Takesaki corollary 3.6

Solution 1:

I have to admit that I don't really follow what Takesaki's logic there. So here is how I know to do it.

Let $S_0\in M'$. For any $x\in\mathscr M$, $$\tag1 \pi(S_0)\pi(x)=\pi(S_0x)=\pi(xS_0)=\pi(x)\pi(S_0). $$ So $\pi(S_0)\in\pi(\mathscr M)'$.

Now let $T\in\pi(\mathscr M)''$. By Proposition 3.5, $T=\pi(T_0)$ for some $T_0\in\mathscr L(\mathfrak H)$. For any $S_0\in\mathscr M'$, $$ \pi(T_0S_0)=T\pi(S_0)=\pi(S_0)T=\pi(S_0T_0). $$ As $\pi$ is faithful, $T_0S_0=S_0T_0$. So $T_0\in\mathscr M''$. This shows that $\pi(\mathscr M)''\subset\pi(\mathscr M'')$.

If we now apply the above to $\mathscr M'$, we get $\pi(\mathscr M')''\subset\pi(\mathscr M')$, which immediately imply $$\tag2 \pi(\mathscr M')''=\pi(\mathscr M'). $$ And applying $(2)$ to $\mathscr M'$, we obtain $$\tag3 \pi(\mathscr M'')''=\pi(\mathscr M''). $$

It now remains to show that "hard" inclusion $\pi(\mathscr M'')\subset\pi(\mathscr M)''$. For this it is good to get some intuition of what's happening. It probably helps to do the case $|I|=2$ explicitly. We have $$ \pi(x)=\begin{bmatrix} x&0\\0&x\end{bmatrix} . $$ One can then check explicitly $$ \pi(\mathscr M)'=\Big\{\begin{bmatrix} a&b\\ c&d\end{bmatrix}:\ a,b,c,d\in\mathscr M'\Big\}, $$ and $$ \pi(\mathscr M)''=\Big\{\begin{bmatrix} a&\\ &a\end{bmatrix}:\ a\in\mathscr M''\Big\}=\pi(\mathscr M''). $$ To actually do the general case, it is not hard to show that $$\tag4 \pi(\mathscr M)'=\big\{X\in\mathscr L(\tilde{\mathfrak H}):\ X_{kj}\in\mathscr M'\ \text{ for all }k,j\big\}. $$ As $\mathscr M'''=\mathscr M'$, applying $(4)$ to $\mathscr M''$ we get $$\tag5 \pi(\mathscr M'')'=\pi(\mathscr M)'. $$ Taking commutants, $$\tag6 \pi(\mathscr M'')''=\pi(\mathscr M)''. $$ And now $(3)$ gives us $$\tag7 \pi(\mathscr M'')=\pi(\mathscr M)''. $$