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Symmetries of a curve on the unit sphere
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Let $mathbbS^2 subseteq mathbbR^3$ be the unit sphere. Let $gamma : [0, l] to mathbbS^2$ be a smooth embedded closed curve. Let $vecnu$ be one of the two continuous unit normal vector field on $gamma$ (with a little abuse, I am denoting with $gamma$ also the image of the curve). Just to be clear, $vecnu$ is tangent to the sphere.
Let $f: gamma to mathbbR$ be a smooth, given function, with only a finite number of zeroes. Let assume also that the following property holds: for every $K in mathcalKill(mathbbS^2)$
beginequationlabelasd
int_gamma f langle nu, Krangle , dsigma = 0.
endequation
With $mathcalKill(mathbbS^2)$ I mean the Lie algebra of Killing vector fields on the sphere.
Question: what can I say about $gamma$ and/or $f$?
For instance, it should be true that if $gamma$ is a circle and $f$ is constant, then $int_gamma f langle nu, Krangle , dsigma = 0$ holds for every $K in mathcalKill(mathbbS^2)$.
Is it true the converse? I feel that the converse might not be true, but maybe it is possible to prove some weaker kind of symmetry for $gamma$ and/or $f$?
differential-geometry riemannian-geometry curves symmetry
edited Jul 17 at 2:24
John Ma
37.5k93669
asked Jul 16 at 16:16
Onil90
1,494521
 |Â
show 1 more comment
up vote
3
down vote
favorite
Let $mathbbS^2 subseteq mathbbR^3$ be the unit sphere. Let $gamma : [0, l] to mathbbS^2$ be a smooth embedded closed curve. Let $vecnu$ be one of the two continuous unit normal vector field on $gamma$ (with a little abuse, I am denoting with $gamma$ also the image of the curve). Just to be clear, $vecnu$ is tangent to the sphere.
Let $f: gamma to mathbbR$ be a smooth, given function, with only a finite number of zeroes. Let assume also that the following property holds: for every $K in mathcalKill(mathbbS^2)$
beginequationlabelasd
int_gamma f langle nu, Krangle , dsigma = 0.
endequation
With $mathcalKill(mathbbS^2)$ I mean the Lie algebra of Killing vector fields on the sphere.
Question: what can I say about $gamma$ and/or $f$?
For instance, it should be true that if $gamma$ is a circle and $f$ is constant, then $int_gamma f langle nu, Krangle , dsigma = 0$ holds for every $K in mathcalKill(mathbbS^2)$.
Is it true the converse? I feel that the converse might not be true, but maybe it is possible to prove some weaker kind of symmetry for $gamma$ and/or $f$?
differential-geometry riemannian-geometry curves symmetry
edited Jul 17 at 2:24
John Ma
37.5k93669
asked Jul 16 at 16:16
Onil90
1,494521
@JohnMa Thanks for the comment and for the edit! mmh.. can you elaborate? It seems to me that it must be zero.. Let $gamma$ be a great circle and $K$ an infinitesimal generation of a rotation of axis $L$, for some line $L$ passing through the origin. Then take $pi$ the plane containing $L$ and perpendicular to $gamma$. It seems to me that $langle nu, K rangle $ is an "odd symmetric" function w.r.t. $pi$. So the stuff on one side of the plane cancels out with the stuff on the other side. Am I wrong?
– Onil90
Jul 17 at 6:36
2
You're right, and you don't even need $gamma$ to be a circle: for any simple closed curve $gamma$ we have $int_gamma langle nu , K rangle d sigma = int_Omega mathcal L_K omega$ where $Omega$ is the region enclosed by $gamma$ and $omega$ is the area form. Since $K$ is Killing we have $mathcal L_K omega = 0.$ (See this related post.)
– Anthony Carapetis
Jul 17 at 8:33
@AnthonyCarapetis Ah, right! Good point!!
– Onil90
Jul 17 at 8:40
When $f$ is not constant, you should be able to show that the condition becomes $int_Omega df(K) omega = 0$ for all $K.$ Not sure if this tells you anything about $f$ or $gamma$; at most I think it would tell you some relationship between them.
– Anthony Carapetis
Jul 17 at 8:46
@AnthonyCarapetis So with "$f$" you mean any smooth extension of $f$ on the inside of $Omega$?
– Onil90
Jul 17 at 8:53
 |Â
show 1 more comment
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Let $mathbbS^2 subseteq mathbbR^3$ be the unit sphere. Let $gamma : [0, l] to mathbbS^2$ be a smooth embedded closed curve. Let $vecnu$ be one of the two continuous unit normal vector field on $gamma$ (with a little abuse, I am denoting with $gamma$ also the image of the curve). Just to be clear, $vecnu$ is tangent to the sphere.
Let $f: gamma to mathbbR$ be a smooth, given function, with only a finite number of zeroes. Let assume also that the following property holds: for every $K in mathcalKill(mathbbS^2)$
beginequationlabelasd
int_gamma f langle nu, Krangle , dsigma = 0.
endequation
With $mathcalKill(mathbbS^2)$ I mean the Lie algebra of Killing vector fields on the sphere.
Question: what can I say about $gamma$ and/or $f$?
For instance, it should be true that if $gamma$ is a circle and $f$ is constant, then $int_gamma f langle nu, Krangle , dsigma = 0$ holds for every $K in mathcalKill(mathbbS^2)$.
Is it true the converse? I feel that the converse might not be true, but maybe it is possible to prove some weaker kind of symmetry for $gamma$ and/or $f$?
differential-geometry riemannian-geometry curves symmetry
edited Jul 17 at 2:24
John Ma
37.5k93669
asked Jul 16 at 16:16
Onil90
1,494521
Let $mathbbS^2 subseteq mathbbR^3$ be the unit sphere. Let $gamma : [0, l] to mathbbS^2$ be a smooth embedded closed curve. Let $vecnu$ be one of the two continuous unit normal vector field on $gamma$ (with a little abuse, I am denoting with $gamma$ also the image of the curve). Just to be clear, $vecnu$ is tangent to the sphere.
Let $f: gamma to mathbbR$ be a smooth, given function, with only a finite number of zeroes. Let assume also that the following property holds: for every $K in mathcalKill(mathbbS^2)$
beginequationlabelasd
int_gamma f langle nu, Krangle , dsigma = 0.
endequation
With $mathcalKill(mathbbS^2)$ I mean the Lie algebra of Killing vector fields on the sphere.
Question: what can I say about $gamma$ and/or $f$?
For instance, it should be true that if $gamma$ is a circle and $f$ is constant, then $int_gamma f langle nu, Krangle , dsigma = 0$ holds for every $K in mathcalKill(mathbbS^2)$.
Is it true the converse? I feel that the converse might not be true, but maybe it is possible to prove some weaker kind of symmetry for $gamma$ and/or $f$?
differential-geometry riemannian-geometry curves symmetry
edited Jul 17 at 2:24
John Ma
37.5k93669
asked Jul 16 at 16:16
Onil90
1,494521
edited Jul 17 at 2:24
John Ma
37.5k93669
edited Jul 17 at 2:24
John Ma
37.5k93669
edited Jul 17 at 2:24
John Ma
37.5k93669
37.5k93669
asked Jul 16 at 16:16
Onil90
1,494521
asked Jul 16 at 16:16
Onil90
1,494521
asked Jul 16 at 16:16
Onil90
1,494521
1,494521
@JohnMa Thanks for the comment and for the edit! mmh.. can you elaborate? It seems to me that it must be zero.. Let $gamma$ be a great circle and $K$ an infinitesimal generation of a rotation of axis $L$, for some line $L$ passing through the origin. Then take $pi$ the plane containing $L$ and perpendicular to $gamma$. It seems to me that $langle nu, K rangle $ is an "odd symmetric" function w.r.t. $pi$. So the stuff on one side of the plane cancels out with the stuff on the other side. Am I wrong?
– Onil90
Jul 17 at 6:36
2
You're right, and you don't even need $gamma$ to be a circle: for any simple closed curve $gamma$ we have $int_gamma langle nu , K rangle d sigma = int_Omega mathcal L_K omega$ where $Omega$ is the region enclosed by $gamma$ and $omega$ is the area form. Since $K$ is Killing we have $mathcal L_K omega = 0.$ (See this related post.)
– Anthony Carapetis
Jul 17 at 8:33
@AnthonyCarapetis Ah, right! Good point!!
– Onil90
Jul 17 at 8:40
When $f$ is not constant, you should be able to show that the condition becomes $int_Omega df(K) omega = 0$ for all $K.$ Not sure if this tells you anything about $f$ or $gamma$; at most I think it would tell you some relationship between them.
– Anthony Carapetis
Jul 17 at 8:46
@AnthonyCarapetis So with "$f$" you mean any smooth extension of $f$ on the inside of $Omega$?
– Onil90
Jul 17 at 8:53
 |Â
show 1 more comment
@JohnMa Thanks for the comment and for the edit! mmh.. can you elaborate? It seems to me that it must be zero.. Let $gamma$ be a great circle and $K$ an infinitesimal generation of a rotation of axis $L$, for some line $L$ passing through the origin. Then take $pi$ the plane containing $L$ and perpendicular to $gamma$. It seems to me that $langle nu, K rangle $ is an "odd symmetric" function w.r.t. $pi$. So the stuff on one side of the plane cancels out with the stuff on the other side. Am I wrong?
– Onil90
Jul 17 at 6:36
2
You're right, and you don't even need $gamma$ to be a circle: for any simple closed curve $gamma$ we have $int_gamma langle nu , K rangle d sigma = int_Omega mathcal L_K omega$ where $Omega$ is the region enclosed by $gamma$ and $omega$ is the area form. Since $K$ is Killing we have $mathcal L_K omega = 0.$ (See this related post.)
– Anthony Carapetis
Jul 17 at 8:33
@AnthonyCarapetis Ah, right! Good point!!
– Onil90
Jul 17 at 8:40
When $f$ is not constant, you should be able to show that the condition becomes $int_Omega df(K) omega = 0$ for all $K.$ Not sure if this tells you anything about $f$ or $gamma$; at most I think it would tell you some relationship between them.
– Anthony Carapetis
Jul 17 at 8:46
@AnthonyCarapetis So with "$f$" you mean any smooth extension of $f$ on the inside of $Omega$?
– Onil90
Jul 17 at 8:53
@JohnMa Thanks for the comment and for the edit! mmh.. can you elaborate? It seems to me that it must be zero.. Let $gamma$ be a great circle and $K$ an infinitesimal generation of a rotation of axis $L$, for some line $L$ passing through the origin. Then take $pi$ the plane containing $L$ and perpendicular to $gamma$. It seems to me that $langle nu, K rangle $ is an "odd symmetric" function w.r.t. $pi$. So the stuff on one side of the plane cancels out with the stuff on the other side. Am I wrong?
– Onil90
Jul 17 at 6:36
@JohnMa Thanks for the comment and for the edit! mmh.. can you elaborate? It seems to me that it must be zero.. Let $gamma$ be a great circle and $K$ an infinitesimal generation of a rotation of axis $L$, for some line $L$ passing through the origin. Then take $pi$ the plane containing $L$ and perpendicular to $gamma$. It seems to me that $langle nu, K rangle $ is an "odd symmetric" function w.r.t. $pi$. So the stuff on one side of the plane cancels out with the stuff on the other side. Am I wrong?
– Onil90
Jul 17 at 6:36
2
2
You're right, and you don't even need $gamma$ to be a circle: for any simple closed curve $gamma$ we have $int_gamma langle nu , K rangle d sigma = int_Omega mathcal L_K omega$ where $Omega$ is the region enclosed by $gamma$ and $omega$ is the area form. Since $K$ is Killing we have $mathcal L_K omega = 0.$ (See this related post.)
– Anthony Carapetis
Jul 17 at 8:33
You're right, and you don't even need $gamma$ to be a circle: for any simple closed curve $gamma$ we have $int_gamma langle nu , K rangle d sigma = int_Omega mathcal L_K omega$ where $Omega$ is the region enclosed by $gamma$ and $omega$ is the area form. Since $K$ is Killing we have $mathcal L_K omega = 0.$ (See this related post.)
– Anthony Carapetis
Jul 17 at 8:33
@AnthonyCarapetis Ah, right! Good point!!
– Onil90
Jul 17 at 8:40
@AnthonyCarapetis Ah, right! Good point!!
– Onil90
Jul 17 at 8:40
When $f$ is not constant, you should be able to show that the condition becomes $int_Omega df(K) omega = 0$ for all $K.$ Not sure if this tells you anything about $f$ or $gamma$; at most I think it would tell you some relationship between them.
– Anthony Carapetis
Jul 17 at 8:46
When $f$ is not constant, you should be able to show that the condition becomes $int_Omega df(K) omega = 0$ for all $K.$ Not sure if this tells you anything about $f$ or $gamma$; at most I think it would tell you some relationship between them.
– Anthony Carapetis
Jul 17 at 8:46
@AnthonyCarapetis So with "$f$" you mean any smooth extension of $f$ on the inside of $Omega$?
– Onil90
Jul 17 at 8:53
@AnthonyCarapetis So with "$f$" you mean any smooth extension of $f$ on the inside of $Omega$?
– Onil90
Jul 17 at 8:53
 |Â
show 1 more comment
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@JohnMa Thanks for the comment and for the edit! mmh.. can you elaborate? It seems to me that it must be zero.. Let $gamma$ be a great circle and $K$ an infinitesimal generation of a rotation of axis $L$, for some line $L$ passing through the origin. Then take $pi$ the plane containing $L$ and perpendicular to $gamma$. It seems to me that $langle nu, K rangle $ is an "odd symmetric" function w.r.t. $pi$. So the stuff on one side of the plane cancels out with the stuff on the other side. Am I wrong?
– Onil90
Jul 17 at 6:36
2
You're right, and you don't even need $gamma$ to be a circle: for any simple closed curve $gamma$ we have $int_gamma langle nu , K rangle d sigma = int_Omega mathcal L_K omega$ where $Omega$ is the region enclosed by $gamma$ and $omega$ is the area form. Since $K$ is Killing we have $mathcal L_K omega = 0.$ (See this related post.)
– Anthony Carapetis
Jul 17 at 8:33
@AnthonyCarapetis Ah, right! Good point!!
– Onil90
Jul 17 at 8:40
When $f$ is not constant, you should be able to show that the condition becomes $int_Omega df(K) omega = 0$ for all $K.$ Not sure if this tells you anything about $f$ or $gamma$; at most I think it would tell you some relationship between them.
– Anthony Carapetis
Jul 17 at 8:46
@AnthonyCarapetis So with "$f$" you mean any smooth extension of $f$ on the inside of $Omega$?
– Onil90
Jul 17 at 8:53