Can we have $n+1$ many orthogonal projections in a type $I_n$ von Neumann algebra?

Question

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

1. Can we have $n +1$ many orthogonal projections in $\mathscr{R}$?
2. 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).

Answer 1

I have found an argument for (2):

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 $\textit{non-zero}$!

Answer 2

Surely not the smoothest path, but here's an argument.

​Let $\mathscr A=E_1\mathscr R E_1$. This is an abelian von Neumann subalgebra. Let $V_1,\ldots,V_n$ be partial isometries with $V_j^*V_j=E_1$, $V_jV^*_j=E_j$. Define $$ F_{kj}=V_kV_j^*,\qquad k,j=1,\ldots,n. $$ It is easy to check that these form a system of matrix units. Then we define $\gamma:M_n(\mathscr A)\to \mathscr R$ as $$ [X_{kj}]\stackrel\gamma\longmapsto\sum_{k,j}F_{k1}X_{kj}F_{1j}. $$ It is straightforward to check that $\gamma$ is a normal $*$-monomorphism. Given $T\in\mathscr R$, we have $$ T=\sum_{k,j}F_kTF_j=\sum_{k,j}F_{k 1}\,F_{1k}TF_{j 1}\,F_{1j}=\gamma([F_{1k}TF_{j 1}]). $$​​ So $\gamma$ is an isomorphism and we can assume $\mathscr R=M_n(\mathscr A)$.

An easy consequence is that $Z(\mathscr R)=\mathscr A\otimes I_n$. And another fact we need is that $\mathscr A\simeq L^\infty(X,\mu)$ for some measure space (see Corollary 7.20 here). So we can assume that $\mathscr A=L^\infty(X,\mu)$.

Now let $P_1,\ldots,P_{n+1}\in\mathscr R$ be pairwise orthogonal. If $E$ is the common central carrier, we can work on $E\mathscr R E$. As $E$ is central, $E=e\otimes I_n$ with $e$ a projection in $\mathscr A$, and $E\mathscr RE=M_n(e\mathscr A)$. So we can work on $M_n(e\mathscr A)$, where the projections are still pairwise orthogonal, now with central carrier the identity.

We have $$ P_h=\sum_{k,j=1}^nf_{hkj}\otimes E_{kj}, $$ with $f_{hkj}\in L^\infty(X,\mu)$. If we fix representatives, we may assume that all the $f_{hkj}$ are functions on $X$. Then, for any $x\in X$, $$ P_h(x)=\sum_{k,j=1}^nf_{hkj}(x)\, E_{kj}\in M_n(\mathbb C). $$ The functions $P_h$ are measurable, since each $f_{hkj}$ is and the rest is basically adding and forming the tuple, which are measurable. For each $x$, $P_1(x),\ldots,P_{n+1}(x)$ are $n+1$ pairwise orthogonal projections in $M_n(\mathbb C)$; so at least one of them is zero. This means that $$\tag1 X=\bigcup_{h=1}^{n+1}X_h,$$ where $$ X_h=\big\{x:\ P_h(x)=0\big\},\quad h=1,\ldots,n+1, $$ are measurable sets.For each $h=1,\ldots,n+1$, let $g_h=1_{X_h}$ and $G_h=g_h\otimes I_n$. Then $I-G_h$ is a central projection in $\mathscr R$ with $(I-G_h)P_m=P_m$; thus $I-G_h=I$, that is $G_h=0$. This means that $X_h$ is a nullset for all $h$, contradicting $(1)$.