commit 68188395cea0fea2c0342eddcf352029677dae97
parent 70f5cb89959791561a5c5c54816b6d23a719ee20
Author: miksa234 <milutin@popovic.xyz>
Date: Wed, 24 Mar 2021 13:52:54 +0100
added box around exercises and solutions
Diffstat:
6 files changed, 256 insertions(+), 130 deletions(-)
diff --git a/pdfs/week1.pdf b/pdfs/week1.pdf
Binary files differ.
diff --git a/pdfs/week2.pdf b/pdfs/week2.pdf
Binary files differ.
diff --git a/pdfs/week5.pdf b/pdfs/week5.pdf
Binary files differ.
diff --git a/src/week1.tex b/src/week1.tex
@@ -14,24 +14,67 @@
\usepackage[colorlinks=true,naturalnames=true,plainpages=false,pdfpagelabels=true]{hyperref}
\usepackage[parfill]{parskip}
+\usepackage{tikz}
+\usetikzlibrary{patterns,decorations.pathmorphing,positioning}
+
+\usepackage[framemethod=TikZ]{mdframed}
+
+\tikzstyle{titlered} =
+ [draw=black, thick, fill=white,%
+ text=black, rectangle,
+ right, minimum height=.7cm]
+
+\newcounter{exercise}
+
+\renewcommand*\theexercise{Exercise~\arabic{exercise}}
+
+\makeatletter
+\mdfdefinestyle{exercisestyle}{%
+ outerlinewidth=1em,%
+ outerlinecolor=white,%
+ leftmargin=-1em,%
+ rightmargin=-1em,%
+ middlelinewidth=1.2pt,%
+ roundcorner=5pt,%
+ linecolor=black,%
+ backgroundcolor=blue!5,
+ innertopmargin=1.2\baselineskip,
+ skipabove={\dimexpr0.5\baselineskip+\topskip\relax},
+ skipbelow={-1em},
+ needspace=3\baselineskip,
+ frametitlefont=\sffamily\bfseries,
+ settings={\global\stepcounter{exercise}},
+ singleextra={%
+ \node[titlered,xshift=1cm] at (P-|O) %
+ {~\mdf@frametitlefont{\theexercise}~};},%
+ firstextra={%
+ \node[titlered,xshift=1cm] at (P-|O) %
+ {~\mdf@frametitlefont{\theexercise}~};},
+}
+\makeatother
+
+\newenvironment{MyExercise}%
+{\begin{mdframed}[style=exercisestyle]}{\end{mdframed}}
+
\theoremstyle{definition}
\newtheorem{definition}{Definition}
\theoremstyle{definition}
\newtheorem{question}{Question}
+\theoremstyle{definition}
+\newtheorem{example}{Example}
+
\theoremstyle{theorem}
\newtheorem{theorem}{Theorem}
\theoremstyle{theorem}
-\newtheorem{exercise}{Exercise}
-
-\theoremstyle{definition}
-\newtheorem{solution}{Solution}
+\newtheorem{lemma}{Lemma}
\newtheorem*{idea}{Proof Idea}
+
\title{Notes on \\ Noncommutative Geometry and Particle Physics}
\author{Popovic Milutin}
\date{Week 1: 05.02 - 12.02}
@@ -134,11 +177,12 @@ Under the pointwise product:
Or I didn't understand this correctly?
\end{question}
-\begin{exercise}
+\begin{MyExercise}
+ \textbf{
Show that $\phi :X_1\ \rightarrow \ X_2$ is injective (surjective) map of finite spaces iff
$\phi ^* :C(X_2)\ \rightarrow \ C(X_1)$ is surjective (injective).
-\end{exercise}
-\begin{solution}
+}\newline
+
Consider $X_1$ with $n$ points and $X_2$ with $m$ points. Then there are three cases:
\begin{enumerate}
\item $n=m$ \\
@@ -151,7 +195,7 @@ Under the pointwise product:
\item $n \langle m $ \\
analogous
\end{enumerate}
-\end{solution}
+\end{MyExercise}
\subsubsection{Matrix Algebras}
\begin{definition}
@@ -237,7 +281,8 @@ Yes
\end{question}
More on that in later chapters.
-\begin{exercise}
+\begin{MyExercise}
+ \textbf{
Given $(H, \pi)$ of a $*$-algebra $A$, the \textbf{commutant} $\pi (A)'$ of $\pi (A)$ is defined as a set
of operators in $L(H)$ that commute with all $\pi (a)$
\begin{align*}
@@ -248,8 +293,8 @@ More on that in later chapters.
\item Show that a representation $(H, \pi)$ of $A$ is irreducible iff the commutant $\pi (A)'$
consists of multiples of the identity
\end{enumerate}
-\end{exercise}
-\begin{solution}
+}
+
1. To show that $\pi (A)'$ is a $*$-algebra we have to show that it is unital, associative and involute.
And note that $\pi (a) \in L(H)\ \forall a \in A$.
Unitarity is given by the unital operator of the $*$-algebra of operators $L(H)$, which exists by definition
@@ -258,9 +303,10 @@ More on that in later chapters.
with a map $*: L(H) \mapsto L(H)$ only for $T$ that commute with $\pi (a)$.
\\
2.?
-\end{solution}
+\end{MyExercise}
-\begin{exercise}
+\begin{MyExercise}
+ \textbf{
\begin{enumerate}
\item If $A$ is a unital $*$-algebra, show that the $n \times n$ matrices $M_n(A)$ with entries
in $A$ form a unital $*$-algebra.
@@ -271,9 +317,7 @@ More on that in later chapters.
\item Let $\tilde{\pi}:M_n(A) \rightarrow L(H^n)$ be a $*$ algebra representation of $M_n(A)$.
Show that $\pi: A \rightarrow L(H^n)$ is a representation of $A$.
\end{enumerate}
-\end{exercise}
-
-\begin{solution}
+}
1. We know $A$ is a $*$ algebra. Unitary operaton in $M_n(A)$ is given by the identity Matrix, which
has to exists because every entry in $M_n(A)$ has to behave like in $A$. Associativity is given by
matrix multiplication. Involutnes is given by the conjugate transpose.\\
@@ -282,7 +326,7 @@ More on that in later chapters.
$U=\mathbbm{1}_n$:\\
$\pi (a) = \mathbbm{1}_n^*\ \tilde{\pi}((a_{ij}))\ \mathbbm{1}_n = \tilde{\pi}((a_{ij})) = \pi (a_{ij})
\Rightarrow a_{ij} = a\mathbbm{1}_n$.
-\end{solution}
+\end{MyExercise}
\subsection{Commutative Matrix Algebras}
\begin{itemize}
@@ -337,11 +381,12 @@ Remark on the notation
\item ${}_A E_B$ $A$-$B$-bimodule $E$;
\end{itemize}
-\begin{exercise}
+\begin{MyExercise}
+ \textbf{
Check that a representation of $\pi : A \rightarrow L(H)$ of a $*$-algebra A turns H into a
left module ${}_A H$.
-\end{exercise}
-\begin{solution}
+}\newline
+
Not quite sure but \\
$a \in A$, $h_1, h_2 \in H$, we know $\pi (a) = T \in L(H)$ than
\begin{align*}
@@ -354,14 +399,14 @@ Remark on the notation
(\pi(a_1)\pi(a_2))h = \pi(a_1)(\pi(a_2) h) = (T_1T_2) h = T_1 (T_2 h)
\end{align*}
For $T_1, T_2 \in L(H)$, which operate naturally from the left on $h$.
-\end{solution}
+\end{MyExercise}
-\begin{exercise}
+\begin{MyExercise}
+ \textbf{
Show that $A$ is a bimodule ${}_A A_A$ with itself.
-\end{exercise}
+}\newline
-\begin{solution}
$\gamma: A\times A\times A \rightarrow A$ which is given by the inner product of the $*$-algebra.
-\end{solution}
+\end{MyExercise}
\end{document}
diff --git a/src/week2.tex b/src/week2.tex
@@ -14,6 +14,48 @@
\usepackage[colorlinks=true,naturalnames=true,plainpages=false,pdfpagelabels=true]{hyperref}
\usepackage[parfill]{parskip}
+\usepackage{tikz}
+\usetikzlibrary{patterns,decorations.pathmorphing,positioning}
+
+\usepackage[framemethod=TikZ]{mdframed}
+
+\tikzstyle{titlered} =
+ [draw=black, thick, fill=white,%
+ text=black, rectangle,
+ right, minimum height=.7cm]
+
+\newcounter{exercise}
+
+\renewcommand*\theexercise{Exercise~\arabic{exercise}}
+
+\makeatletter
+\mdfdefinestyle{exercisestyle}{%
+ outerlinewidth=1em,%
+ outerlinecolor=white,%
+ leftmargin=-1em,%
+ rightmargin=-1em,%
+ middlelinewidth=1.2pt,%
+ roundcorner=5pt,%
+ linecolor=black,%
+ backgroundcolor=blue!5,
+ innertopmargin=1.2\baselineskip,
+ skipabove={\dimexpr0.5\baselineskip+\topskip\relax},
+ skipbelow={-1em},
+ needspace=3\baselineskip,
+ frametitlefont=\sffamily\bfseries,
+ settings={\global\stepcounter{exercise}},
+ singleextra={%
+ \node[titlered,xshift=1cm] at (P-|O) %
+ {~\mdf@frametitlefont{\theexercise}~};},%
+ firstextra={%
+ \node[titlered,xshift=1cm] at (P-|O) %
+ {~\mdf@frametitlefont{\theexercise}~};},
+}
+\makeatother
+
+\newenvironment{MyExercise}%
+{\begin{mdframed}[style=exercisestyle]}{\end{mdframed}}
+
\theoremstyle{definition}
\newtheorem{definition}{Definition}
@@ -27,17 +69,10 @@
\newtheorem{theorem}{Theorem}
\theoremstyle{theorem}
-\newtheorem{exercise}{Exercise}
-
-\theoremstyle{theorem}
\newtheorem{lemma}{Lemma}
-\theoremstyle{definition}
-\newtheorem{solution}{Solution}
-
\newtheorem*{idea}{Proof Idea}
-
\title{Notes on \\ Noncommutative Geometry and Particle Physics}
\author{Popovic Milutin}
\date{Week 2: 12.02 - 19.02}
@@ -81,13 +116,14 @@ $\langle \cdot,\cdot\rangle_E$ needs to satisfy the following for $e, e_1, e_2 \
We denote $KK_f(A,B)$ the set of all \textit{Hilbert bimodules} of $(A,B)$.
-\begin{exercise}
+\begin{MyExercise}
+ \textbf{
Check that a representation $\pi:\ A \ \rightarrow L(H)$ of a matrix algebra $A$ turns $H$ into
a Hilbert bimodule for $(A, \mathbb{C})$.
\label{ex: bimodule}
-\end{exercise}
+}\newline
+
-\begin{solution}
We check if the representation of $a \in A$, $\pi(a)=T \in L(H)$ fulfills
the conditions on the $\mathbb{C}$-valued inner product for $h_1, h_2 \in H$:
\begin{itemize}
@@ -98,24 +134,26 @@ We denote $KK_f(A,B)$ the set of all \textit{Hilbert bimodules} of $(A,B)$.
$\mathbb{C}$-valued inner product
\item $\langle h_1, h_2\rangle \ge 0$, $\mathbb{C}$-valued inner product.
\end{itemize}
-\end{solution}
+\end{MyExercise}
-\begin{exercise}
+\begin{MyExercise}
+ \textbf{
Show that the $A-A$ bimodule given by $A$ is in $KK_f(A,A)$ by taking 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
\end{align*}
\label{exercise: inner-product}
-\end{exercise}
-\begin{solution}
+}\newline
+
+
We check again the conditions on $\langle \cdot, \cdot\rangle _A$, let $a, a_1, a_2 \in A$:
\begin{itemize}
\item $\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 $
\item $\langle a_1, a_2 \cdot a\rangle _A = a^*_1 (a_2\cdot a) = (a^*a_2)\cdot a = \langle a_1, a_2\rangle _A a$
\item $\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{itemize}
-\end{solution}
+\end{MyExercise}
\begin{example}
Consider a $*$ homomorphism between two matrix algebras $\phi:A\rightarrow B$.
@@ -152,7 +190,8 @@ We denote $KK_f(A,B)$ the set of all \textit{Hilbert bimodules} of $(A,B)$.
%What is the meaning of `associative up to isomorphism'? Isomorphism of $F \circ E$ or of $A, B$ or $D$?
%\end{question}
-\begin{exercise}
+ \begin{MyExercise}
+ \textbf{
Show that the association $\phi \leadsto E_\phi$ (from the previous Example) is natural
in the sense
\begin{enumerate}
@@ -164,32 +203,27 @@ We denote $KK_f(A,B)$ the set of all \textit{Hilbert bimodules} of $(A,B)$.
E_{\psi \circ \phi} \in KK_f(A,C)
\end{align*}
\end{enumerate}
-\end{exercise}
-
-\begin{solution}
+}
\begin{enumerate}
- \
\item $\text{id}_A: A \rightarrow A$.\\
To construct $E_{\phi}\in KK_f(A,A)$, we let $E_{\phi}$ be $A$ with a natural right
representation, so $\Rightarrow E_{\phi}\simeq A$.\\
With an inner product, acting on $A$ from the left with $\phi$, $a', a\in A$\\
$a'a = (\phi(a') a) \in A $, which is satisfied by $\text{id}_A$, so $\phi = \text{id}_A$.
- \item $a \cdot b \cdot c = \psi(\phi(a) \cdot b) \cdot c$ for $a \in A$, $b\in B$, and $c\in C$
+ \item $a \cdot b \cdot c = \psi(\phi (a) \cdot b) \cdot c$ for $a \in A$, $b\in B$, and $c\in C$
which is $\psi \circ \phi$
\end{enumerate}
-\end{solution}
+\end{MyExercise}
-\begin{exercise}
+\begin{MyExercise}
+ \textbf{
In the definition of Morita equivalence:
\begin{enumerate}
\item Check that $E \otimes _B F$ is a $A-D$ bimodule
\item Check that $\langle \cdot,\cdot\rangle _{E\oplus _B F}$ defines a $D$ valued inner product
\item Check that $\langle a^*(e_1 \otimes f_1), e_2 \otimes f_2\rangle _{E \otimes _B F} = \langle e_1 \otimes f_1, a(e_2 \otimes f_2)\rangle _{E \otimes _B F}$.
\end{enumerate}
-\end{exercise}
-
-\begin{solution}
- \
+}
\begin{enumerate}
\item $E \otimes _B F = E \otimes F / \{\sum_i e_i b_i \otimes f_i - e_i \otimes b_i f_i;
e_i \in E_i, b_i \in B, f_i \in F\}$ the last part takes out all tensor product elements of
@@ -203,7 +237,7 @@ We denote $KK_f(A,B)$ the set of all \textit{Hilbert bimodules} of $(A,B)$.
$g_2 = e_2 \otimes f_2$ for some $e_1, e_2 \in E$ and $f_1, f_2 \in F$, or else
$G \notin KK_f(A,C)$ which would violate the Kasparov product
\end{enumerate}
-\end{solution}
+ \end{MyExercise}
\begin{definition}
Let $A$, $B$ be \textit{matrix algebras}. They are called \textit{Morita equivalent} if there
@@ -275,17 +309,16 @@ We denote $KK_f(A,B)$ the set of all \textit{Hilbert bimodules} of $(A,B)$.
These maps are each others inverses, thus $\hat{A} \simeq \hat{B}$
\end{proof}
-\begin{exercise}
+\begin{MyExercise}
+ \textbf{
Fill in the gaps in the above proof:
\begin{enumerate}
\item show that the representation of $\pi _A$ defined is irreducible iff $\pi _B$ is.
\item Show that the association of the class $[\pi _A]$ to $[\pi _B]$ is independent
of the choice of representatives $\pi _A$ and $\pi _B$
\end{enumerate}
-\end{exercise}
+}
-\begin{solution}
- \
\begin{enumerate}
\item $(\pi _B, H)$ is irreducible means $H \neq \emptyset$ and only $\emptyset$ or $H$
is invariant under the Action of $B$ on $H$.
@@ -295,7 +328,7 @@ We denote $KK_f(A,B)$ the set of all \textit{Hilbert bimodules} of $(A,B)$.
hence any choice of representation is irreducible, because the structure space denotes all unitary
equivalence classes of irreducible representations.
\end{enumerate}
-\end{solution}
+\end{MyExercise}
\begin{lemma}
The matrix algebra $M_n(\mathbb{C})$ has a unique irreducible representation (up to isomorphism)
diff --git a/src/week5.tex b/src/week5.tex
@@ -17,6 +17,45 @@
\usepackage{tikz}
\usetikzlibrary{patterns,decorations.pathmorphing,positioning}
+\usepackage[framemethod=TikZ]{mdframed}
+
+\tikzstyle{titlered} =
+ [draw=black, thick, fill=white,%
+ text=black, rectangle,
+ right, minimum height=.7cm]
+
+\newcounter{exercise}
+
+\renewcommand*\theexercise{Exercise~\arabic{exercise}}
+
+\makeatletter
+\mdfdefinestyle{exercisestyle}{%
+ outerlinewidth=1em,%
+ outerlinecolor=white,%
+ leftmargin=-1em,%
+ rightmargin=-1em,%
+ middlelinewidth=1.2pt,%
+ roundcorner=5pt,%
+ linecolor=black,%
+ backgroundcolor=blue!5,
+ innertopmargin=1.2\baselineskip,
+ skipabove={\dimexpr0.5\baselineskip+\topskip\relax},
+ skipbelow={-1em},
+ needspace=3\baselineskip,
+ frametitlefont=\sffamily\bfseries,
+ settings={\global\stepcounter{exercise}},
+ singleextra={%
+ \node[titlered,xshift=1cm] at (P-|O) %
+ {~\mdf@frametitlefont{\theexercise}~};},%
+ firstextra={%
+ \node[titlered,xshift=1cm] at (P-|O) %
+ {~\mdf@frametitlefont{\theexercise}~};},
+}
+\makeatother
+
+\newenvironment{MyExercise}%
+{\begin{mdframed}[style=exercisestyle]}{\end{mdframed}}
+
\theoremstyle{definition}
\newtheorem{definition}{Definition}
@@ -32,12 +71,6 @@
\theoremstyle{theorem}
\newtheorem{lemma}{Lemma}
-\theoremstyle{theorem}
-\newtheorem{exercise}{Exercise}
-
-\theoremstyle{definition}
-\newtheorem{solution}{Solution}
-
\newtheorem*{idea}{Proof Idea}
@@ -52,10 +85,11 @@
\section{Noncommutative Geometric Spaces }
\subsection{Exercises}
-\begin{exercise}
- Make the proof of the last theorem (see week4.pdf) explicit for $N=3$
-\end{exercise}
-\begin{solution}
+\begin{MyExercise}
+\textbf{
+ Make the proof of the last theorem (see week4.pdf) explicit for $N=3$.
+}\newline
+
For the C* algebra we have $A=\mathbb{C}^3$
For $H$ we have $H = (\mathbb{C}^2)^{\oplus 3} = H_2 \oplus H_2^1 \oplus H_2^2$.
The symmetric operator $D$ acting on $H$ and the representation $\pi (a)$:
@@ -101,19 +135,19 @@
\end{align}
Then the norm of the commutator would be the largest eigenvalue
\begin{align}
- &||[D, \pi(a)]|| = ||D\pi(a) - \pi(a)D||\nonumber \\
- &=
- \left|\left|
- \setlength{\arraycolsep}{0.1cm}
- \renewcommand{\arraystretch}{0.1}
- \begin{pmatrix}
- 0 & x_1(a(2)-a(1)) & 0 & 0 & 0 & 0 \\
- -x_1(a(2)-a(1)) & 0 & 0 & 0 & 0 & 0 \\
- 0 & 0 & 0 & x_2(a(3)-a(1)) & 0 & 0 \\
- 0 & 0 & -x_2(a(3)-a(1)) & 0 & 0 & 0 \\
- 0 & 0 & 0 & 0 & 0 & x_3(a(3)-a(2)) \\
- 0 & 0 & 0 & 0 & -x_3(a(2)-a(3)) & 0 \\
- \end{pmatrix}\right|\right| \label{skew matrix}
+ &||[D, \pi(a)]|| = ||D\pi(a) - \pi(a)D||\nonumber
+ % &=
+ % \left|\left|
+ % \setlength{\arraycolsep}{0.1cm}
+ % \renewcommand{\arraystretch}{0.1}
+ % \begin{pmatrix}
+ % 0 & x_1(a(2)-a(1)) & 0 & 0 & 0 & 0 \\
+ % -x_1(a(2)-a(1)) & 0 & 0 & 0 & 0 & 0 \\
+ % 0 & 0 & 0 & x_2(a(3)-a(1)) & 0 & 0 \\
+ % 0 & 0 & -x_2(a(3)-a(1)) & 0 & 0 & 0 \\
+ % 0 & 0 & 0 & 0 & 0 & x_3(a(3)-a(2)) \\
+ % 0 & 0 & 0 & 0 & -x_3(a(2)-a(3)) & 0 \\
+ % \end{pmatrix}\right|\right| \label{skew matrix}
\end{align}
The matrix in Equation \ref{shew matrix} is a skew symmetric matrix its eigenvalues
are $i\lambda_1, i\lambda_2, i\lambda_3, i\lambda_4$, where the $\lambda$'s are on the
@@ -123,9 +157,10 @@ the larges eigenvalues:
\begin{align}
||[D, \pi(a)]|| = \max_{a\in A}\{x_i|a(j)-a(k)|\}
\end{align}
-\end{solution}
+\end{MyExercise}
-\begin{exercise}
+\begin{MyExercise}
+ \textbf{
Compute the 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
@@ -138,8 +173,8 @@ the larges eigenvalues:
\end{pmatrix}
\end{align*}
for some $d \in \mathbb{R}$
-\end{exercise}
-\begin{solution}
+}\newline
+
We have $A=\mathbb{C}^3$, $H=\mathbb{C}^3$ and $D$ from above, then
\begin{align}
@@ -151,8 +186,10 @@ the larges eigenvalues:
\end{pmatrix} \right|\right| \\
&= d^{-1} |a(2) - a(1)|
\end{align}
-\end{solution}
-\begin{exercise}
+\end{MyExercise}
+
+\begin{MyExercise}
+ \textbf{
Show that $d_{ij}$ from Equation \ref{ext metric} is a metric on $\hat{A}$ by
establishing that:
\begin{align}
@@ -163,8 +200,8 @@ the larges eigenvalues:
\begin{equation} \label{ext metric}
d_{ij} = \sup_{a\in A}\big\{|\text{Tr}(a(i)) - \text{Tr}((a(j))|: ||[D, a]|| \leq 1\big\}
\end{equation}
-\end{exercise}
-\begin{solution}
+}\newline
+
For Equation \ref{metric 1} set $i=j$ in \ref{ext metric}.
\begin{align*}
d_{ii} &= \sup_{a \in A}\{|\text{Tr}(a(i)) - \text{Tr}((a(i))|: ||[D, a]|| \leq
@@ -177,7 +214,7 @@ addition.
For Equation \ref{metric 3}, for $k=j$ then $d_{kj} = 0$ and the equality
holds. For $i = k$ then $d_{ik} = 0$ and equality holds. Else set $d_{ik} =
1$ and $d_{kj} = 1$ then $d_{ij} = 1 \leq d_{ik} + d_{kj} = 2$
-\end{solution}
+\end{MyExercise}
\subsection{Properties of Matrix Algebras}
\begin{lemma}
@@ -227,14 +264,15 @@ exercises.
differential operators` given $A$, that act on $H$?
\end{question}
Then there is a map $d:A\rightarrow \Omega _D ^1 (A)$, $d = [D, \cdot]$.
-\begin{exercise}
+\begin{MyExercise}
+ \textbf{
Verify that 'd` is a derivation of the C* algebra
\begin{align*}
d(ab) = d(a)b + ad(b) \\
d(a^*) = -d(a)^*
\end{align*}
-\end{exercise}
-\begin{solution}
+}\newline
+
For the record $d(\cdot) = [D, \cdot]$, then we have
\begin{enumerate}
\item
@@ -249,14 +287,16 @@ Then there is a map $d:A\rightarrow \Omega _D ^1 (A)$, $d = [D, \cdot]$.
&= -d(a)^*
\end{align*}
\end{enumerate}
-\end{solution}
-\begin{exercise}
+\end{MyExercise}
+\begin{MyExercise}
+ \textbf{
Verify that $\Omega _D^1 (A)$ is an $A$-bimodule by rewriting
+ }
\begin{align*}
a(a_k[D, b_k]b = \sum_k a'_k[D, b'_k] \;\;\;\; a'_k, b'_k \in A
\end{align*}
-\end{exercise}
-\begin{solution}
+ \newline
+
First off we know the algebra is associative then we know that elements
in $A$ can be represented faithfully on a Hilbert space $H$. Because of
the Hilbert Basis $\{\textbf{n}_i\}_{i\in \mathbb{N}}$ of the Hilbert space we can decompose these elements
@@ -271,7 +311,7 @@ Then there is a map $d:A\rightarrow \Omega _D ^1 (A)$, $d = [D, \cdot]$.
[D, b_k] b = d(b_k)b = d(b_kb) - b_kd(b)\\
\end{align*}
I don't think this is correct I'll try it again
-\end{solution}
+\end{MyExercise}
\begin{lemma}
Let $(A, H, D) = (M_n(\mathbb{C}, \mathbb{C}^n, D)$, with $D$ a hermitian
@@ -293,19 +333,21 @@ I don't think this is correct I'll try it again
(A) \subset L(\mathbb{C}^n) = H \simeq M_n(\mathbb{C}) = A$
\end{proof}
-\begin{exercise}
+\begin{MyExercise}
+ \textbf{
Consider $(A=\mathbb{C}^2, H=\mathbb{C}^2,
D = \begin{pmatrix} 0 & \lambda \\ \bar{\lambda} & 0
\end{pmatrix})$ with $\lambda \neq 0$. Show that $\Omega _D^1(A)
\simeq M_2(\mathbb{C})$
-\end{exercise}
-\begin{solution}
+ }
+\newline
+
Because of the Hilbert Basis $D$ can be extended in terms of
the basis of $M_2(\mathbb{C})$, plugging this into Equation
\ref{basis} will get us the same cyclic result, thus
$\Omega _D^1(A) \simeq M_2(\mathbb{C})$
\
-\end{solution}
+\end{MyExercise}
\subsection{Morphisms Between Finite Spectral Triples}
\begin{definition}
@@ -331,12 +373,12 @@ Some remarks
$\Omega _D^1 (A)$.
\end{itemize}
-\begin{exercise}
+\begin{MyExercise}
+ \textbf{
Show that the unitary equivalence between finite spectral
triples is a equivalence relation
-\end{exercise}
+}\newline
-\begin{solution}
An equivalence relation needs to satisfy reflexivity, symmetry
transitivity.
Let $(A_1, H_1, D_1)$, $(A_2, H_2, D_2)$ and $(A_3, H_3, D_3)$
@@ -377,7 +419,7 @@ Some remarks
D_2 U_23^* \\
&= D_3
\end{align*}
-\end{solution}
+\end{MyExercise}
Extending the this relation we look again at the notion of equivalence from
Morita equivalence of Matrix Algebras.
@@ -436,12 +478,13 @@ With the information thus far we can prove the following theorem
\end{align*}
\end{proof}
-\begin{exercise}
+\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
$E \rightarrow E\otimes _A \Omega _D^1(A)$
-\end{exercise}
-\begin{solution}
+}\newline
+
Both $\nabla$ and $\nabla'$ need to satisfy the Leiblitz rule, so
let's see if $\nabla - \nabla'$ does.
@@ -454,9 +497,10 @@ With the information thus far we can prove the following theorem
&=\bar{\nabla}(ea)
\end{align*}
For some $\bar{\nabla}=\nabla-\nabla'$.
-\end{solution}
+\end{MyExercise}
-\begin{exercise}
+\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
@@ -474,11 +518,11 @@ With the information thus far we can prove the following theorem
difference operator $D'$ with the connection on $A$ given by
$\nabla = d + \omega$
\end{enumerate}
-\end{exercise}
-\begin{solution}
+}
+
I did some notes on this one, but they are not really correct. I'll try
it again next session.
-\end{solution}
+\end{MyExercise}
\subsection{Graphing Finite Spectral Triples}
\begin{definition}
@@ -501,19 +545,20 @@ With the information thus far we can prove the following theorem
\end{tikzpicture}
\caption{A simple graph with three vertices and three edges}
\end{figure}
-\begin{exercise}
+\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 $H$.
-\end{exercise}
-\begin{solution}
+}\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{solution}
+\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
@@ -576,13 +621,14 @@ $D_{ij}$
D = D_e + D_e^*)$}
\end{figure}
-\begin{exercise}
+\begin{MyExercise}
+ \textbf{
Draw a $\Lambda$ decorated graph corresponding to the spectral triple
$(A=\mathbb{C}^3, H=\mathbb{C}^3, D=\begin{pmatrix}0 & \lambda & 0\\
\bar{\lambda} &0 &0 \\ 0&0&0\end{pmatrix})$
-\end{exercise}
-\begin{figure}[h!]
- \centering
+}\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}},]
@@ -593,14 +639,15 @@ $D_{ij}$
\draw[style=thick, -] (1.1,1.7) -- (-1.1,1.7);
\draw[style=thick, -] (1.1,1.3) -- (-1.1,1.3);
\end{tikzpicture}
- \caption{Solution}
-\end{figure}
-\begin{exercise}
+ % \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
$A=\mathbb{C}\oplus M_2(\mathbb{C})$
-\end{exercise}
-\begin{figure}[h!]
+}\newline
+
\centering
\begin{tikzpicture}[
mass/.style = {draw,circle, minimum size=0.4cm, inner sep=0pt, thick},
@@ -625,9 +672,10 @@ $D_{ij}$
\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}
- \caption{Solution $A=M_3(\mathbb{C})$}
-\end{figure}
+
+\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
