Hairy Ball Theorem

[Imaginary Number]

I've heard of this mathematical theorem but could someone explain it to me, in very, very short words and simple sentences?

[ajb]

Take a look at Wiki.

 

It states that there are no non-vanishing continuous vector fields on n-spheres for n even. It is a consequence of the tangent bundle of even dimensional spheres are non-trivial. It has no non-zero global sections.

 

If you try to comb a "hairy ball" you will always get a tuft.

Edited October 3, 2010 by ajb

[Imaginary Number]

Take a look at Wiki.

 

It states that there are no non-vanishing continuous vector fields on n-spheres for n even. It is a consequence of the tangent bundle of even dimensional spheres are non-trivial. It has no non-zero global sections.

 

If you try to comb a "hairy ball" you will always get a tuft.

 

What do you mean by "non-zero global sections?"

What are they?

[ajb]

Look up vector bundles.

 

In this context we mean no non-zero globally defined vectors.

 

In other words, there does not exist an everywhere non-zero tangent vector field on even dimensional spheres. That is at some point(s) on the sphere the vector field will vanish.

[Imaginary Number]

Look up vector bundles.

 

In this context we mean no non-zero globally defined vectors.

 

In other words, there does not exist an everywhere non-zero tangent vector field on even dimensional spheres. That is at some point(s) on the sphere the vector field will vanish.

 

 

So if I am understanding you correctly: There has to be somewhere on the sphere where one of the vectors is zero. Is that right?

I know that a vector is magnitude AND direction, so is the magnitude zero and the direction also zero?

[ajb]

So if I am understanding you correctly: There has to be somewhere on the sphere where one of the vectors is zero. Is that right?

I know that a vector is magnitude AND direction, so is the magnitude zero and the direction also zero?

 

A vector field can be thought of as the assignment of a vector to every point on a manifold in a smooth way.

 

 

The hairy ball theorem says that every vector field on an even dimensional sphere will vanishing at least at one point.

[the tree]

I noticed from the graphics on the wiki that a combed torus was 'quite easy' but saw no expansion on that point. Do you know if other surfaces have been explored in much detail? I'd guess that not every n-torus could be combed.

[ajb]

I noticed from the graphics on the wiki that a combed torus was 'quite easy' but saw no expansion on that point. Do you know if other surfaces have been explored in much detail? I'd guess that not every n-torus could be combed.

 

You need the Euler characteristic [math]\chi[/math] of the surface to be zero in order to "comb the hair flat". The number of possible points in which the "hair" would stick up us given by this number.

 

The Euler characteristic of a closed orientable surface is given by its genus ("number of holes"):

 

[math]\chi = 2-2g[/math].

 

So for [math]g = 1[/math] i.e. the torus we have no problem points.

 

 

The problem here is that the hairy ball theorem does not generalise to surfaces of arbitrary higher genus (>2). However, I am sure you can tackle this question though.

[the tree]

Okay, I don't really know the language to describe this (like, at all). But if you had a dougnut where the hairs were say, pointing down on the inside and up on the outside - that should be well combed.

 

Then another dougnut where the hairs were the other way around, the two could be intersected and then you'd have a surface with genus 2 that would still be well combed (does that make any amount of sense?).

 

If any of that made sense then it should be possible to iterate that action for any genus. Am I at all on the right track?

[ajb]

I think you mean the connected sum. You can do this in the category of smooth manifolds, so thinking about vector fields here is fine.

 

I'd have to think about the hairy tori theorem

[Imaginary Number]

To clear this up, lets say that I have a furry sphere with all the hairs combed in the same direction, almost all, will I have a vector that looks like (0)

(0)

 

Actually would the same rules apply to "four vectors?"

Edited October 18, 2010 by Imaginary Number

[the tree]

@ajb, yeah that's what I meant.

@im.no. according to the wiki article, the theorem applies to 2n dimensional sphere, so no.