Timeline for When do tensor products of C*-algebras commute with colimits?
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
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| Nov 29, 2011 at 15:05 | vote | accept | Fabian Lenhardt | ||
| Nov 29, 2011 at 3:17 | answer | added | Leonel Robert | timeline score: 7 | |
| Nov 28, 2011 at 22:49 | comment | added | Yemon Choi | Sergio,the question is about the category of all C*-algebras, not just the abelian ones | |
| Nov 28, 2011 at 22:38 | comment | added | Buschi Sergio | error above: $C^ast:=C^{\ast}$ | |
| Nov 28, 2011 at 22:37 | comment | added | Buschi Sergio | For algebras like $C(X)$ (resp. $C^ast(X))$ a colimit correspond to a limit of realcompact (resp. compact) spaces (this colimits can do in in Topological spaces category, because are full reflexive subcategories, then the inclusion create limits). THen if we have a representation theorem "each $C^ast$ algebra (of a some specific type) is isomorphic to some funtion spaces $C_X$ on some topological space $X$" and there exist a isomorphism $C_X\otimes C_Y\cong C(X\times Y)$ the question become topological. (also for disprove it) | |
| Nov 28, 2011 at 21:36 | comment | added | Yemon Choi | My guess (but I don't have time to check the details right now, hence this is merely a comment) is that the maximal tensor product of C*-algebras should behave nicely, or at least better, with colimits. Then when $N$ is nuclear the max and min tensor products coincide, so $N \otimes$ would behave nicely. | |
| Nov 28, 2011 at 21:32 | comment | added | Yemon Choi | @Ulrich Pennig: all nuclear C*-algebras are exact... | |
| Nov 28, 2011 at 21:31 | comment | added | Yemon Choi | Chris, what tensor product are you/Guichardet using? I seem to recall that Guichardet defines a tensor which gives decent SMC properties but isn't the spatial t.p. And at C*-level the whole issue of tensor products is MUCH more complicated | |
| Nov 28, 2011 at 21:12 | history | edited | Martin Brandenburg |
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| Nov 28, 2011 at 21:11 | comment | added | Ulrich Pennig | You definitely want $N$ to be an exact $C^*$-algebra, not just a nuclear one, for something like that to be true. | |
| Nov 28, 2011 at 16:43 | comment | added | Chris Heunen | By the way, preserving filtered colimits is the same as preserving colimits of chains [Adamek&Rosicky, Cambridge Univ Press, 1994, Cors 1.5 & 1.7]. So you may assume that I is a total order. If all the maps $A_i \to A_j$ are inclusions, doesn't that make the colimit just the closure of the union of all the $A_i$, leading to a trivial proof? | |
| Nov 28, 2011 at 16:15 | comment | added | Chris Heunen | In the category of von Neumann algebras, the functor $- \otimes N$ preserves coequalizers [Guichardet, Bull Sci Math 90:41-64, 1966, Prop 8.3]. I would hope the same holds for C*-algebras, so that it would suffice to concentrate on coproducts. But also, in the category of von Neumann algebras, colimits do not preserve flatness [Guichardet, Remark 8.2], which is a bad sign. | |
| Nov 28, 2011 at 15:30 | history | edited | Matthew Daws |
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| Nov 28, 2011 at 15:19 | history | asked | Fabian Lenhardt | CC BY-SA 3.0 |