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 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 $\phi_2$.

reallywant. Or, to follow Polya, what is the simplest case where you donotknow these maps explicitly? $\endgroup$ – Yemon Choi Feb 19 '17 at 20:01