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Problem with Milnor´s proof of invariance of index by diffeomorphisms
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I am reading Milnor's ''Topology from the differentiable viewpoint'' and there is a proof (Lemma 1 in section 6 if you have acces to the book) of this lemma:
Suppose that the vector field $v$ in $U$ corresponds to $v'=df circ v circ f^-1 $ on $U'$, under a diffeomorphism $f: U to U'$. Then the index of $v$ at an isolated zero is equal to the index of $v'$ at $f(z)$
So, he proves this for orientation preserving diffeomorphisms, and to prove it for orientation reversing diffeomorphisms it starts:
'' To consider diffeomorphisms that reverse orientation, it is sufficient to consider the special case of a reflection $rho$.''
And then carries on completing the proof using $rho$.
Why is it that you can just consider a reflection? It probably is very easy to see, since I find this a really good book, and he only avoids explaining trivial stuff. But I cannot seem to see it.
differential-topology vector-fields
asked Jul 31 at 14:37
Bajo Fondo
376213
add a comment |Â
up vote
2
down vote
favorite
I am reading Milnor's ''Topology from the differentiable viewpoint'' and there is a proof (Lemma 1 in section 6 if you have acces to the book) of this lemma:
Suppose that the vector field $v$ in $U$ corresponds to $v'=df circ v circ f^-1 $ on $U'$, under a diffeomorphism $f: U to U'$. Then the index of $v$ at an isolated zero is equal to the index of $v'$ at $f(z)$
So, he proves this for orientation preserving diffeomorphisms, and to prove it for orientation reversing diffeomorphisms it starts:
'' To consider diffeomorphisms that reverse orientation, it is sufficient to consider the special case of a reflection $rho$.''
And then carries on completing the proof using $rho$.
Why is it that you can just consider a reflection? It probably is very easy to see, since I find this a really good book, and he only avoids explaining trivial stuff. But I cannot seem to see it.
differential-topology vector-fields
asked Jul 31 at 14:37
Bajo Fondo
376213
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I am reading Milnor's ''Topology from the differentiable viewpoint'' and there is a proof (Lemma 1 in section 6 if you have acces to the book) of this lemma:
Suppose that the vector field $v$ in $U$ corresponds to $v'=df circ v circ f^-1 $ on $U'$, under a diffeomorphism $f: U to U'$. Then the index of $v$ at an isolated zero is equal to the index of $v'$ at $f(z)$
So, he proves this for orientation preserving diffeomorphisms, and to prove it for orientation reversing diffeomorphisms it starts:
'' To consider diffeomorphisms that reverse orientation, it is sufficient to consider the special case of a reflection $rho$.''
And then carries on completing the proof using $rho$.
Why is it that you can just consider a reflection? It probably is very easy to see, since I find this a really good book, and he only avoids explaining trivial stuff. But I cannot seem to see it.
differential-topology vector-fields
asked Jul 31 at 14:37
Bajo Fondo
376213
I am reading Milnor's ''Topology from the differentiable viewpoint'' and there is a proof (Lemma 1 in section 6 if you have acces to the book) of this lemma:
Suppose that the vector field $v$ in $U$ corresponds to $v'=df circ v circ f^-1 $ on $U'$, under a diffeomorphism $f: U to U'$. Then the index of $v$ at an isolated zero is equal to the index of $v'$ at $f(z)$
So, he proves this for orientation preserving diffeomorphisms, and to prove it for orientation reversing diffeomorphisms it starts:
'' To consider diffeomorphisms that reverse orientation, it is sufficient to consider the special case of a reflection $rho$.''
And then carries on completing the proof using $rho$.
Why is it that you can just consider a reflection? It probably is very easy to see, since I find this a really good book, and he only avoids explaining trivial stuff. But I cannot seem to see it.
differential-topology vector-fields
asked Jul 31 at 14:37
Bajo Fondo
376213
asked Jul 31 at 14:37
Bajo Fondo
376213
asked Jul 31 at 14:37
Bajo Fondo
376213
asked Jul 31 at 14:37
Bajo Fondo
376213
376213
add a comment |Â
add a comment |Â
1 Answer
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1
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You can assume that $U'subset mathbbR^n$. Let $r$ be a reflection such that $ r(f(x))=f(x) $. If $f$ does not preserve the orientation $rcirc f$ does and $ind(rcirc f)'(v) _r(f(x)) =ind(v_x)$ since $rcirc f$ preserves the orientation. We also have $ind(rcirc f)'(v) _r(f(x)) =ind(r'(f'(v))_r(f(x)) =ind(f'(v))_f(x) $. Since $r$ is a reflection.
answered Jul 31 at 15:16
Tsemo Aristide
50.8k11143
I'm sorry but I do not undertand your notation. What is $ind(f)(v)_x$? Thanks in advanced.
– Bajo Fondo
Jul 31 at 15:30
$ind(v_x) $ is the index of $v$ at $x$.
– Tsemo Aristide
Jul 31 at 16:20
Ok... I get it you have $v''=r circ df circ v circ f^-1circ r^-1$, then $v''$has the same index as $v$ (because $r circ f$ preserves orientation). Then because $v''=r circ v' circ r^-1$, you have that $v'$ and $v''$ have the same index, hence $v$ and $v'$ also share the same index.
– Bajo Fondo
Jul 31 at 18:07
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
You can assume that $U'subset mathbbR^n$. Let $r$ be a reflection such that $ r(f(x))=f(x) $. If $f$ does not preserve the orientation $rcirc f$ does and $ind(rcirc f)'(v) _r(f(x)) =ind(v_x)$ since $rcirc f$ preserves the orientation. We also have $ind(rcirc f)'(v) _r(f(x)) =ind(r'(f'(v))_r(f(x)) =ind(f'(v))_f(x) $. Since $r$ is a reflection.
answered Jul 31 at 15:16
Tsemo Aristide
50.8k11143
I'm sorry but I do not undertand your notation. What is $ind(f)(v)_x$? Thanks in advanced.
– Bajo Fondo
Jul 31 at 15:30
$ind(v_x) $ is the index of $v$ at $x$.
– Tsemo Aristide
Jul 31 at 16:20
Ok... I get it you have $v''=r circ df circ v circ f^-1circ r^-1$, then $v''$has the same index as $v$ (because $r circ f$ preserves orientation). Then because $v''=r circ v' circ r^-1$, you have that $v'$ and $v''$ have the same index, hence $v$ and $v'$ also share the same index.
– Bajo Fondo
Jul 31 at 18:07
add a comment |Â
up vote
1
down vote
accepted
You can assume that $U'subset mathbbR^n$. Let $r$ be a reflection such that $ r(f(x))=f(x) $. If $f$ does not preserve the orientation $rcirc f$ does and $ind(rcirc f)'(v) _r(f(x)) =ind(v_x)$ since $rcirc f$ preserves the orientation. We also have $ind(rcirc f)'(v) _r(f(x)) =ind(r'(f'(v))_r(f(x)) =ind(f'(v))_f(x) $. Since $r$ is a reflection.
answered Jul 31 at 15:16
Tsemo Aristide
50.8k11143
I'm sorry but I do not undertand your notation. What is $ind(f)(v)_x$? Thanks in advanced.
– Bajo Fondo
Jul 31 at 15:30
$ind(v_x) $ is the index of $v$ at $x$.
– Tsemo Aristide
Jul 31 at 16:20
Ok... I get it you have $v''=r circ df circ v circ f^-1circ r^-1$, then $v''$has the same index as $v$ (because $r circ f$ preserves orientation). Then because $v''=r circ v' circ r^-1$, you have that $v'$ and $v''$ have the same index, hence $v$ and $v'$ also share the same index.
– Bajo Fondo
Jul 31 at 18:07
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
You can assume that $U'subset mathbbR^n$. Let $r$ be a reflection such that $ r(f(x))=f(x) $. If $f$ does not preserve the orientation $rcirc f$ does and $ind(rcirc f)'(v) _r(f(x)) =ind(v_x)$ since $rcirc f$ preserves the orientation. We also have $ind(rcirc f)'(v) _r(f(x)) =ind(r'(f'(v))_r(f(x)) =ind(f'(v))_f(x) $. Since $r$ is a reflection.
answered Jul 31 at 15:16
Tsemo Aristide
50.8k11143
You can assume that $U'subset mathbbR^n$. Let $r$ be a reflection such that $ r(f(x))=f(x) $. If $f$ does not preserve the orientation $rcirc f$ does and $ind(rcirc f)'(v) _r(f(x)) =ind(v_x)$ since $rcirc f$ preserves the orientation. We also have $ind(rcirc f)'(v) _r(f(x)) =ind(r'(f'(v))_r(f(x)) =ind(f'(v))_f(x) $. Since $r$ is a reflection.
answered Jul 31 at 15:16
Tsemo Aristide
50.8k11143
edited Jul 31 at 15:22
answered Jul 31 at 15:16
Tsemo Aristide
50.8k11143
answered Jul 31 at 15:16
Tsemo Aristide
50.8k11143
answered Jul 31 at 15:16
Tsemo Aristide
50.8k11143
50.8k11143
I'm sorry but I do not undertand your notation. What is $ind(f)(v)_x$? Thanks in advanced.
– Bajo Fondo
Jul 31 at 15:30
$ind(v_x) $ is the index of $v$ at $x$.
– Tsemo Aristide
Jul 31 at 16:20
Ok... I get it you have $v''=r circ df circ v circ f^-1circ r^-1$, then $v''$has the same index as $v$ (because $r circ f$ preserves orientation). Then because $v''=r circ v' circ r^-1$, you have that $v'$ and $v''$ have the same index, hence $v$ and $v'$ also share the same index.
– Bajo Fondo
Jul 31 at 18:07
add a comment |Â
I'm sorry but I do not undertand your notation. What is $ind(f)(v)_x$? Thanks in advanced.
– Bajo Fondo
Jul 31 at 15:30
$ind(v_x) $ is the index of $v$ at $x$.
– Tsemo Aristide
Jul 31 at 16:20
Ok... I get it you have $v''=r circ df circ v circ f^-1circ r^-1$, then $v''$has the same index as $v$ (because $r circ f$ preserves orientation). Then because $v''=r circ v' circ r^-1$, you have that $v'$ and $v''$ have the same index, hence $v$ and $v'$ also share the same index.
– Bajo Fondo
Jul 31 at 18:07
I'm sorry but I do not undertand your notation. What is $ind(f)(v)_x$? Thanks in advanced.
– Bajo Fondo
Jul 31 at 15:30
I'm sorry but I do not undertand your notation. What is $ind(f)(v)_x$? Thanks in advanced.
– Bajo Fondo
Jul 31 at 15:30
$ind(v_x) $ is the index of $v$ at $x$.
– Tsemo Aristide
Jul 31 at 16:20
$ind(v_x) $ is the index of $v$ at $x$.
– Tsemo Aristide
Jul 31 at 16:20
Ok... I get it you have $v''=r circ df circ v circ f^-1circ r^-1$, then $v''$has the same index as $v$ (because $r circ f$ preserves orientation). Then because $v''=r circ v' circ r^-1$, you have that $v'$ and $v''$ have the same index, hence $v$ and $v'$ also share the same index.
– Bajo Fondo
Jul 31 at 18:07
Ok... I get it you have $v''=r circ df circ v circ f^-1circ r^-1$, then $v''$has the same index as $v$ (because $r circ f$ preserves orientation). Then because $v''=r circ v' circ r^-1$, you have that $v'$ and $v''$ have the same index, hence $v$ and $v'$ also share the same index.
– Bajo Fondo
Jul 31 at 18:07
add a comment |Â
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