Does countable decomposable von Neumann algebra will necessary imply that hilbert space has to be separable where the von Neumann algebra acts, if not give a counterexample.
Countable decomposable von Neumann algebra
operator-algebrasvon-neumann-algebras
Related Solutions
I'm not very familiar with this part of Takesaki's book. I think the answer to your question is Corollary 8.11. It's not immediately obvious to me what Radon measure to use, but here is a possibility: take a countable dense $\{a_n\}\subset A$; for each $a_n$, get a state $\varphi_n$ with $\varphi(a_n)=1$. Construct Radon measures $\mu_n$ as in the proof of Proposition I.4.5, and form $\mu=\sum_n2^{-n}\mu_n$. Note that the quasi-state space is weak$^*$-compact.
For $x\in G,$ $x\neq e,$ the conjugacy class of $x$ is the subset of $G$ defined by $$C_x=\{gxg^{-1}\:\, g\in G\}$$ The conjugacy classes do not change under the action of inner automorpisms. i.e. $hC_xh^{-1}=C_x$ for every $h\in G.$ According to Dietrich Burde comment
The von Neumann algebra $VN(G)$ of the group $G$ is a factor (the center of $VN(G)$ is trivial) if and only if for every $x\neq e$ the set $C_x$ is infinite.
For the proof of $\Rightarrow$ direction assume by contradiction that there is $x_0\in G,$ $x_0\neq e,$ such the set $C_{x_0}$ is finite. Consider the operator $$A=\sum_{x\in C_{x_0}}\lambda_x$$ Clearly $A$ belongs to the algebra generated by $\lambda_g,$ $g\in G,$ as well as to its strong operator closure, i.e. to $VN(G).$ The operator $A$ commutes with translations $\lambda_h$ for every $h\in G.$ Indeed $$\lambda_h A(\lambda_h)^{-1}=\lambda_h A\lambda_{h^{-1}}= \sum_{x\in C_{x_0}}\lambda_{hxh^{-1}}\underset{y=hxh^{-1}} {=}\sum_{y\in C_{x_0}}\lambda_y=A$$ The operator $A$ is nontrivial as $$\langle A\delta_e,\delta_{x_0}\rangle_{\ell^2(G)}=1$$ This completes the proof of $\Rightarrow$ direction.
For $\Leftarrow $ direction, assume that $VN(G)$ contains an operator $A,$ which commutes with all operators in $VN(G),$ hence it commutes with left translations $\lambda_g$ for every $g\in G.$ We are going to show that $A=0.$ Let $A\delta_e=\sum_{x\in G}a(x)\delta_x.$ The operator $A,$ restricted to the functions with finite support, is of the form $$A=\sum_{x\in G}a(x)\lambda_x$$ as it commutes with right translations $\rho_g.$ Indeed $$A\delta_y=A\rho_y(\delta_e)=\rho_y A\delta_e=\rho_y\left (\sum_{x\in G}a(x)\delta_x\right )=\sum_{x\in G} a(x)\delta_{xy}=\left (\sum_{x\in G} a(x)\lambda_x\right )\delta_y$$ As $A$ commutes with left translations we get $$\sum_{x\in G}a(x)\lambda_x=A=\lambda_{g^{-1}}A\lambda_g=\sum_{x\in G}a(x)\lambda_{g^{-1}xg}=\sum_{x\in G} a(gxg^{-1})\lambda_x$$ Therefore the function $a(x)$ is constant on each conjugacy class. Since $a(x)=A\delta_e\in \ell^2(G),$ the series $\sum_{x\in G} |a(x)|^2$ is convergent. As each conjugacy class is infinite then $a(x)\equiv 0,$ i.e $A=0.$
Best Answer
Counterexample: $\mathbb C\,I\subset B(\ell^2(\mathbb R))$. It is a one-dimensional subalgebra, so not only it is countably decomposable, it is finitely decomposable, even "one-decomposable".
This example is non-degenerate.