Mixing
Can a smooth map between two embedded submanifolds be (locally) smoothly extended?
$begingroup$
Consider $cal M$ and $cal M'$, smooth embedded submanifolds of two linear manifolds $cal E$ and $cal E'$ (respectively). Let $F colon cal M to cal M'$ be a smooth map.
From Lee's textbook (2012, Intro to smooth manifolds), Lemma 5.34, we know that for the special case where $cal M' = mathbb{R}$ we can smoothly extend $F$, at least locally:
There exists an open neighborhood $U$ of $cal M$ in $cal E$ and a smooth function $bar F colon U to mathbb{R}$ such that $F$ is the restriction of $bar F$ to $cal M$, that is, $F = bar F|_{cal M}$.
My question: for the more general case where $cal M'$ is not simply equal to $mathbb{R}$ but can be any embedded submanifold of a linear manifold $cal E'$, can I also have such a smooth extension with a map $bar F colon cal E to cal E'$?
differential-geometry smooth-manifolds smooth-functions
asked Jan 24 at 23:09
Nicolas BoumalNicolas Boumal
14317
$endgroup$
add a comment |
$begingroup$
Consider $cal M$ and $cal M'$, smooth embedded submanifolds of two linear manifolds $cal E$ and $cal E'$ (respectively). Let $F colon cal M to cal M'$ be a smooth map.
From Lee's textbook (2012, Intro to smooth manifolds), Lemma 5.34, we know that for the special case where $cal M' = mathbb{R}$ we can smoothly extend $F$, at least locally:
There exists an open neighborhood $U$ of $cal M$ in $cal E$ and a smooth function $bar F colon U to mathbb{R}$ such that $F$ is the restriction of $bar F$ to $cal M$, that is, $F = bar F|_{cal M}$.
My question: for the more general case where $cal M'$ is not simply equal to $mathbb{R}$ but can be any embedded submanifold of a linear manifold $cal E'$, can I also have such a smooth extension with a map $bar F colon cal E to cal E'$?
differential-geometry smooth-manifolds smooth-functions
asked Jan 24 at 23:09
Nicolas BoumalNicolas Boumal
14317
$endgroup$
$begingroup$
Do you want a map defined on all of $mathcal E$ or will a neighborhood $U$ of a point in $M$ do? Remember that maps to $mathcal E'$ are vector-valued functions.
$endgroup$
– Ted Shifrin
Jan 24 at 23:15
$begingroup$
I expect a smooth extension to all of $cal E$ won't be possible, because already for $cal M' = mathbb{R}$ this requires $cal M$ to be properly embedded (Lee, Lemma 5.34 and Exercise 5-18). So a smooth extension to a neighborhood of $cal M$ in $cal E$ will have to do. My difficulty is with the co-domain $cal M' subseteq cal E'$. Would it suffice to simply consider $F$ as a map into $cal E' approxeq mathbb{R}^d$, and to study its individual components?
$endgroup$
– Nicolas Boumal
Jan 24 at 23:32
$begingroup$
(Also, I suspect that if such smooth extensions exist, they would exist regardless of the linear structure of $cal E'$.)
$endgroup$
– Nicolas Boumal
Jan 24 at 23:33
1
$begingroup$
Closely related: math.stackexchange.com/questions/1893383/…
$endgroup$
– Eric Wofsey
Jan 25 at 6:31
add a comment |
$begingroup$
Consider $cal M$ and $cal M'$, smooth embedded submanifolds of two linear manifolds $cal E$ and $cal E'$ (respectively). Let $F colon cal M to cal M'$ be a smooth map.
From Lee's textbook (2012, Intro to smooth manifolds), Lemma 5.34, we know that for the special case where $cal M' = mathbb{R}$ we can smoothly extend $F$, at least locally:
There exists an open neighborhood $U$ of $cal M$ in $cal E$ and a smooth function $bar F colon U to mathbb{R}$ such that $F$ is the restriction of $bar F$ to $cal M$, that is, $F = bar F|_{cal M}$.
My question: for the more general case where $cal M'$ is not simply equal to $mathbb{R}$ but can be any embedded submanifold of a linear manifold $cal E'$, can I also have such a smooth extension with a map $bar F colon cal E to cal E'$?
differential-geometry smooth-manifolds smooth-functions
asked Jan 24 at 23:09
Nicolas BoumalNicolas Boumal
14317
$endgroup$
Consider $cal M$ and $cal M'$, smooth embedded submanifolds of two linear manifolds $cal E$ and $cal E'$ (respectively). Let $F colon cal M to cal M'$ be a smooth map.
From Lee's textbook (2012, Intro to smooth manifolds), Lemma 5.34, we know that for the special case where $cal M' = mathbb{R}$ we can smoothly extend $F$, at least locally:
There exists an open neighborhood $U$ of $cal M$ in $cal E$ and a smooth function $bar F colon U to mathbb{R}$ such that $F$ is the restriction of $bar F$ to $cal M$, that is, $F = bar F|_{cal M}$.
My question: for the more general case where $cal M'$ is not simply equal to $mathbb{R}$ but can be any embedded submanifold of a linear manifold $cal E'$, can I also have such a smooth extension with a map $bar F colon cal E to cal E'$?
differential-geometry smooth-manifolds smooth-functions
differential-geometry smooth-manifolds smooth-functions
asked Jan 24 at 23:09
Nicolas BoumalNicolas Boumal
14317
asked Jan 24 at 23:09
Nicolas BoumalNicolas Boumal
14317
asked Jan 24 at 23:09
Nicolas BoumalNicolas Boumal
14317
asked Jan 24 at 23:09
Nicolas BoumalNicolas Boumal
14317
asked Jan 24 at 23:09
Nicolas BoumalNicolas Boumal
14317
14317
$begingroup$
Do you want a map defined on all of $mathcal E$ or will a neighborhood $U$ of a point in $M$ do? Remember that maps to $mathcal E'$ are vector-valued functions.
$endgroup$
– Ted Shifrin
Jan 24 at 23:15
$begingroup$
I expect a smooth extension to all of $cal E$ won't be possible, because already for $cal M' = mathbb{R}$ this requires $cal M$ to be properly embedded (Lee, Lemma 5.34 and Exercise 5-18). So a smooth extension to a neighborhood of $cal M$ in $cal E$ will have to do. My difficulty is with the co-domain $cal M' subseteq cal E'$. Would it suffice to simply consider $F$ as a map into $cal E' approxeq mathbb{R}^d$, and to study its individual components?
$endgroup$
– Nicolas Boumal
Jan 24 at 23:32
$begingroup$
(Also, I suspect that if such smooth extensions exist, they would exist regardless of the linear structure of $cal E'$.)
$endgroup$
– Nicolas Boumal
Jan 24 at 23:33
1
$begingroup$
Closely related: math.stackexchange.com/questions/1893383/…
$endgroup$
– Eric Wofsey
Jan 25 at 6:31
add a comment |
$begingroup$
Do you want a map defined on all of $mathcal E$ or will a neighborhood $U$ of a point in $M$ do? Remember that maps to $mathcal E'$ are vector-valued functions.
$endgroup$
– Ted Shifrin
Jan 24 at 23:15
$begingroup$
I expect a smooth extension to all of $cal E$ won't be possible, because already for $cal M' = mathbb{R}$ this requires $cal M$ to be properly embedded (Lee, Lemma 5.34 and Exercise 5-18). So a smooth extension to a neighborhood of $cal M$ in $cal E$ will have to do. My difficulty is with the co-domain $cal M' subseteq cal E'$. Would it suffice to simply consider $F$ as a map into $cal E' approxeq mathbb{R}^d$, and to study its individual components?
$endgroup$
– Nicolas Boumal
Jan 24 at 23:32
$begingroup$
(Also, I suspect that if such smooth extensions exist, they would exist regardless of the linear structure of $cal E'$.)
$endgroup$
– Nicolas Boumal
Jan 24 at 23:33
1
$begingroup$
Closely related: math.stackexchange.com/questions/1893383/…
$endgroup$
– Eric Wofsey
Jan 25 at 6:31
$begingroup$
Do you want a map defined on all of $mathcal E$ or will a neighborhood $U$ of a point in $M$ do? Remember that maps to $mathcal E'$ are vector-valued functions.
$endgroup$
– Ted Shifrin
Jan 24 at 23:15
$begingroup$
Do you want a map defined on all of $mathcal E$ or will a neighborhood $U$ of a point in $M$ do? Remember that maps to $mathcal E'$ are vector-valued functions.
$endgroup$
– Ted Shifrin
Jan 24 at 23:15
$begingroup$
I expect a smooth extension to all of $cal E$ won't be possible, because already for $cal M' = mathbb{R}$ this requires $cal M$ to be properly embedded (Lee, Lemma 5.34 and Exercise 5-18). So a smooth extension to a neighborhood of $cal M$ in $cal E$ will have to do. My difficulty is with the co-domain $cal M' subseteq cal E'$. Would it suffice to simply consider $F$ as a map into $cal E' approxeq mathbb{R}^d$, and to study its individual components?
$endgroup$
– Nicolas Boumal
Jan 24 at 23:32
$begingroup$
I expect a smooth extension to all of $cal E$ won't be possible, because already for $cal M' = mathbb{R}$ this requires $cal M$ to be properly embedded (Lee, Lemma 5.34 and Exercise 5-18). So a smooth extension to a neighborhood of $cal M$ in $cal E$ will have to do. My difficulty is with the co-domain $cal M' subseteq cal E'$. Would it suffice to simply consider $F$ as a map into $cal E' approxeq mathbb{R}^d$, and to study its individual components?
$endgroup$
– Nicolas Boumal
Jan 24 at 23:32
$begingroup$
(Also, I suspect that if such smooth extensions exist, they would exist regardless of the linear structure of $cal E'$.)
$endgroup$
– Nicolas Boumal
Jan 24 at 23:33
$begingroup$
(Also, I suspect that if such smooth extensions exist, they would exist regardless of the linear structure of $cal E'$.)
$endgroup$
– Nicolas Boumal
Jan 24 at 23:33
1
1
$begingroup$
Closely related: math.stackexchange.com/questions/1893383/…
$endgroup$
– Eric Wofsey
Jan 25 at 6:31
$begingroup$
Closely related: math.stackexchange.com/questions/1893383/…
$endgroup$
– Eric Wofsey
Jan 25 at 6:31
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You should be able to do this using the tubular neighborhood theorem of Riemannian geometry. Let $pi:NM to M$ be the basepoint map from the normal bundle of M to M, and let $V$ be a sufficiently small neighborhood of (the canonical image of) $M$ in $NM$, such that there is a diffeomorphism $psi: V to U$ onto some tubular neighborhood of $M$ (by the TNT). Then set $widetilde{F} = F circ pi circ psi^{-1}: U to M'$. This is your extension (and it takes its values in $M'$).
answered Jan 25 at 6:20
Stephen MStephen M
361
$endgroup$
$begingroup$
This is great, thanks Stephen! Further composing the map you propose with the inclusion map from M' into E' gives the final answer as a map from (a neighborhood of M in) E to E', and illustrates the role of having M' embedded in E' (so that the inclusion would be smooth).
$endgroup$
– Nicolas Boumal
Jan 25 at 16:15
$begingroup$
Yes, good catch!
$endgroup$
– Stephen M
Jan 25 at 19:39
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%2f3086498%2fcan-a-smooth-map-between-two-embedded-submanifolds-be-locally-smoothly-extende%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
$begingroup$
You should be able to do this using the tubular neighborhood theorem of Riemannian geometry. Let $pi:NM to M$ be the basepoint map from the normal bundle of M to M, and let $V$ be a sufficiently small neighborhood of (the canonical image of) $M$ in $NM$, such that there is a diffeomorphism $psi: V to U$ onto some tubular neighborhood of $M$ (by the TNT). Then set $widetilde{F} = F circ pi circ psi^{-1}: U to M'$. This is your extension (and it takes its values in $M'$).
answered Jan 25 at 6:20
Stephen MStephen M
361
$endgroup$
$begingroup$
This is great, thanks Stephen! Further composing the map you propose with the inclusion map from M' into E' gives the final answer as a map from (a neighborhood of M in) E to E', and illustrates the role of having M' embedded in E' (so that the inclusion would be smooth).
$endgroup$
– Nicolas Boumal
Jan 25 at 16:15
$begingroup$
Yes, good catch!
$endgroup$
– Stephen M
Jan 25 at 19:39
add a comment |
$begingroup$
You should be able to do this using the tubular neighborhood theorem of Riemannian geometry. Let $pi:NM to M$ be the basepoint map from the normal bundle of M to M, and let $V$ be a sufficiently small neighborhood of (the canonical image of) $M$ in $NM$, such that there is a diffeomorphism $psi: V to U$ onto some tubular neighborhood of $M$ (by the TNT). Then set $widetilde{F} = F circ pi circ psi^{-1}: U to M'$. This is your extension (and it takes its values in $M'$).
answered Jan 25 at 6:20
Stephen MStephen M
361
$endgroup$
$begingroup$
This is great, thanks Stephen! Further composing the map you propose with the inclusion map from M' into E' gives the final answer as a map from (a neighborhood of M in) E to E', and illustrates the role of having M' embedded in E' (so that the inclusion would be smooth).
$endgroup$
– Nicolas Boumal
Jan 25 at 16:15
$begingroup$
Yes, good catch!
$endgroup$
– Stephen M
Jan 25 at 19:39
add a comment |
$begingroup$
You should be able to do this using the tubular neighborhood theorem of Riemannian geometry. Let $pi:NM to M$ be the basepoint map from the normal bundle of M to M, and let $V$ be a sufficiently small neighborhood of (the canonical image of) $M$ in $NM$, such that there is a diffeomorphism $psi: V to U$ onto some tubular neighborhood of $M$ (by the TNT). Then set $widetilde{F} = F circ pi circ psi^{-1}: U to M'$. This is your extension (and it takes its values in $M'$).
answered Jan 25 at 6:20
Stephen MStephen M
361
$endgroup$
You should be able to do this using the tubular neighborhood theorem of Riemannian geometry. Let $pi:NM to M$ be the basepoint map from the normal bundle of M to M, and let $V$ be a sufficiently small neighborhood of (the canonical image of) $M$ in $NM$, such that there is a diffeomorphism $psi: V to U$ onto some tubular neighborhood of $M$ (by the TNT). Then set $widetilde{F} = F circ pi circ psi^{-1}: U to M'$. This is your extension (and it takes its values in $M'$).
answered Jan 25 at 6:20
Stephen MStephen M
361
answered Jan 25 at 6:20
Stephen MStephen M
361
answered Jan 25 at 6:20
Stephen MStephen M
361
answered Jan 25 at 6:20
Stephen MStephen M
361
361
$begingroup$
This is great, thanks Stephen! Further composing the map you propose with the inclusion map from M' into E' gives the final answer as a map from (a neighborhood of M in) E to E', and illustrates the role of having M' embedded in E' (so that the inclusion would be smooth).
$endgroup$
– Nicolas Boumal
Jan 25 at 16:15
$begingroup$
Yes, good catch!
$endgroup$
– Stephen M
Jan 25 at 19:39
add a comment |
$begingroup$
This is great, thanks Stephen! Further composing the map you propose with the inclusion map from M' into E' gives the final answer as a map from (a neighborhood of M in) E to E', and illustrates the role of having M' embedded in E' (so that the inclusion would be smooth).
$endgroup$
– Nicolas Boumal
Jan 25 at 16:15
$begingroup$
Yes, good catch!
$endgroup$
– Stephen M
Jan 25 at 19:39
$begingroup$
This is great, thanks Stephen! Further composing the map you propose with the inclusion map from M' into E' gives the final answer as a map from (a neighborhood of M in) E to E', and illustrates the role of having M' embedded in E' (so that the inclusion would be smooth).
$endgroup$
– Nicolas Boumal
Jan 25 at 16:15
$begingroup$
This is great, thanks Stephen! Further composing the map you propose with the inclusion map from M' into E' gives the final answer as a map from (a neighborhood of M in) E to E', and illustrates the role of having M' embedded in E' (so that the inclusion would be smooth).
$endgroup$
– Nicolas Boumal
Jan 25 at 16:15
$begingroup$
Yes, good catch!
$endgroup$
– Stephen M
Jan 25 at 19:39
$begingroup$
Yes, good catch!
$endgroup$
– Stephen M
Jan 25 at 19:39
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%2f3086498%2fcan-a-smooth-map-between-two-embedded-submanifolds-be-locally-smoothly-extende%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
$begingroup$
Do you want a map defined on all of $mathcal E$ or will a neighborhood $U$ of a point in $M$ do? Remember that maps to $mathcal E'$ are vector-valued functions.
$endgroup$
– Ted Shifrin
Jan 24 at 23:15
$begingroup$
I expect a smooth extension to all of $cal E$ won't be possible, because already for $cal M' = mathbb{R}$ this requires $cal M$ to be properly embedded (Lee, Lemma 5.34 and Exercise 5-18). So a smooth extension to a neighborhood of $cal M$ in $cal E$ will have to do. My difficulty is with the co-domain $cal M' subseteq cal E'$. Would it suffice to simply consider $F$ as a map into $cal E' approxeq mathbb{R}^d$, and to study its individual components?
$endgroup$
– Nicolas Boumal
Jan 24 at 23:32
$begingroup$
(Also, I suspect that if such smooth extensions exist, they would exist regardless of the linear structure of $cal E'$.)
$endgroup$
– Nicolas Boumal
Jan 24 at 23:33
1
$begingroup$
Closely related: math.stackexchange.com/questions/1893383/…
$endgroup$
– Eric Wofsey
Jan 25 at 6:31