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Let $p,q$ be two projections of a $C^*$ algebra. A projection $l$ is called a bisector projection to $(p,q)$ if $$|pl-l|=|ql-l|$$ The motivation comes from the geometric intuition of "Bisector" in plane geometry.

Assume that two projections $p,q$ are similar or Mourray von Neumann equivalent. Does they admit a bisector projection? What is the answer to this question for the particular case $A=M_2(C(S^2))$ with the projections $p=\begin{pmatrix} 1&0\\0&0\end{pmatrix}$ and $q=1/2\begin{pmatrix} 1+z& x+yi\\ x-yi& 1-z \end{pmatrix}$?

Remark: One can prove that for two homotopic projections $p,q$ there always exist a bisector projection $l$. To prove this we may assume that $pq\neq q$ otherwise $l=1-p$ is a bisector projection to $p,q$. Let $\gamma (t)$ be a curve of projections with $\gamma(0)=p,\; \gamma(1)=q$. Now we apply the intermediate value theorem to the continuous function $$\phi (t)= |p\gamma(t)-\gamma(t)|-|q\gamma (t)-\gamma(t)|$$ Observe that $\phi(0)\phi(1)<0$. This completes the proof.

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$l=0$ works always, according to your definition, as $pl-l=0$ and so is $ql-l$. Also, one can take $l= p \cap q$, is the infimum of $p$ and $q$ as well.

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    $\begingroup$ To be honest I mean non trivial projections otherwise both $l=0$ or $l=1$ works. $\endgroup$ Dec 13, 2019 at 15:15

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