What is the difference between a chart and a tangent space?
$begingroup$
To my lay-person mind, a chart is a one-to-one function that maps an area on a manifold to a euclidean space of equal dimension.
Then I understand a tangent space to be the space of vectors that are parallel to the manifold at a specific point.
So does one use the appropriate charts to map points to the tangent space?
These seem like similar things to me (even though I am fairly certain they are not). Is the difference that any point on the manifold can be mapped to that specific tangent space whereas the chart only covers a subset of points in the manifold?
Or do I have two completely different ideas conflated?
Is the mapping of points from the manifold even related to the manifolds charts?
differential-geometry manifolds
asked Dec 11 '18 at 10:36
bidbybidby
977
$endgroup$
add a comment |
$begingroup$
To my lay-person mind, a chart is a one-to-one function that maps an area on a manifold to a euclidean space of equal dimension.
Then I understand a tangent space to be the space of vectors that are parallel to the manifold at a specific point.
So does one use the appropriate charts to map points to the tangent space?
These seem like similar things to me (even though I am fairly certain they are not). Is the difference that any point on the manifold can be mapped to that specific tangent space whereas the chart only covers a subset of points in the manifold?
Or do I have two completely different ideas conflated?
Is the mapping of points from the manifold even related to the manifolds charts?
differential-geometry manifolds
asked Dec 11 '18 at 10:36
bidbybidby
977
$endgroup$
add a comment |
$begingroup$
To my lay-person mind, a chart is a one-to-one function that maps an area on a manifold to a euclidean space of equal dimension.
Then I understand a tangent space to be the space of vectors that are parallel to the manifold at a specific point.
So does one use the appropriate charts to map points to the tangent space?
These seem like similar things to me (even though I am fairly certain they are not). Is the difference that any point on the manifold can be mapped to that specific tangent space whereas the chart only covers a subset of points in the manifold?
Or do I have two completely different ideas conflated?
Is the mapping of points from the manifold even related to the manifolds charts?
differential-geometry manifolds
asked Dec 11 '18 at 10:36
bidbybidby
977
$endgroup$
To my lay-person mind, a chart is a one-to-one function that maps an area on a manifold to a euclidean space of equal dimension.
Then I understand a tangent space to be the space of vectors that are parallel to the manifold at a specific point.
So does one use the appropriate charts to map points to the tangent space?
These seem like similar things to me (even though I am fairly certain they are not). Is the difference that any point on the manifold can be mapped to that specific tangent space whereas the chart only covers a subset of points in the manifold?
Or do I have two completely different ideas conflated?
Is the mapping of points from the manifold even related to the manifolds charts?
differential-geometry manifolds
differential-geometry manifolds
asked Dec 11 '18 at 10:36
bidbybidby
977
asked Dec 11 '18 at 10:36
bidbybidby
977
asked Dec 11 '18 at 10:36
bidbybidby
977
asked Dec 11 '18 at 10:36
bidbybidby
977
asked Dec 11 '18 at 10:36
bidbybidby
977
977
add a comment |
add a comment |
1 Answer
1
active
oldest
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$begingroup$
Each point on a differentiable manifold has its own tangent space. Informally, you can think of the tangent space at a point as the space of tangent vectors to the surface at that point. But this informal picture only works if the manifold is embedded in some higher dimensional space.
A more abstract and more formal point of view is that tangent space at a point $P$ of a differentiable manifold $M$ consists of the derivatives of all curves on $M$ that pass through $P$. The derivatives exist because each map between $M$ and a chart is differentiable and invertible. So we can map a curve on $M$ to a chart, carry out calculus in the chart and "lift" the result back to $M$. The result is independent of the choice of chart, so it is a property of $M$ itself (at $P$).
It is not entirely obvious that we can create a vector space from the derivatives of the curves through $P$, but we can because the derivative is a linear operator. This vector space is then the tangent space at $P$, which we denote by $T_P$. The set of tangent spaces $T_P$ for all points $P$ on $M$ forms a vector bundle over $M$.
answered Dec 11 '18 at 11:50
gandalf61gandalf61
8,801725
$endgroup$
$begingroup$
What is the difference between mapping the curve on M to a chart as you said, and mapping it to a tangent space at the same point?
$endgroup$
– bidby
Dec 11 '18 at 11:56
$begingroup$
Am I correct in my understanding if I say that one maps all the curves through a point P to Euclidean space using any chart that includes P. Once I am in this space I can differentiate all of these curves and I get a bunch of vectors, which form the tangent space/plane at P once I have mapped back to M using the inverse of the chart I originally used. While these tangent vectors may have looked different in the Euclidean space, depending on my choice of chart, the resulting tangent space/plane at P in M is independent of which chart I chose?
$endgroup$
– bidby
Dec 11 '18 at 13:38
1
$begingroup$
@bidby Yes, your second comment is correct. Going back to your first comment, each point in the tangent space $T_P$ represents the derivative at $P$ of a set of curves on $M$ (many curves may have the same derivative at $P$). A point in $T_P$ does not represent some other point on $M$. So you cannot map points on $M$ to $T_P$ - apart from $P$ itself, which you can say maps to the origin in $T_P$ (because the constant "curve" $f(t)=P space forall t$ has derivative $0$).
$endgroup$
– gandalf61
Dec 11 '18 at 13:46
$begingroup$
Ah ok! So is the exponential mapping of point A at point B to the tangent space $T_b$ really just the vector in the tangent space that is parallel with the geodesic between A and B?
$endgroup$
– bidby
Dec 11 '18 at 14:05
$begingroup$
@bidby Yes, but you have to add extra structure such as a Riemannian metric to the manifold in order to define geodesics and exponential maps.
$endgroup$
– gandalf61
Dec 11 '18 at 14:20
add a comment |
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$begingroup$
Each point on a differentiable manifold has its own tangent space. Informally, you can think of the tangent space at a point as the space of tangent vectors to the surface at that point. But this informal picture only works if the manifold is embedded in some higher dimensional space.
A more abstract and more formal point of view is that tangent space at a point $P$ of a differentiable manifold $M$ consists of the derivatives of all curves on $M$ that pass through $P$. The derivatives exist because each map between $M$ and a chart is differentiable and invertible. So we can map a curve on $M$ to a chart, carry out calculus in the chart and "lift" the result back to $M$. The result is independent of the choice of chart, so it is a property of $M$ itself (at $P$).
It is not entirely obvious that we can create a vector space from the derivatives of the curves through $P$, but we can because the derivative is a linear operator. This vector space is then the tangent space at $P$, which we denote by $T_P$. The set of tangent spaces $T_P$ for all points $P$ on $M$ forms a vector bundle over $M$.
answered Dec 11 '18 at 11:50
gandalf61gandalf61
8,801725
$endgroup$
$begingroup$
What is the difference between mapping the curve on M to a chart as you said, and mapping it to a tangent space at the same point?
$endgroup$
– bidby
Dec 11 '18 at 11:56
$begingroup$
Am I correct in my understanding if I say that one maps all the curves through a point P to Euclidean space using any chart that includes P. Once I am in this space I can differentiate all of these curves and I get a bunch of vectors, which form the tangent space/plane at P once I have mapped back to M using the inverse of the chart I originally used. While these tangent vectors may have looked different in the Euclidean space, depending on my choice of chart, the resulting tangent space/plane at P in M is independent of which chart I chose?
$endgroup$
– bidby
Dec 11 '18 at 13:38
1
$begingroup$
@bidby Yes, your second comment is correct. Going back to your first comment, each point in the tangent space $T_P$ represents the derivative at $P$ of a set of curves on $M$ (many curves may have the same derivative at $P$). A point in $T_P$ does not represent some other point on $M$. So you cannot map points on $M$ to $T_P$ - apart from $P$ itself, which you can say maps to the origin in $T_P$ (because the constant "curve" $f(t)=P space forall t$ has derivative $0$).
$endgroup$
– gandalf61
Dec 11 '18 at 13:46
$begingroup$
Ah ok! So is the exponential mapping of point A at point B to the tangent space $T_b$ really just the vector in the tangent space that is parallel with the geodesic between A and B?
$endgroup$
– bidby
Dec 11 '18 at 14:05
$begingroup$
@bidby Yes, but you have to add extra structure such as a Riemannian metric to the manifold in order to define geodesics and exponential maps.
$endgroup$
– gandalf61
Dec 11 '18 at 14:20
add a comment |
$begingroup$
Each point on a differentiable manifold has its own tangent space. Informally, you can think of the tangent space at a point as the space of tangent vectors to the surface at that point. But this informal picture only works if the manifold is embedded in some higher dimensional space.
A more abstract and more formal point of view is that tangent space at a point $P$ of a differentiable manifold $M$ consists of the derivatives of all curves on $M$ that pass through $P$. The derivatives exist because each map between $M$ and a chart is differentiable and invertible. So we can map a curve on $M$ to a chart, carry out calculus in the chart and "lift" the result back to $M$. The result is independent of the choice of chart, so it is a property of $M$ itself (at $P$).
It is not entirely obvious that we can create a vector space from the derivatives of the curves through $P$, but we can because the derivative is a linear operator. This vector space is then the tangent space at $P$, which we denote by $T_P$. The set of tangent spaces $T_P$ for all points $P$ on $M$ forms a vector bundle over $M$.
answered Dec 11 '18 at 11:50
gandalf61gandalf61
8,801725
$endgroup$
$begingroup$
What is the difference between mapping the curve on M to a chart as you said, and mapping it to a tangent space at the same point?
$endgroup$
– bidby
Dec 11 '18 at 11:56
$begingroup$
Am I correct in my understanding if I say that one maps all the curves through a point P to Euclidean space using any chart that includes P. Once I am in this space I can differentiate all of these curves and I get a bunch of vectors, which form the tangent space/plane at P once I have mapped back to M using the inverse of the chart I originally used. While these tangent vectors may have looked different in the Euclidean space, depending on my choice of chart, the resulting tangent space/plane at P in M is independent of which chart I chose?
$endgroup$
– bidby
Dec 11 '18 at 13:38
1
$begingroup$
@bidby Yes, your second comment is correct. Going back to your first comment, each point in the tangent space $T_P$ represents the derivative at $P$ of a set of curves on $M$ (many curves may have the same derivative at $P$). A point in $T_P$ does not represent some other point on $M$. So you cannot map points on $M$ to $T_P$ - apart from $P$ itself, which you can say maps to the origin in $T_P$ (because the constant "curve" $f(t)=P space forall t$ has derivative $0$).
$endgroup$
– gandalf61
Dec 11 '18 at 13:46
$begingroup$
Ah ok! So is the exponential mapping of point A at point B to the tangent space $T_b$ really just the vector in the tangent space that is parallel with the geodesic between A and B?
$endgroup$
– bidby
Dec 11 '18 at 14:05
$begingroup$
@bidby Yes, but you have to add extra structure such as a Riemannian metric to the manifold in order to define geodesics and exponential maps.
$endgroup$
– gandalf61
Dec 11 '18 at 14:20
add a comment |
$begingroup$
Each point on a differentiable manifold has its own tangent space. Informally, you can think of the tangent space at a point as the space of tangent vectors to the surface at that point. But this informal picture only works if the manifold is embedded in some higher dimensional space.
A more abstract and more formal point of view is that tangent space at a point $P$ of a differentiable manifold $M$ consists of the derivatives of all curves on $M$ that pass through $P$. The derivatives exist because each map between $M$ and a chart is differentiable and invertible. So we can map a curve on $M$ to a chart, carry out calculus in the chart and "lift" the result back to $M$. The result is independent of the choice of chart, so it is a property of $M$ itself (at $P$).
It is not entirely obvious that we can create a vector space from the derivatives of the curves through $P$, but we can because the derivative is a linear operator. This vector space is then the tangent space at $P$, which we denote by $T_P$. The set of tangent spaces $T_P$ for all points $P$ on $M$ forms a vector bundle over $M$.
answered Dec 11 '18 at 11:50
gandalf61gandalf61
8,801725
$endgroup$
Each point on a differentiable manifold has its own tangent space. Informally, you can think of the tangent space at a point as the space of tangent vectors to the surface at that point. But this informal picture only works if the manifold is embedded in some higher dimensional space.
A more abstract and more formal point of view is that tangent space at a point $P$ of a differentiable manifold $M$ consists of the derivatives of all curves on $M$ that pass through $P$. The derivatives exist because each map between $M$ and a chart is differentiable and invertible. So we can map a curve on $M$ to a chart, carry out calculus in the chart and "lift" the result back to $M$. The result is independent of the choice of chart, so it is a property of $M$ itself (at $P$).
It is not entirely obvious that we can create a vector space from the derivatives of the curves through $P$, but we can because the derivative is a linear operator. This vector space is then the tangent space at $P$, which we denote by $T_P$. The set of tangent spaces $T_P$ for all points $P$ on $M$ forms a vector bundle over $M$.
answered Dec 11 '18 at 11:50
gandalf61gandalf61
8,801725
answered Dec 11 '18 at 11:50
gandalf61gandalf61
8,801725
answered Dec 11 '18 at 11:50
gandalf61gandalf61
8,801725
answered Dec 11 '18 at 11:50
gandalf61gandalf61
8,801725
8,801725
$begingroup$
What is the difference between mapping the curve on M to a chart as you said, and mapping it to a tangent space at the same point?
$endgroup$
– bidby
Dec 11 '18 at 11:56
$begingroup$
Am I correct in my understanding if I say that one maps all the curves through a point P to Euclidean space using any chart that includes P. Once I am in this space I can differentiate all of these curves and I get a bunch of vectors, which form the tangent space/plane at P once I have mapped back to M using the inverse of the chart I originally used. While these tangent vectors may have looked different in the Euclidean space, depending on my choice of chart, the resulting tangent space/plane at P in M is independent of which chart I chose?
$endgroup$
– bidby
Dec 11 '18 at 13:38
1
$begingroup$
@bidby Yes, your second comment is correct. Going back to your first comment, each point in the tangent space $T_P$ represents the derivative at $P$ of a set of curves on $M$ (many curves may have the same derivative at $P$). A point in $T_P$ does not represent some other point on $M$. So you cannot map points on $M$ to $T_P$ - apart from $P$ itself, which you can say maps to the origin in $T_P$ (because the constant "curve" $f(t)=P space forall t$ has derivative $0$).
$endgroup$
– gandalf61
Dec 11 '18 at 13:46
$begingroup$
Ah ok! So is the exponential mapping of point A at point B to the tangent space $T_b$ really just the vector in the tangent space that is parallel with the geodesic between A and B?
$endgroup$
– bidby
Dec 11 '18 at 14:05
$begingroup$
@bidby Yes, but you have to add extra structure such as a Riemannian metric to the manifold in order to define geodesics and exponential maps.
$endgroup$
– gandalf61
Dec 11 '18 at 14:20
add a comment |
$begingroup$
What is the difference between mapping the curve on M to a chart as you said, and mapping it to a tangent space at the same point?
$endgroup$
– bidby
Dec 11 '18 at 11:56
$begingroup$
Am I correct in my understanding if I say that one maps all the curves through a point P to Euclidean space using any chart that includes P. Once I am in this space I can differentiate all of these curves and I get a bunch of vectors, which form the tangent space/plane at P once I have mapped back to M using the inverse of the chart I originally used. While these tangent vectors may have looked different in the Euclidean space, depending on my choice of chart, the resulting tangent space/plane at P in M is independent of which chart I chose?
$endgroup$
– bidby
Dec 11 '18 at 13:38
1
$begingroup$
@bidby Yes, your second comment is correct. Going back to your first comment, each point in the tangent space $T_P$ represents the derivative at $P$ of a set of curves on $M$ (many curves may have the same derivative at $P$). A point in $T_P$ does not represent some other point on $M$. So you cannot map points on $M$ to $T_P$ - apart from $P$ itself, which you can say maps to the origin in $T_P$ (because the constant "curve" $f(t)=P space forall t$ has derivative $0$).
$endgroup$
– gandalf61
Dec 11 '18 at 13:46
$begingroup$
Ah ok! So is the exponential mapping of point A at point B to the tangent space $T_b$ really just the vector in the tangent space that is parallel with the geodesic between A and B?
$endgroup$
– bidby
Dec 11 '18 at 14:05
$begingroup$
@bidby Yes, but you have to add extra structure such as a Riemannian metric to the manifold in order to define geodesics and exponential maps.
$endgroup$
– gandalf61
Dec 11 '18 at 14:20
$begingroup$
What is the difference between mapping the curve on M to a chart as you said, and mapping it to a tangent space at the same point?
$endgroup$
– bidby
Dec 11 '18 at 11:56
$begingroup$
What is the difference between mapping the curve on M to a chart as you said, and mapping it to a tangent space at the same point?
$endgroup$
– bidby
Dec 11 '18 at 11:56
$begingroup$
Am I correct in my understanding if I say that one maps all the curves through a point P to Euclidean space using any chart that includes P. Once I am in this space I can differentiate all of these curves and I get a bunch of vectors, which form the tangent space/plane at P once I have mapped back to M using the inverse of the chart I originally used. While these tangent vectors may have looked different in the Euclidean space, depending on my choice of chart, the resulting tangent space/plane at P in M is independent of which chart I chose?
$endgroup$
– bidby
Dec 11 '18 at 13:38
$begingroup$
Am I correct in my understanding if I say that one maps all the curves through a point P to Euclidean space using any chart that includes P. Once I am in this space I can differentiate all of these curves and I get a bunch of vectors, which form the tangent space/plane at P once I have mapped back to M using the inverse of the chart I originally used. While these tangent vectors may have looked different in the Euclidean space, depending on my choice of chart, the resulting tangent space/plane at P in M is independent of which chart I chose?
$endgroup$
– bidby
Dec 11 '18 at 13:38
1
1
$begingroup$
@bidby Yes, your second comment is correct. Going back to your first comment, each point in the tangent space $T_P$ represents the derivative at $P$ of a set of curves on $M$ (many curves may have the same derivative at $P$). A point in $T_P$ does not represent some other point on $M$. So you cannot map points on $M$ to $T_P$ - apart from $P$ itself, which you can say maps to the origin in $T_P$ (because the constant "curve" $f(t)=P space forall t$ has derivative $0$).
$endgroup$
– gandalf61
Dec 11 '18 at 13:46
$begingroup$
@bidby Yes, your second comment is correct. Going back to your first comment, each point in the tangent space $T_P$ represents the derivative at $P$ of a set of curves on $M$ (many curves may have the same derivative at $P$). A point in $T_P$ does not represent some other point on $M$. So you cannot map points on $M$ to $T_P$ - apart from $P$ itself, which you can say maps to the origin in $T_P$ (because the constant "curve" $f(t)=P space forall t$ has derivative $0$).
$endgroup$
– gandalf61
Dec 11 '18 at 13:46
$begingroup$
Ah ok! So is the exponential mapping of point A at point B to the tangent space $T_b$ really just the vector in the tangent space that is parallel with the geodesic between A and B?
$endgroup$
– bidby
Dec 11 '18 at 14:05
$begingroup$
Ah ok! So is the exponential mapping of point A at point B to the tangent space $T_b$ really just the vector in the tangent space that is parallel with the geodesic between A and B?
$endgroup$
– bidby
Dec 11 '18 at 14:05
$begingroup$
@bidby Yes, but you have to add extra structure such as a Riemannian metric to the manifold in order to define geodesics and exponential maps.
$endgroup$
– gandalf61
Dec 11 '18 at 14:20
$begingroup$
@bidby Yes, but you have to add extra structure such as a Riemannian metric to the manifold in order to define geodesics and exponential maps.
$endgroup$
– gandalf61
Dec 11 '18 at 14:20
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