commit 446272af324ed22f48abbb532b4db4209aa60552
parent c6e56fff3821a9a1048b668390053ce3a4441c7f
Author: miksa <milutin@popovic.xyz>
Date: Thu, 18 Feb 2021 09:46:40 +0100
to do: write up the solutions
Diffstat:
| M | week2.pdf | | | 0 | |
| M | week2.tex | | | 118 | +++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++-- |
2 files changed, 115 insertions(+), 3 deletions(-)
diff --git a/week2.pdf b/week2.pdf
Binary files differ.
diff --git a/week2.tex b/week2.tex
@@ -20,12 +20,18 @@
\theoremstyle{definition}
\newtheorem{question}{Question}
+\theoremstyle{definition}
+\newtheorem{example}{Example}
+
\theoremstyle{theorem}
\newtheorem{theorem}{Theorem}
\theoremstyle{theorem}
\newtheorem{exercise}{Exercise}
+\theoremstyle{theorem}
+\newtheorem{lemma}{Lemma}
+
\theoremstyle{definition}
\newtheorem{solution}{Solution}
@@ -95,10 +101,23 @@ We denote $KK_f(A,B)$ the set of all \textit{Hilbert bimodules} of $(A,B)$.
\begin{align*}
\langle a, a\rangle_A = a^*a' \;\;\;\; a,a'\in A
\end{align*}
+ \label{exercise: inner-product}
\end{exercise}
\begin{solution}
\end{solution}
+\begin{example}
+ Consider a $*$ homomorphism between two matrix algebras $\phi:A\rightarrow B$.
+ From it we can construct a Hilbert bimodule $E_{\phi} \in KK_f(A, B)$ in the following way.
+ We let $E_{\phi}$ be $B$ in the vector space sense and an inner product from the above
+ Exercise \ref{exercise: inner-product}, with $A$ acting on the left with $\phi$.
+ \begin{align*}
+ a\cdot b = \phi(a)b \;\;\;\; a\in A, b\in E_{\phi}
+ \end{align*}
+\end{example}
+
+
+
\subsubsection{Kasparov Product and Morita Equivalence}
\begin{definition}
Let $E \in KK_f(A, B)$ and $F \in KK_F(B, D)$ the \textit{Kasparov product} is defined as
@@ -159,6 +178,31 @@ What is the meaning of 'associative up to isomorphism? Isomorphism of $F \circ E
Why are $E$ and $F$ each others inverse in the Kasparov Product?
\end{question}
+\begin{example}
+ \
+ \begin{itemize}
+ \item Hilber bimodule of $(A,A)$ is $A$
+ \item Let $E \in KK_f(A,B)$, we take $E \circ A = A\oplus _A E \simeq E$
+ \item we conclude, that $_A A_A$ is the identity in the Kasparov product (up to isomorphism)
+ \end{itemize}
+\end{example}
+
+\begin{example}
+ Let $E = \mathbb{C}^n$, which is a $(M_n(\mathbb{C}), \mathbb{C})$ Hilbert bimodule with the
+ standard $\mathbb{C}$ inner product.\\
+ On the other hand let $F = \mathbb{C}^n$, which is a $(\mathbb{C}, M_n(\mathbb{C}))$ Hilbert
+ bimodule by right matrix multiplication with $M_n(\mathbb{C})$ valued inner product:
+ \begin{align*}
+ <v_1, v_2>=\bar{v_1}v_2^t \;\; \in M_n(\mathbb{C})
+ \end{align*}
+ Now we take the Kasparov product of $E$ and $F$:
+ \begin{itemize}
+ \item $F\circ E\ =\ E\otimes _{\mathbb{C}}F\ \;\;\;\;\;\; \simeq \ M_n(\mathbb{C})$
+ \item $E\circ F\ =\ F\otimes _{M_n(\mathbb{C})}E\ \simeq\ \mathbb{C}$
+ \end{itemize}
+ $M_n(\mathbb{C})$ and $\mathbb{C}$ are Morita equivalent
+\end{example}
+
\begin{theorem}
Two matrix algebras are Morita Equivalent iff their their Structure spaces
are isomorphic as discreet spaces (have the same cardinality / same number of elements)
@@ -169,13 +213,81 @@ What is the meaning of 'associative up to isomorphism? Isomorphism of $F \circ E
E \otimes _B F \simeq A \;\;\; \text{and} \;\;\; F \otimes _A E \simeq B
\end{align*}
Consider $[(\pi _B, H)] \in \hat{B}$ than we construct a representation of $A$,
- $\pi _A \rightarrow L(E \otimes _B H)$ with $\pi _A(a) (e \otimes v) = a e \otimes w$
+ \begin{align*}
+ \pi _A \rightarrow L(E \otimes _B H)\;\;\; \text{with} \;\;\; \pi _A(a) (e \otimes v) = a e \otimes w
+ \end{align*}
\begin{question}
Is $E \simeq H$ and $F \simeq W$?
\end{question}
- \textit{vice versa}, consider $[(\pi _A, W)] \in \hat{A} \Rightarrow
- \pi _B: B \rightarrow L(F \otimes _A W)$ and $\pi _B(b) (f\otimes w) = bf\otimes w$
+ \textit{vice versa}, consider $[(\pi _A, W)] \in \hat{A}$ we can construct $\pi _B$
+ \begin{align*}
+ \pi _B: B \rightarrow L(F \otimes _A W) \;\;\; \text{and}\;\;\; \pi _B(b) (f\otimes w) = bf\otimes w
+ \end{align*}
These maps are each others inverses, thus $\hat{A} \simeq \hat{B}$
\end{proof}
+\begin{exercise}
+ Fill in the gaps in the above proof:
+ \begin{itemize}
+ \item show that the representation of $\pi _A$ defined is irreducible iff $\pi _B$ is.
+ \item Show that the association of the class $[\pi _A]$ to $[\pi _B]$ is independent
+ of the choice of representatives $\pi _A$ and $\pi _B$
+ \end{itemize}
+\end{exercise}
+
+\begin{solution}
+\end{solution}
+
+\begin{lemma}
+ The matrix algebra $M_n(\mathbb{C})$ has a unique inrreducible representation (up to isomorphism)
+ given by the defining representation on $\mathbb{C}^n$.
+\end{lemma}
+\begin{proof}
+ We know $\mathbb{C}^n$ is a irreducible representation of $A= M_n(\mathbb{C})$. Let $H$ be irreducible
+ and of dimension $k$, then we define a map
+ \begin{align*}
+ \phi : A\oplus...\oplus A &\rightarrow H^* \\
+ (a_1,...,a_k) &\mapsto e^1\circ a_1^t+...+e^k\circ a_k^t
+ \end{align*}
+ With $\{e^1,...,e^k\}$ being the basis of the dual space $H^*$ and $(\circ)$ being the pre-composition
+ of elements in $H^*$ and $A$ acting on $H$. This forms a morphism of $M_n(\mathbb{C})$ modules,
+ provided a matrix $a \in A$ acts on $H^*$ with $v\mapsto v\circ a^t$ ($v\in H^*$).
+ Furthermore this morphism is surjective, thus making the pullback $\phi ^*:H\mapsto (A^k)^*$ injective.
+ Now identify $(A^k)^*$ with $A^k$ as a $A$-module and note that
+ $A=M_n(\mathbb{C}) \simeq \oplus ^n \mathbb{C}^n$ as a n A module.
+ It follows that $H$ is a submodule of $A^k \simeq \oplus ^{nk}\mathbb{C}$. By irreducibility
+ $H \simeq \mathbb{C}$.
+\end{proof}
+
+\begin{example}
+ Consider two matrix algebras $A$, and $B$.
+ \begin{align*}
+ A = \bigoplus ^N_{i=1} M_{n_i}(\mathbb{C}) \;\;\; B = \bigoplus ^M_{j=1} M_{m_j}(\mathbb{C})
+ \end{align*}
+ Let $\hat{A} \simeq \hat{B}$ that implies $N=M$ and define $E$ with $A$ acting by block-diagonal
+ matrices on the first tensor and B acting in the same way on the second tensor. Define $F$ vice versa.
+ \begin{align*}
+ E:= \bigoplus _{i=1}^N \mathbb{C}^{n_i} \otimes \mathbb{C}^{m_i} \;\;\;
+ F:= \bigoplus _{i=1}^N \mathbb{C}^{m_i} \otimes \mathbb{C}^{n_i}
+ \end{align*}
+ Then we calculate the Kasparov product.
+ \begin{align*}
+ E \otimes _B F &\simeq \bigoplus _{i=1}^N (\mathbb{C}^{n_i}\otimes\mathbb{C}^{m_i})
+ \otimes _{M_{m_i}(\mathbb{C})} (\mathbb{C}^{m_i}\otimes\mathbb{C}^{n_i}) \\
+ &\simeq \bigoplus _{i=1}^N \mathbb{C}^{n_i}\otimes
+ \left(\mathbb{C}^{m_i}\otimes _{M_{m_i}(\mathbb{C})}\mathbb{C}^{m_i}\right)
+ \oplus \mathbb{C}^{n_i} \\
+ &\simeq \bigoplus _{i=1}^N \mathbb{C}^{m_i}\otimes\mathbb{C}^{n_i} \simeq A
+ \end{align*}
+ and from $F \otimes _A E \simeq B$.
+\end{example}
+
+We conclude that.
+\begin{itemize}
+ \item There is a duality between finite spaces and Morita equivalence classes of matrix algebras.
+ \item By replacing $*$-homomorphism $A\rightarrow B$ with Hilbert bimodules $(A,B)$ we introduce
+ a richer structure of morphism between matrix algebras.
+\end{itemize}
+
+
\end{document}
