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I'm trying to understand the definition of a differential form on $M$ in the context of Fréchet spaces or, more generally, locally convex spaces. The standard procedure defines a k-form as a map $\alpha: M \rightarrow L^k_{\text{alt}}(TM)$ such that the coordinate representation $U \times E^k \rightarrow \mathbb{R}$ is smooth. Here, $L^k_{\text{alt}}(TM)$ denotes the space of k-multilinear, skew-symmetric and continuous maps on $TM$.

Another method ([1], [2]) endows the spaces $L^k_{\text{alt}}(T_mM)$ with the strong topology, constructs a vector bundle structure for $\Lambda^k M := \bigcup_m L^k_{\text{alt}}(T_mM)$ and finally defines differential forms as smooth sections of this bundle.

I have two questions about this later construction:

1) How is the smooth structure of $\Lambda^k M$ defined?

Background: In [3, Remark II.3.5 p. 18], Neeb declares that one can endow the cotangent bundle with the structure of a (topological?) vector bundle, but not with a smooth manifold structure. The smoothness of the transition functions would require that maps of the form $U \times E' \rightarrow E'$, $(x,\alpha) \rightarrow \alpha \circ d(\rho \circ \kappa^{-1})_x$ are smooth for all chart transitions. This seams only to hold for Banach manifolds (Why?). But this no-go-fact would us prevent from talking about smooth sections of the exterior product.

2) Wurzbacher [2] says that $\Lambda^k M$ is a vector bundle because every continuous linear map $T: E \rightarrow E$ maps bounded to bounded sets. Where is this fact needed?

[1]: Kriegl, Michor: Convenient setting of global anlysis (p. 336 ff.)

[2]: Wurzbacher: Fermionic Second Quantization

[3]: Neeb: Infinite-Dimensional Lie Groups

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If evaluation $E\times E'\to \mathbb R$ is jointly continuous for any pair of lcs topologies which are compatible with duality, then $E$ has to be normable: Namely, if $ev(U\times U')\subset (-1,1)$ for open $U$ and $U'$, then $U$ is contained in the polar of an open set, thus is bounded. So $E$ is normable. This extends to the composition of mappings. If you insist that all smooth maps are continuous for the lcd topologies (as Neeb does), then you have problems here. If you allow finer topologies (look for $c^\infty$-topology in [1]), then this difficulty disappear. Then evaluation from above is smooth, even if it is not continuous for the lcs topologies.

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Thanks for this quick reply. Ok, I was suspecting something along this lines and now see the advantage of the convenient setting. In other words, Wurzbacher's statement "obviously $\Lambda^k M$ is a [smooth] vector bundle in the category of l.c. spaces" is faulty or do I miss something? – Tobias Diez Feb 11 '13 at 20:31
It is not faulty, it depends on the calculus you are using. You can use convenient calculus on lc spaces. – Peter Michor Feb 11 '13 at 20:36
He uses the same calculus as Neeb (see p. 336 of Wurzbacher)... – Tobias Diez Feb 11 '13 at 21:28

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