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Let $A$ be a $C^*$-algebra with self-adjoint part $A_{\operatorname{sa}}$. Then $A$ is called monotone complete if every increasing norm bounded net in $A_{\operatorname{sa}}$ has a supremum (with respect to the usual order on $A_{\operatorname{sa}}$ that is induced by the cone of positive semidefinite elements). Since the positive part of the unit ball in $A_{\operatorname{sa}}$ is, when indexed over itself, an increasing net and an approximate unit, it is not difficult to show that every monotone complete $C^*$-algebra is unital.

If one wants to discuss a related completeness property in non-unital $C^*$-algebras, the following property seems natural to consider:

$(*)$ Every increasing net in $A_{\operatorname{sa}}$ that is bounded above by an element from $A_{\operatorname{sa}}$ has a supremum in $A_{\operatorname{sa}}$.

The sequence space $c_0$ and the space of compact linear operators on an infinite-dimensional Hilbert space are simple examples of non-unital $C^*$-algebras that have property $(*)$.

Question. Has property $(*)$ been systematically studied in $C^*$-algebra theory? (And does it have a name which one can use to find literature about it?)

I did an online search and browsed through several papers and books that treat monotone complete $C^*$-algebras to find some hints, but to no avail.

Motivation. A brief explanation of what motivated this question: If $A$ satisfies $(*)$ and is, in addition, separable, then every increasing net in $A_{\operatorname{sa}}$ that is bounded from above by an element of $A_{\operatorname{sa}}$ can be shown to be even norm convergent; this is a special case of a result on ordered Banach spaces (Theorem 3.1 in a recent preprint I wrote). So I thought that it might be possible to develop a structure theory for separable $C^*$-algebras that satisfy $(*)$ (but I haven't actually tried to do it; my background in $C^*$-algebras is quite weak), and then I started wondering whether there might even be a structure theory available for non-separable $C^*$-algebras that satisfy $(*)$.

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    $\begingroup$ A suggestion: I find it plausible that the condition is equivalent to the multiplier algebra $M(A)$ being monotone complete. $M(A)$ being monotone complete clearly implies $(\ast)$. For the converse, I’m thinking along the following line: let $(m_\lambda)$ be an increasing net in $M(A)_{sa}$ bounded above by $1$. Then for any $x \in A$, $(x^\ast m_\lambda x)$ is an increasing net in $A$ bounded above by $x^\ast x$, so it has a supremum $S(x)$. Use polarization identity to get a conjugate-symmetric sesquilinear form $F(x, y)$ with $S(x) = F(x, x)$. Then $L(x) = \lim_\alpha F(u_\alpha, x)$ and… $\endgroup$
    – David Gao
    Commented Aug 21 at 12:20
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    $\begingroup$ $R(x) = \lim_\alpha F(x^\ast, u_\alpha)$ should define a multiplier on $A$, i.e., an element of $M(A)$, which seems to be the supremum of $(m_\lambda)$. (I couldn’t formalize this argument yet, though. The main technical difficulties are going from $S$ to $F$, then to $(L, R)$, and verifying all these things are well-defined.) $\endgroup$
    – David Gao
    Commented Aug 21 at 12:24
  • $\begingroup$ @DavidGao: Interesting idea. I'm quite confident that this is true in the commutative case, but the details seem to be very messy in the non-commutative case. During the last hour or so I've tried to prove it by representing $M(A)$ as a subalgebra of the univeral enveloping von Neumann algebra of $A$, but no luck so far. $\endgroup$
    – Jochen Glueck
    Commented Aug 21 at 15:37
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    $\begingroup$ Let me report what I have been able to accomplish regarding this question: I’ve managed to shown that my idea, up to and including the definition of $F$ as a sesquilinear form, works, and that $F$ behaves “nicely”. Unfortunately, because of this, what I ended up really proving is about the quasimultiplier $QM(A)$ (which is not even, a priori, a $C^\ast$-algebra, but at least it is an operator system, so talking about monotone completeness still makes sense)… $\endgroup$
    – David Gao
    Commented Oct 12 at 1:13
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    $\begingroup$ … I just ended up showing that $A$ satisfies $(\ast)$ iff $QM(A)$ is monotone complete. This is enough to imply my conjectured equivalence when $A$ is commutative, since in that case $M(A) = QM(A)$. For the general case, I don’t see any way to prove $\lim_\alpha F(u_\alpha, x)$ converges, so I haven’t been able to convert the result to $M(A)$. I do have a different idea that hinges on using polarization identity again to get a $C^\ast$-algebra structure on $QM(A)$, but haven’t succeeded yet. $\endgroup$
    – David Gao
    Commented Oct 12 at 1:16

1 Answer 1

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Let $\tilde{A}$ denote the (unique up to isomorphism) unitization of a non-unital $C^*$-algebra $A$.

Proposition 2.1.11 in Saitô and Wright's monograph states the following:

Let $A$ be a $C^*$-algebra without a unit element. Let $S$ be an upward directed set in $A_{sa}$ with supremum $s$ in $A_{sa}$. Then $S$ has a supremum $s$ in $\tilde{A}_{sa}$.

The authors give a proof on p. 14.

In light of this, it sounds to me like you are describing the non-unital $C^*$-algebras whose unitizations are monotone complete $C^*$-algebras. So the study of $C^*$-algebras satisfying your property should just be the study of monotone complete $C^*$-algebras (after unitization).

Edit: As Jochen points out, a non-unital $C^*$-algebra satisfying property $(\ast)$ can have unitization which is not monotone complete.

The closest thing I can find in the literature is on p. 29 of the same monograph.

Call a $C^*$-algebra pseudo monotone $\sigma$-complete if each upper bounded, monotone increasing sequence has a least upper bound.

The authors point out that, for unital algebras, each pseudo monotone $\sigma$-complete algebra is monotone $\sigma$-complete, but that this need not be the case for non-unital algebras.

Following Saitô and Wright, your property could (should?) reasonably be called something like "pseudo monotone complete". However, this term does not seem to appear anywhere in the literature. Moreover, their term "pseudo monotone $\sigma$-complete" also does not seem to appear anywhere, besides briefly in the monograph. So, it would seem that property $(\ast)$ has most likely not been systematically studied.

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    $\begingroup$ Thank you for your answer. However, Proposition 2.1.11 does not imply that $\tilde A$ is monotone complete if $A$ satisfies $(*)$. For instance, the sequence space $c_0$ satisfies $(*)$, but its unitization $c$ (the space of all convergent sequences) is not monotone complete. $\endgroup$
    – Jochen Glueck
    Commented Mar 15 at 6:33
  • $\begingroup$ @JochenGlueck Hmm, I think I must have misread / misunderstood your question initially. I will update my answer shortly, but as far as I can tell, this property has not really been studied (or at least nothing has been published about it). $\endgroup$
    – Alec Gow
    Commented Aug 20 at 19:06
  • $\begingroup$ Thanks for the updated anwers! $\endgroup$
    – Jochen Glueck
    Commented Aug 20 at 19:58
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    $\begingroup$ @AlecGow That doesn’t seem right. $c_0$ satisfies $(\ast)$, but its unitization $c$ is not monotone complete, but the multiplier algebra of $c_0$ is $\ell^\infty$, which is monotone complete. (The same holds for the noncommutative version, namely $K(\ell^2)$, $\widetilde{K(\ell^2)}$, and $B(\ell^2)$.) $\endgroup$
    – David Gao
    Commented Sep 30 at 22:36
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    $\begingroup$ @AlecGow I mean, that is the correct understanding, as far as I can tell. I was saying your conclusion from that understanding is wrong. Say, if $M(A)$ is monotone complete, the theorem only says the completion of $\tilde{A}$ is $M(A)$. That certainly does not mean $\tilde{A}$ is monotone complete. $\endgroup$
    – David Gao
    Commented Sep 30 at 23:51

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