Submanifold of codimension 1 orientable iff there exists unit normal vector field.
$begingroup$
Suppose I have a submanifold $M subset mathbb{R}^{n}$, of dimension $n-1$. Where a unit normal vector field is a section $nu$ of the normal bundle $ TM^{bot} to M$. So the fibers are all the vectors that are perpendicular to the tangent space of the same base point. With the addition that $| nu(p) | =1$, for all $p$ in $M$. Apparently it's orientable if and only if there exists a unit normal vector field on $M$. And I tried to prove this, but I just get stuck everytime. I tried looking at the local flatness definition of a submanifold, and I know I need to do something with the orientation of $mathbb{R}^n$ and the fact that $nu$ is smooth. But at this point I'm just stuck and do not know how to prove it.
differential-geometry manifolds vector-bundles vector-fields orientation
asked Dec 2 '18 at 8:36
AkatsukiMalikiAkatsukiMaliki
313110
$endgroup$
add a comment |
$begingroup$
Suppose I have a submanifold $M subset mathbb{R}^{n}$, of dimension $n-1$. Where a unit normal vector field is a section $nu$ of the normal bundle $ TM^{bot} to M$. So the fibers are all the vectors that are perpendicular to the tangent space of the same base point. With the addition that $| nu(p) | =1$, for all $p$ in $M$. Apparently it's orientable if and only if there exists a unit normal vector field on $M$. And I tried to prove this, but I just get stuck everytime. I tried looking at the local flatness definition of a submanifold, and I know I need to do something with the orientation of $mathbb{R}^n$ and the fact that $nu$ is smooth. But at this point I'm just stuck and do not know how to prove it.
differential-geometry manifolds vector-bundles vector-fields orientation
asked Dec 2 '18 at 8:36
AkatsukiMalikiAkatsukiMaliki
313110
$endgroup$
1
$begingroup$
When you pull the tangent bunde of $mathbb{R}^n$ back to $M$, it splits into the direct sum of the tangent bundle of $M$ and the normal bundle to $M$. The tangent bundle of $R^n$ is orientable hence so is the pullback, and the existence of a unit normal vector field is equivalent to the normal bundle being trivial (which is equivalent to orientability for line bundles). Now you could argue that a sum of a bundle with an orientable bundle is orientable iff the bundle is orientable (a fast argument would be with the first Stiefel-Whitney class, which is additive and 0 if the bundle is trivial).
$endgroup$
– Aleksandar Milivojevic
Dec 2 '18 at 20:25
add a comment |
$begingroup$
Suppose I have a submanifold $M subset mathbb{R}^{n}$, of dimension $n-1$. Where a unit normal vector field is a section $nu$ of the normal bundle $ TM^{bot} to M$. So the fibers are all the vectors that are perpendicular to the tangent space of the same base point. With the addition that $| nu(p) | =1$, for all $p$ in $M$. Apparently it's orientable if and only if there exists a unit normal vector field on $M$. And I tried to prove this, but I just get stuck everytime. I tried looking at the local flatness definition of a submanifold, and I know I need to do something with the orientation of $mathbb{R}^n$ and the fact that $nu$ is smooth. But at this point I'm just stuck and do not know how to prove it.
differential-geometry manifolds vector-bundles vector-fields orientation
asked Dec 2 '18 at 8:36
AkatsukiMalikiAkatsukiMaliki
313110
$endgroup$
Suppose I have a submanifold $M subset mathbb{R}^{n}$, of dimension $n-1$. Where a unit normal vector field is a section $nu$ of the normal bundle $ TM^{bot} to M$. So the fibers are all the vectors that are perpendicular to the tangent space of the same base point. With the addition that $| nu(p) | =1$, for all $p$ in $M$. Apparently it's orientable if and only if there exists a unit normal vector field on $M$. And I tried to prove this, but I just get stuck everytime. I tried looking at the local flatness definition of a submanifold, and I know I need to do something with the orientation of $mathbb{R}^n$ and the fact that $nu$ is smooth. But at this point I'm just stuck and do not know how to prove it.
differential-geometry manifolds vector-bundles vector-fields orientation
differential-geometry manifolds vector-bundles vector-fields orientation
asked Dec 2 '18 at 8:36
AkatsukiMalikiAkatsukiMaliki
313110
asked Dec 2 '18 at 8:36
AkatsukiMalikiAkatsukiMaliki
313110
asked Dec 2 '18 at 8:36
AkatsukiMalikiAkatsukiMaliki
313110
asked Dec 2 '18 at 8:36
AkatsukiMalikiAkatsukiMaliki
313110
asked Dec 2 '18 at 8:36
AkatsukiMalikiAkatsukiMaliki
313110
313110
1
$begingroup$
When you pull the tangent bunde of $mathbb{R}^n$ back to $M$, it splits into the direct sum of the tangent bundle of $M$ and the normal bundle to $M$. The tangent bundle of $R^n$ is orientable hence so is the pullback, and the existence of a unit normal vector field is equivalent to the normal bundle being trivial (which is equivalent to orientability for line bundles). Now you could argue that a sum of a bundle with an orientable bundle is orientable iff the bundle is orientable (a fast argument would be with the first Stiefel-Whitney class, which is additive and 0 if the bundle is trivial).
$endgroup$
– Aleksandar Milivojevic
Dec 2 '18 at 20:25
add a comment |
1
$begingroup$
When you pull the tangent bunde of $mathbb{R}^n$ back to $M$, it splits into the direct sum of the tangent bundle of $M$ and the normal bundle to $M$. The tangent bundle of $R^n$ is orientable hence so is the pullback, and the existence of a unit normal vector field is equivalent to the normal bundle being trivial (which is equivalent to orientability for line bundles). Now you could argue that a sum of a bundle with an orientable bundle is orientable iff the bundle is orientable (a fast argument would be with the first Stiefel-Whitney class, which is additive and 0 if the bundle is trivial).
$endgroup$
– Aleksandar Milivojevic
Dec 2 '18 at 20:25
1
1
$begingroup$
When you pull the tangent bunde of $mathbb{R}^n$ back to $M$, it splits into the direct sum of the tangent bundle of $M$ and the normal bundle to $M$. The tangent bundle of $R^n$ is orientable hence so is the pullback, and the existence of a unit normal vector field is equivalent to the normal bundle being trivial (which is equivalent to orientability for line bundles). Now you could argue that a sum of a bundle with an orientable bundle is orientable iff the bundle is orientable (a fast argument would be with the first Stiefel-Whitney class, which is additive and 0 if the bundle is trivial).
$endgroup$
– Aleksandar Milivojevic
Dec 2 '18 at 20:25
$begingroup$
When you pull the tangent bunde of $mathbb{R}^n$ back to $M$, it splits into the direct sum of the tangent bundle of $M$ and the normal bundle to $M$. The tangent bundle of $R^n$ is orientable hence so is the pullback, and the existence of a unit normal vector field is equivalent to the normal bundle being trivial (which is equivalent to orientability for line bundles). Now you could argue that a sum of a bundle with an orientable bundle is orientable iff the bundle is orientable (a fast argument would be with the first Stiefel-Whitney class, which is additive and 0 if the bundle is trivial).
$endgroup$
– Aleksandar Milivojevic
Dec 2 '18 at 20:25
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
});
}
});
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%2f3022403%2fsubmanifold-of-codimension-1-orientable-iff-there-exists-unit-normal-vector-fiel%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.
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%2f3022403%2fsubmanifold-of-codimension-1-orientable-iff-there-exists-unit-normal-vector-fiel%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
1
$begingroup$
When you pull the tangent bunde of $mathbb{R}^n$ back to $M$, it splits into the direct sum of the tangent bundle of $M$ and the normal bundle to $M$. The tangent bundle of $R^n$ is orientable hence so is the pullback, and the existence of a unit normal vector field is equivalent to the normal bundle being trivial (which is equivalent to orientability for line bundles). Now you could argue that a sum of a bundle with an orientable bundle is orientable iff the bundle is orientable (a fast argument would be with the first Stiefel-Whitney class, which is additive and 0 if the bundle is trivial).
$endgroup$
– Aleksandar Milivojevic
Dec 2 '18 at 20:25