How to find the integral submanifold? [duplicate]
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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.
differential-geometry manifolds
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marked as duplicate by Ted Shifrin
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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.
add a comment |
$begingroup$
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.
differential-geometry manifolds
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marked as duplicate by Ted Shifrin
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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.
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What do you mean "compute"? Do you want to find a formula for a parameterization of an integral submanifold?
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– AnonymousCoward
Dec 6 '18 at 0:15
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If a parametrization is possible.
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– Rikka
Dec 6 '18 at 0:21
add a comment |
$begingroup$
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.
differential-geometry manifolds
$endgroup$
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
differential-geometry manifolds
edited Dec 6 '18 at 0:09
Rikka
asked Dec 6 '18 at 0:04
RikkaRikka
1779
1779
marked as duplicate by Ted Shifrin
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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
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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
add a comment |
$begingroup$
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
add a comment |
1 Answer
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$begingroup$
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.
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add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
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.
answered Dec 6 '18 at 4:23
AnonymousCowardAnonymousCoward
3,1122435
3,1122435
add a comment |
add a comment |
$begingroup$
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