# Arveson's extension for normal completely positive maps

My question deals with a version of Arveson's extension theorem (for the standard version, see, e.g., Paulsen's book Completely Bounded Maps and Operator Algebras). Let $\mathcal A$ be a von Neumann algebra, $\mathcal R\subset\mathcal A$ an operator system, and $\mathcal H$ a Hilbert space. If $\Phi_0:\mathcal R\to\mathcal L(\mathcal H)$ is a linear normal completely positive unital map, does there exist a linear normal completely positive unital extension $\Phi:\mathcal A\to\mathcal L(\mathcal H)$ for $\Phi_0$ (complete positivity for a linear map on an operator system defined as in Paulsen's book)?

I am aware that if one gives certain restrictions for the operator system, the extension can be carried out. E.g., when $\mathcal A$ is a type-I factor and $\mathcal R$ is contained in the ultraweak closure of the set of its compact operators, the Arveson-like extension result for normal maps holds. Does anyone know whether the normal extension exists in general or whether there are some weaker versions of Arveson's theorem for normal maps? I am mostly interested in the case where $\mathcal A$ is a type-I factor. Thank you in advance.

• @MatthewDaws Thanks for your answer. Normal here means continuous with respect to the relative topology generated by the ultraweak topology of $\mathcal A$. Indeed, a typical example in my mind is where $\mathcal R\subset\mathcal L(\mathcal H)$ is a commutative von Neumann algebra, so I guess there will be problems in general. But are you aware of some weaker results other than the one I meantioned in the question? Jul 3 '14 at 8:25

The answer is NO and here's a counterexample.

Let $\prod_{n\in{\mathbb N}}{\mathbb M}_n$ be the $\ell_\infty$-direct sum of the full matrix algebras ${\mathbb M}_n$, and let $X\subset\prod_{n\in{\mathbb N}}{\mathbb M}_n$ be the weak$^*$-closed subspace defined by $$X=\{ (x_n)_{n=1}^\infty : \Theta_{m,n}(x_m)=x_n \mbox{ for all }m>n\},$$ where $\Theta_{m,n}\colon{\mathbb M}_m\to{\mathbb M}_n$ is the compression to the upper-left corner. Then, $\Phi\colon X\to{\mathbb B}(\ell_2)$, $\Phi((x_n)_{n=1}^\infty) = \lim_n x_n$ is a normal unital completely positive map (in fact, an weak$^*$-homeomorphic complete order isomorphism). The map $\Phi$ does not extends to a normal ucp map on $\prod_{n\in{\mathbb N}}{\mathbb M}_n$. (In case one wants a type I factor, use the embedding $\prod_{n\in{\mathbb N}}{\mathbb M}_n\subset{\mathbb B}(\bigoplus\ell_2^n)$.)

Indeed, suppose $\Psi\colon\prod_{n\in{\mathbb N}}{\mathbb M}_n\to{\mathbb B}(\ell_2)$ is any normal ucp map which maps the closed unit ball $(\prod_{n\in{\mathbb N}}{\mathbb M}_n)_1$ onto the closed unit ball $({\mathbb B}(\ell_2))_1$. Then, $x \in (\prod_{n\in{\mathbb N}}{\mathbb M}_n)_1$ such that $\Psi(x)$ is unitary is in the multiplicative domain $\mathrm{mult}(\Psi)$ of $\Psi$. Since ${\mathbb B}(\ell_2)$ is spanned by unitary elements, $\Psi$ restricted to the $\mathrm{C}^*$-subalgebra $\mathrm{mult}(\Psi)$ is a surjective $*$-homomorphism. If $\Psi$ were normal, $\mathrm{mult}(\Psi)$ would become a von Neumann subalgebra of $\prod_{n\in{\mathbb N}}{\mathbb M}_n$ and $\Psi$ would become a normal surjective $*$-homomorphism from $\mathrm{mult}(\Psi)$ onto ${\mathbb B}(\ell_2)$, which is absurd. (Although it's overkill, there's also a result of Sukochev and Haagerup--Rosenthal--Sukochev saying that the predual ${\mathbb B}(\ell_2)_*$ does not Banach embed to the predual $(\prod_{n\in{\mathbb N}}{\mathbb M}_n)_*$.)

• Very nice example! Jul 3 '14 at 13:57

Edit: This is correct, except for the actual application to the question at hand, which is wrong; see below for details.

MR2526788 (2010m:46098)
Haagerup, Uffe(DK-SU-CS); Musat, Magdalena(1-MEMP)
Classification of hyperfinite factors up to completely bounded isomorphism of their preduals.
J. Reine Angew. Math. 630 (2009), 141–176.

See pages 149--150. Let $M,N$ be von Neumann algebras. Recall that any function $\phi\in M^*$ has a unique decomposition as $\phi=\phi_s+\phi_n$ where $\phi_n\in M_*$ and $\phi_s$ is "singular". For $T:M\rightarrow N$ a bounded linear map, we can similarly decompose $T=T_n+T_s$ where $T_n,T_s:M\rightarrow N$ are bounded linear maps with $$\phi\circ T_n = (\phi\circ T)_n, \quad \phi\circ T_s = (\phi\circ T)_s \qquad (\phi\in N_*).$$ It's easy to see that if $T$ is (C)P then so are $T_s,T_n$. An original reference for this claim is:

MR0107825 (21 #6547)
Tomiyama, Jun
On the projection of norm one in W∗-algebras. III.
Tôhoku Math. J. (2) 11 1959 125–129.
http://projecteuclid.org/euclid.tmj/1178244633

So, for your question, let $\Phi_1:\mathcal{A}\rightarrow \mathcal L(H)$ be any UCP extension, and let $\Phi = (\Phi_1)_n$. So $\Phi$ is normal and UCP. As $\Phi_0$ is normal, $(\Phi_0)_n = \Phi_0$ (if there is a problem, it's at this point: what precisely is your definition of "normal" for a UCP map from an operator system?) and so $\Phi$ extends $\Phi_0$ as required.

Edit: This is nonsense as Taka points out. I was claiming that $(\Phi_1)_n=\Phi_0$ on $\mathcal R$, or equivalently that $$\langle \omega, (\Phi_1)_n(a) \rangle = \langle \omega, \Phi_0(a) \rangle \qquad (a\in \mathcal R,\omega\in\mathcal{B}(H)_*),$$ which is in turn equivalent to $$\langle (\omega\circ\Phi_1)_n, a \rangle = \langle \omega\circ (\Phi_1)_n, a \rangle = \langle \omega\circ\Phi_0 , a \rangle \qquad (a\in \mathcal R,\omega\in\mathcal{B}(H)_*).$$ That $\Phi_0$ is normal might be taken to mean that $\omega\circ\Phi_0 \in \mathcal{R}_*$ (if $\mathcal R$ is weak$^*$-closed, say). But all we know about $\Phi_1$ is that $\Phi_1(a)=\Phi(a)$ for $a\in\mathcal R$, which seems to tell us nothing about $(\omega\circ\Phi_1)_n\in\mathcal{A}_*$ restricted to $\mathcal R$.

• That $\Phi$ is singular on $\mathcal A$ does not mean it is singular on the weak$^*$-closed subspace $\mathcal R$. In fact, if ${\mathcal R}\subset{\mathcal B}(H)$ is an injective von Neumann algebra which is not discrete (e.g., $L^\infty[0,1]$), then there is a conditional expectation $\Phi$ from ${\mathcal B}(H)$ onto $\mathcal R$. It is identity on $\mathcal R$, but singular on ${\mathcal B}(H)$. Actually the answer to the original question is NO in general, but I could not come up with a simple example. Jul 2 '14 at 23:49
• After all, there was a very easy example. Consider the weak$^*$-homeomorphic complete order isomorphic embedding $\Theta\colon{\mathbb B}(\ell_2({\mathbb N}))\to\prod_{n\in{\mathbb N}}{\mathbb M}_n$, given by $\Theta(x)=(P_n x P_n)_n$. Here $P_n$ is the rank $n$ projection onto $\ell_2(\{1,\ldots,n\})\subset\ell_2({\mathbb N})$ and $P_n {\mathbb B}(\ell_2({\mathbb N})) P_n$ is identified with the full matrix algebra ${\mathbb M}_n$. Then, the identity map on ${\mathbb B}(\ell_2({\mathbb N}))$ does not admit a normal ucp extension on $\prod_{n\in{\mathbb N}}{\mathbb M}_n$. Jul 3 '14 at 8:42
• Taka, why don't you add your example as a separate answer? Jul 3 '14 at 12:07