commit 27787c7460f0d200134fc99bc5a953c70521b2d0
parent 446272af324ed22f48abbb532b4db4209aa60552
Author: miksa <milutin@popovic.xyz>
Date: Thu, 18 Feb 2021 12:22:39 +0100
finished week2.tex added a solution to week1.tex last exercise
Diffstat:
4 files changed, 88 insertions(+), 23 deletions(-)
diff --git a/week1.pdf b/week1.pdf
Binary files differ.
diff --git a/week1.tex b/week1.tex
@@ -89,7 +89,7 @@ $\lambda \in \mathbb{C}$ and $x \in X$ they provide the following structures:
The author indicates that $\mathbb{C}$-valued functions on $X$ are automatically continuous.
\begin{idea}
- CAN WE USE THE METRIC? >> NO!
+ CAN WE USE THE METRIC? NO!
We know that $X$ is a \textit{finite discrete space}, meaning in an $\epsilon$-$\delta$ approach
for each $x \in X$ the only $y \in X$, that is small enough is $x$ by itself, which implies
$\epsilon$ is always bigger than zero, thus every function $f:X\ \rightarrow\ \mathbb{C}$ is continuous.
@@ -143,12 +143,12 @@ Under the pointwise product:
\begin{enumerate}
\item $n=m$ \\
Obviously $\phi$ is bijective and $\phi ^*$ too.
- \item $n > m$ \\
- $\phi$ assigns $n$ points to $m$ points when $n > m$,
+ \item $n \rangle m$ \\
+ $\phi$ assigns $n$ points to $m$ points when $n \rangle m$,
which is by definition surjective. \\
- $\phi ^*$ assigns $m$ points to $n$ points when $n > m$,
+ $\phi ^*$ assigns $m$ points to $n$ points when $n \rangle m$,
which is by definition injective. \\
- \item $n < m $ \\
+ \item $n \langle m $ \\
analogous
\end{enumerate}
\end{solution}
@@ -345,8 +345,15 @@ Remark on the notation
Not quite sure but \\
$a \in A$, $h_1, h_2 \in H$, we know $\pi (a) = T \in L(H)$ than
\begin{align*}
- <\pi (a) h_1, \pi (a) h_2> = <T h_1, T h_2> = <T^*T h_1, h_2> = <h_1, h_2>
+ \langle \pi (a) h_1, \pi (a) h_2\rangle = \langle T h_1, T h_2\rangle = \langle T^*T h_1, h_2\rangle = \langle h_1, h_2\rangle
\end{align*}
+ Or maybe this \\
+ If $_A H$ than $(a_1a_2) h = a_1 (a_2 h)$ for $a_1, a_2 \in A$ and $h \in H$.\\
+ Then we take the representation of an $a \in A$, $\pi (a)$:
+ \begin{align*}
+ (\pi(a_1)\pi(a_2))h = \pi(a_1)(\pi(a_2) h) = (T_1T_2) h = T_1 (T_2 h)
+ \end{align*}
+ For $T_1, T_2 \in L(H)$, which operate naturally from the left on $h$.
\end{solution}
\begin{exercise}
diff --git a/week2.pdf b/week2.pdf
Binary files differ.
diff --git a/week2.tex b/week2.tex
@@ -55,10 +55,13 @@
Let $A$ be an algebra, $E$ be a \textit{right} $A$-module and $F$ be a \textit{left} $A$-module.
The \textit{balanced tensor product} of $E$ and $F$ forms a $A$-bimodule.
\begin{align*}
- E \otimes _A F := E \otimes F /\left\{\sum _i e_i a_i \otimes f_i - e_i \otimes a_i f_i : \;\;\;
+ E \otimes _A F := E \otimes F / \left\{\sum _i e_i a_i \otimes f_i - e_i \otimes a_i f_i : \;\;\;
a_i \in A,\ e_i \in E,\ f_i \in F \right\}
\end{align*}
\end{definition}
+\begin{question}
+ Does $/$ denote the complement, because one usually writes $\setminus$.
+\end{question}
In other words the balanced tensor product forms only elements of
\begin{itemize}
\item $E$ that preserver the \textit{left} representation of $A$ and
@@ -93,6 +96,16 @@ We denote $KK_f(A,B)$ the set of all \textit{Hilbert bimodules} of $(A,B)$.
\end{exercise}
\begin{solution}
+ We check if the representation of $a \in A$, $\pi(a)=T \in L(H)$ fulfills
+ the conditions on the $\mathbb{C}$-valued inner product for $h_1, h_2 \in H$:
+ \begin{itemize}
+ \item $\langle h_1, \pi(a) h)2\rangle _\mathbb{C} = \langle h_1, T h_2\rangle _\mathbb{C} =
+ \langle T^* h_1, h_2\rangle _\mathbb{C}$, $T^*$ given by the adjoint
+ \item $\langle h_1, h_2 \pi(a)\rangle _\mathbb{C} = \langle h_1, h_2 T\rangle _\mathbb{C} = \langle h_1, h_2\rangle _\mathbb{C}$, $T$ acts from the left
+ \item $\langle h_1, h_2\rangle _\mathbb{C}^* = \langle h_2,h_1\rangle _\mathbb{C}$, hermitian because of the
+ $\mathbb{C}$-valued inner product
+ \item $\langle h_1, h_2\rangle \ge 0$, $\mathbb{C}$-valued inner product.
+ \end{itemize}
\end{solution}
\begin{exercise}
@@ -104,6 +117,12 @@ We denote $KK_f(A,B)$ the set of all \textit{Hilbert bimodules} of $(A,B)$.
\label{exercise: inner-product}
\end{exercise}
\begin{solution}
+ We check again the conditions on $\langle \cdot, \cdot\rangle _A$, let $a, a_1, a_2 \in A$:
+ \begin{itemize}
+ \item $\langle a_1, a\cdot a_2\rangle _A = a^*\ a\cdot a_2 = (a^*a_1)^* a_2 = \langle a^*a_1, a_2\rangle $
+ \item $\langle a_1, a_2 \cdot a\rangle _A = a^*_1 (a_2\cdot a) = (a^*a_2)\cdot a = \langle a_1, a_2\rangle _A a$
+ \item $\langle a_1, a_2\rangle _A^* = (a_1^* a_2)^* = a_2^*(a_1^*)^* = a_2^* a_1 = \langle a_2, a_1\rangle $
+ \end{itemize}
\end{solution}
\begin{example}
@@ -127,18 +146,23 @@ We denote $KK_f(A,B)$ the set of all \textit{Hilbert bimodules} of $(A,B)$.
\end{align*}
Such that $F\circ E \in KK_f(A,D)$ with a $D$-valued inner product.
\begin{align*}
- <e_1 \otimes f_1, e_2 \otimes f_2>_{E\otimes _B F} = <f_1,<e_1, e_2>_E f_2>_F
+ \langle e_1 \otimes f_1, e_2 \otimes f_2\rangle _{E\otimes _B F} = \langle f_1,\langle e_1, e_2\rangle _E f_2\rangle _F
\end{align*}
\end{definition}
\begin{question}
-What is the meaning of 'associative up to isomorphism? Isomorphism of $F \circ E$ or of $A, B$ or $D$?
+ How do we go from $\langle e_1 \otimes f_1, e_2 \otimes f_2\rangle _{E\otimes _B F}$ to $
+ \langle f_1,\langle e_1, e_2\rangle _E f_2\rangle _F$ \label{q: tensorproduct}
+\end{question}
+
+\begin{question}
+What is the meaning of `associative up to isomorphism'? Isomorphism of $F \circ E$ or of $A, B$ or $D$?
\end{question}
\begin{exercise}
- Show that the association $\phi \leadsto E_\phi$ (from the previous Example) is natrual
+ Show that the association $\phi \leadsto E_\phi$ (from the previous Example) is natural
in the sense
- \begin{itemize}
+ \begin{enumerate}
\item $E_{\text{id}_A} \simeq A \in KK_f(A,A)$
\item for $*$-algebra homomorphism $\phi: A \rightarrow B$ and $\psi: B \rightarrow C$ we have
an isomorphism
@@ -146,26 +170,50 @@ What is the meaning of 'associative up to isomorphism? Isomorphism of $F \circ E
E_{\psi} \circ E_{\phi}\ \equiv\ E_{\phi} \otimes _B E_{\psi}\ \simeq\
E_{\psi \circ \phi} \in KK_f(A,C)
\end{align*}
- \end{itemize}
+ \end{enumerate}
\end{exercise}
\begin{solution}
+ \begin{enumerate}
+ \
+ \item $\text{id}_A: A \rightarrow A$.\\
+ To construct $E_{\phi}\in KK_f(A,A)$, we let $E_{\phi}$ be $A$ with a natural right
+ representation, so $\Rightarrow E_{\phi}\simeq A$.\\
+ With an inner product, acting on $A$ from the left with $\phi$, $a', a\in A$\\
+ $a'a = (\phi(a') a) \in A $, which is satisfied by $\text{id}_A$, so $\phi = \text{id}_A$.
+ \item Not sure but: $a \cdot b \cdot c = \psi(\phi(a) \cdot b) \cdot c$ which is in a sense
+ $\psi \circ \phi$
+ \end{enumerate}
\end{solution}
\begin{exercise}
In the definition of Morita equivalence:
- \begin{itemize}
+ \begin{enumerate}
\item Check that $E \otimes _B F$ is a $A-D$ bimodule
- \item Check that $<\cdot,\cdot>_{E\oplus _B F}$ defines a $D$ valued inner product
- \item Check that $<a^*(e_1 \otimes f_1), e_2 \otimes f_2>_{E \otimes _B F} = <e_1 \otimes f_1, a(e_2 \otimes f_2)>_{E \otimes _B F}$.
- \end{itemize}
+ \item Check that $\langle \cdot,\cdot\rangle _{E\oplus _B F}$ defines a $D$ valued inner product
+ \item Check that $\langle a^*(e_1 \otimes f_1), e_2 \otimes f_2\rangle _{E \otimes _B F} = \langle e_1 \otimes f_1, a(e_2 \otimes f_2)\rangle _{E \otimes _B F}$.
+ \end{enumerate}
\end{exercise}
\begin{solution}
+ \
+ \begin{enumerate}
+ \item $E \otimes _B F = E \otimes F / \{\sum_i e_i b_i \otimes f_i - e_i \otimes b_i f_i;
+ e_i \in E_i, b_i \in B, f_i \in F\}$ the last part takes out all tensor product elements of
+ $E$ and $F$ that don't preserver the left/right representation.
+ \item $\langle e_1, e_2\rangle _E \in B$ and $\langle f_1, f_2\rangle _F \in C$ by definition. So let $\langle e_1, e_2\rangle _E =b$. \\
+ Then $\langle e_1 \otimes f_1, e_2 \otimes f_2\rangle _{E\otimes _B F} = \langle f_1, \langle e_1, e_2\rangle _E f_2\rangle _F =
+ \langle f_1, b f_2\rangle _F \in C$
+ \item Check Question \ref{q: tensorproduct}.\\
+ But let $G := E\otimes _B F \in KK_f(A,C)$ then $\forall g_1, g_2 \in G$ and $a \in A$ we need
+ by definition $\langle g_1, ag_2\rangle _G = \langle a^*g_1, g_2\rangle _G$ and we set $g_1 = e_1 \otimes f_1$ and
+ $g_2 = e_2 \otimes f_2$ for some $e_1, e_2 \in E$ and $f_1, f_2 \in F$, or else
+ $G \notin KK_f(A,C)$ which would violate the Kasparov product
+ \end{enumerate}
\end{solution}
\begin{definition}
- Let $A$, $B$ be \textit{matrix algebras}. They are called \textit{Morita eqivalent} if there
+ Let $A$, $B$ be \textit{matrix algebras}. They are called \textit{Morita equivalent} if there
exists an $E \in KK_f(A, B)$ and an $F \in KK_f(B, A)$ such that:
\begin{align*}
E \otimes _B F \simeq A \;\;\; \text{and} \;\;\; F \otimes _A E \simeq B
@@ -193,10 +241,10 @@ What is the meaning of 'associative up to isomorphism? Isomorphism of $F \circ E
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})
+ \langle v_1, v_2\rangle =\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}
+ \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}
@@ -228,18 +276,28 @@ What is the meaning of 'associative up to isomorphism? Isomorphism of $F \circ E
\begin{exercise}
Fill in the gaps in the above proof:
- \begin{itemize}
+ \begin{enumerate}
\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{enumerate}
\end{exercise}
\begin{solution}
+ \
+ \begin{enumerate}
+ \item $(\pi _B, H)$ is irreducible means $H \neq \emptyset$ and only $\emptyset$ or $H$
+ is invariant under the Action of $B$ on $H$.
+ Than $E\otimes _B H$ cannot be empty, because also $E$ preserves left representation of $A$
+ and also $E\otimes _B H \simeq A$.
+ \item The important thing is that $[\pi _A] \in \hat{A}$ respectively $[\pi _B] \in \hat{B}$,
+ hence any choice of representation is irreducible, because the structure space denotes all unitary
+ equivalence classes of irreducible representations.
+ \end{enumerate}
\end{solution}
-\begin{lemma}
- The matrix algebra $M_n(\mathbb{C})$ has a unique inrreducible representation (up to isomorphism)
+ \begin{lemma}
+ The matrix algebra $M_n(\mathbb{C})$ has a unique irreducible representation (up to isomorphism)
given by the defining representation on $\mathbb{C}^n$.
\end{lemma}
\begin{proof}
