Smoothness of a vector field defined by a system of differential equations
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 $TM$
differential-equations riemannian-geometry vector-fields
asked Dec 11 '18 at 21:29
User12239
412216
|
show 1 more comment
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 $TM$
differential-equations riemannian-geometry vector-fields
asked Dec 11 '18 at 21:29
User12239
412216
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?
– LutzL
Dec 11 '18 at 21:41
@LutzL that system of differential equations defines a collection of smooth vector fields only along curves, not on any open set necessarily.
– User12239
Dec 11 '18 at 22:05
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 $TMtimes TM$.
– LutzL
Dec 11 '18 at 22:10
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
– User12239
Dec 11 '18 at 22:17
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.
– LutzL
Dec 11 '18 at 22:37
|
show 1 more comment
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 $TM$
differential-equations riemannian-geometry vector-fields
asked Dec 11 '18 at 21:29
User12239
412216
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 $TM$
differential-equations riemannian-geometry vector-fields
differential-equations riemannian-geometry vector-fields
asked Dec 11 '18 at 21:29
User12239
412216
asked Dec 11 '18 at 21:29
User12239
412216
asked Dec 11 '18 at 21:29
User12239
412216
asked Dec 11 '18 at 21:29
User12239
412216
asked Dec 11 '18 at 21:29
User12239
412216
412216
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?
– LutzL
Dec 11 '18 at 21:41
@LutzL that system of differential equations defines a collection of smooth vector fields only along curves, not on any open set necessarily.
– User12239
Dec 11 '18 at 22:05
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 $TMtimes TM$.
– LutzL
Dec 11 '18 at 22:10
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
– User12239
Dec 11 '18 at 22:17
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.
– LutzL
Dec 11 '18 at 22:37
|
show 1 more comment
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?
– LutzL
Dec 11 '18 at 21:41
@LutzL that system of differential equations defines a collection of smooth vector fields only along curves, not on any open set necessarily.
– User12239
Dec 11 '18 at 22:05
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 $TMtimes TM$.
– LutzL
Dec 11 '18 at 22:10
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
– User12239
Dec 11 '18 at 22:17
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.
– LutzL
Dec 11 '18 at 22:37
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?
– LutzL
Dec 11 '18 at 21:41
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?
– LutzL
Dec 11 '18 at 21:41
@LutzL that system of differential equations defines a collection of smooth vector fields only along curves, not on any open set necessarily.
– User12239
Dec 11 '18 at 22:05
@LutzL that system of differential equations defines a collection of smooth vector fields only along curves, not on any open set necessarily.
– User12239
Dec 11 '18 at 22:05
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 $TMtimes TM$.
– LutzL
Dec 11 '18 at 22:10
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 $TMtimes TM$.
– LutzL
Dec 11 '18 at 22:10
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
– User12239
Dec 11 '18 at 22:17
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
– User12239
Dec 11 '18 at 22:17
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.
– LutzL
Dec 11 '18 at 22:37
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.
– LutzL
Dec 11 '18 at 22:37
|
show 1 more comment
0
active
oldest
votes
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3035852%2fsmoothness-of-a-vector-field-defined-by-a-system-of-differential-equations%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3035852%2fsmoothness-of-a-vector-field-defined-by-a-system-of-differential-equations%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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?
– LutzL
Dec 11 '18 at 21:41
@LutzL that system of differential equations defines a collection of smooth vector fields only along curves, not on any open set necessarily.
– User12239
Dec 11 '18 at 22:05
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 $TMtimes TM$.
– LutzL
Dec 11 '18 at 22:10
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
– User12239
Dec 11 '18 at 22:17
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.
– LutzL
Dec 11 '18 at 22:37