Let $\mathcal{M}$ be a unital von Neumann algebra over $\mathcal{H}$ Hilbert space.
Let's say that we have an isometry $U \in \mathcal{M}$ such that $U^\ast U = I$ and $UU^\ast $ is an orthogonal projection. As a projection we have that $I-UU^\ast > 0 $ so that if we take a faithful trace $\phi$ on $\mathcal{M}$ this should satisfy:
$\phi (I-UU^\ast)=\phi(I) - \phi(UU^\ast) >0$ , but $\phi(I)=\phi(U^\ast U) = \phi(UU^\ast)$ because of the trace property. So there seems to be a contradiction with the fact that the trace should be faithful. Where am I wrong?
There are three possible places to go wrong:
1) You assume your von Neumann algebra has a non-unitary isometry. But this does happen (for example when $\mathcal M= B(H)$), so nothing wrong with this.
2) You assume your von Neumann algebra has a faithful trace. But this happens (for example when $\mathcal M$ is any II$_1$-factor), so nothing wrong with this.
3) You assume your von Neumann algebra has a non-unitary isometry and a faithful trace. This cannot happen, for precisely the proof you give.