If $(e_1, ldots, e_n)$ and $(f_1, ldots, f_n)$ are co-oriented, then so are $(e_1, ldots, e_n, u)$ and $(f_1,...
$begingroup$
Suppose that $S$ is a smooth $n$-submanifold of $M$ where $dim(M) = n+1$. Suppose also that $S$ and $M$ are both Riemannian and oriented.
Suppose that $(e_1, ldots, e_n)$ and $(f_1, ldots, f_n)$ are both positively oriented orthonormal bases for $T_pS$ and that $(e_1, ldots, e_n, u)$ is positively oriented for $T_pM$ where $u$ is orthonormal with respect to $(e_1, ldots, e_n)$.
Is it necessarily true that $(f_1, ldots, f_n, u)$ is also positively
oriented?
It seems to me possible that $(f_1, ldots, f_n, -u)$ is positively oriented. But I am reading a proof that seems to be relying on the fact that it cannot be.
Any help is appreciated.
linear-algebra differential-geometry riemannian-geometry smooth-manifolds orientation
asked Dec 6 '18 at 4:52
CuriousKid7CuriousKid7
1,676717
$endgroup$
add a comment |
$begingroup$
Suppose that $S$ is a smooth $n$-submanifold of $M$ where $dim(M) = n+1$. Suppose also that $S$ and $M$ are both Riemannian and oriented.
Suppose that $(e_1, ldots, e_n)$ and $(f_1, ldots, f_n)$ are both positively oriented orthonormal bases for $T_pS$ and that $(e_1, ldots, e_n, u)$ is positively oriented for $T_pM$ where $u$ is orthonormal with respect to $(e_1, ldots, e_n)$.
Is it necessarily true that $(f_1, ldots, f_n, u)$ is also positively
oriented?
It seems to me possible that $(f_1, ldots, f_n, -u)$ is positively oriented. But I am reading a proof that seems to be relying on the fact that it cannot be.
Any help is appreciated.
linear-algebra differential-geometry riemannian-geometry smooth-manifolds orientation
asked Dec 6 '18 at 4:52
CuriousKid7CuriousKid7
1,676717
$endgroup$
add a comment |
$begingroup$
Suppose that $S$ is a smooth $n$-submanifold of $M$ where $dim(M) = n+1$. Suppose also that $S$ and $M$ are both Riemannian and oriented.
Suppose that $(e_1, ldots, e_n)$ and $(f_1, ldots, f_n)$ are both positively oriented orthonormal bases for $T_pS$ and that $(e_1, ldots, e_n, u)$ is positively oriented for $T_pM$ where $u$ is orthonormal with respect to $(e_1, ldots, e_n)$.
Is it necessarily true that $(f_1, ldots, f_n, u)$ is also positively
oriented?
It seems to me possible that $(f_1, ldots, f_n, -u)$ is positively oriented. But I am reading a proof that seems to be relying on the fact that it cannot be.
Any help is appreciated.
linear-algebra differential-geometry riemannian-geometry smooth-manifolds orientation
asked Dec 6 '18 at 4:52
CuriousKid7CuriousKid7
1,676717
$endgroup$
Suppose that $S$ is a smooth $n$-submanifold of $M$ where $dim(M) = n+1$. Suppose also that $S$ and $M$ are both Riemannian and oriented.
Suppose that $(e_1, ldots, e_n)$ and $(f_1, ldots, f_n)$ are both positively oriented orthonormal bases for $T_pS$ and that $(e_1, ldots, e_n, u)$ is positively oriented for $T_pM$ where $u$ is orthonormal with respect to $(e_1, ldots, e_n)$.
Is it necessarily true that $(f_1, ldots, f_n, u)$ is also positively
oriented?
It seems to me possible that $(f_1, ldots, f_n, -u)$ is positively oriented. But I am reading a proof that seems to be relying on the fact that it cannot be.
Any help is appreciated.
linear-algebra differential-geometry riemannian-geometry smooth-manifolds orientation
linear-algebra differential-geometry riemannian-geometry smooth-manifolds orientation
asked Dec 6 '18 at 4:52
CuriousKid7CuriousKid7
1,676717
asked Dec 6 '18 at 4:52
CuriousKid7CuriousKid7
1,676717
edited Dec 6 '18 at 5:54
CuriousKid7
asked Dec 6 '18 at 4:52
CuriousKid7CuriousKid7
1,676717
asked Dec 6 '18 at 4:52
CuriousKid7CuriousKid7
1,676717
asked Dec 6 '18 at 4:52
CuriousKid7CuriousKid7
1,676717
1,676717
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If $A$ is the invertible matrix that sends each $e_i$ to $f_i$, then the the matrix that sends $(e_1,dots,e_n,u)$ to $(f_1,dots,f_n,u)$ has block form $begin{pmatrix} A& 0 \ 0 & 1end{pmatrix}$ so its determinant equals the determinant of $A$. Hence your assumptions imply that also $(f_1,dots,f_n,u)$ is positively oriented.
answered Dec 6 '18 at 11:38
Andreas CapAndreas Cap
11.1k923
$endgroup$
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%2f3028050%2fif-e-1-ldots-e-n-and-f-1-ldots-f-n-are-co-oriented-then-so-are%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
$begingroup$
If $A$ is the invertible matrix that sends each $e_i$ to $f_i$, then the the matrix that sends $(e_1,dots,e_n,u)$ to $(f_1,dots,f_n,u)$ has block form $begin{pmatrix} A& 0 \ 0 & 1end{pmatrix}$ so its determinant equals the determinant of $A$. Hence your assumptions imply that also $(f_1,dots,f_n,u)$ is positively oriented.
answered Dec 6 '18 at 11:38
Andreas CapAndreas Cap
11.1k923
$endgroup$
add a comment |
$begingroup$
If $A$ is the invertible matrix that sends each $e_i$ to $f_i$, then the the matrix that sends $(e_1,dots,e_n,u)$ to $(f_1,dots,f_n,u)$ has block form $begin{pmatrix} A& 0 \ 0 & 1end{pmatrix}$ so its determinant equals the determinant of $A$. Hence your assumptions imply that also $(f_1,dots,f_n,u)$ is positively oriented.
answered Dec 6 '18 at 11:38
Andreas CapAndreas Cap
11.1k923
$endgroup$
add a comment |
$begingroup$
If $A$ is the invertible matrix that sends each $e_i$ to $f_i$, then the the matrix that sends $(e_1,dots,e_n,u)$ to $(f_1,dots,f_n,u)$ has block form $begin{pmatrix} A& 0 \ 0 & 1end{pmatrix}$ so its determinant equals the determinant of $A$. Hence your assumptions imply that also $(f_1,dots,f_n,u)$ is positively oriented.
answered Dec 6 '18 at 11:38
Andreas CapAndreas Cap
11.1k923
$endgroup$
If $A$ is the invertible matrix that sends each $e_i$ to $f_i$, then the the matrix that sends $(e_1,dots,e_n,u)$ to $(f_1,dots,f_n,u)$ has block form $begin{pmatrix} A& 0 \ 0 & 1end{pmatrix}$ so its determinant equals the determinant of $A$. Hence your assumptions imply that also $(f_1,dots,f_n,u)$ is positively oriented.
answered Dec 6 '18 at 11:38
Andreas CapAndreas Cap
11.1k923
answered Dec 6 '18 at 11:38
Andreas CapAndreas Cap
11.1k923
answered Dec 6 '18 at 11:38
Andreas CapAndreas Cap
11.1k923
answered Dec 6 '18 at 11:38
Andreas CapAndreas Cap
11.1k923
11.1k923
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%2f3028050%2fif-e-1-ldots-e-n-and-f-1-ldots-f-n-are-co-oriented-then-so-are%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
