I am reading Jones famous paper "index for subfactors" recently. And I met some questions in the reading. Here is the link for the paper:https://link.springer.com/article/10.1007/bf01389127.
I got stuck on Lemma 4.1.7 right now. To be more specific, I am not so sure how he get that $ue_iu^*=e_j$ and $ue_ju^*=e_i$ at the same time. We know that if $p\sim q$ in a finite von Neumann algebra $M$, then $p=uqu^*$ for some $u\in U(M)$. However, this will only give us $ue_iu^*=e_j$ and $u^*e_ju=e_i$. So I think we may need something more. I notice that in this case we always have $e_ie_j=e_je_i$. Hence, I guess we may may need this in the proof.
My attempt:
I am trying to prove the following proposition: If $p$, $q\in M$ are projections, $p\sim q$ and $pq=qp$, can we have a self adjoint unitary element $u\in U(M)_{sa}$ such that $p=uqu^*$. If the proposition is true, then Lemma 4.1.7 follows, but I couldn't prove the proposition right now.
My question is, Is the proposition above true? If it is true, how do we prove it? If it is not true, How do we prove Lemma 4.1.7 in the original paper?
Any help will be truly grateful!
The proposition is true for finite $M$.
From the fact that $pq=qp$ we get that $pq$ is a projection.
The following lemma is a particular case of Exercise 6.9.26 in Kadison-Ringrose II.
Lemma. Let $M$ be a finite von Neumann algebra, and $p,q$ be equivalent projections such that $pq=qp$. Then $p-pq$ is equivalent to $q-qp$.
By the Lemma, there exists a partial isometry $w$, with $w^*w=p-pq$, $ww^*=q-qp$. Note that $p-pq$ and $q-qp$ are orthogonal to each other. Now let $$ u=w+w^*+[1-(p+ q-2pq)]. $$ Since $w^2=0$ and $w^*(1-p-q+2pq)=w(1-p-q+2pq)=0$, we have $$ u^*u=u^2=ww^*+w^*w+1-p-q+2pq=1. $$ As $M$ is finite, $u^*u=1$ implies that $u$ is a unitary (or one computes $uu^*$ directly). Finally, using that $w^*p=w^*(q-qp)p=0$, $$ up=wp+pqp=w(p-pq)+pq=(q-pq)w+pq=qu. $$
Proof of the Lemma. If $p-pq\not\preceq q-pq$ by comparison there exists $z\in Z(M)$ with $z(q-pq)\prec z(p-pq)$. So there exists a proper subprojection $g\leq p-pq$ with $$z(q-pq)\sim g\leq (p-pq).$$ Now $$ zp\sim zq=z(q-pq)+zpq\sim g+zpq\leq z(p-pq)+zpq=zp. $$ This shows that $zp\sim g+zpq$. As $g+zpq$ is a proper subprojection of $zp$ and $M$ is finite, we get a contradiction. So $z$ cannot exist, showing that $p-pq\preceq q-pq$. Exchanging roles we get the equivalence.