$\prod_{\mathcal U} L^p( B(\ell^2 \Gamma)   \otimes  
{\mathcal{M}}_\alpha)$ with a net $({\mathcal{M}}_\alpha,\tau_\alpha)$ of semi-finite von Neumann algebras.

To keep the notation simple, we consider the case $p=4$ and write $j$ for $j_4$. The same proof works for every $p \in 2\N$.

Fix $\gamma \in \Gamma$, and let $(\Xi_\alpha)$ be a representative of $j(\lambda(\gamma))$. Express that $j \circ T_\phi^4 = (\prod_{{\mathcal{U}}} M_\phi^{p} \otimes id) \circ j$ for $\phi = \delta_\gamma$ a Dirac mass point. This implies that $j(\lambda(\gamma))=((M_{\delta_\gamma}^4 \otimes id)
(\Xi_\alpha))_{\alpha}$, so that we can assume that $(M_{\delta_\gamma} \otimes id) (\Xi_\alpha) = \Xi_\alpha$ for all $\alpha$. In other words, if we see $\Xi_\alpha$ as an infinite matrix with values in ${\mathcal{S}}_p({\mathcal{H}})$, its $(s,t)$ entry is zero unless $st^{-1}=\gamma$. Equivalently, there is an element $\xi \in
\ell^p(\Gamma;{\mathcal{S}}_p({\mathcal{H}}))\subset {\mathcal{S}}_p(\ell^2 \Gamma \otimes {\mathcal{H}})$ such that $\Xi_\alpha = (\lambda(\gamma) \otimes id) \xi$. Let us denote this $\xi$ by $\xi_\gamma^{(\alpha)}$. Using that $1 =
\|\lambda(\gamma)\|_{L^4({\mathcal{L}}(G))} = \lim_{\alpha, \mathcal U}
\|\xi_\gamma^{(\alpha)}\|_{\ell^4(\Gamma; {\mathcal{S}}_4({\mathcal{H}}))}$, we can as well assume that $\|\xi_\gamma^{(\alpha)}\|_{\ell^4(\Gamma; {\mathcal{S}}_4({\mathcal{H}}))}=1$. Since $\gamma \in\Gamma$ was arbitrary, we can choose such a representative for all $\gamma$.

Let $\eta_\gamma^{(\alpha)}$ be the unit vector in $\ell^4 \Gamma$ defined by $\eta_\gamma^{(\alpha)} =
(\|\xi_\gamma^{(\alpha)}(s)\|_{{\mathcal{S}}_4({\mathcal{H}})})_{s \in \Gamma}$. We claim that, $$\label{eq=invariant}
\forall \gamma,s \in \Gamma, 