We know that center of a factor von Neumann algebra $\mathcal{A} $ is trivial. Let $P_1$ be a projection in $\mathcal{A} $ such that $P_1\neq I,0$ . undoubtedly there exist another projection like $P_2$ in $\mathcal{A}$ such that $P_1P_2\neq P_2P_1$. Since if for every projection $P$, $P_1P=PP_1$, then $P_1$ is in the center of $\mathcal{A}$. so $P_1=I$ and it is contradiction.
Now my question is that can we prove that there is more than one projection in $\mathcal{A}$ (called $P_2$)such that $P_1P_2\neq P_2P_1$? Can we say that there are infinite projections in $\mathcal{A} $ with this property?
thank you for your help.
Yes and yes.
Consider first $\mathcal A=M_2(\mathbb C)$, $P_1=E_{11}$. Then you can construct $$ P_t=\begin{bmatrix}t&\sqrt{t-t^2}\\\sqrt{t-t^2}&1-t\end{bmatrix},\ \ \ \ t\in(0,1) $$ and $P_1P_t\ne P_tP_1$. So uncountably many.
If now $\mathcal A$ is any factor, there exists nonzero $V\in\mathcal A$ with $V^*V\leq P_1$, $VV^*\leq I-P_1$. Now $\mathcal B=\text{span}\,\{V^*V, V, V^*, VV^*\}$ is a copy of $M_2(\mathbb C)$, so we can again play the game above.