commit ca51a9ea241136bf2426523d924a11dd9feb6047
parent 2b11e4d44fc696ca2dbb8048cc280bd250de0d0b
Author: miksa234 <milutin@popovic.xyz>
Date: Fri, 26 Mar 2021 08:37:17 +0100
added some exercise
Diffstat:
3 files changed, 32 insertions(+), 21 deletions(-)
diff --git a/pdfs/week5.pdf b/pdfs/week5.pdf
Binary files differ.
diff --git a/src/week5.tex b/src/week5.tex
@@ -325,20 +325,20 @@ Then there is a map $d:A\rightarrow \Omega _D ^1 (A)$, $d = [D, \cdot]$.
\end{align*}
\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
- 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
+ % 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{MyExercise}
\begin{lemma}
diff --git a/src/week6.tex b/src/week6.tex
@@ -144,7 +144,7 @@ An opposite-algebra $A^\circ$ of a $A$ is defined to be equal to $A$ as a
vector space with the opposite product
\begin{align}
&a\circ b := ba\\
- &\Rihtarrow a^\circ = Ja^* J^{-1} \;\;\; \text{defines the left
+ &\Rightarrow a^\circ = Ja^* J^{-1} \;\;\; \text{defines the left
representation of $A^\circ$ on $H$}
\end{align}
\end{definition}
@@ -158,6 +158,17 @@ vector space with the opposite product
Then we define $\gamma (a) = a$ and $J(a) = a^*$ with $a\in H$.
Since $D$ mus be odd with respect to $\gamma$ it vanishes identically.
\end{example}
+\begin{MyExercise}
+ \textbf{
+ In the previous example, show that the right action on $M_N(\mathbb{C})$
+ on $H = M_N(\mathbb{C})$ as defined by $a \mapsto a^\circ$
+ is given by right matrix multiplication.
+}\newline
+
+ \begin{align*}
+ a^\circ \xi = J a^* J^{-1} = Ja^* \xi^* = J\xi a=\xi^* a
+ \end{align*}
+\end{MyExercise}
\subsection{Morphisms Between Finite Real Spectral Triples}
Extend unitary equivalence of finite spectral triples to real ones (with $J$
@@ -174,8 +185,8 @@ and $\gamma$)
&U\gamma _1 U^* = \gamma _2\\
&UJ_1 U^* = J_2
\end{align}
-\end{definiton}
-\begin{definiton}
+\end{definition}
+\begin{definition}
Let $E$ be a $B$-$A$ bimodule. The \textit{conjugate Module} $E^\circ$ is
given by the $A$-$B$-bimodule.
\begin{align}
@@ -186,7 +197,7 @@ and $\gamma$)
a \cdot \bar{e} \cdot b = b^* \bar{e} a^* \;\;\;\; \forall a\in A, b \in
B
\end{align}
-\end{definiton}
+\end{definition}
$E^\circ$ is not a Hilbert bimodule for $(A, B)$ because it doesn't have a
natural $B$-valued inner product. But there is a $A$-valued inner product on
the left $A$-module $E^\circ$ with
@@ -195,10 +206,10 @@ the left $A$-module $E^\circ$ with
\;\;\;\; e_1, e_2 \in E
\end{align}
and linearity in $A$:
-\begin{aling}
+\begin{align}
\langle a \bar{e}_1, \bar{e}_2 \rangle = a \langle \bar{e}_1, \bar{e}_2
\rangle \;\;\;\; \forall a \in A.
-\end{aling}
+\end{align}
\subsubsection{Construction of a Finite Real Spectral Triple from a Finite
Real Spectral Triple}
Given a Hilbert bimodule $E$ for $(B, A)$ we construct a spectral triple
@@ -267,7 +278,7 @@ Finally for the grading
and for $\epsilon '$
\begin{align*}
J'D'(e_1 \otimes \xi \otimes \bar{e}_2)&=J'((\nabla e_1) \xi \otimes
- \bar{e_2} + e_1 \otims D\xi \otimes \bar{e}_2 + e_1 \otimes \xi (\tau
+ \bar{e_2} + e_1 \otimes D\xi \otimes \bar{e}_2 + e_1 \otimes \xi (\tau
\nabla e_2))\\
&= \epsilon' D'(e_2 \otimes J\xi \otimes \bar{e}_2)\\
&= \epsilon' D'J'(e_1 \otimes \xi \bar{e}_2)
