Are Homotopic Maps of the n-Sphere Antipodal?

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Discussion Overview

The discussion revolves around the homotopy of maps from the n-sphere to itself, specifically addressing the condition that these maps are never antipodal. Participants explore various approaches to demonstrate that such maps are homotopic, engaging in technical reasoning and mathematical exploration.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant suggests working with the n-cube and its boundary to lift maps from Sn to In, noting that In is contractible and could provide a homotopy between the lifted maps.
  • Another participant proposes that the non-antipodal condition implies a unique length-minimizing great circle connecting the images of the maps.
  • A participant questions the continuity of homotopies between points, indicating that having homotopies for corresponding points does not necessarily yield a homotopy between the maps themselves.
  • It is mentioned that moving along great circles can be achieved by connecting points in Euclidean space and projecting onto the sphere, maintaining the non-antipodal condition.
  • A later reply presents a specific formula for a homotopy, indicating a successful approach to the problem.

Areas of Agreement / Disagreement

Participants express differing views on the nature of homotopies and the implications of the non-antipodal condition. There is no consensus on a definitive method or conclusion regarding the homotopy of the maps.

Contextual Notes

Some participants express uncertainty about the application of their ideas, particularly regarding the continuity of homotopies and the implications of the non-antipodal condition. The discussion includes various assumptions and mathematical steps that remain unresolved.

alexfloo
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"Let f,g:Sn→Sn be maps so that f(x) and g(x) are not antipodal for any x. Show that f and g are homotopic."

Here's my initial approach: I figured it would be easier to work in In instead, so I note that Sn is the quotient of the n-cube with its boundary. Therefore, each map Sn→Sn can be lifted to a map In→Sn which is constant on the boundary in Rn.

It may also be possible to use a covering space to lift the other end of the map, getting In→Rn. I played around with this in the n=1 case (the only case where I'm certain it'd work) and it didn't seem to help.

In any case, In is contractible, so we have a homotopy F between f' and g' (our lifted maps). I just can't figure out how to use the fact that f and g are never antipodal to prove that F can be deformed into a homotopy which is constant on the boundary at ever stage. I understand intuitively that that restriction prevents one map from ever "wrapping around" more than the other, but I'm not certain how to apply it.
 
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Not sure but if the two maps are never anitpodal then there is a unique length minimizing great circle connecting them.
 
alexfloo said:
"Let f,g:Sn→Sn be maps so that f(x) and g(x) are not antipodal for any x. Show that f and g are homotopic."

Here's my initial approach: I figured it would be easier to work in In instead, so I note that Sn is the quotient of the n-cube with its boundary. Therefore, each map Sn→Sn can be lifted to a map In→Sn which is constant on the boundary in Rn.

It may also be possible to use a covering space to lift the other end of the map, getting In→Rn. I played around with this in the n=1 case (the only case where I'm certain it'd work) and it didn't seem to help.

In any case, In is contractible, so we have a homotopy F between f' and g' (our lifted maps). I just can't figure out how to use the fact that f and g are never antipodal to prove that F can be deformed into a homotopy which is constant on the boundary at ever stage. I understand intuitively that that restriction prevents one map from ever "wrapping around" more than the other, but I'm not certain how to apply it.

Consider a homotopy between the two points in R^n and then rescale so that the

homotopy is moved to Sn
 
A homotopy between which two points? Having a homotopy between f(x) and g(x) for each fixed x doesn't give me a homotopy between f and g, since those homotopies need not vary continuously with x.
 
If homotopies between corresponding points in paths constituted a homotopy of those paths, then every path-connected space would be simply connected.
 
You can move along the great circles by moving along the line segment in Euclidean space connecting f(x) to g(x) then projecting onto the sphere. The line segment will not pass through the origin because f(s) and g(x) are not antipodal.
 
I meant a homotopy between f(x) and g(x) in general; the (arguably) simplest homotopy.
 
That worked perfectly:

F(z,t) = \frac{tf(z) + (1-t)g(z)}{|tf(z) + (1-t)g(z)|}

is my homotopy. Thanks a lot!
 
Glad it helped.
 

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