Is $pq+qp$ ever a projection?

Question

There is a question over on Mathoverflow to which I added this answer.

Is it an answer? i.e. if $p$ and $q$ are Hilbert space projections, $p=p^2=p^*$ and $q=q^2=q^*$, is $pq+qp$ ever a non-zero Hilbert space projection?

Answer 1

No.

Suppose that $p,q,r:=pq+qp$ are all projections in a vector space, i.e. $pp=p$, $qq=q$, $(pq+qp)^2 = pq+qp$ (We do not need to require that they are orthogonal or that we are on a hilbert space!).

part 1: Let $s$, $t$ be arbitrary operators with $sq = tp$. Then we have $$ \begin{align} sq = tp &\implies sq = tpp = sqp \\ &\implies sq = sqq = sqpq \\ &\implies sq (qp + pq) = sqp + sqpq = 2sq \\ &\implies 2sq = sq (qp+pq) = sq(qp+pq)^2 = 2sq (qp+pq) = 4sq \\ &\implies sq = 0. \end{align} $$

part 2: By assumption, we have $(pq+qp)(pq+qp)=pq+qp$, which is equivalent to $((pq+qp)p -p)q = (q - (pq+qp)q)p$. Applying part 1 yields $((pq+qp)p - p)q=0$. Then we have $$ \begin{align} pq = (pq+qp) pq = pqpq + qpq &\implies qpq = pq - pqpq \\ &\implies qpq = pq - p (pq - pqpq) = pqpq \\ &\implies pq = (pq+qp) pq = pqpq + qpq = 2qpq \\ &\implies 2qpq = pq = 2q(2qpq ) = 4qpq \\ &\implies qpq = 0. \end{align} $$ By exchanging $p$ and $q$ we also get $pqp =0$. In summary, we have $$ pq+qp = (pq + qp)(pq+qp) = pqpq + pqp + qpq + qpqp = 0, $$ which is what we wanted to show.

(Note: A previous version of this answer was incorrect.

The above uses a large number of elementary operations and only relies on $rr=r$-type substitution, which makes me think that there is probably a simpler proof).