# Homotopy between identity and antipodal map

1. Apr 11, 2012

### jgens

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

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

2. Relevant equations

N/A

3. 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
$$h:\left(\sum_{i=1}^{2k+2}x_ie_i,t\right) \mapsto \sum_{i=1}^{k+1}\left(x_{2i-1}\cos{\pi t}-x_{2i}\sin{\pi t}\right)e_{2i-1} + \sum_{i=1}^{k+1}\left(x_{2i-1}\sin{\pi t}+x_{2i}\cos{\pi t}\right)e_{2i}$$
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.

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
$$H:\left(\sum_{i=1}^{2k+2}x_ie_i,t\right) \mapsto \sum_{i=1}^{k+1}\left(x_{2i-1}\cos{\pi t}-x_{2i}\sin{\pi t}\right)e_{2i-1} + \sum_{i=1}^{k+1}\left(x_{2i-1}\sin{\pi t}+x_{2i}\cos{\pi t}\right)e_{2i}$$
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. Apr 11, 2012

### 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.