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I am trying to construct a variety of manifolds which have a topological group structure. I feel that it is possible to do so by taking two manifolds with group structures and defining a bundle of one over the other. For example, can we make an object such as a Klein bottle a manifold with group structure in the following way? Identify the Klein bottle $K$ with $[0,1]\times [0,1]/$~ where ~ denotes the usual quotient (i.e. $(x,0)$~$(x,1)$ and $(0,y)$~$(1,1-y)$). We can define addition of two points mod 1 and then take the quotient. It is my intuition that this would be a well defined group structure. If I am thinking of this correctly it seems that the group would be isomorphic to the torus group. Does the construction I describe make sense? Can it be generalized to fibre bundles?

Joseph Zambrano
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  • Thank you! Is there a well defined way to construct manifolds with group structure given two manifolds with group structures? (besides a simple product) – Joseph Zambrano Mar 01 '14 at 17:29
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    If $G$ is a topological group and $p\colon H\to G$ is a covering space projection, then there is an induced topological group structure on $H$ which makes $p$ into a continuous surjective group homomorphism. See for instance http://math.stackexchange.com/questions/5391/covering-of-a-topological-group-is-a-topological-group. In this case, I guess the fiber is a discrete topological group so $H$ can be seen to be a semdirect product of $G$ and the fiber of the map $p$ above the identity. – Dan Rust Mar 01 '14 at 17:40
  • Thanks again. I'm not sure if this is correct, but putting together the link you posted and the fact that a mapping torus for a given homeomorphism of some surface is a fiber bundle over $S^1$ (i.e a topological group). If the condition that the image of the fundamental group $\pi_1(H\times H,(h,h))$ is a subset of $\pi_1(H,h)$ is satisfied, where $H$ is the mapping torus, then does the mapping torus have the structure of a topological group. Forgive me if this does not make sense. – Joseph Zambrano Mar 01 '14 at 21:16
  • I'd need to brush up on my notes as it's been a while but it seems reasonable on the face of it. It might be good to post that as a separate question in my opinion if you want a better chance at a definitive answer. Sorry I can't be of more help. – Dan Rust Mar 01 '14 at 21:42
  • Thank you! I was afraid to post as a separate question as I am relatively new on stackexchange and was unsure of the appropriateness of asking a similar question. – Joseph Zambrano Mar 01 '14 at 21:44
  • I think it's sufficiently different to your original question that it warrants a different answer and so should be made into a new question (if you want to ask it). As an aside, welcome to math.se :). the community is generally welcoming of well-thought and carefully constructed questions like yours, especially when they show an intent to learn and a previously made effort to answer your own question. – Dan Rust Mar 01 '14 at 21:54

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The quotient also has to respect the group structure which yours does not.

For instance $(0,\frac{1}{4})\sim (1,\frac{3}{4})$ but

$(0,\frac{1}{4})+(\frac{1}{2},0)=(\frac{1}{2},\frac{1}{4})$ and

$(1,\frac{3}{4})+(\frac{1}{2},0)=(\frac{1}{2},\frac{3}{4})\nsim(\frac{1}{2},\frac{1}{4})$.

So you do not have a well defined group structure on the quotient.

To be more specific, if $[x]$ is the equivalence class of the point $x$, then for all $y$ and all $x'\in[x]$ you need $[xy]=[x'y]$.

Dan Rust
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