On some Hilbert space $\mathcal{H}$, I have two orthogonal projections $P,Q:\mathcal{H}\rightarrow \mathcal{H}$. In fact, $Q=UPU^*$ for some unitary $U$ (Note that $Q$ is indeed another orthogonal projection).
I have shown that the following operator
\begin{align}
A:=P-PQP=P-PUPU^*P\,,
\end{align}
is trace-class. However, in this particular setting, I am very interested in whether or not this operator is trace-class,
\begin{align}
B:=P-Q = P-UPU^*\,.
\end{align}
Notice how $A = PBP$. It might not be true at all that $B$ is trace-class, but I was wondering if there are any results out there about trace-class products of operators that I can exploit to investigate whether or not $B$ is trace-class. Intuitively, I think the trace-class property of $A$ must have something to do with $B$, because in my case the projection $P$ is the projection of $L^2(\mathbb{R}^2)$ onto $L^2(\mathbb{R}_+\times \mathbb{R})$, that is, it restricts functions to the upper half-plane (I defined this projection $P$ by means of multiplication operators). I have showed that an orthogonal projection is only trace-class if the dimension of its range is finite, and this really doesn't seem to be the case for $P$, so my guts are telling me that it must have something to do with $B$.
Any links to material are appreciated, or arguments for either cases. Thanks.
Best Answer
Consider $\mathcal{H}= \mathcal{K}\oplus \mathcal{K}$, where $\mathcal{K}$ is an arbitrary Hilbert space and let $P$ be the orthogonal projection onto the first component and $Q$ on the second component. Let $U: \mathcal{H}\to \mathcal{H}$ be the unitary that switches components. Then $$Q = UPU^* = UPU$$ Note further that $PQ = QP = 0$. Thus $B= P-Q$ is also a projection. However, if $\mathcal{K}$ has infinite-dimension, then $B$ does not have finite-dimensional range and thus $B$ is not trace-class by the observation you have made.
Thus, in general, $B$ not not be trace-class.