A net version of dominated convergence theorem (Wrong)

Lemma 1. Suppose \(E\) is a Riesz space, \(F\) is an order ideal of \(E\) and \((f_i)_{i\in I}\) is a net in \(F\). Then \(f_i\) converges (in order) to an element \(f\) in \(F\) iff and only if \(f_i\) converges to \(f\) in \(E\) and \(|f_i|\leq \bar f\) for some \(\bar f\in F\) when \(i\) is sufficiently large.

Proof. Suppose \(f\) is the order limit of \((f_i)\) in \(F\), then we can find a decreasing net \((g_j)_{j\in J}\) in \(F\) with \(\inf_{j\in J} g_j=0\) such that for any \(j\in J\) there exists some \(i_0\in I\) satisfying \(|f_i-f|\leq g_j\) for all \(i\geq i_0.\) Pick \(g\in E\) that \(g\leq g_j\) for all \(j\in J.\) Since \(F\) is an ideal, \(g\) is an element of \(F\) and thus less than \(0\). Therefore, the infimum of \((g_j)_{j\in J}\) in \(E\) is also \(0\). As a result, \(f_i\) converges to \(f\) in \(E\) too. The proof for the only if part ends by taking \(\bar f=|f|+g_{i_0}\).

Conversely, we suppose \(f_i\) converges to \(f\) in \(E\) and \(|f_i|\leq \bar f\) for some \(\bar f\in F\) when \(i\) is sufficiently large. Pick a decreasing net \((g_j)_{j\in J}\) with in \(E\) with \(\inf_{j\in J} g_j=0\) such that for any \(j\in J\) there exists some \(i_0\in I\) satisfying \(|f_i-f|\leq g_j\) for all \(i\geq i_0.\) We may assume \(g_j\leq |f|+\bar f,\) otherwise, replace \(g_j\) with \(g_j\wedge(|f|+\bar f)\). It follows from \(F\) is an ideal of \(E\) that \(g_j\in F\). Since \(F\) is a subset of \(E\), any \(g\in F\) satisfying \(g\leq g_j\) for all \(j\in J\) satisfies \(g\leq 0\). Consequently, the infimum of \((g_j)_{j\in J}\) in \(F\) is also \(0\) and hence \(f_i\) converges to \(f\) in \(F\). ◻

The following theorem is an abstract statement of dominated convergence theorem in Banach lattice.

Theorem 2. Suppose \(E\) is a Riesz space, \(F\) is an order ideal of \(E\) equipped with an order continuous norm. Let \((f_i)\) be a net order convergent to some \(f\) in \(E\) and \(|f_i|\leq \bar f\) for some \(\bar f\in F\). Then \(f_i\) is norm convergent to \(f\) in \(F\).

Proof. By the above lemma, \(f_i\) is order convergent to \(f\) in \(F\). The order continuity of the norm on \(F\) yields that \(f_i\) is norm convergent to \(f\). ◻

The following is not ture!

Now, we can give a net version of the canonical Dominated Convergence Theorem.

Corollary 3. Let \((\Omega,\mathcal{A},\mu)\) be a measure space, \((f_i)\) a net of \({\mathcal {A}}\)-measurable functions. Assume the net \((f_i)\) converges \(\mu\)-almost everywhere to an \(\mathcal {A}\)-measurable function \(f\), and is dominated by a \(g \in L^1(\Omega,\mathcal{A},\mu)\) i.e., for each \(i\) we have: \(|f_i| \leq g\), \(\mu\)-almost everywhere. Then f is in \(L^1(\Omega,\mathcal{A},\mu)\) and \[\lim_{i} \int_\Omega |f_i-f|\,d\mu = 0\] which also implies \[\lim_i\int_\Omega f_i\,d\mu = \int_\Omega f\,d\mu.\]

Proof. Let \(E\) be the linear space of all \(\mathcal{A}\)-measurable functions. It is an Riesz space, with respect to the order defined by \(f\leq g\) if \(f(x)\leq g(x)\) \(\mu\)-a.e., in which a net \((f_i)\) is order convergent if and only if \((f_i(x))\) is convergent \(\mu\)-a.e.. By Theorem 2.4.2, (Meyer-Nieberg 2012), the canonical norm on \(L^1(\Omega,\mathcal{A},\mu)\) is order continuous. Applying Theorem 2 with \(F=L^1(\Omega,\mathcal{A},\mu)\), we finishes the proof. ◻

Meyer-Nieberg, Peter. 2012. Banach Lattices. Springer Science &Business Media.

Half exactness of $K_0$ functor

Theorem. Let $\require{amscd}$ $\begin{CD} 0 @> >> A @>i>> [email protected]>\pi>>[email protected]>>>0 \end{CD}$ be an exact sequence of $\mathbb{C}$-algebras, then $\begin{CD}K_0(A)@>{i_*}>>K_0(B)@>{\pi_*}>>K_0(C)\end{CD}$ is exact in the middle.

Key point. For any invertible element $u$ in $M_m(B)$, $$\begin{pmatrix}u & \\ & -u^{-1}\end{pmatrix}=\begin{pmatrix}& 1 \\ 1& \end{pmatrix}\begin{pmatrix}1 & \\u & 1\end{pmatrix}\begin{pmatrix}1 & -u^{-1}\\ & 1\end{pmatrix}\begin{pmatrix}1 & \\u & 1\end{pmatrix}.$$ Since each factor on the right hand side lifts to a invertible element in $M_{2m}(B^+)$, $\begin{pmatrix}u & \\ & -u^{-1}\end{pmatrix}$ lifts to an invertible element in $M_{2m}(B^+)$.

Proof. Let $[e']-[e'_0]$ be an element in the kernel of $\pi_*$, where $e'\in M_{n'}(B^+)$ and $e'_0\in M_{n'}(\mathbb{C})$ satisfies $e'- e'_0\in M_{n'}(B)$. Then $[\pi_* e']=[\pi_* e'_0]$ and thus there is a unitary $w$ in $M_{n}(C)$ such that $w (\pi_* e) w^{-1}=\pi_* e_0=e_0$, where $n=2n', e= \begin{pmatrix}e & \\ & 0\end{pmatrix}, e_0=\begin{pmatrix}e_0 & \\ & 0\end{pmatrix}.$

Pick a lift $v\in M_{2n}(B^+)$ of $\begin{pmatrix}w& \\ & -w^{-1}\end{pmatrix}$, we have $$\pi_*(v\begin{pmatrix}e & \\ & 0\end{pmatrix}v^{-1}) =\begin{pmatrix}w& \\ & -w^{-1}\end{pmatrix} \begin{pmatrix}\pi_*e& \\ & 0\end{pmatrix} \begin{pmatrix}w^{-1}& \\ & -w\end{pmatrix} =\begin{pmatrix}e_0& \\ & 0\end{pmatrix}.$$ Since $\begin{CD} A @>i>> [email protected]>\pi>>C \end{CD}$ is exact, there is some $d\in M_{2n}(A^+)$ such that $i_* d=v\begin{pmatrix}e & \\ & 0\end{pmatrix}v^{-1}.$ Moreover, $d=d^2$ due to that $i$ is an injection. A direct computation $$i_*([d]-[e_0])=[v\begin{pmatrix}e & \\ & 0\end{pmatrix}v^{-1}]-[e_0]=[e]-[e_0]$$ yields that $\begin{CD}K_0(A)@>{i_*}>>K_0(B)@>{\pi_*}>>K_0(C)\end{CD}$ is exact in the middle.

elements of $K_0$ group

Let $e$ be an idempotent in a unital algebra $A$ and $u=\begin{pmatrix}1-e & e\\ e & 1-e\end{pmatrix}$, then $u$ is invertible and $u\begin{pmatrix}1-e & 0\\ 0 & e\end{pmatrix}=\begin{pmatrix}1 & 0\\ 0 & 0\end{pmatrix}u.$ So, $\begin{pmatrix}e & 0\\ 0 & 1-e\end{pmatrix}$ is Murrary-von Neumann equivalent to $\begin{pmatrix}1 & 0\\ 0 & 1\end{pmatrix}$ in $M_2(A)$.

Suppose $g=[e]-[f]$ is an element of a $K_0$ group, then \begin{align*}g=&[e]+[1-f]-[1-f]-[f]=[\begin{pmatrix}e & 0\\ 0 & 1-f\end{pmatrix}]-[\begin{pmatrix}1-f & 0\\ 0 & f\end{pmatrix}]\\=&[\begin{pmatrix}e & 0\\ 0 & 1-f\end{pmatrix}]-[\begin{pmatrix}1 & 0\\ 0 & 0\end{pmatrix}].\end{align*}

On Homotopy

Lemma. Suppose $A$ is a unital Banach algebra and $\gamma$ is a continuous path in the idempotents of $A$. Then there is a continuous path $u$ in invertible elements of $A$ such that $u(t)\gamma(t) u^{-1}(t)=\gamma(0)$.

Proof. For any $s\in [0,1]$, let $$u_s(t)=\gamma(s)\gamma(t)+(1-\gamma(s))(1-\gamma(t)),$$ then $u_s(t)\gamma(t)=\gamma(s)u_s(t)$ and $\|u_s(t)-1\|=\|(2\gamma(s)-1)(\gamma(t)-\gamma(s))\|\leq \|2\gamma(s)-1\|\|\gamma(s)-\gamma(t)\|.$ Let $r_s>0$ be such that $\|\gamma(s)-\gamma(t)\|<\frac{1}{\|2\gamma(s)-1\|}$ for all $t\in (s-r_s, s+r_s)$, then $u_s$ is invertible on the open set $(s-r_s, s+r_s)$. Since $[0,1]$ is compact, we can find finite points $s_1< s_2<\cdots<s_n$ such that $\{(s_i-r_{s_i}, s_i+r_{s_i}): i=1,2,\cdots,n\}$ is a cover of $[0,1]$, $(s_i-r_{s_i}, s_i+r_{s_i}) \cap(s_{i+1}-r_{s_{i+1}}, s_{i+1}+r_{s_{i+1}})\neq\emptyset$, $0\in (s_1-r_1, s_1+r_1)$ and $1\in (s_n-r_{n}, s_n+r_{s_n})$.

Pick $s^i\in (s_i-r_{s_i}, s_i+r_{s_i})\cap (s_{i+1}-r_{s_{i+1}}, s_{i+1}+r_{s_{i+1}})$ such that $s_i<s^i<s_{i+1}$, $i=1,\cdots,n-1$ and s^0=0, s^n=1. For each $i\in \{1,2,\cdots, n\}$, define $$v_i(t)=\begin{cases}u_{s_i}(s^{i-1})^{-1}u_{s_i}(t), t\in [s^{i-1}, s^i]\\ 1, t\in [0,s^{i-1}]\\ u_{s_i}(s^{i-1})^{-1}u_{s_i}(s^i), t\in [s^i,1]\end{cases}$$ and let $u=v_1v_2\cdots v_n$, then $u$ is a continuous path in invertible elements of $A$ and $u(t) \gamma(t) u^{-1}(t)=\gamma(0)$.

Examples of Idempotents

If $e$ is an idempotent in an algebra, so is $ueu^{-1}$ whenever $u$ is invertible. So, if $e$ and $f$ are homotopic in $M_\infty(A)$ if they are homtopic in some $M_n(A)$ where $A$ is a C*-algebra. Note that $\begin{pmatrix} I & \\ & 0\end{pmatrix}e_t \begin{pmatrix} I & \\ & 0\end{pmatrix}$ is a homotopy if $e_t:[0,1]\to M_\infty(A)$ is.

Examples of idempotents in $M_\infty(\mathbb{C})$. (a)\begin{pmatrix} & & & d\\ & & & c\\ & \Huge 0 & & b \\ & & & a \\ & & & 1\end{pmatrix} (b)$\frac{1}{1-t^2}\begin{pmatrix} -t^2 & t\\ -t &1\end{pmatrix}(t\in(1,\infty))$