4
$\begingroup$

Suppose we have the following:

  • A C*-algebra $A$ and a von Neumann algebra $M$ (we can assume that $M$ is $\mathcal B(H)$).

  • A sequence of *-homomorphisms $\phi_i\colon A\to M$

  • an ultrafilter $\mathcal U\in\beta\mathbb N\setminus\mathbb N$

Define $\phi(a)=\lim_{i\in\mathcal U}\phi_i(a)$ for $a\in A$ where the limit is taken in the ultraweak topology of $M$.

Is $\phi$ a $*$-homomorphism?

If $A$ is unital and every $\phi_i$ is unital, it would be enough to have that $\phi$ is order zero, since then $\phi(1)=1$. So, is it true that given a sequence of cpc order zero maps from $A$ to $M$ and an ultrafilter, its point-ultraweak limit along the ultrafilter is order zero?

More generally, if we have two sequences of positive operators $S_i$ and $T_i$ such that $S_iT_i=0$ for every $i$, and $S=\lim_{\mathcal U}S_i$, $T=\lim_{\mathcal U}T_i$, are $S$ and $T$ orthogonal?

$\endgroup$

1 Answer 1

6
$\begingroup$

No.

Set $M:=\mathcal B(\mathcal H)$ and let $(e_j)$ be an ONB for $\mathcal H$. Set $$ S_i(e_j) := \begin{cases} e_0, \quad &j=0; \\ -e_0,\quad &i=j; \\ 0, \quad &\text{otherwise} \end{cases} $$ and $$ T_i(e_j) := \begin{cases} e_0+e_i, \quad &j=0; \\ 0, \quad &\text{otherwise} \end{cases} $$

Then $S_iT_i=0$. The ultraweak limit of $(S_i)$ is the nonzero projection $S$ defined by $$ S(e_j) := \begin{cases} e_0, \quad &j=0; \\ 0, \quad &\text{otherwise} \end{cases} $$ and $(T_i)$ has the same limit.

This answers your "more generally" question, but also your original question: Let $A$ be the universal C*-algebra generated by a 2 elements, $s,t$ of norm at most 2, satisfying $st=0$, then define $\phi_i(s):=S_i$ and $\phi_i(t):=T_i$. If $\phi$ is a point-ultraweak limit of the $\phi_i$ then $\phi(s)=S=\phi(t)$, so $\phi$ can't be multiplicative.

$\endgroup$
3
  • $\begingroup$ Naive question: for which $A$ would Alessandro's question have a positive answer? (There is something in the original question which reminds me of AMNM, although it isn't actually the same) $\endgroup$
    – Yemon Choi
    Sep 16, 2015 at 16:22
  • $\begingroup$ Uhg.You're right. And if I need them to be positive I can just take the projections onto $e_0+e_i$ and $e_0-e_i$. So even if $A$ is $\mathbb C^2$ that wouldn't work (answering Yemon). Thanks! Naively, what if there are operators $U$ an $V$, and $\epsilon>0$ (small enough) such that for all $i$ we have $||S_i-U||, ||T_i-V||<\epsilon$ (and everything has norm $1$!) $\endgroup$ Sep 16, 2015 at 17:19
  • $\begingroup$ Alessandro: adding that hypothesis doesn't help. Let $U$ be the projection onto $e_1$ and $V$ the projection onto $e_2$ (so that $U,V$ are orthogonal to $S_i,T_i$ for $i>2$), and then define $S_i' := U + \epsilon S_i$ and $T_i':= V + \epsilon T_i$. $\endgroup$ Sep 16, 2015 at 20:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.