It is known that for a completely bounded map $\psi:A\to B(H)$ there exist completely positive maps $\phi\_1,\phi\_2:A\to B(H)$ such that 
$$\Vert \phi\_i\Vert\_{cb}=\Vert \psi\Vert\_{cb},$$
and the map $\Phi:M\_2(A)\to B(H\oplus H)$, given by
$$\Phi\left(\left[\begin{array}{ll}a&b\\c&d\end{array} \right]\right)=\left[ \begin{array}{ll}\phi\_1(a) &\psi(b)\\\psi(c)^\* &\phi\_2(d) \end{array}\right],$$
is completely positive. 
The maps $\phi\_1$, $\phi\_2$ are, in general, obtained via the the Arveson extension theorem, which is a version of Hahn-Banach.


**Question:** Are there non-trivial examples of $A$ and $\psi$ where there maps $\phi\_1$, $\phi\_2$ are explicit and known? 


Any examples, references etc. where these maps are known and written down explicitly would be greatly appreciated!
A case of interest is when $A$ is a $C^\*$-completion of the group ring $\mathbb{C}G$ of a discrete group $G$.


**Second question:** Additionally, if $\psi:A\to B\subseteq B(H)$, where $B$ is a (proper) $C^\*$-subalgebra of $B(H)$, is it possible (or: when is it possible?) to arrange for $\phi\_1$, $\phi\_2$ to also take values in $B$? This is the case when $B$ is injective, but this seems to be to restrictive. Again, the examples where this would be interesting is $A=B=C^\*\_r(G)$, the reduced group $C^\*$-algebra.



Extensions of completely positive maps
It's always worth exploring results and proofs to see if you can get more from them. By pushing the result you claim a bit further, we arrive at the standard representation theorem for cb maps (produced using Stinespring on the $2\times 2$ matrix map; see Effros + Ruan section 5.3 or Paulsen's book, for example). Namely,
$$ \psi(x) = S^\* \pi(x) T \qquad (x\in A) $$
where $\pi:A\rightarrow B(K)$ is a $\*$-representation, and $T,S:H\rightarrow K$ satisfy $\|S\| \|T\| = \|\psi\|\_{cb}$.


But now notice we can reverse this argument. The map $M\_2(A) \rightarrow M\_2(B(K)) = B(K\oplus K)$ given by 
$$ \pi\_2\Big( \begin{pmatrix} a & b \\ c & d \end{pmatrix} \Big) = 
\begin{pmatrix} \pi(a) & \pi(b) \\ \pi(c) & \pi(d) \end{pmatrix} $$
is a $\*$-homomorphism, and let 
$$V:H\rightarrow K\oplus K; \xi \mapsto \begin{pmatrix} S(\xi) \\ T(\xi)
\end{pmatrix}. $$
Then
$$ V^\*\pi\_2\Big( \begin{pmatrix} a & b \\ c & d \end{pmatrix} \Big)V
= \begin{pmatrix} S^\*\pi(a)S & S^\*\pi(b)T \\ T^\*\pi(c)S & T^\*\pi(d)T
\end{pmatrix} $$
and so $\phi\_1(a) = S^\*\pi(a)S$ etc. (I think, if you are careful, you can even get some better control of $\|V\|$, see Paulsen's book, though I don't have a copy to check. **Edit:** This is a bit inaccurate! Notice that $S^\*\pi(1)S$ will rarely be (a multiple of) the identity, whereas we'd normally construct $\phi\_1$ to be unital, up to some constant. So moving back and forth between these viewpoints is not quite reversible.)


So basically that gives a "standard form" for $\phi\_i$.


If $A=C^\*\_r(G)$, can we say more, when $\psi$ is a multiplier? Yes! There's a lovely paper by Jolissaint, "A characterization of completely bounded multipliers of Fourier algebras" in Colloquium Math: [Link to PDF](http://matwbn.icm.edu.pl/ksiazki/cm/cm63/cm63218.pdf) where it's shown that for a cb multiplier, there is a continuous function $\tilde\psi$ say which on $c\_{00}(G) \subseteq C^\*\_r(G)$ acts as multiplication, and that $\tilde\psi$ has the form that there are continuous bounded functions $f,g:G\rightarrow K$ for some Hilbert space $K$ with
$$ \tilde\psi(t^{-1}s) = \langle f(s) | g(t) \rangle \qquad (t,s\in G). $$
Then define $F:L^2(G) \rightarrow L^2(G)\otimes K = L^2(G;K)$ by
$$ F(\xi) = \big( s \mapsto \xi(s)f(s^{-1}) \big)\_{s\in G} \in L^2(G;K). $$
Similarly define $G$ from $g$. Then if $\lambda(s)\in C^\*\_r(G)$ is the translation operator,
$$ \langle G^\*(\lambda(s)\otimes 1\_K) F\xi | \eta \rangle
= \int\_G \langle \xi(s^{-1}t) f((s^{-1}t)^{-1}) | \eta(t) g(t^{-1}) \rangle \ dt
= \tilde\psi(s) \langle \lambda(s)\xi|\eta\rangle. $$
So $G^\*(\lambda(s)\otimes 1\_K)F = \psi(\lambda(s))$ and hence we can take $\phi\_1(\cdot) = G^\*(\cdot \otimes 1\_K)G$ and similarly for $\phi\_2$.


