 Solution

A closed subspace invariant under a star-algebra is reducing, since it is invariant under every adjoint as well. Thus the orthogonal projections $P_1,P_2$ onto $H_1,H_2$ commute with $M$. Consider the bounded intertwinerIts kernel is $H_1\cap H_2$, so its initial polar space isIts closed range isFor the inclusion from left to right, $(I-P_2)h_1=h_1-P_2{h}_1$ lies in $H_1+H_2$ and is orthogonal to $H_2$. Conversely, approximate any $z$ in the displayed $K_2$ by $h_{1,n}+h_{2,n}$ and apply $I-P_2$. The images converge to $z$ and equal $T{h}_{1,n}$, proving the reverse inclusion after closure. The overline on the sum is present in the original PDF and is essential for potentially nonclosed sums.

Take the polar decomposition $T=V|T|$ between these Hilbert spaces. Because $T$ intertwines the two actions of $M$, its adjoint does also, so $T^{\ast }T$ commutes with the action on $H_1$. Its positive square root therefore commutes as well. The identity $V|T|\xi =T\xi$ shows on its dense initial range that $V$ intertwines, and continuity extends this to $K_1$. It is an isometry onto $K_2$. ConsequentlyThe unitary here is a unitary between the indicated Hilbert space modules, not an assertion that the unclosed algebraic sum is a Hilbert space.