Torsion-freeness of a connection and anti-symmetrization
$begingroup$
Let $M$ be an Hermite manifold, and $nabla$ be the Levi-Civita connection on $TM$ and extend it to $Lambda^*_{mathbb{C}}(M)$. Then $nabla$ is torsion-free by definition. But I read from a paper that the torsion-freeness implies $dtheta_i = text{Alt}(nablatheta_i)$ where ${theta_i}$ is a local frame of $Lambda^{1,0}(M)$. As far as I know, the torsion-fressness means the torsion tensor $T(X,Y) = nabla_X Y - nabla_Y X - [X,Y]equiv 0$, so how can we deduce $dtheta_i = text{Alt}(nablatheta_i)$ from this?
differential-geometry connections
asked Nov 24 '18 at 12:10
J. XUJ. XU
431214
$endgroup$
add a comment |
$begingroup$
Let $M$ be an Hermite manifold, and $nabla$ be the Levi-Civita connection on $TM$ and extend it to $Lambda^*_{mathbb{C}}(M)$. Then $nabla$ is torsion-free by definition. But I read from a paper that the torsion-freeness implies $dtheta_i = text{Alt}(nablatheta_i)$ where ${theta_i}$ is a local frame of $Lambda^{1,0}(M)$. As far as I know, the torsion-fressness means the torsion tensor $T(X,Y) = nabla_X Y - nabla_Y X - [X,Y]equiv 0$, so how can we deduce $dtheta_i = text{Alt}(nablatheta_i)$ from this?
differential-geometry connections
asked Nov 24 '18 at 12:10
J. XUJ. XU
431214
$endgroup$
add a comment |
$begingroup$
Let $M$ be an Hermite manifold, and $nabla$ be the Levi-Civita connection on $TM$ and extend it to $Lambda^*_{mathbb{C}}(M)$. Then $nabla$ is torsion-free by definition. But I read from a paper that the torsion-freeness implies $dtheta_i = text{Alt}(nablatheta_i)$ where ${theta_i}$ is a local frame of $Lambda^{1,0}(M)$. As far as I know, the torsion-fressness means the torsion tensor $T(X,Y) = nabla_X Y - nabla_Y X - [X,Y]equiv 0$, so how can we deduce $dtheta_i = text{Alt}(nablatheta_i)$ from this?
differential-geometry connections
asked Nov 24 '18 at 12:10
J. XUJ. XU
431214
$endgroup$
Let $M$ be an Hermite manifold, and $nabla$ be the Levi-Civita connection on $TM$ and extend it to $Lambda^*_{mathbb{C}}(M)$. Then $nabla$ is torsion-free by definition. But I read from a paper that the torsion-freeness implies $dtheta_i = text{Alt}(nablatheta_i)$ where ${theta_i}$ is a local frame of $Lambda^{1,0}(M)$. As far as I know, the torsion-fressness means the torsion tensor $T(X,Y) = nabla_X Y - nabla_Y X - [X,Y]equiv 0$, so how can we deduce $dtheta_i = text{Alt}(nablatheta_i)$ from this?
differential-geometry connections
differential-geometry connections
asked Nov 24 '18 at 12:10
J. XUJ. XU
431214
asked Nov 24 '18 at 12:10
J. XUJ. XU
431214
asked Nov 24 '18 at 12:10
J. XUJ. XU
431214
asked Nov 24 '18 at 12:10
J. XUJ. XU
431214
asked Nov 24 '18 at 12:10
J. XUJ. XU
431214
431214
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
This is a fact from Riemannian geometry, yes, but has nothing to do with the Hermitian structure. The key point is that you can rephrase the torsion-free condition by working with an orthonormal coframe $theta_i$ and saying
$$dtheta_i = sum omega_{ij}wedgetheta_j$$
with $omega_{ij}=-omega_{ji}$. (Ordinarily, you'd have $dtheta_i = sumomega_{ij}wedgetheta_j + tau_i$, where $tau$ gives the torsion.) Here $omega_{ij}$ gives the connection form.
Then you can check that $nabla theta_i = sum omega_{ij}otimestheta_j$, and the rest is immediate, since $dtheta_i = sumomega_{ij}wedgetheta_j = sum omega_{ij}otimestheta_j - theta_jotimesomega_{ij}$.
answered Nov 24 '18 at 19:29
Ted ShifrinTed Shifrin
63.1k44489
$endgroup$
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%2f3011480%2ftorsion-freeness-of-a-connection-and-anti-symmetrization%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
This is a fact from Riemannian geometry, yes, but has nothing to do with the Hermitian structure. The key point is that you can rephrase the torsion-free condition by working with an orthonormal coframe $theta_i$ and saying
$$dtheta_i = sum omega_{ij}wedgetheta_j$$
with $omega_{ij}=-omega_{ji}$. (Ordinarily, you'd have $dtheta_i = sumomega_{ij}wedgetheta_j + tau_i$, where $tau$ gives the torsion.) Here $omega_{ij}$ gives the connection form.
Then you can check that $nabla theta_i = sum omega_{ij}otimestheta_j$, and the rest is immediate, since $dtheta_i = sumomega_{ij}wedgetheta_j = sum omega_{ij}otimestheta_j - theta_jotimesomega_{ij}$.
answered Nov 24 '18 at 19:29
Ted ShifrinTed Shifrin
63.1k44489
$endgroup$
add a comment |
$begingroup$
This is a fact from Riemannian geometry, yes, but has nothing to do with the Hermitian structure. The key point is that you can rephrase the torsion-free condition by working with an orthonormal coframe $theta_i$ and saying
$$dtheta_i = sum omega_{ij}wedgetheta_j$$
with $omega_{ij}=-omega_{ji}$. (Ordinarily, you'd have $dtheta_i = sumomega_{ij}wedgetheta_j + tau_i$, where $tau$ gives the torsion.) Here $omega_{ij}$ gives the connection form.
Then you can check that $nabla theta_i = sum omega_{ij}otimestheta_j$, and the rest is immediate, since $dtheta_i = sumomega_{ij}wedgetheta_j = sum omega_{ij}otimestheta_j - theta_jotimesomega_{ij}$.
answered Nov 24 '18 at 19:29
Ted ShifrinTed Shifrin
63.1k44489
$endgroup$
add a comment |
$begingroup$
This is a fact from Riemannian geometry, yes, but has nothing to do with the Hermitian structure. The key point is that you can rephrase the torsion-free condition by working with an orthonormal coframe $theta_i$ and saying
$$dtheta_i = sum omega_{ij}wedgetheta_j$$
with $omega_{ij}=-omega_{ji}$. (Ordinarily, you'd have $dtheta_i = sumomega_{ij}wedgetheta_j + tau_i$, where $tau$ gives the torsion.) Here $omega_{ij}$ gives the connection form.
Then you can check that $nabla theta_i = sum omega_{ij}otimestheta_j$, and the rest is immediate, since $dtheta_i = sumomega_{ij}wedgetheta_j = sum omega_{ij}otimestheta_j - theta_jotimesomega_{ij}$.
answered Nov 24 '18 at 19:29
Ted ShifrinTed Shifrin
63.1k44489
$endgroup$
This is a fact from Riemannian geometry, yes, but has nothing to do with the Hermitian structure. The key point is that you can rephrase the torsion-free condition by working with an orthonormal coframe $theta_i$ and saying
$$dtheta_i = sum omega_{ij}wedgetheta_j$$
with $omega_{ij}=-omega_{ji}$. (Ordinarily, you'd have $dtheta_i = sumomega_{ij}wedgetheta_j + tau_i$, where $tau$ gives the torsion.) Here $omega_{ij}$ gives the connection form.
Then you can check that $nabla theta_i = sum omega_{ij}otimestheta_j$, and the rest is immediate, since $dtheta_i = sumomega_{ij}wedgetheta_j = sum omega_{ij}otimestheta_j - theta_jotimesomega_{ij}$.
answered Nov 24 '18 at 19:29
Ted ShifrinTed Shifrin
63.1k44489
edited Nov 25 '18 at 1:30
answered Nov 24 '18 at 19:29
Ted ShifrinTed Shifrin
63.1k44489
answered Nov 24 '18 at 19:29
Ted ShifrinTed Shifrin
63.1k44489
answered Nov 24 '18 at 19:29
Ted ShifrinTed Shifrin
63.1k44489
63.1k44489
add a comment |
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%2f3011480%2ftorsion-freeness-of-a-connection-and-anti-symmetrization%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
