I'm struggeling with the proof of a theorem in [BO08]. The first part before the line is what I think I understood. The part after that I don't understand at all.

Let $\Gamma$ be a discrete group and $\mathbb{C}[\Gamma]$ the group ring of $\Gamma$. Let $C_\lambda^\ast(\Gamma)$ be the reduced group $C^\ast$-algebra, i.e. the completion of $\mathbb{C}[\Gamma]$ wrt. the norm $\|x\|=\|\lambda(x)\|$ where λ is the left regular representation of $\Gamma$ on $\mathbb{B}(\ell^2(\Gamma)$. Let $$L(\Gamma) = C_\lambda^\ast(\Gamma)'' \subset \mathcal{B}(\ell^2(\Gamma))$$ be the so called group von Neumann algebra. The full group $C^\ast$-algebra is defined as $C^*(\Gamma) = \overline{\mathbb{C}(\Gamma)}^{\|\cdot\|_u}$ with $\| x\|_u = \sup\{ \|\pi(x)\| \; | \; \pi \text{ is a unitary representation of } \Gamma\}$.

For two $C^\ast$-algebras $A$ and $B$ we say that a map $\varphi_A\to B$ is u.c.p if $\varphi$ is unital and $\varphi_n:M_n(A)→M_n(B)$ defined by $\varphi_n([a_{i,j}])=[\varphi(a_{i,j})]$ maps positive matrices to positive matrices for all $n\in \mathbb{N}$.

$\varphi$ is called normal if it is continuous wrt. ultra-weak topology.

A function $\varphi:\Gamma\to \mathbb{C}$ is called positive definite if for every finite subset $F:=\{s_1, \dots, s_n\}\subseteq \Gamma$ the matrix $$[\varphi(s_i^{-1}s_j)]_{i,j}\in M_n(\mathbb{C}) $$ is positive semidefinite.

For a function $\varphi:\Gamma\to\mathbb{C}$ set \begin{equation*} \begin{split} w_\varphi: \mathbb{C}[\Gamma] &\to \mathbb{C}\\ \sum_{t\in\Gamma}\alpha_t t &\mapsto \sum_{t\in\Gamma} \alpha_t \varphi(t) \end{split} \end{equation*} and \begin{equation*} \begin{split} m_\varphi: \mathbb{C}[\Gamma] &\to \mathbb{C}[\Gamma]\\ \sum_{t\in\Gamma}\alpha_t t &\mapsto \sum_{t\in\Gamma} \alpha_t \varphi(t)t. \end{split} \end{equation*}

The map $m_\varphi$ is also called the Schur-multiplier.

Theorem(2.5.11, [BO08])Let $\varphi:\Gamma\to\mathbb{C}$ be a function with $\varphi(e) =1$. The following conditions are equivalent:

  • The functional $w_\varphi$ extends to a positive functional on $C^\ast(\Gamma)$.
  • The Schur multiplier $m_\varphi$ extends to a u.c.p. map on the two group $C^\ast$-algebras $C^\ast(\Gamma)$ and $C^\ast_\lambda (\Gamma)$ and to a normal u.c.p map on $L(\Gamma)$.

Proof:Since $w_\varphi$ extends to a functional on $C^\ast(\Gamma)$ it follows that $w_\varphi$ extends to a state on $C^\ast(\Gamma)$, which will also be denoted by $w_\varphi$. Embedding $C^\ast(\Gamma)\subseteq \mathcal{B}(H_u)$ where $H_u$ is the Hilbert space of the universal representation $(\pi_u, H_u)$ of $C^\ast(\Gamma)$, we can regard $w_\varphi$ as a state on $\mathcal{B}(H_u)$. We define a map \begin{equation*} \begin{split} \phi: C_\lambda^\ast(\Gamma)\otimes 1 \cong C_\lambda^\ast(\Gamma) &\to C_\lambda^\ast(\Gamma)\otimes C^\ast(\Gamma) \subseteq C_\lambda^\ast(\Gamma)\otimes \mathcal{B}(H_u)\\ \sum_{t\in\Gamma}\alpha_t\lambda(t)\otimes 1 &\mapsto \sum_{t\in\Gamma}\alpha_t(\lambda(t)\otimes t) \end{split} \end{equation*} which is a $\ast$-homomorphism: By Fell's absorption there is a unitary operator $U$ such that $U^\ast(\lambda \otimes \pi_u)(x)U = (\lambda\otimes 1)(x)$ for $x\in \mathbb{C}[\Gamma]$ with \begin{equation*} \begin{split} \|\phi(\sum_{t\in \Gamma} \alpha_t (\lambda(t)\otimes 1))\| &= \|\sum_{t\in \Gamma} \alpha_t (\lambda(t)\otimes t)\| = \|(\lambda\otimes \pi_u)(\sum_{t\in \Gamma} \alpha_t t)\|\\ &=\|U^\ast (\lambda\otimes1)(\sum_{t\in\Gamma} \alpha_t t)U\| = \|\sum_{t\in\Gamma} \alpha_t(\lambda(t)\otimes 1)\|. \end{split} \end{equation*}

We conclude that $\phi$ extends to a normal $\ast$-homomorphism $\phi:L(\Gamma) \to L(\Gamma)\overline{\otimes} B(H_u)$(Why?).\ Checking that $(id_{L[\Gamma)} \otimes w_\varphi)\circ \phi$ coincides with $m_\varphi$ (identifying $L(\Gamma)\overline{\otimes} 1 \cong L(\Gamma)$) concludes the group von Neumann algebra case. The reduced group $C^\ast$-algebra case follows directly.(Why?)

I would be very thankful if someone could give me a hint or some explanations. Thank you very much in advance!

[BO08] Brown and Ozawa, "C∗-Algebras and Fnite-Dimensional Approximations"

  • $\begingroup$ In the comment below the answer, I asked a question Maybe you can help as well :) $\endgroup$ – user745578 Oct 11 '20 at 19:00

It says in [BO08] that "Fell's principle is spatially implemented". That is, $$ \phi:\sum_t\alpha_t\lambda_t\longmapsto \sum_t\alpha_t(\lambda_t\otimes1)=U^*\left(\sum_t\alpha_t(\lambda_t\otimes \pi_u)\right)\,U\longmapsto \sum_t\alpha_t(\lambda_t\otimes \pi_u). $$ As mentioned in the book, both maps in the composition above are spatial, so they are sot continuous; so in particular $\phi$ is normal.

The argument given in book works for the reduced C$^*$-algebra case if you just don't extend to $L(\Gamma)$.

  • $\begingroup$ Sorry to bother you again after such a long time. But could you shortly explain how it follows that this map is SOT-continous? I'm still confused by that. To my understanding today, it directly follows that it is normal, since for a bounded, upward directed set of self-adjoint operators $a_i$, the operators $Ua_iU^{\ast} $ also form a upward directed set of selfadjoint operators that is bounded by $UaU ^{\ast} $ where $a$ denotes the upper bound of the original set of operators, $vand $U$ is a unitary. $\endgroup$ – Opalgal May 5 '19 at 22:41
  • $\begingroup$ Well, that's basically it. If $a_j\to a$ sot, then $Ua_jU^*\to UaU^*$ sot. Is that not clear to you? I can try to explain more if needed. $\endgroup$ – Martin Argerami May 5 '19 at 22:58
  • $\begingroup$ I am stressing myself out right now and even less is clear to me than it usually. I thought what you wrote was clear to me but when I tried to check it by calculation I just didn't know where I was going with it. So an elaboration would be helpful. Sorry. $\endgroup$ – Opalgal May 5 '19 at 23:03
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    $\begingroup$ It's already stated above in the question: $$ \|\sum_{t\in \Gamma} \alpha_t (\lambda(t)\otimes t)\| = \|(\lambda\otimes \pi_u)(\sum_{t\in \Gamma} \alpha_t t)\|\leq \|\sum_{t\in \Gamma} \alpha_t \|.$$ $\endgroup$ – Martin Argerami Oct 12 '20 at 11:37
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    $\begingroup$ Oh, that I understand. I'm unsure very often :D $\endgroup$ – Martin Argerami Oct 12 '20 at 16:34

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