commit 6410158b96d6b52ecbdc7285c54f986fd25af4fd
parent 08728ff7db193b88e1a312f16cd0404357269f9e
Author: miksa234 <milutin@popovic.xyz>
Date: Wed, 17 Mar 2021 12:30:11 +0100
finishing up the worked exercises
Diffstat:
| M | week5.tex | | | 111 | +++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++-- |
1 file changed, 109 insertions(+), 2 deletions(-)
diff --git a/week5.tex b/week5.tex
@@ -227,6 +227,51 @@ 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}
+ 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}
+ For the record $d(\cdot) = [D, \cdot]$, then we have
+ \begin{enumerate}
+ \item
+ \begin{align*}
+ d(ab) &= [D, ab] = [D, a]b + a[D,b]\\
+ &= d(a)b + ad(b)
+ \end{align*}
+ \item
+ \begin{align*}
+ d(a^*) &= [D, a^*] = Da^* - a^*D \\
+ &=-(D^*a - aD^*) = -[D^*, a] \\
+ &= -d(a)^*
+ \end{align*}
+ \end{enumerate}
+\end{solution}
+\begin{exercise}
+ 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}
+ 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
+ in therms of the basis elements.
+ \begin{align*}
+ aa_k &= \sum _{\textbf{n}}(\langle a, \textbf{n} \rangle) a_k \\
+ &= \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:
+ \begin{align*}
+ [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}
\begin{lemma}
Let $(A, H, D) = (M_n(\mathbb{C}, \mathbb{C}^n, D)$, with $D$ a hermitian
@@ -240,14 +285,28 @@ Then there is a map $d:A\rightarrow \Omega _D ^1 (A)$, $d = [D, \cdot]$.
Assume $D = \sum _i \lambda _i e_{ii}$ (diagonal), $\lambda _i \in \mathbb{R}$ and
$\{e_{ij}\}$ the basis of $M_n(\mathbb{C}$. For fixed $i$, $j$ choose $k$
such that $\lambda _k \neq \lambda _j$ then
- \begin{align*}
+ \begin{align} \label{basis}
\left(\frac{1}{\lambda _k - \lambda _j} e_{ik}\right) [D, e_{kj}] =
e_{ij}
- \end{align*}
+ \end{align}
$e_{ij}\in \Omega _D ^1 (A)$ by the above definition. And $\Omega _D ^1
(A) \subset L(\mathbb{C}^n) = H \simeq M_n(\mathbb{C}) = A$
\end{proof}
+\begin{exercise}
+ 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}
+ 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}
+
\subsection{Morphisms Between Finite Spectral Triples}
\begin{definition}
two finite spectral tripes $(A_1, H_1, D_1)$ and $(A_2, H_2, D_2)$ are
@@ -272,6 +331,54 @@ Some remarks
$\Omega _D^1 (A)$.
\end{itemize}
+\begin{exercise}
+ Show that the unitary equivalence between finite spectral
+ triples is a equivalence relation
+\end{exercise}
+
+\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 $(A_3, H_3, D_3)$
+ 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{solution}
+
Extending the this relation we look again at the notion of equivalence from
Morita equivalence of Matrix Algebras.
\newline
