Question about equivalent representations of von Neumann algebras in Kadison's book

Question

When I read Kadison's book Fundamentals of The Theory of Operator Algebras, I meat a qeustion on page 460, at line 14, I can not understand that $A\mapsto AE'$ and $A\mapsto \Phi_n(A)F'$ are unitarily equivalent.

Thanks a lot to any one who can tell me why these two representations are equivalent or give me a hint!

Answer 1

You already have from before in the proof, using Proposition 4.5.3 that there exist unitaries $U_a$ with $U_a\Phi_n(A)F_a'U_a^*=AE_a'$ for all $A\in\mathscr R$, and all $a$.

Now take $U=\sum_a E_a'U_aF_a'$, and show that it is a unitary and that it conjugates your representations.

Edit: To see that $U$ is a unitary, only a straightforward computation is needed. Taking $A=I$ in the formula above, we get $U_aF_a'U_a^*=E_a'$. Then $$ U^*U=\sum_{a,b}F_a'U_a^*E_a'E_b'U_bF_b'=\sum_{a}F_a'U_a^*E_a'U_aF_a'=\sum_aF_a'=I. $$ The computation for $UU^*=I$ is completely similar.