Suppose $ dim(M)<dim(N)$, can there be a diffeomorphism from manifold $M$ to a submanifold of the manifold...
$begingroup$
Diffeomorphisms are necessarily between manifolds of the same dimension. Imagine $f$ going from dimension $n$ to dimension $k$. If $n < k$ then $Df_x$ could never be surjective; and if $n > k$ then $Df_x$ could never be injective. In both cases, therefore, $Df_x$ fails to be a bijection.
The definition is-
$f : M → N$ is called a diffeomorphism if, in coordinate charts, it satisfies the definition above. More precisely: Pick any cover of $M$ by compatible coordinate charts and do the same for $N$. Let $φ$ and $ψ$ be charts on, respectively, $M$ and $N$, with $U$ and $V$ as, respectively, the images of $φ$ and $ψ$. The map $ψfφ^{-1} : U → V$ is then a diffeomorphism as in the definition above, whenever $f(φ^{-1}(U)) ⊂ ψ^{-1}(V).$
$ underline{text{My question:}}$
$(1)$ Can there exists a diffeomorphism between two manifolds of different dimensions?
$(2)$ Suppose $dim(M)<dim(N)$, can there be a diffeomorphism from manifold $M$ to a submanifold of the manifold $N$?
Help me
differential-geometry diffeomorphism
$endgroup$
add a comment |
$begingroup$
Diffeomorphisms are necessarily between manifolds of the same dimension. Imagine $f$ going from dimension $n$ to dimension $k$. If $n < k$ then $Df_x$ could never be surjective; and if $n > k$ then $Df_x$ could never be injective. In both cases, therefore, $Df_x$ fails to be a bijection.
The definition is-
$f : M → N$ is called a diffeomorphism if, in coordinate charts, it satisfies the definition above. More precisely: Pick any cover of $M$ by compatible coordinate charts and do the same for $N$. Let $φ$ and $ψ$ be charts on, respectively, $M$ and $N$, with $U$ and $V$ as, respectively, the images of $φ$ and $ψ$. The map $ψfφ^{-1} : U → V$ is then a diffeomorphism as in the definition above, whenever $f(φ^{-1}(U)) ⊂ ψ^{-1}(V).$
$ underline{text{My question:}}$
$(1)$ Can there exists a diffeomorphism between two manifolds of different dimensions?
$(2)$ Suppose $dim(M)<dim(N)$, can there be a diffeomorphism from manifold $M$ to a submanifold of the manifold $N$?
Help me
differential-geometry diffeomorphism
$endgroup$
2
$begingroup$
1: Are you not satisfied with the explanation in your first paragraph as to why you can't have diffeomorphisms between manifolds of different dimensions? 2: Consider $M = mathbb R$, $N = mathbb R^2$ and $f : M to N$ sending $f(x) = (x,0)$.
$endgroup$
– Kenny Wong
Jan 4 at 11:18
$begingroup$
@KennyWong, Yes I am satisfied but I still has the confusion regarding my question $(2)$. The map $f(x)=(x,0)$ is a diffeomorphism. So here be a diffeomorphism from manifold $M$ to a submanifold of the manifold $N$. Am I right?
$endgroup$
– M. A. SARKAR
Jan 4 at 11:31
1
$begingroup$
@M.A.SARKAR Yes you're right (assuming you have $f: mathbb{R} to mathbb{R}^2$ or something similar).
$endgroup$
– 0x539
Jan 4 at 12:00
add a comment |
$begingroup$
Diffeomorphisms are necessarily between manifolds of the same dimension. Imagine $f$ going from dimension $n$ to dimension $k$. If $n < k$ then $Df_x$ could never be surjective; and if $n > k$ then $Df_x$ could never be injective. In both cases, therefore, $Df_x$ fails to be a bijection.
The definition is-
$f : M → N$ is called a diffeomorphism if, in coordinate charts, it satisfies the definition above. More precisely: Pick any cover of $M$ by compatible coordinate charts and do the same for $N$. Let $φ$ and $ψ$ be charts on, respectively, $M$ and $N$, with $U$ and $V$ as, respectively, the images of $φ$ and $ψ$. The map $ψfφ^{-1} : U → V$ is then a diffeomorphism as in the definition above, whenever $f(φ^{-1}(U)) ⊂ ψ^{-1}(V).$
$ underline{text{My question:}}$
$(1)$ Can there exists a diffeomorphism between two manifolds of different dimensions?
$(2)$ Suppose $dim(M)<dim(N)$, can there be a diffeomorphism from manifold $M$ to a submanifold of the manifold $N$?
Help me
differential-geometry diffeomorphism
$endgroup$
Diffeomorphisms are necessarily between manifolds of the same dimension. Imagine $f$ going from dimension $n$ to dimension $k$. If $n < k$ then $Df_x$ could never be surjective; and if $n > k$ then $Df_x$ could never be injective. In both cases, therefore, $Df_x$ fails to be a bijection.
The definition is-
$f : M → N$ is called a diffeomorphism if, in coordinate charts, it satisfies the definition above. More precisely: Pick any cover of $M$ by compatible coordinate charts and do the same for $N$. Let $φ$ and $ψ$ be charts on, respectively, $M$ and $N$, with $U$ and $V$ as, respectively, the images of $φ$ and $ψ$. The map $ψfφ^{-1} : U → V$ is then a diffeomorphism as in the definition above, whenever $f(φ^{-1}(U)) ⊂ ψ^{-1}(V).$
$ underline{text{My question:}}$
$(1)$ Can there exists a diffeomorphism between two manifolds of different dimensions?
$(2)$ Suppose $dim(M)<dim(N)$, can there be a diffeomorphism from manifold $M$ to a submanifold of the manifold $N$?
Help me
differential-geometry diffeomorphism
differential-geometry diffeomorphism
edited Jan 4 at 11:39
Davide Giraudo
126k16150261
126k16150261
asked Jan 4 at 10:49
M. A. SARKARM. A. SARKAR
2,1911619
2,1911619
2
$begingroup$
1: Are you not satisfied with the explanation in your first paragraph as to why you can't have diffeomorphisms between manifolds of different dimensions? 2: Consider $M = mathbb R$, $N = mathbb R^2$ and $f : M to N$ sending $f(x) = (x,0)$.
$endgroup$
– Kenny Wong
Jan 4 at 11:18
$begingroup$
@KennyWong, Yes I am satisfied but I still has the confusion regarding my question $(2)$. The map $f(x)=(x,0)$ is a diffeomorphism. So here be a diffeomorphism from manifold $M$ to a submanifold of the manifold $N$. Am I right?
$endgroup$
– M. A. SARKAR
Jan 4 at 11:31
1
$begingroup$
@M.A.SARKAR Yes you're right (assuming you have $f: mathbb{R} to mathbb{R}^2$ or something similar).
$endgroup$
– 0x539
Jan 4 at 12:00
add a comment |
2
$begingroup$
1: Are you not satisfied with the explanation in your first paragraph as to why you can't have diffeomorphisms between manifolds of different dimensions? 2: Consider $M = mathbb R$, $N = mathbb R^2$ and $f : M to N$ sending $f(x) = (x,0)$.
$endgroup$
– Kenny Wong
Jan 4 at 11:18
$begingroup$
@KennyWong, Yes I am satisfied but I still has the confusion regarding my question $(2)$. The map $f(x)=(x,0)$ is a diffeomorphism. So here be a diffeomorphism from manifold $M$ to a submanifold of the manifold $N$. Am I right?
$endgroup$
– M. A. SARKAR
Jan 4 at 11:31
1
$begingroup$
@M.A.SARKAR Yes you're right (assuming you have $f: mathbb{R} to mathbb{R}^2$ or something similar).
$endgroup$
– 0x539
Jan 4 at 12:00
2
2
$begingroup$
1: Are you not satisfied with the explanation in your first paragraph as to why you can't have diffeomorphisms between manifolds of different dimensions? 2: Consider $M = mathbb R$, $N = mathbb R^2$ and $f : M to N$ sending $f(x) = (x,0)$.
$endgroup$
– Kenny Wong
Jan 4 at 11:18
$begingroup$
1: Are you not satisfied with the explanation in your first paragraph as to why you can't have diffeomorphisms between manifolds of different dimensions? 2: Consider $M = mathbb R$, $N = mathbb R^2$ and $f : M to N$ sending $f(x) = (x,0)$.
$endgroup$
– Kenny Wong
Jan 4 at 11:18
$begingroup$
@KennyWong, Yes I am satisfied but I still has the confusion regarding my question $(2)$. The map $f(x)=(x,0)$ is a diffeomorphism. So here be a diffeomorphism from manifold $M$ to a submanifold of the manifold $N$. Am I right?
$endgroup$
– M. A. SARKAR
Jan 4 at 11:31
$begingroup$
@KennyWong, Yes I am satisfied but I still has the confusion regarding my question $(2)$. The map $f(x)=(x,0)$ is a diffeomorphism. So here be a diffeomorphism from manifold $M$ to a submanifold of the manifold $N$. Am I right?
$endgroup$
– M. A. SARKAR
Jan 4 at 11:31
1
1
$begingroup$
@M.A.SARKAR Yes you're right (assuming you have $f: mathbb{R} to mathbb{R}^2$ or something similar).
$endgroup$
– 0x539
Jan 4 at 12:00
$begingroup$
@M.A.SARKAR Yes you're right (assuming you have $f: mathbb{R} to mathbb{R}^2$ or something similar).
$endgroup$
– 0x539
Jan 4 at 12:00
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%2f3061514%2fsuppose-dimm-dimn-can-there-be-a-diffeomorphism-from-manifold-m-to-a%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%2f3061514%2fsuppose-dimm-dimn-can-there-be-a-diffeomorphism-from-manifold-m-to-a%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
2
$begingroup$
1: Are you not satisfied with the explanation in your first paragraph as to why you can't have diffeomorphisms between manifolds of different dimensions? 2: Consider $M = mathbb R$, $N = mathbb R^2$ and $f : M to N$ sending $f(x) = (x,0)$.
$endgroup$
– Kenny Wong
Jan 4 at 11:18
$begingroup$
@KennyWong, Yes I am satisfied but I still has the confusion regarding my question $(2)$. The map $f(x)=(x,0)$ is a diffeomorphism. So here be a diffeomorphism from manifold $M$ to a submanifold of the manifold $N$. Am I right?
$endgroup$
– M. A. SARKAR
Jan 4 at 11:31
1
$begingroup$
@M.A.SARKAR Yes you're right (assuming you have $f: mathbb{R} to mathbb{R}^2$ or something similar).
$endgroup$
– 0x539
Jan 4 at 12:00