"Project" an operator outside of a von Neumann Algebra into it

Question

Suppose $W$ is a proper von Neumann Algebra contained in $B(H)$ and the identity in $W$ is the identity mapping of $H$ (namely, $W$ does not have non-trivial null space).

1. Given a self-adjoint $T\in W$, does there exists a projection $P$ that is not in $W$ such that one of $TP, PT, PTP$ is not in $W$? (we definitely do not want all of them to be zero) Can we find a sufficient condition for this to happen?
2. Conversely, given a self-adjoint $T$ that is not in $W$, does there exists a projection $Q$ that is in $W$ such that one of $TQ, QT, QTQ$ is in $W$? (again we do not want all of them to be zero) Can we find a sufficient condition for this to happen?

In general, if $T$ is seld-adjoint, then there exists a projection in $W^*(T)$ (the von Neumann Algebra generated by $T$) such that the range of that projection is the closure of the range of $T$ and that projection corresponds to the characteristic function of $\sigma(T)$. So far, this is the only thing I know about this question. Besides, given $T$ self-adjoint and not in $W$, can $W^*(T)\cap W\neq\emptyset$? If we have, say $f(T)\in W^*(T)\cap W$, can we always recover the entire $W^*(T)$ using $f(T)$ (even when $f$ is not the characteristic function of $\sigma(T)$)?

Answer 1

1. Yes, such a $P$ always exists. If $PT\in W$ for all projections $P$, then as the projections span all of $B(H)$ we get that $ST\in W$ for all $S\in B (H)$. Taking adjoints we also get that $TS\in W$ for all $S$. With $Q$ the range projection of $T$, this also gives us $SQ$, $QS$ in $W$ for all $S$. Given $x,y,z\in H$ with $z=Qz$, we can take $S_1$ the rank-one operator that maps $z\mapsto y$, $S_2$ the rank-one operator that maps $x\mapsto z$, and $S_1QS_2\in W$ is the rank-one operator $x\mapsto y$. Thus $W$ contains all the rank-one operators and so $W=B (H)$. It follows that if $W\ne B (H)$ there exist a projection $P$ with $PT,TP\not\in W$.

2. No. Consider $W=B (H)\otimes 1\subset B (H\otimes H)$, and let $T=1\otimes S$ for any nontrivial $S$. For any projection $Q=Q_1\otimes1\in W$ we have $$ QT=TQ=QTQ=Q_1\otimes S\not \in W. $$