On one property of infimum of two projections

Question

Suppose $M$ is von Neumann algebra and $p,q\in M$ are projections. One can show that $r=p\wedge q + p\wedge (1-q)$ is also projection, where $p\wedge q$ means the $\inf\{p,q\}$. Indeed, $ (p\wedge q)\cdot (p\wedge (1-q)) = (p\wedge q)\cdot q\cdot (1-q)\cdot (p\wedge (1-q)) = 0$. Is it true that $r=p$ and how to prove?

Answer 1

I would prefer an algebraic argument, but this is the first thing that comes to mind.

Think $M\subset B(H)$. The subspaces $PH\cap QH$ and $PH\cap (1-Q)H$ are orthogonal to each other. Your projection $r$ is the orthogonal projection onto $PH\cap QH+PH\cap (1-Q)H$. And now you can see the problem: nothing prevents the summands from being zero.

For instance, in $M_2(\mathbb C)$, take $$p=\begin{bmatrix}1&0\\0&0\end{bmatrix},\ \ \ q=\begin{bmatrix}1/2&1/2\\1/2&1/2\end{bmatrix}.$$ Then $p\wedge q=p\wedge(1-q)=0.$