Smoothness of a vector field defined by a system of differential equations
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There is a lemma stating that there exists a unique vector field $G$ on the tangent bundle $TM$ of a manifold $M$ whose integral curves are of the form $tmapsto (gamma(t),gamma’(t))$ where $gamma$ is a geodesic.
The proof using a system of differential equations first shows that if we assume existence, then its uniqueness follows. However, to show existence the proof defines a vector field locally using the same system of differential equations but why must it define a smooth vector field?
The system of differential equations is $x_k’(t)=y_k, y_k’(t)=-sum_{i,j}Gamma_{ij}^k y_i y_j$, taking $(x_1,...,x_n,y_1,...,y_n)$ as the coordinate of a curve in $TM$
ordinary-differential-equations riemannian-geometry vector-fields
asked Dec 11 '18 at 21:29
User12239User12239
453216
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show 1 more comment
$begingroup$
There is a lemma stating that there exists a unique vector field $G$ on the tangent bundle $TM$ of a manifold $M$ whose integral curves are of the form $tmapsto (gamma(t),gamma’(t))$ where $gamma$ is a geodesic.
The proof using a system of differential equations first shows that if we assume existence, then its uniqueness follows. However, to show existence the proof defines a vector field locally using the same system of differential equations but why must it define a smooth vector field?
The system of differential equations is $x_k’(t)=y_k, y_k’(t)=-sum_{i,j}Gamma_{ij}^k y_i y_j$, taking $(x_1,...,x_n,y_1,...,y_n)$ as the coordinate of a curve in $TM$
ordinary-differential-equations riemannian-geometry vector-fields
asked Dec 11 '18 at 21:29
User12239User12239
453216
$endgroup$
$begingroup$
The manifold is smooth, which means that the chart is smooth. The Christoffel symbols $Γ^k_{ij}$ are linear combinations of the partial derivatives of the metric tensor, thus smooth. Where do you see the obstacle to the vector field $(x',y')$ being smooth?
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– LutzL
Dec 11 '18 at 21:41
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@LutzL that system of differential equations defines a collection of smooth vector fields only along curves, not on any open set necessarily.
$endgroup$
– User12239
Dec 11 '18 at 22:05
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No, it defines a vector field on $TM$, resp on the open set $Utimes Bbb R^n$ corresponding to the chart and the tangent bundle above it. To each $(x,y)$ where $xin M$ and $yin T_xM$ it assigns a vector $(x',y')in T(TM)$. Which looks like $TMtimes TM$.
$endgroup$
– LutzL
Dec 11 '18 at 22:10
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In fact $G$ is defined by $(q,v)mapsto (x_1’(t),...,x_n’(t),y_1’(t),...,y_n’(t))$ where $x_i,y_i$ are as above with the initial condition $x(0)=q,y(0)=v$. So this vector field is smooth on a collection of curves partitioning the coordinate system’s neighborhood, but not necessarily smooth on the whole neighborhood @LutzL
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– User12239
Dec 11 '18 at 22:17
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You are getting confused by the prime, one should find better notations. Set $xi_k(x,y)=y_k$, $eta_k(x,y)=-sum Γ^k_{ij}y_iy_j$, then $(xi(x,y), eta(x,y))$ is a vector field on the chart, not only along curves.
$endgroup$
– LutzL
Dec 11 '18 at 22:37
|
show 1 more comment
$begingroup$
There is a lemma stating that there exists a unique vector field $G$ on the tangent bundle $TM$ of a manifold $M$ whose integral curves are of the form $tmapsto (gamma(t),gamma’(t))$ where $gamma$ is a geodesic.
The proof using a system of differential equations first shows that if we assume existence, then its uniqueness follows. However, to show existence the proof defines a vector field locally using the same system of differential equations but why must it define a smooth vector field?
The system of differential equations is $x_k’(t)=y_k, y_k’(t)=-sum_{i,j}Gamma_{ij}^k y_i y_j$, taking $(x_1,...,x_n,y_1,...,y_n)$ as the coordinate of a curve in $TM$
ordinary-differential-equations riemannian-geometry vector-fields
asked Dec 11 '18 at 21:29
User12239User12239
453216
$endgroup$
There is a lemma stating that there exists a unique vector field $G$ on the tangent bundle $TM$ of a manifold $M$ whose integral curves are of the form $tmapsto (gamma(t),gamma’(t))$ where $gamma$ is a geodesic.
The proof using a system of differential equations first shows that if we assume existence, then its uniqueness follows. However, to show existence the proof defines a vector field locally using the same system of differential equations but why must it define a smooth vector field?
The system of differential equations is $x_k’(t)=y_k, y_k’(t)=-sum_{i,j}Gamma_{ij}^k y_i y_j$, taking $(x_1,...,x_n,y_1,...,y_n)$ as the coordinate of a curve in $TM$
ordinary-differential-equations riemannian-geometry vector-fields
ordinary-differential-equations riemannian-geometry vector-fields
asked Dec 11 '18 at 21:29
User12239User12239
453216
asked Dec 11 '18 at 21:29
User12239User12239
453216
asked Dec 11 '18 at 21:29
User12239User12239
453216
asked Dec 11 '18 at 21:29
User12239User12239
453216
asked Dec 11 '18 at 21:29
User12239User12239
453216
453216
$begingroup$
The manifold is smooth, which means that the chart is smooth. The Christoffel symbols $Γ^k_{ij}$ are linear combinations of the partial derivatives of the metric tensor, thus smooth. Where do you see the obstacle to the vector field $(x',y')$ being smooth?
$endgroup$
– LutzL
Dec 11 '18 at 21:41
$begingroup$
@LutzL that system of differential equations defines a collection of smooth vector fields only along curves, not on any open set necessarily.
$endgroup$
– User12239
Dec 11 '18 at 22:05
$begingroup$
No, it defines a vector field on $TM$, resp on the open set $Utimes Bbb R^n$ corresponding to the chart and the tangent bundle above it. To each $(x,y)$ where $xin M$ and $yin T_xM$ it assigns a vector $(x',y')in T(TM)$. Which looks like $TMtimes TM$.
$endgroup$
– LutzL
Dec 11 '18 at 22:10
$begingroup$
In fact $G$ is defined by $(q,v)mapsto (x_1’(t),...,x_n’(t),y_1’(t),...,y_n’(t))$ where $x_i,y_i$ are as above with the initial condition $x(0)=q,y(0)=v$. So this vector field is smooth on a collection of curves partitioning the coordinate system’s neighborhood, but not necessarily smooth on the whole neighborhood @LutzL
$endgroup$
– User12239
Dec 11 '18 at 22:17
$begingroup$
You are getting confused by the prime, one should find better notations. Set $xi_k(x,y)=y_k$, $eta_k(x,y)=-sum Γ^k_{ij}y_iy_j$, then $(xi(x,y), eta(x,y))$ is a vector field on the chart, not only along curves.
$endgroup$
– LutzL
Dec 11 '18 at 22:37
|
show 1 more comment
$begingroup$
The manifold is smooth, which means that the chart is smooth. The Christoffel symbols $Γ^k_{ij}$ are linear combinations of the partial derivatives of the metric tensor, thus smooth. Where do you see the obstacle to the vector field $(x',y')$ being smooth?
$endgroup$
– LutzL
Dec 11 '18 at 21:41
$begingroup$
@LutzL that system of differential equations defines a collection of smooth vector fields only along curves, not on any open set necessarily.
$endgroup$
– User12239
Dec 11 '18 at 22:05
$begingroup$
No, it defines a vector field on $TM$, resp on the open set $Utimes Bbb R^n$ corresponding to the chart and the tangent bundle above it. To each $(x,y)$ where $xin M$ and $yin T_xM$ it assigns a vector $(x',y')in T(TM)$. Which looks like $TMtimes TM$.
$endgroup$
– LutzL
Dec 11 '18 at 22:10
$begingroup$
In fact $G$ is defined by $(q,v)mapsto (x_1’(t),...,x_n’(t),y_1’(t),...,y_n’(t))$ where $x_i,y_i$ are as above with the initial condition $x(0)=q,y(0)=v$. So this vector field is smooth on a collection of curves partitioning the coordinate system’s neighborhood, but not necessarily smooth on the whole neighborhood @LutzL
$endgroup$
– User12239
Dec 11 '18 at 22:17
$begingroup$
You are getting confused by the prime, one should find better notations. Set $xi_k(x,y)=y_k$, $eta_k(x,y)=-sum Γ^k_{ij}y_iy_j$, then $(xi(x,y), eta(x,y))$ is a vector field on the chart, not only along curves.
$endgroup$
– LutzL
Dec 11 '18 at 22:37
$begingroup$
The manifold is smooth, which means that the chart is smooth. The Christoffel symbols $Γ^k_{ij}$ are linear combinations of the partial derivatives of the metric tensor, thus smooth. Where do you see the obstacle to the vector field $(x',y')$ being smooth?
$endgroup$
– LutzL
Dec 11 '18 at 21:41
$begingroup$
The manifold is smooth, which means that the chart is smooth. The Christoffel symbols $Γ^k_{ij}$ are linear combinations of the partial derivatives of the metric tensor, thus smooth. Where do you see the obstacle to the vector field $(x',y')$ being smooth?
$endgroup$
– LutzL
Dec 11 '18 at 21:41
$begingroup$
@LutzL that system of differential equations defines a collection of smooth vector fields only along curves, not on any open set necessarily.
$endgroup$
– User12239
Dec 11 '18 at 22:05
$begingroup$
@LutzL that system of differential equations defines a collection of smooth vector fields only along curves, not on any open set necessarily.
$endgroup$
– User12239
Dec 11 '18 at 22:05
$begingroup$
No, it defines a vector field on $TM$, resp on the open set $Utimes Bbb R^n$ corresponding to the chart and the tangent bundle above it. To each $(x,y)$ where $xin M$ and $yin T_xM$ it assigns a vector $(x',y')in T(TM)$. Which looks like $TMtimes TM$.
$endgroup$
– LutzL
Dec 11 '18 at 22:10
$begingroup$
No, it defines a vector field on $TM$, resp on the open set $Utimes Bbb R^n$ corresponding to the chart and the tangent bundle above it. To each $(x,y)$ where $xin M$ and $yin T_xM$ it assigns a vector $(x',y')in T(TM)$. Which looks like $TMtimes TM$.
$endgroup$
– LutzL
Dec 11 '18 at 22:10
$begingroup$
In fact $G$ is defined by $(q,v)mapsto (x_1’(t),...,x_n’(t),y_1’(t),...,y_n’(t))$ where $x_i,y_i$ are as above with the initial condition $x(0)=q,y(0)=v$. So this vector field is smooth on a collection of curves partitioning the coordinate system’s neighborhood, but not necessarily smooth on the whole neighborhood @LutzL
$endgroup$
– User12239
Dec 11 '18 at 22:17
$begingroup$
In fact $G$ is defined by $(q,v)mapsto (x_1’(t),...,x_n’(t),y_1’(t),...,y_n’(t))$ where $x_i,y_i$ are as above with the initial condition $x(0)=q,y(0)=v$. So this vector field is smooth on a collection of curves partitioning the coordinate system’s neighborhood, but not necessarily smooth on the whole neighborhood @LutzL
$endgroup$
– User12239
Dec 11 '18 at 22:17
$begingroup$
You are getting confused by the prime, one should find better notations. Set $xi_k(x,y)=y_k$, $eta_k(x,y)=-sum Γ^k_{ij}y_iy_j$, then $(xi(x,y), eta(x,y))$ is a vector field on the chart, not only along curves.
$endgroup$
– LutzL
Dec 11 '18 at 22:37
$begingroup$
You are getting confused by the prime, one should find better notations. Set $xi_k(x,y)=y_k$, $eta_k(x,y)=-sum Γ^k_{ij}y_iy_j$, then $(xi(x,y), eta(x,y))$ is a vector field on the chart, not only along curves.
$endgroup$
– LutzL
Dec 11 '18 at 22:37
|
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$begingroup$
The manifold is smooth, which means that the chart is smooth. The Christoffel symbols $Γ^k_{ij}$ are linear combinations of the partial derivatives of the metric tensor, thus smooth. Where do you see the obstacle to the vector field $(x',y')$ being smooth?
$endgroup$
– LutzL
Dec 11 '18 at 21:41
$begingroup$
@LutzL that system of differential equations defines a collection of smooth vector fields only along curves, not on any open set necessarily.
$endgroup$
– User12239
Dec 11 '18 at 22:05
$begingroup$
No, it defines a vector field on $TM$, resp on the open set $Utimes Bbb R^n$ corresponding to the chart and the tangent bundle above it. To each $(x,y)$ where $xin M$ and $yin T_xM$ it assigns a vector $(x',y')in T(TM)$. Which looks like $TMtimes TM$.
$endgroup$
– LutzL
Dec 11 '18 at 22:10
$begingroup$
In fact $G$ is defined by $(q,v)mapsto (x_1’(t),...,x_n’(t),y_1’(t),...,y_n’(t))$ where $x_i,y_i$ are as above with the initial condition $x(0)=q,y(0)=v$. So this vector field is smooth on a collection of curves partitioning the coordinate system’s neighborhood, but not necessarily smooth on the whole neighborhood @LutzL
$endgroup$
– User12239
Dec 11 '18 at 22:17
$begingroup$
You are getting confused by the prime, one should find better notations. Set $xi_k(x,y)=y_k$, $eta_k(x,y)=-sum Γ^k_{ij}y_iy_j$, then $(xi(x,y), eta(x,y))$ is a vector field on the chart, not only along curves.
$endgroup$
– LutzL
Dec 11 '18 at 22:37