Is velocity field extendible to a vector field on the whole manifold?
In DoCarmo’s Riemannian Geometry book, an affine connection $nabla$ on a smooth manifold $M$ is defined to be a mapping :$Xi(M)times Xi(M)rightarrow Xi(M)$, where $Xi(M)$ is the set of all smooth vector fields on $M$.
Now DoCarmo wants to clarify this definition with a theorem containing the following fact: ... c) if $V$ is a vector field along the curve $c$, and if $V$ is induced by a vector field $Yin Xi(M)$, then the covariant derivative of $V$ along $c$ denoted by $frac{DV}{dt}$ equals $nabla(frac{dc}{dt},Y)$, where $frac{dc}{dt}$ is the velocity field of the curve $c$.
But the domain of $nabla$ does not include $(frac{dc}{dt},Y)$, since $frac{dc}{dt}$ does not belong to $Xi(M)$. What is the justification?
differential-geometry riemannian-geometry smooth-manifolds vector-fields
asked Nov 23 at 12:56
User12239
405215
add a comment |
In DoCarmo’s Riemannian Geometry book, an affine connection $nabla$ on a smooth manifold $M$ is defined to be a mapping :$Xi(M)times Xi(M)rightarrow Xi(M)$, where $Xi(M)$ is the set of all smooth vector fields on $M$.
Now DoCarmo wants to clarify this definition with a theorem containing the following fact: ... c) if $V$ is a vector field along the curve $c$, and if $V$ is induced by a vector field $Yin Xi(M)$, then the covariant derivative of $V$ along $c$ denoted by $frac{DV}{dt}$ equals $nabla(frac{dc}{dt},Y)$, where $frac{dc}{dt}$ is the velocity field of the curve $c$.
But the domain of $nabla$ does not include $(frac{dc}{dt},Y)$, since $frac{dc}{dt}$ does not belong to $Xi(M)$. What is the justification?
differential-geometry riemannian-geometry smooth-manifolds vector-fields
asked Nov 23 at 12:56
User12239
405215
You can extend $frac{dc}{dt}$ to a smooth vector field on the manifold, and the value of the directional derivative along $c$ is independent of that choice.
– Charlie Frohman
Nov 23 at 13:05
DoCarmo immediately after defining $frac{dc}{dt}$ states that vector fields along a curve do not necessarily extend to the whole manifold @CharlieFrohman
– User12239
Nov 23 at 13:09
math.stackexchange.com/questions/1865661/…. According to this, the velocity field is generally not a vector field on the whole manifold
– User12239
Nov 23 at 13:14
The connection is defined globally on sections of the tangent bundle. Then one can show that the value of the covariant derivative $nabla_XY$ at a point $p$ actually only depends on the value $X_p$ and $Y$ in a neighborhood of $p$. This allows for the $frac{D}{dt}$ to be well-defined.
– Matt
Nov 23 at 14:02
Thanks, I got it completely now @CharlieFrohman and Matt
– User12239
Nov 23 at 15:17
add a comment |
In DoCarmo’s Riemannian Geometry book, an affine connection $nabla$ on a smooth manifold $M$ is defined to be a mapping :$Xi(M)times Xi(M)rightarrow Xi(M)$, where $Xi(M)$ is the set of all smooth vector fields on $M$.
Now DoCarmo wants to clarify this definition with a theorem containing the following fact: ... c) if $V$ is a vector field along the curve $c$, and if $V$ is induced by a vector field $Yin Xi(M)$, then the covariant derivative of $V$ along $c$ denoted by $frac{DV}{dt}$ equals $nabla(frac{dc}{dt},Y)$, where $frac{dc}{dt}$ is the velocity field of the curve $c$.
But the domain of $nabla$ does not include $(frac{dc}{dt},Y)$, since $frac{dc}{dt}$ does not belong to $Xi(M)$. What is the justification?
differential-geometry riemannian-geometry smooth-manifolds vector-fields
asked Nov 23 at 12:56
User12239
405215
In DoCarmo’s Riemannian Geometry book, an affine connection $nabla$ on a smooth manifold $M$ is defined to be a mapping :$Xi(M)times Xi(M)rightarrow Xi(M)$, where $Xi(M)$ is the set of all smooth vector fields on $M$.
Now DoCarmo wants to clarify this definition with a theorem containing the following fact: ... c) if $V$ is a vector field along the curve $c$, and if $V$ is induced by a vector field $Yin Xi(M)$, then the covariant derivative of $V$ along $c$ denoted by $frac{DV}{dt}$ equals $nabla(frac{dc}{dt},Y)$, where $frac{dc}{dt}$ is the velocity field of the curve $c$.
But the domain of $nabla$ does not include $(frac{dc}{dt},Y)$, since $frac{dc}{dt}$ does not belong to $Xi(M)$. What is the justification?
differential-geometry riemannian-geometry smooth-manifolds vector-fields
differential-geometry riemannian-geometry smooth-manifolds vector-fields
asked Nov 23 at 12:56
User12239
405215
asked Nov 23 at 12:56
User12239
405215
edited Nov 23 at 13:50
asked Nov 23 at 12:56
User12239
405215
asked Nov 23 at 12:56
User12239
405215
asked Nov 23 at 12:56
User12239
405215
405215
You can extend $frac{dc}{dt}$ to a smooth vector field on the manifold, and the value of the directional derivative along $c$ is independent of that choice.
– Charlie Frohman
Nov 23 at 13:05
DoCarmo immediately after defining $frac{dc}{dt}$ states that vector fields along a curve do not necessarily extend to the whole manifold @CharlieFrohman
– User12239
Nov 23 at 13:09
math.stackexchange.com/questions/1865661/…. According to this, the velocity field is generally not a vector field on the whole manifold
– User12239
Nov 23 at 13:14
The connection is defined globally on sections of the tangent bundle. Then one can show that the value of the covariant derivative $nabla_XY$ at a point $p$ actually only depends on the value $X_p$ and $Y$ in a neighborhood of $p$. This allows for the $frac{D}{dt}$ to be well-defined.
– Matt
Nov 23 at 14:02
Thanks, I got it completely now @CharlieFrohman and Matt
– User12239
Nov 23 at 15:17
add a comment |
You can extend $frac{dc}{dt}$ to a smooth vector field on the manifold, and the value of the directional derivative along $c$ is independent of that choice.
– Charlie Frohman
Nov 23 at 13:05
DoCarmo immediately after defining $frac{dc}{dt}$ states that vector fields along a curve do not necessarily extend to the whole manifold @CharlieFrohman
– User12239
Nov 23 at 13:09
math.stackexchange.com/questions/1865661/…. According to this, the velocity field is generally not a vector field on the whole manifold
– User12239
Nov 23 at 13:14
The connection is defined globally on sections of the tangent bundle. Then one can show that the value of the covariant derivative $nabla_XY$ at a point $p$ actually only depends on the value $X_p$ and $Y$ in a neighborhood of $p$. This allows for the $frac{D}{dt}$ to be well-defined.
– Matt
Nov 23 at 14:02
Thanks, I got it completely now @CharlieFrohman and Matt
– User12239
Nov 23 at 15:17
You can extend $frac{dc}{dt}$ to a smooth vector field on the manifold, and the value of the directional derivative along $c$ is independent of that choice.
– Charlie Frohman
Nov 23 at 13:05
You can extend $frac{dc}{dt}$ to a smooth vector field on the manifold, and the value of the directional derivative along $c$ is independent of that choice.
– Charlie Frohman
Nov 23 at 13:05
DoCarmo immediately after defining $frac{dc}{dt}$ states that vector fields along a curve do not necessarily extend to the whole manifold @CharlieFrohman
– User12239
Nov 23 at 13:09
DoCarmo immediately after defining $frac{dc}{dt}$ states that vector fields along a curve do not necessarily extend to the whole manifold @CharlieFrohman
– User12239
Nov 23 at 13:09
math.stackexchange.com/questions/1865661/…. According to this, the velocity field is generally not a vector field on the whole manifold
– User12239
Nov 23 at 13:14
math.stackexchange.com/questions/1865661/…. According to this, the velocity field is generally not a vector field on the whole manifold
– User12239
Nov 23 at 13:14
The connection is defined globally on sections of the tangent bundle. Then one can show that the value of the covariant derivative $nabla_XY$ at a point $p$ actually only depends on the value $X_p$ and $Y$ in a neighborhood of $p$. This allows for the $frac{D}{dt}$ to be well-defined.
– Matt
Nov 23 at 14:02
The connection is defined globally on sections of the tangent bundle. Then one can show that the value of the covariant derivative $nabla_XY$ at a point $p$ actually only depends on the value $X_p$ and $Y$ in a neighborhood of $p$. This allows for the $frac{D}{dt}$ to be well-defined.
– Matt
Nov 23 at 14:02
Thanks, I got it completely now @CharlieFrohman and Matt
– User12239
Nov 23 at 15:17
Thanks, I got it completely now @CharlieFrohman and Matt
– User12239
Nov 23 at 15:17
add a comment |
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You can extend $frac{dc}{dt}$ to a smooth vector field on the manifold, and the value of the directional derivative along $c$ is independent of that choice.
– Charlie Frohman
Nov 23 at 13:05
DoCarmo immediately after defining $frac{dc}{dt}$ states that vector fields along a curve do not necessarily extend to the whole manifold @CharlieFrohman
– User12239
Nov 23 at 13:09
math.stackexchange.com/questions/1865661/…. According to this, the velocity field is generally not a vector field on the whole manifold
– User12239
Nov 23 at 13:14
The connection is defined globally on sections of the tangent bundle. Then one can show that the value of the covariant derivative $nabla_XY$ at a point $p$ actually only depends on the value $X_p$ and $Y$ in a neighborhood of $p$. This allows for the $frac{D}{dt}$ to be well-defined.
– Matt
Nov 23 at 14:02
Thanks, I got it completely now @CharlieFrohman and Matt
– User12239
Nov 23 at 15:17