Let $(\mathcal{A},\varphi)$ be a free probability space, where $\mathcal{A}$ is a von Neumann algebra and $\varphi$ a finite and faithful trace. Let furthermore $p\in\mathcal{A}$ be a projection. Consider a $A\in\mathcal{A}$ and assume that $p$ and $A$ are free. This implies $$\varphi(p A p)=\varphi(p)\varphi(A)\varphi(p).$$ But this is equal to $$\varphi(p)\varphi(Ap)$$ and since $p$ is a projection and $\varphi$ tracial, we have $$\varphi(p)\varphi(pAp).$$ We therefore have $$\varphi(pAp)=\varphi(p)\varphi(pAp),$$ which implies that $$p=1.$$ I am pretty sure I am missing something since I don't think this is true...
Please help :)
Freness is not multiplicativity. It works for a product of two elements, because $\varphi(x-\varphi(x) I)=0$ and $\varphi(y-\varphi(y)I)=0$, so $$ 0=\varphi((x-\varphi(x)I)(y-\varphi(y)I)=\varphi(xy)-\varphi(x)\varphi(y). $$ But if you have three or more elements, the above does not give you multiplicativity. The way to express $\varphi(xyz)$ in terms of values of products of less degree requires free cumulants.