Cup Product and Representability

[Jamma]

Hello all,

I was just wondering, is there a canonical relationship between the two, in the following sense (I only need to deal with products of elements of the first cohomology group here):

We know that elements of the first simplicial cohomology group can be represented as equivalence classes of maps into some representing space. I'm only bothered about H^1, for now, so we have that H^1(X) = [X,S^1].

We also know that there is a cup product on cohomology definied in a reasonable simple way on cochains.

My question is: given two functions f,g: X ->S^1 which define elements of H^1(X), is there a canonical way of defining the cup product of f and g in terms only of the function f and g? It seems that there should be, but I don't see a canonical way of doing so. Is there some way I should be embedding the torus into the infinite dimensional complex projective space?

 

[Jamma]

Hrmm, I think there may be a partial answer to my question here:
http://en.wikipedia.org/wiki/Cohomology_operation

 

[Jamma]

Actually, no, that only covers 1-ary operations, but I'm sure there must be a generalisation to 2-ary, or possibly higher.

 

[Jamma]

Ahha, found an answer to my own question (feel free to delete thread):
http://www.math.sunysb.edu/~basu/notes/GSS4.pdf

(I don't always talk to myself like this :P)

 

[quasar987]

Nice link. The other pdf file from this guy look equally interresting.

 

[Jamma]

Yes, it's nicely written, quite informative. I don't quite get the last sentence though:

"Armed with this observation, we can conclude that the fundamental
group of any surface (or of any aspherical manifold) has no torsion."

Does this make sense? The Klein bottle is a surface, and that has torsion in its fundamental group (I think). Does he mean orientable surface? I don't understand how it follows from the rest either, although admittedly I haven't had time to read it all properly yet.

 

[quasar987]

Maybe he meant any finitely generated abelian G. That way, G=Z^rx Z_t1 x ... x Z_ts so K(G,1)=(S¹)^xr x L_t1 x ... x L_ts (where L_k = K(Z_k,1) is an infinite lens space)

But here I'm confused, because the homology of this, as given by Kunneth's formula, looks complicated. Can we argue nevertheless that it is nonzero in infinitely many dimensions?

 

[lavinia]

Yes, it's nicely written, quite informative. I don't quite get the last sentence though:

"Armed with this observation, we can conclude that the fundamental
group of any surface (or of any aspherical manifold) has no torsion."

Does this make sense? The Klein bottle is a surface, and that has torsion in its fundamental group (I think). Does he mean orientable surface? I don't understand how it follows from the rest either, although admittedly I haven't had time to read it all properly yet.

- The Klein bottle has no torsion in its fundamental group. The group can be described as as split exact sequence of groups.0 -> Z -> K -> Z - > 0An a spherical manifold must be a K(pi,1). But since every compact manifold is a finite CW-complex - by Morse Theory - an ashperical compact manifold can not have torsion in its fundamental group.- Flat manifolds are aspherical and are covered by Euclidean space be the action of a torsion free group of isometries. Such groups satisfy an exact sequence,0 -> n dimensional lattice,L, -> G -> G/L ->1 where G/L is a finite group.

For the Klein bottle this sequence is0 -> ZxZ -> K -> Z/2 ->0If for instance, if one starts with the standard lattice on R^2 and adds the transformation (x,y,) -> (x+1/2,-y) this group is isomorphic the the fundamental group of the Klein bottle.

Notice that the square of the translated reflection is (x,y) -> (x + 1, y) which is an element of the standard lattice.

 

[lavinia]

Hello all,

I was just wondering, is there a canonical relationship between the two, in the following sense (I only need to deal with products of elements of the first cohomology group here):

We know that elements of the first simplicial cohomology group can be represented as equivalence classes of maps into some representing space. I'm only bothered about H^1, for now, so we have that H^1(X) = [X,S^1].

We also know that there is a cup product on cohomology definied in a reasonable simple way on cochains.

My question is: given two functions f,g: X ->S^1 which define elements of H^1(X), is there a canonical way of defining the cup product of f and g in terms only of the function f and g? It seems that there should be, but I don't see a canonical way of doing so. Is there some way I should be embedding the torus into the infinite dimensional complex projective space?

Is it easy to why the first Z cohomology is homotopy classes of maps into the circle?

 

[Jamma]

Well, it's not that easy, although it's fairly well known that there is a natural bijection between H^n(X,G) and [X,K(G,n)] (i.e. between the n'th cohomology group with coefficients in G and homotopy classes of maps between X and the n'th Eilenberg-MacLane space of G). See Brown's representability theorem. Actually, just see the link I posted for a nice exposition to this. K(Z,1)=S^1.

I'm not 100% on the torsion in the fundamental group of the Klein bottle (I think I have calculated it before and found a presentation of it), although there certainly is torsion in the fundamental group of the real projective plane (it's fundamental group is the cyclic group of order 2).

 

[lavinia]

Well, it's not that easy, although it's fairly well known that there is a natural bijection between H^n(X,G) and [X,K(G,n)] (i.e. between the n'th cohomology group with coefficients in G and homotopy classes of maps between X and the n'th Eilenberg-MacLane space of G). See Brown's representability theorem. Actually, just see the link I posted for a nice exposition to this. K(Z,1)=S^1.

I'm not 100% on the torsion in the fundamental group of the Klein bottle (I think I have calculated it before and found a presentation of it), although there certainly is torsion in the fundamental group of the real projective plane (it's fundamental group is the cyclic group of order 2).

As I said above, the fundamental group of the Klein bottle is a split extension

0 -> Z -> K -> Z -> 0

If a and b generate the two infinite cyclic subgroups then ab(a^-1) = -b where b generates the kernel of the projection of K onto Z.

 

[quasar987]

An a spherical manifold must be a K(pi,1). But since every compact manifold is a finite CW-complex - by Morse Theory - an ashperical compact manifold can not have torsion in its fundamental group.

Could you offer more details on this? Surely you're not suggesting that a finite CW-complex has no torsion in its pi_1? (Think RP²)

 

[lavinia]

Could you offer more details on this? Surely you're not suggesting that a finite CW-complex has no torsion in its pi_1? (Think RP²)

no but if a finite CW complex is a K(pi,1) then it can not have torsion in its fundamental group - this by the comment at the end of the paper in the link.

I made an mistake in my post above. Every flat manifold like the Klein bottle has a fundamental group as I described and they are all K(pi,1)s. But there are other manifolds that are covered by contractible spaces and so are K(pi,1)'s as well. I corrected my post above.

 

[quasar987]

no but if a finite CW complex is a K(pi,1) then it can not have torsion in its fundamental group - this by the comment at the end of the paper in the link.

Err.. but I thought the validity of the comment at the end of the paper in the link is precisely the topic under discussion. We (see post #6 and #7) don't understand why the comment is true. In fact, it seems to us that RP², whose pi_1 is Z_2, is an obvious counterexample (to the part of the statement concerning surfaces at least (because RP^2 is not aspherical)).

 

[Jamma]

no but if a finite CW complex is a K(pi,1) then it can not have torsion in its fundamental group - this by the comment at the end of the paper in the link.

I made an mistake in my post above. Every flat manifold like the Klein bottle has a fundamental group as I described and they are all K(pi,1)s. But there are other manifolds that are covered by contractible spaces and so are K(pi,1)'s as well. I corrected my post above.

My definition of a K(pi,1) is a CW complex which has fundamental group pi and all other homotopy groups trivial. So a K(pi,1) has fundamental group pi, which, well, does have torsion if pi does?

Are we all referring to K(pi,1)'s as different objects? I'm pretty sure that the Klein bottle will not have trivial higher homotopy groups.

 

[Jamma]

[STRIKE]Can I just clear up what "aspherical" means? Are we saying something is aspherical iff it has all trivial homotopy groups for n>1 (it's very hard to think of such objects, except graphs and some of the more obvious K(pi,1)'s anyway)?[/STRIKE]
[Edit:] Just looked it up, and aspherical basically just means an Eilenberg-MacLane space with homotopy concentrated in degree 1, as I thought.

Then the statement at the end of the article partially makes sense, although I still don't understand then what he means by surface- I would usually consider RP^2 as a surface, although it being a clear counterexample, the author clearly can not.

 

[lavinia]

Can I just clear up what "aspherical" means? Are we saying something is aspherical iff it has all trivial homotopy groups for n>1 (it's very hard to think of such objects, except graphs and K(pi,1)'s anyway)? Then the statement at the end of the article makes sense, although I still don't understand then what he means by surface- I would usually consider RP^2 as a surface, although it being a clear counterexample, the author clearly can not.

RP^2 is not aspherical

I reread the quote and it is wrong. The author is referring only to K(pi,1)'s but had a mental slip. The sphere and the projective plane are surfaces but are not aspherical. But all orientable surfaces of genus greater than zero are covered by contractible spaces and so are aspherical. The Klein bottle is also aspherical. So the author is almost right.

Here is the quote at the end of the paper.

"Then it follows that the homology of K(Zn; 1) is Zn in each odd dimension. Consequently, no finite CW
complexes can model K(Zn; 1). More generally, if a group G has torsion then K(G; 1) is never a
finite dimensional CW complex. Armed with this observation, we can conclude that the fundamental
group of any surface (or of any aspherical manifold) has no torsion."His point is that a K(pi,1) with torsion in its fundamental group can not be a finite dimensional CW complex. But every manifold is a finite dimensional CW complex. So if a manifold is a K(pi,1) it can not have torsion in its fundamental group.

 

[lavinia]

My definition of a K(pi,1) is a CW complex which has fundamental group pi and all other homotopy groups trivial. So a K(pi,1) has fundamental group pi, which, well, does have torsion if pi does?

Are we all referring to K(pi,1)'s as different objects? I'm pretty sure that the Klein bottle will not have trivial higher homotopy groups.

The Klein bottle is covered by the Euclidean plane. Use the exact homotopy sequence of a fibration to see that it has no higher homotopy groups.

More generally any flat Riemannian manifold is covered by Euclidean space. I think one proof is just that the exponential mapping has no singularities.

 

[Jamma]

Ah, yes, you are right, it has Euclidean space as a universal cover. So, in a sense, this is another proof that the Klein bottle has no torsion in it's fundamental group.

I realized that RP^2 is not aspherical, but I would consider it a surface. I agree that this is probably just a slip by the author (albeit a very weird one- surface theory has nothing to do with anything else he had been talking about!).

 

[Jamma]

Out of curiosity, where does the more general statement with "for any G with torsion" come from? Is it true that if H<G, then the EM-space for G has the EM-space for H as a subcomplex?

 

[Jamma]

I realized that RP^2 is not aspherical, but I would consider it a surface. I agree that this is probably just a slip by the author (albeit a very weird one- surface theory has nothing to do with anything else he had been talking about!).

Ahha, this entry from wikipedia may answer where this slip came from:

"Using the second of above definitions we easily see that all orientable compact surfaces of genus greater than 0 are aspherical (as they have either the Euclidean plane or the hyperbolic plane as a universal cover)"

So he simply forgot to mention "orientable with genus greater than 0". I think that clears up that question.

[and we did well to notice the real projective plane :) -
"It follows that all non-orientable surfaces, except the real projective plane, are aspherical as well, as they can be covered by an orientable surface genus 1 or higher."]

 

[lavinia]

Ah, yes, you are right, it has Euclidean space as a universal cover. So, in a sense, this is another proof that the Klein bottle has no torsion in it's fundamental group.

By this do you mean that a diffeomorphism of Euclidean space of finite order must have a fixed point? Is that true? Seems that is must be.

 

[Jamma]

Well, I was just rephrasing what you had already said really, the Klein bottle has Euclidean space as a universal cover and hence has trivial homotopy past degree one by the reasons you gave.

As for a diffeomorphism of Euclidean space with finite order having a fixed point, I also feel that it must be true.

 

[quasar987]

What's a diffeomorphism of Euclidean space with finite order?

 

[lavinia]

What's a diffeomorphism of Euclidean space with finite order?

finite order means that a finite repetition of the map returns to the identity map.

From the paper I think it follows that no such map can exist without fixed points because otherwise Euclidean space would cover a finite CW complex with a finite cyclic group of covering transformations.

 

[Jamma]

Hmm, I don't see how that follows, how does an arbitrary diffeomorphism lead to a well defined covering of a surface?

This is a nice question though. It seems that there should be a simple trick to solve it, such using finiteness in some way coupled with the fixed point result for closed ball, but I can't think of a thorough reasoning.

 

[quasar987]

Hmm, I don't see how that follows, how does an arbitrary diffeomorphism lead to a well defined covering of a surface?

This is a nice question though. It seems that there should be a simple trick to solve it, such using finiteness in some way coupled with the fixed point result for closed ball, but I can't think of a thorough reasoning.

If f:R^n-->R^n, is a diffeo of order k, it leads to a G=Z_k action on R^n by a.x=f^a(x). If it is free and proper, than the quotient R^n/G is a manifold and the projection R^n-->R^n/G is a covering map with covering transformation group G. If R^n/G is compact, it is homotopy equivalent to a finite CW-complex whose pi_1 is G=Z_k. (contradicting the fact stated in the text)

-Since G is finite, the action is proper. (check)
-Is it sufficient that f be w/o fixed points for the action to be free? Couldn't it be that f^m has a fixed point for some m?
-What about compactness? When is R^n/G compact?I, for one, still don't understand why the sentence "if a group G has torsion then K(G; 1) is never a finite dimensional CW complex" in the text is true.

 

[lavinia]

Hmm, I don't see how that follows, how does an arbitrary diffeomorphism lead to a well defined covering of a surface?

This is a nice question though. It seems that there should be a simple trick to solve it, such using finiteness in some way coupled with the fixed point result for closed ball, but I can't think of a thorough reasoning.

No. A covering transformation can not have fixed points unless it is the identity map.

But more than that is needed. Will have to think about it.

 

[Jamma]

Hmm, really can't think of anything. Maybe something roughly like this would be plausible (at least for dimension 2)?

Say h has order n and pretend it has no fixed point. Pick any point x. The orbit of x defines some points. Make a path from x to h(x). Now, this path also determines a path from h(x) to h(h(x)) by applying h again. Apply h yet again for another path from h(h(x)) to h(h(h(x))) and so on.

The image of all these paths forms a loop of some sort (which may cross itself :S). The loop with some sort of interior filled in must form an invariant subset of the homeomorphism. But because this subset is a topological disk (or wedge of disks which don't form some sort of loop :S), it must have a fixed point.This is clearly not a proof and there are tonnes of things wrong with it, but it seems to be on the right lines.

 

[Jamma]

Ok, so if we accept that the homeomorphism must be proper, we just need to check that it doesn't matter if it is free or not.

Suppose it's not, i.e. it has a fixed point of order k<n (take maximum such k). Then simply replace the homeo h with h^k. The action is still proper and now free and from what you've said this defines a surface with finite covering transformation group, a contradiction.

Compactness- surely it can never be compact if we mod out by a finite group.