# Parallelizability vs. hairy ball theorem

1. Apr 21, 2008

### mma

Usually the hairy ball theorem is cited for proving that $$S^2$$ is not parallelizable. However, hairy ball theorem is too strong for this. This theorem states that there isn't a nowhere vanishing continuous vector field on $$S^2$$. Unparalellizable property means only that there aren't two linearly independent vector fields on $$S^2$$. Could somebody tell an example of a nonparallelizable n-dimensional manifold on which hairy ball theorem is false, i.e. on which there is continuous nowhere vanishing vector field (but because of the nonparallelizability, n linearly independent aren't)?

2. Apr 22, 2008

### mathwonk

isn't that trivial? (take products.)

3. Apr 22, 2008

### mma

Yes it's trivial. Sorry for the stupid question and thanks.

4. Apr 24, 2008

### mathwonk

sorry for the smartass answer. i have been enjoying slinging around the word "trivial" for a few weeks.

and is it really trivial? is it obvious the product of a torus and a sphere is not parallelizable?

5. Apr 25, 2008

### mma

Yes, it seems to be obvious. If it were parallelizable then there would be 4 continuous vector fields having 4 linearly independent vectors above each (t,s) point of the product. These vectors would span the tangent space of the product space above (t,s). But this tangent space is a pair of tangent spaces: $$(T_t,T_s)$$, $$T_t$$ is of the torus above t and $$T_s$$ of the sphere above s. So the 4 vectors would span both $$T_t$$ and $$T_s$$. But this would contradict to the unparallelizable property of the sphere.

Last edited: Apr 25, 2008
6. Apr 25, 2008

### mathwonk

dont you need just two of them to span the tangent space of the sphere at every point, to get that?

7. Apr 26, 2008

### mma

I'm afraid that you are kidding. Yes, 2 from $$T_s$$ is enough to span $$T_s$$. Two from $$T_p$$ don't span $$T_s$$. And one from $$T_p$$ and one from $$T_s$$ also don't. But 2 from $$T_s$$ we don't have at each point because of the Hedgehog Theorem.

8. Apr 26, 2008

### mma

Sorry, I missed the indexes. p means t (a point of the torus)

9. Apr 26, 2008

### gel

Well, the Klein bottle and Mobius strip are not parallelizable, because they aren't orientable. However it is not too hard to find a nowhere vanishing vector field.

10. Apr 26, 2008

### gel

Also, Sn has a nowhere vanishing vector field whenever n is odd.
To construct one, consider Sn as the elements of Rn+1 with norm 1. Let A be an antisymmeric (n+1)x(n+1) non-singular matrix. As n+1 is even, this can be done by building it out of (n+1)/2 copies of the matrix
$$\left( \begin{array}{rr} 0 & 1 \\ -1 & 0 \end{array} \right).$$
Then, the vector field is v = Ax at each point x of Sn.

However, except for the cases n=1,3,7, Sn is not parallelizable for any odd n (mentioned http://en.wikipedia.org/wiki/Parallelizable" [Broken], although I expect that it is very difficult to prove).

Last edited by a moderator: May 3, 2017
11. Apr 26, 2008

### gel

It is parallelizable.
torus x sphere = circle x circle x sphere.
And, the product of a circle and sphere is parallelizable, which you can see by looking at it as a quotient space of R3-{0}

12. Apr 26, 2008

### gel

the what?

13. Apr 26, 2008

### mma

Hedgehog Theorem = Hairy Ball Theorem.

By the way, what error do you see in my proof?

14. Apr 29, 2008

### mma

Then, finally, is the product of a sphere and a torus parallelizable or not?

15. Apr 29, 2008

### gel

Yes. That's what I was saying in my previous post.

16. Apr 30, 2008

### mma

Yes, but on the one hand I didn't understand that (I don't know what quotient space do you mean), and on the other hand you didn't show the error in my proof.

Perhaps I am wrong and the continuous vector fields of a product space aren't the Cartesian products of the continuous vector fields of the component spaces?

17. Apr 30, 2008

### gel

That's right, but I didn't follow how that was supposed to imply that the component spaces are parallelizable.
If you project linearly independent vectors onto a subspace, they no longer have to be linearly independent or even non-zero.

18. May 1, 2008

### mma

Maybe it's trivial but I don't know what qoutient space dou you mean. Could you explain this or give a reference for the explanation?

19. May 1, 2008

### gel

The circle can be expressed as the real line R quotiented out by the group of translations x |-> x+n (for integer n). That is, you identify x,y in R if x-y is an integer. This expresses R as the covering space of the circle, S1.

You can map R x S2 to R3-{0} by (x,y) |-> exp(x) y, which takes the group of translations of R to the group x |-> exp(n) x on R3-{0}, for integer n.

So, S1 x S2 is diffeomorphic to R3-{0} with points x, exp(n)x identified (integer n).

If ei is a basis for R3 as a vector space, then you have the basis vi(x) = |x| ei for the tangent space of R3-{0} at each point x. This is invariant under the action of the group x |-> exp(n)x, so it maps to a basis for the quotient space S1 x S2.

20. May 2, 2008

### mathwonk

the point is that there are a lot of vectors tangent to the product which are not tangent to either factor.

i.e. consider a family of vectors tangent to the product R^3, and project those vectors onto the sphere. notice that some of these vectors project to zero on the sphere.

Last edited: May 2, 2008
21. May 6, 2008

### mma

Now I understand. On one hand R x S2 is clearly diffeomorfic to R3-{0} bacause the radial half-lines are diffeomorfic to R. On the other hand the x=x+n equivalence relation on R is mapped to an equivalence relation x=exp(n)x equivalence relation on this half-lines, and this generates an equivalence relation on R x S2 that leaves the given basises unchanged. Nice!Thank you!

22. May 7, 2008

### mma

Yes, I see already. The pont is that we can give 3 continuous vector fields on the sphere such that they are generators of the tangent space at each pont, but we can't select 2 of them (from the vector fields) bacause each of them vanishes somewhere (because of the hairy ball theorem). But at such points the vectors of the remaining two can be the basis. Such 3 vector fields can be e.g. the projection of the natural basis of the tangent space of R3 (i.e. of the tangents of the Cartesian coordinate lines of a R3) on the tangent planes of the sphere .

What I still don't understand that why did you mentioned product spaces in post #2. Was it only a joke?

23. May 18, 2008

### mathwonk

i dont remember either but i had the wrong idea at first about this. the mobius band example however shows that a twisted product does give a correct counterexample.

24. May 19, 2008

### mma

Yes, Möbius band is exactly the example what I wanted. But I've never heard the "twisted product" phrase before. I know that if the Möbius band is embedded in R^3 then there is a twist on the strip (we must twist a paper strip before we glue together its ends when making a Möbius band from a paper strip), but I'm afraid that the exact notion of twisted product would be too sophisticated for me. Really I don't understand even the difference between the product topology amd box topology. So it would be a real challenge to explain it for me. (I hope that you like challenges

Last edited: May 19, 2008
25. May 19, 2008

### mathwonk

well it is a union of products, i.e. of rectangles, but the rectangles are twisted before the final gluing, just as intuition suggests.

in fact all tangent bundles are twisted products, since they are unions of tangent bundles of coordinate charts, and the tangent bundle of a coordinate chart is a product.