Talk:Tangent bundle

message: You mean tangent bundle is a term used in auto mechanics, Michael Hardy? Phys
:Not that I've ever heard of, but you can get from cars to tangent bundles in three easy steps: Automobile → differential → derivative → tangent bundle. Yes, I'm bored :) Fropuff 00:54, 2004 Mar 15 (UTC)
This is a very nice and modern description. Note that the purely algebraic description of curvature and especially its group theoretical invariance discussion leads via tangent bundling to examples of cosmoligical models, which general relativity needs for an global approach. Hannes Tilgner
Should have some import of sections, the hairy ball theorem, etc. Jake 22:07, 2 Jun 2005 (UTC)

message: This article is part of a series of closely related articles for which I would like to clarify the interrelations. Please contribute your ideas at Wikipedia talk:WikiProject Mathematics/related articles. --MarSch 14:08, 12 Jun 2005 (UTC)
thread_title:  discussion at [[Wikipedia talk:WikiProject Mathematics/related articles]] 

message: Oleg, when I read your example I was led to visualize the tangent space as the subspace in the plane consisting of a line tangent to the given point on the circle. I do not know whether this is what you intended, but this does not strike me as a good way of looking at things. Instead, I think of a vector space attached to each smooth point, representing all possible tangent vectors to this point. I don't really think about the space being embedding in anything, which is perhaps why I was disturbed by this. I changed it to how I see things for now. Also, I thought the cylinder construction could be clarified... it seems vague right now. Thoughts? - Gauge 05:05, 19 September 2005 (UTC)
replies: text: Hi Gauge. You are the differential geometry guy, so I will defer to you in everything in that region. Let me explain my motivation however. The reason I created that example, and the reason I still prefer it as originally stated, is because I found the tangent bundle page very hard to understand for anybody who is not proficient in differential geometry. That's why I have the silly example with tangent lines to the circle, lines which need to be rotated in 3D so that they don't intersect and so that they align smoothly, eventually having no choice than to settle into a cyllinder. I know this is not rigurous, but I have a question: can one at least intuitively think of the tangent bundle that way? Oleg Alexandrov 15:25, 19 September 2005 (UTC)
replies: text: The tangent bundle to a circle is (isomorphic to) a cylinder, not just intuitively, but in actuality.  Viz. TS^1=S^1 \times \mathbb{R}  (using the product topology, I suppose, if one wanted to get fancy.) linas 00:50, 20 September 2005 (UTC)

text: Yes, every tangent bundle is locally a product, but not globally. About combing the hair, how about the hairy ball theorem? Oleg Alexandrov 04:23, 20 September 2005 (UTC)
replies: text: Yep that's the one, but I see that article is less than illuminating as well. Sigh. Parts of WP are a mess.  linas 04:25, 20 September 2005 (UTC)
thread_title: Description in the introduction

message: The state of this article was deplorable, so I rewrote it. Still to do:
*Expound upon the simplest nontrivial example, T(S2).
*Comment on the conditions for the tangent bundle to be trivial (existence of a global frame) and the definition of parallizability.
*Give the relationship between the tangent bundle of M and the frame bundle of M.
*Mention functorality of the tangent bundle construction and the pushforward map.
I may not get around to these soon, so I invite others to help. -- Fropuff 06:51, 20 September 2005 (UTC)
replies: text: The rewrite made the article much better, thanks! It is still not accessible for people not knowing anything at all about differential geometry, but that is to be expected in a sense, as this is not a trivial concept. Oleg Alexandrov 15:21, 20 September 2005 (UTC)
thread_title: To do

message: why don't you use \bigcup instead of \coprod for disjoint union
I didn't do it myself, as I why would rather ask first (is this the right way to handle such a question?)
:\coprod is the standard symbol for a disjoint union. See that article for reasons. -- Fropuff 19:48, 22 December 2005 (UTC)
replies: text: The idea is tbat the tangent spaces at two different points of the manifold may contain the same tangent vector. In this case, it has to be included twice in the tangent bundle, once as a tangent vector at the first point and once as a tangent vector at the second point. -- Jitse Niesen (talk) 22:10, 22 December 2005 (UTC)
thread_title:  union 

message: It's taking me a while to figure all of this bundle business out. Suppose I have an idealized particle somewhere in space. Its position can be modeled by a point in an affine space. Is it correct then that its complete state (position and velocity) represents a point in (member of the) tangent bundle of that affine space? Similarly, would a bound vector (a vector with a defined tail position as well as direction and magnitude) really be a member of a vector bundle? —Ben FrantzDale 06:20, 28 January 2007 (UTC)
replies: text: Yes and yes. -- Fropuff 16:27, 28 January 2007 (UTC)
replies: text:  Great. Thanks. —Ben FrantzDale 21:46, 28 January 2007 (UTC)
thread_title:  Examples 

message: TopologyExpert deleted the higher order tangent bundle section saying it was completely incorrect.  What exactly do you see as incorrect, because it looks largely correct to me and the people who created it.  The article clearly states the tangent bundle is an object that exists for a smooth manifold, moreover, later in the article it clearly states that the tangent bundle is itself a smooth manifold, so the concept of the tangent bundle of the tangent bundle is perfectly well-defined.   The bit about the tensor is a little uninformative but it was otherwise totally on the mark.  So what do you perceive as totally wrong? Rybu (talk) 04:24, 3 October 2008 (UTC)
replies: text: I brought higher order tangent bundles back.  I moved them ahead of the canonical vector field on the tangent bundle since the double tangent bundle is needed to make sense of that section.  I also added a "role" section where it's explained that the point of the tangent bundle is to serve as domain and range for the derivative of a smooth map of manifolds. Rybu (talk) 03:28, 4 October 2008 (UTC)
replies: text:  Yes, it is true that the total space of the tangent bundle of a smooth manifold is a smooth manifold. I deleted that section because at the time there were no reasons as to why these 'higher order tangent bundles' are of interest. Also, the section claimed (wrongly) that higher order tangent bundles are like 'higher order derivatives' which is meaningless. Of course, higher order derivatives of a smooth function between smooth manifolds can be induced on higher order tangent bundles. But the section is good know so I have no intentions of removing it again.
Topology Expert (talk) 03:34, 9 October 2008 (UTC)
thread_title:  Higher order Tangent bundle deleted by TopologyExpert 

message: The previous section was very helpful in clarifying some problems.  Here are a few items that the discussion has prompted:
1. The upside-down pi symbol is used here to stand for "disjoint union".  However, the disjoint union article, which this article references, uses the \bigsqcup, not the coproduct symbol \coprod.  I think it would help to change to the bigsqcup, unless coproduct is intended. (A previous talk section "union" suggested the plain bigcup symbol, which the other article would argue against, though I do see bigcup used that way in many other web sources.)
2. Footnote note 1 is a bit mangled on the way to almost explaining a point I'm about to make below. I think the phrase "align them in a plane perpendicular plane" doesn't make sense.  Possibly it should be "align them perpendicular to the plane of the circle".
3. From the "Request for clarification" discussion of the last few days, and my finally understanding the notation of the "TM=..." algebra, I think I see where the intro got me way off track, and how the intro could be improved.  
Not understanding the notation, I tried to make it make sense as a statement of how to calculate the arrangement of the tangent bundles in space. I.e.: if we were to apply that algebra somehow to the circle example, it would "calculate" the answer: a set of "vertically" arranged 1-D spaces.  Instead, I now see that the algebra is merely a statement of the requirements: a set of vector spaces, one vector space for each point on the manifold, with each vector space disjoint from the others (and make them collectively contiguous and smooth? -- though I don't see that in the requirements).  The actual solution -- 1-D vector spaces perpendicular to the circle's plane, and arranged around the circle so as to form a cylinder -- was devised by  separate thinking not shown here, and is "the answer" because it satisfies the requirements.
Now, if I'm right about this, it would be helpful for the intro to point this out: That having stated "Tangent bundle" in terms of requirements, we are looking for solutions for those requirements, not calculating the shape of the tangent bundle. 
Thanks again for the discussion Gwideman (talk) 12:54, 9 December 2009 (UTC)
thread_title:  Request for clarification, cont'd 

message: Clearly, "the tangent space of a vector space W is naturally a product, TW \cong W \times W" should read "the tangent bundle of a vector space...", which raises the question why we can apply this product structure to the tangent bundle T(T_x M) at each point. How do we get T(T_x M) from T(TM)? Also, is it obvious that the given "informal" definition gives a coordinate-independent vector field? If not, what is the point of this lengthy explanation? Doesn't the map (x, v) \mapsto (x, v, 0, v) given below achieve the same less cryptically? I would also like to see an explanation of the phrase "splitting the map via". Theowoll (talk) 00:07, 12 March 2018 (UTC)
thread_title:  Canonical vector field: tangent space is naturally a product