Homotopy between identity and antipodal map

[jgens]

Homework Statement

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

Homework Equations

N/A

The Attempt at a Solution

My attempt:
Fix k \in \mathbb{Z}_{\geq 0} and let \{e_i\}_{i=1}^{2k+2} be the standard basis for \mathbb{R}^{2k+2}. Define the map h:S^{2k+1} \times \mathbb{R} \rightarrow S^{2k+1} by setting
<br /> 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}<br />
This map satisfies h(x,0) = x and h(x,1) = -x. To complete the proof it only needs to be shown that h is a smooth map.

Let B^{2k+1} denote the open unit ball in \mathbb{R}^{2k+1} and for each i \in \{1,\dots,2k+2\} define the open hemispheres U_i^{\pm} = \{(x_1,\dots,x_{2k+2}) \in S^{2k+1}:\mathrm{sgn}(x_i) = \pm 1\}. Define for each i \in \{1,\dots,2k+2\} the maps \phi_i^{\pm}:U_i^{\pm} \rightarrow B^{2k+1} such that (x_1,\dots,x_{2k+2}) \mapsto (x_1,\dots,\hat{x_i},\dots,x_{2k+2}). Then the collection \{(U_i^{\pm},\phi_i^{\pm})\} is a smooth atlas for S^{2k+1} and the collection \{(U_i^{\pm} \times \mathbb{R},\phi_i^{\pm} \times \mathrm{id}_{\mathbb{R}})\} is a smooth atlas for S^{2k+1} \times \mathbb{R}. Notice that for each i \in \{1,\dots,2k+2\} a simple computation show that the local representation \phi_i^{\pm} \circ h \circ (\phi_i^{\pm} \times \mathrm{id}_{\mathbb{R}})^{-1} is smooth. This establishes that h 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 h smooth follows directly from the fact that the mapping H:\mathbb{R}^{2k+2} \times \mathbb{R} \rightarrow \mathbb{R}^{2k+2} given by
<br /> 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}<br />
is clearly smooth. If my reasoning is right here, then I think this renders the whole defining atlases thing completely unnecessary.

 

[Dick]

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.