vector field of a sphere
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I have a problem to understand vector fields on manifolds. Theoretically I get it, but I cannot find a concrete example with numbers.
For example, I have the $S^2$ sphere and I take the classic polar parametriazation ($theta$,$phi$)$to$ r($theta$,$phi$) in order to get the vector fields r$theta$,r$phi$.What are those vector fields?Are they a base of the tangent space?Obviously they are, but why?
Are they the same with the canonical base $frac{partial}{partial theta}$, $frac{partial}{partial phi}$?
differential-geometry manifolds
asked Jan 20 at 8:29
janejane
264
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add a comment |
$begingroup$
I have a problem to understand vector fields on manifolds. Theoretically I get it, but I cannot find a concrete example with numbers.
For example, I have the $S^2$ sphere and I take the classic polar parametriazation ($theta$,$phi$)$to$ r($theta$,$phi$) in order to get the vector fields r$theta$,r$phi$.What are those vector fields?Are they a base of the tangent space?Obviously they are, but why?
Are they the same with the canonical base $frac{partial}{partial theta}$, $frac{partial}{partial phi}$?
differential-geometry manifolds
asked Jan 20 at 8:29
janejane
264
$endgroup$
add a comment |
$begingroup$
I have a problem to understand vector fields on manifolds. Theoretically I get it, but I cannot find a concrete example with numbers.
For example, I have the $S^2$ sphere and I take the classic polar parametriazation ($theta$,$phi$)$to$ r($theta$,$phi$) in order to get the vector fields r$theta$,r$phi$.What are those vector fields?Are they a base of the tangent space?Obviously they are, but why?
Are they the same with the canonical base $frac{partial}{partial theta}$, $frac{partial}{partial phi}$?
differential-geometry manifolds
asked Jan 20 at 8:29
janejane
264
$endgroup$
I have a problem to understand vector fields on manifolds. Theoretically I get it, but I cannot find a concrete example with numbers.
For example, I have the $S^2$ sphere and I take the classic polar parametriazation ($theta$,$phi$)$to$ r($theta$,$phi$) in order to get the vector fields r$theta$,r$phi$.What are those vector fields?Are they a base of the tangent space?Obviously they are, but why?
Are they the same with the canonical base $frac{partial}{partial theta}$, $frac{partial}{partial phi}$?
differential-geometry manifolds
differential-geometry manifolds
asked Jan 20 at 8:29
janejane
264
asked Jan 20 at 8:29
janejane
264
asked Jan 20 at 8:29
janejane
264
asked Jan 20 at 8:29
janejane
264
asked Jan 20 at 8:29
janejane
264
264
add a comment |
add a comment |
1 Answer
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The basis of the tangent space is indeed $fracpartial{partialtheta},fracpartial{partialphi}$. This is an abstracted form, which is also sometimes written as $partial_theta,partial_phi$.
We can associate the points on the sphere with the coordinates in the map $(theta,phi)$ and then 'read' the basis vectors 'through the coordinate map'. If we do, we get:
$$fracpartial{partialtheta}(theta,phi)=(1,0)quadtext{and}quadfracpartial{partialphi}(theta,phi)=(0,1)$$
So we see that they span indeed a 2-dimensional tangent space. The differential map of the coordinate map transforms them to the unit vectors in its coordinate space.
For comparison, if we look at the embedding of $S^2$ in $mathbb R^3$ with its canonical cartesian coordinate map $(x,y,z)$, we get:
$$fracpartial{partialtheta}(cosphisintheta, sinphisintheta, costheta)=(cosphicostheta, cosphicostheta, -sintheta)$$
And if we 'read through the cartesian coordinate map' $(x,y)$ of $S^2$, we get:
$$fracpartial{partialtheta}(cosphisintheta, sinphisintheta)=(cosphicostheta, cosphicostheta)$$
which is the representation of $fracpartial{partialtheta}$ in the cartesian coordinate map.
answered Jan 20 at 8:58
I like SerenaI like Serena
4,2221722
$endgroup$
$begingroup$
I am a bit confused now.Is that wright: r$theta$:$R^2$ $to$ $R^3$ as ($theta$,$phi$) $to$ ($frac{partial x}{partial theta}$,$frac{partial y}{partial theta}$,$frac{partial z}{partial theta}$) and this is the same as e1?
$endgroup$
– jane
Jan 20 at 9:34
$begingroup$
@jane, yes, the resulting vector is the representation of $mathbf e_1=fracpartial{partialtheta}$ in the embedding given by $f(theta,phi)=(x(theta,phi), y(theta,phi), z(theta,phi))$.
$endgroup$
– I like Serena
Jan 20 at 10:46
add a comment |
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1 Answer
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1 Answer
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active
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active
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$begingroup$
The basis of the tangent space is indeed $fracpartial{partialtheta},fracpartial{partialphi}$. This is an abstracted form, which is also sometimes written as $partial_theta,partial_phi$.
We can associate the points on the sphere with the coordinates in the map $(theta,phi)$ and then 'read' the basis vectors 'through the coordinate map'. If we do, we get:
$$fracpartial{partialtheta}(theta,phi)=(1,0)quadtext{and}quadfracpartial{partialphi}(theta,phi)=(0,1)$$
So we see that they span indeed a 2-dimensional tangent space. The differential map of the coordinate map transforms them to the unit vectors in its coordinate space.
For comparison, if we look at the embedding of $S^2$ in $mathbb R^3$ with its canonical cartesian coordinate map $(x,y,z)$, we get:
$$fracpartial{partialtheta}(cosphisintheta, sinphisintheta, costheta)=(cosphicostheta, cosphicostheta, -sintheta)$$
And if we 'read through the cartesian coordinate map' $(x,y)$ of $S^2$, we get:
$$fracpartial{partialtheta}(cosphisintheta, sinphisintheta)=(cosphicostheta, cosphicostheta)$$
which is the representation of $fracpartial{partialtheta}$ in the cartesian coordinate map.
answered Jan 20 at 8:58
I like SerenaI like Serena
4,2221722
$endgroup$
$begingroup$
I am a bit confused now.Is that wright: r$theta$:$R^2$ $to$ $R^3$ as ($theta$,$phi$) $to$ ($frac{partial x}{partial theta}$,$frac{partial y}{partial theta}$,$frac{partial z}{partial theta}$) and this is the same as e1?
$endgroup$
– jane
Jan 20 at 9:34
$begingroup$
@jane, yes, the resulting vector is the representation of $mathbf e_1=fracpartial{partialtheta}$ in the embedding given by $f(theta,phi)=(x(theta,phi), y(theta,phi), z(theta,phi))$.
$endgroup$
– I like Serena
Jan 20 at 10:46
add a comment |
$begingroup$
The basis of the tangent space is indeed $fracpartial{partialtheta},fracpartial{partialphi}$. This is an abstracted form, which is also sometimes written as $partial_theta,partial_phi$.
We can associate the points on the sphere with the coordinates in the map $(theta,phi)$ and then 'read' the basis vectors 'through the coordinate map'. If we do, we get:
$$fracpartial{partialtheta}(theta,phi)=(1,0)quadtext{and}quadfracpartial{partialphi}(theta,phi)=(0,1)$$
So we see that they span indeed a 2-dimensional tangent space. The differential map of the coordinate map transforms them to the unit vectors in its coordinate space.
For comparison, if we look at the embedding of $S^2$ in $mathbb R^3$ with its canonical cartesian coordinate map $(x,y,z)$, we get:
$$fracpartial{partialtheta}(cosphisintheta, sinphisintheta, costheta)=(cosphicostheta, cosphicostheta, -sintheta)$$
And if we 'read through the cartesian coordinate map' $(x,y)$ of $S^2$, we get:
$$fracpartial{partialtheta}(cosphisintheta, sinphisintheta)=(cosphicostheta, cosphicostheta)$$
which is the representation of $fracpartial{partialtheta}$ in the cartesian coordinate map.
answered Jan 20 at 8:58
I like SerenaI like Serena
4,2221722
$endgroup$
$begingroup$
I am a bit confused now.Is that wright: r$theta$:$R^2$ $to$ $R^3$ as ($theta$,$phi$) $to$ ($frac{partial x}{partial theta}$,$frac{partial y}{partial theta}$,$frac{partial z}{partial theta}$) and this is the same as e1?
$endgroup$
– jane
Jan 20 at 9:34
$begingroup$
@jane, yes, the resulting vector is the representation of $mathbf e_1=fracpartial{partialtheta}$ in the embedding given by $f(theta,phi)=(x(theta,phi), y(theta,phi), z(theta,phi))$.
$endgroup$
– I like Serena
Jan 20 at 10:46
add a comment |
$begingroup$
The basis of the tangent space is indeed $fracpartial{partialtheta},fracpartial{partialphi}$. This is an abstracted form, which is also sometimes written as $partial_theta,partial_phi$.
We can associate the points on the sphere with the coordinates in the map $(theta,phi)$ and then 'read' the basis vectors 'through the coordinate map'. If we do, we get:
$$fracpartial{partialtheta}(theta,phi)=(1,0)quadtext{and}quadfracpartial{partialphi}(theta,phi)=(0,1)$$
So we see that they span indeed a 2-dimensional tangent space. The differential map of the coordinate map transforms them to the unit vectors in its coordinate space.
For comparison, if we look at the embedding of $S^2$ in $mathbb R^3$ with its canonical cartesian coordinate map $(x,y,z)$, we get:
$$fracpartial{partialtheta}(cosphisintheta, sinphisintheta, costheta)=(cosphicostheta, cosphicostheta, -sintheta)$$
And if we 'read through the cartesian coordinate map' $(x,y)$ of $S^2$, we get:
$$fracpartial{partialtheta}(cosphisintheta, sinphisintheta)=(cosphicostheta, cosphicostheta)$$
which is the representation of $fracpartial{partialtheta}$ in the cartesian coordinate map.
answered Jan 20 at 8:58
I like SerenaI like Serena
4,2221722
$endgroup$
The basis of the tangent space is indeed $fracpartial{partialtheta},fracpartial{partialphi}$. This is an abstracted form, which is also sometimes written as $partial_theta,partial_phi$.
We can associate the points on the sphere with the coordinates in the map $(theta,phi)$ and then 'read' the basis vectors 'through the coordinate map'. If we do, we get:
$$fracpartial{partialtheta}(theta,phi)=(1,0)quadtext{and}quadfracpartial{partialphi}(theta,phi)=(0,1)$$
So we see that they span indeed a 2-dimensional tangent space. The differential map of the coordinate map transforms them to the unit vectors in its coordinate space.
For comparison, if we look at the embedding of $S^2$ in $mathbb R^3$ with its canonical cartesian coordinate map $(x,y,z)$, we get:
$$fracpartial{partialtheta}(cosphisintheta, sinphisintheta, costheta)=(cosphicostheta, cosphicostheta, -sintheta)$$
And if we 'read through the cartesian coordinate map' $(x,y)$ of $S^2$, we get:
$$fracpartial{partialtheta}(cosphisintheta, sinphisintheta)=(cosphicostheta, cosphicostheta)$$
which is the representation of $fracpartial{partialtheta}$ in the cartesian coordinate map.
answered Jan 20 at 8:58
I like SerenaI like Serena
4,2221722
edited Jan 20 at 9:04
answered Jan 20 at 8:58
I like SerenaI like Serena
4,2221722
answered Jan 20 at 8:58
I like SerenaI like Serena
4,2221722
answered Jan 20 at 8:58
I like SerenaI like Serena
4,2221722
4,2221722
$begingroup$
I am a bit confused now.Is that wright: r$theta$:$R^2$ $to$ $R^3$ as ($theta$,$phi$) $to$ ($frac{partial x}{partial theta}$,$frac{partial y}{partial theta}$,$frac{partial z}{partial theta}$) and this is the same as e1?
$endgroup$
– jane
Jan 20 at 9:34
$begingroup$
@jane, yes, the resulting vector is the representation of $mathbf e_1=fracpartial{partialtheta}$ in the embedding given by $f(theta,phi)=(x(theta,phi), y(theta,phi), z(theta,phi))$.
$endgroup$
– I like Serena
Jan 20 at 10:46
add a comment |
$begingroup$
I am a bit confused now.Is that wright: r$theta$:$R^2$ $to$ $R^3$ as ($theta$,$phi$) $to$ ($frac{partial x}{partial theta}$,$frac{partial y}{partial theta}$,$frac{partial z}{partial theta}$) and this is the same as e1?
$endgroup$
– jane
Jan 20 at 9:34
$begingroup$
@jane, yes, the resulting vector is the representation of $mathbf e_1=fracpartial{partialtheta}$ in the embedding given by $f(theta,phi)=(x(theta,phi), y(theta,phi), z(theta,phi))$.
$endgroup$
– I like Serena
Jan 20 at 10:46
$begingroup$
I am a bit confused now.Is that wright: r$theta$:$R^2$ $to$ $R^3$ as ($theta$,$phi$) $to$ ($frac{partial x}{partial theta}$,$frac{partial y}{partial theta}$,$frac{partial z}{partial theta}$) and this is the same as e1?
$endgroup$
– jane
Jan 20 at 9:34
$begingroup$
I am a bit confused now.Is that wright: r$theta$:$R^2$ $to$ $R^3$ as ($theta$,$phi$) $to$ ($frac{partial x}{partial theta}$,$frac{partial y}{partial theta}$,$frac{partial z}{partial theta}$) and this is the same as e1?
$endgroup$
– jane
Jan 20 at 9:34
$begingroup$
@jane, yes, the resulting vector is the representation of $mathbf e_1=fracpartial{partialtheta}$ in the embedding given by $f(theta,phi)=(x(theta,phi), y(theta,phi), z(theta,phi))$.
$endgroup$
– I like Serena
Jan 20 at 10:46
$begingroup$
@jane, yes, the resulting vector is the representation of $mathbf e_1=fracpartial{partialtheta}$ in the embedding given by $f(theta,phi)=(x(theta,phi), y(theta,phi), z(theta,phi))$.
$endgroup$
– I like Serena
Jan 20 at 10:46
add a comment |
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