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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 $pq\noteq q$ otherwise $l=1-p$ is a bisector projection. 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.

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.

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 homotopic or 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 $q=1/2\begin{pmatrix} 1+z& x+yi\\ x-yi& 1-z \end{pmatrix}$?

I tried this for homotopic projections but I can not finish the argument that every two homotopic projections $p,q$ have a bisector projection. Here is what I tried : Let $\gamma (t)$ be a curve of projections with $\gamma(0)=p,\; \gamma(1)=q$. Now we would like to apply the intermediate value theorem to the continuous function $$\phi (t)= |p\gamma(t)-\gamma(t)|-|q\gamma (t)-\gamma(t)|$$ In fact we wish to prove $\phi(0)\phi(1)<0$. For this we need to prove that if $pq=q$ leads us to a contradiction but I do not know how to prove it.

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 $pq\noteq q$ otherwise $l=1-p$ is a bisector projection. 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.

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 $q=1/2\begin{pmatrix} 1+z& x+yi\\ x-yi& 1-z \end{pmatrix}$?

It can be easily shown that every two homotopic projections $p,q$ have a bisector projection. Here is a proof: 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 $$\phi (t)= |p\gamma(t)-\gamma(t)|-|q\gamma (t)-\gamma(t)|$$ In fact we have $\phi(0)\phi(1)<0$. For if $pq=q$ then $qp=q$ then $e=p-q$ would be an idempotent with $e^*=-e$, a contradiction.

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 homotopic or 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 $q=1/2\begin{pmatrix} 1+z& x+yi\\ x-yi& 1-z \end{pmatrix}$?

I tried this for homotopic projections but I can not finish the argument that every two homotopic projections $p,q$ have a bisector projection. Here is what I tried : Let $\gamma (t)$ be a curve of projections with $\gamma(0)=p,\; \gamma(1)=q$. Now we would like to apply the intermediate value theorem to the continuous function $$\phi (t)= |p\gamma(t)-\gamma(t)|-|q\gamma (t)-\gamma(t)|$$ In fact we wish to prove $\phi(0)\phi(1)<0$. For this we need to prove that if $pq=q$ leads us to a contradiction but I do not know how to prove it.