Let $A$, $B$ be two $C^\ast$-algebras and $\mathcal{F}(A,B)$ be the operator ideal of all completely bounded operators $T:A \to B$ for which there are uniformly bounded nets of completely bounded maps $(\psi_k)_k$ and $(\phi_k)_k$ $$ A \xrightarrow{\phi_k} M_{d_k}(\mathbb{C}) \xrightarrow{\psi_k} B $$ such that $\psi_k \circ \phi_k$ tends to $T$ in the pointwise norm topology. We can define a norm for $T$ as $$ \|T\|_{\mathcal{F}(A,B)} = \inf \big\{ \sup_{k} \| \phi_k \|_{cb} \, \| \psi \|_{cb} : \psi_k \text{ and } \phi_k \text{ as above } \big\} $$

Question 1: Has this set of maps been studyied? Is it known to be equivalent to some other "natural" operator ideal?

Question 2: Is the property $id \in \mathcal{F}(A,A)$ equivalent to any known approximation property?

Background. We have that $$ \text{Nuclearity} \subset \{id \in \mathcal{F}(A,A)\} \subset \text{CBAP}. $$ By [Smith] if we assume $id \in \mathcal{F}(A,A)$ and that both $\phi_k$ and $\psi_k$ are complete contractions then $A$ is nuclear. Because of this observation it can not be true that $$ \| id \|_{\mathcal{F}(A)} = \Lambda_{cb}(A), $$ where $\Lambda_{cb}(A)$ is the CBAP constant of $A$. Indeed, take a non-nuclear algebra $A$ with $\Lambda_{cb}(A) = 1$ and assume there are $\phi_k$, $\psi_k$ as above with $\| \psi_k \|_{cb} \| \phi_k \|_{cb} = 1$. If those maps are not contractions then $\| \psi_k \|_{cb} = \lambda$ and $\| \phi_k \|_{cb} =\lambda^{-1} $ and renormalization will give two contractions, which will force $A$ to be nuclear. Therefore, it does not seem likely that the approximation property of Question 2 above is equivalent to CBAP.

In a similar spirit, if we strengthen the condition so that $\psi_k$ is decomposable (i.e.: sum of cp maps) then $T$ is in the class of operators satisfying that $$ \| (id \otimes T): C \otimes_\min A \to C \otimes_\max B \| < \infty. $$ See [Pisier, Corollary 12.6].

[Pisier] Pisier, Gilles, Introduction to operator space theory, London Mathematical Society Lecture Note Series 294. Cambridge: Cambridge University Press (ISBN 0-521-81165-1/pbk). vii, 478 p. (2003). ZBL1093.46001.

[Smith] Smith, R.R., Completely contractive factorizations of $C^*$-algebras, J. Funct. Anal. 64, 330-337 (1985). ZBL0585.46050.

  • 2
    $\begingroup$ The answer to the 2nd question is 'yes, nuclearity.' This was shown by Pisier. It is Theorem 2.9 in his 1996 Memoir "The operator Hilbert space OH, complex interpolation and tensor norms." I don't know about the first question. $\endgroup$ Jan 19 '18 at 0:49
  • $\begingroup$ Hi @CalebEckhardt, I didn't know that this condition on id is equivalent to CPAP. Is there a quick sketch of how/why Theorem 2.9 in Pisier's 1996 memoir works, or is it too technical? $\endgroup$
    – Yemon Choi
    Jan 19 '18 at 11:13
  • $\begingroup$ Thanks for the reference, @CalebEckhardt, i will look that up. $\endgroup$ Jan 19 '18 at 12:26
  • $\begingroup$ @YemonChoi If there is a quick sketch I don't know it $\endgroup$ Jan 19 '18 at 14:06

I don't have access to Pisier's memoir, but I think I can provide a rough sketch of the proof.

First step consists of showing that the assumptions imply that there exists a completely bounded projection $P:B(H) \to A^{\ast\ast}$. Then, and this is the hard part, one has to prove that it implies that $A^{\ast\ast}$ is injective, hence $A$ is nuclear. In fact, the assumptions imply that $A$ is exact and ijectivity of $A^{\ast\ast}$ immediately implies that $A$ has the WEP (weak expectation property), and it is not hard to show that combination of WEP and exactness (understood as existence of approximate factorisation of the embedding $A\subset B(H)$ through matix algebras) gives nuclearity.

In order to do that, we will work with maps $\phi_k^{\ast\ast}: A^{\ast\ast}\to M_{d_k}$ and $\psi_k: M_{d_k} \to A^{\ast\ast}$, whose composition converges to identity in point-ultraweak topology; it is not entirely clear to me that this is actually true. However, there is a way to produce a sequence that will tend to identity in point-ultraweak topology, using local reflexivity. Indeed, the assumptions say that $A$ has a strong form of completely bounded approximation property, therefore it is exact, hence locally reflexive. In particular, it means that whenever $X \subset A^{\ast\ast}$ is a finite dimensional subspace and $(\omega_1, \dots, \omega_n)\subset A^{\ast}$, then for any $\varepsilon > 0$ there is a completely contractive map $P: X \to A$ such that $|\omega_i(x-Px)| \leqslant \varepsilon \|x\|$. Suppose that we have $A\subset A^{\ast\ast} \subset B(H)$. If we compose such a $P$ with $T_k:= \psi_k \circ \phi_k$, then by by Arveson-Wittstock theorem these maps can be extended to maps from $B(H)$ to $A$, since $phi_k\circ P$ has values in $M_{d_k}$. These extensions may be viewied as completely bounded maps from $B(H)$ to $A^{\ast\ast}$, and we have $CB(B(H), (A^{\ast})^{\ast})\simeq B(H)\widehat{\otimes} A^{\ast}$, where $\widehat{\otimes}$ is the projective tensor product (of operator spaces). Therefore any bounded net would have a weak* cluster point. If we are careful enough, we may construct a net whose cluster point must be the identity on $A^{\ast\ast}$.

I will first describe the averaging approach of Christensen and Sinclair (see On von Neumann algebras which are complemented subspaces of $B(H)$, J. Funct. Anal. 122 (1994), no. 1, 91--102). The key is the following: suppose that $P:B(H) \to M$ is a completely bounded projection and $(u_i)$ is a sequence of elements in $M$ such that $\sum u_i^{\ast} u_i = \mathrm{Id}$ strongly. Then the ultraweakly closed convex hull of the maps of the form $x\mapsto \sum P(xu_{i}^{\ast}) u_i$ (indexed by such families of $u_i$'s) contains a completely bounded projection $Q$ which is a right $M$-module map. Doing it from the other side, we obtain an $M$-bimodule projection; this is enough by the results presented by Bunce and Paschke in Quasi expectations and amenable von Neumann algebras, Proc. Amer. Math. Soc. 71 (1978), 232--236. This is actually not very hard. If we assume that $M$ is a finite von Neumann algebra with trace $\tau$ then let $\varphi$ be a functional on $B(H)$ defined by $\varphi(x) :=\tau(Q(x))$, where $Q$ is a bimodular projection. Then $\varphi$ is unitarily invariant. We may assume that $\varphi$ is self-adjoint and then perform the Hahn-Jordan decompostion into positive and neqative parts, and uniqueness of this decomposition implies that both parts are unitarily invariant. If we take the positive part then we obtain a hypertrace on $B(H)$, which is sufficient for proving that $M$ is injective. The general case is handled by standard techniques, the semifinite case by approximation, and the type III case using the crossed product decompostion.

Pisier's (and Haagerup's?) approach is completely different, and he actually requires slightly less than complete boundedness, he requires that the projections works nicely with row and column spaces (see Projections from a von Neumann algebra onto a subalgebra, Bull. Soc. Math. France 123 (1995), 139--153). The main result is: let $M$ be a von Neumann algebra, and let $X_0$ ($X_1$) be $M^n$ equipped with the norm $\|(x_i)\|_{0}:=\|\sum x_i^{\ast}x_i\|_M^{\frac{1}{2}}$ ($\|(x_i)\|_{1}:=\|\sum x_i x_i^{\ast}\|_M^{\frac{1}{2}}$). The interpolation result is that $\|(x_i)\|_{\frac{1}{2}} = \|\sum \overline{x_i}\otimes x_i\|^{\frac{1}{2}}_{\overline{M}\otimes_{\mathrm{max}} M}$. In particular, even though general completely bounded maps do not interact nicely with the maximal tensor product, they do so if we restrict to ''positive definite'' tensors.

Suppose now that $N\subset M$ is a pair of von Neumann algebras such that there exists a completely bounded projection $P:M \to N$. By the interpolation result we have $\|\sum \overline{x_i}\otimes x_i\|_{\overline{M}\otimes_{\mathrm{max}} M} \leqslant \|P\|_{cb}^2 \|\sum \overline{x_i} \otimes x_i\|_{\overline{N} \otimes_{\mathrm{max}} N}$, for any tuple $(x_i)\subset N$. Note that if we denote $t=\sum \overline{x_i}\otimes x_i$, then $(t^{\ast}t)^m$ is a positive definite tensor as well for any $m$, and by $C^{\ast}$-identity we get the above inequality with constant $\|P\|_{cb}^{\frac{1}{m}}$ for any $m$, i.e. with constant $1$; it means that the inclusion $N \subset M$ is $\mathrm{max}^{+}$-injective. Then one can use the self-polar forms to show that in such case there exists a norm one projection from $M$ onto $N$. It would take some time to discuss it, but if anyone is interested I can elaborate on that.

The second approach seems way harder, but it gives more, e.g. Haagerup's characterisation of WEP algebras: a $C^{\ast}$-algebra $A$ is WEP iff the max and min norms coincide on positive definite tensors in $\overline{A}\otimes A$; this shows in particular that WEP depends only on the operator space structure of $A$.

Maybe a comment is in order; it seems to me that all of these results when something a priori completely bounded can be replaced by something completely contractive (or even completely positive), really come from interpreting various properties in terms of tensor products and then using the rigidity of the norm structure of a $C^{\ast}$-algebra, i.e. whenever we have equivalent norms they have to be equal.

I hope that this answer is helpful to someone; writing it certainly clarified matters a lot for me.

  • 1
    $\begingroup$ This is a fantastic answer (in terms of explanation and context) $\endgroup$
    – Yemon Choi
    Feb 19 '18 at 21:24
  • $\begingroup$ @Yemon, thank you very much. Unfortunately, I have to admit that there are serious issues with this answer. In fact, I am unable to show that the assumptions give a cb projection onto $A^{\ast\ast}$, I only get sort of a weak expectation onto $A$. But Pisier's approach still works; it shows that $A$ has WEP and, since the assumptions give CBAP, hence exactness, $A$ has to be nuclear. I will rewrite the answer tomorrow. $\endgroup$ Feb 21 '18 at 4:05
  • $\begingroup$ Turns out that it wasn't too bad after all, although I have a feeling that I am missing something obvious. $\endgroup$ Feb 21 '18 at 18:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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