commit 3041a37a31d5b07d7d6ff27ee6d41ed9ee0bf096
parent fd580db32388d4d35d9c95c07828b876a69cdaa5
Author: miksa234 <milutin@popovic.xyz>
Date: Mon, 26 Jul 2021 10:45:03 +0200
checkpoint
Diffstat:
6 files changed, 125 insertions(+), 118 deletions(-)
diff --git a/src/thesis/back/title.tex b/src/thesis/back/title.tex
@@ -20,7 +20,7 @@
\vspace*{1.5cm}
{\fontsize{12}{0} \selectfont submitted by}\\
-\vspace*{0.4cm}
+\vspace*{0.3cm}
{ \fontsize{14}{0} \selectfont Popović Milutin}\\
@@ -44,7 +44,7 @@
\fontsize{10}{0} \selectfont\\
\fontsize{10}{0} \selectfont degree
- programme a0 it appears on / & \fontsize{10}{0} \selectfont Physics \\
+ programme as it appears on / & \fontsize{10}{0} \selectfont Physics \\
\fontsize{10}{0} \selectfont the student record sheet:\vspace*{0.4cm} &
\fontsize{10}{0} \selectfont \\
diff --git a/src/thesis/chapters/electroncg.tex b/src/thesis/chapters/electroncg.tex
@@ -63,9 +63,9 @@ interchange particles with antiparticles by the following equations
&J_F e_L = \bar{e_L}, \\
\nonumber \\
&\gamma _F e_R = -e_R,\\
- &\gamma_F e_L = e_L \\
+ &\gamma_F e_L = e_L,
\end{align}
-where $J_F$ and $gamma_F$ have to following properties
+where $J_F$ and $\gamma_F$ have to following properties
\begin{align}
&J_F^2 = 1,\\
& J_F \gamma_F = - \gamma_F J_F.
@@ -99,7 +99,7 @@ $\{e_L, e_R, \bar{e}_L, \bar{e}_R\}$ is represented by
\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 \ref{eq:leftrightrepr}. Note
+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}
@@ -149,8 +149,8 @@ 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 \ref{eq:fedfail} and $D_F$
-from equation \ref{eq:feddirac}.
+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
@@ -288,7 +288,8 @@ anti commuting, that is for Grassmann variables $\theta _i, \theta _j$ we have
\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))\\
+ \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}
@@ -298,10 +299,10 @@ anti commuting, that is for Grassmann variables $\theta _i, \theta _j$ we have
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}\label{eq:trheatkernel}
+ \begin{align}
\text{Tr}(f(\frac{D_\omega}{\Lambda})) \simeq& \ 2f_4 \Lambda ^4
- a_0(D_\omega ^2)+ 2f_2\Lambda^2 a_2(D_\omega^2) \\&+ f(0) a_4(D_\omega^4)
- +O(\Lambda^{-1}).
+ 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
@@ -310,8 +311,8 @@ anti commuting, that is for Grassmann variables $\theta _i, \theta _j$ we have
\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
+ 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}
@@ -355,20 +356,20 @@ anti commuting, that is for Grassmann variables $\theta _i, \theta _j$ we have
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) \\
+ \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 \ref{eq:trheatkernel}.
+ heat kernel expansion in equation \eqref{eq:trheatkernel}.
\end{proof}
\subsubsection{Fermionic Action}
-We remind ourselves the definition of the fermionic action in
+We remind ourselves the definition of the fermionic action in definition
\ref{def:fermionic action} and the manifold we are dealing with in equation
-\ref{eq:almost commutative manifold}. The Hilbertspace $H_F$ is separated
+\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
@@ -402,31 +403,31 @@ for a $\xi \in H^+$. Then the straight forward calculation gives \begin{align}
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 \ref{eq:fermionic1} we calculate
+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)+
+ \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)+
+ \\&+\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 \ref{eq:fermionic2} we have
+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)+\\
+ -\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)=\\
+ \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 \ref{eq:fermionic3} we can write
+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)+\\
+ +\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)=\\
+ +\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
diff --git a/src/thesis/chapters/heatkernel.tex b/src/thesis/chapters/heatkernel.tex
@@ -22,7 +22,7 @@ gauge field, the heat kernel reads then
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 \ref{eq:standard} as $t \rightarrow 0$ thus the
+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
@@ -30,75 +30,75 @@ expansion reads
\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
+%%----------------------- 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}
@@ -139,7 +139,8 @@ 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) =\\
+ 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}
@@ -156,7 +157,7 @@ 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),\\
+ 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}
@@ -167,7 +168,7 @@ 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}} = \\
+ \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
@@ -185,7 +186,7 @@ $M_1$. 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))=\\
+ \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}
@@ -206,18 +207,18 @@ equations
e^{-2\varepsilon f}D) =
0\label{eq:var3}.
\end{align}
-Let us explain the equations above. To get the first equation \ref{eq:var1}
+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 \ref{eq:var1}. Equation \ref{eq:var2} is derived similarly.
+then we expand both sides in $t$ and get \eqref{eq:var1}. Equation \eqref{eq:var2} is derived similarly.
-For equation \ref{eq:var3} we consider the following operator
+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 \ref{eq:var1} and we get
+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,
@@ -227,12 +228,12 @@ 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)) =\\
+ D(\varepsilon,\delta)) =\nonumber\\
&=\frac{d}{d\varepsilon}\bigg|_{\varepsilon=0}\frac{d}{d\delta}\bigg|_{\delta=0}a_n(1,
- D(\varepsilon,\delta)) =\\
+ D(\varepsilon,\delta)) =\nonumber\\
&=a_{n-2} ( e^{-2\varepsilon f}F, e^{-2\varepsilon f}D),
\end{align}
-which gives us equation \ref{eq:var3}.
+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
@@ -242,7 +243,7 @@ $u^I$, and by that the first three heat kernel coefficients read
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} + \\
+ \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}
@@ -252,7 +253,7 @@ 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
-\ref{eq:var2}, the coefficient $k = 2$ is
+\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),
@@ -264,7 +265,7 @@ which means $\alpha_1 = 6$. Looking at the coefficient $k=4$ we have
\end{align}
thus $\alpha_4 = 60\alpha_2$ and $\alpha_5 = 180$.
-By applying \ref{eq:var3} to $n=4$ we get
+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.
@@ -276,9 +277,10 @@ 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)
- +a_2(1,-\Delta_1) a_2(1,-\Delta_2) \\&+ a_0(1,-\Delta_1)
- a_4(1,-\Delta_2),
+ 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 =
@@ -287,9 +289,9 @@ curvature of $M_{1/2}$) we obtain $\frac{2}{360}\alpha_7 =
For $n=6$ we get
\begin{align}
0 &= \text{Tr}_V(\int_Md^nx\sqrt{g}
- (F(-2\alpha_3-10\alpha_4+4\alpha_5)f_{,kk}E +\\
- &+(2\alpha_3 + 10\alpha_6)f_{,iijj}+\\
- &+(2\alpha_4 -2\alpha_6 - 20\alpha_7 -2\alpha_8)f_{,ii}R\\
+ (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 $\alpha_9 = 2$
@@ -307,6 +309,6 @@ 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})).
\end{align}
diff --git a/src/thesis/chapters/twopointspace.tex b/src/thesis/chapters/twopointspace.tex
@@ -87,7 +87,7 @@ distance formula on $M\times F_X$ with
To calculate the distance between two points on the Two-Point space $X= \{x,
y\}$, between $x$ and $y$, we consider an $a \in \mathbb{C}^2 = C(X)$, which is
specified by two complex numbers $a(x)$ and $a(y)$. Then we simplify the
-commutator inequality in \ref{eq:commutator inequality}
+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,\\
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
@@ -10,6 +10,10 @@
\newpage
+\tableofcontents
+
+\newpage
+
\input{back/abstract}
%------------------- INTRO -------------------------
