In this blog post, Terry tao states that "It is easy to see that $f$ is continuous, homogeneous of degree zero, and odd, and so its restriction to $S^n$ is an antipodal map."

Why is this the case? It may only be the case that it is true in the specific case of this $f$, as defined in the post, but I can't see why these conditions would lead to this conclusion of antipodality.

  • $\begingroup$ Since $f$ is odd we have that $f(-x) =-f(x)$ which is what you need for it to be the antipodal map. I don't see why the other conditions are required. $\endgroup$ – R_D Jan 25 '16 at 15:53
  • $\begingroup$ @Rise it need not be the antipodal map. It need not even be continuous. $\endgroup$ – Matt Samuel Jan 25 '16 at 15:59
  • $\begingroup$ @MattSamuel, I did not know there can be other antipodal maps. What would an example? Thank you. $\endgroup$ – R_D Jan 26 '16 at 4:28
  • 1
    $\begingroup$ @Rise I didn't either, but I'm trusting Tao. In any case, $f(x)=ix$ on $S^3$ considered as the group of unit quaternions is continuous and odd, and can be extended to $\mathbb R^4$ as homogenous of degree 0, but is not "the" antipodal map. $\endgroup$ – Matt Samuel Jan 26 '16 at 4:32
  • $\begingroup$ @MattSamuel, good point. Thank you. $\endgroup$ – R_D Jan 26 '16 at 4:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.