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

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Hi All,
I need to figure out the definition of the oft-called "Standard" action of the circle ##S^1 ## ( as a Topological/Lie group) ,on ##S^n##, the n-sphere ( I guess seen as ##\{z : |z|=1\}## in Euclidean n-space). My searches returned an action of ##S^1## on ##S^3 ## given by ##A(z_1, z_2): z --> (zz_1, zz_2) ## and ##z-->(zz_1, z^{_})## , for z^ the conjugate of z, but no general definition for the action of ##S^1 ## on all ## S^n ##.
 
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@WWGD I have never heard of a standard action of the circle on spheres. Can you give some references?
 
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I will look more carefully, but it is used as a counterexample to show that ## \mathbb RP^n ## , n odd does not have the fixed point property, i.e., there are continuous self-maps f that do not fix any element, :cont functions with ##f(x) \neq x \forall x##. EDIT: This is from a homework problem from a while back.
 
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Well if ##n## is odd, then you can view ##S^n=S^{2k-1}## as the the set of ##k##-tuples of complex numbers ##(z_1,\ldots,z_k)## such that ##\sum_{i=1}^k|z_i|^2=1##. Then ##S^1## acts on the sphere by ##e^{i\phi}\cdot (z_1,\ldots,z_k)=(e^{i\phi}z_1,\ldots,e^{i\phi}z_k)##. This action takes antipodal points to antipodal points, so any element of ##S^1## gives a self-map on ##\mathbb{P}^n##. This map doesn't take a point to itself or its antipode unless ##\phi## is a multiple of ##\pi## so the map on projective space doesn't have fixed points.
 
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Exactly where did you encounter this @WWGD ?
 
jim mcnamara said:
Exactly where did you encounter this @WWGD ?
I am not sure, I am searching for the source. I saw it at one point used to show a result that Projective Odd-dimensional Real space does not have the fixed point property. The map described by Infrared, the action, does not have a fixed point when you "factor through" (pass to the quotient by identifying a point with its antipode) this map into a map from projective (2n+1)-space to itself.
EDIT: Any map that sends pairs of antipodes to pairs of antipodes will work too, so it comes down to showing these exists, or, better, like @Infrared , produce one such map.
 
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Expanding a bit, we say the map f factors through the map p if there is a third map h with f= hop , with o being composition. The name comes from the analogy with factoring numbers as , e.g., 15=3(5), but in our case, composition plays the role of multiplication in numbers. This factoring is helpful in that it works well with many functors ( from Category Theory; such as fundamental group or homology) _* as in , e.g. (fog)_* =f_*o g_* , which helps find the solution of otherwise hairy problems.
 
WWGD said:
I will look more carefully, but it is used as a counterexample to show that ## \mathbb RP^n ## , n odd does not have the fixed point property, i.e., there are continuous self-maps f that do not fix any element, :cont functions with ##f(x) \neq x \for all x##.

@WWGD I am feeling a little dumb here so help me out. What is the fixed point property?Also the action of the unit complex numbers on an odd dimensional sphere as described in post #4 does not project to a fixed point free action on projective space. While on the sphere coordinate wise multiplication by ##e^{iπ}## has no fixed points on the sphere (It is the antipodal map), it projects to the identity map on projective space. So the action on projective space as described is not fixed point free.
 
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lavinia said:
@WWGD I am feeling a little dumb here so help me out. What is the fixed point property?

The antipodal map is fixed point free on every sphere in every dimension including dimension zero.

Also the action of the unit complex numbers on an odd dimensional sphere as described in post #4 does not project to a fixed point free action on projective space. While on the sphere coordinate wise multiplication by ##e^{iπ}## has no fixed points on the sphere (It is the antipodal map), it projects to the identity map on projective space. So the action on projective space as described is not fixed point free.
Don't , Lavinia, too much terminology and same term has different names. Fixed point property for space X is that every continuous self-map f: X-->X, has a fixed point. I think Infrared wrote ##e^{i\phi}##
 
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  • #10
lavinia said:
@WWGD
Also the action of the unit complex numbers on an odd dimensional sphere as described in post #4 does not project to a fixed point free action on projective space...it projects to the identity map on projective space

No, projecting to the identity map would mean that every point on the sphere it taken to itself or its antipode. This is not the case when ##\phi## is not a multiple of ##\pi##.

Edit: What I meant by saying that it takes antipodes to antipodes is that if ##p,q## are antipodal before acting on them, then they are still antipodal after acting by an element of ##S^1##. This is just saying that we have a well-defined map on projective space.
 
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  • #11
Infrared said:
No, projecting to the identity map would mean that every point on the sphere it taken to itself or its antipode. This is not the case when ##\phi## is not a multiple of ##\pi##.

Edit: What I meant by saying that it takes antipodes to antipodes is that if ##p,q## are antipodal before acting on them, then they are still antipodal after acting by an element of ##S^1##. This is just saying that we have a well-defined map on projective space.
OK. I see so all the other angles give a fixed point free map. I thought one wanted a fixed point free action - which may also be true but not the projected action.
 
  • #12
Infrared said:
No, projecting to the identity map would mean that every point on the sphere it taken to itself or its antipode. This is not the case when ##\phi## is not a multiple of ##\pi##.

Edit: What I meant by saying that it takes antipodes to antipodes is that if ##p,q## are antipodal before acting on them, then they are still antipodal after acting by an element of ##S^1##. This is just saying that we have a well-defined map on projective space.
My bad, I meant descends to the quotient to a self-map in Projective space.
 
  • #13
WWGD said:
I thought one wanted a fixed point free action - which may also be true but not the projected action.

The easy way to do this is to look at the cyclic subgroup generated by any element of ##S^1## that is not a root of unity. Maybe the more interesting question is if this can be done via a Lie group action.

Edit: @WWGD Sorry, what exactly are you referring to?
 
  • #14
@lavinia I think the following action ##S^1## action works as long as ##k>1## (the action is free):

##e^{i\phi}\cdot (z_1,\ldots,z_k)=(e^{i\phi\alpha_1}z_1,\ldots,e^{i\phi\alpha_k}z_k)##

where the ##\alpha_i## are real numbers that are ##\mathbb{Z}##-independent.

Edit: nevermind, this is not well-defined.
 
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  • #15
Infrared said:
The easy way to do this is to look at the cyclic subgroup generated by any element of ##S^1## that is not a root of unity. Maybe the more interesting question is if this can be done via a Lie group action.

Edit: @WWGD Sorry, what exactly are you referring to?
I meant maps that are constant on equivalence classes pass to the quotient in the sense that pof =f^op , i.e., the diagram commutes; f^ is defined on equivalence classes; I don't know how to draw a diagram here.
 
  • #16
I am not really following this in detail, but off the top of my head it seems ( to me) there is an obvious action of S^1 on any S^n. I.e. realize S^n as the result of collapsing the top and the bottom of the cylinder [0,1]xS^(n-1) each to a point, as in homotopy theory (suspension?). Hence the induced action of S^1 seems to be the previously defined one on each slice ≈ S^(n-1), and fixes both poles. hope this is not absurdly wrong. of course you start it off with the "obvious" action of S^1 on S^1, by multiplication of complex numbers of norm one.

a quick google search yields this:

https://mathoverflow.net/questions/18569/circle-action-on-sphereSo I guess a standard action results from using the normal form of orthogonal matrices, decomposing them as made up of rotations. see e.g herstein's algebra book (which I unfortunatey gave away). Then theorem is that every orthogonal transformation is an orthogonal direct sum of actions on invariant subspaces of dimension ≤ 2, on each of which the map is either the identity or a rotation. (The notes on my website say incorrectly "reflection" instead of "rotation"; I might try to claim my spell checker is to blame but the one in my brain is at least as culpable.) So if this is right, the "standard" action could be given in coordinates by a block matrix with as many 2x2 blocks as possible, with sins and cosines giving a standard 2x2 rotation in the 2x2 blocks, and the odd 1x1 has the identity.
 
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  • #17
mathwonk said:
I am not really following this in detail, but off the top of my head it seems ( to me) there is an obvious action of S^1 on any S^n. I.e. realize S^n as the result of collapsing the top and the bottom of the cylinder [0,1]xS^(n-1) each to a point, as in homotopy theory (suspension?). Hence the induced action of S^1 seems to be the previously defined one on each slice ≈ S^(n-1), and fixes both poles. hope this is not absurdly wrong. of course you start it off with the "obvious" action of S^1 on S^1, by multiplication of complex numbers of norm one.

a quick google search yields this:

https://mathoverflow.net/questions/18569/circle-action-on-sphereSo I guess a standard action results from using the normal form of orthogonal matrices, decomposing them as made up of rotations. see e.g herstein's algebra book (which I unfortunatey gave away). Then theorem is that every orthogonal transformation is an orthogonal direct sum of actions on invariant subspaces of dimension ≤ 2, on each of which the map is either the identity or a rotation. (The notes on my website say incorrectly "reflection" instead of "rotation"; I might try to claim my spell checker is to blame but the one in my brain is at least as culpable.) So if this is right, the "standard" action could be given in coordinates by a block matrix with as many 2x2 blocks as possible, with sins and cosines giving a standard 2x2 rotation in the 2x2 blocks, and the odd 1x1 has the identity.

I never heard of a standard action either but it seems that the OP was looking for an action that projects to real projective space and has projected elements that act without fixed points. Rotation of a sphere projects to an action on projective space but as you pointed out every element has a fixed point.

For odd spheres multiplication by complex numbers of length 1 gives a fixed point free action of ##S^1## on the sphere and the projection of this action has elements that act without fixed points.

I am wondering whether there is a fixed point free action of the circle on every odd dimensional real projective space. Modding out by the antipodal map folds each orbit circle on itself twice to give a circle in real projective space. I imagine that on these folded circles there is another fixed point free action of ##S^1## but it is certainly not the projection of the action on the sphere. For example if one views ##RP^3## as the tangent circle bundle to the 2 sphere, then given an orientation and Riemannian metric, ##S^1## acts on each tangent circle by rotation. This is a fixed point free action on ##RP^3##.
 
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  • #18
@WWGD I thought you'd be interested that multiplication by ##e^{iπ/2}## projects to a fixed point free involution of the odd dimensional real projective space. (An involution is a map whose square is the identity map.) A manifold with a fixed point free involution is a boundary. That is: it is the boundary of a 1 higher dimensional manifold. The proof is easy.

No even dimensional real projective space is a boundary and thus can not have a fixed point free involution. This rules out a fixed point free action of ##S^1##.
 
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  • #19
@WWGD Another interesting point is that the action on odd spheres that @Infrared describes in post #4 shows that the antipodal map is homotopic to the identity. On even dimensional spheres it is not homotopic to the identity. So if this is the standard action on odd spheres, it has no analogue for even spheres.
 
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  • #20
lavinia said:
@WWGD I thought you'd be interested that multiplication by ##e^{iπ/2}## projects to a fixed point free involution of the odd dimensional real projective space. (An involution is a map whose square is the identity map.) A manifold with a fixed point free involution is a boundary. That is: it is the boundary of a 1 higher dimensional manifold. The proof is easy.

No even dimensional real projective space is a boundary and thus can not have a fixed point free involution. This rules out a fixed point free action of ##S^1##.
Thank you, can you please give me a sketch of a proof or ref for even dimensional Real projective spaces not being boundaries? What is the obstruction?
 
  • #21
In order for a compact manifold to be a boundary, its Stiefel-Whitney numbers must vanish. You can find a proof in Milnor-Stascheff that the Stiefel-Whitney numbers of ##\mathbb{RP}^n## are all zero if and only if ##n## is odd.

Edited: should be right now
 
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  • #22
To show that an even dimensional projective space ##P^{2n}## is not a boundary one can also use an argument involving the Euler characteristic. Every even dimensional projective space has Euler characteristic 1 and this by itself means that it is not a boundary since the Euler characteristic mod 2 is the top Stiefel-Whitney number. In other words a manifold of odd Euler characteristic is not a boundary.

If one assumes by contradiction that ##P=∂M## that ##P## is the boundary of the compact manifold ##M## then by gluing two copies of ##M## together along ##P## one gets an odd dimensional smooth closed manifold without boundary. This manifold has Euler characteristic zero as do all odd dimensional closed manifolds without boundary. The Euler characteristic of the glued manifold ##M##υ##_{P}M## is twice the Euler characteristic of ##M## minus the Euler characteristic of ##P##. So ##χ(M##υ##_{P}M) = 2χ(M)-χ(P) = 2χ(M)-1##. One can see this by extending a triangulation of ##P## to a triangulation of ##M## then noticing that when one counts simplicies,the simplicies of ##P## should be counted only once. So ##0 = 2χ(M)-1## or ##χ(M)=1/2## which is impossible since the Euler characteristic of any simplicial complex is always an integer.

The proof works for any even dimensional manifold with odd Euler characteristic.

Note: This proof uses a lot of machinery. I would love to see a simpler proof. The proof using Stiefel-Whitney numbers though is much more complicated.
 
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  • #23
WWGD said:
Thank you, can you please give me a sketch of a proof or ref for even dimensional Real projective spaces not being boundaries? What is the obstruction?

An even dimensional projective space is not orientable so its first Stiefel-Whitney class ##ω_1## is not zero. The mod 2 cohomology ring of real projective space is a truncated polynomial algebra -truncated above the dimension of the space - in a single generator in dimension 1. For even dimensional projective spaces, ##ω_1## must be this generator. Since the truncation occurs above the dimension ##2n## of the projective space, the Stiefel-Whitney number ##ω_1^{2n}## is not zero.

Note: The expression ##ω_1^{2n}## denotes the ##2n## -fold cup product of ##ω_1## with itself.
 
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  • #24
the multiplication by complex unit length vectors mentioned by infrared is of course the same as the orthogonal direct sum of real rotations mentioned in post #17. Hence one "analog" in odd dimensions is also the othogonal direct sum of rotations plus one identity map in the extra dimension. I.e. both cases are instances of the standard normal form for orthogonal transformations in real space, as a sum of rotations and identities, (as in herstein's topics in algebra, e.g.), so only in even dimensional real space (hence odd dimensional spheres) can there be only rotations and no identities, hence no fixed points.
 
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  • #25
the nice discussions via characteristic classes seem to be treating the case of smooth manifolds. Does it follow that the even diml projective spaces cannot be boundaries of topological manifolds? (presumably there exist topological manifolds with no smooth structure.) Is there a theory of stiefel whitney classes in that generality? They seem to be topological invariants, but is there a topological definition? Perhaps the key issue is the definition of a topological version of the tangent bundle.
 
  • #26
mathwonk said:
the nice discussions via characteristic classes seem to be treating the case of smooth manifolds. Does it follow that the even diml projective spaces cannot be boundaries of topological manifolds? (presumably there exist topological manifolds with no smooth structure.) Is there a theory of stiefel whitney classes in that generality? They seem to be topological invariants, but is there a topological definition? Perhaps the key issue is the definition of a topological version of the tangent bundle.
What do you mean by a topological definition? Aren't they defined in terms of sections, cup products and pullbacks and related concepts? And, yes, re the boundaries, Lavinia laid it out in post #23. It would be nice if Lavinia could do an "Insights" on Characteristic forms, maybe specialized to Whitney-Stiffel or other. Common, let's pile the pressure on Lavinia. EDIT: Ah, you mean we make use of results like Poincare Duality and definiton of Char forms that assume smoothness?
 
  • #27
the definition is topological once you are given the bundle, but the bundle used in the definition is the tangent bundle, which depends on the smooth structure. Indeed it seems there is a theory of topological "micro bundles" due to Milnor, and using it he shows that in fact the tangent bundle of a smooth manifold is not a topological invariant. E.g. there exist different smooth structures on the same topological manifold that have different pontrjagin classes. Thus it is interesting that stiefel whitney classes are apparently topological invariants not just of the tangent bundle but of the underlying topological manifold, which does not have a uniquely determined smooth tangent bundle. I.e. even though you can define non homeomorphic tangent bundles for different smooth structures on the same topological manifold, they will apparently yield the same SW numbers. Nonetheless that does not tell us how to define SW numbers for a topological manifold that does not have a smooth structure, and hence has no definable smooth tangent bundle.
 
  • #28
An interesting ( at least to me) result is that ##\mathbb CP^n ## is never isomorphic to ##\mathbb RP^{2n}##, which I always wondered about. Because Complex projective spaces are always orientable. I also always wondered if we could tell whether a space A was isomorphic to a product space ## B \times C ## by showing that A is a trivial bundle for some other manifold. Maybe just random musings without much importance.
 
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  • #30
@WWGD

I always thought projective space was non orientable and hence i too was puzzled when I read that complex projective space was always orientable, (the proof I first saw was in Chern's notes on complex manifolds and used only the cauchy riemann equations to compute a chart change determinant as positive). Of course it seems I also missed the case of real complex 1 space, i.e. the circle. I just thought the real projective plane was typical for projective spaces, since that was the only case I had looked at.
 
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  • #31
@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.
 
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  • #32
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.)
 
  • #33
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
 
  • #35
mathwonk said:
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.
 
  • #36
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.
 
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  • #37
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.
 
  • #38
mathwonk said:
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.
 
  • #39
WWGD said:
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?
 
  • #40
lavinia said:
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.
 
  • #41
@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±...)
 
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  • #42
mathwonk said:
@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.
 
  • #43
@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.
 
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  • #44
@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.
 
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  • #45
@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.
 
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  • #46
WWGD said:
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.
 
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  • #47
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.
 
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  • #48
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?
 
  • #49
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.
 
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  • #50
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.
 
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