# When is a group von Neumann algebra a factor?

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?

• 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. Mar 28 at 12:03
• @Ryszard Szwarc You said you were interested in answering this question. Mar 28 at 12:04
• 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)." Mar 28 at 12:19
• @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. Mar 29 at 5:30

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 $$A=0.$$
• 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. Mar 30 at 13:25