^n$, $A$ acts as matrix multiplication\\
+ $H$ is irreducible.
+ \item $A = M_n(\mathbb{C})$, 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}
+
+\begin{definition}
+ Let $(H_1, \pi _1)$ and $(H_2, \pi _2)$ be representations of a $*$-algebra $A$. They 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}
+
+\begin{question}
+ In matrix representation this is diagonalization condition? (unitary diagonalization)
+\end{question}
+Yes
+
+\begin{definition}
+ $A$ a $*$-algebra then, $\hat{A}$ is called the structure space of all \textit{unitary equivalence classes
+ of irreducible representations of A}
+\end{definition}
+
+\begin{question}
+ Gelfand duality and the spectrum of $\hat{A}$, examples Fourier-Transform and Laplace-Transform
+ for simple spaces.
+\end{question}
+More on that in later chapters.
+
+\begin{MyExercise}
+ \textbf{
+ Given $(H, \pi)$ 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)' = \{T \in L(H):\pi (a)T = T\pi (a) \;\;\; \forall a\in A\}
+ \end{align}
+ \begin{enumerate}
+ \item Show that $\pi (A)'$ is a $*$-algebra.
+ \item Show that a representation $(H, \pi)$ of $A$ is irreducible iff the commutant $\pi (A)'$
+ consists of multiples of the identity
+ \end{enumerate}
+}
+
+ 1. To show that $\pi (A)'$ is a $*$-algebra we have to show that it is unital, associative and involute.
+ And note that $\pi (a) \in L(H)\ \forall a \in A$.
+ Unitarity 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 $*$-algebra of $L(H)$, $L(H) \times L(H) \mapsto L(H)$,
+ which is associative by definition. Involutnes is also given by the $*$-algebra $L(H)$
+ with a map $*: L(H) \mapsto L(H)$ only for $T$ that commute with $\pi (a)$.
+ \\
+ 2.?
+\end{MyExercise}
+
+\begin{MyExercise}
+ \textbf{
+ \begin{enumerate}
+ \item If $A$ is a unital $*$-algebra, show that the $n \times n$ matrices $M_n(A)$ with entries
+ in $A$ form a unital $*$-algebra.
+ \item Let $\pi :A\ \rightarrow \ L(H)$ be a representation of a $*$-algebra $A$ and set
+ $H^n = H \oplus ... \oplus H$, $n$ times. Show that $\tilde{\pi}:M_n(A) \rightarrow L(H^n)$
+ of $M_n(A)$ with\\
+ $\tilde{\pi}((a_{ij})) = (\tilde{\pi}(a_{ij})) \in M_n(A)$.
+ \item Let $\tilde{\pi}:M_n(A) \rightarrow L(H^n)$ be a $*$ algebra representation of $M_n(A)$.
+ Show that $\pi: A \rightarrow L(H^n)$ is a representation of $A$.
+ \end{enumerate}
+}
+ 1. We know $A$ is a $*$ algebra. Unitary operaton in $M_n(A)$ is given by the identity Matrix, which
+ has to exists because every entry in $M_n(A)$ has to behave like in $A$. Associativity is given by
+ matrix multiplication. Involutnes is given by the conjugate transpose.\\
+ 2. $A \simeq M_n(A)$ and $H \simeq H^n$ meaning $\tilde{\pi}$ is a valid reducible representation.\\
+ 3. $\tilde{\pi}$ and $\pi$ are unitary equivalent, there is a map $U: H^n \rightarrow H^n$ given by
+ $U=\mathbbm{1}_n$:\\
+ $\pi (a) = \mathbbm{1}_n^*\ \tilde{\pi}((a_{ij}))\ \mathbbm{1}_n = \tilde{\pi}((a_{ij})) = \pi (a_{ij})
+ \Rightarrow a_{ij} = a\mathbbm{1}_n$.
+\end{MyExercise}
+
+\subsubsection{Commutative Matrix Algebras}
+\begin{itemize}
+ \item Commutative matrix algebras can be used to reconstruct a discrete space given
+ a matrix \textit{commutative} matrix algebra.
+ \item The structure space $\hat{A}$ is used for this. Because $A \simeq \mathbb{C}^N$ we all any
+ irreducible representation are of the form
+ $\pi _i:(\lambda_1,...,\lambda_N)\in \mathbb{C}^N \mapsto \lambda_i \in \mathbb{C}$ \\
+ for $i = 1,...,N \Rightarrow \hat{A} \simeq \{1,...,N\}.$
+ \item Conclusion is that there is a duality between discrete spaces and commutative matrix algebra
+ this duality is called the \textit{finite dimensional Gelfand duality}
+\end{itemize}
+
+\subsubsection{Noncommutative Matrix Algebras}
+Aim is to construct duality between finite dimensional spaces and \textit{equivalence classes}
+of matrix algebras, to preserve general non-commutivity of matrices.
+\begin{itemize}
+ \item Equivalence classes are described by a generalized notion of ispomorphisms between matrix
+ algebras (\textit{Morita Equivalence})
+\end{itemize}
+
+\subsubsection{Algebraic Modules}
+\begin{definition}
+ 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_1a_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 $\exists$ a bilinear map $\gamma: F \times B \rightarrow F$ with
+ \begin{align}
+ f \cdot (b_1b_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{definition}
+Notion of 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 (ae) = a \phi (e); \;\;\; a \in A, e \in E.
+\end{align}
+Remark on the notation
+\begin{itemize}
+ \item ${}_A E$ left $A$-module $E$;
+ \item ${}_A E_B$ right $B$-module $F$;
+ \item ${}_A E_B$ $A$-$B$-bimodule $E$;
+\end{itemize}
+
+\begin{MyExercise}
+ \textbf{
+ Check that a representation of $\pi : A \rightarrow L(H)$ of a $*$-algebra A turns H into a
+ left module ${}_A H$.
+}\newline
+
+ Not quite sure but \\
+ $a \in A$, $h_1, h_2 \in H$, we know $\pi (a) = T \in L(H)$ than
+ \begin{align}
+ \langle \pi (a) h_1, \pi (a) h_2\rangle = \langle T h_1, T h_2\rangle = \langle T^*T h_1, h_2\rangle = \langle h_1, h_2\rangle
+ \end{align}
+ Or maybe this \\
+ If $_A H$ than $(a_1a_2) h = a_1 (a_2 h)$ for $a_1, a_2 \in A$ and $h \in H$.\\
+ Then we take the representation of an $a \in A$, $\pi (a)$:
+ \begin{align}
+ (\pi(a_1)\pi(a_2))h = \pi(a_1)(\pi(a_2) h) = (T_1T_2) h = T_1 (T_2 h)
+ \end{align}
+ For $T_1, T_2 \in L(H)$, which operate naturally from the left on $h$.
+\end{MyExercise}
+
+\begin{MyExercise}
+ \textbf{
+ Show that $A$ is a bimodule ${}_A A_A$ with itself.
+}\newline
+
+ $\gamma: A\times A\times A \rightarrow A$ which is given by the inner product of the $*$-algebra.
+\end{MyExercise}
diff --git a/src/thesis/front/conclusion.tex b/src/thesis/chapters/conclusion.tex
diff --git a/src/thesis/chapters/diffgeo.tex b/src/thesis/chapters/diffgeo.tex
@@ -0,0 +1,49 @@
+\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}.
diff --git a/src/thesis/chapters/electroncg.tex b/src/thesis/chapters/electroncg.tex
@@ -0,0 +1,755 @@
+\subsection{Noncommutative Geometry of Electrodynamics}
+\subsubsection{The Two-Point Space}
+One of the basics forms of a noncommutative space is the the two point space $X
+:= \{x, y\}$, it can be represented by the following spectral triple
+\begin{align}
+ F_x := (C(X) = \mathbb{C}^2, H_F, D_F; J_F, \gamma _f).
+\end{align}
+There are three properties of $F_x$ that stand out, first of all the action of
+$C(X)$ on $H_F$ is faithful for $dim(H_F) \geq 2$. Thus we can make a simple
+choice for the Hilbertspace, $H_F = \mathbb{C}^2$. Furthermore $\gamma_F$ is
+the $\mathbb{Z}_2$ grading, which allows us to decompose $H_F$ into
+\begin{align}
+ H_F = H_F^+ \otimes H_F^- = \mathbb{C} \otimes \mathbb{C},
+\end{align}
+where
+\begin{align}
+ H_F^\pm = \{\psi \in H_F |\; \gamma_F\psi = \pm \psi\},
+\end{align}
+are two eigenspaces. And lastly the Dirac operator $D_F$ lets us
+interchange between $H_F^\pm$,
+\begin{align}
+ D_F =
+ \begin{pmatrix}0 & t \\ \bar{t} & 0\end{pmatrix}, \;\;\;\;\;
+ \text{with} \;\; t\in\mathbb{C}.
+\end{align}
+
+ The Two-Point space $F_x$ can only have a real structure if the Dirac
+ operator vanishes, i.e. $D_F = 0$, in that case we have KO-dimension of 0,
+ 2 or 6.
+
+ To elaborate on this, we know that there are two diagram representations of
+ $F_x$ at $\underbrace{\mathbb{C} \oplus \mathbb{C}}_{C(X)}$ on
+ $\underbrace{\mathbb{C} \oplus\mathbb{C}}_{H_F}$, which are:
+ \begin{figure}[h!] \centering
+ \begin{tikzpicture}[
+ dot/.style = {draw, circle, inner sep=0.06cm},
+ no/.style = {},
+ ]
+ \node[no](a) at (0,0) [label=left:$\textbf{1}^\circ$] {};
+ \node[no](b) at (0, -1) [label=left:$\textbf{1}^\circ$] {};
+ \node[no](c) at (1, 0.5) [label=above:$\textbf{1}$] {};
+ \node[no](d) at (2, 0.5) [label=above:$\textbf{1}$] {};
+ \node[dot](d0) at (2,0) [] {};
+ \node[dot](d0) at (1,-1) [] {};
+
+ \node[no](a1) at (6,0) [label=left:$\textbf{1}^\circ$] {};
+ \node[no](b2) at (6, -1) [label=left:$\textbf{1}^\circ$] {};
+ \node[no](c2) at (7, 0.5) [label=above:$\textbf{1}$] {};
+ \node[no](d2) at (8, 0.5) [label=above:$\textbf{1}$] {};
+ \node[dot](d0) at (7,0) [] {};
+ \node[dot](d0) at (8,-1) [] {};
+ \end{tikzpicture}
+ \end{figure}\newline
+If the Two-Point space $F_x$ would be a real spectral triple then $D_F$ can
+only go vertically or horizontally. This would mean that $D_F$ vanishes.
+
+The diagram on the left has KO-dimension 2 and 6, the diagram on the
+right has KO-dimension 0 and 4. Yet KO-dimension 4 is not ruled out because
+$dim(H_F^\pm) = 1$ (see Lemma 3.8 Book), which ultimately means $J_F^2 = -1$ is
+not allowed.
+\subsubsection{The product Space}
+By Extending the Two-Point space with a four dimensional Riemannian spin
+manifold, we get an almost commutative manifold $M\times F_x$, given by
+\begin{align}
+ M\times F_x = (C^\infty(M), \mathbb{C}^2), L^2(S)\otimes \mathbb{C}^2,
+ D_M\otimes 1 ; J_M\otimes J_F, \gamma_M \otimes \gamma_F),
+\end{align}
+where
+\begin{align}
+ C^\infty(M, \mathbb{C}^2) \simeq C^\infty(M) \oplus C^\infty(M).
+\end{align}
+According to Gelfand duality the algebra $C^\infty(M, \mathbb{C}^2)$ of the
+spectral triple corresponds to the space
+\begin{align}
+ N:= M\otimes X \simeq M\sqcup X
+\end{align}
+Keep in mind that we still need to find an appropriate real structure on the
+Riemannian spin manifold, $J_M$.
+\newline
+The total Hilbertspace can be decomposed into $H = L^2(S) \oplus L^2(S)$, such
+that for
+\newline
+$\underbrace{a,b\in C^\infty(M)}_{(a, b) \in C^\infty(N)}$
+and $\underbrace{\psi, \phi \in L^2(S)}_{(\psi, \phi) \in H}$ we have
+\begin{align}
+ (a, b)(\psi, \phi) = (a\psi, b\phi)
+\end{align}
+Along with the decomposition of the total Hilbertspace we can consider a
+distance formula on $M\times F_x$ with
+\begin{align}\label{eq:commutator inequality}
+ d_{D_F}(x,y) = \sup\left\{ |a(x) - a(y)|:a\in A_F, ||[D_F, a]|| \leq
+ 1 \right\}.
+\end{align}
+To calculate the distance between two points on the Two-Point space $X= \{x,
+y\}$, between $x$ and $y$, we consider an $a \in \mathbb{C}^2 = C(X)$, which is
+specified by two complex numbers $a(x)$ and $a(y)$. Then we simplify the
+commutator inequality in \ref{eq:commutator inequality}
+\begin{align}
+ &||[D_F , a]|| = ||(a(y) - a(x))\begin{pmatrix}0 &t\\\bar{t} &0
+ \end{pmatrix}|| \leq 1,\\
+ &\Rightarrow |a(y) - a(x)|\leq \frac{1}{|t|},
+\end{align}
+and the supremum gives us the distance
+\begin{align}
+ d_{D_F} (x,y) = \frac{1}{|t|}.
+\end{align}
+An interesting observation here is that, if the Riemannian spin manifold can be
+represented by a real spectral triple then a real structure $J_M$ exists,
+then it follows that $t=0$ and the distance becomes infinite. This is a purely
+mathematical observation and has no physical meaning.
+
+Now let $p \in M$, then take two points on $N=M\times X$, $(p, x)$ and
+$(p,y)$ and $a \in C^\infty(N)$ is determined by $a_x(p):=a(p, x)$ and
+$a_y(p):=a(p, y)$. The distance between these two points is then
+\begin{align}
+ d_{D_F\otimes 1}(n_1, n_2) = \sup \left\{ |a(n_1) - a(n_2)|: a\in
+ A, ||[D\otimes 1, a]||\right\}
+\end{align}
+\textbf{Remark}: If $n_1 = (p,x)$ and $n_2 = (q, x)$ for $p,q \in M$ then
+\begin{align}
+ d_{D_M \otimes 1} (n_1, n_2) = |a_x(p) - a_x(q)| \;\;\; a_x\in
+ C^\infty(M) \;\; \text{with} \;\; ||[D\otimes 1, a_x]|| \leq 1
+\end{align}
+The distance turns to the geodestic distance formula
+\begin{align}
+ d_{D_M\otimes1}(n_1, n_2) = d_g(p, q)
+\end{align}
+
+However if $n_1 = (p, x)$ and $n_2 = (q, y)$ then the two conditions are
+$||[D_M, a_x]|| \leq 1$ and $||[D_M, a_y|| \leq 1$. They have no
+restriction which results in the distance being infinite! And $N =
+M\times X$ is given by two disjoint copies of M which are separated by
+infinite distance
+
+\textbf{Note}: distance is only finite if $[D_F, a] \neq 1$. The
+commutator
+generates a scalar field say $\phi$ and the finiteness of the distance is
+related to the existence of scalar fields.
+\subsubsection{$U(1)$ Gauge Group}
+Here we determine the Gauge theory corresponding to the almost
+commutative
+Manifold $M\times F_x$.
+
+\textbf{Gauge Group of a Spectral Triple}:
+\begin{align}
+ \mathfrak{B}(A, H; J) := \{ U = uJuJ^{-1} | u\in U(A)\}
+\end{align}
+\begin{definition}
+ A *-automorphism of a *-algebra $A$ is a linear invertible
+ map
+ \begin{align}
+ &\alpha:A \rightarrow A\;\;\; \text{with}\\
+ \nonumber\\
+ &\alpha(ab) = \alpha(a)\alpha(b)\\
+ &\alpha(a)^* = \alpha(a^*)
+ \end{align}
+ The \textbf{Group of automorphisms of the *-Algebra $A$} is
+ $(A)$.\newline
+ The automorphism $\alpha$ is called \textbf{inner} if
+ \begin{align}
+ \alpha(a) = u a u^* \;\;\; \text{for} \;\; U(A)
+ \end{align}
+ where $U(A)$ is
+ \begin{align}
+ U(A) = \{ u\in A|\;\; uu^* = u^*u=1\} \;\;\;
+ \text{(unitary)}
+ \end{align}
+\end{definition}
+The Gauge group is given by the quotient $U(A)/U(A_J)$.
+We want a nontrivial Gauge group so we need to choose $U(A_J) \neq
+U(A)$ which is the same as $U((A_F)_{J_F}) \neq
+U(A_F)$.
+We consider $F_x$ to be
+\begin{align}
+ F_x := \left(\mathbb{C}^2,\mathbb{C}^2, D_F =\begin{pmatrix}
+ 0&0\\0&0\end{pmatrix}; J_f =\begin{pmatrix}
+ 0&C\\C&0\end{pmatrix},
+ \gamma_F = \begin{pmatrix}1&0\\0&-1\end{pmatrix}\right).
+\end{align}
+Here $C$ is the complex conjugation, and $F_X$ is a real even finite
+ spectral triple (space) with $KO-dim=6$
+
+\begin{proposition}
+ The Gauge group $\mathfrak{B}(F)$ of the two point space is given by
+ $U(1)$.
+\end{proposition}
+\begin{proof}
+ Note that $U(A_F) = U(1) \times U(1)$. We need to show that
+ $U(\mathcal{A}_F)
+ \cap U(A_F)_{J_F}) \simeq U(1)$, such that $\mathfrak{B}(F)
+ \simeq U(1)$.\newline
+
+ So for $a \in \mathbb{C}^2$ to be in $(A_F)_{J_F}$ it has
+ to satisfy $J_F a^* J_F = a$.
+ \begin{align}
+ J_F a^* J^{-1} =
+ \begin{pmatrix}0&C\\C&0\end{pmatrix}
+ \begin{pmatrix}\bar{a}_1&0\\0&\bar{a}_2\end{pmatrix}
+ \begin{pmatrix}0&C\\C&0\end{pmatrix}
+ =
+ \begin{pmatrix}a_2&0\\0&a_1\end{pmatrix}
+ \end{align}
+ Which is only the case if $a_1 = a_2$. So we have
+ $(A_F)_{J_F} \simeq \mathbb{C}$, whose unitary elements
+ from $U(1)$ are contained in the diagonal subgroup of
+ $U(\mathcal{A}_F)$.
+\end{proof}
+
+Now we need to find the exact from of the field $B_\mu$ to calculate the
+spectral action of a spectral triple. Since $(A_F)_{J_F} \simeq
+\mathbb{C}$ we find that $\mathfrak{h}(F) = \mathfrak{u}((A_F)_{J_F})
+\simeq i\mathbb{R}$. Where $\mathfrak{h}(F)$ is the Lie Algebra on $F$
+and $\mathfrak{u}((A_F)_{J_F})$ is the Lie algebra of the unitary group
+$(A_F)_{J_F}$.\newline
+
+An arbitrary hermitian field $A_\mu = -ia\partial _\mu b$ is given by
+two
+$U(1)$ Gauge fields $X_\mu^1, X_\mu^2 \in C^\infty(M, \mathbb{R})$.
+However $A_\mu$ appears in combination $A_\mu - J_F A_\mu J_F^{-1}$:
+\begin{align}
+ B_\mu = A_\mu - J_F A_\mu J_F^{-1} =
+ \begin{pmatrix}X_\mu^1&0\\0&X_\mu^2 \end{pmatrix}
+ -
+ \begin{pmatrix}X_\mu^2&0\\0&X_\mu^1 \end{pmatrix}
+ =:
+ \begin{pmatrix}Y_\mu&0\\0&-Y_\mu \end{pmatrix}=
+ Y_\mu \otimes \gamma _F
+\end{align}
+where $Y_\mu$ the $U(1)$ Gauge field is defined as
+\begin{align}
+ Y_\mu := X_\mu^1 - X_\mu^2 \in C^\infty(M, \mathbb{R}) = C^\infty(M,
+ i\ u(1)).
+\end{align}
+
+\begin{proposition}
+ The inner fluctuations of the almost-commutative manifold $M\times
+ F_x$ described above are parametrized by a $U(1)$-gauge field $Y_\mu$
+ as
+ \begin{align}
+ D \mapsto D' = D + \gamma ^\mu Y_\mu \otimes \gamma_F
+ \end{align}
+ The action of the gauge group $\mathfrak{B}(M\times F_X) \simeq
+ C^\infty (M, U(1))$ on $D'$ is implemented by
+ \begin{align}
+ Y_\mu \mapsto Y_\mu - i\ u\partial_\mu u^*; \;\;\;\;\; (u\in
+ \mathfrak{B}(M\times F_X)).
+ \end{align}
+\end{proposition}
+
+\subsection{Electrodynamics}
+Now we use the almost commutative Manifold and the abelian gauge group
+$U(1)$ to describe Electrodynamics. We arrive at a unified description of
+gravity and electrodynamics although in the classical level.
+\newline
+
+The almost commutative Manifold $M\times F_X$ describes a local gauge group
+$U(1)$. The inner fluctuations of the Dirac operator describe $Y_\mu$ the
+gauge field of $U(1)$. There arise two Problems:
+\newline
+(1): With $F_X$, $D_F$ must vanish, however this implies that the electrons
+are massless (this we do not want)
+\newline
+
+(2): The Euclidean action for a free Dirac field is
+\begin{align}
+ S = - \int i \bar{\psi}(\gamma ^\mu\partial _\mu - m) \psi d^4x,
+\end{align}
+$\psi,\ \bar{\psi}$ must be considered as independent variables, which means
+$S_F$ need two independent Dirac Spinors. We write $\{e, \bar{e}\}$ for the
+ONB of $H_F$, where $\{e\}$ is the ONB of $H_F^+$ and $\{\bar{e}\}$ the ONB
+of $H_F^-$ with the real structure this gives us 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}
+The total Hilbertspace is $H = L^2(S) \otimes H_F$, with $\gamma _F$ we can
+decompose $L^2(S) = L^2(S)^+ \oplus L^2(S)^-$, so with $\gamma = \gamma _M
+\otimes \gamma _F$ we can obtain the positive eigenspace $H^+$
+\begin{align}
+ H^+ = L^2(S)^+ \otimes H_F^+ \oplus L^(S)^- \otimes H_F^-.
+\end{align}
+For a $\xi \i H^+$ 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}. This is
+too much restriction for $F_X$.
+\subsubsection{The Finite Space}
+Here we solve the two problems by enlarging(doubling) the Hilbertspace. This
+is done by introducing multiplicities in Krajewski Diagrams which will also
+allow us to choose a nonzero Dirac operator which will connect the two
+vertices (next chapter).
+\newline
+
+We start of with the same algebra $C^\infty(M, \mathbb{C}^2)$, corresponding
+to space $N= M\times X \simeq M\sqcup M$.
+\newline
+
+The Hilbertspace will describe four particles,
+\begin{itemize}
+ \item left handed electrons
+ \item right handed positrons
+\end{itemize}
+Thus we have $\{ \underbrace{e_R, e_L}_{\text{left-handed}},
+\underbrace{\bar{e}_R, \bar{e}_L}_{\text{right-handed}}\}$ the ONB for $H_F
+\mathbb{C}^4$.
+\newline
+Then with $J_F$ we interchange particles with antiparticles we have the
+following properties
+\begin{align}
+ &J_F e_R = \bar{e}_R \;\;\;\;\; &J_F e_L = \bar{e_L} \\
+ &\gamma _F e_R = -e_R \;\;\;\;\; &\gamma_F e_L = e_L \\
+ \text{and}& \nonumber \\
+ &J_F^2 = 1 \;\;\;\;\; & J_F \gamma_F = - \gamma_F J_F
+\end{align}
+This corresponds to KO-dim$= 6$. Then $\gamma_F$ allows us to can decompose
+$H$
+\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}
+Alternatively we can decompose $H$ into the eigenspace of particles and their
+antiparticles (electrons and positrons) 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}
+Now 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}
+ a =
+ \begin{pmatrix}a_1 & a_2 \end{pmatrix} \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.
+\begin{proposition}
+ The data
+ \begin{align}
+ \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}
+ defines a real even spectral triple of KO-dimension 6.
+\end{proposition}
+This spectral triple 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}
+Add a non-zero Dirac Operator to $F_{ED}$. From the Krajewski Diagram, we see
+that edges only exist between the multiple vertices. So we construct a Dirac
+operator mapping between the two vertices.
+\begin{align}\label{dirac}
+ 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 consider 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 $\gamma_F$ like before, $D_F$ like above.
+\subsubsection{The almost-commutative Manifold}
+The almost commutative manifold $M\times F_{ED}$ has KO-dim$=2$, it is the
+following spectral triple
+\begin{align}
+ M\times F_{ED} := \left(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\right)
+\end{align}
+
+The algebra decomposition is like before
+\begin{align}
+ C^\infty(M, \mathbb{C}^2) = C^\infty (M) \oplus C^\infty (M)
+\end{align}
+
+The Hilbertspace decomposition is
+\begin{align}
+ H = (L^2(S) \otimes H_e ) \oplus (L^2(S) \otimes H_{\bar{e}}).
+\end{align}
+Here we have the one component of the algebra acting on $L^2(S) \otimes H_e$,
+and the other one acting on $L^2(S) \otimes H_{\bar{e}}$
+\newline
+
+The derivation of the gauge theory is the same for $F_{ED}$ as for $F_X$, we
+have $\mathfrak{B}(F) \simeq U(1)$ and for $B_\mu = A_\mu - J_F A_\mu
+J_F^{-1}$
+\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}
+
+Our space $N = M\times X \simeq M\sqcup M$ consists of two compies of $M$.
+If $D_F = 0$ we have infinite distance between the two copies. Now we have $D_F$
+nonzero but $[D_F, a] = 0$ $\forall a \in A$ which still yields infinite
+distance.
+\begin{question}
+ What does this imply (physically, mathematically)? Why can we continue
+ even thought we have infinite distance between the same manifold? What do
+ we get if we fix this?
+\end{question}
+\subsubsection{The Spectral Action}
+Here we calculate the Lagrangian of the almost commutative Manifold $M\times
+F_{ED}$, which corresponds to the Lagrangian of Electrodynamics on a curved
+background Manifold (+ gravitational Lagrangian). It consists of the spectral
+action $S_b$ (bosonic) and of the fermionic action $S_f$.
+
+The simples spectral action of a spectral triple $(A, H, D)$ is given by the
+trace of some function of $D$, we also allow 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}
+ 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}f(\frac{D_\omega}{\Lambda})
+ \end{align}
+ where $\Lambda$ is a real cutoff parameter. The minimal condition on $f$
+ is that $f(\frac{D_\omega}{\Lambda})$ is a traclass operator, which mean
+ that it should be compact operator with well defined finite trace
+ independent of the basis. The subscript $b$ of $S_b$ refers to bosonic,
+ because in physical applications $\omega$ will describe bosonic fields.
+
+ Furthermore there is a topological spectral action, defined with the
+ grading $\gamma$
+ \begin{align}
+ S_{\text{top}}[\omega] := \text{Tr}(\gamma\
+ f(\frac{D_\omega}{\Lambda})).
+ \end{align}
+\end{definition}
+\begin{definition}
+ The fermionic action is defined by
+ \begin{align}
+ S_f[\omega, \psi] = (J\tilde{\psi}, D_\omega \tilde{\psi})
+ \end{align}
+ with $\tilde{\psi} \in H_{cl}^+ := \{\tilde{\psi}: \psi \in H^+\}$.
+ $H_{cl}^+$ is the set of Grassmann variables in $H$ in the +1-eigenspace
+ of the grading $\gamma$.
+\end{definition}
+The grasmann variables are a set of Basis vectors of a vector space, they
+form a unital algebra over a vector field say $V$ where the generators are anti commuting, that is for
+$\theta _i, \theta _j$ some Grassmann variables 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}
+\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}
+ with
+ \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}
+ where $N=4$ and $\mathcal{L}_M$ is the Lagrangian of the spectral triple
+ $(C^\infty(M) , L^2(S), D_M)$
+ \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
+ $R_{\nu\sigma} = g^{\mu\varrho} R_{\mu\nu \varrho\sigma}$.
+
+
+ Furthermore $\mathcal{L}_B$ describes the kinetic term of the gauge field
+ \begin{align}
+ \mathcal{L}_B(B_\mu) := \frac{f(0)}{24\pi^2}
+ \text{Tr}(F_{\mu\nu}F^{\mu\nu}).
+ \end{align}
+ Last $\mathcal{L}_\phi$ is the scalar-field Lagrangian with a boundary
+ term.
+ \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))\\
+ &+ \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 our manifold $m$ is $\dim(m) = \text{tr}(id) =4 $. let us
+ take a $x \in m$, we have an asymtotic expansion of
+ $\text{tr}(f(\frac{d_\omega}{\lambda}))$ as $\lambda \rightarrow \infty$
+ \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) \\&+ f(0) a_4(d_\omega^4)
+ +o(\lambda^{-1}).
+ \end{align}
+ note that the heat kernel coefficients are zero for uneven $k$,
+ furthermore 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 we note that $n:= \text{tr}\mathbbm{1_{h_f}})$
+ \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 need to extend in terms of coefficients of $f$, 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}
+ now for the cross terms of $\omega_{\mu\nu}^e\omega^{e\mu\nu}$ the trace
+ vanishes because of the anti-symmetric properties of the riemannian
+ curvature tensor
+ \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 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}
+ and the trace of the whole term is
+ \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}
+ plugging the results into $a_4$ and simplifying we can write
+ \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) \\
+ &+ \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 plug in the heat kernel coefficients into the
+ heat kernel expansion above.
+\end{proof}
+
+Here on we go and calculate the spectral action of $M\times F_{ED}$
+\begin{proposition}
+ The Spectral action of $M\times F_{ED}$ is
+ \begin{align}
+ \text{Tr}(f\frac{D_\omega}{\Lambda}) \sim \int_M \mathcal{L}(g_{\mu\nu},
+ Y_\mu) \sqrt{g}\ d^4x + O(\Lambda^{-1})
+ \end{align}
+ where the Lagrangian is
+ \begin{align}
+ \mathcal{L}(g_{\mu\nu}, Y_\mu) =
+ 4\mathcal{L}_M(g_{\mu\nu})+
+ \mathcal{L}_Y(Y_\mu)+
+ \mathcal{L}_\phi(g_{\mu\nu}, d)
+ \end{align}
+ here the $d$ in $\mathcal{L}_\phi$ is from $D_F$ in equation
+ \ref{dirac}. The Lagrangian $\mathcal{L}_M$ is like in equation
+ \ref{lagr}. The Lagrangian $\mathcal{L}_Y$ is the kinetic term of the
+ $U(1)$ gauge field $Y_\mu$
+ \begin{align}
+ \mathcal{L}_Y(Y_\mu):= \frac{f(0)}{6\pi^2}
+ Y_{\mu\nu}Y^{\mu\nu}\;\;\;\;\;\;\;\;\text{with}\;\;\; Y_{\mu\nu} =
+ \partial_\mu Y_\nu -
+ \partial_\nu Y_\mu.
+ \end{align}
+ Then there is $\mathcal{L}_\phi$, which has two constant terms
+ (disregarding the boundary term) that add up to the Cosmological Constant
+ and a term that for the Einstein-Hilbert action
+ \begin{align}
+ \mathcal{L}_\phi(g_{\mu\nu}, d) := \frac{2f_2 \Lambda ^2}{\pi^2}
+ |d|^2 + \frac{f(0)}{2\pi^2} |d|^4 + \frac{f(0)}{12\pi ^2} s |d|^2.
+ \end{align}
+\end{proposition}
+\begin{proof}
+ The Trace of $\mathbb{C}^4$ (the Hilbertspace) gives $N=4$. With $B_\mu$
+ like in equation \ref{field} we have $\text{Tr}(F_{\mu\nu}
+ F^{\mu\nu})=4Y_{\mu\nu}Y^{\mu\nu}$. This provides $\mathcal{L}_Y$.
+ Furthermore we have $\Phi^2 = D_F^2 = |d|^2$ and $\mathcal{L}_\phi$ only
+ give numerical contributions to the cosmological constant and the
+ Einstein-Hilbert action.
+
+ The proof is relying itself on just plugging the terms into the previous
+ proposition, for which I didn't write the proof for.
+\end{proof}
+
+
+\subsection{fermionic action}
+a quick reminder with what we are dealing with, the fermionic action is defined
+in the following way.
+\begin{definition}
+ the fermionic action is defined by
+ \begin{align}
+ s_f[\omega, \psi] = (j\tilde{\psi}, d_\omega \tilde{\psi})
+ \end{align}
+ with $\tilde{\psi} \in h_{cl}^+ := \{\tilde{\psi}: \psi \in h^+\}$.
+ $h_{cl}^+$ is the set of grassmann variables in $h$ in the +1-eigenspace
+ of the grading $\gamma$.
+\end{definition}
+
+the almostcommutative manifold we are dealing with is the following
+\begin{align}
+ &m\times f_{ed} := \left(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\right).\\
+ \nonumber\\
+ &\text{where:} \nonumber \\
+ &c^\infty(m,\mathbb{c}^2) = c^\infty(m) \otimes c^\infty(m)
+ &\mathcal{h} = \mathcal{h}^+ \otimes \mathcal{h}^-\\
+ &\mathcal{h} = l^2(s)^+ \otimes h_f^+ \oplus l^2(s)^- \otimes h_f^-.
+\end{align}
+where $h_f$ is separated into the particle-anitparticle states with onb $\{e_r,
+e_l, \bar{e}_r, \bar{e}_l\}$. the onb of $h_f^+$ is $\{e_l, \bar{e}_r\}$ and
+for $h_f^-$ we have $\{e_r, \bar{e}_l\}$. furthermore we can decompose a spinor
+$\psi \in l^2(s)$ for each of the eigenspaces $h_f^\pm$, $\psi = \psi_r
+\psi_l$. thus we can write for an arbitrary $\psi \in \mathcal{h}^+$
+\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}
+for $\chi_l, \psi_l \in l^2(s)^+$ and $\chi_r, \psi_r \in l^2(s)^-$.
+\begin{proposition}
+ we can define the action of the fermionic art of $m\times f_{ed}$ in the
+ following way
+ \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}
+\end{proposition}
+\begin{proof}
+ we take the fluctuated dirac operator
+ \begin{align}
+ d_\omega = d_m \otimes i + \gamma^\mu \otimes b_\mu + \gamma_m \otimes
+ d_f
+ \end{align}
+\end{proof}
+the fermionic action is $s_f = (j\tilde{\xi}, d_\omega\tilde{\xi})$ for a $\xi
+\in \mathcal{h}^+$, we can begin to calculate (note that we add the constant
+$\frac{1}{2}$ to the action)
+\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:1}\\
+ &+\frac{1}{2}(j\tilde{\xi}, (\gamma^\mu \otimes b_\mu)
+ \tilde{\xi})\label{eq:2}\\
+ &+\frac{1}{2}(j\tilde{\xi}, (\gamma_m\otimes
+ d_f)\tilde{\xi})\label{eq:3}.
+\end{align}
+for equation \ref{eq:1} 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)+
+ \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)+
+ \frac{1}{2}(j_m\tilde{\chi}_r,d_m\tilde{\chi}_l)\\
+ &= (j_m\tilde{\chi},d_m\tilde{\chi}).
+\end{align}
+for equation \ref{eq:2} 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)+\\
+ &+\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)=\\
+ &= -(j_m\tilde{\chi}, \gamma^\mu y_\mu\tilde{\psi}).
+\end{align}
+for equation \ref{eq:3} we have
+\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)+\\
+ &+\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)=\\
+ &= i(j_m\tilde{\chi}, m\tilde{\psi})
+\end{align}
+note that we obtain a complex mass parameter $d$, so we write $d:=im$ for $m\in \mathbb{r}$,
+which stands for the real mass and we obtain a nice result
+
+\begin{theorem}
+ the full lagrangian of $m\times f_{ed}$ 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}
+
+\end{theorem}
diff --git a/src/thesis/chapters/heatkernel.tex b/src/thesis/chapters/heatkernel.tex
@@ -0,0 +1,373 @@
+
+\subsection{Heat Kernel Expansion}
+\subsubsection{The Heat Kernel}
+The heat kernel $K(t; x, y; D)$ is the fundamental solution of the heat
+equation. It depends on the operator $D$ of Laplacian type.
+\begin{align}
+ (\partial _t + D_x)K(t;x, y;D) =0
+\end{align}
+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}
+we have 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
+guage field, the heat kernel reads then
+\begin{align}
+ K(t;x,y;D) = \langle x|e^{-tD}|y\rangle.
+\end{align}
+We can expand it it in terms of $D_0$ and we still have the
+singularity from the equation \ref{eq:standard} as $t\rightarrow 0$ thus the
+expansion gives
+\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 $b_k(x,y)$ are regular in $y \rightarrow x$. They are called the heat
+kernel coefficients.
+
+\subsubsection{Example}
+Now let us consider a propagator $D^{-1}(x,y)$ defined through the heat kernel
+in an integral representation
+\begin{align}
+ D^{-1} (x,y) = \int_0^\infty dt K(t;x,y;D).
+\end{align}
+We can integrate the expression formally if we assume the heat kernel vanishes
+for $t\rightarrow \infty$ we 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}
+it solves the 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 integral approximation of the propagator we conclude
+that the singularities of $D^{-1}$ coincide with the singularities of the heat
+kernel coefficients.
+We consider now a generating functional in terns of $\det(D)$ which is called
+the one-loop effective action (quantum fields theory)
+\begin{align}
+ W = \frac{1}{2}\ln(\det D)
+\end{align}
+we can relate $W$ with the heat kernel. 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 depent on
+$\lambda$, because of this we can ignore it. Further more we use $\ln(\det D) =
+\text{Tr}(\ln D)$ and therefor we can write for $W$
+\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 is now 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 are just ignored. The divergences at $t\rightarrow 0$
+are cutoff at $t=\Lambda^{-2}$, thus we write
+\begin{align}
+ W_\Lambda = -\frac{1}{2} \int_{\Lambda^{-2}}^\infty dt \frac{K(t, D)}{t}.
+\end{align}
+We can calculate $W_\Lambda$ at 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 $b_2(x,x)$ with $k\leq n$. Now we compute the limit
+$\Lambda \rightarrow \infty$
+\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}
+here $\Gamma$ is the gamma function.
+\subsubsection{Differential Geometry and 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}
+
+\subsubsection{Spectral Functions}
+Manifolds without $M$ boundary condition for the operator $e^{-tD}$ for $t>0$ is a
+trace class operator on $L^2(V)$, this means that for any smooth function $f$
+on $M$ we can define
+\begin{align}
+ K(t,f,D) = \text{Tr}_{L^2}(fe^{-tD})
+\end{align}
+and we can rewrite
+\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 terms of the Heat kernel $K(t;x,y;D)$ in the regular limit $y\rightarrow y$.
+We can write the Heat Kernel in terms of the spectrum of $D$. Say
+$\{\phi_\lambda\}$ is a ONB of eigenfunctions of $D$ corresponding to the
+eigenvalue $\lambda$
+\begin{align}
+ K(t;x,y;D) = \sum_\lambda \phi^\dagger_\lambda(x)
+ \phi_\lambda(y)e^{-t\lambda}.
+\end{align}
+We have an asymtotic expansion at $t \rightarrow 0$ for the trace
+\begin{align}
+ 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}
+We consider a compact Riemmanian Manifold $M$ without boundary condition, a
+vector bundle $V$ over $M$ to define functions which carry discrete (spin or
+gauge) indices. An Laplace style operator $D$ over $V$ and smooth function $f$
+on $M$. There is an asymtotic expansion where the heat kernel coefficients
+\begin{enumerate}
+ \item with odd index $k=2j+1$ vanish
+ $a_{2j+1}(f,D) = 0$
+ \item with even index are locally computable in terms of geometric
+ invariants
+\end{enumerate}
+\begin{align}
+ a_k(f,D) &= \text{Tr}_V\left(\int_M d^n x\sqrt{g}(f(x)a_k(x;D)\right) =\\
+ &=\sum_I \text{Tr}_V\left(\int_M d^nx \sqrt{g}(fu^I \mathcal{A}^I_k(D))\right)
+\end{align}
+here $\mathcal{K}^I_k$ are all possible independent invariants of dimension
+$k$, constructed from $E, \Omega, R_{\mu\nu\varrho\sigma}$ and their
+derivatives, $u^I$ are some constants.
+
+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 everything, i.e.
+\begin{align}
+ e^{-tD} &= e^{-tD_1} \otimes e^{-tD_2}\\
+ f(x_1, x_2) &= f_1(x_1)f_2(x_2)\\
+ a_k(x;D) &= \sum_{p+q=k} a_p(x_1; D_1)a_q(x_2;D_2)
+\end{align}
+Say the spectrum of $D_1$ is known, $l^2, l\in \mathbb{Z}$. We obtain the heat
+kernel asymmetries with the Poisson Summation formula
+\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}} = \\
+ &\simeq \sqrt{\frac{\pi}{t}} + \mathcal{O}(e^{-1/t}).
+\end{align}
+Note that 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 we can write
+\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}
+
+On the other had all geometric invariants associated with $D$ are in the $D_2$
+part. Thus all invariants are independent of $x_1$, so we can choose for $M_1$.
+Say $M_1 = S^1$ with $x\in (0, 2\pi)$ and $D_1=-\partial_{x_1}^2$ we may
+rewrite the heat kernel coefficients in
+\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))=\\
+ &= 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}|_{\varepsilon=0}a_k(1, e^{-2\varepsilon f}D) =
+ (n-k) a_k(f, D),\label{eq:var1}\\
+ &\frac{d}{d\varepsilon}|_{\varepsilon=0}a_k(1, D-\varepsilon F) =
+ a_{k-2}(F,D),\label{eq:var2}\\
+ &\frac{d}{d\varepsilon}|_{\varepsilon=0}a_k(e^{-2\varepsilon f}F,
+ e^{-2\varepsilon f}D) =
+ 0\label{eq:var3}.
+\end{align}
+To prove the equation \ref{eq:var1} we differentiate
+\begin{align}
+ \frac{d}{d\varepsilon}|_{\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 \ref{eq:var1}. Equation \ref{eq:var2}
+is derived similarly. For equation \ref{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 \ref{eq:var1} and we get
+\begin{align}
+ \frac{d}{d\varepsilon}|_{\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}|_{\delta=0}\frac{d}{d\varepsilon}|_{\varepsilon=0}a_n(1,
+ D(\varepsilon,\delta)) =
+ \frac{d}{d\varepsilon}|_{\varepsilon=0}\frac{d}{d\delta}|_{\delta=0}a_n(1,
+ D(\varepsilon,\delta)) =\\
+ &=a_{n-2} ( e^{-2\varepsilon f}F, e^{-2\varepsilon f}D)
+\end{align}
+which proves equation \ref{eq:var3}. With this we calculate the constants $u^I$
+and we can write the first three heat kernel coefficients as
+\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 RE + \alpha_5 E^2
+ \alpha_6 R_{,kk} + \\
+ &+\alpha_7 R^2 + \alpha_8 R_{ij}R_{ij} + \alpha_9
+ R_{ijkl}R_{ijkl} +\alpha_{10} \Omega_{ij}\Omega{ij})).
+\end{align}
+The constants $\alpha_I$ do not depend on the dimension $n$ of the Manifold and
+we can compute them with our variational identities.
+
+The first coefficient $\alpha_0$ can be seen from the heat kernel expanion of
+the Laplacian on $S^1$ (above), $\alpha_0 = 1$. For $\alpha_1$ we use
+\ref{eq:var2}, for $k = 2$
+\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}
+thus we conclude that $\alpha_1 = 6$. Now we take $k=4$
+\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 FE + \alpha_2 FR),
+\end{align}
+thus $\alpha_4 = 60\alpha_2$ and $\alpha_5 = 180$.
+
+Furthermore we apply \ref{eq:var3} to $n=4$
+\begin{align}
+ \frac{d}{d\varepsilon}|_{\varepsilon=0} a_2(e^{-2\varepsilon f}F,
+ e^{-2\varepsilon f}D) = 0.
+\end{align}
+By 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 $M_1, M_2$, then we can decompose the heat
+kernel coefficients for $k=4$
+\begin{align}
+ a_4(1,-\Delta_1-\Delta_2) =& a_4(1, -\Delta_1) a_0(1, -\Delta_2)
+ +a_2(1,-\Delta_1) a_2(1,-\Delta_2) \\&+ 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 +\\
+ &+(2\alpha_3 + 10\alpha_6)f_{,iijj}+\\
+ &+(2\alpha_4 -2\alpha_6 - 20\alpha_7 -2\alpha_8)f_{,ii}R\\
+ &+(-8\alpha_8 -8\alpha_6)f_{,ij}R_{ij}))
+\end{align}
+we obtain $\alpha_3 = 60$, $\alpha_6=12$, $\alpha_8 = -2$ and $\alpha_9 = 2$
+
+For $\alpha_{10}$ we use the Gauss-Bonnet theorem to get $\alpha_{10}=30$,
+which is left out because it is a lengthy computation.
+
+Summarizing we get for the 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/front/intro.tex b/src/thesis/chapters/intro.tex
diff --git a/src/thesis/chapters/main_sec.tex b/src/thesis/chapters/main_sec.tex
@@ -0,0 +1 @@
+\section{Main Section}
diff --git a/src/thesis/front/main_sec.tex b/src/thesis/front/main_sec.tex
@@ -1,2 +0,0 @@
-\section{Main Section}
-\lipsum[8]
diff --git a/src/thesis/main.pdf b/src/thesis/main.pdf
Binary files differ.
diff --git a/src/thesis/main.tex b/src/thesis/main.tex
@@ -8,15 +8,28 @@
\newpage
+%-------------------- BACKHAND ---------------------
+
\input{back/abstract}
-\input{front/intro}
+\input{chapters/intro}
+
+%----------------- MAIN SECTION --------------------
+
+\input{chapters/main_sec}
+
+%\input{chapters/basics}
+%
+%\input{chapters/heatkernel}
+
+\input{chapters/electroncg}
+
-\input{front/main_sec}
+%------------------- BACKHAND ---------------------
-\input{front/conclusion}
+\input{chapters/conclusion}
-\input{front/acknowledgment}
+\input{chapters/acknowledgment}
\input{back/refs}
diff --git a/src/thesis/uni.bib b/src/thesis/thesis.bib
