Comparing minimal projections in two finite type $I$ subfactors

Question

Let $R$ be a von Neumann algebra containing two finite type $I_{n}$ subfactors $M$ and $N$ with matrix units $\{E_{rs}\}$ and $\{F_{rs}\}$, respectively. My question is: is there necessarily a partial isometry $V\in R$ with initial projection $E_{1,1}$ and final projection $F_{1,1}$? If $R$ is finite this is true, but I am wondering if this is true for an arbitrary von Neumann algebra $R$.

Answer 1

This can fail very spectacularly, even in finite dimension, if the embeddings are not required to be unital. You can take $R=M_2(\mathbb C)\oplus M_2(\mathbb C)$ and the canonical matrix units of the two summands are not comparable.

If you require the embeddings to be unital, the answer is yes for any von Neumann algebra. You have two families of pairwise equivalent projections $\{E_{1},\ldots,E_{n}\}$ and $\{F_{1},\ldots,F_{n}\}$, such that each family adds to the identity.

Suppose first that $E_1$ is finite. Then $I=\sum_kE_k$ is finite. Thus $F_1,\ldots,F_n$ are finite. Using comparison, consider a central projection $P$ such that $PE_1\preceq PF_1$. Then $PE_k\preceq PF_k$ for all $k$ and so there exist projections $F_k'$ with $PE_k\sim F_k'\leq PF_k$. Then $$ \sum_kF_k'\sim\sum_kPE_k=\sum_kPF_k=P. $$ As $P$ is finite, this implies that $\sum_kF_k'=\sum_kPF_k$, and then $F_k'=PF_k$ for all $k$, which gives us $PE_k\sim PF_k$. Since $(I-P)F_k\preceq (I-P)E_k$ the argument can be repeated to get $(I-P)E_k\sim (I-P)F_k$, and then $E_k\sim F_k$ for all $k$.

Next suppose that $E_1$ is properly infinite. Then we can halve it, that is there exists a projection $R$ such that $E_1\sim R\sim E_1-R$. Then $$ E_2\sim E_1\sim E_1-R,\qquad\text{ and so }\quad E_1+E_2\sim R+(E_1-R)=E_1. $$ Continuing this way, $E_1\sim\sum_kE_k=I$. If $Q$ is a central projection such that $QF_1$ is finite, then $Q=\sum_kQF_k$ is finite (since $QF_k\sim QF_1$ for all $k$). As $Q=\sum_kQE_k$, this implies that $QE_1\leq Q$ is finite; the only possibility is that $Q=0$, since $E_1$ is properly infinite. It follows that $F_1$ is properly infinite, and by repeating the argument above we get $$ F_1\sim\sum_kF_k=I\sim E_1. $$

Finally, in the general case, there exists a central projection $P$ such that $(I-P)E_1$ is finite, and $PE_1$ is zero or properly infinite. By the two cases above, we have $(I-P)E_1\sim (I-P)F_1$ and $PE_1\sim PF_1$. Thus $$ E_1=(I-P)E_1+PE_1\sim (I-P)F_1+PF_1=F_1. $$

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On top of what was asked, more can be shown: there exists a unitary that conjugates one family to the other. Having partial isometries $V_1,\ldots,V_n$ such that $V_k^*V_k=E_k$ and $V_kV_k^*=F_k$, we can define $$ U=\sum_kV_k. $$ Because both sets of projections are pairwise orthogonal, we have $$\tag1V_j^*V_k=V_j^*F_jF_kV_k=\delta_{k,j}\,E_k$$ and $$\tag2V_jV_k^*=V_jE_jE_kV_k^*=\delta_{k,j}\,F_k.$$ A direct computation, using $(1)$ and $(2)$ and the fact that each set of projections adds to the identity, then gives that $U$ is a unitary in the von Neumann algebra and $$ UE_kU^*=F_k,\qquad\qquad k=1,\ldots,n. $$