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Why is the second mixed partial not belong to tangent space $partial^2_ij notin T_pM$
Understanding the Definition of a Differential Form of Degree $k$How to induce a connection on a submanifold?Tangent spaces at different points and the concept of connectionWhat's the advantage of using a connection instead of embedding it to Euclidean space?On an informal explanation of the tangent space to a manifoldMap Laplacian in terms of covariant derivativesSeeing that the second fundamental form is the orthogonal component of the LaplacianTwo definitions of covariant derivatives in the setting of surfaces in $mathbbR^3$Levi-Civita connections from metrics on the orthogonal frame bundleAre Christoffel symbols structure coefficients?
$begingroup$
So I was just reading on wikipedia about covariant derivatives
And it is mentioned that $fracpartial^2partial_ipartial_j$ is not tangential to $M$, so that means it won't be in the tangent space of $M$.
But isn't there a canonical isomorphism of $T_0(T_pM) approx T_pM$?
Or is there a more simple geometric reason that the rate of change of the velocity field must not be in the tangent space?
Every picture (Da Carmo, or Presley) shows that $dv/dt$ is a vector pointing away from the manifold $M$ and tangent space $T_pM$ with it orthogonal decomposition. How do we know $dv/dt$ can't lie in the spaces?
calculus differential-geometry
$endgroup$
add a comment |
$begingroup$
So I was just reading on wikipedia about covariant derivatives
And it is mentioned that $fracpartial^2partial_ipartial_j$ is not tangential to $M$, so that means it won't be in the tangent space of $M$.
But isn't there a canonical isomorphism of $T_0(T_pM) approx T_pM$?
Or is there a more simple geometric reason that the rate of change of the velocity field must not be in the tangent space?
Every picture (Da Carmo, or Presley) shows that $dv/dt$ is a vector pointing away from the manifold $M$ and tangent space $T_pM$ with it orthogonal decomposition. How do we know $dv/dt$ can't lie in the spaces?
calculus differential-geometry
$endgroup$
add a comment |
$begingroup$
So I was just reading on wikipedia about covariant derivatives
And it is mentioned that $fracpartial^2partial_ipartial_j$ is not tangential to $M$, so that means it won't be in the tangent space of $M$.
But isn't there a canonical isomorphism of $T_0(T_pM) approx T_pM$?
Or is there a more simple geometric reason that the rate of change of the velocity field must not be in the tangent space?
Every picture (Da Carmo, or Presley) shows that $dv/dt$ is a vector pointing away from the manifold $M$ and tangent space $T_pM$ with it orthogonal decomposition. How do we know $dv/dt$ can't lie in the spaces?
calculus differential-geometry
$endgroup$
So I was just reading on wikipedia about covariant derivatives
And it is mentioned that $fracpartial^2partial_ipartial_j$ is not tangential to $M$, so that means it won't be in the tangent space of $M$.
But isn't there a canonical isomorphism of $T_0(T_pM) approx T_pM$?
Or is there a more simple geometric reason that the rate of change of the velocity field must not be in the tangent space?
Every picture (Da Carmo, or Presley) shows that $dv/dt$ is a vector pointing away from the manifold $M$ and tangent space $T_pM$ with it orthogonal decomposition. How do we know $dv/dt$ can't lie in the spaces?
calculus differential-geometry
calculus differential-geometry
asked Mar 22 at 3:37
HawkHawk
5,5751140110
5,5751140110
add a comment |
add a comment |
2 Answers
2
active
oldest
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$begingroup$
They just mean "not necessarily tangential", i.e., you cannot hope to use that they are tangential in any general argument. But sometimes that $vec n$ will be $0$.
Here's a simple geometric example. Consider the circle $S^1subset mathbbR^2$ embedded in the plane, and consider the constant speed path $gamma:mathbbR to S^1$ given by $thetamapsto (cos(theta),sin(theta))$. The velocity vector is always tangent to the circle, but the acceleration is always radial, perpendicular to the circle.
But it's not that these are never in the tangent space. For example, embed $mathbbR$ inside $mathbbR^2$ as the $x$-axis. Any path in this copy of $mathbbR$ necessarily has acceleration which is tangent to the line, because it is straight. But you could also embed $mathbbR$ into $mathbbR^2$ as a parabola by $tmapsto (t,t^2)$, which will have some acceleration vectors which are not tangent to the parabola.
$endgroup$
add a comment |
$begingroup$
$fracpartial^2partial_ipartial_j$ is not a tangent vector because it does not satisfy the product rule:
$$fracpartial^2partial_ipartial_j(x_ix_j)
=1neq 0
=(fracpartial^2partial_ipartial_jx_i)cdot x_j+x_icdot (fracpartial^2partial_ipartial_jx_j)$$
$endgroup$
add a comment |
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2 Answers
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active
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2 Answers
2
active
oldest
votes
active
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active
oldest
votes
$begingroup$
They just mean "not necessarily tangential", i.e., you cannot hope to use that they are tangential in any general argument. But sometimes that $vec n$ will be $0$.
Here's a simple geometric example. Consider the circle $S^1subset mathbbR^2$ embedded in the plane, and consider the constant speed path $gamma:mathbbR to S^1$ given by $thetamapsto (cos(theta),sin(theta))$. The velocity vector is always tangent to the circle, but the acceleration is always radial, perpendicular to the circle.
But it's not that these are never in the tangent space. For example, embed $mathbbR$ inside $mathbbR^2$ as the $x$-axis. Any path in this copy of $mathbbR$ necessarily has acceleration which is tangent to the line, because it is straight. But you could also embed $mathbbR$ into $mathbbR^2$ as a parabola by $tmapsto (t,t^2)$, which will have some acceleration vectors which are not tangent to the parabola.
$endgroup$
add a comment |
$begingroup$
They just mean "not necessarily tangential", i.e., you cannot hope to use that they are tangential in any general argument. But sometimes that $vec n$ will be $0$.
Here's a simple geometric example. Consider the circle $S^1subset mathbbR^2$ embedded in the plane, and consider the constant speed path $gamma:mathbbR to S^1$ given by $thetamapsto (cos(theta),sin(theta))$. The velocity vector is always tangent to the circle, but the acceleration is always radial, perpendicular to the circle.
But it's not that these are never in the tangent space. For example, embed $mathbbR$ inside $mathbbR^2$ as the $x$-axis. Any path in this copy of $mathbbR$ necessarily has acceleration which is tangent to the line, because it is straight. But you could also embed $mathbbR$ into $mathbbR^2$ as a parabola by $tmapsto (t,t^2)$, which will have some acceleration vectors which are not tangent to the parabola.
$endgroup$
add a comment |
$begingroup$
They just mean "not necessarily tangential", i.e., you cannot hope to use that they are tangential in any general argument. But sometimes that $vec n$ will be $0$.
Here's a simple geometric example. Consider the circle $S^1subset mathbbR^2$ embedded in the plane, and consider the constant speed path $gamma:mathbbR to S^1$ given by $thetamapsto (cos(theta),sin(theta))$. The velocity vector is always tangent to the circle, but the acceleration is always radial, perpendicular to the circle.
But it's not that these are never in the tangent space. For example, embed $mathbbR$ inside $mathbbR^2$ as the $x$-axis. Any path in this copy of $mathbbR$ necessarily has acceleration which is tangent to the line, because it is straight. But you could also embed $mathbbR$ into $mathbbR^2$ as a parabola by $tmapsto (t,t^2)$, which will have some acceleration vectors which are not tangent to the parabola.
$endgroup$
They just mean "not necessarily tangential", i.e., you cannot hope to use that they are tangential in any general argument. But sometimes that $vec n$ will be $0$.
Here's a simple geometric example. Consider the circle $S^1subset mathbbR^2$ embedded in the plane, and consider the constant speed path $gamma:mathbbR to S^1$ given by $thetamapsto (cos(theta),sin(theta))$. The velocity vector is always tangent to the circle, but the acceleration is always radial, perpendicular to the circle.
But it's not that these are never in the tangent space. For example, embed $mathbbR$ inside $mathbbR^2$ as the $x$-axis. Any path in this copy of $mathbbR$ necessarily has acceleration which is tangent to the line, because it is straight. But you could also embed $mathbbR$ into $mathbbR^2$ as a parabola by $tmapsto (t,t^2)$, which will have some acceleration vectors which are not tangent to the parabola.
edited Mar 22 at 13:40
answered Mar 22 at 13:26
cspruncsprun
2,795211
2,795211
add a comment |
add a comment |
$begingroup$
$fracpartial^2partial_ipartial_j$ is not a tangent vector because it does not satisfy the product rule:
$$fracpartial^2partial_ipartial_j(x_ix_j)
=1neq 0
=(fracpartial^2partial_ipartial_jx_i)cdot x_j+x_icdot (fracpartial^2partial_ipartial_jx_j)$$
$endgroup$
add a comment |
$begingroup$
$fracpartial^2partial_ipartial_j$ is not a tangent vector because it does not satisfy the product rule:
$$fracpartial^2partial_ipartial_j(x_ix_j)
=1neq 0
=(fracpartial^2partial_ipartial_jx_i)cdot x_j+x_icdot (fracpartial^2partial_ipartial_jx_j)$$
$endgroup$
add a comment |
$begingroup$
$fracpartial^2partial_ipartial_j$ is not a tangent vector because it does not satisfy the product rule:
$$fracpartial^2partial_ipartial_j(x_ix_j)
=1neq 0
=(fracpartial^2partial_ipartial_jx_i)cdot x_j+x_icdot (fracpartial^2partial_ipartial_jx_j)$$
$endgroup$
$fracpartial^2partial_ipartial_j$ is not a tangent vector because it does not satisfy the product rule:
$$fracpartial^2partial_ipartial_j(x_ix_j)
=1neq 0
=(fracpartial^2partial_ipartial_jx_i)cdot x_j+x_icdot (fracpartial^2partial_ipartial_jx_j)$$
answered Mar 22 at 16:18
triitrii
83817
83817
add a comment |
add a comment |
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