A Ref: "Standard" Action of S^1 on S^n ?

[mathwonk]

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.

 

[lavinia]

@WWGD @mathwonk Another proof of the hairy ball theorem, one which is suggestive of further insight but is pretty opaque, is based on the second homology and second homotopy groups of the unit tangent circle bundle. Using a Riemannian metric an everywhere non-zero section can be adjusted to have norm 1 at all points and becomes a section of the tangent circle bundle. If one knows that this bundle is diffeomorphic to $SO(3)$ then one can use either that the second cohomology group of $SO(3)$ with real coefficients is zero, or that $π_2(SO(3))=0$ the second homotopy group of $SO(3)$ is zero. The cohomology argument is just that the pull back $π^{*}(V)$ of the volume element of the 2 sphere to $SO(3)$ is a coboundary $dF$ since $H^2(SO(3);R) = 0$. If there were a non-zero section then the volume element $V$ would equal $s^{*}(dF)$ and would be an exact 2 form.

 

[mathwonk]

so i presume the fact that composing a section with projection equals the identity map of the sphere contradicts the fact that the section is homotopic to zero? of course, using the sequence of maps from π2 sphere to π2 O(3) to π2sphere? of course this is an ignorant question, but is it possible this fact that π2(O(3)) = 0, is more difficult than (or equivalent to) the fact we are trying to deduce?

 

[lavinia]

so i presume the fact that composing a section with projection equals the identity map of the sphere contradicts the fact that the section is homotopic to zero? of course, using the sequence of maps from π2 sphere to π2 O(3) to π2sphere? of course this is an ignorant question, but is it possible this fact that π2(O(3)) = 0, is more difficult than (or equivalent to) the fact we are trying to deduce?

I would think so. It follows from the homotopy exact sequence of a fibration. But that is a lot of machinery.

If one views $SO(3)$ as a three dimensional ball with antipodal points on its boundary sphere identified then an embedded 2 sphere can be shrunk to any point in the interior of the ball this because it cannot touch two antipodal points on the boundary sphere. This eliminates the homotopy sequence of the fibration.

For the Hopf fibration a non-zero section is null homotopic because it is not surjective and the sphere minus a point is contractible. The identity map on the 2 sphere is not null homotopic since its degree is 1. This seems to decrease the amount of machinery in this case.

And this seems to be a little more general than just the Hopf fibration. The argument says that the identity map on the 2 sphere can not be expressed as a composition $S^2→S^3→S^2$. For an embedded $S^2$ in $S^3$ this says that the 2 sphere is not a retract of $S^3$.

 

[lavinia]

Now being guided by the case of riemann surfaces, but without reading anything from Milnor - Stasheff, I conjecture that the method of passing from a bundle to a map to a grassmannian can be done as follows: Given an n plane bundle E over a space X, we want to map X into a grassmannian of n planes in some larger r diml euclidean space.

Consider the vector space of all global sections of E, presumably an infinite dimensional vector space. In that space choose (using a partition of unity to construct them) a sequence s1,...,sr of global sections such that at each point p of X, the values of these sections s1(p),...,sr(p), span the space Ep, the fiber of the bundle at p.

Let V = the r dimensional vector subspace they span in the infinite dimensional space of all sections. Now for each point p, we have a map V-->Ep, evaluation at p. taking each section s to its value in Ep. By construction of the space V, these maps are all surjective, hence all have kernels of dimension r-n. Thus if we send each point p of X to the kernel of the evaluation map V-->Ep, we get a map from X to the grassmannian of r-n planes in V, i.e. in r space. So we have a map X-->Gr(r-n,r) ≈ Gr(n,r). That should do it, but probably Milnor makes it look more elementary and concrete than that.

Ok I looked at Milnor and indeed he makes it look more concrete and elementary but in fact it is almost the same construction as I gave, but where I said one can use partitions of unity to construct certain sections, he explicitly uses the partitions of unity to write down local trivializations, which is how I would construct my sections, and then he defines an explicit bundle map into euclidean space by them. So its essentially the same but looks completely different and more concrete but less intrinsic. It's wonderful how much easier it is to read something when you have already come up with the idea on your own.

If for each local trivialization one projects onto $R^{r}$ then this gives a linear map on each fiber into $R^{r}$. On two overlapping domains, one gets two linear maps of the fiber into $R^{r}$. What do you do with them?

 

[mathwonk]

Milnor takes the direct sum of the maps, summing over open sets of the cover. This is similar to my taking the space of global sections spanned by the ones constructed locally.

And yes now that I understand your argument with SO(3), I like it a lot. Indeed Artin explains in his algebra book the story of the double cover of SO(3) by SU(2) ≈ S^3, hence SO(3) ≈ RP^3 has the same π2 as S^3, namely zero. So this seems more elementary now.

This may be nuts, but it seems that one could start from the Hopf fibration and construct a sequence of other circle bundles over S^2 just by identifying opposite points of each circle. The first such construction gives RP^3 as a circle bundle over S^2, i.e. SO(3). Does indeed Z/2 act again on SO(3)? I guess you would need some metric to define "opposite". The minus map seems to act on SU(2) but it changes the sign of the determinant in O(3). So this seems fishy. Indeed this seems unlikely. The circles are cosets of the subgroup of diagonal matrices in SU(2), but although the minus map seems to act on all such cosets, other multipliers by non real numbers of norm one, e.g. i, probably do not.

... time out for research:
Well what do you know, Steenrod says there is such an action of Z/n on S^3 for every n >0, and the resulting quotient spaces are all ≈ S^2, and the ones we have seen correspond to n = 1,2. The others are called "lens spaces".

oh yes, a coset of the diagonal subgroup is an orbit of the action of the subgroup by translation. within any such orbit we can consider the orbit of the action of the subgroup of diagonal matrices whose diagonal entries are nth roots of 1. I guess this works. whoopee! but I never was too good at algebra. (I actually wrote something down and now it looks better. Thats why algebra is so hard for me, I hate to write stuff down, hence losing entirely the benefit of algebraic computation. I.e. visibly, a subgroup does act on its left cosets by right multiplication.)

 

[lavinia]

Milnor takes the direct sum of the maps, summing over open sets of the cover. This is similar to my taking the space of global sections spanned by the ones constructed locally.

I see now. very nice.

 

[mathwonk]

yes but somehow my approach is dual to his. also in algebraic geometry you can choose a basis s1,...,sr for the space V of global sections of a line bundle, and define a projective mapping to P(V) (=lines in V) using the homogeneous coordinates [s1,...,sr], or you can do it more intrinsically, and map a point to the hyperplane in V of all those sections that vanish at a point, getting a mapping instead to P*(V). These things always confuse me. Of course I guess a choice of basis always defines an isomorphism of a space with its dual. But it seems as if the dual approach, mapping to a hyperplane, avoids choice of a basis, hence seems more intrinsic, but more abstract.

 

[mathwonk]

Remark: The Hopf action given in Artin, Algebra, for S^1 acting on S^3, is almost the same as infrared's double rotation of the pair of complex numbers representing a point of S^3, except Artin conjugates the second complex multiplier. I.e. representing a point of S^3 as a complex 2x2 matrix in SU(2) with rows [a b], [-b* a*], where * denotes complex conjugation, he let's S^1 be the diagonal matrices in SU(2) with rows [c 0], [0 c*], acting by matrix multiplication. This seems to rotate the first complex number a by an angle t = arg(c) and the second one by -t = arg(c*).

I wonder what the quotient space of S^3 is by infrared's action? Is it also S^2? Is it perhaps the S^1 bundle over S^2 associated to the integer -1 rather than 1? or vice versa? It says in Steenrod that these two bundles (the ones corresponding to n and -n, in π1(S^1) ≈ Z) are not equivalent under SO(3) but that they are equivalent under O(3).

...later: well using the exact homotopy sequence of a circle bundle with total space S^3, it seems the base space must be a simply connected surface with π2 ≈ Z, so presumably the quotient space is indeed S^2. ?

 

[lavinia]

This is how I think the two different actions of the circle on $S^3$ are related to the Chern class and to the Hopf fibration - up to mistakes.

The action of $S^1$ on the three sphere can be defined in two opposite ways - in @Infrared's way - multiplication of coordinates by $e^{iθ}$ or in the opposite way - multiplication of coordinates by $e^{-iθ}$. These are the restrictions to $S^3$ of two different complex structures on the canonical complex line bundle over the 2 sphere. The quotient space by these two actions is $S^2$. In both cases this is the Hopf fibration. The difference is in the complex structure not the bundle.

Given a complex structure on a complex vector bundle, there is always another structure called the conjugate structure. It replaces scalar multiplication by a complex number by multiplication by its complex conjugate. Multiplication of the 3 sphere by $e^{- iθ}$ instead of by $e^{iθ}$ is just the restriction of the conjugate structure on the canonical line bundle to the 3 sphere.

The Chern class of a complex line bundle is the Euler class of the underlying real two plane bundle when given a naturally determined orientation. (Euler classes are defined for oriented bundles and the same bundle with opposite orientation has the negative Euler class.)

As Milnor describes, a basis for a complex vector space $x_1 ...x_{n}$ naturally determines an orientation of the underlying real vector space as $(x_1,ix_1, ... x_{j},ix_{j}, ... x_{n},ix_{n})$. For a complex line this is $(x_1,ix_1)$. For the conjugate structure this is $(x_1,-ix_1)$. The orientation is independent of the choice of basis.

For line bundles, a complex structure and its conjugate have oppostite orientation so their Euler/Chern classes are negatives of each other. While the underlying real bundles are isomorphic, the complex bundles are not since the identity map is conjugate linear rather than complex linear.

More generally, the conjugate action on an odd dimensional complex vector space determines the opposite orientation, and on an even dimensional vector space the same orientation.

 

[mathwonk]

That is very interesting and I have not grasped it fully yet. I wonder if I misunderstood something. Rather than the choice of multiplication by e^it or e^-it, it seems that Artin's action does both, i.e. infrareds action has two multiplications, and Artin seems to use e^it as one multiplication but e^-it as then other. So is it a mixture of the two cases you discuss?

I.e. Artin's apparently sends (a,b) to (a.e^it, be^-it). I assume this just by multiplying the matrices he gives on pages 272 and 275, in equations (2.4) and (2.13). I must be misunderstanding something.

well i just noticed it depends on whether i let my matrix act from the left or fom the right! one sends (a,b) to (a.e^it, b.e^it) and the other gives what I wrote 3 lines above. Obviously I need to do some calculations and stop just speculating.

 

[WWGD]

@WWGD: I added an explicit link now, in post #41, that (may) take a lot less time than reading the book, or not. The point is if you know the (total) chern classes of projective space, then you can compute the chern classes of hypersurfaces in projective space too.

If you have the inclination, you might like my notes on RRT there. I have enjoyed studying it all my career and tried to lay out all I learned there, admittedly not everything there is. Well honestly I only studied the curve case mostly all those years, but did enjoy working up these notes on the surface case and 3-fold case.

Seems link is dead, leads to a forbidden error 403

 

[mathwonk]

@WWGD: does this work?https://www.math.uga.edu/sites/default/files/inline-files/rrt.pdf

 

[WWGD]

@WWGD: does this work?https://www.math.uga.edu/sites/default/files/inline-files/rrt.pdf

Yes, thanks so much, Wonk.

 

[mathwonk]

my pleasure. it seems our webpages get rearranged occasionally, i.e. every ten years or so.