Mixing
General case of a result in differential geometry
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Let $W subset Bbb R^n$ open and $F:W to Bbb R$, $F$ is $C^infty$ and $alt b in F(W)$ so that $L=F^-1([a,b])$ is compact and $nabla F(p) ne 0$ for every $p in L$
By taking the field $X = frac nabla F^2 $ then the flow $Æ^X_b-a$ gives us a diffeomormphism between $F^-1(a)$ and $F^-1(b)$.
My question is if we can generalise this for $Bbb R^k$ instead of $Bbb R$?
differential-geometry morse-theory
edited Aug 2 at 17:51
John Ma
37.5k93669
asked Aug 2 at 16:18
Amontillado
433311
add a comment |Â
up vote
2
down vote
favorite
Let $W subset Bbb R^n$ open and $F:W to Bbb R$, $F$ is $C^infty$ and $alt b in F(W)$ so that $L=F^-1([a,b])$ is compact and $nabla F(p) ne 0$ for every $p in L$
By taking the field $X = frac nabla F^2 $ then the flow $Æ^X_b-a$ gives us a diffeomormphism between $F^-1(a)$ and $F^-1(b)$.
My question is if we can generalise this for $Bbb R^k$ instead of $Bbb R$?
differential-geometry morse-theory
edited Aug 2 at 17:51
John Ma
37.5k93669
asked Aug 2 at 16:18
Amontillado
433311
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $W subset Bbb R^n$ open and $F:W to Bbb R$, $F$ is $C^infty$ and $alt b in F(W)$ so that $L=F^-1([a,b])$ is compact and $nabla F(p) ne 0$ for every $p in L$
By taking the field $X = frac nabla F^2 $ then the flow $Æ^X_b-a$ gives us a diffeomormphism between $F^-1(a)$ and $F^-1(b)$.
My question is if we can generalise this for $Bbb R^k$ instead of $Bbb R$?
differential-geometry morse-theory
edited Aug 2 at 17:51
John Ma
37.5k93669
asked Aug 2 at 16:18
Amontillado
433311
Let $W subset Bbb R^n$ open and $F:W to Bbb R$, $F$ is $C^infty$ and $alt b in F(W)$ so that $L=F^-1([a,b])$ is compact and $nabla F(p) ne 0$ for every $p in L$
By taking the field $X = frac nabla F^2 $ then the flow $Æ^X_b-a$ gives us a diffeomormphism between $F^-1(a)$ and $F^-1(b)$.
My question is if we can generalise this for $Bbb R^k$ instead of $Bbb R$?
differential-geometry morse-theory
edited Aug 2 at 17:51
John Ma
37.5k93669
asked Aug 2 at 16:18
Amontillado
433311
edited Aug 2 at 17:51
John Ma
37.5k93669
edited Aug 2 at 17:51
John Ma
37.5k93669
edited Aug 2 at 17:51
John Ma
37.5k93669
37.5k93669
asked Aug 2 at 16:18
Amontillado
433311
asked Aug 2 at 16:18
Amontillado
433311
asked Aug 2 at 16:18
Amontillado
433311
433311
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
No.
For example, think of a regular curve (so $nabla gamma neq 0$) $gamma : (-epsilon, epsilon) to mathbb R^2$ which has a self intersection at $a=(0,0)in mathbb R^2$. Then $gamma^-1(a)$ has more than one points while $gamma^-1(b)$ has only one point in general.
answered Aug 2 at 17:54
John Ma
37.5k93669
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
No.
For example, think of a regular curve (so $nabla gamma neq 0$) $gamma : (-epsilon, epsilon) to mathbb R^2$ which has a self intersection at $a=(0,0)in mathbb R^2$. Then $gamma^-1(a)$ has more than one points while $gamma^-1(b)$ has only one point in general.
answered Aug 2 at 17:54
John Ma
37.5k93669
add a comment |Â
up vote
1
down vote
accepted
No.
For example, think of a regular curve (so $nabla gamma neq 0$) $gamma : (-epsilon, epsilon) to mathbb R^2$ which has a self intersection at $a=(0,0)in mathbb R^2$. Then $gamma^-1(a)$ has more than one points while $gamma^-1(b)$ has only one point in general.
answered Aug 2 at 17:54
John Ma
37.5k93669
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
No.
For example, think of a regular curve (so $nabla gamma neq 0$) $gamma : (-epsilon, epsilon) to mathbb R^2$ which has a self intersection at $a=(0,0)in mathbb R^2$. Then $gamma^-1(a)$ has more than one points while $gamma^-1(b)$ has only one point in general.
answered Aug 2 at 17:54
John Ma
37.5k93669
No.
For example, think of a regular curve (so $nabla gamma neq 0$) $gamma : (-epsilon, epsilon) to mathbb R^2$ which has a self intersection at $a=(0,0)in mathbb R^2$. Then $gamma^-1(a)$ has more than one points while $gamma^-1(b)$ has only one point in general.
answered Aug 2 at 17:54
John Ma
37.5k93669
answered Aug 2 at 17:54
John Ma
37.5k93669
answered Aug 2 at 17:54
John Ma
37.5k93669
answered Aug 2 at 17:54
John Ma
37.5k93669
37.5k93669
add a comment |Â
add a comment |Â
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