Talk:Cotangent bundle

message: Apparently, nobody wants to define the cotangent bundle.  Silly rabbit 08:04, 10 November 2005 (UTC)
replies: text: Maybe we should go on about co-vector fields being dual to vector fields and do it components before we lose most of our readers by mentioning a sheaf? Billlion 05:26, 19 May 2006 (UTC)

text: T^*(M) = \coprod_{x\in M}T^*_x(M) = \left\{(v,x) \colon x \in M \mbox{ and } v \in T^*_x M\right\}
That is, ordered pairs of a point in the manifold with cotangent vectors at that point.  Very simple. --Chris Foster 05:55, 30 July 2006 (UTC)

text: Well yes as a set, and a vector space at each point, but this misses out the topology let alone the differntiable structure.Billlion 20:44, 30 July 2006 (UTC)
replies: text: You're quite right, but perhaps we can attach the topological and smooth structure to the simple definition, as is done in the article on the tangent bundle?  It seems to me that the balance between redundency and explanatory power lies in having a simple definition at the top of the article.  Of course, the mathematically sophisticated version is always nice if you can understand without reading twenty other pages of definitions.  It can be moved further down the page for those who understand the deeper structure.  --Chris Foster 12:06, 7 August 2006 (UTC)

text: Well for any (finite dimensional) manifold the tangent bundle and cotangent bundle are isomorphic as bundles, so of course the total spaces are diffeomorphic.  However there is not a natural isomorphism between them, unless you have some additional structure such as Riemannian metric.Billlion 11:55, 30 December 2006 (UTC)

message: "The one-form assigns to a vector in the tangent bundle of the cotangent bundle the application of the element† in the cotangent bundle (a linear functional) to the projection‡ of the vector into the tangent bundle (the differential of the projection of the cotangent bundle to the original manifold)."
†Which element? The vector?
‡How do you apply an element to a projection? Rather, how is a linear functional an application of an element to a transformation? Or is it the projection that's the linear functional? What is this a description of? As for breaking the sentence up I suggest leading with:
"The differential of the projection of the cotangent bundle to the manifold is also the projection of a vector into the tangent bundle."
And, if the linear functional is the application of the element, saying:
"The one-form assigns a linear functional, the application of the element to that projection of the vector, to a vector in the tangent bundle of the cotangent bundle." ᛭ LokiClock (talk) 19:34, 10 December 2010 (UTC)
replies: text: I changed it. How is it now? Ozob (talk) 23:26, 10 December 2010 (UTC)
replies: text:  Almost crystal clear. But how is the vector in the tangent bundle being projected into the tangent bundle? Or is it that it's being projected from the tangent bundle of the cotangent bundle to the tangent bundle of the manifold? ᛭ LokiClock (talk) 10:51, 11 December 2010 (UTC)
replies: text: It's being projected from the tangent bundle of the cotangent bundle down to the tangent bundle.  I added a couple words to make this more explicit.  Ozob (talk) 12:49, 11 December 2010 (UTC)
replies: text:  That's more information, but I still don't see it. Which tangent bundle? Is there a difference between the tangent bundle you're talking about and the tangent bundle of the original manifold? ᛭ LokiClock (talk) 20:17, 11 December 2010 (UTC)
replies: text: I identified the domain and codomain of dπ.  Is it clear now? Ozob (talk) 17:52, 12 December 2010 (UTC)
replies: text:  Yes. You are wonderful. ᛭ LokiClock (talk) 16:35, 13 December 2010 (UTC)
replies: text: Thank you! Ozob (talk) 00:00, 14 December 2010 (UTC)
thread_title:  A thick set of language 

message: Referring this edit of yours:
* Why is the function g there? The definition only says that the fiber at x\in M is the dual space T^*_xM to the tangent space T_xM. Nowhere does the definition say anything about some extraneous function g\in C^\infty(\mathbb{R}^n). The way you wrote this implies that, for every v^* \in T^*_xM, there is g\in C^\infty(\mathbb{R}^n) such that g(M)=0 and  v^* = dg_x.  First, it's not immediately clear whether this is true. Even if it is true, why use it here?  Why all the extra complication?
* Where did the requirement that g(M)=0 come from? What if I change g to g+c, for some constant c? g(M) will not be 0 anymore, but dg_x is still the same. So, why 0, rather than some other value?
* v is undefined in your example. StrokeOfMidnight (talk) 17:32, 7 August 2020 (UTC)
thread_title:  Examples 

message: While the definition given in #Formal Definition is certainly valid, and may be the best one for those interested in Algebraic geometry, the average reader might be better served by a simpler definition from a different perspective. Also, it is important to note in the article that there are alternate equivalent definitions.
There is a simpler definitions using germs of functions on M directly rather than as pullbacks of germs on M \times M. And, of course, there are slightly longer definitions not using the concept of germs. I'm not sure which is clearest to the general reader. --Shmuel (Seymour J.) Metz Username:Chatul (talk) 19:52, 13 March 2022 (UTC)
thread_title:  Formal Definition