$R^{nabla}(X,Y)Z = Kbig(langle Y,Z rangle X - langle X, Z rangle Ybig)$ for some $K in C^{infty}(M)$
$begingroup$
Let $(M, langle , rangle )$ be a $2$-dimensional Riemannian manifold, $nabla$ it's Levi-Civita connection. Show that there is a function $K in C^{infty}(M)$ such that:
$R^{nabla}(X,Y)Z = Kbig(langle Y,Z rangle X - langle X, Z rangle Ybig)$ for all $X,Y,Z in Gamma(TM)$
I don't really have a strategy, thus far I've only used the definition and tried to use that $nabla$ is metric and torsion-free, but didn't get anything that was helpful. I also don't understand the significance of $M$ being a $2$-dimensional Riemannian manifold. Can someone maybe give a hint and say what role the dimension of $M$ plays here please?
differential-geometry riemannian-geometry
asked Dec 17 '18 at 14:07
eager2learneager2learn
1,24311530
$endgroup$
add a comment |
$begingroup$
Let $(M, langle , rangle )$ be a $2$-dimensional Riemannian manifold, $nabla$ it's Levi-Civita connection. Show that there is a function $K in C^{infty}(M)$ such that:
$R^{nabla}(X,Y)Z = Kbig(langle Y,Z rangle X - langle X, Z rangle Ybig)$ for all $X,Y,Z in Gamma(TM)$
I don't really have a strategy, thus far I've only used the definition and tried to use that $nabla$ is metric and torsion-free, but didn't get anything that was helpful. I also don't understand the significance of $M$ being a $2$-dimensional Riemannian manifold. Can someone maybe give a hint and say what role the dimension of $M$ plays here please?
differential-geometry riemannian-geometry
asked Dec 17 '18 at 14:07
eager2learneager2learn
1,24311530
$endgroup$
add a comment |
$begingroup$
Let $(M, langle , rangle )$ be a $2$-dimensional Riemannian manifold, $nabla$ it's Levi-Civita connection. Show that there is a function $K in C^{infty}(M)$ such that:
$R^{nabla}(X,Y)Z = Kbig(langle Y,Z rangle X - langle X, Z rangle Ybig)$ for all $X,Y,Z in Gamma(TM)$
I don't really have a strategy, thus far I've only used the definition and tried to use that $nabla$ is metric and torsion-free, but didn't get anything that was helpful. I also don't understand the significance of $M$ being a $2$-dimensional Riemannian manifold. Can someone maybe give a hint and say what role the dimension of $M$ plays here please?
differential-geometry riemannian-geometry
asked Dec 17 '18 at 14:07
eager2learneager2learn
1,24311530
$endgroup$
Let $(M, langle , rangle )$ be a $2$-dimensional Riemannian manifold, $nabla$ it's Levi-Civita connection. Show that there is a function $K in C^{infty}(M)$ such that:
$R^{nabla}(X,Y)Z = Kbig(langle Y,Z rangle X - langle X, Z rangle Ybig)$ for all $X,Y,Z in Gamma(TM)$
I don't really have a strategy, thus far I've only used the definition and tried to use that $nabla$ is metric and torsion-free, but didn't get anything that was helpful. I also don't understand the significance of $M$ being a $2$-dimensional Riemannian manifold. Can someone maybe give a hint and say what role the dimension of $M$ plays here please?
differential-geometry riemannian-geometry
differential-geometry riemannian-geometry
asked Dec 17 '18 at 14:07
eager2learneager2learn
1,24311530
asked Dec 17 '18 at 14:07
eager2learneager2learn
1,24311530
edited Dec 17 '18 at 15:03
user593746
asked Dec 17 '18 at 14:07
eager2learneager2learn
1,24311530
asked Dec 17 '18 at 14:07
eager2learneager2learn
1,24311530
asked Dec 17 '18 at 14:07
eager2learneager2learn
1,24311530
1,24311530
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Hint Fix $p in M$. By definition, $R_p(X, Y) {,cdot,} = -R_p(Y, X){,cdot,}$ (for all $X, Y in T_p M$), so we can view $R_p$ as an element of $bigwedge^2 T_p^*M otimes T_p M otimes T^*_p M$. On the other hand, since $nabla g = 0$, unwinding definitions gives that the endomorphism $R_p(X, Y){,cdot,}$ is $g$-skew. So, we can view $R_p$ as an element of $bigwedge^2 T_p^*M otimes operatorname{End}_{operatorname{skew}}(T_p^* M)$. But since $dim M = 2$ this is a $1$-dimensional vector space and so is spanned by a single element.
answered Dec 17 '18 at 17:22
TravisTravis
60.4k767147
$endgroup$
add a comment |
$begingroup$
Hint I: For any n-dimensional Riemannian manifold $(M,g)$, the curvature tensor $R$ has $frac{n^2(n^2-1)}{12}$ indepedent non-zero components.
Hint II: Do you know what sectional curvature is?
answered Dec 17 '18 at 15:31
Luis LopezLuis Lopez
714
$endgroup$
$begingroup$
Thanks for the first hint. We haven't done sectional curvature yet.
$endgroup$
– eager2learn
Dec 17 '18 at 15:49
add a comment |
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%2f3043984%2fr-nablax-yz-k-big-langle-y-z-rangle-x-langle-x-z-rangle-y-big-f%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint Fix $p in M$. By definition, $R_p(X, Y) {,cdot,} = -R_p(Y, X){,cdot,}$ (for all $X, Y in T_p M$), so we can view $R_p$ as an element of $bigwedge^2 T_p^*M otimes T_p M otimes T^*_p M$. On the other hand, since $nabla g = 0$, unwinding definitions gives that the endomorphism $R_p(X, Y){,cdot,}$ is $g$-skew. So, we can view $R_p$ as an element of $bigwedge^2 T_p^*M otimes operatorname{End}_{operatorname{skew}}(T_p^* M)$. But since $dim M = 2$ this is a $1$-dimensional vector space and so is spanned by a single element.
answered Dec 17 '18 at 17:22
TravisTravis
60.4k767147
$endgroup$
add a comment |
$begingroup$
Hint Fix $p in M$. By definition, $R_p(X, Y) {,cdot,} = -R_p(Y, X){,cdot,}$ (for all $X, Y in T_p M$), so we can view $R_p$ as an element of $bigwedge^2 T_p^*M otimes T_p M otimes T^*_p M$. On the other hand, since $nabla g = 0$, unwinding definitions gives that the endomorphism $R_p(X, Y){,cdot,}$ is $g$-skew. So, we can view $R_p$ as an element of $bigwedge^2 T_p^*M otimes operatorname{End}_{operatorname{skew}}(T_p^* M)$. But since $dim M = 2$ this is a $1$-dimensional vector space and so is spanned by a single element.
answered Dec 17 '18 at 17:22
TravisTravis
60.4k767147
$endgroup$
add a comment |
$begingroup$
Hint Fix $p in M$. By definition, $R_p(X, Y) {,cdot,} = -R_p(Y, X){,cdot,}$ (for all $X, Y in T_p M$), so we can view $R_p$ as an element of $bigwedge^2 T_p^*M otimes T_p M otimes T^*_p M$. On the other hand, since $nabla g = 0$, unwinding definitions gives that the endomorphism $R_p(X, Y){,cdot,}$ is $g$-skew. So, we can view $R_p$ as an element of $bigwedge^2 T_p^*M otimes operatorname{End}_{operatorname{skew}}(T_p^* M)$. But since $dim M = 2$ this is a $1$-dimensional vector space and so is spanned by a single element.
answered Dec 17 '18 at 17:22
TravisTravis
60.4k767147
$endgroup$
Hint Fix $p in M$. By definition, $R_p(X, Y) {,cdot,} = -R_p(Y, X){,cdot,}$ (for all $X, Y in T_p M$), so we can view $R_p$ as an element of $bigwedge^2 T_p^*M otimes T_p M otimes T^*_p M$. On the other hand, since $nabla g = 0$, unwinding definitions gives that the endomorphism $R_p(X, Y){,cdot,}$ is $g$-skew. So, we can view $R_p$ as an element of $bigwedge^2 T_p^*M otimes operatorname{End}_{operatorname{skew}}(T_p^* M)$. But since $dim M = 2$ this is a $1$-dimensional vector space and so is spanned by a single element.
answered Dec 17 '18 at 17:22
TravisTravis
60.4k767147
answered Dec 17 '18 at 17:22
TravisTravis
60.4k767147
answered Dec 17 '18 at 17:22
TravisTravis
60.4k767147
answered Dec 17 '18 at 17:22
TravisTravis
60.4k767147
60.4k767147
add a comment |
add a comment |
$begingroup$
Hint I: For any n-dimensional Riemannian manifold $(M,g)$, the curvature tensor $R$ has $frac{n^2(n^2-1)}{12}$ indepedent non-zero components.
Hint II: Do you know what sectional curvature is?
answered Dec 17 '18 at 15:31
Luis LopezLuis Lopez
714
$endgroup$
$begingroup$
Thanks for the first hint. We haven't done sectional curvature yet.
$endgroup$
– eager2learn
Dec 17 '18 at 15:49
add a comment |
$begingroup$
Hint I: For any n-dimensional Riemannian manifold $(M,g)$, the curvature tensor $R$ has $frac{n^2(n^2-1)}{12}$ indepedent non-zero components.
Hint II: Do you know what sectional curvature is?
answered Dec 17 '18 at 15:31
Luis LopezLuis Lopez
714
$endgroup$
$begingroup$
Thanks for the first hint. We haven't done sectional curvature yet.
$endgroup$
– eager2learn
Dec 17 '18 at 15:49
add a comment |
$begingroup$
Hint I: For any n-dimensional Riemannian manifold $(M,g)$, the curvature tensor $R$ has $frac{n^2(n^2-1)}{12}$ indepedent non-zero components.
Hint II: Do you know what sectional curvature is?
answered Dec 17 '18 at 15:31
Luis LopezLuis Lopez
714
$endgroup$
Hint I: For any n-dimensional Riemannian manifold $(M,g)$, the curvature tensor $R$ has $frac{n^2(n^2-1)}{12}$ indepedent non-zero components.
Hint II: Do you know what sectional curvature is?
answered Dec 17 '18 at 15:31
Luis LopezLuis Lopez
714
answered Dec 17 '18 at 15:31
Luis LopezLuis Lopez
714
answered Dec 17 '18 at 15:31
Luis LopezLuis Lopez
714
answered Dec 17 '18 at 15:31
Luis LopezLuis Lopez
714
714
$begingroup$
Thanks for the first hint. We haven't done sectional curvature yet.
$endgroup$
– eager2learn
Dec 17 '18 at 15:49
add a comment |
$begingroup$
Thanks for the first hint. We haven't done sectional curvature yet.
$endgroup$
– eager2learn
Dec 17 '18 at 15:49
$begingroup$
Thanks for the first hint. We haven't done sectional curvature yet.
$endgroup$
– eager2learn
Dec 17 '18 at 15:49
$begingroup$
Thanks for the first hint. We haven't done sectional curvature yet.
$endgroup$
– eager2learn
Dec 17 '18 at 15:49
add a comment |
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.
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%2f3043984%2fr-nablax-yz-k-big-langle-y-z-rangle-x-langle-x-z-rangle-y-big-f%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
