Another proof that a vector field on the sphere must have a zero?

1. May 20, 2010

lavinia

The proofs I have seen that a vector field on the 2-sphere must have a zero rely on the general theorem that the index of any vector field on a manifold equals the manifold's Euler characteristic.

The tangent circle bundle of the 2 sphere is homeomorphic to real projective 3 space.
A non-zero vector field would be a map from

$$S^2 -> RP^3$$

which has a left inverse, p, where p is just the bundle projection map.

Otherwise put,

p o v = identity on $$S^2$$

But the second real homology of projective 3 space is zero so p o v must equal zero on the fundamental cycle of the 2 sphere. This contradicts the equation p o v = identity.

2. Jan 28, 2011

Landau

I learned this as an easy application of homology theory (used to define the degree of a map S^n->S^n):

Theorem: If f,g:S^n->S^n are continuous maps such that f(x)=/g(x) for all x, then f is homotopic equivalent to -g (i.e. the composition of g and the antipodal map).

Corollary: If f is as above with n even, then there exists x such that f(x)=x or f(x)=-x.
(proof: otherwise f would be homotopic equivalent to both the identity and the antipodal map, so would have degree both equal to 1 and (-1)^{n+1}=-1)

Corollary: Any vector field v on a 2k-sphere must have a zero.
Proof: otherwise v/|v| does not take any x to either x or -x, in contradiction to the previous result.

3. Jan 28, 2011

lavinia

Right. I know this proof. It is lovely.

4. Jan 28, 2011

quasar987

Fun proof lavinia, but could you tell me why the tangent cicle bundle of S^2 is homeomorphic to RP^3?

Thx

5. Jan 29, 2011

mathwonk

my favorite is lefschetz's proof. chose any point on the sphere where the vector is not zero. remove a small disk about that point. a bit of thinking will convince you that the degree of the map defined by the vector field on the boundary of the complementary disc is 2, hence on the complement of the disc around that point, there are 2 zeroes counting multiplicities. this takes a little imagination to visualize it but it is very elementary.

6. Jan 30, 2011

lavinia

SO(3) acts on the sphere by rotations. The derivatives of these rotations define a smooth action of SO(3) on the tangent circle bundle of the 2 sphere. This action transitive and is without fixed points. Thus the tangent circle bundle of the 2 sphere is diffeomorphic to SO(3).

7. Jan 30, 2011

quasar987

I see, thanks.

8. Jan 30, 2011

lavinia

Another proof which I would like to try - maybe we can do it together - would be to split the tangent circle bundle along the equator of the 2 sphere into two solid tori then study the attaching map. This could be figured out from the coordinate transformation z -> 1/z.

I think we would be looking to see that twice the homology class of one of the circles on one bounding torus is sent the identity loop in the other. This would mean that the fundamental group of the tangent circle bundle contains a copy of Z/2Z and that would mean that there could not be a non-zero vector field.

9. Jan 30, 2011

quasar987

That is a very interesing idea but I don't understand the main argument. Does the exstence of a nowhere vanishing vector field on S² implies the contractibility of the tangent circle bundle?

10. Jan 30, 2011

lavinia

Not contractibility - but triviality. Since the sphere is orientable one non-zero vector field automatically gives two by rotating the first 90 degrees in the positive direction.

If the tangent circle bundle of the sphere is trivial, its first homology can not have any torsion.

11. Jan 30, 2011

lavinia

It would also be interesting to see what other manifolds can be obtained from different attaching maps. Certainly the 3 sphere is one since it fibers over the 2 sphere by the Hopf fibration.

12. Jan 31, 2011

lavinia

Here's where I got to.

the coordinate transformation z ->1/z can be written in coordinates as

u = x/(x^2 + y^2) v = -y/(x^2 + y^2)

The Jacobian for points on the unit circle is

d/dx = (y^2 - x^2)d/du - 2xyd/dv

d/dy = 2xyd/du + (y^2 - x^2)d/dv

This is the a rotation through an angle whose sine is 2xy.
Note that if (x,y) rotates around the unit circle once, the image of d/dx circles twice around the unit circle in the (d/du,d/dv) plane.

The tangent circle bundle of the 2 sphere falls into two solid tori that are glued together along their boundary tori above the unit circle. The Jacobian is the gluing map,

The two generators of the fundamental group of the bounding torus in the xy plane are
d/dx is held constant and (x,y,) is held constant. The first is null homotopic in the solid torus, the second is not. If their homology class are a and b respectively ( a is the null homotopic circle) and the corresponding classes in the other solid torus are c and d then the Jacobian maps a to c + 2d. So 2d is null homotopic which seems to says that the first homology of the tangent circle bundle is Z/2Z. Needs more detail.

Last edited: Jan 31, 2011
13. Jan 31, 2011

lavinia

I think I may have a sign error here. I think the signs of 2xy need to be reversed.

I think a similar argument should work for the Hopf fibration. Here though one already knows that the circle bundle is S^3. still the solid tori gluing should show that the resulting manifold is simply connected. I wonder if a is identified with c + d.

14. Feb 1, 2011

lavinia

Here is another thought.

The sphere has a connection of constant postitive curvature,1.

The exterior derivative of the connection one form on the tangent circle bundle equals minus the pull back (under the bundle projection map ) of the volume form of the sphere.

Therefore, if there were a non zero vector field the exterior derivative of the connection one form would pull back to the volume element of the sphere. This is not possible because the volume element of the sphere is not exact.

By the same reasoning the curvature of every connection on the torus must have points where it is zero because the tangent bundle of the torus is trivial.

So it seems that the Gauss Bonnet theorem is overkill if you just want to prove that the sphere has no non-zero vector field. You do not need to know that the index of every vector field is the Euler characterisitic.

Last edited: Feb 1, 2011
15. Feb 6, 2011

lavinia

One can also see the attaching map in complex coordinates as d/dz -> (-1/z^2) d/dw

where w = 1/z are the coordinates around infinity. On the unit circle this is

d/dz -> -(e^-i2a)d/dw where z = e^ia so the double rotation is evident.

Equivalently, choose any tangent vector at the north pole of the sphere and parallel translate it along great circles to a small circle near the south pole.The vector field along the boundary has index 2 around the south pole

16. Feb 7, 2011

lavinia

Continuing the parallel translation all the way to the south pole sweeps out a projective plane whose Z boundary is twice the fiber circle over the south pole. This shows that the tangent circle bundle is not trivial for if it were then no multiple of any fiber would be a boundary.

17. Feb 8, 2011

lavinia

What if I take the vector field that tangent to geodesic flow from the north to the south pole. There are two singularities each with degree one. I think you mean the vector field obtained by parallel translating a fixed vector along great circles.

18. Feb 10, 2011

mathwonk

I do not understand. in your example there are indeed two zeroes counting multiplicities, i.e. you have two zeroes eachg with multipliciy one, and 1+1 = 2.

The argument is to look near any point where the vector is not zero, hence nearby the vectors are all parallel roughly to that one. assume all these vectors are roughly horizontal, and point say to the right. Now the complement of a small disc about that point is also a disc and has a non zero vector at each boundary point.

Then the boundary circle, viewed as the boundary circle of the complementary disc, has both vectors at top and bottom of the disc still parallel. but on the sides, the vector on the left points out of the complementary disc, while on the right side the vector points into the complementary disc. If you think about it, this means the vector field on the boundary circle has revolved 360 degrees on each half circle, hence has winding number 2 on the complementary boundary circle. that implies the sum of the degrees at the zeroes of the vector field on the complementary disc is 2, if there is only a finite number of zeros.

sorry if this is not clear, but it has nothing to do with parallel translation as far as I can see.

19. Feb 10, 2011

zhentil

You can get Z of them, I believe :) Look at the degree of the map from the equator to U(1).

20. Feb 10, 2011

lavinia

I guess I didn't understand. Sorry. I just thought that each zero only winds once on each 2 disc not twice. It will wind twice on the only complementary 2 disc if there is only one zero.
Parallel translation of a single vector would create a single zero with multiplicity 2 and defines a null homotopy of the doubly winding curve.

21. Feb 10, 2011

mathwonk

yes, that's exactly it, the idea is to look at the complement of a disc on which there are NO zeroes. you might enjoy reading lefschetz's better description than mine.

you have to have an argument that works without assuming any zeroes exist.

22. Feb 11, 2011

lavinia

On the other hand, on the boundary of an open ball around a zero, the vector field winds around the surface of a solid torus. It one thinks of the fiber at the zero as the central meridian of the solid torus, then the index of the vector field at the zero is the winding number of the vector field around the meridian.

23. Feb 11, 2011

lavinia

In my example there is only one zero with multiplicity 2.

ok

This is confusing. Can you explain it again?

agreed

24. Feb 11, 2011

lavinia

As this thread unfolds I am learning more that I originally hoped for about the 2 sphere.

I just read a proof in Milner's Topology from the Differentiable Viewpoint that the sum of the indices of a vector field on a manifold is the same for all vector fields.

He embeds the manifold in Euclidean space and the extends the vector field to a vector field on a tubular neighborhood of the embedded manifold that points outward along the boundary and has the same zeros with the same indexes as the original vector field. The construction is straight forward.

By cutting out small balls around the zeros then normalizing the vector field to have unit length, he gets a map of the tube minus the open balls into the unit sphere.

The degree of this map on the boundary must be zero so the sum of the indices of the vector field equals the degree of the map of the boundary of the tube into the sphere.

For a hypersurface of Euclidean space, the tube has two components and the map into the sphere on each component has the same degree. This means that the Euler charateristic of a manifold that can be embedded as a hypersurface of Euclidean space is even. So every orientable 2 dimensional surface has even Euler characteristic. For the 2 sphere these maps have degree 2.

25. Feb 12, 2011

mathwonk

When I said you had two zeroes, each of degree one, I was quoting your own statement in #17.

Cut out a small disc from the sphere, centered at a point where there is a non zero tangent vector, and on which all vectors are roughly horizontal, pointing to the right. Now look at the complement, a large sphere with a small missing disc.

In order to view that large punctured sphere as a disc, think of it as rubber, and let the rubber contract. The large outside part of the sphere snaps into place inside the small boundary circle. As it does this, some of the vectors on the boundary circle, in order to remain tangent to the surface flip over.

The vectors at top and bottom stay fixed pointing right, but the vectors at 3 o'clock and 9 o'clock flip 180 degrees and now point left. The other vectors flip differently according to the angle they make with the boundary circle. I.e. we are "reflecting" the part of the sphere outside the boundary circle to the inside. So vectors tangent to the boundary circle are left fixed, and the others are reflected in the tangent line to the boundary circle.

Thus the more perpendicular they are to the boundary circle the more they change direction. Since the angle between the previously horizontal vectors and the tangent lines to the circle, go through two full cycles as you go around the circle, the winding number, i.e. the total degree, of the vector field on the boundary circle of the new disc, is two. Hence either there is an infinite number of zeroes inside the new disc, or there are a finite number with the sum of the degrees equaling 2.

Either way there is a zero inside the new disc, i.e. outside the old disc.