# Connes fusion and the composition of completely positive maps

Let $N$ be a type $II_{1}$-factor with trace $\tau$.

An $N-N$ correspondence is a Hilbert $N$-bimodule $H$ where the left and right actions are both ultraweakly continuous. Equivalently, a correspondence is given by a *-homomorphism of $N$ into an amplification of $N$. (See Popa's correspondences INCREST preprint for more detail.)

The Connes fusion $H \otimes_{N} K$ of correspondences $H$ and $K$ is also called the composition of $H$ and $K$, since it in the "amplification picture" described above, $H \otimes_{N} K$ is isomorphic to the composition of an amplification of the *-homomorphism associated to $K$ with the *-homomorphism associated to $H$.

Furthermore, the Stinespring construction allows us to associate a (cyclic) correspondence $H_{\phi}$ to a completely positive map $\phi$ on $N$, and all cyclic correspondences are of this form, i.e. we can recover the completely positive map from a (unit) cyclic vector. If $\phi_{\xi}$ (resp. $\phi_{\eta}$) denotes the c.p. map associated to a cyclic vector $\xi$ (resp. $\eta$), then it is straightforward to prove that $H_{\phi_{\eta} \circ \phi_{\xi}} \cong H_{\phi_{\xi \otimes_{N} \eta}}$.

It is not true, in general, that $H_{\phi}\otimes_{N} H_{\psi} \cong H_{\psi \circ \phi}$ for normal completely positive maps $\phi$ and $\psi$ on $N$.

Question: Precisely when is $H_{\phi}\otimes_{N} H_{\psi} \cong H_{\psi \circ \phi}$?

It is true if the maps are unital *-homomorphisms, but is false when both maps are $\tau$. Are there any other cases where the statement is true?

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I'm not familiar with the Stinespring construction. Could you add a reference (or explanation) please? –  André Henriques Mar 24 '11 at 19:11
@André: Essentially, it is a GNS analogue where you use the CP map $\phi$ instead of a state. In our case, equip the algebraic tensor product N \otimes N with the sesquilinear form $\langle x_{1} \otimes y_{1}, x_{2} \otimes y_{2} \rangle=\tau(y_{2}^{*}\phi(x_{2}^{*}x_{1})y_{1})$ separate and complete, prove things are bounded, etc... This all can be found in Popa's preprint. A nicer explanation of the general construction is located in Paulsen's book: Completely bounded maps and operator algebras. Let me know if this did the trick. Thanks for thinking about my problem! –  Jon Bannon Mar 24 '11 at 19:18
BTW: It may be that I am asking precisely when $\xi \otimes_{N} \eta$ is a cyclic vector, but I hope there is something more I'm not seeing. –  Jon Bannon Mar 24 '11 at 19:20
It may be revealing to look at the non-factor case $N\cong L^\infty[0,1]$. Then your normal c.p. map $\phi$ is determined by a measure $\eta_\phi$ on $[0,1]^2$ with the property that the push-forwards of $\eta_\phi$ onto the two coordinate directions are Lebesgue-abs. continuous. If I am not mistaken, then $\xi\otimes_N \eta$ is cyclic iff either $\xi$ or $\eta$ (densely) generates $H_\phi$ (resp., $H_\psi$) as a left (right) module. If you disintegrate $d\eta_\phi(x,y)=d\mu_x(y) dx$, the condition is that $\mu_x$ is atomic x-a.e. It may be more fun to ask for isom. up to multiplicity. –  Dima Shlyakhtenko Mar 25 '11 at 2:49
No, $H_{\phi_\xi} \otimes H_{\phi_\eta}$ just about never has a cyclic vector even in this abelian case $N=L^\infty[0,1]$. What I was saying is that the condition on cyclicity of $\xi\otimes\eta$ can be more readily seen in the abelian case, where there is a nice picture for the correspondences. Very roughly, you need either $H_\phi$ or $H_\psi$ to be of dimension <=1 as left (right) $N$-modules, which translates into the condition that the module is the compression of a module associated to a homomorphism. –  Dima Shlyakhtenko Mar 25 '11 at 22:27