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Cup Product and Representability 
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#19
Jun1511, 05:49 AM

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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 realised 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!). 


#20
Jun1511, 05:55 AM

P: 429

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 EMspace for G has the EMspace for H as a subcomplex?



#21
Jun1511, 05:58 AM

P: 429

"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 nonorientable surfaces, except the real projective plane, are aspherical as well, as they can be covered by an orientable surface genus 1 or higher."] 


#22
Jun1511, 06:30 AM

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#23
Jun1511, 08:41 AM

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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. 


#24
Jun1511, 08:56 AM

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



#25
Jun1511, 09:08 AM

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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. 


#26
Jun1511, 09:37 AM

P: 429

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. 


#27
Jun1511, 10:30 AM

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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. 


#28
Jun1511, 10:36 AM

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


#29
Jun1511, 10:59 AM

P: 429

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. 


#30
Jun1511, 11:13 AM

P: 429

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. 


#31
Jun1611, 08:38 AM

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I don't see how to generalize this. One idea might be to identify the second Z/2 cohomology of a group,G, with group extensions 0 > Z/2 > H > G > 0 and try to show that if the group,G, has torsion then there is always an extension that is not split and which restricts to a nonsplit extension over one of its finite cyclic subgroups. There must be a general construction for a K(pi,1) as a CW complex that has cell in unbounded dimensions when the group has torsion. I will try to find one. I know for Z/2 it is the infinite dimensional real projective space which I think has a cell in every dimension. A K(pi,1) is the universal classifying space for vector bundles whose structure group is the group,pi, with the discrete topology. These are so called flat bundles. Maybe one could show that if pi has torsion then in any dimension there is a manifold with a flat vector bundle over it whose structure group is,pi, and whose Euler class is not zero. 


#32
Jun1611, 09:44 AM

P: 429

The paper goes through the construction for the EM1spaces for the cyclic groups at the end they are the infinite lens spaces and indeed their homology is non zero in infinitely many dimensions. It seems to go from this to arbitrary groups quite suddenly I don't know if this is intentional to imply that not much more work needs to be done, or it is simply mentioning the generalisation. However, it seems that the K(Zn,1)'s having to be infinite dimensional should somehow imply the same for any K(G,1) for which G has torsion. Your group extension idea is good though I'm sure more can be done with that. 


#33
Jun1611, 09:51 AM

P: 429

Ahha, I seem to remember something along the lines of this:
If we have a group extension such as: 0 > H > G > G/H > 0 Then this leads canonically to a fibre bundle: K(H,1) > K(G,1) > K(G/H) I'm not 100% sure on this, will check up on it when I get time. I don't even know if this helps perhaps there is some sort of spectral sequence argument, although it would seem that we should know something about K(G/H,1). 


#34
Jun1611, 09:59 AM

P: 429

Or perhaps I'm overcomplicating things in the above another way of defining group cohomology is to take the simplicial cohomology of the classifying space (which is a K(Z,1) in the discrete case). So maybe there is an argument in here with exact sequences.



#35
Jun1611, 10:05 AM

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