Well-definedness of assinging a value to a point of a manifold with respect to parallelism
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By definition as in DoCarmo’s Riemannian Geometry book, in a smooth manifold $M$ a vector field $V$ along a smooth curve $c:(a,b)rightarrow M$ is a mapping $tmapsto V(t)in T_{c(t)}M$ such that $tmapsto V(t)f$ is a smooth map for all smooth maps $fin C^{infty}(M)$.
Suppose that $c_1,c_2$ are two smooth curves in $M$ such that $c_1(t_0)=c_2(t_0)$, and $c_1(t_1)=c_2(t_1)$ where $t_0$ and $t_1$ are two distinct points in the (same) domain of $c_1,c_2$. Suppose that $V_0in T_{c_1(t_0)}M$. There exists a unique parallel vector field $V_1$ along $c_1$ such that $V_1(t_0)=V_0$, and similarly a unique $V_2$ along $c_2$ such that $V_2(t_0)=V_0$. Does it follow that $V_1(t_1)=V_2(t_1)$?
differential-geometry riemannian-geometry smooth-manifolds vector-fields
asked Dec 5 at 13:38
User12239
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By definition as in DoCarmo’s Riemannian Geometry book, in a smooth manifold $M$ a vector field $V$ along a smooth curve $c:(a,b)rightarrow M$ is a mapping $tmapsto V(t)in T_{c(t)}M$ such that $tmapsto V(t)f$ is a smooth map for all smooth maps $fin C^{infty}(M)$.
Suppose that $c_1,c_2$ are two smooth curves in $M$ such that $c_1(t_0)=c_2(t_0)$, and $c_1(t_1)=c_2(t_1)$ where $t_0$ and $t_1$ are two distinct points in the (same) domain of $c_1,c_2$. Suppose that $V_0in T_{c_1(t_0)}M$. There exists a unique parallel vector field $V_1$ along $c_1$ such that $V_1(t_0)=V_0$, and similarly a unique $V_2$ along $c_2$ such that $V_2(t_0)=V_0$. Does it follow that $V_1(t_1)=V_2(t_1)$?
differential-geometry riemannian-geometry smooth-manifolds vector-fields
asked Dec 5 at 13:38
User12239
405215
add a comment |
up vote
1
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up vote
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down vote
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By definition as in DoCarmo’s Riemannian Geometry book, in a smooth manifold $M$ a vector field $V$ along a smooth curve $c:(a,b)rightarrow M$ is a mapping $tmapsto V(t)in T_{c(t)}M$ such that $tmapsto V(t)f$ is a smooth map for all smooth maps $fin C^{infty}(M)$.
Suppose that $c_1,c_2$ are two smooth curves in $M$ such that $c_1(t_0)=c_2(t_0)$, and $c_1(t_1)=c_2(t_1)$ where $t_0$ and $t_1$ are two distinct points in the (same) domain of $c_1,c_2$. Suppose that $V_0in T_{c_1(t_0)}M$. There exists a unique parallel vector field $V_1$ along $c_1$ such that $V_1(t_0)=V_0$, and similarly a unique $V_2$ along $c_2$ such that $V_2(t_0)=V_0$. Does it follow that $V_1(t_1)=V_2(t_1)$?
differential-geometry riemannian-geometry smooth-manifolds vector-fields
asked Dec 5 at 13:38
User12239
405215
By definition as in DoCarmo’s Riemannian Geometry book, in a smooth manifold $M$ a vector field $V$ along a smooth curve $c:(a,b)rightarrow M$ is a mapping $tmapsto V(t)in T_{c(t)}M$ such that $tmapsto V(t)f$ is a smooth map for all smooth maps $fin C^{infty}(M)$.
Suppose that $c_1,c_2$ are two smooth curves in $M$ such that $c_1(t_0)=c_2(t_0)$, and $c_1(t_1)=c_2(t_1)$ where $t_0$ and $t_1$ are two distinct points in the (same) domain of $c_1,c_2$. Suppose that $V_0in T_{c_1(t_0)}M$. There exists a unique parallel vector field $V_1$ along $c_1$ such that $V_1(t_0)=V_0$, and similarly a unique $V_2$ along $c_2$ such that $V_2(t_0)=V_0$. Does it follow that $V_1(t_1)=V_2(t_1)$?
differential-geometry riemannian-geometry smooth-manifolds vector-fields
differential-geometry riemannian-geometry smooth-manifolds vector-fields
asked Dec 5 at 13:38
User12239
405215
asked Dec 5 at 13:38
User12239
405215
edited Dec 5 at 13:57
asked Dec 5 at 13:38
User12239
405215
asked Dec 5 at 13:38
User12239
405215
asked Dec 5 at 13:38
User12239
405215
405215
add a comment |
add a comment |
1 Answer
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Not necessarily, unless the space is flat. This is actually a definition; a space is flat if and only if the property you ask about holds. The Euclidean space is flat.
These things are the starting point of the concept of curvature.
The following picture is taken from Wikipedia and it shows how the choice of a path can produce different parallel trasported vectors, in the case of the sphere.
answered Dec 5 at 14:15
Giuseppe Negro
17.3k330122
Thanks for this intuitive illustration
– User12239
Dec 5 at 14:26
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Not necessarily, unless the space is flat. This is actually a definition; a space is flat if and only if the property you ask about holds. The Euclidean space is flat.
These things are the starting point of the concept of curvature.
The following picture is taken from Wikipedia and it shows how the choice of a path can produce different parallel trasported vectors, in the case of the sphere.
answered Dec 5 at 14:15
Giuseppe Negro
17.3k330122
Thanks for this intuitive illustration
– User12239
Dec 5 at 14:26
add a comment |
up vote
1
down vote
accepted
Not necessarily, unless the space is flat. This is actually a definition; a space is flat if and only if the property you ask about holds. The Euclidean space is flat.
These things are the starting point of the concept of curvature.
The following picture is taken from Wikipedia and it shows how the choice of a path can produce different parallel trasported vectors, in the case of the sphere.
answered Dec 5 at 14:15
Giuseppe Negro
17.3k330122
Thanks for this intuitive illustration
– User12239
Dec 5 at 14:26
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Not necessarily, unless the space is flat. This is actually a definition; a space is flat if and only if the property you ask about holds. The Euclidean space is flat.
These things are the starting point of the concept of curvature.
The following picture is taken from Wikipedia and it shows how the choice of a path can produce different parallel trasported vectors, in the case of the sphere.
answered Dec 5 at 14:15
Giuseppe Negro
17.3k330122
Not necessarily, unless the space is flat. This is actually a definition; a space is flat if and only if the property you ask about holds. The Euclidean space is flat.
These things are the starting point of the concept of curvature.
The following picture is taken from Wikipedia and it shows how the choice of a path can produce different parallel trasported vectors, in the case of the sphere.
answered Dec 5 at 14:15
Giuseppe Negro
17.3k330122
answered Dec 5 at 14:15
Giuseppe Negro
17.3k330122
answered Dec 5 at 14:15
Giuseppe Negro
17.3k330122
answered Dec 5 at 14:15
Giuseppe Negro
17.3k330122
17.3k330122
Thanks for this intuitive illustration
– User12239
Dec 5 at 14:26
add a comment |
Thanks for this intuitive illustration
– User12239
Dec 5 at 14:26
Thanks for this intuitive illustration
– User12239
Dec 5 at 14:26
Thanks for this intuitive illustration
– User12239
Dec 5 at 14:26
add a comment |
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