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I know that an operator algebra $A$ has a "minimal" $C^*$-algebra $C$ containing $A$, which is known as the $C^*$ envelope of $A$. The existence of such a minimal $C^*$-algebra generated by $A$ (a minimal $C^*$ cover for $A$) is well known.

On the other hand, any dual operator algebra $A$ generates $W^*$-algebras. For example, there is a maximal $W^*$ cover of $A$, which I've seen denoted by $W^{*}_{max}(A)$. My question is: is it known that a minimal $W^*$ cover exist? That is, does a dual operator algebra has a $W^*$ envelope?

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  • $\begingroup$ Are you conidering $A$ here a sub-algebra of operators of a given Hilbert space, or do you have in mind an "abstract" Banach algebra and you wish to define a ctegorical envelope considering all possible representations? $\endgroup$
    – Uri Bader
    Commented Apr 28, 2016 at 7:04
  • $\begingroup$ I'm talking about an abstract dual operator algebra $A$ (that is, there is a Hilbert space H and a w$^{*}$-continuous completely isometric isomorphism $\pi:A\to B(H)$ ). I want to know if anybody has proven that there is a minimal $W^{*}$-algebra generated by $A$ (yes, considering all representations, just like in the case of the $C^{*}$-envelope of an operator algebra). $\endgroup$
    – epsilon
    Commented Apr 28, 2016 at 22:30
  • $\begingroup$ epsilon, I erased an answer I previously posted because I realized it is was of your point. $\endgroup$
    – Uri Bader
    Commented Apr 29, 2016 at 13:40
  • $\begingroup$ I guess the answer to my question is: It's not known. It's an open problem. $\endgroup$
    – epsilon
    Commented May 3, 2016 at 2:16
  • $\begingroup$ Epsilon, I think it is hasty to say "the answer is not known" - not every operator algebraist reads MathOverflow, and those who do might not have seen your question. $\endgroup$
    – Yemon Choi
    Commented May 9, 2016 at 19:58

2 Answers 2

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This can be done in two ways, algebraically by taking its bicommutant (if is is realised as an algebra of operators on a Hilbert space), or topologically by taking its completion in the bounded weak topology, i.e., the finest locally convex topology which agrees with the weak topology on the unit ball.

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  • $\begingroup$ It doesn't sound like you are familiar with the $C^{*}$ envelope of an operator algebra. In the case of a dual operator algebra ,we have to consider the $W^{*}$ algebras generated by ALL weak$^{*}$ continuous completely isometric homomorphisms $j:A\to B$ such that $B$ is generated as a $W^{*}$-algebra by $j(A)$. From all these $W^{*}$-algebras, is there a minimal one? $\endgroup$
    – epsilon
    Commented Apr 28, 2016 at 22:47
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The obvious attempt to construct a W$^*$-envelope like the C$^*$-envelope fails because if $\mathcal{M} \subseteq \mathcal{B}(\mathcal{H})$ is an injective von Neumann algebra there may not be a normal contraction from $\mathcal{B}(\mathcal{H})$ onto $\mathcal{M}$, e.g. if $\mathcal{M}$ is not purely atomic.

I think there are pretty strong obstructions to the usefulness of such an object even in the commutative case, although I can't immediately see how to use them to rule out a W$^*$-envelope altogether.

If $\mathcal{A}$ is a commutative operator algebra whose C$^*$-envelope is $\mathrm{C}([0, 1])$, then its injective envelope is the Dixmier algebra of bounded Borel functions on $[0, 1]$ modulo the ideal of functions with meagre support. See this paper of Blecher and Magajna for a proof. The Dixmier algebra is an AW$^*$-algebra that is not a W$^*$-algebra, and it has no normal linear functionals.

Since any commutative W$^*$-algebra is injective, any W$^*$-envelope of a commutative operator algebra whose C$^*$-envelope is $\mathrm{C}([0, 1])$ has to contain the Dixmier algebra and every embedding of a dual operator algebra in its W$^*$-envelope would have to factor through the Dixmier algebra.

The pure state space of the Dixmier algebra is separable, so it has many faithful representations on separable Hilbert spaces, but it certainly seems that there is no canonical representation.

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