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

[WWGD]

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 $S^n$.

 

[lavinia]

@WWGD I have never heard of a standard action of the circle on spheres. Can you give some references?

 

[WWGD]

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.

 

[Infrared]

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.

 

[jim mcnamara]

Exactly where did you encounter this @WWGD ?

 

[WWGD]

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.

 

[WWGD]

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.

 

[lavinia]

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.

 

[WWGD]

@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 $e^{i\phi}$

 

[Infrared]

@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.

 

[lavinia]

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.

 

[WWGD]

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.

 

[Infrared]

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?

 

[Infrared]

@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.

 

[WWGD]

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.

 

[mathwonk]

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.

 

[lavinia]

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 $RP^3$.

 

[lavinia]

@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 $S^1$.

 

[lavinia]

@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.

 

[WWGD]

@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?

 

[Infrared]

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

 

[lavinia]

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.

 

[lavinia]

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.

 

[mathwonk]

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.

 

[mathwonk]

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.

 

[WWGD]

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?

 

[mathwonk]

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.

 

[WWGD]

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.

 

[mathwonk]

Apparently it is known that if a smooth manifold bounds a topological manifold, then it also bounds a smooth manifold, which would answer my question in some sense. It is also asserted that the fact that rational pontrjagin classes can be defined for topological manifolds is relevant. here is a link to some recent comments on math overflow:

https://mathoverflow.net/questions/208501/topological-cobordisms-between-smooth-manifolds

 

[mathwonk]

@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.

 

[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.