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It is well-known that a von Neumann algebra (on a separable Hilbert space) can be written as a direct integrals of factors, i.e., von Neumann algebras with center $\mathbb C I$. As such, factors play a central role in the structure theory of von Neumann algebras.

A prime example of von Neumann algebras are group von Neumann algebras, constructed as follows: Given a countable discrete group $G$, let $$ \lambda\colon G\to B(\ell^2(G)),\,\lambda_g\delta_h=\delta_{gh}. $$ The group von Neumann algebra $L(G)$ is the von Neumann algebra generated by $\lambda(G)$.

Question: When is the group von Neumann algebra a factor?

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  • $\begingroup$ This question was asked before (math.stackexchange.com/questions/4411178/…) as a (now closed) homework question without context. But I think the question itself is a good fit for MSE. Of course the answer is well-known to the experts, but the proof is short enough for a post on MSE and quite instructive in my opinion. $\endgroup$
    – MaoWao
    Mar 28 at 12:03
  • $\begingroup$ @Ryszard Szwarc You said you were interested in answering this question. $\endgroup$
    – MaoWao
    Mar 28 at 12:04
  • $\begingroup$ See this MO-post for an answer and references: "The group von Neumann algebra $L\Gamma$ is a factor if and only if the group $\Gamma$ is ICC (i.e. infinite conjugacy class property)." $\endgroup$ Mar 28 at 12:19
  • $\begingroup$ @MaoWao I have just found the question, as I expected that the previous one would be re-edited. I will post an answer today, hopefully instructive and possibly correct. $\endgroup$ Mar 29 at 5:30

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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 $A=0.$

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  • $\begingroup$ Thank you for your answer. I think in the second part one has to be a little careful because the sum $A=\sum_x a(x)\lambda_x$ may not make sense in $L(G)$. You may want to say that $A$ acts like this on the span of the $\delta_y$, which is well-defined and all you need for your argument. $\endgroup$
    – MaoWao
    Mar 30 at 13:25
  • $\begingroup$ Thanks, I will correct that. $\endgroup$ Mar 30 at 14:20

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