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Brower degree and index (Poincare-hopf)
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I'm reading the book of J. Milnor "Topology from a differentiable viewpoint" and I got stuck at chapter 6. I don't have much knowledge about topology, so I apologise if I say something silly.
Poincare-Hopf theorem implies that over a $s^2$-sphere a smooth vector field must have, at least, one zero of index two or two zeros of index one.ÂÂ
Let me assume that we have a smooth field with only one zero of index two at the north pole of the $s^2$.
A) Now by definition, the degree of the field in a loop $c_N $ that enclose this zero must be 2, isn't it?.ÂÂ
B) On the other hand, if I take a small circle , $c_S$, around the south pole, the degree of the field over this circle must be zero, since the field only have regular points and extends smoothly to the interior (the one that contains south pole) of this circle (Lemma 1 chap 5)*. ÂÂ
But, by the same reasoning, taking both loops (around the north and the south poles) as a boundary of a manifold composed only by regular points of the field, by lemma 1 the degrees must be equal on both circles, which certainly contradict A-B. Please any comment on this will be very welcome.ÂÂ
In particular, I'm interested in the degree of the field over the boundary of a 2 dimensional disk with a field with index n inside. ÂÂ
   ÂÂ
 *In what follows I'm assuming that if I explicitly compute the degree of the field over $c_s$ from the other half of the $S^2$, the one that corresponds to the north pole, will also be zero, since the field and the curve are the same. Maybe this is what is wrong.
differential-geometry differential-topology
asked Aug 3 at 16:16
viri
63
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up vote
1
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I'm reading the book of J. Milnor "Topology from a differentiable viewpoint" and I got stuck at chapter 6. I don't have much knowledge about topology, so I apologise if I say something silly.
Poincare-Hopf theorem implies that over a $s^2$-sphere a smooth vector field must have, at least, one zero of index two or two zeros of index one.ÂÂ
Let me assume that we have a smooth field with only one zero of index two at the north pole of the $s^2$.
A) Now by definition, the degree of the field in a loop $c_N $ that enclose this zero must be 2, isn't it?.ÂÂ
B) On the other hand, if I take a small circle , $c_S$, around the south pole, the degree of the field over this circle must be zero, since the field only have regular points and extends smoothly to the interior (the one that contains south pole) of this circle (Lemma 1 chap 5)*. ÂÂ
But, by the same reasoning, taking both loops (around the north and the south poles) as a boundary of a manifold composed only by regular points of the field, by lemma 1 the degrees must be equal on both circles, which certainly contradict A-B. Please any comment on this will be very welcome.ÂÂ
In particular, I'm interested in the degree of the field over the boundary of a 2 dimensional disk with a field with index n inside. ÂÂ
   ÂÂ
 *In what follows I'm assuming that if I explicitly compute the degree of the field over $c_s$ from the other half of the $S^2$, the one that corresponds to the north pole, will also be zero, since the field and the curve are the same. Maybe this is what is wrong.
differential-geometry differential-topology
asked Aug 3 at 16:16
viri
63
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I'm reading the book of J. Milnor "Topology from a differentiable viewpoint" and I got stuck at chapter 6. I don't have much knowledge about topology, so I apologise if I say something silly.
Poincare-Hopf theorem implies that over a $s^2$-sphere a smooth vector field must have, at least, one zero of index two or two zeros of index one.ÂÂ
Let me assume that we have a smooth field with only one zero of index two at the north pole of the $s^2$.
A) Now by definition, the degree of the field in a loop $c_N $ that enclose this zero must be 2, isn't it?.ÂÂ
B) On the other hand, if I take a small circle , $c_S$, around the south pole, the degree of the field over this circle must be zero, since the field only have regular points and extends smoothly to the interior (the one that contains south pole) of this circle (Lemma 1 chap 5)*. ÂÂ
But, by the same reasoning, taking both loops (around the north and the south poles) as a boundary of a manifold composed only by regular points of the field, by lemma 1 the degrees must be equal on both circles, which certainly contradict A-B. Please any comment on this will be very welcome.ÂÂ
In particular, I'm interested in the degree of the field over the boundary of a 2 dimensional disk with a field with index n inside. ÂÂ
   ÂÂ
 *In what follows I'm assuming that if I explicitly compute the degree of the field over $c_s$ from the other half of the $S^2$, the one that corresponds to the north pole, will also be zero, since the field and the curve are the same. Maybe this is what is wrong.
differential-geometry differential-topology
asked Aug 3 at 16:16
viri
63
I'm reading the book of J. Milnor "Topology from a differentiable viewpoint" and I got stuck at chapter 6. I don't have much knowledge about topology, so I apologise if I say something silly.
Poincare-Hopf theorem implies that over a $s^2$-sphere a smooth vector field must have, at least, one zero of index two or two zeros of index one.ÂÂ
Let me assume that we have a smooth field with only one zero of index two at the north pole of the $s^2$.
A) Now by definition, the degree of the field in a loop $c_N $ that enclose this zero must be 2, isn't it?.ÂÂ
B) On the other hand, if I take a small circle , $c_S$, around the south pole, the degree of the field over this circle must be zero, since the field only have regular points and extends smoothly to the interior (the one that contains south pole) of this circle (Lemma 1 chap 5)*. ÂÂ
But, by the same reasoning, taking both loops (around the north and the south poles) as a boundary of a manifold composed only by regular points of the field, by lemma 1 the degrees must be equal on both circles, which certainly contradict A-B. Please any comment on this will be very welcome.ÂÂ
In particular, I'm interested in the degree of the field over the boundary of a 2 dimensional disk with a field with index n inside. ÂÂ
   ÂÂ
 *In what follows I'm assuming that if I explicitly compute the degree of the field over $c_s$ from the other half of the $S^2$, the one that corresponds to the north pole, will also be zero, since the field and the curve are the same. Maybe this is what is wrong.
differential-geometry differential-topology
asked Aug 3 at 16:16
viri
63
edited 18 hours ago
asked Aug 3 at 16:16
viri
63
asked Aug 3 at 16:16
viri
63
asked Aug 3 at 16:16
viri
63
63
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