commit 50d2a7089b2150049fe2207a1ad762a89c8aeb03
parent 6410158b96d6b52ecbdc7285c54f986fd25af4fd
Author: miksa234 <milutin@popovic.xyz>
Date: Thu, 18 Mar 2021 17:45:04 +0100
done with week5
Diffstat:
| M | week5.pdf | | | 0 | |
| M | week5.tex | | | 166 | ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++--- |
2 files changed, 161 insertions(+), 5 deletions(-)
diff --git a/week5.pdf b/week5.pdf
Binary files differ.
diff --git a/week5.tex b/week5.tex
@@ -266,7 +266,7 @@ Then there is a map $d:A\rightarrow \Omega _D ^1 (A)$, $d = [D, \cdot]$.
&= \sum _{k} a'_{k}
\end{align*}
Which would than be the same as the sum of some elements
- $\a'_{k} \in A$. Then we calculate the commutator:
+ $a'_{k} \in A$. Then we calculate the commutator:
\begin{align*}
[D, b_k] b = d(b_k)b = d(b_kb) - b_kd(b)\\
\end{align*}
@@ -359,8 +359,8 @@ Some remarks
&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} \;\;\;
@@ -436,6 +436,50 @@ With the information thus far we can prove the following theorem
\end{align*}
\end{proof}
+\begin{exercise}
+ 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}
+ 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[\prime{D}, a]\\
+ &=\bar{\nabla}(ea)
+ \end{align*}
+ For some $\bar{\nabla}=\nabla-\nabla'$.
+\end{solution}
+
+\begin{exercise}
+ 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}
+\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}
+
\subsection{Graphing Finite Spectral Triples}
\begin{definition}
A \textit{graph} is a ordered pair $(\Gamma ^{(0)}, \Gamma ^{(1)})$.
@@ -457,6 +501,19 @@ 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}
+ 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}
+ $A\simeq \bigoplus_{i=1}^N M_{n_i}(\mathbb{C})$ 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}
\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
@@ -506,8 +563,7 @@ $D_{ij}$
D = \sum _{e \in \Gamma ^{(1)}} D_e + D_e^*
\end{equation}
\end{theorem}
-\begin{example}
-\begin{figure}[h!]
+ \begin{figure}[h!]
\centering
\begin{tikzpicture}[
mass/.style = {draw,circle, minimum size=0.3cm, inner sep=0pt, thick},
@@ -519,5 +575,105 @@ $D_{ij}$
\caption{A $\Lambda$-decorated Graph of $(M_n(\mathbb{C}), \mathbb{C}^n,
D = D_e + D_e^*)$}
\end{figure}
-\end{example}
+
+\begin{exercise}
+ 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
+\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}
+ \caption{Solution}
+\end{figure}
+\begin{exercise}
+ 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!]
+ \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}
+ \caption{Solution $A=M_3(\mathbb{C})$}
+\end{figure}
+\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}
+
+
\end{document}
