Determining the normal functionals on the direct sum of von Neumann algebras Consider a collection $\{M_i: i \in I\}$ of von Neumann algebras and consider their $\ell^\infty$-direct sum $M$. I'm trying to show that a normal functional $\omega$ on $M$ is of the form
$$\omega(m) = \sum_i \omega_i(m_i), m = (m_i)_i \in M$$
where $\omega_i \in (M_i)_*$ and $\sum_i \|\omega_i\|<\infty$.
In another post, user Ruy suggested to note that
$$
  (M_*)^* \cong
  M \cong
  \oplus_i^\infty M_i \cong
  \oplus_i^\infty ({M_i}_*)^* \cong
  (\oplus_i^1{M_i}_*)^*.
  $$
and then invoke uniqueness of the predual of a von Neumann algebra to obtain
$$M_* \cong \bigoplus_i^1 {M_i}_*.$$
However, to prove the above result I guess I need to know explicitely how this isomorphism looks like. How can I see this?
 A: Let me try another perhaps more pedestrian  approach.
So we're given that  $M=\ell ^\infty (M_i)$ (I guess I prefer this notation over ${\oplus}_i^\infty  M_i$), where the $M_i$ are von
Neumann algebras.   Denoting by $M_*$ the space of normal linear functionals, we thus need to prove that
$$
  M_*=\ell ^1({M_i}_*).
  $$
Given $\omega \in M_*$, denote by $\omega _i$ the restriction of $\omega $ to $M_i$, so that $\omega _i\in {M_i}_*$, and
let us prove that
$$
  \|\omega \| = \sum_i\Vert \omega _i\Vert.
  \tag 1
  $$
Fixing  $\rho $ in the interval $(0, 1)$,  and for each $i$,  choose some  $m_i$ in $M_i$ such that $\Vert m_i\Vert =1$, and
$\omega _i(m_i)\geq \rho \Vert \omega _i\Vert $.
If $F$ is a finite subset of the index set $I$, and $m_F\in M$ is defined to have coordinates $m_i$, for $i$ in $F$, and zero
elsewhere, then $\Vert m_F\Vert =1$, and
$$
  \Vert \omega \Vert  \geq  |\omega (m_F)| = \sum_{i\in F}\omega _i(m_i) \geq  \sum_{i\in F} \rho \Vert \omega _i\Vert .
  $$
Since both $F$ and $\rho $ are arbitrary,  it follows that
$\Vert \omega \Vert  \geq \sum_{i\in I} \Vert \omega _i\Vert $.
In order to prove the reverse inequality, let $N$ be the subalgebra of $M$ formed by the elements
with finitely many nonzero coordinates.  Then, for every $m$ in the unit ball of $N$, we have that
$$
  |\omega (m)| =
  \big |\sum_{i\in I}\omega _i(m_i)\big | \leq
  \sum_{i\in I}|\omega _i(m_i) | \leq  $$$$ \leq
  \sup_{i\in I}\|m_i\|\sum_{i\in I}\|\omega _i\| \leq
  \sum_{i\in I}\|\omega _i\|.
  \tag 2
  $$
Using von Neumann's double commutant Theorem, it is easy to see that $N$ is weakly dense in $M$.  Furthermore, by Kaplansky's
density theorem, we have that the unit ball of $N$ is dense in the unit ball of $M$ relative to $\sigma (M,M_*)$, aka the weak$^*$
topology.  Therefore the inequality in (2) is preserved if, instead,  $m$ is taken  in  the unit ball of $M$.  This concludes
the proof of claim (1)
Therefore this defines  an isometric  linear map
$$
  \omega \in M_*\mapsto (\omega _i)_i\in \ell ^1({M_i}_*),
  \tag 3
  $$
and we'll be done once we prove it to be  onto.
To see that this is indeed the case, pick $(\omega _i)_i\in \ell ^1({M_i}_*)$, and
define for each subset $J\subseteq I$, the linear functional $\omega _J$ on $M$ by
$$
  \omega _J(m) =\sum_{i\in F} \omega _i(m_i),\quad\forall m\in M.
  $$
For $J$ finite it is clear that $\omega _J$ is normal,  and hence so is $\omega :=\omega _I$,  since the latter is the norm limit of the
$\omega _F$,  for $F$ finite.
Finally notice that the map in (3) sends $\omega $ to $(\omega _i)_i$, concluding the proof.
