Non vanishing one form on the circle
I am trying to understand this classical example of a nowhere vanishing differential one form from Tu's Introduction to Manifolds
Let $omega = -ydx + xdy$ be restricted to $S^1 subset mathbb{R}^2$. Then $omega(X)$ is nowhere vanishing for every vector field $X in mathfrak{X}(S^1)$.
Now I understand that if $iota: S^1 hookrightarrow mathbb{R}^2$ is the inclusion map and $frac{d}{dt} in mathfrak{X}(S^1)$ is the vector field $frac{d}{dt} = c'(t) = (-sin t, cos t)$, then $iota_*(frac{d}{dt}) = -y partial_x + x partial_y$, and we have $omega(iota_*(frac{d}{dt})) = 1$. But if $z(t) in C^{infty}(S^1, mathbb{R})$ I believe that $z(t) frac{d}{dt} in mathfrak{X}(S^1)$ as well. Now my question, is the pushforward $iota_*(z(t) frac{d}{dt}) = z(t) (-y partial_x + x partial_y)?$ and if it is, then couldn't I set $z(t_0) = 0$ for some $t_0$ and thus $omega$ wouldn't be non vanishing?
differential-geometry differential-forms
asked 2 days ago
Emilio Minichiello
3297
add a comment |
I am trying to understand this classical example of a nowhere vanishing differential one form from Tu's Introduction to Manifolds
Let $omega = -ydx + xdy$ be restricted to $S^1 subset mathbb{R}^2$. Then $omega(X)$ is nowhere vanishing for every vector field $X in mathfrak{X}(S^1)$.
Now I understand that if $iota: S^1 hookrightarrow mathbb{R}^2$ is the inclusion map and $frac{d}{dt} in mathfrak{X}(S^1)$ is the vector field $frac{d}{dt} = c'(t) = (-sin t, cos t)$, then $iota_*(frac{d}{dt}) = -y partial_x + x partial_y$, and we have $omega(iota_*(frac{d}{dt})) = 1$. But if $z(t) in C^{infty}(S^1, mathbb{R})$ I believe that $z(t) frac{d}{dt} in mathfrak{X}(S^1)$ as well. Now my question, is the pushforward $iota_*(z(t) frac{d}{dt}) = z(t) (-y partial_x + x partial_y)?$ and if it is, then couldn't I set $z(t_0) = 0$ for some $t_0$ and thus $omega$ wouldn't be non vanishing?
differential-geometry differential-forms
asked 2 days ago
Emilio Minichiello
3297
Well, of course: If you multiply any $1$-form by the zero function, you end up with the zero $1$-form. But that is no longer the $omega$ you started with!
– Ted Shifrin
2 days ago
But I multiplied the vector field by a smooth function that vanishes at say a single point.
– Emilio Minichiello
2 days ago
1
Of course, if your vector field $X$ vanishes at a point, then so does $omega(X)$. No $1$-form on any manifold can satisfy your requirement. The best you can get is a $1$-form $omega$, such that for any non-vanishing $X$ the contraction $omega(X)$ is non-vanishing. The differential form $omega$ in question indeed satisfies that.
– Amitai Yuval
2 days ago
See that's what I guessed, but Tu never properly defines "nowhere vanishing" so I thought I was mistaken. Thank you!
– Emilio Minichiello
2 days ago
add a comment |
I am trying to understand this classical example of a nowhere vanishing differential one form from Tu's Introduction to Manifolds
Let $omega = -ydx + xdy$ be restricted to $S^1 subset mathbb{R}^2$. Then $omega(X)$ is nowhere vanishing for every vector field $X in mathfrak{X}(S^1)$.
Now I understand that if $iota: S^1 hookrightarrow mathbb{R}^2$ is the inclusion map and $frac{d}{dt} in mathfrak{X}(S^1)$ is the vector field $frac{d}{dt} = c'(t) = (-sin t, cos t)$, then $iota_*(frac{d}{dt}) = -y partial_x + x partial_y$, and we have $omega(iota_*(frac{d}{dt})) = 1$. But if $z(t) in C^{infty}(S^1, mathbb{R})$ I believe that $z(t) frac{d}{dt} in mathfrak{X}(S^1)$ as well. Now my question, is the pushforward $iota_*(z(t) frac{d}{dt}) = z(t) (-y partial_x + x partial_y)?$ and if it is, then couldn't I set $z(t_0) = 0$ for some $t_0$ and thus $omega$ wouldn't be non vanishing?
differential-geometry differential-forms
asked 2 days ago
Emilio Minichiello
3297
I am trying to understand this classical example of a nowhere vanishing differential one form from Tu's Introduction to Manifolds
Let $omega = -ydx + xdy$ be restricted to $S^1 subset mathbb{R}^2$. Then $omega(X)$ is nowhere vanishing for every vector field $X in mathfrak{X}(S^1)$.
Now I understand that if $iota: S^1 hookrightarrow mathbb{R}^2$ is the inclusion map and $frac{d}{dt} in mathfrak{X}(S^1)$ is the vector field $frac{d}{dt} = c'(t) = (-sin t, cos t)$, then $iota_*(frac{d}{dt}) = -y partial_x + x partial_y$, and we have $omega(iota_*(frac{d}{dt})) = 1$. But if $z(t) in C^{infty}(S^1, mathbb{R})$ I believe that $z(t) frac{d}{dt} in mathfrak{X}(S^1)$ as well. Now my question, is the pushforward $iota_*(z(t) frac{d}{dt}) = z(t) (-y partial_x + x partial_y)?$ and if it is, then couldn't I set $z(t_0) = 0$ for some $t_0$ and thus $omega$ wouldn't be non vanishing?
differential-geometry differential-forms
differential-geometry differential-forms
asked 2 days ago
Emilio Minichiello
3297
asked 2 days ago
Emilio Minichiello
3297
edited 2 days ago
asked 2 days ago
Emilio Minichiello
3297
asked 2 days ago
Emilio Minichiello
3297
asked 2 days ago
Emilio Minichiello
3297
3297
Well, of course: If you multiply any $1$-form by the zero function, you end up with the zero $1$-form. But that is no longer the $omega$ you started with!
– Ted Shifrin
2 days ago
But I multiplied the vector field by a smooth function that vanishes at say a single point.
– Emilio Minichiello
2 days ago
1
Of course, if your vector field $X$ vanishes at a point, then so does $omega(X)$. No $1$-form on any manifold can satisfy your requirement. The best you can get is a $1$-form $omega$, such that for any non-vanishing $X$ the contraction $omega(X)$ is non-vanishing. The differential form $omega$ in question indeed satisfies that.
– Amitai Yuval
2 days ago
See that's what I guessed, but Tu never properly defines "nowhere vanishing" so I thought I was mistaken. Thank you!
– Emilio Minichiello
2 days ago
add a comment |
Well, of course: If you multiply any $1$-form by the zero function, you end up with the zero $1$-form. But that is no longer the $omega$ you started with!
– Ted Shifrin
2 days ago
But I multiplied the vector field by a smooth function that vanishes at say a single point.
– Emilio Minichiello
2 days ago
1
Of course, if your vector field $X$ vanishes at a point, then so does $omega(X)$. No $1$-form on any manifold can satisfy your requirement. The best you can get is a $1$-form $omega$, such that for any non-vanishing $X$ the contraction $omega(X)$ is non-vanishing. The differential form $omega$ in question indeed satisfies that.
– Amitai Yuval
2 days ago
See that's what I guessed, but Tu never properly defines "nowhere vanishing" so I thought I was mistaken. Thank you!
– Emilio Minichiello
2 days ago
Well, of course: If you multiply any $1$-form by the zero function, you end up with the zero $1$-form. But that is no longer the $omega$ you started with!
– Ted Shifrin
2 days ago
Well, of course: If you multiply any $1$-form by the zero function, you end up with the zero $1$-form. But that is no longer the $omega$ you started with!
– Ted Shifrin
2 days ago
But I multiplied the vector field by a smooth function that vanishes at say a single point.
– Emilio Minichiello
2 days ago
But I multiplied the vector field by a smooth function that vanishes at say a single point.
– Emilio Minichiello
2 days ago
1
1
Of course, if your vector field $X$ vanishes at a point, then so does $omega(X)$. No $1$-form on any manifold can satisfy your requirement. The best you can get is a $1$-form $omega$, such that for any non-vanishing $X$ the contraction $omega(X)$ is non-vanishing. The differential form $omega$ in question indeed satisfies that.
– Amitai Yuval
2 days ago
Of course, if your vector field $X$ vanishes at a point, then so does $omega(X)$. No $1$-form on any manifold can satisfy your requirement. The best you can get is a $1$-form $omega$, such that for any non-vanishing $X$ the contraction $omega(X)$ is non-vanishing. The differential form $omega$ in question indeed satisfies that.
– Amitai Yuval
2 days ago
See that's what I guessed, but Tu never properly defines "nowhere vanishing" so I thought I was mistaken. Thank you!
– Emilio Minichiello
2 days ago
See that's what I guessed, but Tu never properly defines "nowhere vanishing" so I thought I was mistaken. Thank you!
– Emilio Minichiello
2 days ago
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%2f3060864%2fnon-vanishing-one-form-on-the-circle%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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f3060864%2fnon-vanishing-one-form-on-the-circle%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
Well, of course: If you multiply any $1$-form by the zero function, you end up with the zero $1$-form. But that is no longer the $omega$ you started with!
– Ted Shifrin
2 days ago
But I multiplied the vector field by a smooth function that vanishes at say a single point.
– Emilio Minichiello
2 days ago
1
Of course, if your vector field $X$ vanishes at a point, then so does $omega(X)$. No $1$-form on any manifold can satisfy your requirement. The best you can get is a $1$-form $omega$, such that for any non-vanishing $X$ the contraction $omega(X)$ is non-vanishing. The differential form $omega$ in question indeed satisfies that.
– Amitai Yuval
2 days ago
See that's what I guessed, but Tu never properly defines "nowhere vanishing" so I thought I was mistaken. Thank you!
– Emilio Minichiello
2 days ago