# Normal u.c.p extension of Schur-multiplier

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"

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 $$L(\Gamma)$$.
• 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$is a unitary. – Opalgal May 5 at 22:41 • Well, that's basically it. If$a_j\to a$sot, then$Ua_jU^*\to UaU^*$sot. Is that not clear to you? I can try to explain more if needed. – Martin Argerami May 5 at 22:58 • I am stressing myself out right now and even less is clear to me than it usually. I thought what you wrote was clear to me but when I tried to check it by calculation I just didn't know where I was going with it. So an elaboration would be helpful. Sorry. – Opalgal May 5 at 23:03 • No worries. If$a_j\to a$sot, then $$\|Ua_jU^*x-U^*aUx\|=\|U(a_j-a)(U^*x)\|=\|(a_j-a)(U^*x)\|\to0.$$ So$Ua_jU^*\to UaU^*$. Is that what you are asking? – Martin Argerami May 5 at 23:12 • Ok, well, that was really stupid of me. For whatever reason my panicked self was trying to check that$Ua_jU^\ast$strongly converges to$a\$. Thank you very much for your time. – Opalgal May 5 at 23:19