If $F$ is a Frechet space, is there any locally convex space topology on the dual $F'$, such that for each local diffeomorphism $f$ from an open subset $U$ of $F$ to $F$, the map $U \times F' \longrightarrow F'$ is smooth?
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$\begingroup$ Your question is not precise enough. You should specify how you define "the map $U \times F' \longrightarrow F'$ " and also in what sense you want it to be smooth. $\endgroup$– TaQCommented Jul 24, 2014 at 8:23
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$\begingroup$ Smoothness in the sense of Michal and Bastiani (also known Keller). And the map is defined by $(x, \lambda) \mapsto \lambda \circ df $. Actually, the question is how to define a smooth structure on cotangent Frechet bundle? $\endgroup$– kavehCommented Jul 24, 2014 at 10:56
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This map is not smooth for every vector space topology on the dual space, see Remark I.3.9. in Neeb, K.-H. "Towards a Lie theory of locally convex groups" 2006 for an explicit counterexample.
Thus the cotangent bundle does not carry a (natural) smooth structure for Fréchet manifolds and one has to specify what one understands under smooth differential forms more directly. The usual way is to require the chart representation of the differential $k$-form $\alpha$ to be smooth as as a map $\alpha: U \times F^k \to \mathbb{R}$ (see Definition I.4.1 in the above paper).
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$\begingroup$ 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? $\endgroup$– kavehCommented Jul 24, 2014 at 19:05
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$\begingroup$ @ 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. $\endgroup$– TaQCommented Jul 24, 2014 at 21:14
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$\begingroup$ 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. $\endgroup$ Commented Jul 25, 2014 at 7:35
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$\begingroup$ 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. $\endgroup$– kavehCommented Jul 25, 2014 at 12:29
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$\begingroup$ 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? $\endgroup$ Commented Jul 25, 2014 at 13:02