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.

  • $\begingroup$ @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? $\endgroup$ Jul 3, 2014 at 8:25

2 Answers 2


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 $(\prod_{n\in{\mathbb N}}{\mathbb M}_n)_*$.)

  • $\begingroup$ Very nice example! $\endgroup$
    – Nik Weaver
    Jul 3, 2014 at 13:57

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

I learnt about this technique from:

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.

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 $\mathcal R$.

  • 3
    $\begingroup$ 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. $\endgroup$ Jul 2, 2014 at 23:49
  • 2
    $\begingroup$ 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 $\prod_{n\in{\mathbb N}}{\mathbb M}_n$. $\endgroup$ Jul 3, 2014 at 8:42
  • $\begingroup$ Taka, why don't you add your example as a separate answer? $\endgroup$
    – Nik Weaver
    Jul 3, 2014 at 12:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.