How to find the integral submanifold? [duplicate]












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  • find the general integral manifolds of a given distribution

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Suppose $U subseteq R^3$ is the subset that all three coordinates are positive. Let $D$ be the distribution on $U$ spanned by two vector fields:



$X = yfrac{partial }{partial z}-zfrac{partial }{partial y}$ , $Y=zfrac{partial }{partial x}-xfrac{partial }{partial z}$.



Then how to find the integral submanifold?



I had shown that $[X, Y]_p = frac{y(p)}{z(p)}Y_p+frac{x(p)}{z(p)}X_p$, hence $D$ is involutive. I also know how to calculate the integral curve of $X$ and $Y$. But I dont know how to proceed to calculate the integral submanifold specifically.










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marked as duplicate by Ted Shifrin differential-geometry
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Dec 6 '18 at 20:26


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


















  • $begingroup$
    What do you mean "compute"? Do you want to find a formula for a parameterization of an integral submanifold?
    $endgroup$
    – AnonymousCoward
    Dec 6 '18 at 0:15










  • $begingroup$
    If a parametrization is possible.
    $endgroup$
    – Rikka
    Dec 6 '18 at 0:21
















1












$begingroup$



This question already has an answer here:




  • find the general integral manifolds of a given distribution

    1 answer




Suppose $U subseteq R^3$ is the subset that all three coordinates are positive. Let $D$ be the distribution on $U$ spanned by two vector fields:



$X = yfrac{partial }{partial z}-zfrac{partial }{partial y}$ , $Y=zfrac{partial }{partial x}-xfrac{partial }{partial z}$.



Then how to find the integral submanifold?



I had shown that $[X, Y]_p = frac{y(p)}{z(p)}Y_p+frac{x(p)}{z(p)}X_p$, hence $D$ is involutive. I also know how to calculate the integral curve of $X$ and $Y$. But I dont know how to proceed to calculate the integral submanifold specifically.










share|cite|improve this question











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Dec 6 '18 at 20:26


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


















  • $begingroup$
    What do you mean "compute"? Do you want to find a formula for a parameterization of an integral submanifold?
    $endgroup$
    – AnonymousCoward
    Dec 6 '18 at 0:15










  • $begingroup$
    If a parametrization is possible.
    $endgroup$
    – Rikka
    Dec 6 '18 at 0:21














1












1








1





$begingroup$



This question already has an answer here:




  • find the general integral manifolds of a given distribution

    1 answer




Suppose $U subseteq R^3$ is the subset that all three coordinates are positive. Let $D$ be the distribution on $U$ spanned by two vector fields:



$X = yfrac{partial }{partial z}-zfrac{partial }{partial y}$ , $Y=zfrac{partial }{partial x}-xfrac{partial }{partial z}$.



Then how to find the integral submanifold?



I had shown that $[X, Y]_p = frac{y(p)}{z(p)}Y_p+frac{x(p)}{z(p)}X_p$, hence $D$ is involutive. I also know how to calculate the integral curve of $X$ and $Y$. But I dont know how to proceed to calculate the integral submanifold specifically.










share|cite|improve this question











$endgroup$





This question already has an answer here:




  • find the general integral manifolds of a given distribution

    1 answer




Suppose $U subseteq R^3$ is the subset that all three coordinates are positive. Let $D$ be the distribution on $U$ spanned by two vector fields:



$X = yfrac{partial }{partial z}-zfrac{partial }{partial y}$ , $Y=zfrac{partial }{partial x}-xfrac{partial }{partial z}$.



Then how to find the integral submanifold?



I had shown that $[X, Y]_p = frac{y(p)}{z(p)}Y_p+frac{x(p)}{z(p)}X_p$, hence $D$ is involutive. I also know how to calculate the integral curve of $X$ and $Y$. But I dont know how to proceed to calculate the integral submanifold specifically.





This question already has an answer here:




  • find the general integral manifolds of a given distribution

    1 answer








differential-geometry manifolds






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share|cite|improve this question













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share|cite|improve this question








edited Dec 6 '18 at 0:09







Rikka

















asked Dec 6 '18 at 0:04









RikkaRikka

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marked as duplicate by Ted Shifrin differential-geometry
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Dec 6 '18 at 20:26


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.









marked as duplicate by Ted Shifrin differential-geometry
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Dec 6 '18 at 20:26


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














  • $begingroup$
    What do you mean "compute"? Do you want to find a formula for a parameterization of an integral submanifold?
    $endgroup$
    – AnonymousCoward
    Dec 6 '18 at 0:15










  • $begingroup$
    If a parametrization is possible.
    $endgroup$
    – Rikka
    Dec 6 '18 at 0:21


















  • $begingroup$
    What do you mean "compute"? Do you want to find a formula for a parameterization of an integral submanifold?
    $endgroup$
    – AnonymousCoward
    Dec 6 '18 at 0:15










  • $begingroup$
    If a parametrization is possible.
    $endgroup$
    – Rikka
    Dec 6 '18 at 0:21
















$begingroup$
What do you mean "compute"? Do you want to find a formula for a parameterization of an integral submanifold?
$endgroup$
– AnonymousCoward
Dec 6 '18 at 0:15




$begingroup$
What do you mean "compute"? Do you want to find a formula for a parameterization of an integral submanifold?
$endgroup$
– AnonymousCoward
Dec 6 '18 at 0:15












$begingroup$
If a parametrization is possible.
$endgroup$
– Rikka
Dec 6 '18 at 0:21




$begingroup$
If a parametrization is possible.
$endgroup$
– Rikka
Dec 6 '18 at 0:21










1 Answer
1






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0












$begingroup$

In this case, the result follows by close inspection of the two vector fields: one generates rotation in the $y,z$-plane and the other generates rotation in the $x,z$-plane.



This is a good point to pause and think geometrically about what the integral submanifolds of these two vector fields might be.



...



...



Using this observation, you should be able to ``see'' geometrically that the integral submanifolds for your distribution are given as level sets of the function
$$ f(x,y,z) = x^2 + y^2 + z^2.$$



Let's prove this in detail. The level sets of the function $f$ are all 2-dimensional (except for the fiber $f^{-1}(0) = {(0,0,0)}$, which we ignore for the moment). They are concentric spheres.



The foliation of $mathbb{R}^3$ by level sets of $f$ has an involutive tangent distribution (of dimension 2, since they are regular fibers of a smooth map $mathbb{R}^3 to mathbb{R}^1$). At each point, this distribution is defined as the kernel of the derivative of $f$,
$$ ker(Df_p).$$



You can check that
$$ Xvert_p, Yvert_p in ker(Df_p)$$
(or equivalently, you can check that the Lie derivatives $L_X f = L_Y f = 0$).



Thus, your distribution is contained in the distribution $ker(Df_p)$. But they both have dimension 2, so they coincide.



Thus the integral submanifolds for your distribution are the same as the level sets of $f$: they are the concentric spheres centered at the origin.



If you want a nice parameterization of this foliation, you are in luck, it has been studied a lot. Spherical or cylindrical coordinates are common favourites.






share|cite|improve this answer









$endgroup$




















    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    In this case, the result follows by close inspection of the two vector fields: one generates rotation in the $y,z$-plane and the other generates rotation in the $x,z$-plane.



    This is a good point to pause and think geometrically about what the integral submanifolds of these two vector fields might be.



    ...



    ...



    Using this observation, you should be able to ``see'' geometrically that the integral submanifolds for your distribution are given as level sets of the function
    $$ f(x,y,z) = x^2 + y^2 + z^2.$$



    Let's prove this in detail. The level sets of the function $f$ are all 2-dimensional (except for the fiber $f^{-1}(0) = {(0,0,0)}$, which we ignore for the moment). They are concentric spheres.



    The foliation of $mathbb{R}^3$ by level sets of $f$ has an involutive tangent distribution (of dimension 2, since they are regular fibers of a smooth map $mathbb{R}^3 to mathbb{R}^1$). At each point, this distribution is defined as the kernel of the derivative of $f$,
    $$ ker(Df_p).$$



    You can check that
    $$ Xvert_p, Yvert_p in ker(Df_p)$$
    (or equivalently, you can check that the Lie derivatives $L_X f = L_Y f = 0$).



    Thus, your distribution is contained in the distribution $ker(Df_p)$. But they both have dimension 2, so they coincide.



    Thus the integral submanifolds for your distribution are the same as the level sets of $f$: they are the concentric spheres centered at the origin.



    If you want a nice parameterization of this foliation, you are in luck, it has been studied a lot. Spherical or cylindrical coordinates are common favourites.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      In this case, the result follows by close inspection of the two vector fields: one generates rotation in the $y,z$-plane and the other generates rotation in the $x,z$-plane.



      This is a good point to pause and think geometrically about what the integral submanifolds of these two vector fields might be.



      ...



      ...



      Using this observation, you should be able to ``see'' geometrically that the integral submanifolds for your distribution are given as level sets of the function
      $$ f(x,y,z) = x^2 + y^2 + z^2.$$



      Let's prove this in detail. The level sets of the function $f$ are all 2-dimensional (except for the fiber $f^{-1}(0) = {(0,0,0)}$, which we ignore for the moment). They are concentric spheres.



      The foliation of $mathbb{R}^3$ by level sets of $f$ has an involutive tangent distribution (of dimension 2, since they are regular fibers of a smooth map $mathbb{R}^3 to mathbb{R}^1$). At each point, this distribution is defined as the kernel of the derivative of $f$,
      $$ ker(Df_p).$$



      You can check that
      $$ Xvert_p, Yvert_p in ker(Df_p)$$
      (or equivalently, you can check that the Lie derivatives $L_X f = L_Y f = 0$).



      Thus, your distribution is contained in the distribution $ker(Df_p)$. But they both have dimension 2, so they coincide.



      Thus the integral submanifolds for your distribution are the same as the level sets of $f$: they are the concentric spheres centered at the origin.



      If you want a nice parameterization of this foliation, you are in luck, it has been studied a lot. Spherical or cylindrical coordinates are common favourites.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        In this case, the result follows by close inspection of the two vector fields: one generates rotation in the $y,z$-plane and the other generates rotation in the $x,z$-plane.



        This is a good point to pause and think geometrically about what the integral submanifolds of these two vector fields might be.



        ...



        ...



        Using this observation, you should be able to ``see'' geometrically that the integral submanifolds for your distribution are given as level sets of the function
        $$ f(x,y,z) = x^2 + y^2 + z^2.$$



        Let's prove this in detail. The level sets of the function $f$ are all 2-dimensional (except for the fiber $f^{-1}(0) = {(0,0,0)}$, which we ignore for the moment). They are concentric spheres.



        The foliation of $mathbb{R}^3$ by level sets of $f$ has an involutive tangent distribution (of dimension 2, since they are regular fibers of a smooth map $mathbb{R}^3 to mathbb{R}^1$). At each point, this distribution is defined as the kernel of the derivative of $f$,
        $$ ker(Df_p).$$



        You can check that
        $$ Xvert_p, Yvert_p in ker(Df_p)$$
        (or equivalently, you can check that the Lie derivatives $L_X f = L_Y f = 0$).



        Thus, your distribution is contained in the distribution $ker(Df_p)$. But they both have dimension 2, so they coincide.



        Thus the integral submanifolds for your distribution are the same as the level sets of $f$: they are the concentric spheres centered at the origin.



        If you want a nice parameterization of this foliation, you are in luck, it has been studied a lot. Spherical or cylindrical coordinates are common favourites.






        share|cite|improve this answer









        $endgroup$



        In this case, the result follows by close inspection of the two vector fields: one generates rotation in the $y,z$-plane and the other generates rotation in the $x,z$-plane.



        This is a good point to pause and think geometrically about what the integral submanifolds of these two vector fields might be.



        ...



        ...



        Using this observation, you should be able to ``see'' geometrically that the integral submanifolds for your distribution are given as level sets of the function
        $$ f(x,y,z) = x^2 + y^2 + z^2.$$



        Let's prove this in detail. The level sets of the function $f$ are all 2-dimensional (except for the fiber $f^{-1}(0) = {(0,0,0)}$, which we ignore for the moment). They are concentric spheres.



        The foliation of $mathbb{R}^3$ by level sets of $f$ has an involutive tangent distribution (of dimension 2, since they are regular fibers of a smooth map $mathbb{R}^3 to mathbb{R}^1$). At each point, this distribution is defined as the kernel of the derivative of $f$,
        $$ ker(Df_p).$$



        You can check that
        $$ Xvert_p, Yvert_p in ker(Df_p)$$
        (or equivalently, you can check that the Lie derivatives $L_X f = L_Y f = 0$).



        Thus, your distribution is contained in the distribution $ker(Df_p)$. But they both have dimension 2, so they coincide.



        Thus the integral submanifolds for your distribution are the same as the level sets of $f$: they are the concentric spheres centered at the origin.



        If you want a nice parameterization of this foliation, you are in luck, it has been studied a lot. Spherical or cylindrical coordinates are common favourites.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 6 '18 at 4:23









        AnonymousCowardAnonymousCoward

        3,1122435




        3,1122435















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