at some point in the proof of the thoerem of double commutant of Von Neumann, you have to show that
$$\pi(\mathcal{M})'' = \pi(\mathcal{M}'') $$
I was looking the proof on Theory of Operator Algebras I, by Takesaki and I don't understand why he constructs all this coming stuff to get the above result.
Basically he denotes $\mathcal{H}$ a fixed Hilbert space and $\{\mathcal{H}\}_{i \in I}$ a family of replicas of $\mathcal{H}$. Then we call $$ \tilde{\mathcal{H}} = \sum_{i \in I}^{\bigoplus}\mathcal{H}_i$$
Let now be $U_i$ the isometry of $\mathcal{H}$ onto $\mathcal{H}_i$. For any perator $x \in \mathcal{L(\tilde{H})}$, putting
$$x_{i,j} = U_i^*xU_j \hspace{4mm} i,j \in I,$$
we obtain a matrix $(x_{i,j})$ of bounded operators on $\mathcal{H}$. Clearly the map $$x \in \mathcal{L(\tilde{H})} \rightarrow (x_{i,j})$$
is injective, so we may write $x = (x_{i,j})$.
For each $x \in \mathcal{L(H)}$ we define an operator $\tilde{x} \in \mathcal{L(\tilde{H})}$ by $\tilde{x} \sum_{i \in I}^{\bigoplus}\xi_i = \sum_{i \in I}^{\bigoplus}x\xi_i$.
Then the map
$$\pi:x \in \mathcal{L(H)} \rightarrow \tilde{x} \in \mathcal{L(\tilde{H}})$$ is an isomorphism of $\mathcal{L(H)}$ into $\mathcal{L(\tilde{H})}$
What we have to show now, knowing this preliminaries, is that
For an operator $\tilde{x} \in \mathcal{L(\tilde{H})}$ to be of the form $\tilde{x} = \pi(x)$ for some $x \in \mathcal{L(H)}$ it is necessary and sufficient that $\tilde{x}$ commutes with all $U_iU_j^*$
And especially on the implication ($\Leftarrow$), I'd need some help to formalize it.
Thank you in advance!
Let $$\pi: \mathcal{L}(\mathcal{H}) \to \mathcal{L}(\mathcal{H}^{\oplus n}): T \mapsto \begin{pmatrix} T & & \\ & \ddots &\\ &&T \end{pmatrix}.$$
Then we have for any unital C*-subalgebra $A \subset \mathcal{L}(\mathcal{H})$ that $\pi(A)' = M_n(A')$. (why?).
Where $M_n(C)$ denotes the C*-subalgebra of $\mathcal{L}(\mathcal{H}^{\oplus n})$ consisting of those linear maps such that if we restrict them to the i-th copy of $\mathcal{H}$ and compose them with the projection onto the $j$-th copy of $\mathcal{H}$, we get an element of $C$ (and this for any $1 \leq i,j \leq n$).
Next prove that for any unital C*-subalgebra $B \subset \mathcal{L}(\mathcal{H})$ we have that $M_n(B)' = \pi(B')$.
Combining these two it is easy to conclude that $\pi(A)'' = M_n(A')' = \pi(A'')$.