commit 49238ee2daaba327d0c814407ac67e7518958d1c
parent 68188395cea0fea2c0342eddcf352029677dae97
Author: miksa234 <milutin@popovic.xyz>
Date: Wed, 24 Mar 2021 16:07:22 +0100
boxes around exercise and finished some exercises
Diffstat:
2 files changed, 63 insertions(+), 27 deletions(-)
diff --git a/pdfs/week5.pdf b/pdfs/week5.pdf
Binary files differ.
diff --git a/src/week5.tex b/src/week5.tex
@@ -85,6 +85,7 @@
\section{Noncommutative Geometric Spaces }
\subsection{Exercises}
+
\begin{MyExercise}
\textbf{
Make the proof of the last theorem (see week4.pdf) explicit for $N=3$.
@@ -118,10 +119,10 @@
0 & x_1 \\ x_1 & 0
\end{pmatrix} \oplus
\begin{pmatrix}
- 0 & x_1 \\ x_1 & 0
+ 0 & x_2 \\ x_2 & 0
\end{pmatrix} \oplus
\begin{pmatrix}
- 0 & x_1 \\ x_1 & 0
+ 0 & x_3 \\ x_3 & 0
\end{pmatrix} \nonumber \\
&=
\begin{pmatrix}
@@ -135,19 +136,19 @@
\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
@@ -155,7 +156,18 @@ upper and lower diagonal check \url{https://en.wikipedia.org/wiki/Skew-symmetric
matrix#Skew-symmetrizable_matrix}. The matrix norm of would be the maximum of the norm of
the larges eigenvalues:
\begin{align}
- ||[D, \pi(a)]|| = \max_{a\in A}\{x_i|a(j)-a(k)|\}
+ ||[D, \pi(a)]|| = \max_{a\in A}\{&x_1|a(2)-a(1)|,\\
+ &x_2|(a(3)-a(1))|,\nonumber\\
+ &x_3|(a(3)-a(2))|,\}\nonumber
+\end{align}
+The metric is then:
+\begin{align}
+ d =
+ \begin{pmatrix}
+ 0 & a(1)-a(2) & a(1)-a(3)\\
+ a(2)-a(1) & 0 & a(2)-a(3)\\
+ a(3)-a(1) & a(3)-a(2) & 0
+ \end{pmatrix}
\end{align}
\end{MyExercise}
@@ -183,8 +195,16 @@ the larges eigenvalues:
0 & a(2)-a(1) & 0 \\
-(a(2)-a(1)) & 0 & 0 \\
0 & 0 & 0
- \end{pmatrix} \right|\right| \\
- &= d^{-1} |a(2) - a(1)|
+ \end{pmatrix} \right|\right|
+ \end{align}
+ The metric is then
+ \begin{align}
+ d =
+ \begin{pmatrix}
+ 0 & a(1)-a(2) & a(1) \\
+ a(2)-a(1) & 0 & a(2) \\
+ -a(1) & -a(2) & 0
+ \end{pmatrix}
\end{align}
\end{MyExercise}
@@ -234,6 +254,10 @@ holds. For $i = k$ then $d_{ik} = 0$ and equality holds. Else set $d_{ik} =
What does the author mean when he sais 'acts faithfully on a
Hilbertspace`? Then the representation is fully reducible, or that the
presentation is irreducible?
+ \newline
+
+ For a *-representation 'faithful` if it is injective. For a
+ *-homomorphism 'faithful` means one-to-one correspondance
\end{question}
\begin{example}
@@ -243,14 +267,18 @@ holds. For $i = k$ then $d_{ik} = 0$ and equality holds. Else set $d_{ik} =
\end{example}
$D$ is referred to as a finite Dirac operator as in as its $\infty$
-dimensional on Riemannian Spin manifolds coming in Chapter 4. Now we
-introduce it as
+dimensional on Riemannian Spin manifolds coming in Chapter 4.
+\newline
+
+Now can introduce a 'differential 'geometric structure` on the finite space X
+with the \textbf{devided difference}
\begin{equation}
\frac{a(i)-a(j)}{d_{ij}}
\end{equation}
-for each pair $i$, $j$ $\in X$ the finite dimensional discrete space.
+for each pair $i$, $j$ $\in X$ the finite dimensional discrete space $X$.
This appears in the entries in the commutator $[D, a]$ in the above
exercises.
+
\begin{definition}
Given an finite spectral triple $(A, H, D)$, the $A$-bimodule of
Connes' differential one form is:
@@ -293,7 +321,7 @@ Then there is a map $d:A\rightarrow \Omega _D ^1 (A)$, $d = [D, \cdot]$.
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
+ a(a_k[D, b_k])b = \sum_k a'_k[D, b'_k] \;\;\;\; a'_k, b'_k \in A
\end{align*}
\newline
@@ -493,10 +521,10 @@ With the information thus far we can prove the following theorem
&-(\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}a + e\otimes[D', a]\\
&=\bar{\nabla}(ea)
\end{align*}
- For some $\bar{\nabla}=\nabla-\nabla'$.
+ Therefore $\nabla-\nabla'$ is a linear map.
\end{MyExercise}
\begin{MyExercise}
@@ -519,9 +547,17 @@ With the information thus far we can prove the following theorem
$\nabla = d + \omega$
\end{enumerate}
}
-
- I did some notes on this one, but they are not really correct. I'll try
- it again next session.
+ \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}
\subsection{Graphing Finite Spectral Triples}
