commit 6c1434b922b99f7b363042a50c7da249f4e1ce52
parent 6c74872af07f02ef9d8c0882959368d8b3d2661c
Author: miksa234 <milutin@popovic.xyz>
Date: Tue, 27 Jul 2021 15:57:40 +0200
checkpoint
Diffstat:
6 files changed, 419 insertions(+), 540 deletions(-)
diff --git a/src/thesis/12 b/src/thesis/12
@@ -1,49 +0,0 @@
-\documentclass[12pt]{article}
-
-%-------------------- BACKHAND ---------------------
-
-\input{back/packages}
-
-\begin{document}
-
-\input{back/title}
-
-\newpage
-
-\tableofcontents
-
-\newpage
-
-\input{back/abstract}
-
-%------------------- INTRO -------------------------
-
-\input{chapters/intro}
-
-%----------------- MAIN SECTION --------------------
-
-\input{chapters/main_sec}
-
-%\input{chapters/basics}
-
-\input{chapters/finitencg}
-
-%\input{chapters/realncg}
-
-%\input{chapters/heatkernel}
-%
-%\input{chapters/twopointspace}
-%
-%\input{chapters/electroncg}
-
-%------------------ OUTRO -------------------------
-
-\input{chapters/conclusion}
-
-\input{chapters/acknowledgment}
-
-%------------------- BACKHAND ---------------------
-
-\input{back/refs}
-
-\end{document}
diff --git a/src/thesis/chapters/basics.tex b/src/thesis/chapters/basics.tex
@@ -400,7 +400,6 @@ which is satisfied by $\phi = \text{id}_A$
that $A$ or $B$ is a bimodule by itself.
\end{definition}
-\end{align}
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
diff --git a/src/thesis/chapters/finitencg.tex b/src/thesis/chapters/finitencg.tex
@@ -1,4 +1,4 @@
-
+\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
@@ -45,13 +45,13 @@ metric on $X$ in terms of algebraic data.
(d_{21})^{-1} & 0
\end{pmatrix}
\end{align}
- Then we commpute the commutator
+ 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 the representation $\pi (a)$ reads
+ \oplus H_2^1 \oplus H_2^2$. The representation $\pi (a)$ reads
\begin{align}
\pi((a(1), a(2), a(3)) &=
\begin{pmatrix}
@@ -214,13 +214,13 @@ mathematical structure which encodes finite discrete geometry into algebraic dat
\begin{proof}
The wording 'acting faithfully on a Hilbertspace' means that the
$*$-representation is injective, or for a $*$-homomorphism that means
- one-to-one correspondance. And since $A$ acts faithfully on a Hilbert
+ 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 ilustration would be for an algebra $A = M_n(\mathbb{C})$ and
+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.
@@ -233,15 +233,15 @@ inner product and $D$ on $H$ is a hermitian matrix $n\times n$ matrix.
\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 sence that
+$d$ is a derivation of the $*$-algebra in the sense that
\begin{align}
- d(ab) = d(a)b + ad(b), \\
+ 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)
+ &= d(a)\ b + a\ d(b).
\end{align}
And
\begin{align}
@@ -250,7 +250,7 @@ And
&= -d(a)^*.
\end{align}
Furthermore $\Omega _D^1 (A)$ is an $A$-bimodule, which can be seen by
-rewriting the definition \eqref{eq:connesoneforms} into
+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\\
@@ -300,116 +300,94 @@ rewriting the definition \eqref{eq:connesoneforms} into
\
\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 unitarily equivalent if
- \begin{itemize}
- \item $A_1 = A_2$
- \item $\exists \;\; U: H_1 \rightarrow H_2$, unitary with
- \begin{enumerate}
- \item $U\pi_1(a)U^* = \pi_2(a)$ with $a \in A_1$
- \item $UD_1 U^* = D_2$
- \end{enumerate}
- \end{itemize}
+ 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 $\Omega _D^1 (A)$.
-Some remarks
-\begin{itemize}
- \item the above is an equivalence relation
- \item spectral unitary equivalence is given by the unitaries of the
- matrix algebra itself
- \item for any such $U$ then $(A, H, D) \sim (A, H, UDU^*)$
- \item $UDU^* = D + U[D, U^*]$ of the form of elements in
- $\Omega _D^1 (A)$.
-\end{itemize}
-
-%\begin{MyExercise}
-% \textbf{
-% Show that the unitary equivalence between finite spectral
-% triples is a equivalence relation
-%}\newline
-%
-% An equivalence relation needs to satisfy reflexivity, symmetry
-% transitivity.
-% Let $(A_1, H_1, D_1)$, $(A_2, H_2, D_2)$ and $(A_3, H_3, D_3)$
-% be three finite spectral triples.
-% \newline
-%
-% For reflexivity $(A_1, H_1, D_1) \sim (A_1, H_1, D_1)$. So there
-% exists a $U: H_1 \rightarrow H_1$ unitary, which is the identity
-% and always exists.
-% \newline
-%
-% For symmetry we need
-% \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:
-% \begin{align}
-% &U\pi_1(a)U^* = \pi_2(a) \;\;\; | \cdot U^*\boxdot U \\
-% &U^*U\pi_1(a)U^*U = \pi_1(a) = U^*\pi_2(a)U \\
-% \end{align}
-% The same with the symmetric operator $D$.
-% \newline
-%
-% For transitivity we need
-% \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) \\
-% &\Rightarrow (A_1, H_1, D_1) \sim (A_3, H_3, D_3)
-% \end{align}
-% There are two unitary maps $U_{12}:H_1 \rightarrow H_2$ and
-% $U_{23}: H_2 \rightarrow H_3$ then
-% \begin{align}
-% U_{23}U_{12} \pi_1(a) U^*_{12}U^*_{23} &= U_{23}
-% \pi_2(a) U_23^* \\
-% &= \pi_3(a) \\
-% U_{23}U_{12} D_1U^*_{12}U^*_{23} &= U_{23}
-% D_2 U_23^* \\
-% &= D_3
-% \end{align}
-%\end{MyExercise}
-
-Extending the this relation we look again at the notion of equivalence from
-Morita equivalence of Matrix Algebras.
-\newline
+%-------------- 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 $ab \in I$ for $a \in A$ and $b\in I$ (or $ba \in
+ 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
+ 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
+ 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.
+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
+ D'(e\otimes \xi) = e \otimes D\ \xi + \nabla (e)\ \xi, \;\;\;\; e\in
+ E, a\in A,
\end{equation}
-Where $\nabla$ is called the \textit{connection on the right A-module E}
-associated with the derivation $d=[D, \cdot]$ and satisfying the
-\textit{Leibnitz Rule} which is
+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
+ \nabla(ae) = \nabla(e)a + e \otimes [D, a], \;\;\;\;\; e\in E,\; a\in A.
\end{equation}
-Then $D'$ is well defined on $E \otimes _A H$:
+Hence $D'$ is well defined on $E \otimes _A H$
\begin{align}
- D'(ea \otimes \xi - e \otimes a \xi) &= D'(ea \otimes \xi) - D'(e
- \otimes \xi) \\
- &= ea\otimes D\xi + \nabla(ae) \xi - e \otimes D(a\xi ) - \nabla (e)a
- \xi \\
+ 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
@@ -423,75 +401,68 @@ With the information thus far we can prove the following theorem
\end{equation}
\end{theorem}
\begin{proof}
- $E\otimes _A H$ was shown in the previous subsection (text before the
- theorem). The only thing left is to show that $D'$ is a symmetric
+ $E\otimes _A H$ 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 \\
- &= \langle \xi _1, \langle \nabla e_1, e_2\rangle _E \xi _2\rangle _H + \langle \xi _1, d\langle e_1, e_2\rangle _E
- \xi _2\rangle _H \\
- &+ \langle D\xi _1,\langle e_1, e_2\rangle _E \xi _2\rangle _H - \langle \xi _1, [D, \langle e_1, e_2\rangle _E] \xi
- _2 \rangle _H \\
+ \langle \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}
-%\begin{MyExercise}
-% \textbf{
-% Let $\nabla$ and $\nabla'$ be two connections on a right $A$-module
-% $E$. Show that $\nabla - \nabla'$ is a right $A$-linear map
-% $E \rightarrow E\otimes _A \Omega _D^1(A)$
-%}\newline
-%
-% Both $\nabla$ and $\nabla'$ need to satisfy the Leiblitz rule, so
-% let's see if $\nabla - \nabla'$ does.
-%
-% \begin{align}
-% \nabla(ea)-\nabla'(ea)&=\nabla(e) + e\otimes[D, a]\\
-% &-(\nabla'(e)a + e\otimes[D',a])\\
-% &=\bar{\nabla}a + e\otimes(Da-aD-D'a+aD')\\
-% &=\bar{\nabla}a + e\otimes((D-D')a-a(D-D'))\\
-% &=\bar{\nabla}a + e\otimes[D', a]\\
-% &=\bar{\nabla}(ea)
-% \end{align}
-% Therefore $\nabla-\nabla'$ is a linear map.
-%\end{MyExercise}
+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
+$E \rightarrow E\otimes _A \Omega _D^1(A)$.
-%\begin{MyExercise}
-% \textbf{
-% Construct a finite spectral triple $(A, H', D')$ from $(A, H, D)$
-% \begin{enumerate}
-% \item show that 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
-% \item Upon identifying $A\otimes_A H\simeq H$, what is $D'$
-% when the connection is $d(\cdot)$.
-% \item Use 1) and 2) to show that any connection $\nabla:
-% A\rightarrow A\otimes_A \Omega_D^1(A)$ is given by
-% \begin{align}
-% \nabla = d + \omega
-% \end{align}
-% where $\omega \in \Omega_D^1(A)$
-% \item Upon identifying $A\otimes_A H \simeq H$, what is the
-% difference operator $D'$ with the connection on $A$ given by
-% $\nabla = d + \omega$
-% \end{enumerate}
-%}
-% \begin{enumerate}
-% \item $\nabla(e \cdot a) = d(a)$
-% \item
-% $D'(a\xi) = a(D\xi) + (\nabla a) \xi = D(a\xi)$
-% \item Use the identity element $e \in A$\\
-% $\nabla (e\cdot a) = \nabla(e) a + 1 \otimes d(a)=d(a)
-% \nabla(e) a$
-% \item $D'(a\otimes \xi) = D'(a \xi) = a(D\xi) + (\nabla a)\xi =
-% a(D\xi) + \nabla(e \cdot a) \xi \\
-% = D(a\xi) + \nabla(e) (a\xi)$
-% \end{enumerate}
-%\end{MyExercise}
+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}
diff --git a/src/thesis/chapters/realncg.tex b/src/thesis/chapters/realncg.tex
@@ -1,329 +1,287 @@
-%\subsection{Finite Real Noncommutative Spaces}
-%\subsubsection{Finite Real Spectral Triples}
-%Add on to finite real spectral triples a \textit{real structure}. The
-%requirement is that $H$ is a $A$-$A$-bimodule (before only a $A$-left
-%module).
-%\newline
-%
-%For this we introduce a $\mathbb{Z}_2$-grading $\gamma$ with
-%\begin{align}
-% &\gamma ^* = \gamma \\
-% &\gamma ^2 = 1 \\
-% &\gamma D = - D \gamma\\
-% &\gamma a = a \gamma \;\;\;\; a\in A
-%\end{align}
-%
-%\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}
-% They are called the \textit{commutant property}, and mean 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 \\
-% &JD = \epsilon \ DJ\\
-% &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}
-%
-%
-%\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} \;\;\; \text{defines the left
-% representation of $A^\circ$ on $H$}
-%\end{align}
-%\end{definition}
-%
-%
-%\begin{example}
-% 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}
-% Then we define $\gamma (a) = a$ and $J(a) = a^*$ with $a\in H$.
-% Since $D$ mus be odd with respect to $\gamma$ it vanishes identically.
-%\end{example}
-%
-%\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}
-%
-%%\begin{MyExercise}
-%% \textbf{
-%% In the previous example, show that the right action on $M_N(\mathbb{C})$
-%% on $H = M_N(\mathbb{C})$ as defined by $a \mapsto a^\circ$
-%% is given by right matrix multiplication.
-%%}\newline
-%%
-%% \begin{align}
-%% a^\circ \xi = J a^* J^{-1}\xi = Ja^* \xi^* = J\xi a=\xi^* a
-%% \end{align}
-%%\end{MyExercise}
-%%\begin{MyExercise}
-%% \textbf{
-%% Let $A= \bigoplus _i M_{n_i}(\mathbb{C})$, represented on $H = \bigoplus_i \mathbb{C}^{n_i}
-%% \otimes \mathbb{C}^{m_i}$, meaning that the irreducible representation $\textbf{n}_i$ has
-%% multiplicity $m_i$.
-%% \begin{enumerate}
-%% \item Show that the commutant $A'$ of $A$ is $A'\simeq \bigoplus_i M_{m_i} (\mathbb{C})$. As a consequence show $A'' \simeq A$.
-%% \item Show that if $\xi$ is a separating vector for $A$ than it is cyclic for $A'$.
-%% \end{enumerate}
-%% }
-%%
-%%
-%% \begin{enumerate}
-%% \item We know the multiplicity space is $V_i = \mathbb{C}^{m_i}$. We know that
-%% for $T\in H$ and
-%% $a\in A'$ to work we need $aT=Ta$ by laws of matrix multiplication we need
-%% $A' \simeq \oplus _i M_{m_i}(\mathbb{C})$ for this to work since $H = \bigoplus_i
-%% \mathbb{C}^{n_i}
-%% \otimes \mathbb{C}^{m_i}$
-%%
-%% \item Suppose $\xi$ is cyclic for $A$ then $A'\xi = \{0\}$. Under the action of $A$ we
-%% then have $A'A\xi = AA' \xi = 0 \Rightarrow A' = 0$.\\
-%% Suppose now $\xi$ is separating for $A'$, we have $A'\xi = \{0\}$. We can define a
-%% projection in $A'$, $A\xi = P'$. With this projection we have $(1-P')\xi = 0
-%% \Rightarrow 1-P' = 0 \Rightarrow A\xi = H$.
-%% \end{enumerate}
-%%\end{MyExercise}
-%%\begin{MyExercise}
-%% \textbf{ Suppose $(A, H, D = 0)$ is a finite spectral triple such that $H$ possesses a
-%% cyclic and separating vector for $A$.
-%% \begin{enumerate}
-%% \item Show that the formula $S(a \xi) = a* \xi$ defines a anti-linear operator\\
-%% $S: H \rightarrow H$.
-%% \item Show that $S$ is invertible
-%% \item Let $J: H \rightarrow H$ be the operator in $S = J \Delta ^{1/2}$ with
-%% $\Delta = S^*S$. Show that $J$ is anti-unitary
-%% \end{enumerate}
-%% }
-%%
-%%
-%% \begin{enumerate}
-%% \item By composition $S(a\xi) = a*\xi$ this is literally anti-linearity. Does this mean
-%% $S\xi = \xi$?
-%% \item Let $\xi \in H$ be cyclic then: $S(A\xi) = A^*\xi = A\xi = H$. The same has to work
-%% for $S^{-1}$ if not then $\xi$ wouldn't exist. $S^{-1}(A^*\xi) = S^{-1}(H) = H$.
-%% \item Since $S$ is bijective then $\Delta ^{1/2}$ and $J$ need to be bijective.
-%% We also have $J = S \Delta^{-1/2}$ and $\Delta^* = \Delta$\\
-%% Now let $\xi _1 , \xi _2 \in H$ \begin{align}
-%% <J \xi _1 , J \xi _2 > &= < J^*J\xi_1 , \xi_2>^* =\\
-%% &= <(\Delta ^{-1/2})^* S^* S \Delta ^{-1/2} \xi_1, \xi_2>^* = \\
-%% &= <(\Delta^{-1/2})^* \Delta \Delta^{-1/2} \xi_1, \xi_2>^* =\\
-%% &= <\Delta^{-1/2} \Delta^{1/2}\Delta^{1/2} \Delta^{-1/2} \xi_1, \xi_2>^* =\\
-%% &= <\xi _1, \xi_2>^* = <\xi_2 , \xi_1>.
-%% \end{align}
-%% \end{enumerate}
-%%\end{MyExercise}
-%\subsubsection{Morphisms Between Finite Real Spectral Triples}
-%Extend unitary equivalence of finite spectral triples to real ones (with $J$
-%and $\gamma$)
-%
-%\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)\\
-% &UD_1U^*=D_2\\
-% &U\gamma _1 U^* = \gamma _2\\
-% &UJ_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}
-%$E^\circ$ 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 $A$:
-%\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}
-%
-%%\begin{MyExercise}
-%% \textbf{Show that $E^\circ$ is a Hilbert bimodule $(B^{\circ}, A^{\circ})$
-%% }\newline
-%%
-%%
-%% Straightforward show properties of the Hilbert bimodule and its $B^{\circ}$
-%% valued inner product. Let $\bar{e}_1, \bar{e}_2 \in E^{\circ}$ and $a^\circ \in A,
-%% b^\circ \in B$. \\
-%% \begin{align}
-%% <\bar{e}_1, a^\circ \bar{e}_2> &= <\bar{e}_1, Ja^*J^{-1} \bar{e}_2>=\\
-%% &= <\bar{e}_1 , J a^* e_2> = \\
-%% &= <J^{-1} e_1, a^* e_2> =\\
-%% & = <a^* e_1, e_2>= <J^{-1}(a^\circ)^* J e_1, e_2> = \\
-%% & = <J^{-1} (a^\circ)^* \bar{e}_1, e_2> =\\
-%% & = <(a^\circ)^* \bar{e}_1 , \bar{e}_2>.
-%% \end{align}
-%%
-%% Next $<\bar{e}_1, \bar{e}_2 b^\circ> = <\bar{e}_1, \bar{e_2}> b^\circ$.
-%% \begin{align}
-%% <\bar{e}_1, \bar{e}_2 b^\circ> &= <\bar{e}_1, \bar{e}_2 Jb^*J^{-1}> =\\
-%% &= <\bar{e}_1, \bar{e_2}> Jb^*J^{-1} = \\
-%% &= <\bar{e}_1, \bar{e}_2> b^\circ.
-%% \end{align}
-%% Then:
-%% \begin{align}
-%% (<\bar{e}_1, \bar{e}_2)>_{E^\circ})^* &= (<e_2, e_1>_E)^* =\\
-%% &= <e_1, e_2>_E^* = <\bar{e}_2, \bar{e}_2>_{E^\circ}
-%% \end{align}
-%% And of course $<\bar{e}, \bar{e}> = <e, e> \geq 0$
-%%\end{MyExercise}
-%
-%\subsubsection{Construction of a Finite Real Spectral Triple from a Finite
-%Real Spectral Triple}
-%Given a Hilbert bimodule $E$ for $(B, A)$ we construct a spectral triple
-%$(B, H', D'; J', \gamma ')$ from $(A, H, D; J, \gamma)$
-%
-%For the $H'$ 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 $H$.
-%\begin{align}
-% H' := E\otimes _A H \otimes _A E^\circ
-%\end{align}
-%
-%Then the action of $B$ on $H'$ is:
-%\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
-%$E^\circ$
-%\begin{align}
-% J'(e_1 \otimes \xi \otimes \bar{e}_2) = e_2 \otimes J \xi \otimes
-% \bar{e}_1
-%\end{align}
-%with $b^\circ = J' b^* (J')^{-1}$, $b^* \in B$ action on $H'$.
-%\newline
-%
-%
-%\newpage
-%%\begin{MyExercise}
-%% \textbf{ Let $\nabla : E \Rightarrow E \otimes _A \Omega _d^1 (A)$ be a right connection on $E$
-%% 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}
-%% Show that 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:
-%% \begin{equation}
-%% \bar{\nabla}(a\bar{e}) = [D, a] \otimes \bar{e} + a \bar{\nabla}(\bar{e})
-%% \end{equation}
-%% }
-%%
-%% Hagime:
-%% \begin{align}
-%% &\text{For one:}\\
-%% &\tau \circ \nabla(ae) = \bar{\nabla}(a\bar{e}) = \bar{\nabla}(a^* \bar{e})\\
-%% &\text{For two:}\\
-%% &\tau \circ \nabla(ae) = \tau(\nabla(e)a) + \tau \circ(e \otimes d(a))=\\
-%% &=a^*\bar{\nabla}(\bar{e}) - d(a)^* \otimes \bar{e}. \\
-%% &= a^*\bar{\nabla}(\bar{e}) + d(a^*) \otimes \bar{e}.
-%% \end{align}
-%%\end{MyExercise}
-%Then the connections
-%\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}
-%give us the Dirac operator on $H' = E \otimes _A H \otimes _A E^\circ$
-%\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
-%\begin{align}
-% \gamma ' = 1 \otimes \gamma \otimes 1
-%\end{align}
-%
-%\begin{theorem}
-% Suppose $(A, H, D; J, \gamma)$ is a finite spectral triple of
-% $KO$-dimension $k$, let $\nabla$ be like above satisfying the
-% compatibility condition (like with finite spectral triples).
-%
-% Then $(B, H',D'; J', \gamma')$ is a finite spectral triple of
-% $KO$-Dimension $k$. ($H', D', J', \gamma'$ like above)
-%\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}
-% and for $\epsilon '$
-% \begin{align}
-% J'D'(e_1 \otimes \xi \otimes \bar{e}_2)&=J'((\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))\\
-% &= \epsilon' D'(e_2 \otimes J\xi \otimes \bar{e}_2)\\
-% &= \epsilon' D'J'(e_1 \otimes \xi \bar{e}_2)
-% \end{align}
-%\end{proof}
+\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/main.pdf b/src/thesis/main.pdf
Binary files differ.
diff --git a/src/thesis/main.tex b/src/thesis/main.tex
@@ -24,17 +24,17 @@
\input{chapters/main_sec}
-%\input{chapters/basics}
+\input{chapters/basics}
\input{chapters/finitencg}
-%\input{chapters/realncg}
+\input{chapters/realncg}
-%\input{chapters/heatkernel}
-%
-%\input{chapters/twopointspace}
-%
-%\input{chapters/electroncg}
+\input{chapters/heatkernel}
+
+\input{chapters/twopointspace}
+
+\input{chapters/electroncg}
%------------------ OUTRO -------------------------
