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

[lavinia]

@WWGD @mathwonk Stiefel-Whitney classes are the same for homotopy equivalent manifolds.

One can retrieve the Stiefel-Whitney classes from Poincare Duality and the Steenrod algebra and nothing else.
Using Wu classes Stiefel-Whitney classes generalize to all topological manifolds - in fact to spaces that satisfy Poincare Duality.

It follows that under a homotopy equivalence, Stiefel-Whitney classes are mapped bijectively onto each other. The homotopy equivalence need not even be differentiable.(A homotopy equivalence is a pair of mappings $f:M→N$ and $g:N→M$ such that $f \circ g$ and $g \circ f$ are homotopic to the identity map.)

This does not mean that the tangent bundles of two different differentiable structures are bundle isomorphic but I do not know of examples in the case of compact manifolds.

The example of Euler characteristic illustrates the topological invariance of Stiefel-Whitney classes. The top Stiefel-Whitney class evaluated on the fundamental mod 2 cycle is the Euler characteristic mod 2. The Euler characteristic can be calculated from a triangulation or from the ranks of the integer homology groups.

More generally, the Poincare dual of the $i$ 'th Stiefel-Whitney class is the sum of the $n-i$ dimensional simplicies of the first barycentric subdivision of the triangulation.

Here is how one can find the first Stiefel-Whitney class of a smooth manifold without using bundle theory. If one triangulates the manifold (every smooth manifold has a triangulation), one can try to write the fundamental top dimensional cycle as a signed sum of $n$ simplicies choosing signs so that faces of adjacent $n$ simplicies cancel out under the boundary operator. If this works then the manifold is orientable. If not, the manifold is not orientable and some of the faces are counted twice rather than cancelling out. Forgetting signs the doubly counted $n-1$ faces sum up to form a simplicial $n-1$ chain(counting each face only once) and this chain represents an obstruction to orienting the manifold. It is a mod 2 homology cycle (which is not hard to prove) and its Poincare dual is the first Stiefel-Whitney class of the manifold.

Sadly I used Spanier's Algebraic Topology to learn most of these theorems. But I imagine Hatcher's book is more readable and also more modern. For Wu classes a little web searching will produce PDF files with good explanations - or if you are brave, there is Milnor's Characteristic Classes.

 

[mathwonk]

thank you lavinia. if the SW classes are indeed defined for any topological manifold, do they again give necessary and sufficient conditions for a topological manifold to be a boundary? (my google searches have suggested that the signature and the kirby - siebenman invariants are relevant for this, and i have only found results in low dimensions.)

 

[WWGD]

Sorry if this is too far off, but the question on whether a space was isomorphic to a product came about with the question on whether $\mathbb R^{2n+1}$ was a perfect square * , i.e., whether there was some topological space Z so that the product $Z \times Z$ is isomorphic to $\mathbb R^{2n+1}$ ( obviously, $\mathbb R^{2n}$ is iso to $\mathbb R^n \times \mathbb R^n$) The answer is no, not too complicated, using the degree of a map. But I don't remember it now. A Dutchman gave the answer; they are big in pointset and general topology (Think Brouwer)

*Which itself came out of a joke

 

[mathwonk]

At least for CW complexes Z, the (covering) dimension of ZxZ seems to be even, where as the (covering) dimension of R^(2n+1) is presumably odd.

https://projecteuclid.org/download/pdf_1/euclid.tkbjm/1496158375

 

[lavinia]

thank you lavinia. if the SW classes are indeed defined for any topological manifold, do they again give necessary and sufficient conditions for a topological manifold to be a boundary? (my google searches have suggested that the signature and the kirby - siebenman invariants are relevant for this, and i have only found results in low dimensions.)

I don't know cobordism theory well except for smooth manifolds. My guess is that topologically defined Stiefel-Whitney classes do not determine the topological cobordism class.

 

[mathwonk]

I rather like Spanier, but have not read too much of it. Bott recommended it to us in 1964, but that was about all there was out there that was fairly comprehensive. Like Spanier, my later teacher Ed Brown Jr. started us off with simplicial complexes, which I think is a good way to get a hands on feel for algebraic topology. Notice most of us still use some triangulation reasoning when trying to give an elementary calculation or explanation. E.g. the double cover of projective space by the sphere, plus a triangulation, gives the fact that the euler characteristic of projective space is 1 in even dimensions. One also gets quickly the very useful Hurwitz formula for euler characteristics of branched covers of riemann surfaces.

 

[WWGD]

Any chance you can do one-or-two basic mptations of chern or other characteristic classes? I am curious . Sering computations make the seem more approachable.

 

[lavinia]

I rather like Spanier, but have not read too much of it. Bott recommended it to us in 1964, but that was about all there was out there that was fairly comprehensive. Like Spanier, my later teacher Ed Brown Jr. started us off with simplicial complexes, which I think is a good way to get a hands on feel for algebraic topology. Notice most of us still use some triangulation reasoning when trying to give an elementary calculation or explanation. E.g. the double cover of projective space by the sphere, plus a triangulation, gives the fact that the euler characteristic of projective space is 1 in even dimensions. On also gets quickly the very useful Hurwitz formula for euler characteristics of branched covers of riemann surfaces.

You studied with some of the immortals.

I like Spanier but it is dense and abstract. It cures what ails you but you are still left with a headache.

 

[lavinia]

Any chance you can do one-or-two basic mptations of chern or other characteristic classes? I am curious . Sering computations make the seem more approachable.

Sure but what are you looking for?

 

[WWGD]

Sure but what are you looking for?

Anything that is not actually trivial but not too demanding, if it is possible to find that sweet spot.

 

[mathwonk]

@WWGD I do not have lavinia's expertise, but when I needed some chern class computations I found a useful formula in bott-tu at least for hyper surfaces, which i used for my computation of chern classes for surfaces in complex projective 3 space and maybe 3-folds in 4 space. This was included in my notes on riemann roch theorems, available on my website at uga math dept. i will look for a more precise link.

OK look at page 50 of these notes:
http://alpha.math.uga.edu/%7Eroy/rrt.pdfThe point is to use Whitney's formula for the chern class of a direct sum. Then we have that the direct sum of the tangent bundle of a hypersurface with its normal (line) bundle (which is known), equals the tangent bundle of the ambient projective space, which is also known. So by the 3 term principle, we have an equation involving 3 terms of which we know 2 of them. So we can solve for the chern class of a hypersurface in terms of the chern classes of projective space and of the normal bundle to a hypersurface. These "known" guys have chern classes expressible in terms of the class h dual to a hyperplane, namely that of P^r is (1+h)^r, and that of the normal bundle to a hypersurface of degree n is (1+nh).

Thus the (total) chern class of the hypersurface of degree n in P^r equals
(1+h)^r.(1+nh)^(-1) = (1+h)^r.(1 -nh + n^2h^2-n^3h^3±...)

 

[WWGD]

@WWGD I do not have lavinia's expertise, but when I needed some chern class computations I found a useful formula in bott-tu at least for hyper surfaces, which i used for my computation of chern classes for surfaces in complex projective 3 space and maybe 3-folds in 4 space. This was included in my notes on riemann roch theorems, available on my website at uga math dept. i will look for a more precise link.

Thanks, I have been meaning , hoping to go over that book for a long time now.

 

[mathwonk]

@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.

 

[mathwonk]

@WWGD:
Ok, the basic case for chern classes is the chern class of a line bundle. A line bundle is in general equivalent to a hypersurface, which in complex varieties has real codimension 2, hence defines by duality a cohomology class in H^2, which is the chern class of the line bundle. This hypersurface is obtained as the zero locus of any section of the line bundle.

On projective space there is one line bundle for each integer, and for a positive integer n, the associated hypersurface is the zero locus of any homogeneous polynomial of degree n. I.e. the unique line bundle on P^r of degree n, called O(n), has as sections exactly the homogeneous polynomials of degre n, and any ≠0 one of those cuts out a hypersurface dual to the chern class. Now we know the chern classes of all line bundles on projective space.

We find it convenient to work with total chern classes, so if h is the 2 dimensional cohomology class dual to a hyperplane, i.e. the first chern class of O(1), then the total chern class of O(1) is 1+h, where 1 belongs to H^0 and h belongs to H^2, (it seems chern classes only occur in even dimensions). Furthermore, the first chern class of the line bundle O(n) associated to a hypersurface of degree n, is 1+nh, also in H^0 + H^2.

Now I seem to have read in Bott - Tu that the total chern class of P^r is the rth power of the total class of the line bundle O(1), i.e. c(P^r) = (1+h)^r, a class in H^0+H^2+...H^2r. (I forget how this goes, but it probably has to do with pulling back a vector bundle until it splits into a sum of line bundles and then multiplying the chern classes of those line bundles together. So presumably T(P^r) splits into r copies of O(1).)

Now consider a hypersurface X of degree n in P^r and ask what is its normal bundle? That is a line bundle which equals, (what else?), the restriction of O(n) to X. Hence the total chern class of that restriction is also (1+nh), restricted to X.

Now by whitneys formula, the total chern class of X, the degree n hyperaurface, multiplied by the class of the normal bundle, equals the class of P^r, so we get c(X).(1+nh) = (1+h)^r, hence c(X) = (1+h)^r.(1+nh)^(-1).

We invert a chern class using the geometric series. so (1+nh)^(-1) = 1-nh + n^2h^2-n^3h^3...

I carried this out with the results you see in my RRT notes for surfaces in P^3 and 3-folds in P^4. I thought it was fun. Hope you do too. So I guess you will probably enjoy it more if you also do the calculations yourself.

I found the reference to Bott-Tu as pages 280-282.

 

[mathwonk]

@lavinia;
"You studied with some of the immortals. " yes but they are not to blame for the results of my inattention. I only started applying myself years later, as a father and husband needing to provide. Still one finds that their wisdom still lurks in ones brain, once one starts to heed it.

The moral is that these stars have valuable things to say, but you may benefit more from whomever you study with, if you listen and reflect on it, and work judiciously through the books they recommend.

 

[lavinia]

Anything that is not actually trivial but not too demanding, if it is possible to find that sweet spot.

The simplest non-trivial example I know of is the first Stiefel-Whitney class of the Mobius band viewed as a line bundle over the circle, $π:M → S^1$.

For the real line, an orientation is just a choice of a direction $+$ or $-$ . A real line bundle is orientable if there is a consistent continuously varying choice of direction for each of the fibers. Otherwise put, the bundle is orientable if there is a section of the bundle that is everywhere non zero. The Mobius band is not orientable so it has no continuous non-zero section. (A section of a bundle is a continuous map from the base space to the total space that projects back to the identity map on the base.)

Excercise: Use the Intermediate Value Theorem to show that the Mobius band does not have a non-zero section.

In fact, any smooth section that is transverse to the base circle must cross the base circle an odd number of times.

One way to translate this into a cohomology class is to divide the base circle into three segments so that it becomes a simplicial triangle. Try to build a non-zero section in stages by first choosing a direction at each of the three vertices. Next ask whether these choices can be continuously extended over the interiors of each the segments. If it is yes for a segment assign a $0$ to it. If no assign a $1$. This gives a map from the three segments into the group $Z_2$ and as such is a $1$ dimensional cochain. This co-chain is the Stiefel-Whitney class of the Mobius band.

The first Stiefel-Whitney class in this case is cohomologous to zero if an only if a continuous non-zero section can be found so it completely determines whether or not the line bundle is orientable/trivial.

While this example may seem too simple it illustrates the relation of characteristic classes to the existence of sections of vector bundles. For an orientable vector bundle there is another characteristic class called the Euler class. For a $k$ dimensional vector bundle, it is an cohomology class in $H^{k}($base space $;Z)$. The Euler class is an obstruction to a non-zero section of the bundle. This means that if there is a non-zero section then the Euler class must be zero.

The Chern class of a complex line bundle can be defined to be the Euler class of the underlying real bundle - the same bundle but forgetting the complex structure.

 

[mathwonk]

Thank you lavinia for this nice clear description. As a contrast, if one tries to learn SW classes from Milnor Stasheff, see p. 38, axiom 4, then the fact that the SW class of the mobius bundle is one, is an axiom, thus giving no insight at all as to its meaning.

The euler class also is a nice example. As you say, the basic principle is that the zero loci of any two non trivial sections of a line bundle (presumably also of any bundle of rank ≤ dimension of base manifold), are always homologous, hence the homology class of any section has a dual class, the euler class of the bundle. Then by definition, the vanishing of this class is an obstruction to the existence of a never zero section. Thus if we want to use that class to tell us whether such a section exists, we need some other way to calculate the class other than first finding a section.

For a compact connected riemann surface, the fundamental fact is that every non constant meromorphic function has a divisor of "degree zero", i.e. every meromorphic function has the same number of zeroes and poles. This follows from the fact that such a function defines a holomorphic map to the riemann sphere, and by degree theory, the same number of points map to the north pole as to the south pole, so their difference, the degree of the divisor of the meromorphic function, is zero.

Then one notes that given two non trivial meromorphic sections of any line bundle, their quotient is a meromorphic function, hence the two sections have (if holomorphic) the same number of zeroes, and even if meromorphic, their divisors have the same degree. Hence every non trivial section of a holomorphic line bundle has the same degree as any other. This degree then is the (1st) chern class of the line bundle. For riemann surfaces, even if the degree of a line bundle is non zero, it is not obvious that a meromorphic section exists, indeed this is a fundamental existence theorem. But if two sections exist, they have the same degree.

There is another definition of chern class in this case that does not assume the existence of a section, stemming from the exponential sequence: 0-->Z-->O-->O*-->1, the right hand map is exponentiation of a holomorphic function, times 2πi, so the kernel is the integers. The associated long exact equence of cohomology has the segment H^1(O*)-->H^2(Z). Since one can show that every holomorphic line bundle defines , by taking its transition functions, a 1st cech cohomology class with coefficients in O* (never vanishing holomorphic functions), the image of that class via the coboundary map in H^2(Z) is another definition of the 1st chern class of the line bundle.

Another approach I like to characteristic classes uses the classifying space concept, i.e. the grassmannian. First one computes the cohomology of a "grassmannian", i.e. of the space of all linear subspaces of a fixed dimension in a given euclidean space. Then one observes there are certain standard cohomology classes defined for a grassmannian. Now if one can only map ones own manifold into a grassmannian in a natural way, then one can pull those classes back to ones own manifold, and call the pulled back classes the characteristic classes of ones manifold. How to do that?

First embed your compact smooth manifold into Euclidean space, whence there is an induced map of each tangent space to your manifold linearly into Euclidean space. Thus if your manifold had dimension n and the ambient euclidean space had dimension r, we have defined a map of our manifold into the grassmannian of n planes in r space. Now the interesting part is that if we let r go to infinity, and use all possible embeddings of our manifold into these euclidean spaces, apparently the results stabilize. I.e. for large r, the maps to the grassmannian given by any two embeddings are homotopic, and the resulting pull back classes are independent of r!

I first read this stuff in some mimeographed notes on differential topology by Milnor that were handed around in the 60's, and that presumably were codified in the Milnor Stasheff book, but I am not sure.

Yes this discussion is on pages 60-70 of Milnor - Stasheff, where he observes that it is a generalization of the Gauss map from a surface in R^3 to the unit sphere. In fact it appears that the datum of an n plane bundle, up to isomorphism, is equivalent to the datum of a map to an infinite grassmannian, up to homotopy, (Remark, p.70).

I realize now, that in the case of riemann surfaces, all this is just the familiar practice of using a line bundle to define a holomorphic map of the riemann surface to projective space, which is the grassmannian for line bundles, and then the 1st chern class of the line bundle is just the intersection number of the embedded riemann surface, with a hyperplane, the homology class dual to the 1st chern class of projective space. Of course in this setting, not all line bundles define holomorphic embeddings, so there is more detail to worry about. E.g. sometimes there are divisors of form M+B where M is the moving part and B is the fixed part ("base divisor"), and in these cases it can happen that the resulting map to projective space loses the fixed B part, so the intersection number with a hyperplane only measures the chern class of the M part. E.g. suppose we take a plane cubic curve C with a double point p. Then the pencil of all lines through p will define a pencil of divisors on C of degree 3, but all of them contain the point p, the fixed part. The resulting map of C to projective space is the projection along these lines from the cubic to P^1, which gives a map of degree 2, not 3. I.e. the map only recovers the family of the pairs of moving points cut by a line through p, not p itself. (In this case "intersection with a hyperplane" means the pull - back of one point of P^1.)

So in Riemann surface theory there is a whole subject studying which line bundles define holomorphic embeddings into projective space. It follows from Riemann Roch theorem (RRT) that if the genus of the riemann surface is g, then any line bundle of degree more than 2g works I believe. Indeed the RRT is a formula helping compute the number of linearly independent holomorphic sections of a line bundle just from knowing the degree of the bundle, i.e. the number of zeroes of a section. In particular, if the degree d is more than 2g-2, then there are always exactly (1-g + d) independent holomorphic sections. If the degree is ≤ 2g-2, then there could be some holomorphic differentials vanishing on the divisor, and if there are exactly i independent such differentials, then this adds sections to the bundle and there are then exactly (1-g+d+i) independent holomorphic sections. And as remarked above, if d > 2g, then the holomorphic sections suffice to embed the riemann surface in projective space of dimension (d-g).

For a "non hyperelliptic" riemann surface (one that does not possesses a map to P^1 of degree 2), one can do better, and one can embed it in P^(g-1), using the "canonical line bundle", the bundle of degree 2g-2 of holomorphic one forms dual to the tangent bundle, (for which i=1, by definition).

To illustrate further the difference between smooth and holomorphic properties of line bundles on riemann surfaces, I note that the smooth homeomorphism class of a complex line bundle on a compact (connected) riemann surface is completely determined by its degree, i.e. its 1st chern class, (because the sheaf of smooth sections of a line bundle is a "fine" sheaf, i.e. admits partitions of unity), whereas for holomorphic line bundles, for each degree, there is a compact manifold (a torus) of dimension g parametrizing the holomorphic isomorphism classes of that degree. By the Riemann Roch theorem however, even in the holomorphic case, at least for large degree d, the number of independent holomorphic sections is determined by the chern class, i.e. the topology.

The generalized Riemann Roch theorem says that at least the holomorphic euler characteristic, i.e. the alternating sum of the dimensions of the holomorphic cohomology groups, is a topological invariant. In the classical case this holomorphic euler characteristic is the alternating sum: (# sections) - i, which equals the topological invariant 1-g+d. Whenever one can prove "vanishing" i.e. that higher dimensional groups are all zero, one deduces a formula for the number of sections just from the topological computation of the euler characteristic. So in some sense, major results in algebraic geometry consist of deciding when one gets holomorphic information just via topology.

It was this that first drew me into the subject of riemann surfaces, or algebraic geometry, the fact that there is a level of complexity that is not immediately visible to the naked eye, i.e. to topology, but where topology is useful, in fact essential. To someone who found topology appealing and fairly natural, this added mystery was intriguing. I suppose number theory is then a further refinement of algebraic geometry, so just as algebraic geometers need and use topology, so do number theorists depend on algebraic geometry, and extend its reach, by exploring which subfields of the complex numbers may already contain solution sets to their equations.

 

[WWGD]

Thanks, both, excellent posts. One thing I would like is to help me understand the notion of continuity in sections of bundles. I am having trouble having a no technical understanding. I can tell, e.g., the mobius strip as a line bundle over the circle must twist in a discontinuous way at some point, but I don't see a more general sense of what continuity men's in this context. I also have trouble understanding, e.g., what is a continuous vector field.i want to read the proof of the hairy balls theorem but I would like to have a feel for what continuity means in this context. Any ideas, suggestions please?

 

[mathwonk]

continuity is local. Locally a mobius strip is just U x R^1, where U is an open subset of S^1, so locally a section is a function U -->UxR^1, and since it hs to be a section, it is the identity in the U component, so locally section is just a function U-->R^1, hence is continuous exactly when this functionm is continuousn inmthe usual sense. In general since a vector bundle is locally of form UxV, a continuous section is locally a continuous function U-->V.

As for the mobius strip, it is a rectangle [0,1] x R^1, with opposite ends {0}xR^1 and {1}xR^1 identified by reversing them, i.e. identify the points {0}x{t} and {1}x{-t}. so a continuous section of the mobius strip is a continuous function f from [0,1] to R^1, that satisfies f(0) = -f(1). Hence by the IVT, f must take the value 0 somewhere.

I.e. continuous section of the mobius bundle is a continuous curve in the mobius strip that meets each line perpendicular to the equator exactly once. Hence (by IVT argument above) it must cross the equator.

 

[mathwonk]

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.

 

[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.