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A Banach algebra (assumed associative and unital) is precisely a monoid object in the monoidal category of Banach spaces, short linear maps, and the projective tensor product. A Banach coalgebra is then a comonoid object in this monoidal category. It's straightforward to write down what a Banach *-coalgebra is too. It's a little less obvious what a C*-coalgebra is, and I don't know if that term appears in the literature, but I've written down my definition.

Generally, the dual space of a coalgebra is an algebra (but not conversely), and that works here too: the dual of a C*-coalgebra is a C*-algebra. But not every C*-algebra arises in this way; obviously, since all of these C*-algebras have preduals (having been explicitly constructed so), they are W*-algebras. But I don't know what other conditions must be satisfied.

So my question is, and I'll be grateful for incomplete answers: Which W*-algebras arise (up to abstract isomorphism) as duals of C*-coalgebras?

Partial answers: The sequence space $l^\infty$ is the dual of $l^1$, and $l^1$ is a C*-coalgebra. But this doesn't work for $L^\infty(R)$; this is the dual of $L^1(R)$, but I can't make $L^1(R)$ into even a Banach coalgebra (in an appropriate way), essentially because the diagonal in $R^2$ has measure zero. (Unless I've miscalculated something, and I'm trying to do the wrong thing here.) Of course, these are quite limited examples: they're (co)commutative. I'd be grateful for more.

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    $\begingroup$ Warning bells sound in my head when I read your definition of a Banach coalgebra - I think the dead hand of dogma / the wisdom of experience says that one should be using a different tensor product for comonoids. (The problem is that $A\otimes A$ is too small if you take $\otimes$ as the canonical monoidal tensor for that category. Going to Cstar world won't help.) $\endgroup$
    – Yemon Choi
    Commented Aug 26, 2012 at 8:24
  • $\begingroup$ Ah, on rereading, I think you might be OK in the (co)commutative setting, but that only works when your co-algebras are $\ell^1$. (This is all related to ideas people have been trying out for Hopf von Neumann algebras, but you are asking for less algebraic structure so I guess there may be more room to get some extra examples) $\endgroup$
    – Yemon Choi
    Commented Aug 26, 2012 at 8:31
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    $\begingroup$ With regard to Yemon's comment about projective tensor products, read Matt Daws' answer to this question: mathoverflow.net/questions/50302/… $\endgroup$
    – Ollie
    Commented Aug 26, 2012 at 17:36
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    $\begingroup$ With regard to Ollie's commemt, just ask Matt Daws about all this, quite frankly. $\endgroup$
    – Yemon Choi
    Commented Aug 27, 2012 at 9:16
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    $\begingroup$ As already indicated by Ollie above, if we use the extended Haagerup tensor product, then all von Neumann algebras are duals of C*-coalgebras (well, I don't have a proof of the co-C*-identity for preduals of von Neumann algebras, but it seems plausible). I would also like to add that there is a lot of evidence that suggests that in the framework of C*-algebras and von Neumann algebras the “right” tensor product is often not the usual injective or projective tensor product, but rather one of the tensor products coming from the theory of operator spaces, such as the one cited above. $\endgroup$
    – Dmitri Pavlov
    Commented Aug 28, 2012 at 8:33

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Only a partial answer, but I thought I'd stick it up to give other people something to poke holes in or build on. Hopefully I will come back to elaborate or correct.

Let $M$ be the W*-algebra whose predual $A$ is your co-algebra. Then your comultiplication is meant to land in $A\hat{\otimes A}$, I gather. So dualizing you want your multiplication on $M$ to be a well-defined map $(M_*\hat{\otimes} M_*)^*\to M$. But now I think one can invoke machinery from tensor norms of Banach spaces to say that this would imply multiplication in $M$ extends to a well-defined continuous linear map $M\check{\otimes} M \to M$ where $\check{\otimes}$ is Grothendieck's injective tensor product. Then if you want this map to be contractive I believe $M$ has to be commutative. If you want it to be bounded, I think you even need $M$ to be a finite sum of (commutative) $\otimes$ (matrices).

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    $\begingroup$ I am coming to this 8 years too late... But Quigg's paper (mathscinet.ams.org/mathscinet-getitem?mr=806641) shows exactly this, in Section 5. However, this only shows that the coproduct $\Delta$ on $A$ is bounded iff $M$ is as stated. The original question wanted this map to be "short", that is, contractive. Surely then you only get commutative vn algs, but I don't quite see the proof. Similarly, for a full answer, you then need to character which commutative vn algs give $\Delta$ mapping into $A\widehat\otimes A$, and not something larger. $\endgroup$
    – Matthew Daws
    Commented Jun 23, 2020 at 9:17
  • $\begingroup$ @MatthewDaws 8 years is fine, I'm still one day going to solve the problems from my PhD... $\endgroup$
    – Yemon Choi
    Commented Jun 23, 2020 at 18:43
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    $\begingroup$ @MatthewDaws to show contractive forces M to be commutative, maybe show things go wrong for 2 by 2 matrices, and then argue that a noncommutative M contains a copy of ${\bf M}_2$. But as you say it's not clear if M commutative is even sufficient to get $\Delta(A) \subseteq A \hat\otimes A$ $\endgroup$
    – Yemon Choi
    Commented Jun 23, 2020 at 18:46

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