Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 1 / 8 / d

One implication is immediate: a unitary module equivalence makes each space an isometric copy of the entire other space, which is closed and invariant. For the converse, choose isometric intertwiners $u:H_1\to H_2$ and $v:H_2\to H_1$ supplied by the two embeddings. Their ranges are closed reducing subspaces. We construct the unitary, proving the Schröder-Bernstein theorem for Hilbert space modules in this setting.

Put $w=vu$, an isometric intertwiner of $H_1$ into itself, and let $D=H_1\ominus v{H}_2$. The spaces $w^{n}D$, $n\ge 0$, are mutually orthogonal. For $m>n$, applying the isometry relation reduces their inner product to one between $D$ and $w^{m-n}D\subseteq w{H}_1\subseteq v{H}_2$, which is zero. Henceis a closed reducing submodule of $H_1$, and $K=D\oplus wK$. Its orthogonal complement lies in $v{H}_2$, because it is orthogonal to $D$. Combining these decompositions givesApply the inverse isometry $v^{\ast }$ on $v{H}_2$ to obtainThe operatoris therefore an isometry onto $H_2$: it is isometric on each orthogonal summand and their images are the two orthogonal summands just exhibited. Both pieces intertwine the action of $M$, as do the projections onto $K$ and $K^{\perp }$, so $W$ is a unitary module equivalence. This proves the reverse implication with no finiteness assumption on either module.