# A question on a lemma about the product map

Here is a Lemma in the book “C*-algebras and Finite-Dimensional Approximations”:

Lemma 3.8.4. Let $A$ be a C*-algebra, $M\subset B(H)$ be a con Neumann algebra and $\phi: A\rightarrow M$ be a completely positive map. Assume that the product map $$\phi \times \iota_{M}: A\odot M’ \rightarrow B(H),~\phi(\sum\limits_{i}a_{i}\otimes m_{i}’)=\sum\limits_{i}\phi(a_{i})m_{i}’.$$ is continuous with respect to the spatial (or minimal) tensor product norm and let $\pi: M\rightarrow B(K)$ be any normal representation. Then the product map $$(\pi \circ \phi)\times \iota_{\pi(M)’}: A\odot \pi(M)’\rightarrow B(K).$$ is also min-continuous (That is, continuous with respect to the spatial (or minimal) tensor product norm).

Proof. Any normal representation of $M$ can be identified with the cut-down by a projection in the commutant of the representation $M\otimes 1_{K}\subset B(H\otimes K)$. Hence it suffices to show that the product map with the commutant in this particular representation is min-continuous.

Since $(M\otimes 1_{K})'\cap B(H\otimes K)=M'\bar{\otimes} B(K)$ (Here, the $\bar{\otimes}$ denote the tensor product of two von Neumann algebra) -- just think of $B(H\otimes K)$ as matrices with entries in $B(H)$ -- we thus have to show that $$(\phi\otimes1_{B(K)})\times\iota_{M'\bar{\otimes}B(K)}: A\odot(M'\bar{\otimes B(K)}~)\rightarrow B(H\otimes K)$$ is min-continuous. But, excepet for the horrific notation required, this is easy since $(\phi\otimes1_{B(K)})\times\iota_{M'\bar{\otimes}B(K)}$ is a point-strong limit of min-continuous maps (with uniformly bounded norms). More precisely, if $P\in B(K)$ is a finite-rank projection, then the map $$(\phi\otimes1_{B(PK)})\times\iota_{M'\bar{\otimes}B(PK)}: A\odot(M'\bar{\otimes B(PK)}~)\rightarrow B(H\otimes PK)$$ is min-continuous and its norm is bounded by $||\phi\times \iota_{M'}||$ because it can be identified with $$(\phi\times\iota_{M'})\otimes id_{B(PK)}: (A\odot M')\odot B(PK)\rightarrow B(H\otimes PK)$$ (Here, it use the Exercise 3.5.1 in this book). Finally, taking a net $\{P_{\lambda}\}$ of finite-rank projections which converge to $1_{K}$ in the strong operator topology and fixing $$x=\sum a_{i}\otimes T_{i}\in A\odot (M'\bar{\otimes}B(K)),$$ it is easy to check that $$(\phi\otimes 1_{B(P_{\lambda}K)})\times \iota_{M'\otimes B(P_{\lambda}K)}((1_{H}\otimes P_{\lambda})x(1_{H}\otimes P_{\lambda}))\rightarrow (\phi \otimes 1_{B(K)})\times \iota_{M'\bar{\otimes}B(K)}(x).$$ in the strong operator topology. This completes the proof.

I have three questions on the proof above:

1. How to comprehend the first sentence "Any normal representation of $M$ can be identified with the cut-down by a projection in the commutant of the representation $M\otimes 1_{K}\subset B(H\otimes K)$."

2. Why does $(M\otimes 1_{K})'\cap B(H\otimes K)=M'\bar{\otimes} B(K)$ hold?

3. How to check the last srong operator topology $$(\phi\otimes 1_{B(P_{\lambda}K)})\times \iota_{M'\otimes B(P_{\lambda}K)}((1_{H}\otimes P_{\lambda})x(1_{H}\otimes P_{\lambda}))\rightarrow (\phi \otimes 1_{B(K)})\times \iota_{M'\bar{\otimes}B(K)}(x).$$

1. It is a general theorem about von Neumann algebras (I know it from Dixmier's vN algebra book, but it should appear in other places too) that any normal representation $\pi:M\to N$ (for some vN algebra $N$) is of the form $$\pi(x)=V^*[(x\otimes 1_{B(K)})\,P\,]\,V$$ for some Hilbert space $K$ (which is not the same $K$ from the statement), $P\in(M\otimes 1_{B(K)})'$ a projection, and $V$ a unitary. We have $$\pi(M)'=[V^*(M\otimes1_{B(K)})PV]'=V^*P(M\otimes1_{B(K)})'PV=V^*P(M'\otimes B(K))PV.$$ So, for $y\in M'$, $z\in B(K)$, $$(\pi\circ\phi)\times\iota_{\pi(M)'}[x\otimes(V^*P(y\otimes z)PV] =V^*(\phi(x)\otimes1_{B(K)})PVV^*P(y\otimes z)PV =V^*P(\phi(x)\otimes1_{B(K)})(y\otimes z)PV=V^*P[(\phi\otimes1_{B(K)})\times\iota_{M'\otimes B(K)}(x\otimes(y\otimes z)) ]PV.$$ As conjugating with a unitary and with a projection is min-continuous, one only needs to check the min-continuity "inside", which is what the authors do.
2. If $M\subset B(H)$, $N\subset B(K)$, then $(M\otimes N)'=M'\otimes N'$ in $B(H\otimes K)$. This is again a general result about tensor products of von Neumann algebras.
3. It should say $1_A\otimes1_H\otimes P_\lambda$. Then $$(\phi\otimes 1_{B(P_{\lambda}K)})\times \iota_{M'\otimes B(P_{\lambda}K)}((1_A\otimes1_{H}\otimes P_{\lambda})(a\otimes T)(1_A\otimes1_{H}\otimes P_{\lambda})) =(\phi(a)\otimes1_{B(K)}(1_H\otimes P_\lambda)T(1_H\otimes P_\lambda)\\ \longrightarrow(\phi(a)\otimes1_{B(K)})T= (\phi \otimes 1_{B(K)})\times \iota_{M'\bar{\otimes}B(K)}(a\otimes T).$$
• Yes. ${\ \ \ }$ Jul 13 '14 at 14:54