commit 5b023921cca4f36e1f6268fd94ed761dd7dab504
parent bd23e8a8ad7160a22748ebfc347dfa3fab149c5b
Author: miksa234 <milutin@popovic.xyz>
Date: Mon, 9 Aug 2021 21:07:31 +0200
checkpoint introduction 1/2
Diffstat:
26 files changed, 68 insertions(+), 3140 deletions(-)
diff --git a/books/Bachelor-Guide.pdf b/books/Bachelor-Guide.pdf
Binary files differ.
diff --git a/src/thesis/back/abstract.tex b/src/thesis/back/abstract.tex
@@ -1,3 +1,13 @@
\begin{abstract}
- \lipsum[1]
+ Noncommutative geometry is a branch of mathematics that has deep
+ connections to applications in physics. From reconstructing the
+ theory of electrodynamics with minimal coupling to gravity, to deriving
+ the full Lagrangian of the standard model and predicting the Higgs mass.
+ One of the reasons for this is the natural existence of a nontrivial
+ gauge group of a mathematical structure called the spectral triple, which
+ encodes (classical) geometrical data intro algebraic data. Altogether
+ this thesis is based on literature work, mostly from Walter D. Suijlekom's
+ book \cite{ncgwalter}. We summarize enough information to both establish
+ the basic backbone of noncommutative geometry and to further out derive
+ the Lagrangian of electrodynamics.
\end{abstract}
diff --git a/src/thesis/back/title.tex b/src/thesis/back/title.tex
@@ -13,9 +13,10 @@
\fontsize{12}{0} \selectfont Title of the Bachelor's Thesis\\
-\vspace*{0.4cm}
+\vspace*{0.3cm}
-\fontsize{18}{0} \selectfont \textbf{Noncommutative Geomtetry and Physics}\\
+\fontsize{18}{0} \selectfont \textbf{Noncommutative Geomtetry and
+Electrodynamics}\\
\vspace*{1.5cm}
@@ -27,14 +28,14 @@
\vspace*{2cm}
{\fontsize{12}{0} \selectfont in partial fulfilment of the requirements for the degree of}\\
-\vspace*{0.4cm}
+\vspace*{0.3cm}
{ \fontsize{14}{0} \selectfont Bachelor of Science (BSc)}\\
\vspace*{2cm}
{ \fontsize{10}{0} \selectfont Vienna, July 2021}\\
-\vspace*{3.4cm}
+\vspace*{3.0cm}
\begin{tabular}{p{9cm}p{11.25cm}}
\fontsize{10}{0} \selectfont
diff --git a/src/thesis/chapters/backup/acknowledgment.tex b/src/thesis/chapters/backup/acknowledgment.tex
@@ -1,3 +0,0 @@
-
-\section{Acknowledgment}
-\lipsum[1]
diff --git a/src/thesis/chapters/backup/backelectroncg.tex b/src/thesis/chapters/backup/backelectroncg.tex
@@ -1,455 +0,0 @@
-\subsection{Noncommutative Geometry of Electrodynamics}
-In this chapter we describe Electrodynamics with the almost commutative
-manifold $M\times F_X$ and the abelian gauge group $U(1)$.
-We arrive at a unified description of gravity and electrodynamics although in the classical level.
-
-The almost commutative Manifold $M\times F_X$ describes a local gauge group
-$U(1)$. The inner fluctuations of the Dirac operator relate to $Y_\mu$ the
-gauge field of $U(1)$. According to the setup we ultimately arrive at two
-serious problems.
-
-First of all in the Two-Point space $F_X$, the operator $D_F$ must vanish for
-us to have a real structure. However this implies that the electrons
-are massless, which would be absurd.
-
-The second problem arises when looking at the Euclidean action for a free
-Dirac field
-\begin{align}
- S = - \int i \bar{\psi}(\gamma ^\mu\partial _\mu - m) \psi d^4x,
-\end{align}
-where $\psi,\ \bar{\psi}$ must be considered as independent variables, which
-means that the fermionic action $S_f$ needs two independent Dirac spinors.
-Let us try and construct two independent Dirac spinors with our data. To do
-this we take a look at the decomposition of the basis and of the total
-Hilbertspace $H = L^2(S) \otimes H_F$. For the orthonormal basis of $H_F$ we
-can write $\{e, \bar{e}\}$ , where $\{e\}$ is the orthonormal basis of
-$H_F^+$ and $\{\bar{e}\}$ the orthonormal basis of $H_F^-$. Accompanied with
-the real structure we arrive at the following relations
-\begin{align}
- J_F e &= \bar{e} \;\;\;\;\;\; J_F \bar{e} = e, \\
- \gamma_F e &= e \;\;\;\;\;\; \gamma_F \bar{e} = \bar{e}.
-\end{align}
-Along with the decomposition of $L^2(S) = L^2(S)^+ \oplus L^2(S)^-$ and $\gamma = \gamma _M
-\otimes \gamma _F$ we can obtain the positive eigenspace
-\begin{align}
- H^+ = L^2(S)^+ \otimes H_F^+ \oplus L^(S)^- \otimes H_F^-.
-\end{align}
-So, for a $\xi \in H^+$ we can write
-\begin{align}
- \xi = \psi _L \otimes e + \psi _R \otimes \bar{e}
-\end{align}
-where $\psi_L \in L^2(S)^+$ and $\psi _R \in L^2(S)^-$ are the two Wheyl
-spinors. We denote that $\xi$ is only determined by one Dirac spinor $\psi :=
-\psi_L + \psi _R$, \textbf{but we require two independent spinors}. Our
-conclusion is that the definition of the fermionic action gives too much
-restrictions to the Two-Point space $F_X$.
-\subsubsection{The Finite Space}
-To solve the two problems we simply enlarge (double) the Hilbertspace. This
-is visualized by introducing multiplicities in Krajewski Diagrams which will also
-allow us to choose a nonzero Dirac operator that will connect the two
-vertices and preserve real structure making our particles massive and
-bringing anti-particles into the mix.
-
-We start of with the same algebra $C^\infty(M, \mathbb{C}^2)$, corresponding
-to space $N= M\times X$. The Hilbertspace describes four particles, meaning
-it has four orthonormal basis elements. It describes \textbf{left handed
-electrons} and \textbf{right handed positrons}. Pointing this out, we have
-$\{ \underbrace{e_R, e_L}_{\text{left-handed}}, \underbrace{\bar{e}_R,
-\bar{e}_L}_{\text{right-handed}}\}$ the orthonormal basis for $H_F =
-\mathbb{C}^4$. Accompanied with the real structure $J_F$, which allows us to
-interchange particles with antiparticles by the following equations
-\begin{align}
- &J_F e_R = \bar{e}_R, \\
- &J_F e_L = \bar{e_L}, \\
- \nonumber \\
- &\gamma _F e_R = -e_R,\\
- &\gamma_F e_L = e_L,
-\end{align}
-where $J_F$ and $\gamma_F$ have to following properties
-\begin{align}
- &J_F^2 = 1,\\
- & J_F \gamma_F = - \gamma_F J_F.
-\end{align}
-By means of $\gamma_F$ we have two options to decompose the total
-Hilbertspace $H$, firstly into
-\begin{align}
- H_F = \underbrace{H_F^+}_{\text{ONB } \{e_L, \bar{e}_L\}}
- \oplus \underbrace{H_F^-}_{\text{ONB } \{e_R, \bar{e}_R\}},
-\end{align}
-or alternatively into the eigenspace of particles and their
-antiparticles (electrons and positrons) which is preferred in literature and
-which we will use going further
-\begin{align}
- H_F = \underbrace{H_{e}}_{\text{ONB } \{e_L, e_R\}} \oplus
- \underbrace{H_{\bar{e}}}_{\text{ONB } \{\bar{e}_L, \bar{e}_R\}}.
-\end{align}
-Here ONB means orthonormal basis.
-
-The action of $a \in A = \mathbb{C}^2$ on $H$ with respect to the ONB
-$\{e_L, e_R, \bar{e}_L, \bar{e}_R\}$ is represented by
-\begin{align}\label{eq:leftrightrepr}
- a =
- (a_1 , a_2 ) \mapsto
- \begin{pmatrix}
- a_1 &0 &0 &0\\
- 0&a_1 &0 &0\\
- 0 &0 &a_2 &0\\
- 0 &0 &0 &a_2\\
- \end{pmatrix}
-\end{align}
-Do note that this action commutes wit the grading and that
-$[a, b^\circ] = 0$ with $b:= J_F b^*J_F$ because both the left and the right
-action is given by diagonal matrices by equation \eqref{eq:leftrightrepr}. Note
-that we are still left with $D_F = 0$ and the following spectral
-triple
-\begin{align}\label{eq:fedfail}
- \left( \mathbb{C}^2, \mathbb{C}^2, D_F=0; J_F =
- \begin{pmatrix}
- 0 & C \\ C &0
- \end{pmatrix},
- \gamma _F =
- \begin{pmatrix}
- 1 & 0 \\ 0 &-1
- \end{pmatrix}
- \right).
- \end{align}
-It can be represented in the following Krajewski diagram,
-with two nodes of multiplicity two
- \begin{figure}[H] \centering
- \begin{tikzpicture}[
- dot/.style = {draw, circle, inner sep=0.06cm},
- bigdot/.style = {draw, circle, inner sep=0.09cm},
- no/.style = {},
- ]
- \node[no](a) at (0,0) [label=left:$\textbf{1}^\circ$] {};
- \node[no](b) at (0, -1) [label=left:$\textbf{1}^\circ$] {};
- \node[no](c) at (0.5, 0.5) [label=above:$\textbf{1}$] {};
- \node[no](d) at (1.5, 0.5) [label=above:$\textbf{1}$] {};
- \node[dot](d0) at (1.5,0) [] {};
- \node[dot](d0) at (0.5,-1) [] {};
- \node[bigdot](d0) at (1.5,0) [] {};
- \node[bigdot](d0) at (0.5,-1) [] {};
- \end{tikzpicture}
- \end{figure}
-\subsubsection{A noncommutative Finite Dirac Operator}
-To extend our spectral triple with a non-zero Operator, we need to take a
-closer look at the Krajewski diagram above. Notice that edges only exist
-between multiple vertices, meaning we can construct a Dirac operator mapping
-between the two vertices. The operator can be represented by the following matrix
-\begin{align}\label{eq:feddirac}
- D_F =
- \begin{pmatrix}
- 0 & d & 0 & 0 \\
- \bar{d} & 0 & 0 & 0 \\
- 0 & 0 & 0 & \bar{d} \\
- 0 & 0 & d & 0
- \end{pmatrix}
-\end{align}
-We can now define the finite space $F_{ED}$.
-\begin{align}
- F_{ED} := (\mathbb{C}^2, \mathbb{C}^4, D_F; J_F, \gamma_F)
-\end{align}
-where $J_F$ and $\gamma_F$ are like in equation \eqref{eq:fedfail} and $D_F$
-from equation \eqref{eq:feddirac}.
-
-\subsubsection{Almost commutative Manifold of Electrodynamics}
-The almost commutative manifold $M\times F_{ED}$ has KO-dimension 2, and is
-represented by the following spectral triple
-\begin{align}\label{eq:almost commutative manifold}
- M\times F_{ED} := \big(C^\infty(M,\mathbb{C}^2),\ L^2(S)\otimes
- \mathbb{C}^4,\
- D_M\otimes 1 +\gamma _M \otimes D_F;\; J_M\otimes J_F,\ \gamma_M\otimes
- \gamma _F\big)
-\end{align}
-The algebra didn't change, thus we can decompose it like before
-\begin{align}
- C^\infty(M, \mathbb{C}^2) = C^\infty (M) \oplus C^\infty (M)
-\end{align}
-As for the Hilbertspace, we can decomposition it in the following way
-\begin{align}
- H = (L^2(S) \otimes H_e ) \oplus (L^2(S) \otimes H_{\bar{e}}).
-\end{align}
-Note that the one component of the algebra is acting on $L^2(S) \otimes H_e$,
-and the other one acting on $L^2(S) \otimes H_{\bar{e}}$. In other words the components of
-the decomposition of both the algebra and the Hilbertspace match by the action of
-the algebra.
-
-The derivation of the gauge theory is the same for $F_{ED}$ as for the
-Two-Point space $F_X$. We have $\mathfrak{B}(F) \simeq U(1)$ and for an
-arbitrary gauge field $B_\mu = A_\mu - J_F A_\mu J_F^{-1}$ we can write
-\begin{align} \label{field}
- B_\mu =
- \begin{pmatrix}
- Y_\mu & 0 & 0 & 0 \\
- 0 & Y_\mu& 0 & 0 \\
- 0 & 0 & Y_\mu& 0 \\
- 0 & 0 & 0 & Y_\mu
- \end{pmatrix} \;\;\;\;\;\ \text{for} \;\;\ Y_\mu (x) \in \mathbb{R}.
-\end{align}
-We have one single $U(1)$ gauge field $Y_\mu$, carrying the action of the
-gauge group
-\begin{align}
- \text{$\mathfrak{B}$}(M\times F_{ED}) \simeq C^\infty(M, U(1))
-\end{align}
-
-The space $N = M\times X$ consists of two copies of $M$.
-If $D_F = 0$ we have infinite distance between the two copies. Now have
-hacked the spectral triple to have nonzero Dirac operator $D_F$. The new
-Dirac operator still has a commuting relation with the algebra $[D_F, a] = 0$
-$\forall a \in A$, and we should note that the distance between the two
-copies of $M$ is still infinite. This is purely an mathematically abstract
-observation and doesn't affect physical results.
-
-\subsubsection{Spectral Action}
-In this chapter we bring all our results together to establish an
-Action functional to describe a physical system. It turns out that
-the Lagrangian of the almost commutative manifold $M\times F_{ED}$
-corresponds to the Lagrangian of Electrodynamics on a curved
-background manifold (+ gravitational Lagrangian), consisting of the spectral
-action $S_b$ (bosonic) and of the fermionic action $S_f$.
-
-The simplest spectral action of a spectral triple $(A, H, D)$ is given by the
-trace of a function of $D$. We also consider inner fluctuations of the Dirac
-operator $D_\omega = D + \omega + \varepsilon' J\omega J^{-1}$ where $\omega =
-\omega ^* \in \Omega_D^1(A)$.
-\begin{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}\big(f(\frac{D_\omega}{\Lambda})\big)
- \end{align}
- where $\Lambda$ is a real cutoff parameter. The minimal condition on $f$
- is that $f(\frac{D_\omega}{\Lambda})$ is a trace class operator. A trace
- class operator is a compact operator with a well defined finite trace
- independent of the basis. The subscript $b$ in $S_b$ stands for bosonic,
- because in physical applications $\omega$ will describe bosonic fields.
-
- In addition to the bosonic action $S_b$ we can define a topological spectral
- action $S_{top}$. Leaning on the grading $\gamma$ the topological spectral action is
- \begin{align}
- S_{\text{top}}[\omega] := \text{Tr}(\gamma\
- f(\frac{D_\omega}{\Lambda})).
- \end{align}
-\end{definition}
-\begin{definition}\label{def:fermionic action}
- The fermionic action is defined by
- \begin{align}
- S_f[\omega, \psi] = (J\tilde{\psi}, D_\omega \tilde{\psi})
- \end{align}
- with $\tilde{\psi} \in H_{cl}^+ := \{\tilde{\psi}: \psi \in H^+\}$, where
- $H_{cl}^+$ is a set of Grassmann variables in $H$ in the +1-eigenspace
- of the grading $\gamma$.
-\end{definition}
-
-%---------------------- APPENDIX ?????????????--------------------
-\textbf{APPENDIX??}
-Grassmann variables are a set of Basis vectors of a vector space, they
-form a unital algebra over a vector field $V$, where the generators are
-anti commuting, that is for Grassmann variables $\theta _i, \theta _j$ we have
-\begin{align}
- &\theta _i \theta _j = -\theta _j \theta _i \\
- &\theta _i x = x\theta _j \;\;\;\; x\in V \\
- &(\theta_i)^2 = 0 \;\;\; (\theta _i \theta _i = -\theta _i \theta _i)
-\end{align}
-%---------------------- APPENDIX ?????????????--------------------
-\begin{proposition}
- The spectral action of the almost commutative manifold $M$ with $\dim(M)
- =4$ with a fluctuated Dirac operator is
- \begin{align}
- \text{Tr}(f\frac{D_\omega}{\Lambda}) \sim \int_M \mathcal{L}(g_{\mu\nu},
- B_\mu, \Phi) \sqrt{g}\ d^4x + O(\Lambda^{-1}),
- \end{align}
- where
- \begin{align}
- \mathcal{L}(g_{\mu\nu}, B_\mu, \Phi) =
- N\mathcal{L}_M(g_{\mu\nu})
- \mathcal{L}_B(B_\mu)+
- \mathcal{L}_\phi(g_{\mu\nu}, B_\mu, \Phi).
- \end{align}
- The Lagrangian $\mathcal{L}_M$ is of the spectral triple, represented by
- the following term
- $(C^\infty(M) , L^2(S), D_M)$
- \begin{align}\label{lagr}
- \mathcal{L}_M(g_{\mu\nu}) := \frac{f_4 \Lambda ^4}{2\pi^2} -
- \frac{f_2 \Lambda^2}{24\pi ^2}s - \frac{f(0)}{320\pi^2} C_{\mu\nu
- \varrho \sigma}C^{\mu\nu \varrho \sigma},
- \end{align}
- here $C^{\mu\nu \varrho \sigma}$ is defined in terms of the Riemannian
- curvature tensor $R_{\mu\nu \varrho \sigma}$ and the Ricci tensor
- $R_{\nu\sigma} = g^{\mu\varrho} R_{\mu\nu \varrho\sigma}$.
- The kinetic term of the gauge field is described by the Lagrangian
- $\mathcal{L}_B$, which takes the following shape
- \begin{align}
- \mathcal{L}_B(B_\mu) := \frac{f(0)}{24\pi^2}
- \text{Tr}(F_{\mu\nu}F^{\mu\nu}).
- \end{align}
- Lastly $\mathcal{L}_\phi$ is the scalar-field Lagrangian with a boundary
- term, given by
- \begin{align}
- \mathcal{L}_\phi(g_{\mu\nu}, B_\mu, \Phi) :=
- &-\frac{2f_2\Lambda^2}{4\pi^2}\text{Tr}(\Phi^2) + \frac{f(0)}{8\pi^2}
- \text{Tr}(\Phi^4) + \frac{f(0)}{24\pi^2}
- \Delta(\text{Tr}(\Phi^2))\nonumber\\
- &+ \frac{f(0)}{48\pi^2}s\text{Tr}(\Phi^2)
- \frac{f(0)}{8\pi^2}\text{Tr}((D_\mu \Phi)(D^\mu \Phi)).
- \end{align}
-\end{proposition}
-\begin{proof}
- The dimension of the manifold $M$ is $\dim(M) = \text{Tr}(id) =4$. For
- an $x \in M$, we have an asymptotic expansion of the term
- $\text{Tr}(f(\frac{D_\omega}{\Lambda}))$ as $\Lambda$ goes to infinity,
- which can be written as
- \begin{align}
- \text{Tr}(f(\frac{D_\omega}{\Lambda})) \simeq& \ 2f_4 \Lambda ^4
- a_0(D_\omega ^2)+ 2f_2\Lambda^2 a_2(D_\omega^2)\nonumber \\&+ f(0) a_4(D_\omega^4)
- +O(\Lambda^{-1}).\label{eq:trheatkernel}
- \end{align}
- We have to note here that the heat kernel coefficients are zero for uneven $k$,
- and they are dependent on the fluctuated Dirac operator
- $D_\omega$. We can rewrite the heat kernel coefficients in terms of $D_M$,
- for the first two terms $a_0$ and $a_2$ we use $N:=
- \text{Tr}\mathbbm{1_{H_F}})$ and write
- \begin{align}
- a_0(D_\omega^2) &= Na_0(D_M^2),\\
- a_2(D_\omega^2 &= Na_2(D_M^2) - \frac{1}{4\pi^2}\int_M
- \text{Tr}(\Phi^2)\sqrt{g}d^4x.
- \end{align}
- For $a_4$ we extend in terms of coefficients of $F$, \textbf{REWRITE: look week9.pdf
- for the standard version}
- \begin{align}
- &\frac{1}{360}\text{Tr}(60sF)= -\frac{1}{6}S(Ns + 4
- \text{Tr}(\Phi^2))\\
- \nonumber\\
- &F^2 = \frac{1}{16}s^2\otimes 1 + 1\otimes \Phi^4 - \frac{1}{4}
- \gamma^\mu\gamma^\nu \gamma^\varrho\gamma^\sigma F_{\mu\nu}F^{\mu\nu}+\\
- &\;\;\;\;\;\;\;+\gamma^\mu\gamma^\nu\otimes(D_\mu\Phi)(D_\nu
- \Phi)+\frac{1}{2}s\otimes \Phi^2 + \ \text{traceless terms}\\
- \nonumber\\
- &\frac{1}{360}\text{Tr}(180F^2) = \frac{1}{8}s^2N + 2\text{Tr}(\Phi^4)
- + \text{Tr}(F_{\mu\nu}F^{\mu\nu}) +\\
- &\;\;\;\;\;\;\;+2\text{Tr}((D_\mu\Phi)(D^\mu\Phi))
- + s\text{Tr}(\Phi^2)\\
- \nonumber\\
- &\frac{1}{360}\text{Tr}(-60\Delta F)=
- \frac{1}{6}\Delta(Ns+4\text{Tr}(\Phi^2)).
- \end{align}
- The cross terms of the trace in $\Omega_{\mu\nu}^E\Omega^{E\mu\nu}$
- vanishes because of the antisymmetric property of the Riemannian
- curvature tensor, thus we can write
- \begin{align}
- \Omega_{\mu\nu}^E\Omega^{E\mu\nu} = \Omega_{\mu\nu}^S\Omega^{S\mu\nu}
- \otimes 1 - 1\otimes F_{\mu\nu}F^{\mu\nu} + 2i\Omega_{\mu\nu}^S
- \otimes F^{\mu\nu}.
- \end{align}
- The trace of the cross term $\Omega^{S}_{\mu\nu}$ vanishes because
- \begin{align}
- \text{Tr}(\Omega^{S}_{\mu\nu}) = \frac{1}{4}
- R_{\mu\nu\varrho\sigma}\text{Tr}(\gamma^\mu\gamma^\nu) = \frac{1}{4}
- R_{\mu\nu\varrho\sigma}g^{\mu\nu} =0,
- \end{align}
- then the trace of the whole term is given by
- \begin{align}
- \frac{1}{360}\text{Tr}(30\Omega^E_{\mu\nu}\Omega^{E\mu\nu}) =
- \frac{N}{24}R_{\mu\nu\varrho\sigma}R^{\mu\nu\varrho\sigma}
- -\frac{1}{3}\text{Tr}(F_{\mu\nu}F^{\mu\nu}).
- \end{align}
- Finally plugging the results into the coefficient $a_4$ and simplifying we get
- \begin{align}
- a_4(x, D_\omega^4) &= Na_4(x, D_M^2) + \frac{1}{4\pi^2}\bigg(\frac{1}{12} s
- \text{Tr}(\Phi^2) + \frac{1}{2}\text{Tr}(\Phi^4) \nonumber \\
- &+ \frac{1}{4}
- \text{Tr}((D_\mu\Phi)(D^\mu \Phi)) + \frac{1}{6}
- \Delta\text{Tr}(\Phi^2) + \frac{1}{6}
- \text{Tr}(F_{\mu\nu}F^{\mu\nu})\bigg).
- \end{align}
- The only thing left is to substitute the heat kernel coefficients into the
- heat kernel expansion in equation \eqref{eq:trheatkernel}.
-\end{proof}
-
-\subsubsection{Fermionic Action}
-We remind ourselves the definition of the fermionic action in definition
-\ref{def:fermionic action} and the manifold we are dealing with in equation
-\eqref{eq:almost commutative manifold}. The Hilbertspace $H_F$ is separated
-into the particle-antiparticle states with ONB $\{e_R, e_L, \bar{e}_R,
-\bar{e}_L\}$. The orthonormal basis of $H_F^+$ is $\{e_L, \bar{e}_R\}$ and
-consequently for $H_F^-$, $\{e_R, \bar{e}_L\}$. We can decompose a spinor
-$\psi \in L^2(S)$ in each of the eigenspaces $H_F^\pm$, $\psi = \psi_R+
-\psi_L$. That means for an arbitrary $\psi \in H^+$ we can write
-\begin{align}
- \psi = \chi_R \otimes e_R + \chi_L \otimes e_L + \psi_L \otimes
- \bar{e}_R+
- \psi_R \otimes \bar{e}_L,
-\end{align}
-where $\chi_L, \psi_L \in L^2(S)^+$ and $\chi_R, \psi_R \in L^2(S)^-$.
-
-Since the fermionic action yields too much restriction on $F_{ED}$ (modified
-Two-Point space $F_X$) we redefine it by taking account the fluctuated Dirac
-operator
-\begin{align}
- D_\omega = D_M \otimes i + \gamma^\mu \otimes B_\mu + \gamma_M \otimes
- D_F.
-\end{align}
-The Fermionic Action is
-\begin{align}
-S_F = (J\tilde{\xi}, D_\omega\tilde{\xi})
-\end{align}
-for a $\xi \in H^+$. Then the straight forward calculation gives \begin{align}
- \frac{1}{2}(J\tilde{\xi}, D_\omega\tilde{\xi})
- &=\frac{1}{2}(J\tilde{\xi}, (D_M \otimes
- i)\tilde{\xi})\label{eq:fermionic1}\\
- &+\frac{1}{2}(J\tilde{\xi}, (\gamma^\mu \otimes B_\mu)
- \tilde{\xi})\label{eq:fermionic2}\\
- &+\frac{1}{2}(J\tilde{\xi}, (\gamma_M\otimes
- D_F)\tilde{\xi})\label{eq:fermionic3},
-\end{align}
-(note that we add the constant $\frac{1}{2}$ to the action).
-For the term in \eqref{eq:fermionic1} we calculate
-\begin{align}
- \frac{1}{2}(J\tilde{\xi}, (D_M\otimes 1)\tilde{\xi}) &=
- \frac{1}{2}(J_M\tilde{\chi}_R,D_M\tilde{\psi}_L)+\nonumber
- \frac{1}{2}(J_M\tilde{\chi}_L,D_M\tilde{\psi}_R)+
- \\&+\frac{1}{2}(J_M\tilde{\psi}_L,D_M\tilde{\psi}_R)+\nonumber
- \frac{1}{2}(J_M\tilde{\chi}_R,D_M\tilde{\chi}_L)\\
- &= (J_M\tilde{\chi},D_M\tilde{\chi}).
-\end{align}
-For the term in \eqref{eq:fermionic2} we have
-\begin{align}
- \frac{1}{2}(J\tilde{\xi}, (\gamma^\mu \otimes B_\mu)\tilde{\xi})&=
- -\frac{1}{2}(J_M\tilde{\chi}_R, \gamma^\mu Y_\mu\tilde{\psi}_R)
- -\frac{1}{2}(J_M\tilde{\chi}_L, \gamma^\mu Y_\mu\tilde{\psi}_R)+\nonumber\\
- &+\frac{1}{2}(J_M\tilde{\psi}_L, \gamma^\mu Y_\mu\tilde{\chi}_R)+
- \frac{1}{2}(J_M\tilde{\psi}_R, \gamma^\mu Y_\mu\tilde{\chi}_L)=\nonumber\\
- &= -(J_M\tilde{\chi}, \gamma^\mu Y_\mu\tilde{\psi}).
-\end{align}
-And for \eqref{eq:fermionic3} we can write
-\begin{align}
- \frac{1}{2}(J\tilde{\xi}, (\gamma_M\otimes D_F)\tilde{\xi})&=
- +\frac{1}{2}(J_M\tilde{\chi}_R, d\gamma_M\tilde{\chi}_R)
- +\frac{1}{2}(J_M\tilde{\chi}_L, \bar{d}\gamma_M\tilde{\chi}_L)+\nonumber\\
- &+\frac{1}{2}(J_M\tilde{\chi}_L, \bar{d}\gamma_M\tilde{\chi}_L)
- +\frac{1}{2}(J_M\tilde{\chi}_R, d\gamma_M\tilde{\chi}_R)=\nonumber\\
- &= i(J_M\tilde{\chi}, m\tilde{\psi}).
-\end{align}
-A small problem arises, we obtain a complex mass parameter $d$, but we can
-write $d:=im$ for $m\in \mathbb{R}$, which stands for the real mass.
-
-Finally the fermionic action of $M\times F_{ED}$ takes the form
- \begin{align}
- S_f = -i\big(J_M\tilde{\chi}, \gamma(\nabla^S_\mu - i\Gamma_\mu)
- \tilde{\Psi}\big) + \big(S_M\tilde{\chi}_L, \bar{d}\tilde{\psi}_L\big) -
- \big(J_M\tilde{\chi}_R, d \tilde{\psi}_R\big).
- \end{align}
-Ultimately we arrive at the full Lagrangian of $M\times F_{ED}$, which is the
-sum of purely gravitational Lagrangian
- \begin{align}
- \mathcal{L}_{grav}(g_{\mu\nu})=4\mathcal{L}_M(g_{\mu\nu})+
- \mathcal{L}_\phi (g_{\mu\nu}),
- \end{align}
-and the Lagrangian of electrodynamics
- \begin{align}
- \mathcal{L}_{ED} = -i\bigg\langle
- J_M\tilde{\chi},\big(\gamma^\mu(\nabla^S_\mu - iY_\mu) -m\big)\tilde{\psi})
- \bigg\rangle
- +\frac{f(0)}{6\pi^2} Y_{\mu\nu}Y^{\mu\nu}.
- \end{align}
-
diff --git a/src/thesis/chapters/backup/basics.tex b/src/thesis/chapters/backup/basics.tex
@@ -1,553 +0,0 @@
-\subsection{Noncommutative Geometric Spaces}
-\subsubsection{$*$-Algebra}
-To grasp the idea of encoding geometrical data into a spectral triple we
-introduce the first ingredient of a spectral triple, an unital $*$ algebra.
-\begin{mydefinition}
- A \textit{vector space} $A$ over $\mathbb{C}$ is called a \textit{complex, unital Algebra} if, \\
- $\forall a,b \in A$ :
- \begin{align}
- A \times A \rightarrow A\\
- (a,\ b)\ &\mapsto \ a\cdot b,
- \end{align}
- with an identity element:
- \begin{align}
- 1a = a1 =a.
- \end{align}
- Extending the definition, a $*$-algebra is an algebra $A$ with a \textit{conjugate linear map (involution)} $*:A\ \rightarrow A$,
- $\forall a, b \in A$ satisfying
- \begin{align}
- (a\ b)^* &= b^*a^*,
- (a^*)^* &= a.
- \end{align}
-\end{mydefinition}
-In the following all unital algebras are referred to as algebras.
-
-\subsubsection{Finite Discrete Space}
-Let us consider an example of an $*$-algebra of continuous functions $C(X)$
-on a discrete topological space $X$ with $N$ points. Functions of a
-continuous $*$-algebra $C(X)$ assign values to $\mathbb{C}$, thus $f,\ g \in
-C(X)$, $\lambda \in \mathbb{C}$ and $x \in X$ they provide the following structure:
-\begin{itemize}
- \item \textit{pointwise linear} \\
- $(f + g)(x) = f(x) + g(x)$,\\
- $(\lambda\ f)(x) = \lambda (f(x)),$
- \item \textit{pointwise multiplication} \\
- $f\ g\ (x) = f(x)g(x)$,
- \item \textit{pointwise involution} \\
- $f^*(x) = \overline{f(x)}.$
-\end{itemize}
-The $*$-algebra $C(X)$ is \textit{isomorphic} to a $*$-algebra $\mathbb{C}^N$
-with involution ($N$ number of points in $X$), we write $C(X) \simeq
-\mathbb{C}^N$. Isomorphisms are bijective maps that preserve structure and
-don't lose physical information. A function $f:X\ \rightarrow\ \mathbb{C}$
-can be represented with $N \times N$ diagonal matrices, where each diagonal
-value represents the function value at the corresponding $i$-th point for $i
-= 1,...,N$. Because of matrix multiplication and hermitian conjugate of
-matrices we have a preserving structure.
-
-Moreover we can \textit{map} between finite discrete spaces $X_1$ and $X_2$ with a
-function
-\begin{align}
- \phi:\ X_1 \rightarrow\ X_2.
-\end{align}
-For every such map there exists a corresponding map
-\begin{align}
- \phi ^*:C(X_2)\ \rightarrow C(X_1),
-\end{align}
-which `pulls back' values even if $\phi$ is not bijective.
-Note that the pullback doesn't map points back, but maps functions on an $*$-algebra $C(X)$.
-The pullback, in literature often called a $*$-homomorphism or a $*$-algebra map under
-pointwise product has the following properties
-\begin{itemize}
- \item $\phi ^*(f\ g) = \phi ^*(f)\ \phi ^*(g)$,
- \item $\phi ^*(\overline{f}) = \overline{\phi ^*(f)}$,
- \item $\phi ^*(\lambda\ f + g) = \lambda\ \phi ^*(f) + \phi ^*(g)$.
-\end{itemize}
-%------------ Exercise
- The map $\phi :X_1\ \rightarrow \ X_2$ is an injective (surjective) map,
- if only if the corresponding pullback $\phi ^* :C(X_2)\ \rightarrow \
- C(X_1)$ is surjective (injective). Let us say, that $X_1$ has $n$ points and
- $X_2$ with $m$ points. Then there are three different cases, first $n=m$ and
- obviously $\phi$ is bijective and $\phi ^*$ too. Then $n > m$, in this case
- $\phi$ assigns $n$ points to $m$ points when $n > m$, which is by definition
- surjective. On the other hand $\phi ^*$ assigns $m$ points to $n$ points when
- $n > m$, which is by definition injective. Lastly $n < m $, which is
- completely analogous to the case $n > m$.
-%------------ Exercise
-
-\subsubsection{Matrix Algebras}
-\begin{mydefinition}
- A \textit{(complex) matrix algebra} A is a direct sum, for $n_i, N \in
- \mathbb{N}$
- \begin{align}
- A = \bigoplus _{i=1}^{N} M_{n_i}(\mathbb{C}).
- \end{align}
- The involution is the hermitian conjugate, a $*$ algebra with involution is referred to as
- a matrix algebra
-\end{mydefinition}
-From a topological discrete space $X$, we can construct a $*$-algebra
-$C(X)$ which is isomorphic to a matrix algebra $A$. Then the question instantly
-arises, if we can construct $X$ given $A$? For a matrix algebra $A$,
-which in most cases is not commutative, the answer is generally no.
-
-Thus there are two options. We can restrict ourselves to commutative matrix algebras,
-which are the vast minority and not physically interesting.
-Or we can allow more morphisms (isomorphisms) between matrix algebras.
-
-\subsubsection{Finite Inner Product Spaces and Representations}
-Until now we looked at finite topological discrete spaces, moreover we can consider a
-finite dimensional inner product space $H$ (finite Hilbert-spaces), with inner product
-$(\cdot,\cdot)\rightarrow \mathbb{C}$. We denote $L(H)$ as the $*$-algebra of operators on $H$
-equipped with a product given by composition and involution of the adjoint, $T \mapsto T^*$.
-Then $L(H)$ is a \textit{normed vector space} with
-\begin{align}
- \|T\|^2 &= \sup_{h \in H}\big\{(T\ h,\ T\ h): (h,\ h) \leq 1\big|\ T
- \in L(H)\big \},\\
- \|T\| &= \sup\big\{\sqrt{\lambda}:\; \lambda \text{ eigenvalue of } T\big\}.
-\end{align}
-This allows us to define representations of $*$-algebras.
-\begin{mydefinition}
- The \textit{representation} of a finite dimensional $*$-algebra $A$ is a
- pair $(H, \pi)$, where $H$ is a finite dimensional inner product space
- and $\pi$ is a $*$-\textit{algebra map}
- \begin{align}
- \pi:A\ \rightarrow \ L(H).
- \end{align}
- We call the representation $(H, \pi)$ \textit{irreducible} if
- \begin{itemize}
- \item $H \neq \emptyset$,
- \item only $\emptyset$ or $H$ is invariant under the action of $A$ on
- $H$.
- \end{itemize}
-\end{mydefinition}
-Here are some examples of reducible and irreducible representations
-\begin{itemize}
- \item For $A = M_n(\mathbb{C})$ the representation $H=\mathbb{C}^n$, $A$ acts as matrix multiplication\\
- $H$ is irreducible.
- \item For $A = M_n(\mathbb{C})$ the representation $H=\mathbb{C}^n\oplus \mathbb{C}^n$, with $a \in A$ acting
- in block form \\ $\pi: a \mapsto \big(\begin{smallmatrix} a & 0\\ 0 & a \end{smallmatrix}\big)$ is
- reducible.
-\end{itemize}
-Naturally there are also certain equivalences between different
-representations.
-\begin{mydefinition}
-Two representations of a $*$-algebra $A$, $(H_1, \pi _1)$ and
-$(H_2, \pi _2)$ are called \textit{unitary equivalent} if there exists a map
-$U: H_1 \rightarrow H_2$ such that.
- \begin{align}
- \pi _1(a) = U^* \pi _2(a) U
- \end{align}
-\end{mydefinition}
-
-Furthermore we define a mathematical structure called the structure space,
-which will later become important, when speaking of the duality between a spectral
-triple and a space.
-\begin{mydefinition}
- Let $A$ a $*$-algebra then, $\hat{A}$ is called the structure space of all \textit{unitary equivalence classes
- of irreducible representations of A}
-\end{mydefinition}
-%------------- EXERCISE
- Given a representation $(H, \pi)$ of a $*$-algebra $A$, the \textbf{commutant} $\pi (A)'$ of $\pi (A)$ is defined as a set
- of operators in $L(H)$ that commute with all $\pi (a)$
- \begin{align}
- \pi (A)' = \big\{T \in L(H):\ \pi(a)\ T = T\ \pi(a) \;\; \forall a\in
- A\big\}
- \end{align}
- The commutant $\pi (A)'$ is also a $*$-algebra, because it has unital,
- associative and involutive properties.
- We note that $\pi (a) \in L(H)\ \forall a \in A$, unitary property is given
- by the unital operator of the $*$-algebra of operators $L(H)$, which exists
- by definition because H is a inner product space. Associativity is given by
- the $*$-algebra of $L(H)$, where $L(H) \times L(H) \mapsto L(H)$, which is
- associative by definition. The involutive property is also given by the $*$-algebra $L(H)$
- with a map $*: L(H) \mapsto L(H)$ only for a $T$ that commutes with $\pi (a)$.
-%------------- EXERCISE
-
-%------------- EXERCISE
- For a unital algebra $*$-algebra $A$, the matrices $M_n(A)$ with entries
- in $A$ form a unital $*$-algebra, because unitary operation in $M_n(A)$ is given by the identity Matrix, which
- has to exists in every entry in $M_n(A)$, and behaves like in $A$. Associativity is given by
- matrix multiplication. Lastly involution is given by the conjugate transpose.
-
- A representation $\pi :A\ \rightarrow \ L(H)$ of a $*$-algebra $A$, for
- $H^n = H \oplus ... \oplus H$, $n$ times. Then we have the following
- representation $\tilde{\pi}:M_n(A) \rightarrow L(H^n)$ for the Matrix
- Algebra with $\tilde{\pi}((a_{ij})) = (\tilde{\pi}(a_{ij})) \in M_n(A)$.
- We have direct isomorphisms of $A \simeq M_n(A)$ and $H \simeq H^n$
- meaning $\tilde{\pi}$ is a valid reducible representation.
-
- Let $\tilde{\pi}:M_n(A) \rightarrow L(H^n)$ be a $*$ algebra
- representation of $M_n(A)$, then $\pi: A \rightarrow L(H^n)$ is a representation of $A$.
- The fact that $\tilde{\pi}$ and $\pi$ are unitary equivalent, there is
- a map $U: H^n \rightarrow H^n$ given by $U=\mathbbm{1}_n$, thus
- \begin{align}
- \pi (a) &= \mathbbm{1}_n^*\ \tilde{\pi}((a_{ij})), \\
- \mathbbm{1}_n &= \tilde{\pi}((a_{ij})) = \pi (a_{ij})
- \Rightarrow a_{ij} = a\ \mathbbm{1}_n.
- \end{align}
-%------------- EXERCISE
-
-
-A commutative matrix algebra can be used to reconstruct a discrete space.
-The structure space $\hat{A}$ is used for this. Because $A \simeq
-\mathbb{C}^N$ all
-irreducible representation are of the form
-\begin{align}
- \pi _i:(\lambda_1,...,\lambda_N)\in \mathbb{C}^N \mapsto \lambda_i \in
- \mathbb{C}
-\end{align}
-for $i = 1,...,N$ and thus $\hat{A} \simeq \{1,...,N\}$.
-The conclusion is that, there is a duality between discrete spaces and
-commutative matrix algebra this duality is called the \textit{finite
-dimensional Gelfand duality}
-
-Our aim is to construct a duality between finite dimensional spaces and
-\textit{equivalence classes} of matrix algebras, to preserve general
-non-commutativity of matrices. Equivalence classes are described by a
-generalized notion of isomorphisms between matrix algebras (\textit{Morita
-Equivalence})
-
-\subsubsection{Algebraic Modules}
-An important notion for Morita Equivalence are algebraic modules, later
-extended with Hilbert bimodules.
-\begin{mydefinition}
- Let $A$, $B$ be algebras (need not be matrix algebras)
- \begin{enumerate}
- \item \textit{left} A-module is a vector space $E$, that carries a left
- representation of $A$, that is $\exists$ a bilinear map $\gamma: A
- \times E \rightarrow E$ with
- \begin{align}
- (a_1\ a_2)\cdot e = a_1 \cdot (a_2 \cdot e);\;\;\; a_1, a_2 \in
- A, e \in E.
- \end{align}
- \item \textit{right} B-module is a vector space $F$, that carries a
- right representation of $A$, that is there exists a bilinear map
- $\gamma: F \times B \rightarrow F$ with
- \begin{align}
- f \cdot (b_1\ b_2)= (f \cdot b_1) \cdot b_2;\;\;\; b_1, b_2 \in B, f \in F
- \end{align}
- \item \textit{left} A-module and \textit{right} B-module is a
- \textit{bimodule}, a vector space $E$ satisfying
- \begin{align}
- a \cdot (e \cdot b)= (a \cdot e) \cdot b;\;\;\; a \in A, b \in B, e \in E
- \end{align}
- \end{enumerate}
-\end{mydefinition}
-An $A$-\textbf{module homomorphism} as linear map $\phi: E\rightarrow F$ which respects the
-representation of A, e.g.\ for left module.
-\begin{align}
- \phi (a\ e) = a \phi (e); \;\;\; a \in A, e \in E.
-\end{align}
-We will use the notation
-\begin{itemize}
- \item ${}_A E$, for left $A$-module $E$;
- \item ${}_A E_B$, for right $B$-module $F$;
- \item ${}_A E_B$, for $A$-$B$-bimodule $E$, simply bimodule.
-\end{itemize}
-%------------------- EXERCISE
-From a simple observation, we see that an arbitrary representation $\pi : A
-\rightarrow L(H)$ of a $*$-algebra A, turns H into a left module ${}_A H$. If
-$_A H$ than $(a_1\ a_2) h = a_1 (a_2\ h)$ for $a_1, a_2 \in A$ and $h \in H$. We
-take the representation of an $a \in A$, $\pi (a)$, and write
-\begin{align}
- \big(\pi(a_1)\ \pi(a_2)\big)h = \pi(a_1)\big(\pi(a_2)\ h\big) =
- \big(T_1\ T_2\big) h = T_1 \big(T_2\ h\big)
-\end{align}
-For $T_1, T_2 \in L(H)$, which operate naturally from the left on $h$.
-
-%------------------- EXERCISE
-%------------------- EXERCISE
-
-Furthermore notice that that an $*$-algebra $A$ is a bimodule ${}_A A_A$ with
-itself, given by the map
-\begin{align}
- \gamma: A\times A\times A \rightarrow A,
-\end{align}
-which is the inner product of a $*$-algebra.
-%------------------- EXERCISE
-
-\subsubsection{Balanced Tensor Product and Hilbert Bimodules}
-
-\begin{mydefinition}
- Let $A$ be an algebra, $E$ be a \textit{right} $A$-module and $F$ be a
- \textit{left} $A$-module. The \textit{balanced tensor product} of $E$ and
- $F$ forms a $A$-bimodule.
- \begin{align}
- E \otimes _A F := E \otimes F / \left\{\sum _i e_i a_i \otimes f_i -
- e_i \otimes a_i f_i : \;\;\; a_i \in A,\ e_i \in E,\ f_i \in F
- \right\}.
- \end{align}
-\end{mydefinition}
-The $/$ denotes the quotient space. By that the operation $\otimes _A$ takes
-two left/right modules and makes a bimodule with the help the tensor product of
-the two modules and the quotient space that takes out all the elements from the
-tensor product that don't preserver the left/right representation and that are
-duplicates.
-\begin{mydefinition}
- Let $A$, $B$ be \textit{matrix algebras}. The \textit{Hilbert bimodule} for
- $(A, B)$ is given by an $A$-$B$-bimodue $E$ and by an $B$-valued
- \textit{inner product} $\langle \cdot,\cdot\rangle_E: E\times E \rightarrow
- B$, which satisfies the following conditions for $e, e_1, e_2 \in
- E,\ a \in A$ and $b \in B$
-\begin{align}
- \langle e_1,\ a\cdot e_2\rangle_E &= \langle a^*\cdot e_1,\ e_2\rangle_E
- \;\;\;\; & \text{sesquilinear in $A$},\\
- \langle e_1,\ e_2 \cdot b\rangle_E
- &= \langle e_1,\ e_2\rangle_E b \;\;\;\; & \text{scalar in $B$},\\
- \langle e_1,\ e_2\rangle_E &= \langle e_2,\ e_1\rangle^*_E \;\;\;\; &
- \text{hermitian}, \\
- \langle e,\ e\rangle_E &\ge 0 \;\;\;\; & \text{equality
- holds iff $e=0$}.
-\end{align}
-We denote $KK_f(A,\ B)$ as the set of all \textit{Hilbert bimodules} of $(A,\ B)$.
-\end{mydefinition}
-%-------------- EXERCISE
-
-And indeed the Hilbert bimodule extension takes a representation $\pi:\ A \
-\rightarrow L(H)$ of a matrix algebra $A$ and turns $H$ into a Hilbert bimodule for
-$(A, \mathbb{C})$, because the representation of $a \in A$, $\pi(a)=T \in L(H)$ fulfills
-the conditions of the $\mathbb{C}$-valued inner product for $h_1, h_2 \in H$
-\begin{itemize}
- \item $\langle h_1,\ \pi(a)\ h_2\rangle _\mathbb{C} = \langle h_1,\ T\ h_2\rangle _\mathbb{C} =
- \langle T^* h_1, h_2\rangle _\mathbb{C}$, $T^*$ given by the adjoint,
- \item $\langle h_1,\ h_2\ \pi(a)\rangle _\mathbb{C} = \langle h_1,\ h_2\
- T\rangle _\mathbb{C} = \langle h_1,\ h_2\rangle _\mathbb{C}$ , $T$ acts
- from the left,
- \item $\langle h_1,\ h_2\rangle _\mathbb{C}^* = \langle h_2,\ h_1\rangle _\mathbb{C}$, hermitian because of the
- $\mathbb{C}$-valued inner product
- \item $\langle h_1,\ h_2\rangle \ge 0$, $\mathbb{C}$-valued inner product.
-\end{itemize}
-%-------------- EXERCISE
-
-%-------------- EXERCISE
-Take again the $A-A$ bimodule given by an $*$-algebra $A$, it is in
-$KK_f(A,\ A)$. This becomes clear by looking at the following inner product
- $\langle \cdot,\cdot\rangle_A:A \times A \rightarrow A$:
- \begin{align}
- \langle a,\ a\rangle_A = a^*a' \;\;\;\; a,a'\in A. \label{eq:inner-product}
- \end{align}
- Simply checking the conditions in $\langle \cdot, \cdot\rangle _A$ for
- $a, a_1, a_2 \in A$
- \begin{align}
- \langle a_1,\ a\cdot a_2\rangle _A &= a^* a\cdot a_2 =
- (a^*a_1)^*\ a_2 = \langle a^*\ a_1,\ a_2\rangle, \\
- \langle a_1,\ a_2 \cdot a\rangle _A &= a^*_1\ (a_2\cdot a) =
- (a^*a_2)\cdot a = \langle a_1,\ a_2\rangle _A\ a,\\
- \langle a_1,\ a_2\rangle _A^* &= (a_1^*\ a_2)^* = a_2^*\
- (a_1^*)^* = a_2^*\ a_1 = \langle a_2,\ a_1\rangle.
- \end{align}
-
-%-------------- EXERCISE
-
-%-------------- EXAMPLE
-As an exemplar for overview consider a $*$ homomorphism between two matrix
-algebras $\phi:A\rightarrow B$, we can construct a Hilbert bimodule
-$E_{\phi} \in KK_f(A, B)$ in the following way. We let $E_{\phi}$ be $B$ in
-as an vector space and an inner product from above in equation
-\eqref{eq:inner-product}, with $A$ acting on the left with $\phi$.
-\begin{align}
- a\cdot b = \phi(a)\ b
-\end{align}
-for $a\in A, b\in E_{\phi}$.
-%-------------- EXAMPLE
-
-\subsubsection{Kasparov Product and Morita Equivalence}
-\begin{mydefinition}
- Let $E \in KK_f(A, B)$ and $F \in KK_F(B, D)$ the \textit{Kasparov product} is defined as
- with the balanced tensor product
- \begin{align}
- F \circ E := E \otimes _B F.
- \end{align}
- Then $F\circ E \in KK_f(A,D)$ is equipped with a $D$-valued inner product
- \begin{align}
- \langle e_1 \otimes f_1,\ e_2 \otimes f_2\rangle _{E\otimes _B F} =
- \langle f_1,\langle e_1,\ e_2\rangle _E f_2\rangle _F
- \end{align}
-\end{mydefinition}
-
-%-------------- EXERCISE
-The Kasparov product for $*$-algebra homomorphism $\phi: A \rightarrow B$ and
-$\psi: B \rightarrow C$ are isomorphisms in the sense of
-\begin{align}
- E_{\psi} \circ E_{\phi}\ \equiv\ E_{\phi} \otimes _B E_{\psi}\
- \simeq\
- E_{\psi \circ \phi} \in KK_f(A,C).
-\end{align}
-
-In the direct computation for elements $a \in A$, $b\in B$, and $c\in C$ which
-is $\psi \circ \phi$ gives us
-\begin{align}
-a \cdot b \cdot c = \psi(\phi (a) \cdot b) \cdot c
-\end{align}
-An interesting case arises when looking at $E_{\text{id}_A} \simeq A \in KK_f(A,A)$ for
-$\text{id}_A: A \rightarrow A$. This is obvious when we let $E_{\phi}$ be $A$
-with a natural right representation. It follows that $E_{\phi}\simeq A$, thus
-an inner product, acting from the left on $A$ for $\phi$, $a', a\in A$ reads
-\begin{align}
- a'\ a = (\phi(a')\ a) \in A,
-\end{align}
-which is satisfied by $\phi = \text{id}_A$
-
-\begin{mydefinition}
- Let $A$, $B$ be \textit{matrix algebras}. They are called \textit{Morita equivalent} if there
- exists an $E \in KK_f(A, B)$ and an $F \in KK_f(B, A)$ such that
- \begin{align}
- E \otimes _B F \simeq A \;\;\; \text{and} \;\;\; F \otimes _A E \simeq
- B,
- \end{align}
- where $\simeq$ denotes the isomorphism between Hilbert bimodules and note
- that $A$ or $B$ is a bimodule by itself.
-\end{mydefinition}
-
-The modules $E$ and $F$ are each others inverse in regards to the Kasparov
-Product, because we land in the same space as we started. To clarify, in
-the definition we have $E \in KK_f(A, B)$. We start from $A$ and $E \otimes
-_B F$, which lands in $A$. Oppositely we have $F \in KK_f(B, D)$ we start
-from $B$ and $F \otimes _A E$, which lands in $B$.
-
-
-%------------- EXERCISE
-By definition $E \otimes _B F$ is a $A-D$ bimodule. Since
-\begin{align}
- E \otimes _B F = E \otimes F / \bigg\{\sum_i\ e_i\ b_i \otimes f_i - e_i
- \otimes b_i\ f_i\ \big|\;\; e_i \in E_i,\ b_i \in B,\ f_i \in F\bigg\},
-\end{align}
-the last part takes out all tensor product elements of $E$ and $F$ that don't
-preserver the left/right representation and that are duplicates.
-
-Additionally $\langle \cdot,\cdot\rangle _{E\oplus _B F}$ defines a $D$ valued
-inner product, as $\langle e_1,\ e_2\rangle _E \in B$ and $\langle f_1,\ f_2\rangle _F \in C$ by
-definition. So for $\langle e_1,\ e_2\rangle _E =b$ we have
-\begin{align}
- \langle e_1 \otimes f_1,\ e_2 \otimes f_2\rangle _{E\otimes _B F} = \langle
- f_1,\ \langle e_1,\ e_2\rangle _E\ f_2\rangle _F = \langle f_1,\ b\ f_2\rangle _F \in C
-\end{align}
-%------------- EXERCISE
-%------------- EXAMPLE
-Picking up the example of $(A, A)$, the Hilbert bimodule $A$, we can
-consider an $E \in KK_f(A,B)$ for
-\begin{align}
- E \circ A = A\oplus _A E \simeq E.
-\end{align}
-We conclude, that $_A A_A$ is the identity element in the Kasparov product (up
-to isomorphism).
-%------------- EXAMPLE
-%------------- EXAMPLE
-Let us examine another example for $E = \mathbb{C}^n$, which is a
-$(M_n(\mathbb{C}), \mathbb{C})$ Hilbert bimodule with the standard $\mathbb{C}$
-inner product. Further let $F = \mathbb{C}^n$, which is a $(\mathbb{C},
-M_n(\mathbb{C}))$ Hilbert bimodule by right matrix multiplication with
-$M_n(\mathbb{C})$ valued inner product, we can write
- \begin{align}
- \langle v_1, v_2\rangle =\bar{v_1}v_2^t \;\; \in M_n(\mathbb{C}).
- \end{align}
-If we take the Kasparov product of $E$ and $F$
- \begin{align}
- F\circ E\ &=\ E\otimes _{\mathbb{C}}F\ \;\;\;\;\;\; \simeq \
- M_n(\mathbb{C}),\\
- E\circ F\ &=\ F\otimes _{M_n(\mathbb{C})}E\ \simeq\ \mathbb{C},
- \end{align}
-we see that $M_n(\mathbb{C})$ and $\mathbb{C}$ are Morita equivalent!
-%------------- EXAMPLE
-
-\begin{mylemma}
- Two matrix algebras are Morita Equivalent if, and only if their their structure spaces
- are isomorphic as discreet spaces (have the same cardinality / same number
- of elements).
-\end{mylemma}
-\begin{proof}
- Let $A$, $B$ be \textit{Morita equivalent}. Then there exist $_A E_B$ and $_B F_A$ with
- \begin{align}
- E \otimes _B F \simeq A \;\;\; \text{and} \;\;\; F \otimes _A E \simeq
- B.
- \end{align}
- Also consider $[(\pi _B, H)] \in \hat{B}$. We can construct a
- representation of $A$, which reads
- \begin{align}
- \pi _A \rightarrow L(E \otimes _B H)\;\;\; \text{with} \;\;\; \pi _A(a)
- (e \otimes v) = a e \otimes w
- \end{align}
- Vice versa, we have $[(\pi _A, W)] \in \hat{A}$ we can construct $\pi _B$
- as
- \begin{align}
- \pi _B: B \rightarrow L(F \otimes _A W) \;\;\; \text{and}\;\;\; \pi
- _B(b) (f\otimes w) = bf\otimes w.
- \end{align}
- Now we need to show that the representation $\pi _A$ is irreducible if and
- only if $\pi _B$ is irreducible. For $(\pi _B, H)$ to be irreducible, we
- need $H \neq \emptyset$ and only $\emptyset$ or $H$ to be invariant under
- the Action of $B$ on $H$. Than $E\otimes _B H$ and $E\otimes _B H \simeq A$
- cannot be empty, because $E$ preserves left representation of $A$.
-
- Lastly we need to check if the association of the class $[\pi _A]$ to $[\pi
- _B]$ is independent of the choice of representatives $\pi _A$ and $\pi _B$.
- The important thing is that $[\pi _A] \in \hat{A}$ respectively $[\pi _B] \in
- \hat{B}$, hence any choice of representation is irreducible, because the
- structure space denotes all unitary equivalence classes of irreducible
- representations.
-
- Note that the statements $E \simeq H$ and $F \simeq W$ are not particularly
- true, since all infinite dimensional Hilbert spaces are isomorphic. Here
- we are looking at finite dimensional Hilbert spaces. Another thing to keep
- in mind, is that for $[\pi _B, H] \in \hat{B}$ and looking at algebraic
- bimodules, we know that $H$ is a bimodule of $B$, hence $E \otimes _B
- H\simeq A$, and for $[\pi _A, W]$, which is the same.
- Finally we can conclude, that these maps are each others inverses, thus
- $\hat{A} \simeq \hat{B}$.
-\end{proof}
-
-\begin{mylemma}
- The matrix algebra $M_n(\mathbb{C})$ has a unique irreducible
- representation (up to isomorphism) given by the defining representation on
- $\mathbb{C}^n$.
-\end{mylemma}
-\begin{proof}
- We know $\mathbb{C}^n$ is a irreducible representation of $A=
- M_n(\mathbb{C})$. Let $H$ be irreducible and of dimension $k$, then we
- define a map
- \begin{align}
- \phi : A\oplus...\oplus A &\rightarrow H^* \\
- (a_1,...,a_k)&\mapsto e^1\circ a_1^t+...+e^k\circ a_k^t,
- \end{align}
-where $\{e^1,...,e^k\}$ is the basis of the dual space $H^*$ and
-$(\circ)$ being the pre-composition of elements in $H^*$ and $A$ acting on $H$.
-This forms a morphism of $M_n(\mathbb{C})$ modules, provided a matrix $a \in A$
-acts on $H^*$ with $v\mapsto v\circ a^t$ ($v\in H^*$). Furthermore this
-morphism is surjective, thus making the pullback $\phi ^*:H\mapsto (A^k)^*$
-injective. Now identify $(A^k)^*$ with $A^k$ as a $A$-module and note that
-$A=M_n(\mathbb{C}) \simeq \oplus ^n \mathbb{C}^n$ as a n A module. It follows
-that $H$ is a submodule of $A^k \simeq \oplus ^{nk}\mathbb{C}$. By
-irreducibility $H \simeq \mathbb{C}$.
-\end{proof}
-
-%---------------- EXAMPLE
-Let us look at an examples for two matrix algebras $A$, and $B$.
-\begin{align}
- A = \bigoplus ^N_{i=1} M_{n_i}(\mathbb{C}), \;\;\;
- B = \bigoplus ^M_{j=1} M_{m_j}(\mathbb{C}).
-\end{align}
-Let $\hat{A} \simeq \hat{B}$, this implies $N=M$. Further define $E$ with $A$
-acting by block-diagonal matrices on the first tensor and B acting in the same
-manner on the second tensor. Define $F$ vice versa, ultimately reading
-\begin{align}
- E:= \bigoplus _{i=1}^N \mathbb{C}^{n_i} \otimes \mathbb{C}^{m_i}, \;\;\;
- F:= \bigoplus _{i=1}^N \mathbb{C}^{m_i} \otimes \mathbb{C}^{n_i}.
-\end{align}
-When we calculate the Kasparov product we get the following
-\begin{align}
- E \otimes _B F &\simeq \bigoplus _{i=1}^N (\mathbb{C}^{n_i}\otimes\mathbb{C}^{m_i})
- \otimes _{M_{m_i}(\mathbb{C})} (\mathbb{C}^{m_i}\otimes\mathbb{C}^{n_i}) \\
- &\simeq \bigoplus _{i=1}^N \mathbb{C}^{n_i}\otimes
- \left(\mathbb{C}^{m_i}\otimes _{M_{m_i}(\mathbb{C})}\mathbb{C}^{m_i}\right)
- \oplus \mathbb{C}^{n_i} \\
- &\simeq \bigoplus _{i=1}^N
- \mathbb{C}^{m_i}\otimes\mathbb{C}^{n_i} \simeq A.
-\end{align}
-On the other hand we get
-\begin{align}
- F \otimes _A E \simeq B.
-\end{align}
-%---------------- EXAMPLE
-
-To summarize, there is a duality between finite spaces and Morita equivalence classes of matrix algebras. By replacing $*$-homomorphism $A\rightarrow B$ with Hilbert bimodules $(A,B)$ we introduce
-a richer structure of morphism between matrix algebras.
diff --git a/src/thesis/chapters/backup/conclusion.tex b/src/thesis/chapters/backup/conclusion.tex
@@ -1,2 +0,0 @@
-\section{Conclusion}
-\lipsum[1]
diff --git a/src/thesis/chapters/backup/diffgeo.tex b/src/thesis/chapters/backup/diffgeo.tex
@@ -1,128 +0,0 @@
-\subsection{Excurse}
-\textbf{Manifold:} A topological space that is locally Euclidean.
-\newline
-\textbf{Riemannian Manifold:}A Manifold equipped with a Riemannian
-Metric, a
-symmetric bilinear form on Vector Fields $\Gamma(TM)$
-\begin{align}
- &g: \Gamma(TM) \times \Gamma(TM) \rightarrow C(M) \\
- \text{with}& \nonumber\\
- &g(X, Y) \in \mathbb{R} \;\;\; \text{if $X, Y \in \mathbb{R}$}\\
- &\text{$g$ is $C(M)$-bilinear } \forall f\in C(M):\;\; g(fX, Y) =
- g(X,
- fY) = fg(X,Y)\\
- &g(X,X) \begin{cases}\geq 0 \;\;\; \forall X \\ = 0 \;\;\; \forall X
- =0
- \end{cases}
-\end{align}
-$g$ on $M$ gives rise to a distance function on $M$
-\begin{align}
- d_g(x, y) = \inf_\gamma \left\{\int_0^1(\dot{\gamma}(t),
- \dot{\gamma}(t))dt;\;\; \gamma(0) = x, \gamma(1) = y \right\}
-\end{align}
-Riemannian Manifold is called spin$^c$ if there exists a vector bundle $S
-\rightarrow M$ with an algebra bundle isomorphism
-\begin{align}
- \mathbb{C}\text{I}(TM) &\simeq \text{End}(S)\;\;\; &\text{($dim(M)$
- even)}\\
- \mathbb{C}\text{I}(TM)^\circ &\simeq \text{End}(S)\;\;\;
- &\text{($dim(M)$ odd)}\\
-\end{align}
-$(M,S)$ is called the \textbf{spin$^c$ structure on $M$}.
-\newline
-$S$ is called the \textbf{spinor Bundle}.
-\newline
-$\Gamma(S)$ are the \textbf{spinors}.
-
-Riemannian spin$^c$ Manifold is called spin if there exists an
-anti-unitary
-operator $J_M:\Gamma(S) \rightarrow \Gamma(S)$ such that:
-\begin{enumerate}
- \item $J_M$ commutes with the action of real-valued continuous
- functions
- on $\Gamma(S)$.
- \item $J_M$ commutes with $\text{Cliff}^-(M)$ (even case)\\
- $J_M$ commutes with $\text{Cliff}^-(M)^\circ$ (odd case)
-\end{enumerate}
-$(S, J_M)$ is called the \textbf{spin Structure on $M$}
-\newline
-$J_M$ is called the \textbf{charge conjugation}.
-
-\subsection{Operators of Laplace Type}
-Let $M$ be a $n$ dimensional compact Riemannian manifold with $\partial M = 0$.
-Then consider a vector bundle $V$ over $M$ (i.e. there is a vector space to
-each point on $M$), so we can define smooth functions. We want to look at
-arbitrary differential operators $D$ of Laplace type on $V$, they have the general
-from
-\begin{align}
- D = -(g^{\mu\nu} \partial_\mu\partial_\nu + a^\sigma\partial_\sigma +b)
-\end{align}
-where $a^\sigma, b$ are matrix valued functions on $M$ and $g^{\mu\nu}$ is the
-inverse metric on $M$. There is a unique connection on $V$ and a unique
-endomorphism (matrix valued function) $E$ on $V$, then we can rewrite $D$ in
-terms of $E$ and covariant derivatives
-\begin{align}
- D = -(g^{\mu\nu} \nabla_\mu \nabla_\nu +E)
-\end{align}
-Where the covariant derivative consists of $\nabla = \nabla^{[R]} +\omega$ the
-standard Riemannian covariant derivative $\nabla^{[R]}$ and a "gauge" bundle
-$\omega$ (fluctuations). WE can write $E$ and $\omega$ in terms of geometrical
-identities
-\begin{align}
- \omega_\delta &= \frac{1}{2}g_{\nu\delta}(a^\nu
- +g^{\mu\sigma}\Gamma^\nu_{\mu\sigma}I_V)\\
- E &= b - g^{\nu\mu}(\partial_\mu \omega_\nu + \omega_\nu \omega_\mu -
- \omega_\sigma \Gamma^\sigma_{\nu\mu})
-\end{align}
-where $I_V$ is the identity in $V$ and the Christoffel symbol
-\begin{align}
- \Gamma^\sigma_{\mu\nu} = g^{\sigma\varrho} \frac{1}{2} (\partial_\mu
- g_{\nu\varrho} + \partial_\nu g_{\mu\varrho} - \partial_\varrho g_{\mu\nu})
-\end{align}
-Furthermore we remind ourselves of the Riemmanian curvature tensor, Ricci
-Tensor and the Scalar curavture.
-\begin{align}
- R^\mu_{\nu\varrho\sigma} &= \partial_\sigma \Gamma^{\mu}_{\nu\varrho}
- -\partial_\varrho \Gamma^\mu_{\nu\sigma}
- \Gamma^{\lambda}_{\nu\varrho}\Gamma^{\mu}_{\lambda\sigma}
- \Gamma^{\lambda}_{\nu\sigma}\Gamma^{\mu}_{\lambda\varrho}\\
- R_{\mu\nu} &:= R^{\sigma}_{\mu\nu\sigma}\\
- R &:= R^\mu_{\ \mu}
-\end{align}
-
-The we let $\{e_1, \dots, e_n\}$ be the local orthonormal frame of
-$TM$(tangent bundle $M$), which will be noted with flat indices $i,j,k,l
-\in\{1,\dots, n\}$, we use $e^k_\mu, e^\nu_j$ to transform between flat indices
-and curved indices $\mu, \nu, \varrho$.
-\begin{align}
- e^\mu_j e^\nu_k g_{\mu\nu} &= \delta_{jk}\\
- e^\mu_j e^\nu_k \delta^{jk} &= g^{\mu\nu} \\
- e^j_\mu e^\mu_k &= \delta^j_k
-\end{align}
-
-The Riemannian part of the covariant derivative contains the standard
-Levi-Civita connection, so that for a $v_\nu$ we write
-\begin{align}
- \nabla_\mu^{[R]} v_\nu = \partial_\mu v_\nu -
- \Gamma^{\varrho}_{\mu\nu}v_\varrho.
-\end{align}
-The extended covariant derivative reads then
-\begin{align}
- \nabla_\mu v^j = \partial_\mu v^j + \sigma^{jk}_\mu v_k.
-\end{align}
-the condition $\nabla_\mu e^k_\nu = 0$ gives us the general connection
-\begin{align}
- \sigma^{kl}_\mu = e^\nu_l\Gamma^{\varrho}_{\mu\nu}e^k_\varrho - e^\nu_l
- \partial_\mu e^k_\nu
-\end{align}
-The we may define the field strength $\Omega_{\mu\nu}$ of the connection $\omega$
-\begin{align}
- \Omega_{\mu\nu} = \partial_\mu \omega_\nu -\partial_\nu \omega_\mu
- +\omega_\mu \omega_\nu -\omega_\nu\omega_\mu.
-\end{align}
-If we apply the covariant derivative on $\Omega$ we get
-\begin{align}
- \nabla_\varrho\Omega_{\mu\nu} = \partial_\varrho \Omega_{\mu\nu} -
- \Gamma^{\sigma}_{\varrho \mu} \Omega_{\sigma\mu} + [\omega_\varrho,
- \Omega_{\mu\nu}]
-\end{align}
diff --git a/src/thesis/chapters/backup/electroncg.tex b/src/thesis/chapters/backup/electroncg.tex
@@ -1,455 +0,0 @@
-\subsection{Noncommutative Geometry of Electrodynamics}
-In this chapter we describe Electrodynamics with the almost commutative
-manifold $M\times F_X$ and the abelian gauge group $U(1)$.
-We arrive at a unified description of gravity and electrodynamics although in the classical level.
-
-The almost commutative Manifold $M\times F_X$ describes a local gauge group
-$U(1)$. The inner fluctuations of the Dirac operator relate to $Y_\mu$ the
-gauge field of $U(1)$. According to the setup we ultimately arrive at two
-serious problems.
-
-First of all in the Two-Point space $F_X$, the operator $D_F$ must vanish for
-us to have a real structure. However this implies that the electrons
-are massless, which would be absurd.
-
-The second problem arises when looking at the Euclidean action for a free
-Dirac field
-\begin{align}
- S = - \int i \bar{\psi}(\gamma ^\mu\partial _\mu - m) \psi d^4x,
-\end{align}
-where $\psi,\ \bar{\psi}$ must be considered as independent variables, which
-means that the fermionic action $S_f$ needs two independent Dirac spinors.
-Let us try and construct two independent Dirac spinors with our data. To do
-this we take a look at the decomposition of the basis and of the total
-Hilbertspace $H = L^2(S) \otimes H_F$. For the orthonormal basis of $H_F$ we
-can write $\{e, \bar{e}\}$ , where $\{e\}$ is the orthonormal basis of
-$H_F^+$ and $\{\bar{e}\}$ the orthonormal basis of $H_F^-$. Accompanied with
-the real structure we arrive at the following relations
-\begin{align}
- J_F e &= \bar{e} \;\;\;\;\;\; J_F \bar{e} = e, \\
- \gamma_F e &= e \;\;\;\;\;\; \gamma_F \bar{e} = \bar{e}.
-\end{align}
-Along with the decomposition of $L^2(S) = L^2(S)^+ \oplus L^2(S)^-$ and $\gamma = \gamma _M
-\otimes \gamma _F$ we can obtain the positive eigenspace
-\begin{align}
- H^+ = L^2(S)^+ \otimes H_F^+ \oplus L^(S)^- \otimes H_F^-.
-\end{align}
-So, for a $\xi \in H^+$ we can write
-\begin{align}
- \xi = \psi _L \otimes e + \psi _R \otimes \bar{e}
-\end{align}
-where $\psi_L \in L^2(S)^+$ and $\psi _R \in L^2(S)^-$ are the two Wheyl
-spinors. We denote that $\xi$ is only determined by one Dirac spinor $\psi :=
-\psi_L + \psi _R$, \textbf{but we require two independent spinors}. Our
-conclusion is that the definition of the fermionic action gives too much
-restrictions to the Two-Point space $F_X$.
-\subsubsection{The Finite Space}
-To solve the two problems we simply enlarge (double) the Hilbertspace. This
-is visualized by introducing multiplicities in Krajewski Diagrams which will also
-allow us to choose a nonzero Dirac operator that will connect the two
-vertices and preserve real structure making our particles massive and
-bringing anti-particles into the mix.
-
-We start of with the same algebra $C^\infty(M, \mathbb{C}^2)$, corresponding
-to space $N= M\times X$. The Hilbertspace describes four particles, meaning
-it has four orthonormal basis elements. It describes \textbf{left handed
-electrons} and \textbf{right handed positrons}. Pointing this out, we have
-$\{ \underbrace{e_R, e_L}_{\text{left-handed}}, \underbrace{\bar{e}_R,
-\bar{e}_L}_{\text{right-handed}}\}$ the orthonormal basis for $H_F =
-\mathbb{C}^4$. Accompanied with the real structure $J_F$, which allows us to
-interchange particles with antiparticles by the following equations
-\begin{align}
- &J_F e_R = \bar{e}_R, \\
- &J_F e_L = \bar{e_L}, \\
- \nonumber \\
- &\gamma _F e_R = -e_R,\\
- &\gamma_F e_L = e_L,
-\end{align}
-where $J_F$ and $\gamma_F$ have to following properties
-\begin{align}
- &J_F^2 = 1,\\
- & J_F \gamma_F = - \gamma_F J_F.
-\end{align}
-By means of $\gamma_F$ we have two options to decompose the total
-Hilbertspace $H$, firstly into
-\begin{align}
- H_F = \underbrace{H_F^+}_{\text{ONB } \{e_L, \bar{e}_L\}}
- \oplus \underbrace{H_F^-}_{\text{ONB } \{e_R, \bar{e}_R\}},
-\end{align}
-or alternatively into the eigenspace of particles and their
-antiparticles (electrons and positrons) which is preferred in literature and
-which we will use going further
-\begin{align}
- H_F = \underbrace{H_{e}}_{\text{ONB } \{e_L, e_R\}} \oplus
- \underbrace{H_{\bar{e}}}_{\text{ONB } \{\bar{e}_L, \bar{e}_R\}}.
-\end{align}
-Here ONB means orthonormal basis.
-
-The action of $a \in A = \mathbb{C}^2$ on $H$ with respect to the ONB
-$\{e_L, e_R, \bar{e}_L, \bar{e}_R\}$ is represented by
-\begin{align}\label{eq:leftrightrepr}
- a =
- (a_1 , a_2 ) \mapsto
- \begin{pmatrix}
- a_1 &0 &0 &0\\
- 0&a_1 &0 &0\\
- 0 &0 &a_2 &0\\
- 0 &0 &0 &a_2\\
- \end{pmatrix}
-\end{align}
-Do note that this action commutes wit the grading and that
-$[a, b^\circ] = 0$ with $b:= J_F b^*J_F$ because both the left and the right
-action is given by diagonal matrices by equation \eqref{eq:leftrightrepr}. Note
-that we are still left with $D_F = 0$ and the following spectral
-triple
-\begin{align}\label{eq:fedfail}
- \left( \mathbb{C}^2, \mathbb{C}^2, D_F=0; J_F =
- \begin{pmatrix}
- 0 & C \\ C &0
- \end{pmatrix},
- \gamma _F =
- \begin{pmatrix}
- 1 & 0 \\ 0 &-1
- \end{pmatrix}
- \right).
- \end{align}
-It can be represented in the following Krajewski diagram,
-with two nodes of multiplicity two
- \begin{figure}[H] \centering
- \begin{tikzpicture}[
- dot/.style = {draw, circle, inner sep=0.06cm},
- bigdot/.style = {draw, circle, inner sep=0.09cm},
- no/.style = {},
- ]
- \node[no](a) at (0,0) [label=left:$\textbf{1}^\circ$] {};
- \node[no](b) at (0, -1) [label=left:$\textbf{1}^\circ$] {};
- \node[no](c) at (0.5, 0.5) [label=above:$\textbf{1}$] {};
- \node[no](d) at (1.5, 0.5) [label=above:$\textbf{1}$] {};
- \node[dot](d0) at (1.5,0) [] {};
- \node[dot](d0) at (0.5,-1) [] {};
- \node[bigdot](d0) at (1.5,0) [] {};
- \node[bigdot](d0) at (0.5,-1) [] {};
- \end{tikzpicture}
- \end{figure}
-\subsubsection{A noncommutative Finite Dirac Operator}
-To extend our spectral triple with a non-zero Operator, we need to take a
-closer look at the Krajewski diagram above. Notice that edges only exist
-between multiple vertices, meaning we can construct a Dirac operator mapping
-between the two vertices. The operator can be represented by the following matrix
-\begin{align}\label{eq:feddirac}
- D_F =
- \begin{pmatrix}
- 0 & d & 0 & 0 \\
- \bar{d} & 0 & 0 & 0 \\
- 0 & 0 & 0 & \bar{d} \\
- 0 & 0 & d & 0
- \end{pmatrix}
-\end{align}
-We can now define the finite space $F_{ED}$.
-\begin{align}
- F_{ED} := (\mathbb{C}^2, \mathbb{C}^4, D_F; J_F, \gamma_F)
-\end{align}
-where $J_F$ and $\gamma_F$ are like in equation \eqref{eq:fedfail} and $D_F$
-from equation \eqref{eq:feddirac}.
-
-\subsubsection{Almost commutative Manifold of Electrodynamics}
-The almost commutative manifold $M\times F_{ED}$ has KO-dimension 2, and is
-represented by the following spectral triple
-\begin{align}\label{eq:almost commutative manifold}
- M\times F_{ED} := \big(C^\infty(M,\mathbb{C}^2),\ L^2(S)\otimes
- \mathbb{C}^4,\
- D_M\otimes 1 +\gamma _M \otimes D_F;\; J_M\otimes J_F,\ \gamma_M\otimes
- \gamma _F\big)
-\end{align}
-The algebra didn't change, thus we can decompose it like before
-\begin{align}
- C^\infty(M, \mathbb{C}^2) = C^\infty (M) \oplus C^\infty (M)
-\end{align}
-As for the Hilbertspace, we can decomposition it in the following way
-\begin{align}
- H = (L^2(S) \otimes H_e ) \oplus (L^2(S) \otimes H_{\bar{e}}).
-\end{align}
-Note that the one component of the algebra is acting on $L^2(S) \otimes H_e$,
-and the other one acting on $L^2(S) \otimes H_{\bar{e}}$. In other words the components of
-the decomposition of both the algebra and the Hilbertspace match by the action of
-the algebra.
-
-The derivation of the gauge theory is the same for $F_{ED}$ as for the
-Two-Point space $F_X$. We have $\mathfrak{B}(F) \simeq U(1)$ and for an
-arbitrary gauge field $B_\mu = A_\mu - J_F A_\mu J_F^{-1}$ we can write
-\begin{align} \label{field}
- B_\mu =
- \begin{pmatrix}
- Y_\mu & 0 & 0 & 0 \\
- 0 & Y_\mu& 0 & 0 \\
- 0 & 0 & Y_\mu& 0 \\
- 0 & 0 & 0 & Y_\mu
- \end{pmatrix} \;\;\;\;\;\ \text{for} \;\;\ Y_\mu (x) \in \mathbb{R}.
-\end{align}
-We have one single $U(1)$ gauge field $Y_\mu$, carrying the action of the
-gauge group
-\begin{align}
- \text{$\mathfrak{B}$}(M\times F_{ED}) \simeq C^\infty(M, U(1))
-\end{align}
-
-The space $N = M\times X$ consists of two copies of $M$.
-If $D_F = 0$ we have infinite distance between the two copies. Now have
-hacked the spectral triple to have nonzero Dirac operator $D_F$. The new
-Dirac operator still has a commuting relation with the algebra $[D_F, a] = 0$
-$\forall a \in A$, and we should note that the distance between the two
-copies of $M$ is still infinite. This is purely an mathematically abstract
-observation and doesn't affect physical results.
-
-\subsubsection{Spectral Action}
-In this chapter we bring all our results together to establish an
-Action functional to describe a physical system. It turns out that
-the Lagrangian of the almost commutative manifold $M\times F_{ED}$
-corresponds to the Lagrangian of Electrodynamics on a curved
-background manifold (+ gravitational Lagrangian), consisting of the spectral
-action $S_b$ (bosonic) and of the fermionic action $S_f$.
-
-The simplest spectral action of a spectral triple $(A, H, D)$ is given by the
-trace of a function of $D$. We also consider inner fluctuations of the Dirac
-operator $D_\omega = D + \omega + \varepsilon' J\omega J^{-1}$ where $\omega =
-\omega ^* \in \Omega_D^1(A)$.
-\begin{mydefinition}
- Let $f:\mathbb{R} \rightarrow \mathbb{R}$ be a suitable function
- \textbf{positive and even}. The spectral action is then
- \begin{align}
- S_b [\omega] := \text{Tr}\big(f(\frac{D_\omega}{\Lambda})\big)
- \end{align}
- where $\Lambda$ is a real cutoff parameter. The minimal condition on $f$
- is that $f(\frac{D_\omega}{\Lambda})$ is a trace class operator. A trace
- class operator is a compact operator with a well defined finite trace
- independent of the basis. The subscript $b$ in $S_b$ stands for bosonic,
- because in physical applications $\omega$ will describe bosonic fields.
-
- In addition to the bosonic action $S_b$ we can define a topological spectral
- action $S_{top}$. Leaning on the grading $\gamma$ the topological spectral action is
- \begin{align}
- S_{\text{top}}[\omega] := \text{Tr}(\gamma\
- f(\frac{D_\omega}{\Lambda})).
- \end{align}
-\end{mydefinition}
-\begin{mydefinition}\label{def:fermionic action}
- The fermionic action is defined by
- \begin{align}
- S_f[\omega, \psi] = (J\tilde{\psi}, D_\omega \tilde{\psi})
- \end{align}
- with $\tilde{\psi} \in H_{cl}^+ := \{\tilde{\psi}: \psi \in H^+\}$, where
- $H_{cl}^+$ is a set of Grassmann variables in $H$ in the +1-eigenspace
- of the grading $\gamma$.
-\end{mydefinition}
-
-%---------------------- APPENDIX ?????????????--------------------
-\textbf{APPENDIX??}
-Grassmann variables are a set of Basis vectors of a vector space, they
-form a unital algebra over a vector field $V$, where the generators are
-anti commuting, that is for Grassmann variables $\theta _i, \theta _j$ we have
-\begin{align}
- &\theta _i \theta _j = -\theta _j \theta _i \\
- &\theta _i x = x\theta _j \;\;\;\; x\in V \\
- &(\theta_i)^2 = 0 \;\;\; (\theta _i \theta _i = -\theta _i \theta _i)
-\end{align}
-%---------------------- APPENDIX ?????????????--------------------
-\begin{proposition}
- The spectral action of the almost commutative manifold $M$ with $\dim(M)
- =4$ with a fluctuated Dirac operator is
- \begin{align}
- \text{Tr}(f\frac{D_\omega}{\Lambda}) \sim \int_M \mathcal{L}(g_{\mu\nu},
- B_\mu, \Phi) \sqrt{g}\ d^4x + O(\Lambda^{-1}),
- \end{align}
- where
- \begin{align}
- \mathcal{L}(g_{\mu\nu}, B_\mu, \Phi) =
- N\mathcal{L}_M(g_{\mu\nu})
- \mathcal{L}_B(B_\mu)+
- \mathcal{L}_\phi(g_{\mu\nu}, B_\mu, \Phi).
- \end{align}
- The Lagrangian $\mathcal{L}_M$ is of the spectral triple, represented by
- the following term
- $(C^\infty(M) , L^2(S), D_M)$
- \begin{align}\label{lagr}
- \mathcal{L}_M(g_{\mu\nu}) := \frac{f_4 \Lambda ^4}{2\pi^2} -
- \frac{f_2 \Lambda^2}{24\pi ^2}s - \frac{f(0)}{320\pi^2} C_{\mu\nu
- \varrho \sigma}C^{\mu\nu \varrho \sigma},
- \end{align}
- here $C^{\mu\nu \varrho \sigma}$ is defined in terms of the Riemannian
- curvature tensor $R_{\mu\nu \varrho \sigma}$ and the Ricci tensor
- $R_{\nu\sigma} = g^{\mu\varrho} R_{\mu\nu \varrho\sigma}$.
- The kinetic term of the gauge field is described by the Lagrangian
- $\mathcal{L}_B$, which takes the following shape
- \begin{align}
- \mathcal{L}_B(B_\mu) := \frac{f(0)}{24\pi^2}
- \text{Tr}(F_{\mu\nu}F^{\mu\nu}).
- \end{align}
- Lastly $\mathcal{L}_\phi$ is the scalar-field Lagrangian with a boundary
- term, given by
- \begin{align}
- \mathcal{L}_\phi(g_{\mu\nu}, B_\mu, \Phi) :=
- &-\frac{2f_2\Lambda^2}{4\pi^2}\text{Tr}(\Phi^2) + \frac{f(0)}{8\pi^2}
- \text{Tr}(\Phi^4) + \frac{f(0)}{24\pi^2}
- \Delta(\text{Tr}(\Phi^2))\nonumber\\
- &+ \frac{f(0)}{48\pi^2}s\text{Tr}(\Phi^2)
- \frac{f(0)}{8\pi^2}\text{Tr}((D_\mu \Phi)(D^\mu \Phi)).
- \end{align}
-\end{proposition}
-\begin{proof}
- The dimension of the manifold $M$ is $\dim(M) = \text{Tr}(id) =4$. For
- an $x \in M$, we have an asymptotic expansion of the term
- $\text{Tr}(f(\frac{D_\omega}{\Lambda}))$ as $\Lambda$ goes to infinity,
- which can be written as
- \begin{align}
- \text{Tr}(f(\frac{D_\omega}{\Lambda})) \simeq& \ 2f_4 \Lambda ^4
- a_0(D_\omega ^2)+ 2f_2\Lambda^2 a_2(D_\omega^2)\nonumber \\&+ f(0) a_4(D_\omega^4)
- +O(\Lambda^{-1}).\label{eq:trheatkernel}
- \end{align}
- We have to note here that the heat kernel coefficients are zero for uneven $k$,
- and they are dependent on the fluctuated Dirac operator
- $D_\omega$. We can rewrite the heat kernel coefficients in terms of $D_M$,
- for the first two terms $a_0$ and $a_2$ we use $N:=
- \text{Tr}\mathbbm{1_{H_F}})$ and write
- \begin{align}
- a_0(D_\omega^2) &= Na_0(D_M^2),\\
- a_2(D_\omega^2 &= Na_2(D_M^2) - \frac{1}{4\pi^2}\int_M
- \text{Tr}(\Phi^2)\sqrt{g}d^4x.
- \end{align}
- For $a_4$ we extend in terms of coefficients of $F$, \textbf{REWRITE: look week9.pdf
- for the standard version}
- \begin{align}
- &\frac{1}{360}\text{Tr}(60sF)= -\frac{1}{6}S(Ns + 4
- \text{Tr}(\Phi^2))\\
- \nonumber\\
- &F^2 = \frac{1}{16}s^2\otimes 1 + 1\otimes \Phi^4 - \frac{1}{4}
- \gamma^\mu\gamma^\nu \gamma^\varrho\gamma^\sigma F_{\mu\nu}F^{\mu\nu}+\\
- &\;\;\;\;\;\;\;+\gamma^\mu\gamma^\nu\otimes(D_\mu\Phi)(D_\nu
- \Phi)+\frac{1}{2}s\otimes \Phi^2 + \ \text{traceless terms}\\
- \nonumber\\
- &\frac{1}{360}\text{Tr}(180F^2) = \frac{1}{8}s^2N + 2\text{Tr}(\Phi^4)
- + \text{Tr}(F_{\mu\nu}F^{\mu\nu}) +\\
- &\;\;\;\;\;\;\;+2\text{Tr}((D_\mu\Phi)(D^\mu\Phi))
- + s\text{Tr}(\Phi^2)\\
- \nonumber\\
- &\frac{1}{360}\text{Tr}(-60\Delta F)=
- \frac{1}{6}\Delta(Ns+4\text{Tr}(\Phi^2)).
- \end{align}
- The cross terms of the trace in $\Omega_{\mu\nu}^E\Omega^{E\mu\nu}$
- vanishes because of the antisymmetric property of the Riemannian
- curvature tensor, thus we can write
- \begin{align}
- \Omega_{\mu\nu}^E\Omega^{E\mu\nu} = \Omega_{\mu\nu}^S\Omega^{S\mu\nu}
- \otimes 1 - 1\otimes F_{\mu\nu}F^{\mu\nu} + 2i\Omega_{\mu\nu}^S
- \otimes F^{\mu\nu}.
- \end{align}
- The trace of the cross term $\Omega^{S}_{\mu\nu}$ vanishes because
- \begin{align}
- \text{Tr}(\Omega^{S}_{\mu\nu}) = \frac{1}{4}
- R_{\mu\nu\varrho\sigma}\text{Tr}(\gamma^\mu\gamma^\nu) = \frac{1}{4}
- R_{\mu\nu\varrho\sigma}g^{\mu\nu} =0,
- \end{align}
- then the trace of the whole term is given by
- \begin{align}
- \frac{1}{360}\text{Tr}(30\Omega^E_{\mu\nu}\Omega^{E\mu\nu}) =
- \frac{N}{24}R_{\mu\nu\varrho\sigma}R^{\mu\nu\varrho\sigma}
- -\frac{1}{3}\text{Tr}(F_{\mu\nu}F^{\mu\nu}).
- \end{align}
- Finally plugging the results into the coefficient $a_4$ and simplifying we get
- \begin{align}
- a_4(x, D_\omega^4) &= Na_4(x, D_M^2) + \frac{1}{4\pi^2}\bigg(\frac{1}{12} s
- \text{Tr}(\Phi^2) + \frac{1}{2}\text{Tr}(\Phi^4) \nonumber \\
- &+ \frac{1}{4}
- \text{Tr}((D_\mu\Phi)(D^\mu \Phi)) + \frac{1}{6}
- \Delta\text{Tr}(\Phi^2) + \frac{1}{6}
- \text{Tr}(F_{\mu\nu}F^{\mu\nu})\bigg).
- \end{align}
- The only thing left is to substitute the heat kernel coefficients into the
- heat kernel expansion in equation \eqref{eq:trheatkernel}.
-\end{proof}
-
-\subsubsection{Fermionic Action}
-We remind ourselves the definition of the fermionic action in definition
-\ref{def:fermionic action} and the manifold we are dealing with in equation
-\eqref{eq:almost commutative manifold}. The Hilbertspace $H_F$ is separated
-into the particle-antiparticle states with ONB $\{e_R, e_L, \bar{e}_R,
-\bar{e}_L\}$. The orthonormal basis of $H_F^+$ is $\{e_L, \bar{e}_R\}$ and
-consequently for $H_F^-$, $\{e_R, \bar{e}_L\}$. We can decompose a spinor
-$\psi \in L^2(S)$ in each of the eigenspaces $H_F^\pm$, $\psi = \psi_R+
-\psi_L$. That means for an arbitrary $\psi \in H^+$ we can write
-\begin{align}
- \psi = \chi_R \otimes e_R + \chi_L \otimes e_L + \psi_L \otimes
- \bar{e}_R+
- \psi_R \otimes \bar{e}_L,
-\end{align}
-where $\chi_L, \psi_L \in L^2(S)^+$ and $\chi_R, \psi_R \in L^2(S)^-$.
-
-Since the fermionic action yields too much restriction on $F_{ED}$ (modified
-Two-Point space $F_X$) we redefine it by taking account the fluctuated Dirac
-operator
-\begin{align}
- D_\omega = D_M \otimes i + \gamma^\mu \otimes B_\mu + \gamma_M \otimes
- D_F.
-\end{align}
-The Fermionic Action is
-\begin{align}
-S_F = (J\tilde{\xi}, D_\omega\tilde{\xi})
-\end{align}
-for a $\xi \in H^+$. Then the straight forward calculation gives \begin{align}
- \frac{1}{2}(J\tilde{\xi}, D_\omega\tilde{\xi})
- &=\frac{1}{2}(J\tilde{\xi}, (D_M \otimes
- i)\tilde{\xi})\label{eq:fermionic1}\\
- &+\frac{1}{2}(J\tilde{\xi}, (\gamma^\mu \otimes B_\mu)
- \tilde{\xi})\label{eq:fermionic2}\\
- &+\frac{1}{2}(J\tilde{\xi}, (\gamma_M\otimes
- D_F)\tilde{\xi})\label{eq:fermionic3},
-\end{align}
-(note that we add the constant $\frac{1}{2}$ to the action).
-For the term in \eqref{eq:fermionic1} we calculate
-\begin{align}
- \frac{1}{2}(J\tilde{\xi}, (D_M\otimes 1)\tilde{\xi}) &=
- \frac{1}{2}(J_M\tilde{\chi}_R,D_M\tilde{\psi}_L)+\nonumber
- \frac{1}{2}(J_M\tilde{\chi}_L,D_M\tilde{\psi}_R)+
- \\&+\frac{1}{2}(J_M\tilde{\psi}_L,D_M\tilde{\psi}_R)+\nonumber
- \frac{1}{2}(J_M\tilde{\chi}_R,D_M\tilde{\chi}_L)\\
- &= (J_M\tilde{\chi},D_M\tilde{\chi}).
-\end{align}
-For the term in \eqref{eq:fermionic2} we have
-\begin{align}
- \frac{1}{2}(J\tilde{\xi}, (\gamma^\mu \otimes B_\mu)\tilde{\xi})&=
- -\frac{1}{2}(J_M\tilde{\chi}_R, \gamma^\mu Y_\mu\tilde{\psi}_R)
- -\frac{1}{2}(J_M\tilde{\chi}_L, \gamma^\mu Y_\mu\tilde{\psi}_R)+\nonumber\\
- &+\frac{1}{2}(J_M\tilde{\psi}_L, \gamma^\mu Y_\mu\tilde{\chi}_R)+
- \frac{1}{2}(J_M\tilde{\psi}_R, \gamma^\mu Y_\mu\tilde{\chi}_L)=\nonumber\\
- &= -(J_M\tilde{\chi}, \gamma^\mu Y_\mu\tilde{\psi}).
-\end{align}
-And for \eqref{eq:fermionic3} we can write
-\begin{align}
- \frac{1}{2}(J\tilde{\xi}, (\gamma_M\otimes D_F)\tilde{\xi})&=
- +\frac{1}{2}(J_M\tilde{\chi}_R, d\gamma_M\tilde{\chi}_R)
- +\frac{1}{2}(J_M\tilde{\chi}_L, \bar{d}\gamma_M\tilde{\chi}_L)+\nonumber\\
- &+\frac{1}{2}(J_M\tilde{\chi}_L, \bar{d}\gamma_M\tilde{\chi}_L)
- +\frac{1}{2}(J_M\tilde{\chi}_R, d\gamma_M\tilde{\chi}_R)=\nonumber\\
- &= i(J_M\tilde{\chi}, m\tilde{\psi}).
-\end{align}
-A small problem arises, we obtain a complex mass parameter $d$, but we can
-write $d:=im$ for $m\in \mathbb{R}$, which stands for the real mass.
-
-Finally the fermionic action of $M\times F_{ED}$ takes the form
- \begin{align}
- S_f = -i\big(J_M\tilde{\chi}, \gamma(\nabla^S_\mu - i\Gamma_\mu)
- \tilde{\Psi}\big) + \big(S_M\tilde{\chi}_L, \bar{d}\tilde{\psi}_L\big) -
- \big(J_M\tilde{\chi}_R, d \tilde{\psi}_R\big).
- \end{align}
-Ultimately we arrive at the full Lagrangian of $M\times F_{ED}$, which is the
-sum of purely gravitational Lagrangian
- \begin{align}
- \mathcal{L}_{grav}(g_{\mu\nu})=4\mathcal{L}_M(g_{\mu\nu})+
- \mathcal{L}_\phi (g_{\mu\nu}),
- \end{align}
-and the Lagrangian of electrodynamics
- \begin{align}
- \mathcal{L}_{ED} = -i\bigg\langle
- J_M\tilde{\chi},\big(\gamma^\mu(\nabla^S_\mu - iY_\mu) -m\big)\tilde{\psi})
- \bigg\rangle
- +\frac{f(0)}{6\pi^2} Y_{\mu\nu}Y^{\mu\nu}.
- \end{align}
-
diff --git a/src/thesis/chapters/backup/finitencg.tex b/src/thesis/chapters/backup/finitencg.tex
@@ -1,664 +0,0 @@
-\subsection{Finite Spectral Triples}
-\subsubsection{Metric on Finite Discrete Spaces}
-Let us come back to our finite discrete space $X$, we can describe it by a
-structure space $\hat{A}$ of a matrix algebra $A$. To describe distance between
-two points in $X$ (as we would in a metric space) we use an array $\{d_{ij}\}_{i,
-j \in X}$ of \textit{real non-negative} entries in $X$ such that
-\begin{itemize}
- \item $d_{ij} = d_{ji}$ Symmetric
- \item $d_{ij} \leq d_{ik} d_{kj}$ Triangle Inequality
- \item $d_{ij} = 0$ for $i=j$ (the same element)
-\end{itemize}
-
-In the commutative case, the algebra $A$ is commutative and can describe the
-metric on $X$ in terms of algebraic data.
-\begin{theorem}
- Let $d_{ij}$ be a metric on $X$ a finite discrete space with $N$ points, $A = \mathbb{C}^N$
- with elements $a = (a(i))_{i=1}^N$ such that $\hat{A} \simeq X$. Then there exists a
- representation $\pi$ of $A$ on a finite-dimensional inner product space $H$ and a symmetric
- operator $D$ on $H$ such that
- \begin{equation}
- d_{ij} = \sup_{a\in A}\bigg\{\big|a(i)-a(j)\big| : |\big|\big[D,
- \pi(a)]\big|\big| \leq 1\bigg\}
- \end{equation}
-\end{theorem}
-
-\begin{proof}
- We claim that this would follow from the equality:
- \begin{equation}
- \big|\big|[D, \pi(a)]\big|\big| = \max_{k\neq l}
- \bigg\{\frac{1}{d_{kl}}\big|a(k) - a(l)\big|\bigg\}
- \label{induction}
- \end{equation}
- This can be proved with induction. Set $N=2$ then $H=\mathbb{C}^2$, $\pi:A\rightarrow L(H)$ and
- a hermitian matrix $D$.
- \begin{align}
- \pi(a) =
- \begin{pmatrix}
- a(1) & 0 \\
- 0 & a(2)
- \end{pmatrix}
- \;\;\;\;
- D =
- \begin{pmatrix}
- 0 & (d_{12})^{-1} \\
- (d_{21})^{-1} & 0
- \end{pmatrix}
- \end{align}
- Then we compute the commutator
- \begin{align}
- \big|\big|[D, \pi(a)]\big|\big| = (d_{12})^{-1} \big| a(1) - a(2)\big|
- \end{align}
-
- For the case $A=\mathbb{C}^3$, we have $H = (\mathbb{C}^2)^{\oplus 3} = H_2
- \oplus H_2^1 \oplus H_2^2$. The representation $\pi (a)$ reads
- \begin{align}
- \pi((a(1), a(2), a(3)) &=
- \begin{pmatrix}
- a(1) & 0 \\ 0 & a(2)
- \end{pmatrix} \oplus
- \begin{pmatrix}
- a(1) & 0 \\ 0 & a(3)
- \end{pmatrix} \oplus
- \begin{pmatrix}
- a(2) & 0 \\ 0 & a(2)
- \end{pmatrix} \nonumber \\
- & = \text{diag}\big(a(1), a(2), a(1), a(3), a(2),
- a(3)\big)
- \end{align}
- And the operator $D$ takes the form
- \begin{align}
- D &=
- \begin{pmatrix}
- 0 & x_1 \\ x_1 & 0
- \end{pmatrix} \oplus
- \begin{pmatrix}
- 0 & x_2 \\ x_2 & 0
- \end{pmatrix} \oplus
- \begin{pmatrix}
- 0 & x_3 \\ x_3 & 0
- \end{pmatrix} \nonumber \\
- &=
- \begin{pmatrix}
- 0 & x_1 & 0 & 0 & 0 & 0 \\
- x_1 & 0 & 0 & 0 & 0 & 0 \\
- 0 & 0 & 0 & x_2 & 0 & 0 \\
- 0 & 0 & x_2 & 0 & 0 & 0 \\
- 0 & 0 & 0 & 0 & 0 & x_3 \\
- 0 & 0 & 0 & 0 & x_3 & 0 \\
- \end{pmatrix}.
- \end{align}
- Then the norm of the commutator would be the largest eigenvalue
- \begin{align}\label{eq:skew matrix}
- &\big|\big|[D, \pi(a)]\big|\big| = \big|\big|D\pi(a) - \pi(a)D\big|\big|,
- \end{align}
- where the matrix in the norm from equation \eqref{eq:skew matrix} is a
- skew symmetric matrix. Its eigenvalues are $i\lambda_1, i\lambda_2,
- i\lambda_3, i\lambda_4$. The $\lambda$'s are on the upper and lower
- diagonal. The matrix norm would be the maximum of the norm with the
- larges eigenvalues:
- \begin{align}
- \big|\big|[D, \pi(a)]\big|\big| = \max_{a\in A}\bigg\{&x_1\big|a(2)-a(1)\big|,\nonumber\\ &x_2\big|(a(3)-a(1))\big|,\nonumber\\
- &x_3\big|(a(3)-a(2))\big|\bigg\}.
- \end{align}
- Hence the metric turns out to be
- \begin{align}
- d =
- \begin{pmatrix}
- 0 & a(1)-a(2) & a(1)-a(3)\\
- a(2)-a(1) & 0 & a(2)-a(3)\\
- a(3)-a(1) & a(3)-a(2) & 0
- \end{pmatrix}
- \end{align}
-
- Suppose this holds for $N$ with $\pi_N$, $H_N = \mathbb{C}^N$ and $D_N$.
- Then it has to holds for $N+1$ with $H_{N+1} = H_{N} \oplus \bigoplus_{i=1}^N
- H_N^i$, since the representation reads
- \begin{align}
- \pi_{N+1}(a(1),\dots,a(N+1)) &= \pi_N(a(1),\dots,a(N))
- \oplus
- \begin{pmatrix}
- a(1) & 0 \\
- 0 & a(N+1)
- \end{pmatrix} \oplus \nonumber\\
- &\oplus \cdots \oplus
- \begin{pmatrix}
- a(N) & 0 \\
- 0 1 & a(N+1)
- \end{pmatrix}
- \end{align}
- And the operator $D_{N+1}$ is
- \begin{align}
- D_{N+1} &= D_N
- \oplus
- \begin{pmatrix}
- 0 & (d_{1(N+1)})^{-1} \\
- (d_{1(N+1)})^{-1} & 0
- \end{pmatrix}\oplus \nonumber \\
- &\oplus \cdots \oplus
- \begin{pmatrix}
- 0 & (d_{N(N+1)})^{-1} \\
- (d_{N(N+1)})^{-1} & 0
- \end{pmatrix}
- \end{align}
- From this follows equation \eqref{induction}.
- Thus we can continue the proof by setting for fixed $i, j$, $a(k) =
- d_{ik}$, which then gives $|a(i) - a(j)| = d_{ij}$ and thereby it follows
- that
- \begin{align}
- \frac{1}{d_{kl}} \big| a(k) - a(l) \big| = \frac{1}{d_{kl}} \big|
- d_{ik} - d_{il} \big| \leq 1.
- \end{align}
-\end{proof}
-
-%---------------- EXERCISE
-To get a better understanding of the results of the theorem let us compute a
-metric on the space of three points given by $d_{ij} = \sup_{a\in A}\{|a(i) -
-a(j)|: ||[D, \pi(a)]|| \leq 1\}$ for the set of data $A = \mathbb{C}^3$ acting
-in the defining representation $H = \mathbb{C}^3$, and
-\begin{align}
- D =
- \begin{pmatrix}
- 0 & d^{-1} & 0 \\ d^{-1} & 0 & 0 \\ 0 & 0 & 0
- \end{pmatrix},
-\end{align}
-for some $d \in \mathbb{R}$.
-From the data $A=\mathbb{C}^3$, $H=\mathbb{C}^3$ and $D$ we compute the
-commutator
-\begin{align}
- \big|\big|[D, \pi(a)]\big|\big| &= d^{-1}\left|\left|
-\begin{pmatrix}
- 0 & a(2)-a(1) & 0 \\
- -(a(2)-a(1)) & 0 & 0 \\
- 0 & 0 & 0
-\end{pmatrix} \right|\right|.
-\end{align}
-Hence the metric is
-\begin{align}
-d =
- \begin{pmatrix}
- 0 & a(1)-a(2) & a(1) \\
- a(2)-a(1) & 0 & a(2) \\
- -a(1) & -a(2) & 0
- \end{pmatrix}.
-\end{align}
-%---------------- EXERCISE
-
-The translation of the metric on $X$ into algebraic data assumes commutativity
-in $A$, this can be extended to a noncommutative matrix algebra, by the
-following metric on a structure space $\hat{A}$ of a matrix algebra
-$M_{n_i}(\mathbb{C}$
-\begin{equation}
- d_{ij} = \sup_{a\in A}\big\{|\text{Tr}(a(i)) - \text{Tr}((a(j))|: ||[D,
- a]|| \leq 1\big\}.\label{eq:discretemetric}
-\end{equation}
-Equation \eqref{eq:discretemetric} is special case of the Connes' distance
-formula on a structure space of $A$.
-
-Finally we have all three ingredients to define a finite spectral triple, an
-mathematical structure which encodes finite discrete geometry into algebraic data.
-\begin{definition}
- A \textit{finite spectral triple} is a tripe $(A, H, D)$, where $A$ is a unital $*$-algebra,
- faithfully represented on a finite-dimensional Hilbert space $H$, with a symmetric operator
- $D: H \rightarrow H$. (Note that $A$ is automatically a matrix algebra.)
-\end{definition}
-
-\subsubsection{Properties of Matrix Algebras}
-\begin{lemma}
- If $A$ is a unital C* algebra acting faithfully on a finite
- dimensional Hilbert space, then $A$ is a matrix algebra of the Form:
- \begin{align}
- A \simeq \bigoplus _{i=1}^N M_{n_i}(\mathbb{C})
- \end{align}
-\end{lemma}
-\begin{proof}
- The wording 'acting faithfully on a Hilbertspace' means that the
- $*$-representation is injective, or for a $*$-homomorphism that means
- one-to-one correspondence. And since $A$ acts faithfully on a Hilbert
- space, this means that $A$ is a $*$ subalgebra of a matrix algebra $L(H) = M_{\dim
- (H)}(\mathbb{C}$. Hence it follows, that $A$ is isomorphic to a matrix
- algebra.
-\end{proof}
-
-A simple illustration would be for an algebra $A = M_n(\mathbb{C})$ and
-$H=\mathbb{C}^n$. Since $A$ acts on $H$ with matrix multiplication and standard
-inner product and $D$ on $H$ is a hermitian matrix $n\times n$ matrix.
-
-\begin{definition}
- Given an finite spectral triple $(A, H, D)$, the $A$-bimodule of
- Connes' differential one-forms is
- \begin{align}\label{eq:connesoneforms}
- \Omega _D ^1 (A) := \left\{ \sum _k a_k[D, b_k]: a_k, b_k \in A
- \right\}.
- \end{align}
-\end{definition}
-Then there is a map $d:A\rightarrow \Omega _D ^1 (A)$, $d = [D, \cdot]$. Where
-$d$ is a derivation of the $*$-algebra in the sense that
-\begin{align}
- d(a\ b) = d(a)b + ad(b), \\
- d(a^*) = -d(a)^*.
-\end{align}
-Since we have $d(\cdot) = [D, \cdot]$, we can easily check the above equations
-\begin{align}
- d(a\ b) &= [D, a\ b] = [D, a]b + a[D,b]\nonumber\\
- &= d(a)\ b + a\ d(b).
-\end{align}
-And
-\begin{align}
- d(a^*) &= [D, a^*] = Da^* - a^*D \nonumber\\
- &=-(D^*\ a - a\ D^*) = -[D^*, a] \nonumber\\
- &= -d(a)^*.
-\end{align}
-Furthermore $\Omega _D^1 (A)$ is an $A$-bimodule, which can be seen by
-rewriting the defining equation \eqref{eq:connesoneforms} into
-\begin{align}
- a\ (a_k[D, b_k])\ b &= a\ a_k(D\ b_k - b_k\ D)\ b = \nonumber\\
- &= a\ a_k(D\ b_k\ b - b_k\ D\ b)=\nonumber\\
- &= a\ a_k(D\ b_k\ b - b_k\ D\ b - b_k\
- b\ D +b_k\ b\ D)=
- \nonumber\\
- &= a\ a_k(D\ b_k\ b-b_k\ b\ D + b_k\ b\ D - b_k\ D\ b) = \nonumber \\
- &= a\ a_k [D, b_k\ b] + a\ a_k\ b [D, b]=\nonumber\\
- &= \sum _k\ a_k'\ [D, b_k']
-\end{align}
-
-\begin{lemma}
- Let $\big(A, H, D\big) = \big(M_n(\mathbb{C}), \mathbb{C}^n, D\big)$, where
- $D$ is a hermitian $n\times n$ matrix. If $D$ is not a multiple of the
- identity then
- \begin{align}
- \Omega _D ^1 (A) \simeq M_n(\mathbb{C}) = A
- \end{align}
-\end{lemma}
-\begin{proof}
- Assume $D = \sum _i \lambda _i e_{ii}$ is diagonal, $\lambda _i \in \mathbb{R}$ and
- $\{e_{ij}\}$ is the basis of $M_n(\mathbb{C})$. Then for fixed $i$, $j$ choose $k$
- such that $\lambda _k \neq \lambda _j$, hence we have
- \begin{align} \label{eq:basis}
- \left(\frac{1}{\lambda _k - \lambda _j} e_{ik}\right) [D, e_{kj}] =
- e_{ij},
- \end{align}
- for $e_{ij}\in \Omega _D ^1 (A)$ by the above definition
- \eqref{eq:connesoneforms}. Ultimately we have
- \begin{align}
- \Omega _D ^1
- (A) \subset L(\mathbb{C}^n) = H \simeq M_n(\mathbb{C}) = A
- \end{align}
-\end{proof}
-
- Consider an example
- \begin{align}
- \left(A=\mathbb{C}^2, H=\mathbb{C}^2,
- D = \begin{pmatrix} 0 & \lambda \\ \bar{\lambda} & 0
- \end{pmatrix}\right)
- \end{align}
- with $\lambda \neq 0$. We can show that $\Omega _D^1(A)
- \simeq M_2(\mathbb{C})$. The Hilbert Basis $D$ can be extended in terms of
- the basis of $M_2(\mathbb{C})$, plugging this into Equation
- \eqref{eq:basis} will get us the same cyclic result, thus
- $\Omega _D^1(A) \simeq M_2(\mathbb{C})$.
-\
-
-\subsubsection{Morphisms Between Finite Spectral Triples}
-Next we will define an equivalence relation between finite spectral triples, called
-spectral unitary equivalence, which is given by the unitarity of the
-two matrix algebras themselves, and an additional map $U$ which allows us to associate a
-one operator to another second operator.
-\begin{definition}
- Two finite spectral tripes $(A_1, H_1, D_1)$ and $(A_2, H_2, D_2)$ are
- called unitary equivalent if $A_1 = A_2$ and there exists a map $U:\ H_1
- \rightarrow H_2$ that satisfies
- \begin{align}
- U\ \pi_1(a)\ U^* &= \pi_2(a)\;\;\;\; \text{with} \;\;\; a \in A_1,\\
- U\ D_1\ U^* &= D_2.
- \end{align}
-\end{definition}
-Notice that for any such $U$ we have the relation $(A, H, D) \sim (A, H, UDU^*)$.
-And hence $U\ D\ U^* = D + U[D, U^*]$ are of the form of elements in $\Omega _D^1 (A)$.
-
-%-------------- EXERCISE
-To make it clear that the above definition is an equivalence relation between
-finite spectral triples, we need to see if the relation satisfies
-reflexivity, symmetry and transitivity. Let us look then at three spectral
-triples $(A_1, H_1, D_1)$, $(A_2, H_2, D_2)$ and $(A_3, H_3, D_3)$.
-For reflexivity $(A_1, H_1, D_1) \sim (A_1, H_1, D_1)$. So there
-exists the unitary map $U: H_1 \rightarrow H_1$, which is the identity
-and always exists. On the other hand the symmetry condition requires
-\begin{align}
- (A_1, H_1, D_1) \sim (A_2, H_2, D_2) \Leftrightarrow
- (A_2, H_2, D_2) \sim (A_1, H_1, D_1).
-\end{align}
-Because $U$ is unitary we can rewrite for the representation for $A_1$
-\begin{align}
- &U\pi_1(a)U^* = \pi_2(a) \;\;\; | \cdot U^*\boxdot U\nonumber \\
- &U^*U\pi_1(a)U^*U = \pi_1(a) = U^*\pi_2(a)U.
-\end{align}
-The same relation applies for the symmetric operator $D$.
-Lastly for transitivity the condition is
-\begin{align}
- (A_1, H_1, D_1) &\sim (A_2, H_2, D_2) \;\;\; \text{and} \;\;\;
- (A_2, H_2, D_2) \sim (A_3, H_3, D_3) \nonumber\\
- &\Rightarrow (A_1, H_1, D_1) \sim (A_3, H_3, D_3).
-\end{align}
-Therefore the two unitary maps $U_{12}:H_1 \rightarrow H_2$ and
-$U_{23}: H_2 \rightarrow H_3$ are
-\begin{align}
- U_{23}\ U_{12}\ \pi_1(a)\ U^*_{12}\ U^*_{23} &= U_{23}\
- \pi_2(a)\ U_{23}^*\nonumber \\
- &= \pi_3(a), \\
- U_{23}\ U_{12}\ D_1\ U^*_{12}\ U^*_{23} &= U_{23}\
- D_2 U_{23}^* \nonumber\\
- &= D_3.
-\end{align}
-%-------------- EXERCISE
-
-In order to extend this relation we take a look at Morita equivalence of Matrix Algebras.
-\begin{definition}
- Let $A$ be an algebra. We say that $I \subset A$, as a vector space, is a
- right(left) ideal if $a\ b \in I$ for $a \in A$ and $b\in I$ (or $b\ a \in
- I$, $b\in I$, $a\in A$). We call a left-right ideal simply an ideal.
-\end{definition}
-
-Given a Hilbert bimodule $E \in KK_f(B, A)$ and $(A, H, D)$ we construct
-a finite spectral triple on $B$, $(B, H', D')$
-\begin{equation}
- H' = E \otimes _A H.
-\end{equation}
-We might define $D'$ with $D'(e \otimes \xi) = e\otimes D\xi$, thought this
-would not satisfy the ideal defining the balanced tensor product over $A$,
-which is generated by elements of the form
-\begin{align}
- e\ a \otimes \xi - e\otimes a\ \xi, \;\;\;\;\; e\in E, a\in A, \xi \in H.
-\end{align}
-This inherits the left action on $B$ from $E$ and has a $\mathbb{C}$
-valued inner product space. $B$ also satisfies the ideal
-\begin{equation}
- D'(e\otimes \xi) = e \otimes D\ \xi + \nabla (e)\ \xi, \;\;\;\; e\in
- E, a\in A,
-\end{equation}
-where $\nabla$ is called the \textit{connection on the right A-module E}
-associated with the derivation $d=[D, \cdot]$. The connection needs to
-satisfy the \textit{Leibnitz Rule}
-\begin{equation}
- \nabla(ae) = \nabla(e)a + e \otimes [D, a], \;\;\;\;\; e\in E,\; a\in A.
-\end{equation}
-Hence $D'$ is well defined on $E \otimes _A H$
-\begin{align}
- D'(e\ a \otimes \xi - e \otimes a\ \xi) &= D'(e\ a \otimes \xi) - D'(e
- \otimes \xi) \nonumber\\
- &= e\ a\otimes D\ \xi + \nabla(a\ e)\ \xi - e \otimes D(a\ \xi) - \nabla
- (e)\ a\ \xi \nonumber\\
- &= 0.
-\end{align}
-With the information thus far we can prove the following theorem
-\begin{theorem}
- If $(A, H, D)$ a finite spectral triple, $E \in KK_f(B, A)$.
- Then $(V, E\otimes _A H, D')$ is a finite spectral triple, provided that
- $\nabla$ satisfies the compatibility condition
- \begin{equation}
- \langle e_1, \nabla e_2 \rangle _E - \langle \nabla e_1, e_2
- \rangle _E = d\langle e_1, e_2 \rangle _E \;\;\;\; e_1, e_2 \in E
- \end{equation}
-\end{theorem}
-\begin{proof}
- $E\otimes _A H$ was previously. The only thing left is to show that $D'$ is a symmetric
- operator, this we can just compute. Let $e_1, e_2 \in E$ and $\xi _1,
- \xi _2 \in H$ then
- \begin{align}
- \langle e_1 \otimes \xi _1, D'(e_2 \otimes \xi_2)\rangle _{E\otimes _A H} &=
- \langle \xi _1, \langle e_1, \nabla e_2\rangle _E\ \xi _2\rangle
- \langle \xi _1 , \langle e_1, e_2\rangle _E\ D\ \xi_2\rangle_H \nonumber\\
- &= \langle \xi _1, \langle \nabla e_1, e_2\rangle _E\ \xi _2\rangle _H + \langle \xi _1, d\langle e_1, e_2\rangle _E
- \ \xi _2\rangle _H \nonumber\\
- &+ \langle D\ \xi _1,\langle e_1, e_2\rangle _E\ \xi _2\rangle _H -
- \langle \xi _1, [D, \langle e_1, e_2\rangle _E]\ \xi
- _2 \rangle _H \nonumber\\
- &= \langle D'(e_1 \otimes \xi _1), e_2 \otimes \xi _2\rangle _{E \otimes _A H}
- \end{align}
-\end{proof}
-
-Let us examin what happens if we look at difference of connectoins $\nabla$ and
-$\nabla'$ on a right $A$-module $E$. Since both connections need to satisfy
-the Leiblitz rule, the difference should also
- \begin{align}
- \nabla(ea)-\nabla'(ea)&=\nabla(e) + e\otimes[D, a]\nonumber\\
- &-(\nabla'(e)a + e\otimes[D',a])\nonumber\\
- &=\bar{\nabla}a + e\otimes(Da-aD-D'a+aD')\nonumber\\
- &=\bar{\nabla}a + e\otimes((D-D')a-a(D-D'))\nonumber\\
- &=\bar{\nabla}a + e\otimes[D', a]\nonumber\\
- &=\bar{\nabla}(ea).
- \end{align}
-Therefore $\nabla-\nabla'$ is a right $A$-linear map
-$E \rightarrow E\otimes _A \Omega _D^1(A)$.
-
-To get a better grasp of the results let us construct a finite spectral
-triple $(A, H', D')$ from $(A, H, D)$. The derivation $d(\cdot):A \rightarrow
-A\otimes _A \Omega_D^1(A)=\Omega_D^1(A)$ is a connection on $A$ considered a
-right $A$-module
-\begin{align}
- \nabla(e \cdot a) = d(a),
-\end{align}
-hence $A\otimes_A H\simeq H$. Next we can construct the operator $D'$
-for the connection $d(\cdot)$,
-\begin{align}
- D'(a\xi) = a(D\xi) + (\nabla a) \xi = D(a\xi).
-\end{align}
-By using the identity element in the connection relation
-\begin{align}
- \nabla (e\cdot a) = \nabla(e) a + 1 \otimes d(a)=d(a) \nabla(e) a,
-\end{align}
-we see that any connection $\nabla: A\rightarrow A\otimes_A \Omega_D^1(A)$ is
-given by
-\begin{align}
- \nabla = d + \omega,
-\end{align}
-where $\omega \in \Omega_D^1(A)$. Ultimately the the
-difference operator $D'$ with the connection on $A$ is given by
-\begin{align}
- D'(a\otimes \xi) &= D'(a \xi) = a(D\xi) + (\nabla a)\xi \nonumber \\
- &=a(D\xi) + \nabla(e \cdot a) \xi \nonumber\\
- &= D(a\xi) + \nabla(e) (a\xi).
-\end{align}
-So any such connection is of the form
-\begin{align}
- \nabla = d + \omega.
-\end{align}
-
-%\subsubsection{Graphing Finite Spectral Triples}
-%\begin{definition}
-% A \textit{graph} is a ordered pair $(\Gamma ^{(0)}, \Gamma ^{(1)})$.
-% Where $\Gamma ^{(0)}$ is the set of vertices (nodes) and $\Gamma ^{(1)}$
-% a set of pairs of vertices (edges)
-%\end{definition}
-%\begin{figure}[h!]
-% \centering
-%\begin{tikzpicture}[
-% mass/.style = {draw,circle, minimum size=0.2cm, inner sep=0pt, thick},
-% spring/.style = {decorate,decoration={zigzag, pre length=1cm,post length=1cm,segment length=5pt}},]
-% \node[mass] (m1) at (1,1.5) {};
-% \node[mass] (m2) at (-1,1.5) {};
-% \node[mass] (m3) at (0,0) {};
-%
-% \draw (m1) -- (m2);
-% \draw (m1) -- (m3);
-% \draw (m2) -- (m3);
-% \end{tikzpicture}
-% \caption{A simple graph with three vertices and three edges}
-%\end{figure}
-%%\begin{MyExercise}
-%% \textbf{
-%% Show that any finite-dimensional faithful representation $H$ of a matrix
-%% algebra $A$ is completely reducible. To do that show that the complement
-%% $W^{\perp}$ of an $A$-submodule $W\subset H$ is also an $A$-submodule
-%% of $H$.
-%%}\newline
-%%
-%% $A\simeq \bigoplus_{i=1}^N M_{n_i}(\mathbb{C})$ is the matrix algebra
-%% then $H$ is a Hilbert $A$-bimodule and $W$ a submodule of $A$.
-%% Because we have $H = W \cup W^{\perp}$, then $W^{\perp}$ is naturally a
-%% $A$-submodule, because elements in $W^{\perp}$ need to satisfy the
-%% bimodularity.
-%%\end{MyExercise}
-%\begin{definition}
-% A $\Lambda$-decorated graph is given by an ordered pair $(\Gamma,
-% \Lambda)$ of a finite graph $\Gamma$ and a set of positive integers
-% $\Lambda$ with the labeling
-% \begin{itemize}
-% \item of the vetices $v\in \Gamma ^{(0)}$ given by $n(\nu) \in
-% \Lambda$
-% \item of the edges $e = (\nu _1, \nu _2) \in \Gamma ^{(1)}$ by
-% operators
-% \begin{itemize}
-% \item $D_e: \mathbb{C}^{n(\nu _1)} \rightarrow
-% \mathbb{C}^{n(\nu _2)}$
-% \item and $D_e^*: \mathbb{C}^{n(\nu _2)} \rightarrow
-% \mathbb{C}^{n(\nu _1)}$ its conjugate traspose
-% (pullback?)
-% \end{itemize}
-% \end{itemize}
-% such that
-% \begin{equation}
-% n(\Gamma ^{(0)}) = \Lambda
-% \end{equation}
-%\end{definition}
-%\begin{question}
-% Would then $D_e$ be the pullback?
-%\end{question}
-%\begin{question}
-% These graphs are important in the next chapter I should look
-% into it more, I don't understand much here, specific
-% how to construct them with the abstraction of a spectral triple...
-%\end{question}
-%
-%The operator $D_e$ between $\textbf{n}_i$ and $\textbf{n}_j$ add up to
-%$D_{ij}$
-%\begin{align}
-% D_{ij} = \sum\limits_{\substack{e = (\nu _1, \nu _2) \\ n(\nu _1) =
-% \textbf{n}_i \\ n(\nu _2) = \textbf{n}_j}} D_e
-%\end{align}
-%
-%\begin{theorem}
-% There is a on to one correspondence between finite spectral triples
-% modulo unitary equivalence and $\Lambda$-decorated graphs, given by
-% associating a finite spectral triples $(A, H, D)$ to a $\Lambda$ decorated
-% graph $(\Gamma, \Lambda)$ in the following way:
-% \begin{equation}
-% A = \bigoplus _{n\in \Lambda} M_n(\mathbb{C}); \;\;\;
-% H = \bigoplus _{\nu \in \Gamma ^{(0)}} \mathbb{C}^{n(\nu)}; \;\;\;
-% D = \sum _{e \in \Gamma ^{(1)}} D_e + D_e^*
-% \end{equation}
-%\end{theorem}
-% \begin{figure}[h!]
-% \centering
-% \begin{tikzpicture}[
-% mass/.style = {draw,circle, minimum size=0.3cm, inner sep=0pt, thick},
-% ]
-%
-% \node[mass, label={\textbf{n}}] (m1) at (1,0) {};
-% \draw (m1) to [out=330, in=210, looseness=25] node[above] {$D_e$} (m1);
-% \end{tikzpicture}
-% \caption{A $\Lambda$-decorated Graph of $(M_n(\mathbb{C}), \mathbb{C}^n,
-% D = D_e + D_e^*)$}
-%\end{figure}
-%
-%%\begin{MyExercise}
-%% \textbf{
-%% Draw a $\Lambda$ decorated graph corresponding to the spectral triple
-%% $(A=\mathbb{C}^3, H=\mathbb{C}^3, D=\begin{pmatrix}0 & \lambda & 0\\
-%% \bar{\lambda} &0 &0 \\ 0&0&0\end{pmatrix})$
-%%}\newline
-%%
-%%\centering
-%%\begin{tikzpicture}[
-%% mass/.style = {draw,circle, minimum size=0.4cm, inner sep=0pt, thick},
-%% spring/.style = {decorate,decoration={zigzag, pre length=1cm,post length=1cm,segment length=5pt}},]
-%% \node[mass] (m1) at (-1,1.5) {\textbf{1}};
-%% \node[mass] (m2) at (1,1.5) {\textbf{2}};
-%% \node[mass] (m3) at (3,1.5) {\textbf{3}};
-%%
-%% \draw[style=thick, -] (1.1,1.7) -- (-1.1,1.7);
-%% \draw[style=thick, -] (1.1,1.3) -- (-1.1,1.3);
-%% \end{tikzpicture}
-%% % \captionof{figure}{Solution}
-%%\end{MyExercise}
-%%\begin{MyExercise}
-%% \textbf{
-%% Use $\Lambda$-decorated graphs to classify all finite spectral triples
-%% (modulo unitary equivalence) on the matrix algebra
-%% $A=\mathbb{C}\oplus M_2(\mathbb{C})$
-%%}\newline
-%%
-%% \centering
-%%\begin{tikzpicture}[
-%% mass/.style = {draw,circle, minimum size=0.4cm, inner sep=0pt, thick},
-%% spring/.style = {decorate,decoration={zigzag, pre length=1cm,post length=1cm,segment length=5pt}},]
-%% \node[mass] (m1) at (-1,1) {\textbf{1}};
-%% \node[mass] (m2) at (1,1) {\textbf{2}};
-%% \node[mass] (m3) at (3,1) {\textbf{3}};
-%%
-%% \node[mass] (m4) at (-1,0) {\textbf{1}};
-%% \node[mass] (m5) at (1,0) {\textbf{2}};
-%% \node[mass] (m6) at (3,0) {\textbf{3}};
-%%
-%% \node[mass] (m7) at (-1,-1) {\textbf{1}};
-%% \node[mass] (m8) at (1,-1) {\textbf{2}};
-%% \node[mass] (m9) at (3,-1) {\textbf{3}};
-%%
-%% \node[mass] (m10) at (-1,-2) {\textbf{1}};
-%% \node[mass] (m11) at (1,-2) {\textbf{2}};
-%% \node[mass] (m12) at (3,-2) {\textbf{3}};
-%%
-%% \draw[style=thick, -] (1.1,0.2) -- (-1.1,0.2);
-%% \draw[style=thick, -] (1.1,-0.2) -- (-1.1,-0.2);
-%% \draw[style=thick, -] (m7) to [out=330, in=210, looseness=10] node[above] {} (m7);
-%% \draw[style=thick, -] (m10) -- (m11) ;
-%%
-%%\end{tikzpicture}
-%%% \captionof{figure}{Solution $A=M_3(\mathbb{C})$}
-%%\end{MyExercise}
-%\subsubsection{Graph Construction of Finite Spectral Triples}
-%\textbf{Algebra:}We know if a acts on a finite dimensional Hilbert space then
-%this C* algebra is isomorphic to a matrix algebra so $A \simeq
-%\bigoplus_{i=1}^{N}M_{n_i}(\mathbb{C})$. Where $i\in
-%\hat{A}$ represents an equivalence class and runs from $1$ to $N$,
-%thus $\hat{A}\simeq\{1,\dots, N\}$. We label equivalence classes by
-%$\textbf{n}_i$, then $\hat{A}\simeq\{\textbf{n}_1,\dots,\textbf{n}_N\}$.
-%\newline
-%
-%\textbf{Hilbert Space:} Since every Hilbert space that acts faithfully on a
-%C* algebra is completely reducible, it is isomorphic to the composition
-%of irreducible representations. $H \simeq \bigoplus_{i=1}^N\mathbb{C}^{n_i}
-%\otimes V_i$. Where all $V_i$'s are Vector spaces, their dimension is the
-%multiplicity of the representation landed by $\textbf{n}_i$ to $V_i$ itself
-%by the multiplicity space.
-%\newline
-%
-%\textbf{Finite Dirac Operator:} $D_{ij}$ is connecting nodes $\textbf{n}_i$
-%and $\textbf{n}_j$, with a symmetric map $D_{ij}:\mathbb{C}^{n_i}\otimes V_i
-%\rightarrow \mathbb{C}^{n_j}\otimes V_j$
-%\newline
-%
-%To draw a graph, draw nodes in position $\textbf{n}_i\in \hat{A}$.
-%Multiple nodes at the same position represent multiplicities in $H$.
-%Draw lines between nodes to represent $D_{ij}$.
-%
-%\begin{figure}[h!]
-% \centering
-%\begin{tikzpicture}
-% \node[draw, label=above:{$\textbf{n}_1$},circle, thick] at (-3,0) {};
-% \node[label=above:{$\dots$}] at (-2,0) {};
-% \node[draw, label=above:{$\textbf{n}_i$},circle, thick] at (-1,0) {};
-% \node[label=above:{$\dots$}] at (0,0) {};
-% \node[draw, label=above:{$\textbf{n}_j$},circle, thick] at (1,0) {};
-% \node[draw, label=above:{},circle, thick, inner sep=0cm, minimum
-% size=0.2cm] at (1,0) {};
-% \node[label=above:{$\dots$}] at (2,0) {};
-% \node[draw, label=above:{$\textbf{n}_N$},circle, thick] at (3,0) {};
-%
-% \draw[style=thick, -] (-1,-0.2) -- (1,-0.2);
-% \draw[style=thick, -] (-1,0.2) -- (1,0.2);
-% \path[style=thick, -] (-1,-0.2) edge[bend right=15]
-% node[pos=0.5,below] {} (3,-0.2);
-% \end{tikzpicture}
-% \caption{Example}
-%\end{figure}
diff --git a/src/thesis/chapters/backup/heatkernel.tex b/src/thesis/chapters/backup/heatkernel.tex
@@ -1,314 +0,0 @@
-\subsection{Heat Kernel Expansion}
-\subsubsection{The Heat Kernel}
-The heat kernel $K(t; x, y; D)$ is the fundamental solution of the heat
-equation
-\begin{align}
- (\partial _t + D_x)K(t;x, y;D) =0,
-\end{align}
-which depends on the operator $D$ of Laplacian type.
-
-For a flat manifold $M = \mathbb{R}^n$ and $D = D_0 := -\Delta_\mu\Delta^\mu +m^2$ the
-Laplacian with a mass term and the initial condition
-\begin{align}
- K(0;x,y;D) = \delta(x,y),
-\end{align}
-takes the form of the standard fundamental solution
-\begin{align}\label{eq:standard}
- K(t;x,y;D_0) = (4\pi t)^{-n/2}\exp\left(-\frac{(x-y)^2}{4t}-tm^2\right).
-\end{align}
-Let us consider now a more general operator $D$ with a potential term or a
-gauge field, the heat kernel reads then
-\begin{align}
- K(t;x,y;D) = \langle x|e^{-tD}|y\rangle.
-\end{align}
-We can expand the heat kernel in $t$, still having a
-singularity from the equation \eqref{eq:standard} as $t \rightarrow 0$ thus the
-expansion reads
-\begin{align}
- K(t;x,y;D) = K(t;x,y;D_0)\left(1 + tb_2(x,y) + t^2b_4(x,y) + \dots
- \right),
-\end{align}
-where $b_k(x,y)$ become regular as $y \rightarrow x$. These coefficients are called the heat
-kernel coefficients.
-%%----------------------- KANN WEGGELASSEN WERDEN
-%\newline
-%\textbf{KANN WEGELASSEN WERDEN BIS ZUM NÄCHSTEN KAPITEL}
-%Let's turn our attention to a propagator $D^{-1}(x,y)$ defined through the
-%heat kernel, with an integral representation
-%\begin{align}
-% D^{-1} (x,y) = \int_0^\infty dt K(t;x,y;D).
-%\end{align}
-%If we assume the heat kernel vanishes for $t\rightarrow \infty$, we can
-%integrate formally to get
-%\begin{align}
-% D^{-1}(x,y) \simeq
-% 2(4\pi)^{-n/2}\sum_{j=0}\left(\frac{|x-y|}{2m}\right)^{-\frac{n}{2}+j+1}
-% K_{-\frac{n}{2}+j+1}(|x-y|m)b_{2j}(x,y),
-%\end{align}
-%where $b_0 = 1$ and $K_\nu (z)$ is the Bessel function
-%\begin{align}
-% K_\nu(z) = \frac{1}{\pi} \int_0^\pi \cos(\nu\tau-z\sin(\tau))d\tau.
-%\end{align}
-%The Bessel function solves the following differential equation
-%\begin{align}
-% z^2 \frac{d^2K}{dz^2} + z \frac{dK}{dz} + (z^2 - \nu^2)=0.
-%\end{align}
-%By looking at an integral approximation for the propagator we conclude that
-%the singularities of $D^{-1}$ coincide with the singularities of the heat
-%kernel coefficients. Thus we can say, that a generating functional in terms of
-%$\det(D)$ is called the one-loop effective action (quantum field theory)
-%\begin{align}
-% W = \frac{1}{2}\ln(\det D).
-%\end{align}
-%We have a direct relation with one-loop effective action $W$ and the
-%heat kernel. Furthermore notice that for each eigenvalue $\lambda >0$ of $D$
-%we can write the identity.
-%\begin{align}
-% \ln \lambda = -\int_0^\infty \frac{e^{-t\lambda}}{t}dt
-%\end{align}
-%This expression is correct up to an infinite constant which does not depend
-%on the eigenvalue $\lambda$, thus we can ignore it. By substituting
-%$\ln(\det D) = \text{Tr}(\ln D)$ we can rewrite the one-loop effective action
-%$W$ into
-%\begin{align}
-% W = -\frac{1}{2} \int_0^\infty dt \frac{K(t, D)}{t},
-%\end{align}
-%where
-%\begin{align}
-% K(t, D) = \text{Tr}(e^{-tD}) = \int d^n x \sqrt{g}K(t;x,x;D).
-%\end{align}
-%The problem now is that the integral of $W$ is divergent at both limits. Yet
-%the divergences at $t\rightarrow \infty$ are caused by $\lambda \leq 0$ of $D$
-%(infrared divergences) and can be ignored. The divergences at $t\rightarrow 0$
-%are cutoff at $t=\Lambda^{-2}$, simply written as
-%\begin{align}
-% W_\Lambda = -\frac{1}{2} \int_{\Lambda^{-2}}^\infty dt \frac{K(t, D)}{t}.
-%\end{align}
-%We can calculate $W_\Lambda$ up to an order of $\lambda ^0$
-%\begin{align}
-% W_\Lambda &= -(4\pi)^{-n/2} \int d^n x\sqrt{g}\bigg(
-% \sum_{2(j+l)<n}\Lambda^{n-2j-2l}b_{2j}(x,x) \frac{(-m^2)^l l!}{n-2j-2l} +\\
-% &+ \sum_{2(j+l) =n }\ln(\Lambda) (-m^2)^l l! b_{2j}(x,x)
-% \mathcal{O}(\lambda^0) \bigg)
-%\end{align}
-%There is an divergence at $b_2(x,x)$ for $k\leq n$. Computing the limit
-%$\Lambda \rightarrow \infty$ we get
-%\begin{align}
-% -\frac{1}{2}(4\pi)^{n/2}m^n\int d^n x\sqrt{g} \sum_{2j>n}
-% \frac{b_{2j}(x,x)}{m^{2j}}\Gamma(2j-n),
-%\end{align}
-%where $\Gamma$ stands for the gamma function.
-%%----------------------- KANN WEGGELASSEN WERDEN
-
-
-\subsubsection{Spectral Functions}
-Manifolds $M$ with a disappearing boundary condition for the operator $e^{-tD}$ for $t>0$ is a
-trace class operator on $L^2(V)$. Meaning for any smooth function $f$ on $M$
-we can define
-\begin{align}
- K(t,f,D) := \text{Tr}_{L^2}(fe^{-tD}),
-\end{align}
-or alternately write an integral representation
-\begin{align}
- K(t, f, D) = \int_M d^n x \sqrt{g} \text{Tr}_V(K(t;x,x;D)f(x)),
-\end{align}
-in the regular limit $y\rightarrow y$. We can write the Heat Kernel in terms
-of the spectrum of $D$. So for an orthonormal basis $\{\phi_\lambda\}$ of
-eigenfunctions for $D$, which corresponds to the eigenvalue $\lambda$, we
-can rewrite the heat kernel into
-\begin{align}
- K(t;x,y;D) = \sum_\lambda \phi^\dagger_\lambda(x)
- \phi_\lambda(y)e^{-t\lambda}.
-\end{align}
-An asymptotic expansion as $t \rightarrow 0$ for the trace is then
-\begin{align}
- \text{Tr}_{L^2}(fe^{-tD}) \simeq \sum_{k\geq 0}t^{(k-n)/2}a_k(f,D),
-\end{align}
-where
-\begin{align}
- a_k(f,D) = (4\pi)^{-n/2} \int_M d^4x \sqrt{g} b_k(x,x) f(x).
-\end{align}
-\subsubsection{General Formulae}
-Let us summarize what we have obtained in the last chapter, we considered a
-compact Riemannian manifold $M$ without boundary condition, a vector bundle
-$V$ over $M$ to define functions which carry discrete (spin or gauge)
-indices, an operator $D$ of Laplace type over $V$ and smooth function $f$ on
-$M$.
-
-There is an asymptotic expansion where the heat kernel coefficients with an
-odd index $k=2j+1$ vanish $a_{2j+1}(f,D) = 0$. On the other hand coefficients
-with an even index are locally computable in terms of geometric invariants
-\begin{align}
- a_k(f,D) &= \text{Tr}_V\left(\int_M d^n x\sqrt{g}(f(x)a_k(x;D)\right)
- =\nonumber\\
- &=\sum_I \text{Tr}_V\left(\int_M d^nx \sqrt{g}(fu^I
- \mathcal{A}^I_k(D))\right).
-\end{align}
-We denote $\mathcal{A}^I_k$ as all possible independent invariants of
-dimension $k$, and $u^I$ are constants. The invariants are constructed from
-$E, \Omega, R_{\mu\nu\varrho\sigma}$ and their derivatives If $E$ has
-dimension two, then the derivative has dimension one. So if $k=2$ there are
-only two independent invariants, $E$ and $R$. This corresponds to the
-statement $a_{2j+1}=0$.
-
-If we consider $M = M_1 \times M_2$ with coordinates $x_1$ and $x_2$ and a
-decomposed Laplace style operator $D = D_1 \otimes 1 + 1 \otimes D_2$ we can
-separate functions acting on operators and on coordinates linearly by the
-following
-\begin{align}
- e^{-tD} &= e^{-tD_1} \otimes e^{-tD_2},\\
- f(x_1, x_2) &= f_1(x_1)f_2(x_2),
-\end{align}
-thus the heat kernel coefficients are separated by
-\begin{align}
- a_k(x;D) &= \sum_{p+q=k} a_p(x_1; D_1)a_q(x_2;D_2)
-\end{align}
-If we know the eigenvalues of $D_1$ are known, $l^2, l\in \mathbb{Z}$, we
-can obtain the heat kernel asymmetries with the Poisson summation formula
-giving us an approximation in the order of $e^{-1/t}$
-\begin{align}
- K(t, D_1) &= \sum_{l\in\mathbb{Z}} e^{-tl^2} = \sqrt{\frac{\pi}{t}}
- \sum_{l\in\mathbb{Z}} e^{-\frac{\pi^2l^2}{t}} = \nonumber \\
- &\simeq \sqrt{\frac{\pi}{t}} + \mathcal{O}(e^{-1/t}).
-\end{align}
-The exponentially small terms have no effect on the heat kernel
-coefficients and that the only nonzero coefficient is $a_0(1, D_1) =
-\sqrt{\pi}$, therefore the heat coefficients can be written as
-\begin{align}
- a_k(f(x^2), D) = \sqrt{\pi}\int_{M_2}
- d^{n-1}x\sqrt{g}\sum_I\text{Tr}_V\left(f(x^2)u^I_{(n-1)}
- \mathcal{A}^I_n(D_2)\right).
-\end{align}
-
-Because all of the geometric invariants associated with $D$ are in the $D_2$
-part, they are independent of $x_1$. Ultimately meaning we are free to choose
-$M_1$. For $M_1 = S^1$ with $x\in (0, 2\pi)$ and $D_1=-\partial_{x_1}^2$
-we can rewrite the heat kernel coefficients into
-\begin{align}
- a_k(f(x_2), D) &= \int_{S^1\times M_2}d^nx \sqrt{g} \sum_I
- \text{Tr}_V(f(x_2) u_{(n)}^I \mathcal{A}^I_k(D_2))= \nonumber\\
- &= 2\pi \int_{M_2} d^nx\sqrt{g} \sum_I\text{Tr}_V(f(x_2) u_{(n)}^I
- \mathcal{A}^I_k(D_2)).
-\end{align}
-Computing the two equations above we see that
-\begin{align}
- u_{(n)}^I = \sqrt{4\pi} u^I_{(n+1)}
-\end{align}
-
-\subsubsection{Heat Kernel Coefficients}
-To calculate the heat kernel coefficients we need the following variational
-equations
-\begin{align}
- &\frac{d}{d\varepsilon}\bigg|_{\varepsilon=0}a_k(1, e^{-2\varepsilon f}D) =
- (n-k) a_k(f, D),\label{eq:var1}\\
- &\frac{d}{d\varepsilon}\bigg|_{\varepsilon=0}a_k(1, D-\varepsilon F) =
- a_{k-2}(F,D),\label{eq:var2}\\
- &\frac{d}{d\varepsilon}\bigg|_{\varepsilon=0}a_k(e^{-2\varepsilon f}F,
- e^{-2\varepsilon f}D) =
- 0\label{eq:var3}.
-\end{align}
-Let us explain the equations above. To get the first equation \eqref{eq:var1}
-we differentiate \begin{align}
- \frac{d}{d\varepsilon}\bigg|_{\varepsilon=0} \text{Tr}(\exp(-e^{-2\varepsilon
- f}tD) = \text{Tr}(2ftDe^{-tD}) = -2t\frac{d}{dt}\text{Tr}(fe^{-tD}))
-\end{align}
-then we expand both sides in $t$ and get \eqref{eq:var1}. Equation \eqref{eq:var2} is derived similarly.
-
-For equation \eqref{eq:var3} we consider the following operator
-\begin{align}
- D(\varepsilon,\delta) = e^{-2\varepsilon f}(D-\delta F)
-\end{align}
-for $k=n$ we use equation \eqref{eq:var1} and we get
-\begin{align}
- \frac{d}{d\varepsilon}\bigg|_{\varepsilon=0}a_n(1,D(\varepsilon,\delta))
- =0,
-\end{align}
-then we take the variation in terms of $\delta$, evaluated at $\delta =0$ and
-swap the differentiation, allowed by theorem of Schwarz
-\begin{align}
- 0 &=
- \frac{d}{d\delta}\bigg|_{\delta=0}\frac{d}{d\varepsilon}\bigg|_{\varepsilon=0}a_n(1,
- D(\varepsilon,\delta)) =\nonumber\\
- &=\frac{d}{d\varepsilon}\bigg|_{\varepsilon=0}\frac{d}{d\delta}\bigg|_{\delta=0}a_n(1,
- D(\varepsilon,\delta)) =\nonumber\\
- &=a_{n-2} ( e^{-2\varepsilon f}F, e^{-2\varepsilon f}D),
-\end{align}
-which gives us equation \eqref{eq:var3}.
-
-Now that we have established the ground basis, we can calculate the constants
-$u^I$, and by that the first three heat kernel coefficients read
-\begin{align}
- a_0(f, D) &= (4\pi)^{-n/2}\int_Md^n x\sqrt{g} \text{Tr}_V(a_0 f),\\
- a_2(f, D) &= (4\pi)^{-n/2}\frac{1}{6}\int_Md^n
- x\sqrt{g}\text{Tr}_V)(f\alpha _1 E+\alpha _2 R),\\
- a_4(f, D) &= (4\pi)^{-n/2}\frac{1}{360}\int_Md^n
- x\sqrt{g}\text{Tr}_V(f(\alpha_3 E_{,kk} + \alpha_4\ R\ E + \alpha_5 E^2
- \alpha_6 R_{,kk} + \nonumber\\
- &+\alpha_7 R^2 + \alpha_8 R_{ij}R_{ij} + \alpha_9
- R_{ijkl}R_{ijkl} +\alpha_{10} \Omega_{ij}\Omega{ij})),
-\end{align}
-where the comma subscript $,$ denotes the derivative and constants $\alpha_I$
-do not depend on the dimension of the Manifold and we can compute them with
-our variational identities.
-
-The first coefficient $\alpha_0$ can be read from the heat kernel expansion of
-the Laplacian on $S^1$ (above), $\alpha_0 = 1$. For $\alpha_1$ we use
-\eqref{eq:var2}, the coefficient $k = 2$ is
-\begin{align}
- \frac{1}{6} \int_M d^n x\sqrt{g} \text{Tr}_V(\alpha_1F) = \int_M d^n
- x\sqrt{g} \text{Tr}_V(F),
-\end{align}
-which means $\alpha_1 = 6$. Looking at the coefficient $k=4$ we have
-\begin{align}
- \frac{1}{360}\int_Md^n x \sqrt{g}\text{Tr}_V(\alpha_4\ F\ R + 2\alpha_5\ F\ E)
- = \frac{1}{6} \int_Md^n x\sqrt{g}\text{Tr}_V(\alpha_1\ F\ E + \alpha_2\ F\ R),
-\end{align}
-thus $\alpha_4 = 60\alpha_2$ and $\alpha_5 = 180$.
-
-By applying \eqref{eq:var3} to $n=4$ we get
-\begin{align}
- \frac{d}{d\varepsilon}|_{\varepsilon=0} a_2(e^{-2\varepsilon f}F,
- e^{-2\varepsilon f}D) = 0.
-\end{align}
-Collecting the terms with $\text{Tr}_V(\int_Md^nx\sqrt{g}(Ff_{,jj}))$ we
-obtain $\alpha_1 = 6\alpha_2$, that is $\alpha_2 = 1$, so $\alpha_4 = 60$.
-
-Now we let $M=M_1\times M_2$ and split $D = -\Delta_1 -\Delta_2$, where
-$\Delta_{1/2}$ are Laplacians for $M_1, M_2$. This allows us to decompose the heat
-kernel coefficient for $k=4$ into
-\begin{align}
- a_4(1,-\Delta_1-\Delta_2) =& a_4(1, -\Delta_1) a_0(1,
- -\Delta_2)\nonumber+ \\
- &+a_2(1,-\Delta_1) a_2(1,-\Delta_2)\nonumber \\
- &+ a_0(1,-\Delta_1)a_4(1,-\Delta_2),
-\end{align}
-with $E=0$ and $\Omega =0$ and by calculating the terms with $R_1R_2$ (scalar
-curvature of $M_{1/2}$) we obtain $\frac{2}{360}\alpha_7 =
-(\frac{\alpha_2}{6})^2$, thus $\alpha_7 = 5$.
-
-For $n=6$ we get
-\begin{align}
- 0 &= \text{Tr}_V(\int_Md^nx\sqrt{g}
- (F(-2\alpha_3-10\alpha_4+4\alpha_5)f_{,kk}E +\nonumber\\
- &+(2\alpha_3 + 10\alpha_6)f_{,iijj}+\nonumber\\
- &+(2\alpha_4 -2\alpha_6 - 20\alpha_7 -2\alpha_8)f_{,ii}R\nonumber\\
- &+(-8\alpha_8 -8\alpha_6)f_{,ij}R_{ij}))
-\end{align}
-we obtain $\alpha_3 = 60$, $\alpha_6=12$, $\alpha_8 = -2$ and $\alpha_9 = 2$
-
-To get $\alpha_{10}$ we use the Gauss-Bonnet theorem, ultimately giving us
-$\alpha_{10}=30$. We leave out this lengthy calculation and refer to
-\cite{heatkernel} for further reading.
-
-Let us summarize our calculations which ultimately give us the following heat kernel
-coefficients
-\begin{align}
- \alpha_0(f, D) &= (4\pi)^{-n/2}\int_M d^n x \sqrt{g} \text{Tr}_V(f),\\
- \alpha_2(f, D) &= (4\pi)^{-n/2}\frac{1}{6}\int_M d^n x \sqrt{g}
- \text{Tr}_V(f(6E+R)),\\
- \alpha_4(f, D) &= (4\pi)^{-n/2}\frac{1}{360}\int_M d^n x \sqrt{g}
- \text{Tr}_V(f(60E_{,kk}+60RE+ 180E^2 +\\
- &+12R_{,kk} + 5R^2 - 2 R_{ij}R_{ij}
- 2R_{ijkl}R_{ijkl} +30\Omega_{ij}\Omega_{ij})).
-\end{align}
-
diff --git a/src/thesis/chapters/backup/intro.tex b/src/thesis/chapters/backup/intro.tex
@@ -1,2 +0,0 @@
-\section{Introduction}
-\lipsum[2]
diff --git a/src/thesis/chapters/backup/main_sec.tex b/src/thesis/chapters/backup/main_sec.tex
@@ -1 +0,0 @@
-\section{Main Section}
diff --git a/src/thesis/chapters/backup/realncg.tex b/src/thesis/chapters/backup/realncg.tex
@@ -1,287 +0,0 @@
-\subsection{Finite Real Noncommutative Spaces}
-\subsubsection{Finite Real Spectral Triples}
-In this chapter we supplement the finite spectral triples with a \textit{real
-structure}. We additionally require a symmetry condition that that $H$ is a
-$A$-$A$-bimodule rather than only a $A$-left module. This ansatz has tight
-bounds with physical properties such as charge conjugation, which we will
-dive in deeper in later chapters. For this we will need to set a basis
-of definitions to get an overview.
-First we introduce a $\mathbb{Z}_2$-grading $\gamma$ with the following
-properties
-\begin{align}
- \gamma ^* &= \gamma, \\
- \gamma ^2 &= 1, \\
- \gamma D &= - D \gamma,\\
- \gamma a &= a \gamma, \;\;\;\; a\in A.
-\end{align}
-Then we can define a finite real spectral triple.
-\begin{definition}
- A \textit{finite real spectral triple} is given by a finite spectral
- triple $(A, H, D)$ and a anti-unitary operator $J:H\rightarrow H$ called
- the \textit{real structure}, such that
- \begin{align}
- a^\circ := J\ a^*\ J^{-1},
- \end{align}
- is a right representation of $A$ on $H$, that is $(ab)^\circ = b^\circ
- a^\circ$. With two requirements
- \begin{align}
- &[a, b^\circ] = 0,\\
- &[[D, a],\ b^\circ] = 0.
- \end{align}
- The two properties are called the \textit{commutant property}, they
- require that the left action of an element in $A$ and $\Omega _D^1(A)$ commutes with the right
- action on $A$.
-\end{definition}
-\begin{definition}
- The $KO$-dimension of a real spectral triple is determined by the sings
- $\epsilon, \epsilon ' ,\epsilon '' \in \{-1, 1\}$ appearing in
- \begin{align}
- J^2 &= \epsilon, \\
- J\ D &= \epsilon \ D\ J,\\
- J\ \gamma &= \epsilon''\ \gamma\ J.
- \end{align}
-\end{definition}
-\begin{table}[h!]
- \centering
- \caption{$KO$-dimension $k$ modulo $8$ of a real spectral triple}
- \begin{tabular}{ c | c c c c c c c c}
- \hline
- $k$ & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\
- \hline
- $\epsilon$ & 1 & 1 & -1 & -1 & -1 & -1 & 1 & 1 \\
- $\epsilon '$ & 1 & -1 & 1 & 1 & 1 & -1 & 1 & 1 \\
- $\epsilon ''$ & 1 & & -1 & & 1 & & -1 & \\
- \hline
- \end{tabular}
-\end{table}
-\noindent
-Even thought the KO-dimension of a real spectral triple is important, we will
-not be doing in-depth introduction of the KO-dimension, for this we reference
-to \cite{ncgwalter}.
-
-\begin{definition}
-An opposite-algebra $A^\circ$ of a $A$ is defined to be equal to $A$ as a
-vector space with the opposite product
-\begin{align}
- &a\circ b := ba\\
- &\Rightarrow a^\circ = Ja^* J^{-1},
-\end{align}
-which defines the left representation of $A^\circ$ on $H$
-\end{definition}
-
-
-%------------EXAMPLE EXERCISE
-Let us examine an example of a matrix algebra $M_N(\mathbb{C})$ acting on
-$H=M_N(\mathbb{C})$ by left matrix multiplication with the Hilbert Schmidt
-inner product.
-\begin{align}
- \langle a , b \rangle = \text{Tr}(a^* b).
-\end{align}
-We can define $\gamma (a) = a$ and $J(a) = a^*$ with $a\in H$. Since $D$
-must be odd with respect to $\gamma$ it vanishes identically. Furthermore we
-know the multiplicity space is $V_i = \mathbb{C}^{m_i}$, and also we know
-that for $T\in H$ and$a\in A'$ to work we need $a\ T=T\ a$. Thus by laws of
-matrix multiplication we need $A' \simeq \bigoplus _i M_{m_i}(\mathbb{C})$. For
-this to work we naturally need $H = \bigoplus_i \mathbb{C}^{n_i} \otimes
-\mathbb{C}^{m_i}$. Hence the right action of $M_N(\mathbb{C})$ on $H =
-M_N(\mathbb{C})$ as defined by $a \mapsto a^\circ$ is given by right matrix
-multiplication
-\begin{align}
- a^\circ \xi = J a^* J^{-1}\xi = Ja^* \xi^* = J\xi a=\xi^* a
-\end{align}
-
-%------------EXAMPLE EXERCISE
-
-\begin{definition}
- We call $\xi \in H$ \textbf{cyclic vector} in $A$ if:
- \begin{align}
- A\xi := { a\xi:\;\; a\in A} = H
- \end{align}
- We call $\xi \in H$ \textbf{separating vector} in $A$ if:
- \begin{align}
- a\xi = 0\;\; \Rightarrow \;\; a=0;\;\;\; a\in A
- \end{align}
-\end{definition}
-Suppose $(A, H, D = 0)$ is a finite spectral triple such that $H$ possesses a
-cyclic and separating vector for $A$ and let $J: H \rightarrow H$ be the
-operator in $S = J \Delta ^{1/2}$ with $\Delta = S^*S$ . By composition
-$S(a\xi) = a*\xi$ this is literally anti-linearity, then $S(a \xi) = a* \xi$
-defines a anti-linear operator. Furthermore the operator $S$ is invertible
-because, if a $\xi \in H$ is cyclic then we have $S(A\xi) = A^*\xi = A\xi =
-H$. Vice versa the same has to work for $S^{-1}$, otherwise $\xi$ wouldn't
-exist. And hence $S^{-1}(A^*\xi) = S^{-1}(H) = H$. Additionally $J$ is
-anti-unitary because firstly, $S$ is bijective thus $\Delta ^{1/2}$ and $J$ need to be bijective.
-Also have $J = S \Delta^{-1/2}$ and $\Delta^* = \Delta$, so for a $\xi _1 ,
-\xi _2 \in H$ we can write
-\begin{align}
- <J \xi _1 , J \xi _2 > &= < J^*J\xi_1 , \xi_2>^* =\nonumber\\
- &= <(\Delta ^{-1/2})^* S^* S \Delta ^{-1/2} \xi_1, \xi_2>^* =\nonumber \\
- &= <(\Delta^{-1/2})^* \Delta \Delta^{-1/2} \xi_1, \xi_2>^* =\nonumber\\
- &= <\Delta^{-1/2} \Delta^{1/2}\Delta^{1/2} \Delta^{-1/2} \xi_1, \xi_2>^*
- =\nonumber\\
- &= <\xi _1, \xi_2>^* = <\xi_2 , \xi_1>,
-\end{align}
-which concludes the anti-unitarity by definition.
-\subsubsection{Morphisms Between Finite Real Spectral Triples}
-Like the unitary equivalence relation for finite spectral triples, we can it
-to finite real spectral triples.
-\begin{definition}
- We call two finite real spectral triples $(A_1, H_1 ,D_1 ; J_1 , \gamma
- _1)$ and $(A_2, H_2, D_2; J_2, \gamma _2)$ unitarily equivalent if $A_1 =
- A_2$ and if there exists a unitary operator $U: H_1 \rightarrow H_2$ such
- that
- \begin{align}
- U\ \pi_1(a)\ U^* &= \pi _2(a),\\
- U\ D_1\ U^* &= D_2,\\
- U \gamma _1\ U^* &= \gamma _2,\\
- U\ J_1\ U^* &= J_2.
- \end{align}
-\end{definition}
-\begin{definition}
- Let $E$ be a $B$-$A$ bimodule. The \textit{conjugate Module} $E^\circ$ is
- given by the $A$-$B$-bimodule.
- \begin{align}
- E^\circ = \{\bar{e} : e\in E\},
- \end{align}
- with
- \begin{align}
- a \cdot \bar{e} \cdot b = b^*\ \bar{e}\ a^*, \;\;\;\; \forall a\in A, b \in
- B.
- \end{align}
-\end{definition}
-We bear in mind that $E^\circ$ is not a Hilbert bimodule for $(A, B)$ because
-it doesn't have a natural $B$-valued inner product. But there is a $A$-valued
-inner product on the left $A$-module $E^\circ$ with
-\begin{align}
- \langle \bar{e}_1, \bar{e}_2 \rangle = \langle e_2 , e_1 \rangle,
- \;\;\;\; e_1, e_2 \in E.
-\end{align}
-And linearity in $A$ by the terms
-\begin{align}
- \langle a\ \bar{e}_1, \bar{e}_2 \rangle = a \langle \bar{e}_1, \bar{e}_2
- \rangle, \;\;\;\; \forall a \in A.
-\end{align}
-
-%------------- EXERCISE
-With this it becomes obvious that $E^\circ$ is a Hilbert bimodule
-of $(B^{\circ}, A^{\circ})$. A straightforward calculation of the properties of the Hilbert bimodule and its $B^{\circ}$
-valued inner product gives the results. So for $\bar{e}_1, \bar{e}_2 \in E^{\circ}$ and $a^\circ \in A,
-b^\circ \in B$ we write
-\begin{align}
- \langle\bar{e}_1, a^\circ \bar{e}_2\rangle &= \langle\bar{e}_1, Ja^*J^{-1}
- \bar{e}_2\rangle=\nonumber\\
- &= \langle\bar{e}_1 , J a^* e_2\rangle \nonumber \\
- &= \langle J^{-1} e_1, a^* e_2\rangle \nonumber\\
- & = \langle a^* e_1, e_2\rangle= \langle J^{-1}(a^\circ)^* J e_1, e_2\rangle \nonumber\\
- & = \langle J^{-1} (a^\circ)^* \bar{e}_1, e_2\rangle \nonumber\\
- & = \langle (a^\circ)^* \bar{e}_1 , \bar{e}_2\rangle.
-\end{align}
-Next for $\langle\bar{e}_1, \bar{e}_2 b^\circ\rangle = \langle\bar{e}_1,
-\bar{e_2}\rangle b^\circ$ we write
-\begin{align}
- \langle\bar{e}_1, \bar{e}_2 b^\circ\rangle &= \langle\bar{e}_1, \bar{e}_2 Jb^*J^{-1}\rangle
- \nonumber\\
- &= \langle\bar{e}_1, \bar{e_2}\rangle Jb^*J^{-1} \nonumber \\
- &= \langle\bar{e}_1, \bar{e}_2\rangle b^\circ.
-\end{align}
-Additionally we have
-\begin{align}
- (\langle\bar{e}_1, \bar{e}_2)\rangle_{E^\circ})^* &= (\langle e_2, e_1\rangle_E)^*\nonumber\\
- &= \langle e_1, e_2\rangle_E^* \nonumber\\
- &= \langle\bar{e}_2, \bar{e}_2\rangle_{E^\circ}.
-\end{align}
-And finally of course we have
-\begin{align}
- \langle\bar{e}, \bar{e}\rangle = \langle e, e\rangle \geq 0
-\end{align}
-%------------- EXERCISE
-
-Given the results thus far with a Hilbert bimodule $E$ for $(B, A)$, we
-construct a spectral triple $(B, H', D'; J', \gamma ')$ from $(A, H, D; J,
-\gamma)$. For $H'$ we make a $\mathbb{C}$-valued inner product on $H'$ by combining
-the $A$ valued inner product on $E$ and $E^\circ$ with the
-$\mathbb{C}$-valued inner product on $H$ by defining
-\begin{align}
- H' := E\otimes _A H \otimes _A E^\circ.
-\end{align}
-Then the action of $B$ on $H'$ takes the following form
-\begin{align}
- b(e_2 \otimes \xi \otimes \bar{e}_2 ) = (be_1) \otimes \xi \otimes
- \bar{e}_2.
-\end{align}
-The right action of $B$ on $H'$ defined by action on the right component
-$E^\circ$ is
-\begin{align}
- J'(e_1 \otimes \xi \otimes \bar{e}_2) = e_2 \otimes J \xi \otimes
- \bar{e}_1,
-\end{align}
-where $b^\circ = J' b^* (J')^{-1}$, $b^* \in B$ is the action on $H'$.
-Hence the connection reads
-\begin{align}
- &\nabla: E \rightarrow E\otimes _A \Omega _D ^1(A) \\
- &\bar{\nabla}:E^\circ \rightarrow \Omega _D^1(A) \otimes _A E^\circ,
-\end{align}
-which gives us the Dirac operator on $H' = E \otimes _A H \otimes _A
-E^\circ$ as
-\begin{align}
- D'(e_1 \otimes \xi \otimes \bar{e}_2) = (\nabla e_1) \xi \otimes
- \bar{e_2}+ e_1 \otimes D\xi \otimes \bar{e}_2 + e_1 \otimes
- \xi(\bar{\nabla}\bar{e}_2).
-\end{align}
-And the right action of $\omega \in \Omega _D ^1(A)$ on $\xi \in H$ is
-defined by
-\begin{align}
- \xi \mapsto \epsilon' J \omega ^* J^{-1}\xi.
-\end{align}
-Finally for the grading we have
-\begin{align}
- \gamma ' = 1 \otimes \gamma \otimes 1.
-\end{align}
-
-Summarizing we can write down the following theorem
-\begin{theorem}
- Suppose $(A, H, D; J, \gamma)$ is a finite spectral triple of
- $KO$-dimension $k$, let $\nabla$ be a connection satisfying the
- compatibility condition (same as with finite spectral triples).
- Then $(B, H',D'; J', \gamma')$ is a finite spectral triple of
- $KO$-Dimension $k$. ($H', D', J', \gamma'$)
-\end{theorem}
-
-\begin{proof}
- The only thing left is to check if the $KO$-dimension is preserved,
- for this we check if the $\epsilon$'s are the same.
- \begin{align}
- &(J')^2 = 1 \otimes J^2 \otimes 1 = \epsilon,\\
- &J' \gamma '= \epsilon ''\gamma'J'.
- \end{align}
- Lastly for $\epsilon '$ we have
- \begin{align}
- J'D'(e_1 \otimes \xi \otimes \bar{e}_2)&=J'\big((\nabla e_1) \xi \otimes
- \bar{e_2} + e_1 \otimes D\xi \otimes \bar{e}_2 + e_1 \otimes \xi (\tau
- \nabla e_2)\big)\nonumber \\
- &= \epsilon' D'\left(e_2 \otimes J\xi \otimes \bar{e}_2\right)\nonumber\\
- &= \epsilon' D'J'\left(e_1 \otimes \xi \bar{e}_2\right)
- \end{align}
-\end{proof}
-
-Let us take a look at $\nabla : E \Rightarrow E \otimes _A \Omega _d^1 (A)$ right connection on $E$
-and consider the following anti-linear map
-\begin{align}
- \tau : E \otimes_A \Omega _D^1 (A) &\rightarrow \Omega _D^1 (A) \otimes_A E^\circ\\
- e \otimes \omega &\mapsto -\omega ^* \otimes \bar{e}.
-\end{align}
-Interestingly the map $\bar{\nabla} : E^\circ \rightarrow \Omega _D^1(A) \otimes E^\circ$
-with $\bar{\nabla}(\bar{e}) = \tau \circ \nabla(e)$ is a left connection, that means
-show that it satisfied the left Leibniz rule, for one
-\begin{align}
- \tau \circ \nabla(ae) = \bar{\nabla}(a\bar{e}) = \bar{\nabla}(a^*
- \bar{e}).
-\end{align}
-And for two
-\begin{align}
- \tau \circ \nabla(ae) &= \tau(\nabla(e)a) + \tau \circ(e \otimes
- d(a))\nonumber \\
- &=a^*\bar{\nabla}(\bar{e}) - d(a)^* \otimes \bar{e}. \nonumber\\
- &= a^*\bar{\nabla}(\bar{e}) + d(a^*) \otimes \bar{e}.
-\end{align}
-
diff --git a/src/thesis/chapters/backup/twopointspace.tex b/src/thesis/chapters/backup/twopointspace.tex
@@ -1,241 +0,0 @@
-\subsection{Almost-commutative Manifold}
-\subsubsection{Two-Point Space}
-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).
-\end{align}
-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}
- 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.
-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{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 = \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}
- 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, $J_M$. 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
-\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 \eqref{eq:commutator inequality}
-\begin{align}
- &||[D_F , a]|| = ||(a(y) - a(x))\begin{pmatrix}0 &t\\\bar{t} &0
- \end{pmatrix}|| \leq 1,\\
- &\Leftrightarrow |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.
-
-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\}.
-\end{align}
-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)| \;\;\;\text{for}\;\;
- a_x\in
- C^\infty(M) \;\; \text{with} \;\; ||[D\otimes 1, a_x]|| \leq 1
-\end{align}
-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),
-\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
-\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
-
-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}
-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
- \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 denoted by
- $(A)$.\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 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}
- 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) of KO-dimension 6.
-
-\begin{proposition}
-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 $U(A_F) \cap
- U(A_F)_{J_F}) \simeq U(1)$, 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}.
- \end{align}
- 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
- $U(A_F)$.
-\end{proof}
-
-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}
- 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 $M\times
- F_X$ 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}
- 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}
-
diff --git a/src/thesis/chapters/basics.tex b/src/thesis/chapters/basics.tex
@@ -1,4 +1,4 @@
-\subsection{Noncommutative Geometric Spaces}
+\subsection{Noncommutative Geometric Spaces\label{sec:1}}
\subsubsection{$*$-Algebra}
To grasp the idea of encoding geometrical data into a spectral triple we
introduce the first ingredient of a spectral triple, an unital $*$ algebra.
@@ -65,8 +65,8 @@ Note that the pullback does not map points back, but maps functions on an $*$-al
The pullback, in literature often called a $*$-homomorphism or a $*$-algebra map under
pointwise product has the following properties
\begin{align}
- \phi ^*(f\ g) = \phi ^*(f)\ \phi ^*(g),
- \phi ^*(\overline{f}) = \overline{\phi ^*(f)},
+ \phi ^*(f\ g) = \phi ^*(f)\ \phi ^*(g),\\
+ \phi ^*(\overline{f}) = \overline{\phi ^*(f)},\\
\phi ^*(\lambda\ f + g) = \lambda\ \phi ^*(f) + \phi ^*(g).
\end{align}
%------------ Exercise
diff --git a/src/thesis/chapters/conclusion.tex b/src/thesis/chapters/conclusion.tex
@@ -1,2 +1,2 @@
\section{Conclusion}
-\lipsum[1]
+\lipsum
diff --git a/src/thesis/chapters/electroncg.tex b/src/thesis/chapters/electroncg.tex
@@ -1,4 +1,4 @@
-\subsection{Noncommutative Geometry of Electrodynamics}
+\subsection{Noncommutative Geometry of Electrodynamics\label{sec:5}}
In this chapter we go through a derivation Electrodynamics with
the almost commutative manifold $M\times F_X$ and the abelian gauge group
$U(1)$. The conclusion is an unified description of gravity and
@@ -318,27 +318,28 @@ anti commuting, that is for Grassmann variables $\theta _i, \theta _j$ we have
\text{Tr}(\mathbbm{1}_{H_F})$ and one obtains
\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
+ a_2(D_\omega^2) &= Na_2(D_M^2) - \frac{1}{4\pi^2}\int_M
\text{Tr}(\Phi^2)\sqrt{g}d^4x.
\end{align}
- For $a_4$ we extend in terms of coefficients of $F$, \textbf{REWRITE: look week9.pdf
- for the standard version}
+ For $a_4$ we extend in terms of coefficients of $F$ from equation
+ \eqref{eq: a_4}
\begin{align}
- &\frac{1}{360}\text{Tr}(60sF)= -\frac{1}{6}S(Ns + 4
+ &\frac{1}{360}\text{Tr}(60RE)= -\frac{1}{6}S(NR + 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}+\\
+ &E^2 = \frac{1}{16}R^2\otimes 1 + 1\otimes \Phi^4 - \frac{1}{4}
+ \gamma^\mu\gamma^\nu \gamma^\varrho\gamma^\sigma
+ F_{\mu\nu}F^{\mu\nu}+\nonumber\\
&\;\;\;\;\;\;\;+\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}) +\\
+ &\frac{1}{360}\text{Tr}(180E^2) = \frac{1}{8}R^2N + 2\text{Tr}(\Phi^4)
+ + \text{Tr}(F_{\mu\nu}F^{\mu\nu}) +\nonumber\\
&\;\;\;\;\;\;\;+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)).
+ &\frac{1}{360}\text{Tr}(-60\Delta E)=
+ \frac{1}{6}\Delta(NR+4\text{Tr}(\Phi^2)).
\end{align}
The cross terms of the trace in $\Omega_{\mu\nu}^E\Omega^{E\mu\nu}$
vanishes because of the antisymmetric property of the Riemannian
diff --git a/src/thesis/chapters/finitencg.tex b/src/thesis/chapters/finitencg.tex
@@ -1,4 +1,4 @@
-\subsection{Finite Spectral Triples}
+\subsection{Finite Spectral Triples\label{sec:2}}
\subsubsection{Metric on Finite Discrete Spaces}
We can describe our finite discrete space $X$ by a structure space $\hat{A}$
of a matrix algebra $A$. To establish a distance between two points in $X$ (as
diff --git a/src/thesis/chapters/heatkernel.tex b/src/thesis/chapters/heatkernel.tex
@@ -1,4 +1,4 @@
-\subsection{Heat Kernel Expansion}
+\subsection{Heat Kernel Expansion\label{sec:4}}
\subsubsection{The Heat Kernel}
The heat kernel $K(t; x, y; D)$ is the fundamental solution to the heat
equation
@@ -314,6 +314,6 @@ coefficients
\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})).
+ 2R_{ijkl}R_{ijkl} +30\Omega_{ij}\Omega_{ij})).\label{eq: a_4}
\end{align}
diff --git a/src/thesis/chapters/intro.tex b/src/thesis/chapters/intro.tex
@@ -1,2 +1,23 @@
\section{Introduction}
-\lipsum[2]
+Noncommutative geometry is a branch of mathematics that incorporates many
+different mathematical fields, e.g. Functional analysis, K-Theory,
+Differential Geometry, Representation Theory and many more. The origins can
+be dated back to the 1940s where two Russian mathematicians Gelfand and
+Naimark proved a theorem that connects (in the sense of duality) (classical)
+geometry and algebras. From the beginning it was obvious that noncommutative
+geometry has physical applications, explicitly with gauge theories, since a
+nontrivial gauge group arises naturally from the main structure of
+noncommutative geometry called the spectral triple. We will naturally use
+this property to present how to derive the Lagrangian of electrodynamics
+\ref{sec:5}, and additionally get a purely gravitational Lagrangian.
+In regards to this to get to the action principles in terms of geometrical
+invariants, a method called the heat kernel expansion is used.
+
+The aim of this thesis is to give a basic foundation of noncommutative
+geometry and to present a physical application which can be derived from this
+theory. Additionally we emphasize that this thesis is only literature work,
+where chapters \ref{sec:1}-\ref{sec:3} and \ref{sec:5} are from
+the work of Walter D. Suijlekom's book \cite{ncgwalter} and chapter
+\ref{sec:4} from D.V. Vassilevich's paper \cite{heatkernel}.
+
+\textbf{NOW:CHAPTER OVERVIEW}
diff --git a/src/thesis/chapters/realncg.tex b/src/thesis/chapters/realncg.tex
@@ -1,4 +1,4 @@
-\subsection{Finite Real Noncommutative Spaces}
+\subsection{Finite Real Noncommutative Spaces\label{sec:3}}
\subsubsection{Finite Real Spectral Triples}
In this chapter we supplement the finite spectral triples with a \textit{real
structure}. We additionally require a symmetry condition that that $H$ is an
diff --git a/src/thesis/chapters/twopointspace.tex b/src/thesis/chapters/twopointspace.tex
@@ -1,4 +1,4 @@
-\subsection{Almost-commutative Manifold}
+\subsection{Almost-commutative Manifold\label{sec:4}}
\subsubsection{Two-Point Space}
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
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
@@ -24,15 +24,15 @@
\input{chapters/main_sec}
-\input{chapters/basics} % ausgearbeitet ohne exercises, ohne examples
+\input{chapters/basics}
-\input{chapters/finitencg} % ausgearbeitet ohne exercises, ohne examples
+\input{chapters/finitencg}
-\input{chapters/realncg} % ausgearbeitet ohne exercises, ohne examples
+\input{chapters/realncg}
-\input{chapters/heatkernel} % ausgearbeitet ohne exercises, ohne examples
+\input{chapters/heatkernel}
-\input{chapters/twopointspace} % ausgearbeitet ohne exercises, ohne examples
+\input{chapters/twopointspace}
\input{chapters/electroncg}
diff --git a/src/thesis/todo.md b/src/thesis/todo.md
@@ -1,19 +1,19 @@
# NOT FORGET LIST
* rewrite geometrical invariants $E$ in heatkernel.tex into the right one
- from electroncg.tex !
- * figures need caption and numbering
+ from electroncg.tex ! DONE
+ * figures need caption and numbering DONE
# NORMAL TODO
1. go through the main part and rewrite the chapters make sure the context
is followed, the equations have a comma or a dot at the sentence end
etc. DONE
- 2. rethink the chapters
+ 2. rethink the chapters DONE
3. write introduction
4. write conclusion
5. cut out exercises and examples in the main part if necessary, read
through the not cut out and write them up nicely
- 6. write abstract
+ 6. write abstract DONE
7. read through
8. submit
