Exercise: Compute a distribution given the integral manifold












0












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I am new in Stack Mathematics. I need your help in solving the follow exercise.



"Compute a distribution $Delta$ over $mathbb{R}^3$ whose integral manifold is the surface of a sphere (i.e. the set of points at the same distance $r$ from the center)."



Usually, the distribution is given and I have to find, if it exists, the integral manifold, but this time is quite the opposite. Does anyone have any idea how to solve this kind of exercises?



Thank you in advance!






differential-geometry differential-topology nonlinear-system







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  • $begingroup$
    I don't think this is possible---do you mean, perhaps $Bbb R^3 - {C}$, where $C$ is the prescribed center?
    $endgroup$
    – Travis
    Jan 7 at 22:43












  • $begingroup$
    I simply copied the text of the exercise, I think that any assumprions can be made
    $endgroup$
    – Francesco Roscia
    Jan 7 at 23:35
















0












$begingroup$


I am new in Stack Mathematics. I need your help in solving the follow exercise.



"Compute a distribution $Delta$ over $mathbb{R}^3$ whose integral manifold is the surface of a sphere (i.e. the set of points at the same distance $r$ from the center)."



Usually, the distribution is given and I have to find, if it exists, the integral manifold, but this time is quite the opposite. Does anyone have any idea how to solve this kind of exercises?



Thank you in advance!






differential-geometry differential-topology nonlinear-system







share|cite|improve this question









$endgroup$












  • $begingroup$
    I don't think this is possible---do you mean, perhaps $Bbb R^3 - {C}$, where $C$ is the prescribed center?
    $endgroup$
    – Travis
    Jan 7 at 22:43












  • $begingroup$
    I simply copied the text of the exercise, I think that any assumprions can be made
    $endgroup$
    – Francesco Roscia
    Jan 7 at 23:35














0












0








0





$begingroup$


I am new in Stack Mathematics. I need your help in solving the follow exercise.



"Compute a distribution $Delta$ over $mathbb{R}^3$ whose integral manifold is the surface of a sphere (i.e. the set of points at the same distance $r$ from the center)."



Usually, the distribution is given and I have to find, if it exists, the integral manifold, but this time is quite the opposite. Does anyone have any idea how to solve this kind of exercises?



Thank you in advance!






differential-geometry differential-topology nonlinear-system







share|cite|improve this question









$endgroup$




I am new in Stack Mathematics. I need your help in solving the follow exercise.



"Compute a distribution $Delta$ over $mathbb{R}^3$ whose integral manifold is the surface of a sphere (i.e. the set of points at the same distance $r$ from the center)."



Usually, the distribution is given and I have to find, if it exists, the integral manifold, but this time is quite the opposite. Does anyone have any idea how to solve this kind of exercises?



Thank you in advance!






differential-geometry differential-topology nonlinear-system




differential-geometry differential-topology nonlinear-system






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 7 at 22:11









Francesco RosciaFrancesco Roscia

12




12












  • $begingroup$
    I don't think this is possible---do you mean, perhaps $Bbb R^3 - {C}$, where $C$ is the prescribed center?
    $endgroup$
    – Travis
    Jan 7 at 22:43












  • $begingroup$
    I simply copied the text of the exercise, I think that any assumprions can be made
    $endgroup$
    – Francesco Roscia
    Jan 7 at 23:35


















  • $begingroup$
    I don't think this is possible---do you mean, perhaps $Bbb R^3 - {C}$, where $C$ is the prescribed center?
    $endgroup$
    – Travis
    Jan 7 at 22:43












  • $begingroup$
    I simply copied the text of the exercise, I think that any assumprions can be made
    $endgroup$
    – Francesco Roscia
    Jan 7 at 23:35
















$begingroup$
I don't think this is possible---do you mean, perhaps $Bbb R^3 - {C}$, where $C$ is the prescribed center?
$endgroup$
– Travis
Jan 7 at 22:43






$begingroup$
I don't think this is possible---do you mean, perhaps $Bbb R^3 - {C}$, where $C$ is the prescribed center?
$endgroup$
– Travis
Jan 7 at 22:43














$begingroup$
I simply copied the text of the exercise, I think that any assumprions can be made
$endgroup$
– Francesco Roscia
Jan 7 at 23:35




$begingroup$
I simply copied the text of the exercise, I think that any assumprions can be made
$endgroup$
– Francesco Roscia
Jan 7 at 23:35










1 Answer
1






active

oldest

votes


















0












$begingroup$

Hint Without loss of generality take the center to be the origin ${bf 0}$, and consider any point $p := (x, y, z) neq {bf 0}$. Since the sphere $S_{|p|}^2$ of radius $|p|$ centered at $bf 0$ is an integral manifold of the distribution $D$,
$$D_p = T_p S_{|p|}^2 .$$




Additional hint With respect to the Euclidean metric (and invoking the canonical identification of the tangent space to a vector space at a point with the vector space itself), $$T_p S_{|p|}^2 = {p}^{perp} .$$







share|cite|improve this answer









$endgroup$













  • $begingroup$
    Hi Travis. Have you any idea about how calculate the distribution? Let call $lambda(x,y,z)$ the function which describes the points on the surface of the sphere. I thought that I need to find vector fields $tau _i (x,y,z)$ such that $d lambda (x,y,z) tau _i (x,y,z) = 0$. These vectors fields span the distribution $Delta$. Is it correct?
    $endgroup$
    – Francesco Roscia
    Jan 7 at 23:59












  • $begingroup$
    That description works, but only locally: The Hairy Ball Theorem says you can't even choose a single nonvanishing vector field tangent to $S^2$, let alone two linearly independent fields. One option is, as you suggest, to find such a function $lambda$ for which the sphere is the preimage $lambda^{-1}(c)$ (for some regular value $c$ of $lambda$). Then, more or less by construction, the distribution you want is just $D_p := ker (dlambda)_p$.
    $endgroup$
    – Travis
    Jan 8 at 8:12












  • $begingroup$
    Thanks Trevis!!
    $endgroup$
    – Francesco Roscia
    Jan 8 at 16:01










  • $begingroup$
    You're welcome, I hope you found my comments useful!
    $endgroup$
    – Travis
    Jan 9 at 5:40












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1 Answer
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1 Answer
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$begingroup$

Hint Without loss of generality take the center to be the origin ${bf 0}$, and consider any point $p := (x, y, z) neq {bf 0}$. Since the sphere $S_{|p|}^2$ of radius $|p|$ centered at $bf 0$ is an integral manifold of the distribution $D$,
$$D_p = T_p S_{|p|}^2 .$$




Additional hint With respect to the Euclidean metric (and invoking the canonical identification of the tangent space to a vector space at a point with the vector space itself), $$T_p S_{|p|}^2 = {p}^{perp} .$$







share|cite|improve this answer









$endgroup$













  • $begingroup$
    Hi Travis. Have you any idea about how calculate the distribution? Let call $lambda(x,y,z)$ the function which describes the points on the surface of the sphere. I thought that I need to find vector fields $tau _i (x,y,z)$ such that $d lambda (x,y,z) tau _i (x,y,z) = 0$. These vectors fields span the distribution $Delta$. Is it correct?
    $endgroup$
    – Francesco Roscia
    Jan 7 at 23:59












  • $begingroup$
    That description works, but only locally: The Hairy Ball Theorem says you can't even choose a single nonvanishing vector field tangent to $S^2$, let alone two linearly independent fields. One option is, as you suggest, to find such a function $lambda$ for which the sphere is the preimage $lambda^{-1}(c)$ (for some regular value $c$ of $lambda$). Then, more or less by construction, the distribution you want is just $D_p := ker (dlambda)_p$.
    $endgroup$
    – Travis
    Jan 8 at 8:12












  • $begingroup$
    Thanks Trevis!!
    $endgroup$
    – Francesco Roscia
    Jan 8 at 16:01










  • $begingroup$
    You're welcome, I hope you found my comments useful!
    $endgroup$
    – Travis
    Jan 9 at 5:40
















0












$begingroup$

Hint Without loss of generality take the center to be the origin ${bf 0}$, and consider any point $p := (x, y, z) neq {bf 0}$. Since the sphere $S_{|p|}^2$ of radius $|p|$ centered at $bf 0$ is an integral manifold of the distribution $D$,
$$D_p = T_p S_{|p|}^2 .$$




Additional hint With respect to the Euclidean metric (and invoking the canonical identification of the tangent space to a vector space at a point with the vector space itself), $$T_p S_{|p|}^2 = {p}^{perp} .$$







share|cite|improve this answer









$endgroup$













  • $begingroup$
    Hi Travis. Have you any idea about how calculate the distribution? Let call $lambda(x,y,z)$ the function which describes the points on the surface of the sphere. I thought that I need to find vector fields $tau _i (x,y,z)$ such that $d lambda (x,y,z) tau _i (x,y,z) = 0$. These vectors fields span the distribution $Delta$. Is it correct?
    $endgroup$
    – Francesco Roscia
    Jan 7 at 23:59












  • $begingroup$
    That description works, but only locally: The Hairy Ball Theorem says you can't even choose a single nonvanishing vector field tangent to $S^2$, let alone two linearly independent fields. One option is, as you suggest, to find such a function $lambda$ for which the sphere is the preimage $lambda^{-1}(c)$ (for some regular value $c$ of $lambda$). Then, more or less by construction, the distribution you want is just $D_p := ker (dlambda)_p$.
    $endgroup$
    – Travis
    Jan 8 at 8:12












  • $begingroup$
    Thanks Trevis!!
    $endgroup$
    – Francesco Roscia
    Jan 8 at 16:01










  • $begingroup$
    You're welcome, I hope you found my comments useful!
    $endgroup$
    – Travis
    Jan 9 at 5:40














0












0








0





$begingroup$

Hint Without loss of generality take the center to be the origin ${bf 0}$, and consider any point $p := (x, y, z) neq {bf 0}$. Since the sphere $S_{|p|}^2$ of radius $|p|$ centered at $bf 0$ is an integral manifold of the distribution $D$,
$$D_p = T_p S_{|p|}^2 .$$




Additional hint With respect to the Euclidean metric (and invoking the canonical identification of the tangent space to a vector space at a point with the vector space itself), $$T_p S_{|p|}^2 = {p}^{perp} .$$







share|cite|improve this answer









$endgroup$



Hint Without loss of generality take the center to be the origin ${bf 0}$, and consider any point $p := (x, y, z) neq {bf 0}$. Since the sphere $S_{|p|}^2$ of radius $|p|$ centered at $bf 0$ is an integral manifold of the distribution $D$,
$$D_p = T_p S_{|p|}^2 .$$




Additional hint With respect to the Euclidean metric (and invoking the canonical identification of the tangent space to a vector space at a point with the vector space itself), $$T_p S_{|p|}^2 = {p}^{perp} .$$








share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 7 at 22:49









TravisTravis

63.8k769151




63.8k769151












  • $begingroup$
    Hi Travis. Have you any idea about how calculate the distribution? Let call $lambda(x,y,z)$ the function which describes the points on the surface of the sphere. I thought that I need to find vector fields $tau _i (x,y,z)$ such that $d lambda (x,y,z) tau _i (x,y,z) = 0$. These vectors fields span the distribution $Delta$. Is it correct?
    $endgroup$
    – Francesco Roscia
    Jan 7 at 23:59












  • $begingroup$
    That description works, but only locally: The Hairy Ball Theorem says you can't even choose a single nonvanishing vector field tangent to $S^2$, let alone two linearly independent fields. One option is, as you suggest, to find such a function $lambda$ for which the sphere is the preimage $lambda^{-1}(c)$ (for some regular value $c$ of $lambda$). Then, more or less by construction, the distribution you want is just $D_p := ker (dlambda)_p$.
    $endgroup$
    – Travis
    Jan 8 at 8:12












  • $begingroup$
    Thanks Trevis!!
    $endgroup$
    – Francesco Roscia
    Jan 8 at 16:01










  • $begingroup$
    You're welcome, I hope you found my comments useful!
    $endgroup$
    – Travis
    Jan 9 at 5:40


















  • $begingroup$
    Hi Travis. Have you any idea about how calculate the distribution? Let call $lambda(x,y,z)$ the function which describes the points on the surface of the sphere. I thought that I need to find vector fields $tau _i (x,y,z)$ such that $d lambda (x,y,z) tau _i (x,y,z) = 0$. These vectors fields span the distribution $Delta$. Is it correct?
    $endgroup$
    – Francesco Roscia
    Jan 7 at 23:59












  • $begingroup$
    That description works, but only locally: The Hairy Ball Theorem says you can't even choose a single nonvanishing vector field tangent to $S^2$, let alone two linearly independent fields. One option is, as you suggest, to find such a function $lambda$ for which the sphere is the preimage $lambda^{-1}(c)$ (for some regular value $c$ of $lambda$). Then, more or less by construction, the distribution you want is just $D_p := ker (dlambda)_p$.
    $endgroup$
    – Travis
    Jan 8 at 8:12












  • $begingroup$
    Thanks Trevis!!
    $endgroup$
    – Francesco Roscia
    Jan 8 at 16:01










  • $begingroup$
    You're welcome, I hope you found my comments useful!
    $endgroup$
    – Travis
    Jan 9 at 5:40
















$begingroup$
Hi Travis. Have you any idea about how calculate the distribution? Let call $lambda(x,y,z)$ the function which describes the points on the surface of the sphere. I thought that I need to find vector fields $tau _i (x,y,z)$ such that $d lambda (x,y,z) tau _i (x,y,z) = 0$. These vectors fields span the distribution $Delta$. Is it correct?
$endgroup$
– Francesco Roscia
Jan 7 at 23:59






$begingroup$
Hi Travis. Have you any idea about how calculate the distribution? Let call $lambda(x,y,z)$ the function which describes the points on the surface of the sphere. I thought that I need to find vector fields $tau _i (x,y,z)$ such that $d lambda (x,y,z) tau _i (x,y,z) = 0$. These vectors fields span the distribution $Delta$. Is it correct?
$endgroup$
– Francesco Roscia
Jan 7 at 23:59














$begingroup$
That description works, but only locally: The Hairy Ball Theorem says you can't even choose a single nonvanishing vector field tangent to $S^2$, let alone two linearly independent fields. One option is, as you suggest, to find such a function $lambda$ for which the sphere is the preimage $lambda^{-1}(c)$ (for some regular value $c$ of $lambda$). Then, more or less by construction, the distribution you want is just $D_p := ker (dlambda)_p$.
$endgroup$
– Travis
Jan 8 at 8:12






$begingroup$
That description works, but only locally: The Hairy Ball Theorem says you can't even choose a single nonvanishing vector field tangent to $S^2$, let alone two linearly independent fields. One option is, as you suggest, to find such a function $lambda$ for which the sphere is the preimage $lambda^{-1}(c)$ (for some regular value $c$ of $lambda$). Then, more or less by construction, the distribution you want is just $D_p := ker (dlambda)_p$.
$endgroup$
– Travis
Jan 8 at 8:12














$begingroup$
Thanks Trevis!!
$endgroup$
– Francesco Roscia
Jan 8 at 16:01




$begingroup$
Thanks Trevis!!
$endgroup$
– Francesco Roscia
Jan 8 at 16:01












$begingroup$
You're welcome, I hope you found my comments useful!
$endgroup$
– Travis
Jan 9 at 5:40




$begingroup$
You're welcome, I hope you found my comments useful!
$endgroup$
– Travis
Jan 9 at 5:40


















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