Existence of a diffemorphism that maps one curve to another
Consider $1 < nin mathbb {N} $, let $gamma_1,gamma_2 : [0,1] to mathbb{R}^{n}$ be smooth paths such that
$$gamma_1 (0) = gamma_2(0) neq gamma_1(1) = gamma_2(1) $$
and $gamma_1, gamma_2$ are injetive functions.
Question: Does there exist a diffeomorphism $varphi: mathbb{R}^n to mathbb{R}^n$ such that $varphi (gamma_1 ([0,1])) = gamma_2 ([0,1]).$
Does anyone know if this result is true?
This seems true but I do not know how to prove it, can anyone help me?
analysis differential-geometry differential-topology
asked Nov 5 at 18:51
Matheus Manzatto
1,3111523
add a comment |
Consider $1 < nin mathbb {N} $, let $gamma_1,gamma_2 : [0,1] to mathbb{R}^{n}$ be smooth paths such that
$$gamma_1 (0) = gamma_2(0) neq gamma_1(1) = gamma_2(1) $$
and $gamma_1, gamma_2$ are injetive functions.
Question: Does there exist a diffeomorphism $varphi: mathbb{R}^n to mathbb{R}^n$ such that $varphi (gamma_1 ([0,1])) = gamma_2 ([0,1]).$
Does anyone know if this result is true?
This seems true but I do not know how to prove it, can anyone help me?
analysis differential-geometry differential-topology
asked Nov 5 at 18:51
Matheus Manzatto
1,3111523
Is $Bbb R^{n>1}$ the set of functions from the interval ${n >1}$ into $Bbb R$ endowed with uniform convergence on each compact subinterval?
– Will M.
Nov 5 at 18:54
No, it is $mathbb {R}^n$ with $n>1$. I will change.
– Matheus Manzatto
Nov 5 at 18:55
Are the $gamma_i$ $C^1$? $C^infty$? Merely continuous?
– Jason DeVito
Nov 5 at 19:00
It cannot be true as stated because you can take $gamma_1([0,1])$ to be a line segment while $gamma_2([0, 1])$ to be two disconnected line segments. I still think this will be false if you do not have $gamma_i$ ($i=1,2$) to be differentiable with continuity. (To see why this would have to fail if the $gamma_i$ are only continuous, take a square and each $gamma_i$ be two of its sides.)
– Will M.
Nov 5 at 19:00
Smooth functions, sorry
– Matheus Manzatto
Nov 5 at 19:00
add a comment |
Consider $1 < nin mathbb {N} $, let $gamma_1,gamma_2 : [0,1] to mathbb{R}^{n}$ be smooth paths such that
$$gamma_1 (0) = gamma_2(0) neq gamma_1(1) = gamma_2(1) $$
and $gamma_1, gamma_2$ are injetive functions.
Question: Does there exist a diffeomorphism $varphi: mathbb{R}^n to mathbb{R}^n$ such that $varphi (gamma_1 ([0,1])) = gamma_2 ([0,1]).$
Does anyone know if this result is true?
This seems true but I do not know how to prove it, can anyone help me?
analysis differential-geometry differential-topology
asked Nov 5 at 18:51
Matheus Manzatto
1,3111523
Consider $1 < nin mathbb {N} $, let $gamma_1,gamma_2 : [0,1] to mathbb{R}^{n}$ be smooth paths such that
$$gamma_1 (0) = gamma_2(0) neq gamma_1(1) = gamma_2(1) $$
and $gamma_1, gamma_2$ are injetive functions.
Question: Does there exist a diffeomorphism $varphi: mathbb{R}^n to mathbb{R}^n$ such that $varphi (gamma_1 ([0,1])) = gamma_2 ([0,1]).$
Does anyone know if this result is true?
This seems true but I do not know how to prove it, can anyone help me?
analysis differential-geometry differential-topology
analysis differential-geometry differential-topology
asked Nov 5 at 18:51
Matheus Manzatto
1,3111523
asked Nov 5 at 18:51
Matheus Manzatto
1,3111523
edited Nov 24 at 20:59
asked Nov 5 at 18:51
Matheus Manzatto
1,3111523
asked Nov 5 at 18:51
Matheus Manzatto
1,3111523
asked Nov 5 at 18:51
Matheus Manzatto
1,3111523
1,3111523
Is $Bbb R^{n>1}$ the set of functions from the interval ${n >1}$ into $Bbb R$ endowed with uniform convergence on each compact subinterval?
– Will M.
Nov 5 at 18:54
No, it is $mathbb {R}^n$ with $n>1$. I will change.
– Matheus Manzatto
Nov 5 at 18:55
Are the $gamma_i$ $C^1$? $C^infty$? Merely continuous?
– Jason DeVito
Nov 5 at 19:00
It cannot be true as stated because you can take $gamma_1([0,1])$ to be a line segment while $gamma_2([0, 1])$ to be two disconnected line segments. I still think this will be false if you do not have $gamma_i$ ($i=1,2$) to be differentiable with continuity. (To see why this would have to fail if the $gamma_i$ are only continuous, take a square and each $gamma_i$ be two of its sides.)
– Will M.
Nov 5 at 19:00
Smooth functions, sorry
– Matheus Manzatto
Nov 5 at 19:00
add a comment |
Is $Bbb R^{n>1}$ the set of functions from the interval ${n >1}$ into $Bbb R$ endowed with uniform convergence on each compact subinterval?
– Will M.
Nov 5 at 18:54
No, it is $mathbb {R}^n$ with $n>1$. I will change.
– Matheus Manzatto
Nov 5 at 18:55
Are the $gamma_i$ $C^1$? $C^infty$? Merely continuous?
– Jason DeVito
Nov 5 at 19:00
It cannot be true as stated because you can take $gamma_1([0,1])$ to be a line segment while $gamma_2([0, 1])$ to be two disconnected line segments. I still think this will be false if you do not have $gamma_i$ ($i=1,2$) to be differentiable with continuity. (To see why this would have to fail if the $gamma_i$ are only continuous, take a square and each $gamma_i$ be two of its sides.)
– Will M.
Nov 5 at 19:00
Smooth functions, sorry
– Matheus Manzatto
Nov 5 at 19:00
Is $Bbb R^{n>1}$ the set of functions from the interval ${n >1}$ into $Bbb R$ endowed with uniform convergence on each compact subinterval?
– Will M.
Nov 5 at 18:54
Is $Bbb R^{n>1}$ the set of functions from the interval ${n >1}$ into $Bbb R$ endowed with uniform convergence on each compact subinterval?
– Will M.
Nov 5 at 18:54
No, it is $mathbb {R}^n$ with $n>1$. I will change.
– Matheus Manzatto
Nov 5 at 18:55
No, it is $mathbb {R}^n$ with $n>1$. I will change.
– Matheus Manzatto
Nov 5 at 18:55
Are the $gamma_i$ $C^1$? $C^infty$? Merely continuous?
– Jason DeVito
Nov 5 at 19:00
Are the $gamma_i$ $C^1$? $C^infty$? Merely continuous?
– Jason DeVito
Nov 5 at 19:00
It cannot be true as stated because you can take $gamma_1([0,1])$ to be a line segment while $gamma_2([0, 1])$ to be two disconnected line segments. I still think this will be false if you do not have $gamma_i$ ($i=1,2$) to be differentiable with continuity. (To see why this would have to fail if the $gamma_i$ are only continuous, take a square and each $gamma_i$ be two of its sides.)
– Will M.
Nov 5 at 19:00
It cannot be true as stated because you can take $gamma_1([0,1])$ to be a line segment while $gamma_2([0, 1])$ to be two disconnected line segments. I still think this will be false if you do not have $gamma_i$ ($i=1,2$) to be differentiable with continuity. (To see why this would have to fail if the $gamma_i$ are only continuous, take a square and each $gamma_i$ be two of its sides.)
– Will M.
Nov 5 at 19:00
Smooth functions, sorry
– Matheus Manzatto
Nov 5 at 19:00
Smooth functions, sorry
– Matheus Manzatto
Nov 5 at 19:00
add a comment |
1 Answer
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Counterexample in $mathbb R^2$: Define $$gamma_1 (t)=begin{cases} (e^{1/(2t-1)},0)&tin [0,1/2)\0& t=1/2\(0,e^{1/(1-2t)})&tin (1/2,1]end{cases}$$
This is a $C^infty$ injective mapping that that traces out the line segment $[(1/e,0),(0,0)]$ followed by the segment $[(0,0),(0,1/e)].$
Note that $gamma_1$ makes a right angle turn at $(0,0).$ There is no diffeomorphism that can turn $gamma_1$ into the usual straight line segment $gamma_2$ from $(1/e,0)$ to $(0,1/e).$
answered Nov 5 at 20:48
zhw.
71.4k43075
add a comment |
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1 Answer
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1 Answer
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active
oldest
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votes
Counterexample in $mathbb R^2$: Define $$gamma_1 (t)=begin{cases} (e^{1/(2t-1)},0)&tin [0,1/2)\0& t=1/2\(0,e^{1/(1-2t)})&tin (1/2,1]end{cases}$$
This is a $C^infty$ injective mapping that that traces out the line segment $[(1/e,0),(0,0)]$ followed by the segment $[(0,0),(0,1/e)].$
Note that $gamma_1$ makes a right angle turn at $(0,0).$ There is no diffeomorphism that can turn $gamma_1$ into the usual straight line segment $gamma_2$ from $(1/e,0)$ to $(0,1/e).$
answered Nov 5 at 20:48
zhw.
71.4k43075
add a comment |
Counterexample in $mathbb R^2$: Define $$gamma_1 (t)=begin{cases} (e^{1/(2t-1)},0)&tin [0,1/2)\0& t=1/2\(0,e^{1/(1-2t)})&tin (1/2,1]end{cases}$$
This is a $C^infty$ injective mapping that that traces out the line segment $[(1/e,0),(0,0)]$ followed by the segment $[(0,0),(0,1/e)].$
Note that $gamma_1$ makes a right angle turn at $(0,0).$ There is no diffeomorphism that can turn $gamma_1$ into the usual straight line segment $gamma_2$ from $(1/e,0)$ to $(0,1/e).$
answered Nov 5 at 20:48
zhw.
71.4k43075
add a comment |
Counterexample in $mathbb R^2$: Define $$gamma_1 (t)=begin{cases} (e^{1/(2t-1)},0)&tin [0,1/2)\0& t=1/2\(0,e^{1/(1-2t)})&tin (1/2,1]end{cases}$$
This is a $C^infty$ injective mapping that that traces out the line segment $[(1/e,0),(0,0)]$ followed by the segment $[(0,0),(0,1/e)].$
Note that $gamma_1$ makes a right angle turn at $(0,0).$ There is no diffeomorphism that can turn $gamma_1$ into the usual straight line segment $gamma_2$ from $(1/e,0)$ to $(0,1/e).$
answered Nov 5 at 20:48
zhw.
71.4k43075
Counterexample in $mathbb R^2$: Define $$gamma_1 (t)=begin{cases} (e^{1/(2t-1)},0)&tin [0,1/2)\0& t=1/2\(0,e^{1/(1-2t)})&tin (1/2,1]end{cases}$$
This is a $C^infty$ injective mapping that that traces out the line segment $[(1/e,0),(0,0)]$ followed by the segment $[(0,0),(0,1/e)].$
Note that $gamma_1$ makes a right angle turn at $(0,0).$ There is no diffeomorphism that can turn $gamma_1$ into the usual straight line segment $gamma_2$ from $(1/e,0)$ to $(0,1/e).$
answered Nov 5 at 20:48
zhw.
71.4k43075
edited Nov 24 at 17:32
answered Nov 5 at 20:48
zhw.
71.4k43075
answered Nov 5 at 20:48
zhw.
71.4k43075
answered Nov 5 at 20:48
zhw.
71.4k43075
71.4k43075
add a comment |
add a comment |
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Is $Bbb R^{n>1}$ the set of functions from the interval ${n >1}$ into $Bbb R$ endowed with uniform convergence on each compact subinterval?
– Will M.
Nov 5 at 18:54
No, it is $mathbb {R}^n$ with $n>1$. I will change.
– Matheus Manzatto
Nov 5 at 18:55
Are the $gamma_i$ $C^1$? $C^infty$? Merely continuous?
– Jason DeVito
Nov 5 at 19:00
It cannot be true as stated because you can take $gamma_1([0,1])$ to be a line segment while $gamma_2([0, 1])$ to be two disconnected line segments. I still think this will be false if you do not have $gamma_i$ ($i=1,2$) to be differentiable with continuity. (To see why this would have to fail if the $gamma_i$ are only continuous, take a square and each $gamma_i$ be two of its sides.)
– Will M.
Nov 5 at 19:00
Smooth functions, sorry
– Matheus Manzatto
Nov 5 at 19:00