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In reviewing the familiar Poincare-Hopf theorem I come across the following question:

Suppose $x$ an isolated 0 of $V$. Pick up a disk around $x$ in its neighborhood. Calculate the degree of the map $$u:\partial D \rightarrow S^{m-1},u(z)=\frac{V(z)}{|V(z)|}$$

where $V(z)$ is the the map $M\rightarrow TM$ which represents a vector field.

I am confused because while in case 1 and 2 it is clear the degree $S^{1}\rightarrow S^{1}$ must be 1 since they are orientation preserving maps combined with rotation, I do not know how to calculate case 3. The author claimed the index is -1. But how? A hint would be mostly welcome.

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In the last case you "turn around" clockwise. –  M.B. Aug 8 '12 at 22:30
    
I still feel confused. The last case seems quite complicated. With a rotation the top and bottom part has degree 1, but left and right part is an antipodal map. If antipodal maps have degree -1 then case 1 should have degree -1 too. –  Bombyx mori Aug 8 '12 at 22:40

1 Answer 1

up vote 2 down vote accepted

Draw a circle around the $0$. Then, at each point of this circle there will be a vector. As you travel around the circle counterclockwise, the direction of the vector at that point will change. The number of times this vector rotates about the origin, counting orientation, is the degree of the Gauss map.

Try keeping track of the vector in case two as you travel around the $0$. You should notice that the vector rotates clockwise exactly once, and hence the degree is $-1$.

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great. This is absolutely clear. –  Bombyx mori Aug 8 '12 at 23:13
    
A nice explanation! However, it's difficult to find the integral curves of a given vector field (even in dimension 2!). –  amine Apr 24 '13 at 16:09

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