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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?

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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.

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  • $\begingroup$ 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
    Commented 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 $1$!) $\endgroup$
    – Alessandro Vignati
    Commented 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 $T_i':= V + \epsilon T_i$. $\endgroup$
    – Aaron Tikuisis
    Commented Sep 16, 2015 at 20:39

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