TITLE: Which W*-algebras are the duals of C*-coalgebras?
QUESTION [10 upvotes]: 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.

REPLY [2 votes]: 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).