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I want to know of a fixed-point-free map from a 2-sphere minus a point to itself. Please also mention why you thought of your example.

My example: a sphere minus the north pole is the complex plane via stereographic projection, and consider a non-trivial translation of the plane.

Kyle Miller
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user345777
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1 Answers1

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Take a vector field on a 2-sphere which vanishes only at a single point, and a flow of the identity map will give a diffeomorphism with no fixed points in the complement of that single point.

Your example is the case of taking the constant vector field on the plane. The inverse of the stereographic projection of this vector field to the sphere gives a vector field with a single index-2 isolated zero.

The Euler characteristic of a sphere is $2$, so I believe your example should be the only example up to homotopy.

Kyle Miller
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  • Could you elaborate on how the Euler Characteristic leads to the homotopy comment? – Duncan Ramage Sep 10 '17 at 21:33
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    For a vector field with isolated singularities, the sum of the indices of the singularities is equal to the Euler characteristic (Hopf index theorem). This forces the vector field to have an index-2 singularity at the puncture, and by the Lefschetz fixed point theorem the Gauss map is degree 1. Two sphere maps with the same degree are homotopic. Given a map without fixed points, you can obtain a vector field; I imagined doing it from the stereographic projection. (More directly, I think you can just use this vector field to get a homotopy of the fixed-point-free map to the identity.) – Kyle Miller Sep 10 '17 at 21:47