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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.

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    $\begingroup$ Without thinking, I assume that if $A=B(H)$, $\psi(a)=a$ and ${\rm dim} H=2$ then everything is explicit. So I think you need to give some more thought or explanation as to what you really want. Or, to follow Polya, what is the simplest case where you do not know these maps explicitly? $\endgroup$ – Yemon Choi Feb 19 '17 at 20:01
  • $\begingroup$ @YemonChoi An interesting case would be for instance $A$= the reduced group $C^*$-algebra of a discrete group $G$ and $\psi$ a multiplier associated to a function on $G$ $\endgroup$ – user10439561 Feb 19 '17 at 20:07
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    $\begingroup$ For Q2 the buzzword is "decomposable map". See Pisier's book, chapters 11 and 14 (the latter attributed to Kirchberg but unpublished, apparently). $\endgroup$ – Matthew Daws Mar 1 '17 at 22:36
  • $\begingroup$ Thanks, I read through both chapters but I did not find anything that allows to restrict the range of the extension from $B(H)$ to $B$. Any hints? $\endgroup$ – user10439561 Mar 4 '17 at 19:08
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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 $\phi_2$.

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  • $\begingroup$ Really nice answer. $\endgroup$ – Nik Weaver Mar 1 '17 at 23:23
  • $\begingroup$ Shame about the two Gs :) $\endgroup$ – Yemon Choi Mar 3 '17 at 3:15
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    $\begingroup$ Oh blast! Well, I'm a programmer these days (much of the time) so I'm allowed to get away with a bit of overloading... Hopefully it's clear from context. $\endgroup$ – Matthew Daws Mar 3 '17 at 9:34

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