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?



Point-ultraweak limit of *-homomorphisms/cpc order zero maps
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.


