Can we have $n+1$ many orthogonal projections in a type $I_n$ von Neumann algebra?
Suppose $\mathscr{R}$ is a type $I_n$ von Neumann algebra, for $n<\infty$, acting on a Hilbert space $\mathcal{H}$. This means there are equivalent abelian projections $E_1, \dots , E_n$ such that $\sum E_i = I$. My question is
- Can we have $n +1$ many orthogonal projections in $\mathscr{R}$?
- What about $n+1$ many orthogonal projections in $\mathscr{R}$ with same central support?
Actually I can see that for $M_n(\Bbb{C})$ both of $1$ and $2$ are impossible. And this follows immediately because of finite dimensionality. But I cannot see any (dis)proof for the general case. Any help is appreciated. Thanks.
EDIT Okay if $\mathscr{R}=L^\infty[0, 1]$ then $\mathscr{R}$ is type $I_1$ but $\chi_{[0, 1/2)}$ and $\chi_{[1/2, 1]}$ are orthogonal. So $1$ is not "YES" in general. Although this example cannot invalidate (2).
Solution 1:
I have found an argument for (2):
- If $\mathscr{R}$ is of type $I_n$ ($n<\infty$) von Neumann algebra then is it possible for $\mathscr{R}$ to have $n+1$ non-zero orthogonal projections with same central carrier?
Answer.
No, it is impossible!
Suppose $\{E_1, \dots, E_n\}$ be a family of equivalent abelian projection such that $\sum E_i = I$. Assume $F_1, \dots, F_n, F_{n+1}$ are all $\textit{non-zero}$ projection with $C_{E_i}= P$ for all $i = 1, 2, \dots, n+1$.
Now restricting our attention to the (type $I_n$) algebra $\mathscr{R} P$ we can assume WLOG that $C_{E_i}= I$ for all $i = 1, 2, \dots, n+1$. But then since $E_i$ is abelian we have $E_i \precsim F_i$ for all $i=1, 2, \dots, n$. This implies $E_1+\dots+E_n \precsim F_1 +\dots + F_n$ then,
$$I = E_1+\dots+E_n \sim F_0 \le F_1 +\dots + F_n\le I$$
This implies by finiteness of $\mathscr{R}$, $F_0 = I$ and hence $F_1 +\dots + F_n=I$. Thus
$$I = F_1 +\dots + F_n \le F_1 +\dots + F_n +F_{n + 1}\le I.$$
Thus $F_{n+1}=0$, a contradiction to the assumption that all $F_i$'s were $\textit{non-zero}$!