This is a follow up question on Von Neumann Algebra of an Infinite Group.
Let $\mathbb{Z}$ be the additive group of the integers. What is the Von Neumann algebra of $\mathbb{Z}$? Let $M$ be the Von Neumann algebra of $\mathbb{Z}$. I know that we have $M \subset B(\ell^2(\mathbb{Z}))$. But what are the linear maps in $M$ here?
If $\Gamma$ is a discrete group (in your case $\Gamma = \mathbb{Z}$), you can consider the injective $*$-homomorphism $$\lambda: \mathbb{C}[\Gamma]\to B(\ell^2(\Gamma))$$ defined by $\lambda(s)\delta_t = \delta_{st}$ for $s,t \in \Gamma$. Then the von Neumann algebra of $\Gamma$ is exactly the set $$L(\Gamma):= \lambda(\mathbb{C}[\Gamma])'' \subseteq B(\ell^2(\Gamma)).$$ In your case, $\lambda(m)\delta_n = \delta_{m+n}$ for $m,n \in \mathbb{Z}$.
Following on QuantumSpace's answer, since we are working with additive groups we have $\lambda(k)= \lambda(1)^k$.
My question is: what is $\lambda(1) $ here?
Thank you in advance!
Not entirely sure what you are looking for, since what $\lambda(1)$ is appears already in the text you quoted. You have $$ \lambda(1)\delta_n=\delta_{n+1}. $$ That is, $\lambda(1)$ is the bilateral shift corresponding to the canonical basis.