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Math Help - Covering space

  1. #1
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    Covering space

    Give an explicit description of the covering space of the torus T^{2} corresponding to the subgroup of \mathbb{Z}\times\mathbb{Z} generated by (n,0), n>0.
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  2. #2
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    Re: Covering space

    Quote Originally Posted by kierkegaard View Post
    Give an explicit description of the covering space of the torus T^{2} corresponding to the subgroup of \mathbb{Z}\times\mathbb{Z} generated by (n,0), n>0.
    You know that the universal cover of \mathbb{T}^2 is \mathbb{R}^2 and \text{Aut}_D(\mathbb{R}^2) consists of translations by elements of the lattice \mathbb{Z}^2. Your subgroup G corresponds to the subgroup \Gamma_G\subseteq\text{Aut}_D(\mathbb{R}^2) consisting of (x,y)\mapsto (x+n,y). Your covering space \widetilde{\mathbb{T}^2} then corresponds to the quotient \mathbb{R}^2/\Gamma_G, where (as I hope you know) this means that (x,y)\sim (x',y') if they differ by a \Gamma_G-action.
    Last edited by Drexel28; December 1st 2011 at 08:00 PM.
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    Re: Covering space

    Quote Originally Posted by Drexel28 View Post
    You know that the universal cover of \mathbb{T}^2 is \mathbb{R} and \text{Aut}_D(\mathbb{R}^2) consists of translations by elements of the lattice \mathbb{Z}^2. Your subgroup G corresponds to the subgroup \Gamma_G\subseteq\text{Aut}_D(\mathbb{R}^2) consisting of (x,y)\mapsto (x+n,y). Your covering space \widetilde{\mathbb{T}^2} then corresponds to the quotient \mathbb{R}^2/\Gamma_G, where (as I hope you know) this means that (x,y)\sim (x',y') if they differ by a \Gamma_G-action.
    So, and please,correct me if I am wrong, can we say that the covering space corresponding this subgroup is the torus itself?
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    MHF Contributor Drexel28's Avatar
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    Re: Covering space

    Quote Originally Posted by kierkegaard View Post
    So, and please,correct me if I am wrong, can we say that the covering space corresponding this subgroup is the torus itself?
    Ok, well, why would you think that? What is your evidence? Write something out for us.
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    Re: Covering space

    Wait, I'm a little confused. It's a cylinder, isn't it? If we consider the relation (x,y)\mapsto(x+n,y) and the quotient map p:\mathbb{R}\mapsto\mathbb{R}/\sim , the image of p is a cylinder identifying a pair of opposite sides of a square. I hope to be right.
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  6. #6
    MHF Contributor Drexel28's Avatar
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    Re: Covering space

    Quote Originally Posted by kierkegaard View Post
    Wait, I'm a little confused. It's a cylinder, isn't it? If we consider the relation (x,y)\mapsto(x+n,y) and the quotient map p:\mathbb{R}\mapsto\mathbb{R}/\sim , the image of p is a cylinder identifying a pair of opposite sides of a square. I hope to be right.
    Right! Intuitively what we've done is decomposed \pi_1(\mathbb{T}^2)=G_1\oplus G_2 ( G_1=\mathbb{Z}\times\{0\} and G_2=\{0\}\times\mathbb{Z}) which induces a decomposition \text{Aut}_D(\mathbb{R}^2)=\Gamma_{G_1}\oplus \Gamma_{G_2} (where \Gamma_{G_1},\Gamma_{G_2} are the deck transformation (x,y)\mapsto (x+n,y) and (x,y)\mapsto (x,y+m) respectively). Intuitively you can think then of \mathbb{T}^2=\mathbb{R}^2/(\text{Aut}_D(\mathbb{R}^2)) as (\mathbb{R}^2/\Gamma_{G_1})/\Gamma_{G_2} where the identifcation \mathbb{R}^2/\Gamma_{G_1} (our identification) corresponds to taking \mathbb{R}^2 and wrapping it around itself (horizontally) to get a cylinder, and then modding by \Gamma_{G_2} gives you the vertical identifications giving us the torus.



    So, in short, yes. It's the cylinder.
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