Talk:Hairy ball theorem

message: In the main picture for this article why is the hairy sphere described as having 'uncomfortable' tufts....... 
Imagine if you were combing your hair and you have a cowlick. It's not that the tuft is actually uncomfortable, it's that the sight of them makes you uncomfortable. Nobody likes looking at them anyways, right? 24.23.13.3 (talk) 21:47, 22 January 2015 (UTC)
thread_title: Uncomfortable

message: "And the award for funniest name for a mathematical theorem goes to:"
http://forums.fark.com/cgi/fark/comments.pl?IDLink=1600618
-You shouldn't lock this from being edited. I'm sure more people are getting use out of it for humorous reasons than there are people getting use out of it for mathematical reasons. Power to the people; isn't that the point of wikipedia?
:The point is to write an encyclopedia, rather than anything else. Charles Matthews 06:52, 2 August 2005 (UTC)
replies: text: Topologically, a sphere is the same as a football, or a cylinder. And the shape of the comb doesn't matter. This isn't to say you can't comb a sphere; you certainly can, but you can't make the end result in a particular way. Think of it if you had control of all the hairs on a sphere, you couldn't make them lie down flat without having two next to each other have drastically different directions. Quentin mcalmott 05:29, 11 August 2005 (UTC)
replies: text:  I don't think it's true that a cylinder is topologically the same as a sphere.  Don't the edges of the cylinder prevent any attempts at creating a diffeomorphism? Threepounds 04:11, 8 December 2005 (UTC)
replies: text:  (1) It is, because in Top only homeomorphisms count and you can easily construct a homeomorphism from S^2 to the cylinder.  (2)  Surprisingly, if a space X admits a differentiable structure, any space X' homeomorphic to X also admits a differentiable structure.  Therefore, the cylinder admits a differentiable structure and is actually diffeomorphic to the sphere.  For a differentiable structure on a topological manifold, you don't actually need diffeomorphisms from the manifold to the model space in the atlas; you just need homeomorphisms which, when composed on overlaps of chart neighborhoods, are differentiable maps from (subsets of) the model space to (subsets of) the model space.  Heuristically, then, you just need the kinks to composition on the overlapping maps to ``smooth out the kinks.  (You can actually work out an explicit formula for stereographic projection from the cube to \mathbb{R}^2 to turn the cube into a differentiable manifold.)  (I know I am replying to a five-year-old question.  Who cares?  It's interesting material.)  99.186.238.191 (talk) 03:44, 12 December 2010 (UTC)
thread_title:  Theorem linked to from humour site 

message: This is number 32 in the most popular articles on Wikipedia. Getting people to find out about it is (for most wikipedians) a good thing.  The article is interesting also!
POOOOOOOOOOOOOOOOOOOOOOOOOOOOP  —Preceding unsigned comment added by 128.174.114.66 (talk) 22:27, 31 March 2010 (UTC)
replies: text: Hairy balls... I get it. LOL and Happy Holidays from The Doctahedron, 16:01, 25 December 2011 (UTC)
thread_title:  humour has a place 

message: A potential candidate can be found here --146.145.37.154 21:18, 26 June 2006 (UTC)
replies: text: As for the image added recently, it is not very appropriate because vectors are not tangent to the sphere. --Army1987 09:28, 5 July 2006 (UTC)
thread_title:  Image of a ball in a vector field 

message: A google translation of this gives the name "Sentence of the hedgehog".  violet/riga (t) 13:23, 28 August 2006 (UTC)
replies: text: Satz is German for theorem, does mean sentence/proposition. Charles Matthews 14:03, 8 November 2006 (UTC)
thread_title:  Google translation of German article 

message: It would be helpful if someone could add an outline of a proof of the theorem to the article.  As is the article just begs the question of how to prove it. Dugwiki 17:24, 7 November 2006 (UTC)
replies: text: +1, as with any theorem, I'd like to see proof.--Jasper Deng (talk) 22:15, 21 July 2014 (UTC)
thread_title:  Possible to provide proof outline? 

message: The article states that you need to have "is at least one p such that f(p) = 0". Is it possible to generate an f(p) s.t. there is only one p s.t. f(p)=0, or do you need at least two? I can't visualise a situation in which you only have one problem point, but of course that doesn't mean it's not possible. Perhaps, if it is possible to construct a vector field with only one problem point, we should include an explanation of it? AdinaBob 19:00, 23 February 2007 (UTC)
replies: text: To answer my own question, it seems that by the Euler characteristic, the sphere must have at least two "problem points". I think there should be some mention of this in the hairy ball theorem. I'll put it in when I find some time.AdinaBob 19:59, 23 February 2007 (UTC)
replies: text:  You can put the two points together, to get what's called an index 2 zero.  What's it look like? Pick a point p on the sphere and a great arc A going through it.  Now consider all circles on the sphere that go through p and have "centers" on A.  Then let the vector field flow consistently around the circles but always die at p.  This is the vector field with just one zero. 24.59.106.118 16:37, 23 September 2007 (UTC)
thread_title: How many problem points on a sphere?

message: As has been mentioned, this is one of wikipedia's most popular articles.  A brief, yet encyclopedic note on the fact that people find this funny should be in there.--Loodog 04:50, 24 February 2007 (UTC)
thread_title:  Encyclopedic note on humor 

message: gluLookAt(eyeX, eyeY,  eyeZ,  centerX,  centerY,  centerZ,  upX,  upY,  upZ) actually takes two vectors for camera orientation: f = center - eye and up = (upX, upY, upZ). The third vector, which is computed from the function, is simply the dot product of the former two. Therefore gluLookAt is continuous and has nothing to do with the Hairy ball theorem. Hairy ball theorem states that there cannot be continuous function f(v) which returns v' orthogonal to v.
: Firstly you mean 'cross product' not 'dot product'. Secondly, it's a standard problem in computer graphics to generate a vector orthogonal to another. For example if we want a camera at A to look at an object at B we want to generate a normalised 'up' vector for the camera that is orthogonal to the line AB. Animators often want such a function to be continuous. No such function exists because of the hairy ball theorem. This is a classic example of the hairy ball theorem at work. gluLookAt is the compromise - you have to sacrifice continuity and instead supply gluLookAt with an up vector. As a result we end up with a function that, as you rightly point out, is continuous, but unfortunately sometimes goes to zero, as the hairy ball theorem says it must. This is a real world consequence of the hairy ball theorem that computer graphics people have to work around every day and to me it seems relevant to the wikipedia entry. I might not have worded it very well however. Sigfpe 22:13, 27 February 2007 (UTC)
replies: text: I don't get that last statement. There is no everywhere non-zero continuous tangent vector field on the sphere so how can it be true that "Given a non zero continous vector field there is always a continuous non-zero tangent vector field.". I must be misreading it. Sigfpe 16:53, 14 March 2007 (UTC)
replies: text:  There is always an orthogonal* non-zero vector field. Sorry.
Someone please make sense of the computer graphics application. It makes a claim about a vector. I can understand that given a continuous vector field then there is not always a continuous orthogonal vector field (for example take the vectors orthogonal to the surface of a sphere). But for one vector?  —Preceding unsigned comment added by Polfbroekstraat (talk • contribs) 08:48, 2 November 2007 (UTC)

text: I have made some sense of it! -jsnx (talk) 07:12, 12 February 2008 (UTC)
thread_title:  Error in paragraph Application to Computer Graphics 

message: When combing physical hair, I think the vector field representing the direction of the hair does not have to be continuous: there may be lines or spots from which the hair is "combed away". In that sense, "you can't comb a hairy ball flat" is not the precise meaning of this theorem. In fact, I guess I can imagine how a hairy ball could be "combed flat" in an intuitive sense with hair combed away from the poles (with the hair exactly at the pole combed in arbitrary direction) and combed along the equator at the equator, so that the vector field of the hair would have two discontinuities (at the poles). Maybe there should be a short note on that in the article.81.20.159.197 13:57, 2 September 2007 (UTC)
thread_title: Hair combing

message: "you can't comb the hair on a billiard ball"
Um, that's trivially true by virtue of the fact that billiard balls don't have hair...
The way I always heard it was "you can't comb a tennis ball without creating a cowlick"  —Preceding unsigned comment added by 71.199.159.232 (talk) 15:00, 14 October 2007 (UTC)
replies: text: "you can't comb the hair on a billiard ball": This statement simply isn't true. It's trivially possible to comb all the hair on a billiard ball in such a way that no tufts, cusps or swirls are generated. This case corresponds with a vanishing tangent vector field. Shinobu (talk) 14:16, 13 January 2008 (UTC)
thread_title:  billiard ball??? 

message: What do you think about this image? It shows that one "pole" is sufficient. There is an animated version of this ball, too: :File:Hairy ball one pole animated.gif. --RokerHRO (talk) 10:35, 26 October 2009 (UTC)
thread_title:  New images 

message: Now what was I thinking about just then? ResMar 04:30, 28 December 2009 (UTC)
thread_title:  Nice name 

message: To make the statement most useful, the article should connect the technical and nontechnical language. I shall attempt to do so, but am unsure, so am posting my attempt here: The flat hairs are the tangent vectors. To comb them is to make the field continuous. The cowlick is a pole, at which the function vanishes. ᛭ LokiClock (talk) 06:38, 9 July 2011 (UTC)
thread_title:  Elaboration on combing statement 

message: I heard there was a way to prove the Hairy Ball Theorem using Green's Theorem. Does anyone know how this would be done?  — Preceding unsigned comment added by 2607:F470:6:4001:B852:5D3F:AF01:405 (talk) 15:33, 26 April 2022 (UTC)
thread_title:  Green Theorem Proof