Mixing
Is Riemann curvature determined by commuting vectorfields?
0
$begingroup$
Let $E$ be a vector bundle over a smooth manifold $M$ with two connections$nabla$, $nabla'$ and define by $R$, $R'$ the corresponding Riemann curvature tensors.
If $R$ and $R'$ agree on all commuting vectorfields, i.e. $R(X,Y)Z=R'(X,Y)Z$ for all $X,Yin Gamma (TM)$ with $[X,Y]=0$ and $ZinGamma (E)$, does it follow that $R=R'$?
differential-geometry riemannian-geometry
edited Jan 16 at 21:15
Bernard
121k740116
asked Jan 16 at 21:03
triitrii
1315
$endgroup$
add a comment |
0
$begingroup$
Let $E$ be a vector bundle over a smooth manifold $M$ with two connections$nabla$, $nabla'$ and define by $R$, $R'$ the corresponding Riemann curvature tensors.
If $R$ and $R'$ agree on all commuting vectorfields, i.e. $R(X,Y)Z=R'(X,Y)Z$ for all $X,Yin Gamma (TM)$ with $[X,Y]=0$ and $ZinGamma (E)$, does it follow that $R=R'$?
differential-geometry riemannian-geometry
edited Jan 16 at 21:15
Bernard
121k740116
asked Jan 16 at 21:03
triitrii
1315
$endgroup$
$begingroup$
Yes. The curvature is a tensor. This means that the value of $R(X,Y)Z$ at a point $p$ is determined by the values of $X,Y$ and $Z$ at $p$. Every two tangent vectors at a point can be extended to two commuting vector fields at a neighborhood, and the claim follows.
$endgroup$
– Amitai Yuval
Jan 16 at 21:54
$begingroup$
Curvature tensors are, as the name suggests, tensors, so they are completely determined from their evaluation over fibers. that is, as any pair of vectors $(X_p,Y_p)$ at a point $p$ extends to a vector field $X,Y$ s.t $[X,Y]=0$, $R(X,Y)_p=R(X_p,Y_p)$ is completely determined by such a commuting pair of vector fields.
$endgroup$
– cjackal
Jan 16 at 21:56
$begingroup$
Thanks this was very helpfull! I would accept this as an answer.
$endgroup$
– trii
Jan 17 at 18:31
add a comment |
0
0
0
$begingroup$
Let $E$ be a vector bundle over a smooth manifold $M$ with two connections$nabla$, $nabla'$ and define by $R$, $R'$ the corresponding Riemann curvature tensors.
If $R$ and $R'$ agree on all commuting vectorfields, i.e. $R(X,Y)Z=R'(X,Y)Z$ for all $X,Yin Gamma (TM)$ with $[X,Y]=0$ and $ZinGamma (E)$, does it follow that $R=R'$?
differential-geometry riemannian-geometry
edited Jan 16 at 21:15
Bernard
121k740116
asked Jan 16 at 21:03
triitrii
1315
$endgroup$
Let $E$ be a vector bundle over a smooth manifold $M$ with two connections$nabla$, $nabla'$ and define by $R$, $R'$ the corresponding Riemann curvature tensors.
If $R$ and $R'$ agree on all commuting vectorfields, i.e. $R(X,Y)Z=R'(X,Y)Z$ for all $X,Yin Gamma (TM)$ with $[X,Y]=0$ and $ZinGamma (E)$, does it follow that $R=R'$?
differential-geometry riemannian-geometry
differential-geometry riemannian-geometry
edited Jan 16 at 21:15
Bernard
121k740116
asked Jan 16 at 21:03
triitrii
1315
edited Jan 16 at 21:15
Bernard
121k740116
asked Jan 16 at 21:03
triitrii
1315
edited Jan 16 at 21:15
Bernard
121k740116
edited Jan 16 at 21:15
Bernard
121k740116
edited Jan 16 at 21:15
Bernard
121k740116
121k740116
asked Jan 16 at 21:03
triitrii
1315
asked Jan 16 at 21:03
triitrii
1315
asked Jan 16 at 21:03
triitrii
1315
1315
$begingroup$
Yes. The curvature is a tensor. This means that the value of $R(X,Y)Z$ at a point $p$ is determined by the values of $X,Y$ and $Z$ at $p$. Every two tangent vectors at a point can be extended to two commuting vector fields at a neighborhood, and the claim follows.
$endgroup$
– Amitai Yuval
Jan 16 at 21:54
$begingroup$
Curvature tensors are, as the name suggests, tensors, so they are completely determined from their evaluation over fibers. that is, as any pair of vectors $(X_p,Y_p)$ at a point $p$ extends to a vector field $X,Y$ s.t $[X,Y]=0$, $R(X,Y)_p=R(X_p,Y_p)$ is completely determined by such a commuting pair of vector fields.
$endgroup$
– cjackal
Jan 16 at 21:56
$begingroup$
Thanks this was very helpfull! I would accept this as an answer.
$endgroup$
– trii
Jan 17 at 18:31
add a comment |
$begingroup$
Yes. The curvature is a tensor. This means that the value of $R(X,Y)Z$ at a point $p$ is determined by the values of $X,Y$ and $Z$ at $p$. Every two tangent vectors at a point can be extended to two commuting vector fields at a neighborhood, and the claim follows.
$endgroup$
– Amitai Yuval
Jan 16 at 21:54
$begingroup$
Curvature tensors are, as the name suggests, tensors, so they are completely determined from their evaluation over fibers. that is, as any pair of vectors $(X_p,Y_p)$ at a point $p$ extends to a vector field $X,Y$ s.t $[X,Y]=0$, $R(X,Y)_p=R(X_p,Y_p)$ is completely determined by such a commuting pair of vector fields.
$endgroup$
– cjackal
Jan 16 at 21:56
$begingroup$
Thanks this was very helpfull! I would accept this as an answer.
$endgroup$
– trii
Jan 17 at 18:31
$begingroup$
Yes. The curvature is a tensor. This means that the value of $R(X,Y)Z$ at a point $p$ is determined by the values of $X,Y$ and $Z$ at $p$. Every two tangent vectors at a point can be extended to two commuting vector fields at a neighborhood, and the claim follows.
$endgroup$
– Amitai Yuval
Jan 16 at 21:54
$begingroup$
Yes. The curvature is a tensor. This means that the value of $R(X,Y)Z$ at a point $p$ is determined by the values of $X,Y$ and $Z$ at $p$. Every two tangent vectors at a point can be extended to two commuting vector fields at a neighborhood, and the claim follows.
$endgroup$
– Amitai Yuval
Jan 16 at 21:54
$begingroup$
Curvature tensors are, as the name suggests, tensors, so they are completely determined from their evaluation over fibers. that is, as any pair of vectors $(X_p,Y_p)$ at a point $p$ extends to a vector field $X,Y$ s.t $[X,Y]=0$, $R(X,Y)_p=R(X_p,Y_p)$ is completely determined by such a commuting pair of vector fields.
$endgroup$
– cjackal
Jan 16 at 21:56
$begingroup$
Curvature tensors are, as the name suggests, tensors, so they are completely determined from their evaluation over fibers. that is, as any pair of vectors $(X_p,Y_p)$ at a point $p$ extends to a vector field $X,Y$ s.t $[X,Y]=0$, $R(X,Y)_p=R(X_p,Y_p)$ is completely determined by such a commuting pair of vector fields.
$endgroup$
– cjackal
Jan 16 at 21:56
$begingroup$
Thanks this was very helpfull! I would accept this as an answer.
$endgroup$
– trii
Jan 17 at 18:31
$begingroup$
Thanks this was very helpfull! I would accept this as an answer.
$endgroup$
– trii
Jan 17 at 18:31
add a 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
});
}
});
draft saved
draft discarded
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%2f3076291%2fis-riemann-curvature-determined-by-commuting-vectorfields%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
draft saved
draft discarded
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.
draft saved
draft discarded
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%2f3076291%2fis-riemann-curvature-determined-by-commuting-vectorfields%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
$begingroup$
Yes. The curvature is a tensor. This means that the value of $R(X,Y)Z$ at a point $p$ is determined by the values of $X,Y$ and $Z$ at $p$. Every two tangent vectors at a point can be extended to two commuting vector fields at a neighborhood, and the claim follows.
$endgroup$
– Amitai Yuval
Jan 16 at 21:54
$begingroup$
Curvature tensors are, as the name suggests, tensors, so they are completely determined from their evaluation over fibers. that is, as any pair of vectors $(X_p,Y_p)$ at a point $p$ extends to a vector field $X,Y$ s.t $[X,Y]=0$, $R(X,Y)_p=R(X_p,Y_p)$ is completely determined by such a commuting pair of vector fields.
$endgroup$
– cjackal
Jan 16 at 21:56
$begingroup$
Thanks this was very helpfull! I would accept this as an answer.
$endgroup$
– trii
Jan 17 at 18:31