The trace-cyclic theorem says that linear operators commute (or cycle) within the trace, that is
$${\rm Tr}(XY) = {\rm Tr}(YX).$$
Now if $X$ and $Y$ operate on a product space $H_A \times H_B \times ...$, it is also possible to take the partial trace over any combination of those spaces. But for various reasons, the cyclic theorem will not hold in that case.
$${\rm Tr_A}(XY) \ne {\rm Tr_A}(YX) \hspace{4em} \mbox{(except special cases).}$$
My general question is whether there is still some constraint on the commuted trace. That is, does the trace-cyclic theorem generalise in some weakened form.
My specific question involves lots of details:
- We are working in complex Hilbert space (quantum).
- $X = \left|R\rangle\langle{E}\right|$ is rank-1 i.e. an outer product of arbitrarily entangled Hilbert-space vectors.
- $Y = U$ is unitary, but it freely entangles the subspaces.
- There can be more than two subspaces and it is given that $$ {\rm Tr_{\bar S}}(XU) = 0. $$
Where $\bar{S}$ means "all subspaces other than $S$", and $S$ is arbitrary. That is, $X$ is chosen such that when you trace over $XU$ and leave any one subspace alive, you always end up with zero. So what I want to know is, whether:
$$ {\rm Tr_{\bar S}}(UX) = 0. $$
If this is true, I would like to know why. If it isn't true, I would like to know if any related thing is true.