1
$\begingroup$

The following question comes from a statement in Lemma 16.4 in K-theory and $C^{\ast}$-Algebras written by N.E. Wegge-Olsen. Let $A$ be a non-unital $C^*$-algebra, $\{p_n\}_{n\in\mathbb{N}}$ be a sequence of orthogonal rank 1 projection. Suppose $A$ has a countable approximate unit $\{a_n\}_{n\in\mathbb{N}}$ such that $a_{n+1}a_n = a_n$ for each $n\in\mathbb{N}$. It is shown in Lemma 16.1 that $a_n\bigotimes x_n$ converges strictly to zero in $\mathcal{M}(A\bigotimes\mathcal{K}(\ell^2))$ for $\{a_n\}$ being uniformly bounded (the definition of strict convergence can be found here).

My question is: does $\sum_{1\leq k \leq N}a_k\bigotimes p_k$ converges strictly in $\mathcal{M}(A\bigotimes\mathcal{K}(\ell^2))$? It is claimed without details in the proof of Lemma 16.4 but I cannot see how this series converges. While $\{a_n\}$ converges strictly to the identity in $\mathcal{M}(A)$, $\{p_n\}$ converges strictly to zero, it seems like, for each $a\in A$, I can find large $N, M\in\mathbb{N} (N\leq M)$ such that:

$$ 0 < \| \sum_{N\leq n \leq M}a_n a\bigotimes p_n \| = \| \big( \sum_{1\leq n \leq N} a_n\bigotimes p_n - \sum_{1\leq n \leq M} a_n\bigotimes p_n \big)\big( \sum_{N \leq n \leq M}a\bigotimes p_n\big)\| $$ and hence the partial sum is not strictly Cauchy. Any hints will be appreciated.

Improve this question
$\endgroup$
3
  • $\begingroup$ (Your "strict element" against you test your sequence must not depend on the sequence, so you should select different $N,M$. That is your failure.) $\endgroup$
    – Ewrt Wert
    Commented Apr 23, 2023 at 13:42
  • $\begingroup$ I am afraid I did not understand your question. What confused me is if the sequence $\{ \sum_{1\leq k \leq N} a_k\bigotimes p_k \}$ converges and I found out that this sequence is not Cauchy with respect to the strict topology. It seems like without the condition $a_{n+1} a_n = a_n$, such a sequence will not converge. $\endgroup$
    – Sanae Kochiya
    Commented Apr 25, 2023 at 17:33
  • $\begingroup$ This is really just @EwrtWert 's comment, but from a different perspective, which makes more sense in my mind. For strict convergence, you need to test against a member of $A\otimes\mathcal{K}(\ell^2)$, but you seem to be testing against $a\otimes 1$ (not in the algebra) or maybe $1 \otimes \sum_{n=1}^M p_n$ which, as Ewrt Wert points out, depends on $M$. What happens if you test against $a\otimes\theta$ for some compact operator $\theta$? $\endgroup$
    – Matthew Daws
    Commented Apr 30, 2023 at 8:52

1 Answer 1

Reset to default
1
$\begingroup$

(Add to comments: No you did not show Cauchy divergence. Your sequence is $x_N := \sum_{k \le i \le N} a_i \otimes p_i$. You must show $(x_N y)_N$ converges for every fixed $y \in X:= A \otimes K$. But you choosed a $y= y_N$ dependent on $N$, wich is not allowed. Hint: Show that $z_N := x_N y$ converges for every $y_M= \sum_{1 \le i \le M} b_i \otimes e_{i,j}$ ($M$ fixed independent of $N$!), then for every $y \in X$ which is a norm limit of $(y_M)$, so when "$M= \infty$".)

Improve this answer
$\endgroup$
2
  • $\begingroup$ I see your point now and I am sorry for not making my example clear enough. Since $\{a_i\bigotimes p_i\}$ is a family of pairwise orthogonal elements, we will first have: $$\|\sum_{1\leq i \leq N}a_i\bigotimes p_i\| = \max_{1\leq i \leq N}\|a_i\bigotimes p_i\|$$ and then, for $n$ large enough, we will have $\|a a_i\| > \|a\|-\epsilon$ for some $\epsilon>0$ for any $i\geq n$. Then, pick an arbitrary $m > n$, for any large $N \gg m$, we will have: $\endgroup$
    – Sanae Kochiya
    Commented May 1, 2023 at 18:00
  • $\begingroup$ $$ \| \Big( \sum_{n\leq i \leq m} a\bigotimes p_i\Big) \Big( \sum_{1\leq i \leq N}a_i\bigotimes p_i - \sum_{1\leq i \leq n} a_i\bigotimes p_i\Big)\| = \| \Big( \sum_{n\leq i \leq m} a\bigotimes p_i\Big)\Big( \sum_{n\leq i \leq N} a_i\bigotimes p_i\Big)\| = \|\sum_{n\leq i \leq m} aa_i\bigotimes p_i\| > \|a\|-\epsilon $$. If the sequence is strictly Cauchy, then for each $x\in A\bigotimes\mathcal{K}$, we will have: $$ \limsup_{n, m}\| \big( \sum_{n\leq i \leq m}a_i\bigotimes p_i\big)x\|=0$$. $\endgroup$
    – Sanae Kochiya
    Commented May 1, 2023 at 18:24

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .