commit 2b11e4d44fc696ca2dbb8048cc280bd250de0d0b
parent da75f78c681a6deb7b42ee5ccf92cc30ba40cce9
Author: miksa234 <milutin@popovic.xyz>
Date: Thu, 25 Mar 2021 17:41:21 +0100
done writing up not done with exercises
Diffstat:
| M | src/week6.tex | | | 188 | +++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ |
1 file changed, 188 insertions(+), 0 deletions(-)
diff --git a/src/week6.tex b/src/week6.tex
@@ -85,5 +85,193 @@
\section{Finite Real Noncommutative Spaces}
\subsection{Finite Real Spectral Triples}
+Add on to finite real spectral triples a \textit{real structure}. The
+requirement is that $H$ is a $A$-$A$-bimodule (before only a $A$-left
+module).
+\newline
+
+For this we introduce a $\mathbb{Z}_2$-grading $\gamma$ with
+\begin{align}
+ &\gamma ^* = \gamma \\
+ &\gamma ^2 = 1 \\
+ &\gamma D = - D \gamma\\
+ &\gamma a = a \gamma \;\;\;\; a\in A
+\end{align}
+
+\begin{definition}
+ A \textit{finite real spectral triple} is given by a finite spectral
+ triple $(A, H, D)$ and a anti-unitary operator $J:H\rightarrow H$ called
+ the \textit{real structure}, such that
+ \begin{align}
+ a^\circ := J a^* J^{-1}
+ \end{align}
+ is a right representation of $A$ on $H$, that is $(ab)^\circ = b^\circ
+ a^\circ$. With two requirements
+ \begin{align}
+ &[a, b^\circ] = 0\\
+ &[[D, a],b^\circ] = 0.
+ \end{align}
+ They are called the \textit{commutant property}, and mean that the left
+ action of an element in $A$ and $\Omega _D^1(A)$ commutes with the right
+ action on $A$.
+\end{definition}
+\begin{definition}
+ The $KO$-dimension of a real spectral triple is determined by the sings
+ $\epsilon, \epsilon ' ,\epsilon '' \in \{-1, 1\}$ appearing in
+ \begin{align}
+ &J^2 = \epsilon \\
+ &JD = \epsilon \ DJ\\
+ &J\gamma = \epsilon '' \gamma J.
+ \end{align}
+\end{definition}
+\begin{table}[h!]
+ \centering
+ \caption{$KO$-dimension $k$ modulo $8$ of a real spectral triple}
+ \begin{tabular}{ c | c c c c c c c c}
+ \hline
+ $k$ & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\
+ \hline
+ $\epsilon$ & 1 & 1 & -1 & -1 & -1 & -1 & 1 & 1 \\
+ $\epsilon '$ & 1 & -1 & 1 & 1 & 1 & -1 & 1 & 1 \\
+ $\epsilon ''$ & 1 & & -1 & & 1 & & -1 & \\
+ \hline
+ \end{tabular}
+\end{table}
+
+
+\begin{definition}
+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
+ representation of $A^\circ$ on $H$}
+\end{align}
+\end{definition}
+
+\begin{example}
+ Matrix algebra $M_N(\mathbb{C})$ acting on $H=M_N(\mathbb{C})$ by left
+ matrix multiplication with the Hilbert Schmidt inner product.
+ \begin{align}
+ \langle a , b \rangle = \text{Tr}(a^* b)
+ \end{align}
+ 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}
+
+\subsection{Morphisms Between Finite Real Spectral Triples}
+Extend unitary equivalence of finite spectral triples to real ones (with $J$
+and $\gamma$)
+
+\begin{definition}
+ We call two finite real spectral triples $(A_1, H_1 ,D_1 ; J_1 , \gamma
+ _1)$ and $(A_2, H_2, D_2; J_2, \gamma _2)$ unitarily equivalent if $A_1 =
+ A_2$ and if there exists a unitary operator $U: H_1 \rightarrow H_2$ such
+ that
+ \begin{align}
+ &U\pi_1(a) U^* = \pi _2(a)\\
+ &UD_1U^*=D_2\\
+ &U\gamma _1 U^* = \gamma _2\\
+ &UJ_1 U^* = J_2
+ \end{align}
+\end{definiton}
+\begin{definiton}
+ Let $E$ be a $B$-$A$ bimodule. The \textit{conjugate Module} $E^\circ$ is
+ given by the $A$-$B$-bimodule.
+ \begin{align}
+ E^\circ = \{\bar{e} : e\in E\}
+ \end{align}
+ with
+ \begin{align}
+ a \cdot \bar{e} \cdot b = b^* \bar{e} a^* \;\;\;\; \forall a\in A, b \in
+ B
+ \end{align}
+\end{definiton}
+$E^\circ$ 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
+\begin{align}
+ \langle \bar{e}_1, \bar{e}_2 \rangle = \langle e_2 , e_1 \rangle
+ \;\;\;\; e_1, e_2 \in E
+\end{align}
+and linearity in $A$:
+\begin{aling}
+ \langle a \bar{e}_1, \bar{e}_2 \rangle = a \langle \bar{e}_1, \bar{e}_2
+ \rangle \;\;\;\; \forall a \in A.
+\end{aling}
+\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
+$(B, H', D'; J', \gamma ')$ from $(A, H, D; J, \gamma)$
+
+For the $H'$ we make a $\mathbb{C}$-valued inner product on $H'$ by combining
+the $A$ valued inner product on $E$ and $E^\circ$ with the
+$\mathbb{C}$-valued inner product on $H$.
+\begin{align}
+ H' := E\otimes _A H \otimes _A E^\circ
+\end{align}
+
+Then the action of $B$ on $H'$ is:
+\begin{align}
+ b(e_2 \otimes \xi \otimes \bar{e}_2 ) = (be_1) \otimes \xi \otimes
+ \bar{e}_2
+\end{align}
+The right action of $B$ on $H'$ defined by action on the right component
+$E^\circ$
+\begin{align}
+ J'(e_1 \otimes \xi \otimes \bar{e}_2) = e_2 \otimes J \xi \otimes
+ \bar{e}_1
+\end{align}
+with $b^\circ = J' b^* (J')^{-1}$, $b^* \in B$ action on $H'$.
+\newline
+
+Then the connections
+\begin{align}
+ &\nabla: E \rightarrow E\otimes _A \Omega _D ^1(A) \\
+ &\bar{\nabla}:E^\circ \rightarrow \Omega _D^1(A) \otimes _A E^\circ
+\end{align}
+give us the Dirac operator on $H' = E \otimes _A H \otimes _A E^\circ$
+\begin{align}
+ D'(e_1 \otimes \xi \otimes \bar{e}_2) = (\nabla e_1) \xi \otimes
+ \bar{e_2}+ e_1 \otimes D\xi \otimes \bar{e}_2 + e_1 \otimes
+ \xi(\bar{\nabla}\bar{e}_2)
+\end{align}
+
+And the right action of $\omega \in \Omega _D ^1(A)$ on $\xi \in H$ is
+defined by
+\begin{align}
+ \xi \mapsto \epsilon' J \omega ^* J^{-1}\xi
+\end{align}
+
+Finally for the grading
+\begin{align}
+ \gamma ' = 1 \otimes \gamma \otimes 1
+\end{align}
+
+\begin{theorem}
+ Suppose $(A, H, D; J, \gamma)$ is a finite spectral triple of
+ $KO$-dimension $k$, let $\nabla$ be like above satisfying the
+ compatibility condition (like with finite spectral triples).
+
+ Then $(B, H',D'; J', \gamma')$ is a finite spectral triple of
+ $KO$-Dimension $k$. ($H', D', J', \gamma'$ like above)
+\end{theorem}
+
+\begin{proof}
+ The only thing left is to check if the $KO$-dimension is preserved,
+ for this we check if the $\epsilon$'s are the same.
+ \begin{align*}
+ &(J')^2 = 1 \otimes J^2 \otimes 1 = \epsilon\\
+ &J' \gamma '= \epsilon ''\gamma'J'
+ \end{align*}
+ 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
+ \nabla e_2))\\
+ &= \epsilon' D'(e_2 \otimes J\xi \otimes \bar{e}_2)\\
+ &= \epsilon' D'J'(e_1 \otimes \xi \bar{e}_2)
+ \end{align*}
+\end{proof}
\end{document}
