The orientation induced on the boundary of a manifold.
$begingroup$
I just learned about the notion of orientability of a manifold which is difficult and abstract for me. If we consider all basis of a vector space, the matrix that transforms one basis in another basis can have either positive or negative determinant. So we can partition the set of basis in two equivalent classes where in each equivalent class the determinant of the matrix of change of basis in an equivalent class is positive. So the pointwise orientation of a manifold is the choice for each $p in M$ of one of the two equivalence classes. Then an orientation is a pointwise orientation which is continuous (i.e. for each point $p$, there is an open $U$ that contains $p$ and a smooth vector field $(X_1,...,X_n)$ such that the equivalent class $[(X_{1p},...,X_{np})]$ gives the pointwise orientation at $p$. Now if we consider a manifold $M$ with boundary, given an orientation $mathcal{O}$ on it (i.e. $mathcal{O}_p$ is one of the two equivalent classes for each $p$), it defines an orientation on the boundary : In each point $p in partial M$, apparently we can choose $mathcal{O}_p=[(Y_{1p},...,Y_{np})]$ where $Y_{1p}$ is an outward vector(i.e. the pushforward of a chart $phi_*Y_{1p}$ has a négative component along $frac{partial}{partial x_n}rvert_{phi(p)}$) and the $Y_{2p},...,Y_{np}$ are in $Tp(partial M)$, i.e. the subspace of $T_pM$ genereated by $(frac{partial}{partial phi_1}rvert_p,...,frac{partial}{partial phi_{n-1}}rvert_p)$. Then $[(Y_{2p},...,Y_{np})]$ is an orientation on $T_ppartial M$.
Question : How do we know that such an $(Y_{1p},...,Y_{np})$ exists?
For $[(Y_{2p},...,Y_{np})]$ to be an orientation, we need that for $p in partial M$ there is an open $U in partial M$ that contains $p$ and vector fields $(Z_2,...,Z_{n-1})$ that are smooth and when evaluated in p give us the same orientation : $[(Z_{2p},...,Z_{n-1p })]=[(Y_{2p},...,Y_{np})]$,why is this true?
differential-geometry manifolds orientation manifolds-with-boundary
asked Jan 11 at 14:09
roi_saumonroi_saumon
63938
$endgroup$
add a comment |
$begingroup$
I just learned about the notion of orientability of a manifold which is difficult and abstract for me. If we consider all basis of a vector space, the matrix that transforms one basis in another basis can have either positive or negative determinant. So we can partition the set of basis in two equivalent classes where in each equivalent class the determinant of the matrix of change of basis in an equivalent class is positive. So the pointwise orientation of a manifold is the choice for each $p in M$ of one of the two equivalence classes. Then an orientation is a pointwise orientation which is continuous (i.e. for each point $p$, there is an open $U$ that contains $p$ and a smooth vector field $(X_1,...,X_n)$ such that the equivalent class $[(X_{1p},...,X_{np})]$ gives the pointwise orientation at $p$. Now if we consider a manifold $M$ with boundary, given an orientation $mathcal{O}$ on it (i.e. $mathcal{O}_p$ is one of the two equivalent classes for each $p$), it defines an orientation on the boundary : In each point $p in partial M$, apparently we can choose $mathcal{O}_p=[(Y_{1p},...,Y_{np})]$ where $Y_{1p}$ is an outward vector(i.e. the pushforward of a chart $phi_*Y_{1p}$ has a négative component along $frac{partial}{partial x_n}rvert_{phi(p)}$) and the $Y_{2p},...,Y_{np}$ are in $Tp(partial M)$, i.e. the subspace of $T_pM$ genereated by $(frac{partial}{partial phi_1}rvert_p,...,frac{partial}{partial phi_{n-1}}rvert_p)$. Then $[(Y_{2p},...,Y_{np})]$ is an orientation on $T_ppartial M$.
Question : How do we know that such an $(Y_{1p},...,Y_{np})$ exists?
For $[(Y_{2p},...,Y_{np})]$ to be an orientation, we need that for $p in partial M$ there is an open $U in partial M$ that contains $p$ and vector fields $(Z_2,...,Z_{n-1})$ that are smooth and when evaluated in p give us the same orientation : $[(Z_{2p},...,Z_{n-1p })]=[(Y_{2p},...,Y_{np})]$,why is this true?
differential-geometry manifolds orientation manifolds-with-boundary
asked Jan 11 at 14:09
roi_saumonroi_saumon
63938
$endgroup$
add a comment |
$begingroup$
I just learned about the notion of orientability of a manifold which is difficult and abstract for me. If we consider all basis of a vector space, the matrix that transforms one basis in another basis can have either positive or negative determinant. So we can partition the set of basis in two equivalent classes where in each equivalent class the determinant of the matrix of change of basis in an equivalent class is positive. So the pointwise orientation of a manifold is the choice for each $p in M$ of one of the two equivalence classes. Then an orientation is a pointwise orientation which is continuous (i.e. for each point $p$, there is an open $U$ that contains $p$ and a smooth vector field $(X_1,...,X_n)$ such that the equivalent class $[(X_{1p},...,X_{np})]$ gives the pointwise orientation at $p$. Now if we consider a manifold $M$ with boundary, given an orientation $mathcal{O}$ on it (i.e. $mathcal{O}_p$ is one of the two equivalent classes for each $p$), it defines an orientation on the boundary : In each point $p in partial M$, apparently we can choose $mathcal{O}_p=[(Y_{1p},...,Y_{np})]$ where $Y_{1p}$ is an outward vector(i.e. the pushforward of a chart $phi_*Y_{1p}$ has a négative component along $frac{partial}{partial x_n}rvert_{phi(p)}$) and the $Y_{2p},...,Y_{np}$ are in $Tp(partial M)$, i.e. the subspace of $T_pM$ genereated by $(frac{partial}{partial phi_1}rvert_p,...,frac{partial}{partial phi_{n-1}}rvert_p)$. Then $[(Y_{2p},...,Y_{np})]$ is an orientation on $T_ppartial M$.
Question : How do we know that such an $(Y_{1p},...,Y_{np})$ exists?
For $[(Y_{2p},...,Y_{np})]$ to be an orientation, we need that for $p in partial M$ there is an open $U in partial M$ that contains $p$ and vector fields $(Z_2,...,Z_{n-1})$ that are smooth and when evaluated in p give us the same orientation : $[(Z_{2p},...,Z_{n-1p })]=[(Y_{2p},...,Y_{np})]$,why is this true?
differential-geometry manifolds orientation manifolds-with-boundary
asked Jan 11 at 14:09
roi_saumonroi_saumon
63938
$endgroup$
I just learned about the notion of orientability of a manifold which is difficult and abstract for me. If we consider all basis of a vector space, the matrix that transforms one basis in another basis can have either positive or negative determinant. So we can partition the set of basis in two equivalent classes where in each equivalent class the determinant of the matrix of change of basis in an equivalent class is positive. So the pointwise orientation of a manifold is the choice for each $p in M$ of one of the two equivalence classes. Then an orientation is a pointwise orientation which is continuous (i.e. for each point $p$, there is an open $U$ that contains $p$ and a smooth vector field $(X_1,...,X_n)$ such that the equivalent class $[(X_{1p},...,X_{np})]$ gives the pointwise orientation at $p$. Now if we consider a manifold $M$ with boundary, given an orientation $mathcal{O}$ on it (i.e. $mathcal{O}_p$ is one of the two equivalent classes for each $p$), it defines an orientation on the boundary : In each point $p in partial M$, apparently we can choose $mathcal{O}_p=[(Y_{1p},...,Y_{np})]$ where $Y_{1p}$ is an outward vector(i.e. the pushforward of a chart $phi_*Y_{1p}$ has a négative component along $frac{partial}{partial x_n}rvert_{phi(p)}$) and the $Y_{2p},...,Y_{np}$ are in $Tp(partial M)$, i.e. the subspace of $T_pM$ genereated by $(frac{partial}{partial phi_1}rvert_p,...,frac{partial}{partial phi_{n-1}}rvert_p)$. Then $[(Y_{2p},...,Y_{np})]$ is an orientation on $T_ppartial M$.
Question : How do we know that such an $(Y_{1p},...,Y_{np})$ exists?
For $[(Y_{2p},...,Y_{np})]$ to be an orientation, we need that for $p in partial M$ there is an open $U in partial M$ that contains $p$ and vector fields $(Z_2,...,Z_{n-1})$ that are smooth and when evaluated in p give us the same orientation : $[(Z_{2p},...,Z_{n-1p })]=[(Y_{2p},...,Y_{np})]$,why is this true?
differential-geometry manifolds orientation manifolds-with-boundary
differential-geometry manifolds orientation manifolds-with-boundary
asked Jan 11 at 14:09
roi_saumonroi_saumon
63938
asked Jan 11 at 14:09
roi_saumonroi_saumon
63938
asked Jan 11 at 14:09
roi_saumonroi_saumon
63938
asked Jan 11 at 14:09
roi_saumonroi_saumon
63938
asked Jan 11 at 14:09
roi_saumonroi_saumon
63938
63938
add a comment |
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%2f3069863%2fthe-orientation-induced-on-the-boundary-of-a-manifold%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%2f3069863%2fthe-orientation-induced-on-the-boundary-of-a-manifold%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
