Index of a critical point in a vecor field

logarithmic
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Homework Statement


This is from Henle's Combinatorial Introduction to Topology, section 9 question 2d).

Here's the phase portrait: http://img69.imageshack.us/img69/70/phasepor.jpg

Find the index of the point in the center of that picture.

The Attempt at a Solution


I used the fact that index = 1 + (e - h)/2, where e = number of elliptic sectors, and h = number of hyperbolic sectors.

In the picture, only D is elliptic, so e = 1, and B, E, F are hyperbolic, so h = 3, which gives index = 0. (A and C are parabolic sectors and don't count for the purposes of finding the index.)

The correct answer is -2. Which can be confirmed by noting that if we pick "pointing upwards" as a reference direction, only at X and Y does it point upwards, and at both these places there is a clockwise turn (a turn in the negative direction), so index = -2.

What's wrong with the first method?
 
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logarithmic said:
The correct answer is -2. Which can be confirmed by noting that if we pick "pointing upwards" as a reference direction, only at X and Y does it point upwards, and at both these places there is a clockwise turn (a turn in the negative direction), so index = -2.

Are you trying to compute the winding number here? You can't just pick two points on a circle around the critical point and note the direction that the vector is turning to do that. You need to follow the complete circle and keep track of the total angle swept out by the vector field. If you do this for your diagram, you'll find that that the winding number is zero. For example, if you follow the curve from X to Y, you'll find that the angle swept is 0, not 2\pi.
 
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