My question concerns Characterisation of Murray-von Neumann equivalence of subprojections
The OP worked out an argument himself for the seemingly easier direction of the proof. I have concerns because the partial isometry $v$ he obtains is in my view not the same as the original one ($u$).
Can somebody please elaborate the proof of this direction?
Thanks!
What I have been thinking so far:
In the setting of the related question I think $ran(u^*)=ran(p)$ and $ran(u)=ran(q)$ as well as $ran(v^*)=ran(p_\infty)$ and $ran(v)=ran(q_\infty)$
So if now $vp_\infty v^* = q_\infty$ at first I feared that the initial space of $u^*$ is bigger than the initial space of $v^*$ but since $ker(p_\infty)^\bot=ran(p_\infty)=ran(v^*)$ that doesn't really matter as all elements that are annulled by v^* but not by $u^*$ will then be annulled by $p_\infty$ which does the trick too.
But what is still kind of a riddle to me is why not only is the initial space of $u^*$ bigger than that of $v^*$ but also $u^*$ extends $v^*$
I think I have similar problems with the $vp_\infty$-part of $vp_\infty v^*$
If anyone could help me sort out my thoughts I#d be really glad!
As you are reading it, the assertion the question you refer is not true. If you fix $u$ and $p_\infty$, then $up_\infty u^*$ is fixed, but you are still free to choose $q_\infty$.
For instance, choose $p=q=u=I$ in $M_2(\mathbb C)$, and $p_\infty=E_{11}$, $q_{\infty}=E_{22}$. Then $p_\infty\sim q_\infty$, but $up_\infty u^*=p_\infty\perp q_\infty$.