Existence of a diffemorphism that maps one curve to another












4














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







share|cite|improve this question
























  • 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
















4














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







share|cite|improve this question
























  • 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














4












4








4


0





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







share|cite|improve this question















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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 24 at 20:59

























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


















  • 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










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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 $(0,1/e).$






share|cite|improve this 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 $(0,1/e).$






    share|cite|improve this answer




























      4














      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 $(0,1/e).$






      share|cite|improve this 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 $(0,1/e).$






        share|cite|improve this answer














        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 $(0,1/e).$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 24 at 17:32

























        answered Nov 5 at 20:48









        zhw.

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        71.4k43075






























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