Timeline for Topology on the dual of a Frechet space
Current License: CC BY-SA 3.0
7 events
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| Jul 25, 2014 at 15:19 | comment | added | kaveh | Tobias, I agree, it would be better if he put more details; it is a bit hard to follow this work. However, this paper is quit interesting and I think this definition of $\beta$-cotangent bundle is suitable for my purpose which is getting a Finsler structure on a cotangent bundle by using pairing. He says that the $\beta$-exterior bundle is locally trivial if all continuous linear maps, map the bornology into itself, but I dont know how often can it occurs in practice. | |
| Jul 25, 2014 at 13:02 | comment | added | Tobias Diez | It's a long time since I looked at Prof. Wurzbachers work, so take the following with a grain of salt: I wasn't to convince that his definition of a cotangent bundle does not run in the same kind of complications, i.e. it is not locally trivial in a smooth manner (since the above result of Neeb yields a counter example for every vector space topology on the dual). If I remember it correctly, Wurzbacher just says "this bundle is obviously locally trivial" and don't comment on possible issues. Kaveh, do you think his work is rigorous? | |
| Jul 25, 2014 at 12:29 | comment | added | kaveh | Thanks for comments, It seems that there is a replacement for a cotangent bundle with a manifold structure in standard calculus (Michal and Bastiani). If we define the topology of uniform convergence on all compact sets of a cotangent space. If $\beta$ is a bornology on the model space. Then the $\beta$-cotangent bundle which is the union of the cotangent spaces with the above topology has a structure of a manifold, see Wurzbacher, Fermionic Second Quantization. | |
| Jul 25, 2014 at 7:35 | comment | added | Tobias Diez | You can still form the exterior bundle (and in particular the cotangent bundle) in the set-theoretic sense. So the differential $df$ still belongs to the cotangent bundle, however you don't have a smooth structure on it and thus you cannot say that the map $m \mapsto (df)_m$ is smooth in the usual way. | |
| Jul 24, 2014 at 21:14 | comment | added | TaQ | @ Kaveh Eftekharinasab: If instead of the Michal−Bastiani smoothness you choose to use the Frölicher−Kriegl−Michor "Convenient Calculus", then you can construct the cotangent bundle in the usual manner. However, then there is the drawback that a smooth map with domain a nonmetrizable space need not be continuous. | |
| Jul 24, 2014 at 19:05 | comment | added | kaveh | Thanks, it was suggested a replacement for a cotangent bundle which is a sub-bundle of the cotangent bundle with a natural smooth structure (Neeb, Remark II.3.5, Monastir Summer School: Infinite-Dimensional Lie Groups). A question which arise now is that if we have for example a functional f on manifold the differential df is the element of the cotangent bundle, is there a replacement of the cotangent bundle which df belongs to? | |
| Jul 24, 2014 at 11:19 | history | answered | Tobias Diez | CC BY-SA 3.0 |