# Getting U.C.P map on group operator algebras using Fell's absorbtion principle.

I'm struggling a bit with this theorem:

Let $\Gamma$ be a discrete group and $\mathbb{C}\Gamma$ be the group ring of $\Gamma$ i.e. the set of formal sums $\sum_{t \in \Gamma} \alpha_t t$. Furthermore let $\mathbb{C}^*_\lambda(\Gamma) \subseteq B(\ell^2(\Gamma))$ be the completion of $\mathbb{C}\Gamma$ wrt. the norm $||x|| = ||\lambda(x)||_{op}$ where $\lambda$ is the left representation of $\Gamma$ on $B(\ell^2(\Gamma))$. Let $L(\Gamma) = (\mathbb{C}\Gamma)''$ be the completion of $\mathbb{C}\Gamma$ wrt. the weak operator topology. Finally let $C^*(\Gamma)$ be the completion of $\mathbb{C}\Gamma$ wrt. to the norm $||x|| = sup_\pi||\pi(x)||_{op}$ (taken over all unitary representations $\pi$ of $\Gamma$ on $B(\ell^2(\Gamma))$).

If and A and B are $C^*$-algebras, we say that a map $\phi: A \rightarrow B$ is u.c.p if $\phi$ is unital and $\phi_n: M_n(A) \rightarrow M_n(B)$ defined by $\phi_n([a_{i,j}]) = [\phi(a_{i,j})]$ maps positively(maps positive matrices to positive matrices). $\phi$ is called normal if it is continuous wrt. ultra-weak topology.

Theorem:

Let $\phi: \Gamma \Rightarrow \mathbb{C}$ be a function with $\phi(e) = 1$, and define $\psi_\phi: \mathbb{C}\Gamma \rightarrow \mathbb{C}$
$$\psi_\phi(\sum_{t \in \Gamma}\alpha_t t) = \sum_{t \in \Gamma}\alpha_t \phi(t)$$ And $m_\phi: \mathbb{C}\Gamma \rightarrow \mathbb{C}\Gamma$ $$m_\phi(\sum_{t \in \Gamma}\alpha_t t) = \sum_{t \in \Gamma}\alpha_t \phi(t)t$$

Show that if the functional $\psi_\phi$ extends to a state on $C^*(\Gamma)$, then $m_\phi$ extends to a u.c.p map on either $C^*(\Gamma)$ or $C^*_\lambda(\Gamma)$, or extends to a normal u.c.p map on $L(\Gamma)$.

So the proof uses Fell's absorbtion principle, which states that given any representation $\pi:\Gamma \rightarrow H$ on a Hilbert space H, then $\lambda \otimes \pi$ is unitarily equivalent to $\lambda \otimes 1_H$. So there exists a unitary operator V which conjugates $C^*_{\lambda}(\Gamma) \otimes 1$ onto $C^*_{\lambda}(\Gamma) \otimes C^*(\Gamma)$. The map which does the job is $$\sigma(\sum_{t} \alpha_t \lambda_t \otimes 1_H) = \sum_{t} \alpha_t \lambda_t \otimes t$$ Since $C^*_{\lambda}(\Gamma) \otimes 1 \simeq C^*_{\lambda}(\Gamma)$ i assume we can identify $\sigma$ as an operator from $C^*_{\lambda}(\Gamma)$ into $C^*_{\lambda}(\Gamma) \otimes C^*(\Gamma)$ with $$\sigma(\sum_{t} \alpha_t \lambda_t) = \sum_{t} \alpha_t \lambda_t \otimes t$$ which is a $^*$-homomorphism. They now argue(which is the point where i fall off), that this extends to a normal $^*$-homomorphism $\bar{\sigma}$ from $L\Gamma$ into $L\Gamma \otimes B(H)$. By assumption the functional $\psi_\phi$ extends to a state $\widetilde{\psi_\phi}$ on $C^*(\Gamma)$, if we let $\nu = id_{L(\Gamma)} \otimes \widetilde{\psi_\phi}$, then $\nu: L(\Gamma) \otimes C^*(\Gamma) \rightarrow L(\Gamma) \otimes \mathbb{C}$. And $$\nu \circ \bar{\sigma}(\sum_{t} \alpha_t t) = \sum_{t} \alpha_t t \otimes \widetilde{\phi}(t) = \sum_{t} \alpha_t t \widetilde{\phi}(t) \otimes 1_{\mathbb{C}} \in L(\Gamma) \otimes \mathbb{C} \simeq L(\Gamma)$$ So this gives an extension of $m_\phi$ if we consider $\nu \otimes \bar{\sigma}$ as a map from $C(\Gamma)$ into $C(\Gamma)$.

Is this sort of the way that you are supposed to go, or am i completely wrong? Since this is a $^*$-homomorphism it is u.c.p, but why is it normal? Does $$C_\lambda^*(\Gamma) \subseteq L\Gamma$$ automatically imply the reduced case? And how do you show the $C^*(\Gamma)$ case?