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Homotopy between identity and antipodal map

  1. Apr 11, 2012 #1

    jgens

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    Gold Member

    1. The problem statement, all variables and given/known data

    Prove that the identity map [itex]\mathrm{id}_{S^{2k+1}}[/itex] and the antipodal map [itex]-\mathrm{id}_{S^{2k+1}}[/itex] are smoothly homotopic.

    2. Relevant equations

    N/A

    3. The attempt at a solution

    My attempt:
    Fix [itex]k \in \mathbb{Z}_{\geq 0}[/itex] and let [itex]\{e_i\}_{i=1}^{2k+2}[/itex] be the standard basis for [itex]\mathbb{R}^{2k+2}[/itex]. Define the map [itex]h:S^{2k+1} \times \mathbb{R} \rightarrow S^{2k+1}[/itex] by setting
    [tex]
    h:\left(\sum_{i=1}^{2k+2}x_ie_i,t\right) \mapsto \sum_{i=1}^{k+1}\left(x_{2i-1}\cos{\pi t}-x_{2i}\sin{\pi t}\right)e_{2i-1} + \sum_{i=1}^{k+1}\left(x_{2i-1}\sin{\pi t}+x_{2i}\cos{\pi t}\right)e_{2i}
    [/tex]
    This map satisfies [itex]h(x,0) = x[/itex] and [itex]h(x,1) = -x[/itex]. To complete the proof it only needs to be shown that [itex]h[/itex] is a smooth map.

    Let [itex]B^{2k+1}[/itex] denote the open unit ball in [itex]\mathbb{R}^{2k+1}[/itex] and for each [itex]i \in \{1,\dots,2k+2\}[/itex] define the open hemispheres [itex]U_i^{\pm} = \{(x_1,\dots,x_{2k+2}) \in S^{2k+1}:\mathrm{sgn}(x_i) = \pm 1\}[/itex]. Define for each [itex]i \in \{1,\dots,2k+2\}[/itex] the maps [itex]\phi_i^{\pm}:U_i^{\pm} \rightarrow B^{2k+1}[/itex] such that [itex](x_1,\dots,x_{2k+2}) \mapsto (x_1,\dots,\hat{x_i},\dots,x_{2k+2})[/itex]. Then the collection [itex]\{(U_i^{\pm},\phi_i^{\pm})\}[/itex] is a smooth atlas for [itex]S^{2k+1}[/itex] and the collection [itex]\{(U_i^{\pm} \times \mathbb{R},\phi_i^{\pm} \times \mathrm{id}_{\mathbb{R}})\}[/itex] is a smooth atlas for [itex]S^{2k+1} \times \mathbb{R}[/itex]. Notice that for each [itex]i \in \{1,\dots,2k+2\}[/itex] a simple computation show that the local representation [itex]\phi_i^{\pm} \circ h \circ (\phi_i^{\pm} \times \mathrm{id}_{\mathbb{R}})^{-1}[/itex] is smooth. This establishes that [itex]h[/itex] is a smooth map and completes the proof.


    My comments:
    So obviously there a some details missing, in particular, that the local representations are smooth. When I wrote it out, the formula was rather long, so I decided not to include it.That aside, here are my questions ...
    1) Does this work and if so is there a cleaner solution?
    2) If this works, is including the atlases for our manifolds necessary? I mean, strictly speaking, it certainly is as well as a computation that shows that the local representations are smooth. But for the purposes of this proof, I feel that including the atlases and defining the smooth local representations does little to help the presentation.

    Edit:
    Thinking about the problem more, I am pretty sure that [itex]h[/itex] smooth follows directly from the fact that the mapping [itex]H:\mathbb{R}^{2k+2} \times \mathbb{R} \rightarrow \mathbb{R}^{2k+2}[/itex] given by
    [tex]
    H:\left(\sum_{i=1}^{2k+2}x_ie_i,t\right) \mapsto \sum_{i=1}^{k+1}\left(x_{2i-1}\cos{\pi t}-x_{2i}\sin{\pi t}\right)e_{2i-1} + \sum_{i=1}^{k+1}\left(x_{2i-1}\sin{\pi t}+x_{2i}\cos{\pi t}\right)e_{2i}
    [/tex]
    is clearly smooth. If my reasoning is right here, then I think this renders the whole defining atlases thing completely unnecessary.
     
    Last edited: Apr 11, 2012
  2. jcsd
  3. Apr 11, 2012 #2

    Dick

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    Science Advisor
    Homework Helper

    I think you are ok. id:R^2->R^2 is homotopic to -id:R^2->R^2. You just rotate the identity by pi to get -id. And that fixes the sphere S^1. It's clearly smooth. And as you showed in an even dimensional space you can just pick the basis vectors pairwise and do the same thing.
     
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