Mixing
The Image of sphere under the map $f: (x,y,z) to (x^2,y^2,z^2,sqrt{2}yz, sqrt{2}zx, sqrt{2}xy)$ and...
$begingroup$
I'm going to take a course about diferentiable manifolds next semester and I'm preparing for it by solving some of the problems from the book An Introduction to Differential Manifolds by D. Barden and C. Thomas. My understanding of the material is very rudimentary, I'm just memorizing the definitions and the theorems so far without reading their proofs. So, please post only the solutions that use the basic definitions and theorems.
Show that the image of $S^2 subset mathbb{R}^3$ under the map $f: (x,y,z) to (x^2,y^2,z^2,sqrt{2}yz, sqrt{2}zx, sqrt{2}xy)$ is a submanifold of $mathbb{R}^6$. How does it relate to $mathbb{R}P^2$?
Introducing charts on the $mathrm{Im}(f)mid_{S^2}$ seems not trivial. However, I have made an observation that I think handles the first part.
$$Df=begin{bmatrix}
2x & 0 & 0\
0 & 2y & 0\
0 & 0 & 2z\
0 & sqrt{2}z & sqrt{2}y\
sqrt{2}z & 0 & sqrt{2}x\
sqrt{2}y & sqrt{2}x & 0
end{bmatrix}$$
has constant rank $2$ over $S$ because at least one of the coordinates $x,y,z$ must be non-zero and then I can find a $2times 2$ sub-matrix with non-zero determinant. If necessary, I can first exchange two rows to create this sub-matrix.
I think this shows that the tangent to $mathrm{Im}(f)mid_{S^2}$ is always a plane, so it's a $2$-dimensional manifold in $mathbb{R}^6$. But I don't know how to make my reasoning rigorous.
I think that $mathrm{Im}(f)$ is diffeomoprhic to $mathbb{R}P^2$. Well, at least they have the same dimension. Plus, the image of $f$ is given by a homogeneous polynomial. I don't know how to proceed from here.
geometry differential-geometry manifolds differential-topology smooth-manifolds
asked Jan 3 at 22:53
stressed outstressed out
4,2191533
$endgroup$
add a comment |
$begingroup$
I'm going to take a course about diferentiable manifolds next semester and I'm preparing for it by solving some of the problems from the book An Introduction to Differential Manifolds by D. Barden and C. Thomas. My understanding of the material is very rudimentary, I'm just memorizing the definitions and the theorems so far without reading their proofs. So, please post only the solutions that use the basic definitions and theorems.
Show that the image of $S^2 subset mathbb{R}^3$ under the map $f: (x,y,z) to (x^2,y^2,z^2,sqrt{2}yz, sqrt{2}zx, sqrt{2}xy)$ is a submanifold of $mathbb{R}^6$. How does it relate to $mathbb{R}P^2$?
Introducing charts on the $mathrm{Im}(f)mid_{S^2}$ seems not trivial. However, I have made an observation that I think handles the first part.
$$Df=begin{bmatrix}
2x & 0 & 0\
0 & 2y & 0\
0 & 0 & 2z\
0 & sqrt{2}z & sqrt{2}y\
sqrt{2}z & 0 & sqrt{2}x\
sqrt{2}y & sqrt{2}x & 0
end{bmatrix}$$
has constant rank $2$ over $S$ because at least one of the coordinates $x,y,z$ must be non-zero and then I can find a $2times 2$ sub-matrix with non-zero determinant. If necessary, I can first exchange two rows to create this sub-matrix.
I think this shows that the tangent to $mathrm{Im}(f)mid_{S^2}$ is always a plane, so it's a $2$-dimensional manifold in $mathbb{R}^6$. But I don't know how to make my reasoning rigorous.
I think that $mathrm{Im}(f)$ is diffeomoprhic to $mathbb{R}P^2$. Well, at least they have the same dimension. Plus, the image of $f$ is given by a homogeneous polynomial. I don't know how to proceed from here.
geometry differential-geometry manifolds differential-topology smooth-manifolds
asked Jan 3 at 22:53
stressed outstressed out
4,2191533
$endgroup$
add a comment |
$begingroup$
I'm going to take a course about diferentiable manifolds next semester and I'm preparing for it by solving some of the problems from the book An Introduction to Differential Manifolds by D. Barden and C. Thomas. My understanding of the material is very rudimentary, I'm just memorizing the definitions and the theorems so far without reading their proofs. So, please post only the solutions that use the basic definitions and theorems.
Show that the image of $S^2 subset mathbb{R}^3$ under the map $f: (x,y,z) to (x^2,y^2,z^2,sqrt{2}yz, sqrt{2}zx, sqrt{2}xy)$ is a submanifold of $mathbb{R}^6$. How does it relate to $mathbb{R}P^2$?
Introducing charts on the $mathrm{Im}(f)mid_{S^2}$ seems not trivial. However, I have made an observation that I think handles the first part.
$$Df=begin{bmatrix}
2x & 0 & 0\
0 & 2y & 0\
0 & 0 & 2z\
0 & sqrt{2}z & sqrt{2}y\
sqrt{2}z & 0 & sqrt{2}x\
sqrt{2}y & sqrt{2}x & 0
end{bmatrix}$$
has constant rank $2$ over $S$ because at least one of the coordinates $x,y,z$ must be non-zero and then I can find a $2times 2$ sub-matrix with non-zero determinant. If necessary, I can first exchange two rows to create this sub-matrix.
I think this shows that the tangent to $mathrm{Im}(f)mid_{S^2}$ is always a plane, so it's a $2$-dimensional manifold in $mathbb{R}^6$. But I don't know how to make my reasoning rigorous.
I think that $mathrm{Im}(f)$ is diffeomoprhic to $mathbb{R}P^2$. Well, at least they have the same dimension. Plus, the image of $f$ is given by a homogeneous polynomial. I don't know how to proceed from here.
geometry differential-geometry manifolds differential-topology smooth-manifolds
asked Jan 3 at 22:53
stressed outstressed out
4,2191533
$endgroup$
I'm going to take a course about diferentiable manifolds next semester and I'm preparing for it by solving some of the problems from the book An Introduction to Differential Manifolds by D. Barden and C. Thomas. My understanding of the material is very rudimentary, I'm just memorizing the definitions and the theorems so far without reading their proofs. So, please post only the solutions that use the basic definitions and theorems.
Show that the image of $S^2 subset mathbb{R}^3$ under the map $f: (x,y,z) to (x^2,y^2,z^2,sqrt{2}yz, sqrt{2}zx, sqrt{2}xy)$ is a submanifold of $mathbb{R}^6$. How does it relate to $mathbb{R}P^2$?
Introducing charts on the $mathrm{Im}(f)mid_{S^2}$ seems not trivial. However, I have made an observation that I think handles the first part.
$$Df=begin{bmatrix}
2x & 0 & 0\
0 & 2y & 0\
0 & 0 & 2z\
0 & sqrt{2}z & sqrt{2}y\
sqrt{2}z & 0 & sqrt{2}x\
sqrt{2}y & sqrt{2}x & 0
end{bmatrix}$$
has constant rank $2$ over $S$ because at least one of the coordinates $x,y,z$ must be non-zero and then I can find a $2times 2$ sub-matrix with non-zero determinant. If necessary, I can first exchange two rows to create this sub-matrix.
I think this shows that the tangent to $mathrm{Im}(f)mid_{S^2}$ is always a plane, so it's a $2$-dimensional manifold in $mathbb{R}^6$. But I don't know how to make my reasoning rigorous.
I think that $mathrm{Im}(f)$ is diffeomoprhic to $mathbb{R}P^2$. Well, at least they have the same dimension. Plus, the image of $f$ is given by a homogeneous polynomial. I don't know how to proceed from here.
geometry differential-geometry manifolds differential-topology smooth-manifolds
geometry differential-geometry manifolds differential-topology smooth-manifolds
asked Jan 3 at 22:53
stressed outstressed out
4,2191533
asked Jan 3 at 22:53
stressed outstressed out
4,2191533
asked Jan 3 at 22:53
stressed outstressed out
4,2191533
asked Jan 3 at 22:53
stressed outstressed out
4,2191533
asked Jan 3 at 22:53
stressed outstressed out
4,2191533
4,2191533
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You are right, $text{Im}(f)$ and $mathbb{R}P^n$ are diffeomorphic.
Recall that we have a quotient map $p:S^n to mathbb{R}P^n$ which identifies antipodal points. Note that $f$ is constant over the fibers of $p$ so it pass to the quotient, that is, there exists a unique differentiable function $tilde{f} : mathbb{R}P^n to mathbb{R}^6$ such that $tilde{f}circ p = f$.
Note that $tilde{f}$ is injective so its corestriction $g := tilde{f}|^{text{Im}(f)}:mathbb{R}P^n to text{Im}(f)$ is a bijection. Now, because $mathbb{R}P^n$ is compact and manifolds are Hausdorff spaces we have that $g$ is a homeomorphism and then by the Inverse Function Theorem it is a diffeomorphism.
answered Jan 3 at 23:13
Dante GrevinoDante Grevino
96319
$endgroup$
$begingroup$
Thanks. Two questions: is the quotient map smooth? (why?) And doesn't the inverse function theorem give us local diffeomorphism? I don't understand the last part. + how can I make my answer to the first part rigorous? Is it even right?
$endgroup$
– stressed out
Jan 3 at 23:18
$begingroup$
Yes, the quotient map is differentiable by definition because we endow the quotient space with the unique differential structure such that the quotient map is a local diffeomorphism. (Of course, not every quotient of a manifold is a manifold but there are results which says that under very general conditions the quotient is a manifold.)
$endgroup$
– Dante Grevino
Jan 3 at 23:27
$begingroup$
For the second question, well... you have to endow the image of $f$ with a suitable differential structure. For example you can prove that it is a regular (or embedded) submanifold of $mathbb{R}^6$ but this follows from the fact that $f$ is a (topological) embedding and it has maximum constant rank.
$endgroup$
– Dante Grevino
Jan 3 at 23:27
1
$begingroup$
Sorry for the confusion. I can not say that $g$ is a diffeomorphism if its codomain does not have a differential structure. But we have a definition of regular/embedded submanifold as the image of a topological embedding that has maximum constant rank and a result following this definition give us the differential structure of such a submanifold. Let me look for a reference.
$endgroup$
– Dante Grevino
Jan 3 at 23:41
1
$begingroup$
See Proposition 5.2 in John M. Lee, Introduction to smooth manifolds, Springer, 2013, second edition.
$endgroup$
– Dante Grevino
Jan 3 at 23:53
|
show 4 more comments
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%2f3061115%2fthe-image-of-sphere-under-the-map-f-x-y-z-to-x2-y2-z2-sqrt2yz-sqrt%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 are right, $text{Im}(f)$ and $mathbb{R}P^n$ are diffeomorphic.
Recall that we have a quotient map $p:S^n to mathbb{R}P^n$ which identifies antipodal points. Note that $f$ is constant over the fibers of $p$ so it pass to the quotient, that is, there exists a unique differentiable function $tilde{f} : mathbb{R}P^n to mathbb{R}^6$ such that $tilde{f}circ p = f$.
Note that $tilde{f}$ is injective so its corestriction $g := tilde{f}|^{text{Im}(f)}:mathbb{R}P^n to text{Im}(f)$ is a bijection. Now, because $mathbb{R}P^n$ is compact and manifolds are Hausdorff spaces we have that $g$ is a homeomorphism and then by the Inverse Function Theorem it is a diffeomorphism.
answered Jan 3 at 23:13
Dante GrevinoDante Grevino
96319
$endgroup$
$begingroup$
Thanks. Two questions: is the quotient map smooth? (why?) And doesn't the inverse function theorem give us local diffeomorphism? I don't understand the last part. + how can I make my answer to the first part rigorous? Is it even right?
$endgroup$
– stressed out
Jan 3 at 23:18
$begingroup$
Yes, the quotient map is differentiable by definition because we endow the quotient space with the unique differential structure such that the quotient map is a local diffeomorphism. (Of course, not every quotient of a manifold is a manifold but there are results which says that under very general conditions the quotient is a manifold.)
$endgroup$
– Dante Grevino
Jan 3 at 23:27
$begingroup$
For the second question, well... you have to endow the image of $f$ with a suitable differential structure. For example you can prove that it is a regular (or embedded) submanifold of $mathbb{R}^6$ but this follows from the fact that $f$ is a (topological) embedding and it has maximum constant rank.
$endgroup$
– Dante Grevino
Jan 3 at 23:27
1
$begingroup$
Sorry for the confusion. I can not say that $g$ is a diffeomorphism if its codomain does not have a differential structure. But we have a definition of regular/embedded submanifold as the image of a topological embedding that has maximum constant rank and a result following this definition give us the differential structure of such a submanifold. Let me look for a reference.
$endgroup$
– Dante Grevino
Jan 3 at 23:41
1
$begingroup$
See Proposition 5.2 in John M. Lee, Introduction to smooth manifolds, Springer, 2013, second edition.
$endgroup$
– Dante Grevino
Jan 3 at 23:53
|
show 4 more comments
$begingroup$
You are right, $text{Im}(f)$ and $mathbb{R}P^n$ are diffeomorphic.
Recall that we have a quotient map $p:S^n to mathbb{R}P^n$ which identifies antipodal points. Note that $f$ is constant over the fibers of $p$ so it pass to the quotient, that is, there exists a unique differentiable function $tilde{f} : mathbb{R}P^n to mathbb{R}^6$ such that $tilde{f}circ p = f$.
Note that $tilde{f}$ is injective so its corestriction $g := tilde{f}|^{text{Im}(f)}:mathbb{R}P^n to text{Im}(f)$ is a bijection. Now, because $mathbb{R}P^n$ is compact and manifolds are Hausdorff spaces we have that $g$ is a homeomorphism and then by the Inverse Function Theorem it is a diffeomorphism.
answered Jan 3 at 23:13
Dante GrevinoDante Grevino
96319
$endgroup$
$begingroup$
Thanks. Two questions: is the quotient map smooth? (why?) And doesn't the inverse function theorem give us local diffeomorphism? I don't understand the last part. + how can I make my answer to the first part rigorous? Is it even right?
$endgroup$
– stressed out
Jan 3 at 23:18
$begingroup$
Yes, the quotient map is differentiable by definition because we endow the quotient space with the unique differential structure such that the quotient map is a local diffeomorphism. (Of course, not every quotient of a manifold is a manifold but there are results which says that under very general conditions the quotient is a manifold.)
$endgroup$
– Dante Grevino
Jan 3 at 23:27
$begingroup$
For the second question, well... you have to endow the image of $f$ with a suitable differential structure. For example you can prove that it is a regular (or embedded) submanifold of $mathbb{R}^6$ but this follows from the fact that $f$ is a (topological) embedding and it has maximum constant rank.
$endgroup$
– Dante Grevino
Jan 3 at 23:27
1
$begingroup$
Sorry for the confusion. I can not say that $g$ is a diffeomorphism if its codomain does not have a differential structure. But we have a definition of regular/embedded submanifold as the image of a topological embedding that has maximum constant rank and a result following this definition give us the differential structure of such a submanifold. Let me look for a reference.
$endgroup$
– Dante Grevino
Jan 3 at 23:41
1
$begingroup$
See Proposition 5.2 in John M. Lee, Introduction to smooth manifolds, Springer, 2013, second edition.
$endgroup$
– Dante Grevino
Jan 3 at 23:53
|
show 4 more comments
$begingroup$
You are right, $text{Im}(f)$ and $mathbb{R}P^n$ are diffeomorphic.
Recall that we have a quotient map $p:S^n to mathbb{R}P^n$ which identifies antipodal points. Note that $f$ is constant over the fibers of $p$ so it pass to the quotient, that is, there exists a unique differentiable function $tilde{f} : mathbb{R}P^n to mathbb{R}^6$ such that $tilde{f}circ p = f$.
Note that $tilde{f}$ is injective so its corestriction $g := tilde{f}|^{text{Im}(f)}:mathbb{R}P^n to text{Im}(f)$ is a bijection. Now, because $mathbb{R}P^n$ is compact and manifolds are Hausdorff spaces we have that $g$ is a homeomorphism and then by the Inverse Function Theorem it is a diffeomorphism.
answered Jan 3 at 23:13
Dante GrevinoDante Grevino
96319
$endgroup$
You are right, $text{Im}(f)$ and $mathbb{R}P^n$ are diffeomorphic.
Recall that we have a quotient map $p:S^n to mathbb{R}P^n$ which identifies antipodal points. Note that $f$ is constant over the fibers of $p$ so it pass to the quotient, that is, there exists a unique differentiable function $tilde{f} : mathbb{R}P^n to mathbb{R}^6$ such that $tilde{f}circ p = f$.
Note that $tilde{f}$ is injective so its corestriction $g := tilde{f}|^{text{Im}(f)}:mathbb{R}P^n to text{Im}(f)$ is a bijection. Now, because $mathbb{R}P^n$ is compact and manifolds are Hausdorff spaces we have that $g$ is a homeomorphism and then by the Inverse Function Theorem it is a diffeomorphism.
answered Jan 3 at 23:13
Dante GrevinoDante Grevino
96319
answered Jan 3 at 23:13
Dante GrevinoDante Grevino
96319
answered Jan 3 at 23:13
Dante GrevinoDante Grevino
96319
answered Jan 3 at 23:13
Dante GrevinoDante Grevino
96319
96319
$begingroup$
Thanks. Two questions: is the quotient map smooth? (why?) And doesn't the inverse function theorem give us local diffeomorphism? I don't understand the last part. + how can I make my answer to the first part rigorous? Is it even right?
$endgroup$
– stressed out
Jan 3 at 23:18
$begingroup$
Yes, the quotient map is differentiable by definition because we endow the quotient space with the unique differential structure such that the quotient map is a local diffeomorphism. (Of course, not every quotient of a manifold is a manifold but there are results which says that under very general conditions the quotient is a manifold.)
$endgroup$
– Dante Grevino
Jan 3 at 23:27
$begingroup$
For the second question, well... you have to endow the image of $f$ with a suitable differential structure. For example you can prove that it is a regular (or embedded) submanifold of $mathbb{R}^6$ but this follows from the fact that $f$ is a (topological) embedding and it has maximum constant rank.
$endgroup$
– Dante Grevino
Jan 3 at 23:27
1
$begingroup$
Sorry for the confusion. I can not say that $g$ is a diffeomorphism if its codomain does not have a differential structure. But we have a definition of regular/embedded submanifold as the image of a topological embedding that has maximum constant rank and a result following this definition give us the differential structure of such a submanifold. Let me look for a reference.
$endgroup$
– Dante Grevino
Jan 3 at 23:41
1
$begingroup$
See Proposition 5.2 in John M. Lee, Introduction to smooth manifolds, Springer, 2013, second edition.
$endgroup$
– Dante Grevino
Jan 3 at 23:53
|
show 4 more comments
$begingroup$
Thanks. Two questions: is the quotient map smooth? (why?) And doesn't the inverse function theorem give us local diffeomorphism? I don't understand the last part. + how can I make my answer to the first part rigorous? Is it even right?
$endgroup$
– stressed out
Jan 3 at 23:18
$begingroup$
Yes, the quotient map is differentiable by definition because we endow the quotient space with the unique differential structure such that the quotient map is a local diffeomorphism. (Of course, not every quotient of a manifold is a manifold but there are results which says that under very general conditions the quotient is a manifold.)
$endgroup$
– Dante Grevino
Jan 3 at 23:27
$begingroup$
For the second question, well... you have to endow the image of $f$ with a suitable differential structure. For example you can prove that it is a regular (or embedded) submanifold of $mathbb{R}^6$ but this follows from the fact that $f$ is a (topological) embedding and it has maximum constant rank.
$endgroup$
– Dante Grevino
Jan 3 at 23:27
1
$begingroup$
Sorry for the confusion. I can not say that $g$ is a diffeomorphism if its codomain does not have a differential structure. But we have a definition of regular/embedded submanifold as the image of a topological embedding that has maximum constant rank and a result following this definition give us the differential structure of such a submanifold. Let me look for a reference.
$endgroup$
– Dante Grevino
Jan 3 at 23:41
1
$begingroup$
See Proposition 5.2 in John M. Lee, Introduction to smooth manifolds, Springer, 2013, second edition.
$endgroup$
– Dante Grevino
Jan 3 at 23:53
$begingroup$
Thanks. Two questions: is the quotient map smooth? (why?) And doesn't the inverse function theorem give us local diffeomorphism? I don't understand the last part. + how can I make my answer to the first part rigorous? Is it even right?
$endgroup$
– stressed out
Jan 3 at 23:18
$begingroup$
Thanks. Two questions: is the quotient map smooth? (why?) And doesn't the inverse function theorem give us local diffeomorphism? I don't understand the last part. + how can I make my answer to the first part rigorous? Is it even right?
$endgroup$
– stressed out
Jan 3 at 23:18
$begingroup$
Yes, the quotient map is differentiable by definition because we endow the quotient space with the unique differential structure such that the quotient map is a local diffeomorphism. (Of course, not every quotient of a manifold is a manifold but there are results which says that under very general conditions the quotient is a manifold.)
$endgroup$
– Dante Grevino
Jan 3 at 23:27
$begingroup$
Yes, the quotient map is differentiable by definition because we endow the quotient space with the unique differential structure such that the quotient map is a local diffeomorphism. (Of course, not every quotient of a manifold is a manifold but there are results which says that under very general conditions the quotient is a manifold.)
$endgroup$
– Dante Grevino
Jan 3 at 23:27
$begingroup$
For the second question, well... you have to endow the image of $f$ with a suitable differential structure. For example you can prove that it is a regular (or embedded) submanifold of $mathbb{R}^6$ but this follows from the fact that $f$ is a (topological) embedding and it has maximum constant rank.
$endgroup$
– Dante Grevino
Jan 3 at 23:27
$begingroup$
For the second question, well... you have to endow the image of $f$ with a suitable differential structure. For example you can prove that it is a regular (or embedded) submanifold of $mathbb{R}^6$ but this follows from the fact that $f$ is a (topological) embedding and it has maximum constant rank.
$endgroup$
– Dante Grevino
Jan 3 at 23:27
1
1
$begingroup$
Sorry for the confusion. I can not say that $g$ is a diffeomorphism if its codomain does not have a differential structure. But we have a definition of regular/embedded submanifold as the image of a topological embedding that has maximum constant rank and a result following this definition give us the differential structure of such a submanifold. Let me look for a reference.
$endgroup$
– Dante Grevino
Jan 3 at 23:41
$begingroup$
Sorry for the confusion. I can not say that $g$ is a diffeomorphism if its codomain does not have a differential structure. But we have a definition of regular/embedded submanifold as the image of a topological embedding that has maximum constant rank and a result following this definition give us the differential structure of such a submanifold. Let me look for a reference.
$endgroup$
– Dante Grevino
Jan 3 at 23:41
1
1
$begingroup$
See Proposition 5.2 in John M. Lee, Introduction to smooth manifolds, Springer, 2013, second edition.
$endgroup$
– Dante Grevino
Jan 3 at 23:53
$begingroup$
See Proposition 5.2 in John M. Lee, Introduction to smooth manifolds, Springer, 2013, second edition.
$endgroup$
– Dante Grevino
Jan 3 at 23:53
|
show 4 more comments
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%2f3061115%2fthe-image-of-sphere-under-the-map-f-x-y-z-to-x2-y2-z2-sqrt2yz-sqrt%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
