ncg

bachelorthesis in physics
git clone git://popovic.xyz/ncg.git
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commit 258e29d485faf66c87ca5d12b75124d7621b86b0
parent 9c62b5a750a93f4f219109435edd51149c7f239b
Author: miksa234 <milutin@popovic.xyz>
Date:   Wed, 21 Jul 2021 15:04:10 +0200

checkpoint

Diffstat:

Msrc/thesis/back/title.tex | 7++++---


Dsrc/thesis/chapters/1 | 754-------------------------------------------------------------------------------


Msrc/thesis/chapters/electroncg.tex | 975+++++++++++++++++++++++++++++++++++++++----------------------------------------


3 files changed, 491 insertions(+), 1245 deletions(-)

diff --git a/src/thesis/back/title.tex b/src/thesis/back/title.tex
@@ -5,7 +5,7 @@
 \hspace{8.8cm}
 \includegraphics[width=8cm]{pics/uni_logo}
 \end{figure}
-\vspace*{2cm}
+\vspace*{1cm}
 
     \fontsize{22}{0} \fontfamily{lmss}\selectfont \textbf{Bachelor's Thesis}\\
 
@@ -15,7 +15,8 @@
 
 \vspace*{0.4cm}
 
-\fontsize{18}{0} \selectfont \textbf{Noncommutative Geomtetry and Physics}\\
+\fontsize{18}{0} \selectfont \textbf{From Noncommutative Geometry to
+Electrodynamics}\\
 
 \vspace*{1.5cm}
 
@@ -26,7 +27,7 @@
 
 \vspace*{2cm}
 
-    {\fontsize{12}{0} \selectfont  in partial fulfilment of the requirements for the degree of}\\
+    {\fontsize{12}{0} \selectfont  in partial fulfillment of the requirements for the degree of}\\
 \vspace*{0.4cm}
 
     { \fontsize{14}{0} \selectfont Bachelor of Science (BSc)}\\
diff --git a/src/thesis/chapters/1 b/src/thesis/chapters/1
@@ -1,754 +0,0 @@
-\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/chapters/electroncg.tex b/src/thesis/chapters/electroncg.tex
@@ -1,12 +1,12 @@
 \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
+One of the basics forms of noncommutative space is the Two-Point space $X
+:= \{x, y\}$. The Two-Point space 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).
+    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
+Three properties of $F_X$ 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}
@@ -24,12 +24,10 @@ interchange between $H_F^\pm$,
             \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
+    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}[
@@ -51,19 +49,18 @@ interchange between $H_F^\pm$,
         \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
+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
+As for the KO-dimension The diagram on the left has KO-dimension 2 and 6, the diagram on the
+right 0 and 4. Yet KO-dimension 4 is 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}
+\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
+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),
+    M\times F_X = \big(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\big),
 \end{align}
 where
 \begin{align}
@@ -72,21 +69,17 @@ where
 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
+    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
+Riemannian spin manifold, $J_M$. Furthermore total Hilbertspace can be decomposed into $H = L^2(S) \oplus L^2(S)$, such that for
 $\underbrace{a,b\in C^\infty(M)}_{(a, b) \in C^\infty(N)}$
 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
+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\}.
@@ -98,7 +91,7 @@ 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|},
+    &\Leftrightarrow |a(y) - a(x)|\leq \frac{1}{|t|},
 \end{align}
 and the supremum gives us the distance
 \begin{align}
@@ -106,45 +99,57 @@ and the supremum gives us the distance
 \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.
+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
+We can also construct a distance formula on $N$ (in reference to a point $p
+\in M$) between two points on $N=M\times X$, $(p, x)$ and $(p,y)$. Then an $a
+\in C^\infty(N)$ is determined by $a_x(p):=a(p, x)$ and $a_y(p):=a(p, y)$.
+The distance between these two points is
 \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\}
+    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
+On the other hand if we consider $n_1 = (p,x)$ 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
+    d_{D_M \otimes 1} (n_1, n_2) = |a_x(p) - a_x(q)| \;\;\;\text{for}\;\;
+    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
+The distance formula turns to out to be the geodesic distance formula
 \begin{align}
-    d_{D_M\otimes1}(n_1, n_2) = d_g(p, q)
+    d_{D_M\otimes1}(n_1, n_2) = d_g(p, q),
 \end{align}
-
+which is to be expected since we are only looking at the manifold.
 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
+\begin{align}
+    &||[D_M, a_x]|| \leq 1, \;\;\; \text{and}\\
+    &||[D_M, a_y|| \leq 1.
+\end{align}
+These conditions 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
+The distance is only finite if $[D_F, a] < 1$. In this case the commutator
+generates a scalar field 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}:
+\subsubsection{$U(1)$ Gauge Group}
+To get a insight into the physical properties of the almost commutative
+manifold $M\times F_X$, that is to calculate the spectral action, we need to
+determine the corresponding Gauge theory.
+For this we set of with simple definitions and important propositions to
+help us break down and search for the gauge group of the Two-Point $F_X$
+space which we then extend to $M\times F_X$. We will only be diving
+superficially into this chapter, for further reading we refer to
+\cite{ncgwalter}.
+\begin{definition}
+Gauge Group of a real spectral triple is given by
 \begin{align}
     \mathfrak{B}(A, H; J) := \{ U = uJuJ^{-1} | u\in U(A)\}
 \end{align}
+\end{definition}
 \begin{definition}
     A *-automorphism of a *-algebra $A$ is a linear invertible
     map
@@ -154,7 +159,7 @@ Manifold $M\times F_x$.
         &\alpha(ab) = \alpha(a)\alpha(b)\\
         &\alpha(a)^* = \alpha(a^*)
     \end{align}
-    The \textbf{Group of automorphisms of the *-Algebra $A$} is
+    The \textbf{Group of automorphisms of the *-Algebra $A$} is denoted by
     $(A)$.\newline
     The automorphism $\alpha$ is called \textbf{inner} if
     \begin{align}
@@ -166,65 +171,53 @@ Manifold $M\times F_x$.
         \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
+The Gauge group of $F_X$ is given by the quotient $U(A)/U(A_J)$.
+We want a nontrivial Gauge group so we need to choose a $U(A_J) \neq
+U(A)$ and $U((A_F)_{J_F}) \neq U(A_F)$.
+We consider our Two-Point space $F_X$ to be equipped with a real structure,
+which means the operator vanishes, and the spectral triple representation is
 \begin{align}
-    F_x := \left(\mathbb{C}^2,\mathbb{C}^2, D_F =\begin{pmatrix}
+    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 $KO-dim=6$
+spectral triple (space) of KO-dimension 6.
 
 \begin{proposition}
-    The Gauge group $\mathfrak{B}(F)$ of the two point space is given by
-    $U(1)$.
+The Gauge group of the Two-Point space $\mathfrak{B}(F_X)$ is $U(1)$.
 \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 $J_F a^* J_F = a$.
+    Note that $U(A_F) = U(1) \times U(1)$. We need to show that $U(A_F) \cap
+    U(A_F)_{J_F}) \simeq U(1)$, such that $\mathfrak{B}(F) \simeq U(1)$. So
+    for an element $a \in \mathbb{C}^2$ 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}
+            \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
+    This can only be 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
-    $U(\mathcal{A}_F)$.
+    $U(A_F)$.
 \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 $X_\mu^1, X_\mu^2 \in C^\infty(M, \mathbb{R})$.
+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} =
+ 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
+    \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}
@@ -234,8 +227,7 @@ where $Y_\mu$ the $U(1)$ Gauge field is defined as
 
 \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
+    F_X$ are parameterized by a $U(1)$-gauge field $Y_\mu$ as
     \begin{align}
         D \mapsto D' = D + \gamma ^\mu Y_\mu \otimes \gamma_F
     \end{align}
@@ -247,91 +239,99 @@ where $Y_\mu$ the $U(1)$ Gauge field is defined as
     \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
+In this chapter we describe Electrodynamics with the almost commutative
+manifold $M\times F_X$ and the abelian gauge group $U(1)$.
+We arrive at a unified description of gravity and electrodynamics although in the classical level.
 
 The almost commutative Manifold $M\times F_X$ describes a local gauge group
-$U(1)$. The inner fluctuations of the Dirac operator 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
+$U(1)$. The inner fluctuations of the Dirac operator relate to $Y_\mu$ the
+gauge field of $U(1)$. According to the setup we ultimately arrive at two
+serious problems.
+
+First of all in the Two-Point space $F_X$, the operator $D_F$ must vanish for
+us to have a real structure. However this implies that the electrons
+are massless, which would be absurd.
 
-(2): The Euclidean action for a free Dirac field is
+The second problem arises when looking at the Euclidean action for a free
+Dirac field
 \begin{align}
     S = - \int i \bar{\psi}(\gamma ^\mu\partial _\mu - m) \psi d^4x,
 \end{align}
-$\psi,\ \bar{\psi}$ 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
+where $\psi,\ \bar{\psi}$ must be considered as independent variables, which
+means that the fermionic action $S_f$ needs two independent Dirac spinors.
+Let us try and construct two independent Dirac spinors with our data. To do
+this we take a look at the decomposition of the basis and of the total
+Hilbertspace $H = L^2(S) \otimes H_F$. For the orthonormal basis of $H_F$ we
+can write $\{e, \bar{e}\}$ , where $\{e\}$ is the orthonormal basis of
+$H_F^+$ and $\{\bar{e}\}$ the orthonormal basis of $H_F^-$. Accompanied with
+the real structure we arrive at the following relations
 \begin{align}
-    J_F e &= \bar{e} \;\;\;\;\;\; J_F \bar{e} = e \\
+    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 $H^+$
+Along with the decomposition of $L^2(S) = L^2(S)^+ \oplus L^2(S)^-$ and  $\gamma = \gamma _M
+\otimes \gamma _F$ we can obtain the positive eigenspace
 \begin{align}
     H^+ = L^2(S)^+ \otimes H_F^+ \oplus L^(S)^- \otimes H_F^-.
 \end{align}
-For a $\xi \i H^+$ we can write
+So, for a $\xi \in H^+$ we can write
 \begin{align}
     \xi = \psi _L \otimes e + \psi _R \otimes \bar{e}
 \end{align}
-where $\psi _L \in L^2(S)^+$ and $\psi _R \in L^2(S)^-$ are the two Wheyl
+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 $F_X$.
+\psi_L + \psi _R$, \textbf{but we require two independent spinors}. Our
+conclusion is that the definition of the fermionic action gives too much
+restrictions to the Two-Point space $F_X$.
 \subsubsection{The Finite Space}
-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
+To solve the two problems we simply enlarge (double) the Hilbertspace. This
+is visualized by introducing multiplicities in Krajewski Diagrams which will also
+allow us to choose a nonzero Dirac operator that will connect the two
+vertices and preserve real structure making our particles massive and
+bringing anti-particles into the mix.
 
 We start of with the same algebra $C^\infty(M, \mathbb{C}^2)$, corresponding
-to space $N= M\times X \simeq M\sqcup M$.
-\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
+to space $N= M\times X$. The Hilbertspace describes four particles, meaning
+it has four orthonormal basis elements. It describes \textbf{left handed
+electrons} and \textbf{right handed positrons}. Pointing this out, we have
+$\{ \underbrace{e_R, e_L}_{\text{left-handed}}, \underbrace{\bar{e}_R,
+\bar{e}_L}_{\text{right-handed}}\}$ the orthonormal basis for $H_F =
+\mathbb{C}^4$. Accompanied with the real structure $J_F$, which allows us to
+interchange particles with antiparticles by the following equations
+\begin{align}
+    &J_F e_R = \bar{e}_R, \\
+    &J_F e_L = \bar{e_L}, \\
+    \nonumber \\
+    &\gamma _F e_R = -e_R,\\
+    &\gamma_F e_L = e_L \\
+\end{align}
+where $J_F$ and $gamma_F$ have to following properties
 \begin{align}
-    &J_F 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
+    &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
-$H$
+By means of $\gamma_F$ we have two options to decompose the total
+Hilbertspace $H$, firstly into
 \begin{align}
     H_F = \underbrace{H_F^+}_{\text{ONB } \{e_L, \bar{e}_L\}}
-    \oplus \underbrace{H_F^-}_{\text{ONB } \{e_R, \bar{e}_R\}}.
+    \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.
+or alternatively into the eigenspace of particles and their
+antiparticles (electrons and positrons) which is preferred in literature and
+which we will use going further
 \begin{align}
     H_F = \underbrace{H_{e}}_{\text{ONB } \{e_L, e_R\}} \oplus
-    \underbrace{H_{\bar{e}}}_{\text{ONB } \{\bar{e}_L, \bar{e}_R\}}
+    \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
+Here ONB means orthonormal basis.
+
+The action of $a \in A = \mathbb{C}^2$ on $H$ with respect to the ONB
 $\{e_L, e_R, \bar{e}_L, \bar{e}_R\}$ is represented by
-\begin{align}
+\begin{align}\label{eq:leftrightrepr}
     a =
-    \begin{pmatrix}a_1 & a_2 \end{pmatrix} \mapsto
+    (a_1 , a_2 ) \mapsto
         \begin{pmatrix}
             a_1 &0 &0 &0\\
              0&a_1 &0 &0\\
@@ -340,11 +340,11 @@ $\{e_L, e_R, \bar{e}_L, \bar{e}_R\}$ is represented by
         \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}
+$[a, b^\circ] = 0$ with $b:= J_F b^*J_F$ because both the left and the right
+action is given by diagonal matrices by equation \ref{eq:leftrightrepr}. Note
+that we are still left with $D_F = 0$ and the following spectral
+triple
+\begin{align}\label{eq:fedfail}
         \left( \mathbb{C}^2, \mathbb{C}^2, D_F=0; J_F =
         \begin{pmatrix}
             0 & C \\ C &0
@@ -353,13 +353,11 @@ action is given by diagonal matrices.
         \begin{pmatrix}
             1 & 0 \\ 0 &-1
         \end{pmatrix}
-        \right)
+        \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,
+It can be represented in the following Krajewski diagram,
 with two nodes of multiplicity two
-    \begin{figure}[h!] \centering
+    \begin{figure}[H] \centering
     \begin{tikzpicture}[
         dot/.style = {draw, circle, inner sep=0.06cm},
         bigdot/.style = {draw, circle, inner sep=0.09cm},
@@ -376,10 +374,11 @@ with two nodes of multiplicity two
         \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}
+To extend our spectral triple with a non-zero Operator, we need to take a
+closer look at the Krajewski diagram above. Notice that edges only exist
+between multiple vertices, meaning we can construct a Dirac operator mapping
+between the two vertices. The operator can be represented by the following matrix
+\begin{align}\label{eq:feddirac}
     D_F =
     \begin{pmatrix}
     0 & d & 0 & 0 \\
@@ -388,37 +387,38 @@ operator mapping between the two vertices.
     0 & 0 & d & 0
     \end{pmatrix}
 \end{align}
-We can now consider the finite space $F_{ED}$.
+We can now define the finite space $F_{ED}$.
 \begin{align}
     F_{ED} := (\mathbb{C}^2, \mathbb{C}^4, D_F; J_F, \gamma_F)
 \end{align}
-where $J_F$ and $\gamma_F$ like before, $D_F$ like above.
+where $J_F$ and $\gamma_F$ are like in equation \ref{eq:fedfail} and $D_F$
+from equation \ref{eq:feddirac}.
+
 \subsubsection{The almost-commutative Manifold}
-The almost commutative manifold $M\times F_{ED}$ has KO-dim$=2$, it is the
-following spectral triple
+The almost commutative manifold $M\times F_{ED}$ has KO-dimension 2, and is
+represented by the following spectral triple
 \begin{align}
-    M\times F_{ED} := \left(C^\infty(M,\mathbb{C}^2),\ L^2(S)\otimes
+    M\times F_{ED} := \big(C^\infty(M,\mathbb{C}^2),\ L^2(S)\otimes
     \mathbb{C}^4,\
     D_M\otimes 1 +\gamma _M \otimes D_F;\; J_M\otimes J_F,\ \gamma_M\otimes
-    \gamma _F\right)
+    \gamma _F\big)
 \end{align}
-
-The algebra decomposition is like before
+The algebra didn't change, thus we can decompose it like before
 \begin{align}
     C^\infty(M, \mathbb{C}^2) = C^\infty (M) \oplus C^\infty (M)
 \end{align}
-
-The Hilbertspace decomposition is
+As for the Hilbertspace, we can decomposition it in the following way
 \begin{align}
     H = (L^2(S) \otimes H_e ) \oplus (L^2(S) \otimes H_{\bar{e}}).
 \end{align}
-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
+Note that the one component of the algebra is acting on $L^2(S) \otimes H_e$,
+and the other one acting on $L^2(S) \otimes H_{\bar{e}}$. In other words the components of
+the decomposition of both the algebra and the Hilbertspace match by the action of
+the algebra.
 
-The derivation of the gauge theory is the same for $F_{ED}$ as for $F_X$, we
-have $\mathfrak{B}(F) \simeq U(1)$ and for $B_\mu = A_\mu - J_F A_\mu
-J_F^{-1}$
+The derivation of the gauge theory is the same for $F_{ED}$ as for the
+Two-Point space $F_X$. We have $\mathfrak{B}(F) \simeq U(1)$ and for an
+arbitrary gauge field $B_\mu = A_\mu - J_F A_\mu J_F^{-1}$ we can write
 \begin{align} \label{field}
     B_\mu =
     \begin{pmatrix}
@@ -434,322 +434,321 @@ gauge group
    \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}
+Our space $N = M\times X \simeq M\sqcup M$ consists of two copies of $M$.
+If $D_F = 0$ we have infinite distance between the two copies. Now have
+hacked the spectral triple to have nonzero Dirac operator $D_F$. The new
+Dirac operator still has a commuting relation with the algebra $[D_F, a] = 0$
+$\forall a \in A$, and we should note that the distance between the two
+copies of $M$ is still infinite. This is purely an mathematically abstract
+observation and doesn't affect physical results.
+
+%\subsubsection{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}