interpretation of bracket vector field as derivation along trajectories
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In DoCarmo’s Riemannian Geometry book, there is a theorem as follows: $[X,Y](p)=lim_{trightarrow 0} {1/t}times [Y-dphi_tY](phi_t(p))$, where $X$ and $Y$ are smooth vector fields on a smooth manifold $M$, $pin M$, and $phi_t$ is the local flow of $X$ in a neighborhood $U$ of $p$.
Both sides of that equality are functions from $U$ to the tangent bundle $TU$. But how can we take limit of the right side when the metric structure of $TU$ has not been assumed by the author? What does the author mean by that equality?
differential-geometry riemannian-geometry smooth-manifolds
asked Nov 16 at 11:16
Selflearner
377214
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show 4 more comments
up vote
0
down vote
favorite
In DoCarmo’s Riemannian Geometry book, there is a theorem as follows: $[X,Y](p)=lim_{trightarrow 0} {1/t}times [Y-dphi_tY](phi_t(p))$, where $X$ and $Y$ are smooth vector fields on a smooth manifold $M$, $pin M$, and $phi_t$ is the local flow of $X$ in a neighborhood $U$ of $p$.
Both sides of that equality are functions from $U$ to the tangent bundle $TU$. But how can we take limit of the right side when the metric structure of $TU$ has not been assumed by the author? What does the author mean by that equality?
differential-geometry riemannian-geometry smooth-manifolds
asked Nov 16 at 11:16
Selflearner
377214
Better to think of it this way: both sides of the equality are elements of the tangent space $T_pM$. This way, you should be able to make sense of the limit as usual.
– AnonymousCoward
Nov 16 at 11:27
Still, $T_pM$ has not been assumed to be a normed vector space for taking limits to make sense @AnonymousCoward
– Selflearner
Nov 16 at 11:29
Take any choice of inner product, they will all define the same topology.
– AnonymousCoward
Nov 16 at 11:30
I don’t think the author had a norm on $T_pM$ in his mind. By that equality, he most probably means something else
– Selflearner
Nov 16 at 11:33
To put my answer differently: finite dimension topological vector spaces are classified by dimension only. You do not need to worry about a choice of norm for the limit to be well-defined.
– AnonymousCoward
Nov 16 at 11:38
|
show 4 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
In DoCarmo’s Riemannian Geometry book, there is a theorem as follows: $[X,Y](p)=lim_{trightarrow 0} {1/t}times [Y-dphi_tY](phi_t(p))$, where $X$ and $Y$ are smooth vector fields on a smooth manifold $M$, $pin M$, and $phi_t$ is the local flow of $X$ in a neighborhood $U$ of $p$.
Both sides of that equality are functions from $U$ to the tangent bundle $TU$. But how can we take limit of the right side when the metric structure of $TU$ has not been assumed by the author? What does the author mean by that equality?
differential-geometry riemannian-geometry smooth-manifolds
asked Nov 16 at 11:16
Selflearner
377214
In DoCarmo’s Riemannian Geometry book, there is a theorem as follows: $[X,Y](p)=lim_{trightarrow 0} {1/t}times [Y-dphi_tY](phi_t(p))$, where $X$ and $Y$ are smooth vector fields on a smooth manifold $M$, $pin M$, and $phi_t$ is the local flow of $X$ in a neighborhood $U$ of $p$.
Both sides of that equality are functions from $U$ to the tangent bundle $TU$. But how can we take limit of the right side when the metric structure of $TU$ has not been assumed by the author? What does the author mean by that equality?
differential-geometry riemannian-geometry smooth-manifolds
differential-geometry riemannian-geometry smooth-manifolds
asked Nov 16 at 11:16
Selflearner
377214
asked Nov 16 at 11:16
Selflearner
377214
edited Nov 16 at 11:25
asked Nov 16 at 11:16
Selflearner
377214
asked Nov 16 at 11:16
Selflearner
377214
asked Nov 16 at 11:16
Selflearner
377214
377214
Better to think of it this way: both sides of the equality are elements of the tangent space $T_pM$. This way, you should be able to make sense of the limit as usual.
– AnonymousCoward
Nov 16 at 11:27
Still, $T_pM$ has not been assumed to be a normed vector space for taking limits to make sense @AnonymousCoward
– Selflearner
Nov 16 at 11:29
Take any choice of inner product, they will all define the same topology.
– AnonymousCoward
Nov 16 at 11:30
I don’t think the author had a norm on $T_pM$ in his mind. By that equality, he most probably means something else
– Selflearner
Nov 16 at 11:33
To put my answer differently: finite dimension topological vector spaces are classified by dimension only. You do not need to worry about a choice of norm for the limit to be well-defined.
– AnonymousCoward
Nov 16 at 11:38
|
show 4 more comments
Better to think of it this way: both sides of the equality are elements of the tangent space $T_pM$. This way, you should be able to make sense of the limit as usual.
– AnonymousCoward
Nov 16 at 11:27
Still, $T_pM$ has not been assumed to be a normed vector space for taking limits to make sense @AnonymousCoward
– Selflearner
Nov 16 at 11:29
Take any choice of inner product, they will all define the same topology.
– AnonymousCoward
Nov 16 at 11:30
I don’t think the author had a norm on $T_pM$ in his mind. By that equality, he most probably means something else
– Selflearner
Nov 16 at 11:33
To put my answer differently: finite dimension topological vector spaces are classified by dimension only. You do not need to worry about a choice of norm for the limit to be well-defined.
– AnonymousCoward
Nov 16 at 11:38
Better to think of it this way: both sides of the equality are elements of the tangent space $T_pM$. This way, you should be able to make sense of the limit as usual.
– AnonymousCoward
Nov 16 at 11:27
Better to think of it this way: both sides of the equality are elements of the tangent space $T_pM$. This way, you should be able to make sense of the limit as usual.
– AnonymousCoward
Nov 16 at 11:27
Still, $T_pM$ has not been assumed to be a normed vector space for taking limits to make sense @AnonymousCoward
– Selflearner
Nov 16 at 11:29
Still, $T_pM$ has not been assumed to be a normed vector space for taking limits to make sense @AnonymousCoward
– Selflearner
Nov 16 at 11:29
Take any choice of inner product, they will all define the same topology.
– AnonymousCoward
Nov 16 at 11:30
Take any choice of inner product, they will all define the same topology.
– AnonymousCoward
Nov 16 at 11:30
I don’t think the author had a norm on $T_pM$ in his mind. By that equality, he most probably means something else
– Selflearner
Nov 16 at 11:33
I don’t think the author had a norm on $T_pM$ in his mind. By that equality, he most probably means something else
– Selflearner
Nov 16 at 11:33
To put my answer differently: finite dimension topological vector spaces are classified by dimension only. You do not need to worry about a choice of norm for the limit to be well-defined.
– AnonymousCoward
Nov 16 at 11:38
To put my answer differently: finite dimension topological vector spaces are classified by dimension only. You do not need to worry about a choice of norm for the limit to be well-defined.
– AnonymousCoward
Nov 16 at 11:38
|
show 4 more comments
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Better to think of it this way: both sides of the equality are elements of the tangent space $T_pM$. This way, you should be able to make sense of the limit as usual.
– AnonymousCoward
Nov 16 at 11:27
Still, $T_pM$ has not been assumed to be a normed vector space for taking limits to make sense @AnonymousCoward
– Selflearner
Nov 16 at 11:29
Take any choice of inner product, they will all define the same topology.
– AnonymousCoward
Nov 16 at 11:30
I don’t think the author had a norm on $T_pM$ in his mind. By that equality, he most probably means something else
– Selflearner
Nov 16 at 11:33
To put my answer differently: finite dimension topological vector spaces are classified by dimension only. You do not need to worry about a choice of norm for the limit to be well-defined.
– AnonymousCoward
Nov 16 at 11:38