a continuous vector field of tangent vectors on the 2 sphere is just a family of tangent vectors to the sphere, one based at each point of the sphere, so that at nearby points the vectors have almost the same length and direction, except when a vector is zero at a point, then nearby vectors only have to have length near zero, but can have any direction.

my favorite proof that in any continuous tangent vector field on the sphere, some point must have the zero vector, takes a little effort to visualize. You prove first that if you look at a small circle on the sphere, such that no point of the circle has zero vector, then there is a map from that circle to itself defined as follows: given a point of the circle, look at the non zero tangent vector based at that point and shrink it down to have length equal to the radius of the circle, and then parallel translate it to have base at the center of the circle, so that its head now points to a (usually new) point of the circle. E.g. if the vector at the center of the circle is non zero, and the circle is very small, then all vectors at all points of the circle, by continuity, will be nearly the same lemngth and direction as that vector at the center, so this map will take every point of the circle very close to the point on the circle where that center vector points to. In particular the map will miss most points of the circle.

On the other hand, if the vector at the center of the circle is zero, and if the vectors on the circle are arranged radially, i.e. at each point of the circle the tangent vector based at that point is parallel; to the radius of the circle to that point, then the map of the circle to itself defined by these vectors will be th identity, i.e. it will wrap the circle exactly once aroiund itself. In general the wrapping number of this map will equal the number of points inside the circle where the tangent vector based there is zero, but counted with "orientation", i.e. the map can wrap the circle around itself in either the positive or the negatyive direction and if both directiopns occur the wrapping numbers will cancel. Nonetheless oif the wrappiong number is not zero, then there must be some zero tangent vectors based at some points inside the circle.

Ok assuming that, look at a point on the sphere where the tangent vector based there is non zero. if no such points exist of course the vector field is zero everywhere and the theorem is proved. Now draw a small circle around that point and note that all vectors at all points of that circle will popint in essentially the same direction. Thus the wrapping number of the map is zero and indeed there are no zero vectors inside that circle. But we claim then there must be some zero vectors, in fact 2 of them outside the circle. To see that visually imagine the sphere made of rubber and let the outside of the circle snap back inside the circle. Eventually you can see that this changes the directiomns of the tangent vectors based at points of the circle. I.e. those vectors athat are actually tangent not jiust to the sphere biut to th ircle, will not have their directions change, new resulting map of the circle to itself will have wrapping number 2.

This is a proof from a book by the great intuitive topologist Solomon Lefschetz.

There are also proofs based on the stokes theorem, showing that a never zero smooth tangent vector field would define a smooth homotopy from the identity map of the sphere to the antipodal map, but by integrating a spherical angle form, this is impossible. Then I suppose you have to use some smoothing approximation argument to get it for the continuous case.

Of course if you knew the total oriented number of zeroes of any tangent vector field must equal the euler number, you could just compute that as V-E+F = 2. That is sometimes proved by showing first that all general vector fields have the same number of zeroes, and then constructing one nice vector field suited to a given triangulation, having a zero vector at the center of each cell, and orientation equal to (-1)^(dim of the cell). for this see Milnors delighful little book on differential topology, or for perhaps less crystal clear insight, but more elementary step by step arguments, see Guillemin and Pollack's book on the same subject.

But to respond to your question on computable characteristic classes, the most computable one is perhaps this euler class of a surface, namely V-E+F. I always like to point out that this class was computed by riemann when he showed the euler number of the cotangent, i.e. "canonical", bundle of a riemann surface is 2g-2, hence the (usual) euler number of the tangent bundle is 2-2g. This may have been the first theorem counting zeroes of sections of bundles using topological invariants. Of course there was Gauss Bonnet, presumably related.

my favorite proof that in any continuous tangent vector field on the sphere, some point must have the zero vector, takes a little effort to visualize. You prove first that if you look at a small circle on the sphere, such that no point of the circle has zero vector, then there is a map from that circle to itself defined as follows: given a point of the circle, look at the non zero tangent vector based at that point and shrink it down to have length equal to the radius of the circle, and then parallel translate it to have base at the center of the circle, so that its head now points to a (usually new) point of the circle. E.g. if the vector at the center of the circle is non zero, and the circle is very small, then all vectors at all points of the circle, by continuity, will be nearly the same lemngth and direction as that vector at the center, so this map will take every point of the circle very close to the point on the circle where that center vector points to. In particular the map will miss most points of the circle.

On the other hand, if the vector at the center of the circle is zero, and if the vectors on the circle are arranged radially, i.e. at each point of the circle the tangent vector based at that point is parallel; to the radius of the circle to that point, then the map of the circle to itself defined by these vectors will be th identity, i.e. it will wrap the circle exactly once aroiund itself. In general the wrapping number of this map will equal the number of points inside the circle where the tangent vector based there is zero, but counted with "orientation", i.e. the map can wrap the circle around itself in either the positive or the negatyive direction and if both directiopns occur the wrapping numbers will cancel. Nonetheless oif the wrappiong number is not zero, then there must be some zero tangent vectors based at some points inside the circle.

Ok assuming that, look at a point on the sphere where the tangent vector based there is non zero. if no such points exist of course the vector field is zero everywhere and the theorem is proved. Now draw a small circle around that point and note that all vectors at all points of that circle will popint in essentially the same direction. Thus the wrapping number of the map is zero and indeed there are no zero vectors inside that circle. But we claim then there must be some zero vectors, in fact 2 of them outside the circle. To see that visually imagine the sphere made of rubber and let the outside of the circle snap back inside the circle. Eventually you can see that this changes the directiomns of the tangent vectors based at points of the circle. I.e. those vectors athat are actually tangent not jiust to the sphere biut to th ircle, will not have their directions change, new resulting map of the circle to itself will have wrapping number 2.

This is a proof from a book by the great intuitive topologist Solomon Lefschetz.

There are also proofs based on the stokes theorem, showing that a never zero smooth tangent vector field would define a smooth homotopy from the identity map of the sphere to the antipodal map, but by integrating a spherical angle form, this is impossible. Then I suppose you have to use some smoothing approximation argument to get it for the continuous case.

Of course if you knew the total oriented number of zeroes of any tangent vector field must equal the euler number, you could just compute that as V-E+F = 2. That is sometimes proved by showing first that all general vector fields have the same number of zeroes, and then constructing one nice vector field suited to a given triangulation, having a zero vector at the center of each cell, and orientation equal to (-1)^(dim of the cell). for this see Milnors delighful little book on differential topology, or for perhaps less crystal clear insight, but more elementary step by step arguments, see Guillemin and Pollack's book on the same subject.

But to respond to your question on computable characteristic classes, the most computable one is perhaps this euler class of a surface, namely V-E+F. I always like to point out that this class was computed by riemann when he showed the euler number of the cotangent, i.e. "canonical", bundle of a riemann surface is 2g-2, hence the (usual) euler number of the tangent bundle is 2-2g. This may have been the first theorem counting zeroes of sections of bundles using topological invariants. Of course there was Gauss Bonnet, presumably related.

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