Timeline for Which W*-algebras are the duals of C*-coalgebras?
Current License: CC BY-SA 3.0
19 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 8, 2023 at 1:11 | comment | added | Toby Bartels | @YemonChoi may know more. | |
| Dec 8, 2023 at 1:11 | comment | added | Toby Bartels | @AlexM. : If anything, they seem to work better with the injective tensor product on Ban. This isn't closed, and it doesn't make sense (or at least doesn't work as well, I don't remember how badly) if you extend to nonarchimedean fields. But it has more coalgebras. Unfortunately, I haven't really thought about this stuff in a decade. | |
| Dec 6, 2023 at 18:09 | comment | added | Alex M. | @TobyBartels: I understand. So, unlike {Banach, C^*, W^*}-algebras (which are universally agreed-upon concepts), the "co-" analogues have not received a definitive shape yet? | |
| Dec 6, 2023 at 17:22 | comment | added | Toby Bartels | @AlexM. : Oh, I can tell you why only short maps though; that's the category that gets you the correct isomorphisms. The hom-set (V,W) is the set of short linear maps, but the internal hom-object [V,W] is the Banach space of all bounded linear maps, so they're still there. (You can think of the underlying set of a Banach space as the set of elements of its closed unit ball, not the set of all of its elements.) It's a weird perspective, since it means that Ban isn't a concrete category in the usual way, but it works out. | |
| Dec 6, 2023 at 17:19 | comment | added | Toby Bartels | @AlexM. : I forgot to tag you in my previous comment and I don't know if the edit will fix that. | |
| Dec 6, 2023 at 17:18 | comment | added | Toby Bartels | @AlexM. : That nLab page, which has been frozen for the past 11 years, should be updated; but in brief, the definition there is the correct one from a category-theorist's perspective if you think of Ban as a closed category (with the Banach space of bounded linear maps from V to W as [V,W]) and do everything automatically from there. But the problem is that it doesn't have many examples, as shown by Yemon's answer, and there are other tensor products on Ban that, while not closed, have more interesting coalgebras. So all that needs to go in there. | |
| Dec 6, 2023 at 16:27 | comment | added | Alex M. | @TobyBartels, let me please hijack your question: is there any standard text laying down the basics of "coalgebras + functional analysis", i.e. C^*-coalgebras and related stuff? I have visited your nLab link, but there is no bibliography therein, so I wonder whether all this is your own view of the subject, or the general consensus on it? I.e. why are the concepts defined the way they are, why the projective tensor project and not any other one, why short maps etc. | |
| Sep 18, 2012 at 0:12 | comment | added | Toby Bartels | That's what I thought (although Matt mentioned the extended HTP and I'm unclear about the difference). Possibly there is some natural way to equip a Banach space with a default operator-space structure, but then is Matt Daws's response to Dmitri's question even correct using the EHTP of the predual with its default op-sp structure? (instead of the op-sp structure that it gets from the algebra structure on its dual). | |
| Sep 17, 2012 at 17:36 | comment | added | Yemon Choi | Toby: the haagerup tp is only defined in the category of op spaces. | |
| Aug 29, 2012 at 18:58 | comment | added | Toby Bartels | (In fact, it was after reading Matt Daws's answer to your question cited by Ollie above, and having this problem, that I decided to see what one could do with Banach coalgebras (not expecting to get all $W^*$-algebras this way but hopefully some), and the literature that I found on those used the projective tensor product. But that literature was also mostly concerned with nonArchimedean spaces, so it just may not be relevant.) | |
| Aug 29, 2012 at 18:55 | comment | added | Toby Bartels | I'm having trouble figuring out what the extended Haagerup product actually means for plain Banach spaces, that is those without an action of an operator algebra or otherwise with the structure of operator spaces. | |
| Aug 28, 2012 at 8:33 | comment | added | Dmitri Pavlov | 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. | |
| Aug 27, 2012 at 9:16 | comment | added | Yemon Choi | With regard to Ollie's commemt, just ask Matt Daws about all this, quite frankly. | |
| Aug 26, 2012 at 17:36 | comment | added | Ollie | With regard to Yemon's comment about projective tensor products, read Matt Daws' answer to this question: mathoverflow.net/questions/50302/… | |
| Aug 26, 2012 at 8:45 | comment | added | Yemon Choi | BTW, maybe this should be a co-$C^\ast$-algebra rather than a $C^\ast$-coalgebra, but it depends how you feel about clashes with terminology used by people in operator algebras. | |
| Aug 26, 2012 at 8:43 | answer | added | Yemon Choi | timeline score: 3 | |
| Aug 26, 2012 at 8:31 | comment | added | Yemon Choi | 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) | |
| Aug 26, 2012 at 8:24 | comment | added | Yemon Choi | 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.) | |
| Aug 26, 2012 at 8:08 | history | asked | Toby Bartels | CC BY-SA 3.0 |