commit 74dd4465987dfdadea227e83522f04e12fa404e2
parent d67465d0c2bad1ea4b8b7bf477583e6106c48da5
Author: miksa <milutin@popovic.xyz>
Date: Thu, 27 May 2021 17:45:37 +0200
done exercises
Diffstat:
4 files changed, 80 insertions(+), 14 deletions(-)
diff --git a/pdfs/week6.pdf b/pdfs/week6.pdf
Binary files differ.
diff --git a/pdfs/week7.pdf b/pdfs/week7.pdf
Binary files differ.
diff --git a/src/week6.tex b/src/week6.tex
@@ -196,7 +196,7 @@ vector space with the opposite product
\item Show that the commutant $A'$ of $A$ is $A'\simeq \bigoplus_i M_{m_i} (\mathbb{C})$. As a consequence show $A'' \simeq A$.
\item Show that if $\xi$ is a separating vector for $A$ than it is cyclic for $A'$.
\end{enumerate}
- }\newline
+ }
\begin{enumerate}
@@ -222,23 +222,23 @@ vector space with the opposite product
$S: H \rightarrow H$.
\item Show that $S$ is invertible
\item Let $J: H \rightarrow H$ be the operator in $S = J \Delta ^{1/2}$ with
- $\Delta = S*S$. Show that $J$ is anti-unitary
+ $\Delta = S^*S$. Show that $J$ is anti-unitary
\end{enumerate}
- }\newline
+ }
\begin{enumerate}
\item By composition $S(a\xi) = a*\xi$ this is literally anti-linearity. Does this mean
$S\xi = \xi$?
- \item Let $\xi \in H$ be cyclic then: $S(A\xi) = A*\xi = A\xi = H$. The same has to work
- for $S^{-1}$ if not then $\xi$ wouldn't exist. $S^{-1}(A*\xi) = S^{-1}(H) = H$.
- \item Since $S$ is bijective then $\Delta ^{1/2}$ and $J$ need to be bijective.\\
- Now let $\xi _1 , \xi _2 \in H$.\\
- \begin{align}
+ \item Let $\xi \in H$ be cyclic then: $S(A\xi) = A^*\xi = A\xi = H$. The same has to work
+ for $S^{-1}$ if not then $\xi$ wouldn't exist. $S^{-1}(A^*\xi) = S^{-1}(H) = H$.
+ \item Since $S$ is bijective then $\Delta ^{1/2}$ and $J$ need to be bijective.
+ We also have $J = S \Delta^{-1/2}$ and $\Delta^* = \Delta$\\
+ Now let $\xi _1 , \xi _2 \in H$ \begin{align}
<J \xi _1 , J \xi _2 > &= < J^*J\xi_1 , \xi_2>^* =\\
- &= <(\Delta ^{1/2})^* S^* S \Delta ^{1/2} \xi_1, \xi_2>^* = \\
- &= <(SS^*)^{1/2}S^*S(SS^*) \xi_1, \xi_2>^* =\\
- &= <(SS^*SS^*)^{1/2} \xi_1, \xi_2>^* = \\
+ &= <(\Delta ^{-1/2})^* S^* S \Delta ^{-1/2} \xi_1, \xi_2>^* = \\
+ &= <(\Delta^{-1/2})^* \Delta \Delta^{-1/2} \xi_1, \xi_2>^* =\\
+ &= <\Delta^{-1/2} \Delta^{1/2}\Delta^{1/2} \Delta^{-1/2} \xi_1, \xi_2>^* =\\
&= <\xi _1, \xi_2>^* = <\xi_2 , \xi_1>.
\end{align}
\end{enumerate}
@@ -350,14 +350,13 @@ with $b^\circ = J' b^* (J')^{-1}$, $b^* \in B$ action on $H'$.
\tau : E \otimes_A \Omega _D^1 (A) &\rightarrow \Omega _D^1 (A) \otimes_A E^\circ\\
e \otimes \omega &\mapsto -\omega ^* \otimes \bar{e}
\end{align}
- Show that the map $\bar{\nabla} : E^\circ \righarrow \Omega _D^1(A) \otimes E^\circ$
+ Show that the map $\bar{\nabla} : E^\circ \rightarrow \Omega _D^1(A) \otimes E^\circ$
with $\bar{\nabla}(\bar{e}) = \tau \circ \nabla(e)$ is a left connection, that means
show that it satisfied the left Leibniz rule:
\begin{equation}
\bar{\nabla}(a\bar{e}) = [D, a] \otimes \bar{e} + a \bar{\nabla}(\bar{e})
\end{equation}
- }\newline
-
+ }
Hagime:
\begin{align}
diff --git a/src/week7.tex b/src/week7.tex
@@ -468,6 +468,73 @@ such that $(ab)^* = b^*a^*$ and $(a^*)^* \;\;\; \forall a,b\in A$
\simeq M_{2k}(\mathbb{C})$ as complex *algebras.
\end{enumerate}
}
+ 1). Let us take some $a, b \in \mathbb{H}$ with
+ \begin{align}
+ a = \begin{pmatrix}
+ \alpha & \beta \\
+ -\bar{\beta} & \bar{\alpha}
+ \end{pmatrix} \;\;\;\;
+ b = \begin{pmatrix}
+ \gamma & \delta \\
+ -\bar{\delta} & \bar{\gamma}
+ \end{pmatrix}
+ \end{align}
+ where $\alpha, \beta, \gamma, \delta \in \mathbb{C}$. Since
+ $\mathbb{H}$ is represented in standard $2x2$ matrices, the involution
+ is just subsequent from there, the only thing left to show is the
+ closure $ab \in \mathbb{H}$.
+ \begin{align}
+ ab &=
+ \begin{pmatrix}
+ \alpha & \beta \\
+ -\bar{\beta} & \bar{\alpha}
+ \end{pmatrix}
+ \begin{pmatrix}
+ \gamma & \delta \\
+ -\bar{\delta} & \bar{\gamma}
+ \end{pmatrix} =\\
+ &=
+ \begin{pmatrix}
+ \alpha\beta - \beta\bar{\delta}& \alpha\delta + \beta \bar{\gamma}\\
+ -(\bar{\alpha}\bar{\delta} + \bar{\beta}\gamma) &
+ \bar{\alpha}\gamma-\bar{\beta}\delta
+ \end{pmatrix} =
+ \begin{pmatrix}
+ \xi& \psi\\
+ -\bar{\psi} & \bar{\xi}
+ \end{pmatrix} \in \mathbb{H}
+ \end{align}
+ where $\xi, \psi \in \mathbb{C}$ because of closure of complex numbers
+ in regards to multiplication and addition, which is $\mathbb{R}
+ \otimes_{\mathbb{R}}\mathbb{C} \simeq \mathbb{C}$, e.g. $\beta \cdot c
+ \in \mathbb{C}$ with $c \in \mathbb{C}$.
+ \newline
+ 2)For $\mathbb{H}\otimes_\mathbb{R} \mathbb{C}$ we have for some $h \in
+ \mathbb{H}$ and $c \in mathbb{C}$
+ \begin{align}
+ h\otimes c &=
+ \begin{pmatrix}
+ \alpha & \beta \\
+ -\bar{\beta} & \bar{\alpha}
+ \end{pmatrix}\otimes c = \\
+ &=
+ \begin{pmatrix}
+ \alpha c & \beta c \\
+ -\bar{\beta} c & \bar{\alpha} c
+ \end{pmatrix} \simeq M_2(\mathbb{C})
+ \end{align}
+ because again of $\mathbb{R} \otimes_\mathbb{R} \mathbb{C} \simeq
+ \mathbb{C}$.
+ \newline
+ 3)We know that $\mathbb{H}$ is a real subalgebra of $M_2(\mathbb{C})$,
+ so $M_k(\mathbb{H})$ is just an extension and an real subalgebra of
+ $M_{2k}(\mathbb{C})$.
+ \newline
+ 4) Here we use what we have learned
+ \begin{align}
+ M_k(\mathbb{H})\otimes_\mathbb{R} \mathbb{C} \simeq
+ M_k(M_2(\mathbb{C})) = M_{2k}(\mathbb{C})
+ \end{align}
\end{MyExercise}
\begin{definition}
A representation of a finite-dimensional real * algebra $A$ is a pair $(\pi
