norm of tensor product of c.b. maps on Von Neumann algebras $M$ is a Von Neumann algebra and $\phi$ is a c.b. map on $M$. Let $H$ be a Hilbert space. Then consider the map $Id\otimes \phi$ on $B(H)\overline{\otimes} M$. Is $id\otimes \phi$ a c.b. map? what is $\|id\otimes \phi\|_{cb}$? I think it is true that the tensor product of the c.b. map is c.b. on the minimal tensor product of operator spaces. But is it true that the tensor product of von Neumann algebras is same as the minimal tensor product of von Neumann algebras?
 A: If $\phi\colon M\to M$ is a normal cb map, then $\mathrm{id}\otimes \phi$ extends uniquely to a normal cb map on $B(H)\bar\otimes M$ and $\lVert \mathrm{id}\otimes \phi\rVert_{cb}=\lVert \phi\rVert_{cb}$. Indeed, since the algebraic tensor product $B(H)\odot M$ is $\sigma$-weakly dense in $B(H)\bar\otimes M$, there is at most one normal extension.
The existence of this extension follows rather directly from the generalized Stinespring theorem: There exists a Hilbert space $K$, bounded linear operators $V\colon L^2(M)\to K$ and $W\colon K\to L^2(M)$ and a normal representation $\pi$ of $M$ on $B(K)$ such that $\phi(x)=W\pi(x)V$ and $\lVert \phi\rVert_{cb}=\lVert V\rVert \lVert W\rVert$. Then the normal extension of $\mathrm{id}\otimes \phi$ to $B(H)\bar\otimes M$ is given by $(\mathrm{id}\otimes W)(\mathrm{id}\otimes \pi)(\mathrm{id}\otimes V)$, which is a normal cb map. This also implies the norm inequality $\lVert \mathrm{id}\otimes \phi\rVert_{cb}\leq \lVert \phi\rVert_{cb}$. The reverse inequality follows from the embedding $M\cong \mathbb C\otimes M\hookrightarrow B(H)\bar\otimes M$.
