Timeline for The tangent bundle to an infinite-dimensional manifold
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 10, 2017 at 14:25 | comment | added | Thomas Rot | @vanmeri: You can look at the books of Klingenberg | |
| Dec 23, 2016 at 18:01 | comment | added | Pedro Lauridsen Ribeiro | The reason is that the modelling topological vector spaces are spaces of smooth sections with compact support for a certain family of pullback vector bundles over $A$, and for these the topology which yields a smooth manifold topology on $C^\infty(A,B)$ with the atlas written in Michael Coffey's answer is actually finer than the standard inductive limit topology. This is well described in Sections 41 and 42 of the book of Kriegl and Michor. In either case the resulting manifold topology is metrizable if and only if $A$ is compact. | |
| Dec 23, 2016 at 17:49 | comment | added | Pedro Lauridsen Ribeiro | @TheoJohnson-Freyd the compact-open $C^\infty$ topology is the one which turns $C^\infty(A,B)$ into a metrizable topological space (provided $A$ is $\sigma$-compact and $B$ is metrizable, to be more precise), but this topology is not a manifold topology if $A$ is not compact and $B$ is not contractible, since in this case this topology is not locally arcwise connected. The Whitney $C^\infty$ topology described in Fly by Night's answer above solves this problem partially, yielding a topological but not smooth manifold structure. | |
| Dec 23, 2016 at 17:21 | comment | added | vanmeri | This is a clear answer, can anyone recommend a book on the subject, please? Thank you | |
| May 25, 2011 at 14:39 | comment | added | Theo Johnson-Freyd | Oh, I know that it's not that hard. I'm just not very good at real analysis, and so try to avoid it whenever I can mumble some categorical incantations. Note that there's no way to decide which of the $C^r$ or $H^s$ spaces is "the right" one, and for most applications it doesn't matter which you pick. But note that any such choice does not recover the Grothendieck-style definition, because there are more "smooth maps from point" to the analytic choices than to the sheaf --- and if you want the "points" to be just the smooth maps, you're stuck: $C^\infty(A,B)$ is Frechet but not Banach. | |
| May 25, 2011 at 8:39 | comment | added | Michael Coffey | Oh, and then it is not hard to show that we can identify the tangent spaces by [T_f C^r (A,B)= \{v \colon A \to TB | v(a) \in T_{f(a)}B\}]. So in fact $T_{id}C^r(A,A)$ is precisely the space of $C^r$ vector fields on $A$. | |
| May 25, 2011 at 8:18 | history | answered | Michael Coffey | CC BY-SA 3.0 |