Non vanishing one form on the circle
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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 Jan 3 at 18:34
Emilio MinichielloEmilio Minichiello
33417
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add a comment |
$begingroup$
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 Jan 3 at 18:34
Emilio MinichielloEmilio Minichiello
33417
$endgroup$
$begingroup$
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!
$endgroup$
– Ted Shifrin
Jan 3 at 19:28
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But I multiplied the vector field by a smooth function that vanishes at say a single point.
$endgroup$
– Emilio Minichiello
Jan 3 at 20:07
1
$begingroup$
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.
$endgroup$
– Amitai Yuval
Jan 3 at 20:58
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See that's what I guessed, but Tu never properly defines "nowhere vanishing" so I thought I was mistaken. Thank you!
$endgroup$
– Emilio Minichiello
Jan 3 at 21:01
add a comment |
$begingroup$
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 Jan 3 at 18:34
Emilio MinichielloEmilio Minichiello
33417
$endgroup$
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 Jan 3 at 18:34
Emilio MinichielloEmilio Minichiello
33417
asked Jan 3 at 18:34
Emilio MinichielloEmilio Minichiello
33417
edited Jan 3 at 20:08
Emilio Minichiello
asked Jan 3 at 18:34
Emilio MinichielloEmilio Minichiello
33417
asked Jan 3 at 18:34
Emilio MinichielloEmilio Minichiello
33417
asked Jan 3 at 18:34
Emilio MinichielloEmilio Minichiello
33417
33417
$begingroup$
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!
$endgroup$
– Ted Shifrin
Jan 3 at 19:28
$begingroup$
But I multiplied the vector field by a smooth function that vanishes at say a single point.
$endgroup$
– Emilio Minichiello
Jan 3 at 20:07
1
$begingroup$
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.
$endgroup$
– Amitai Yuval
Jan 3 at 20:58
$begingroup$
See that's what I guessed, but Tu never properly defines "nowhere vanishing" so I thought I was mistaken. Thank you!
$endgroup$
– Emilio Minichiello
Jan 3 at 21:01
add a comment |
$begingroup$
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!
$endgroup$
– Ted Shifrin
Jan 3 at 19:28
$begingroup$
But I multiplied the vector field by a smooth function that vanishes at say a single point.
$endgroup$
– Emilio Minichiello
Jan 3 at 20:07
1
$begingroup$
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.
$endgroup$
– Amitai Yuval
Jan 3 at 20:58
$begingroup$
See that's what I guessed, but Tu never properly defines "nowhere vanishing" so I thought I was mistaken. Thank you!
$endgroup$
– Emilio Minichiello
Jan 3 at 21:01
$begingroup$
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!
$endgroup$
– Ted Shifrin
Jan 3 at 19:28
$begingroup$
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!
$endgroup$
– Ted Shifrin
Jan 3 at 19:28
$begingroup$
But I multiplied the vector field by a smooth function that vanishes at say a single point.
$endgroup$
– Emilio Minichiello
Jan 3 at 20:07
$begingroup$
But I multiplied the vector field by a smooth function that vanishes at say a single point.
$endgroup$
– Emilio Minichiello
Jan 3 at 20:07
1
1
$begingroup$
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.
$endgroup$
– Amitai Yuval
Jan 3 at 20:58
$begingroup$
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.
$endgroup$
– Amitai Yuval
Jan 3 at 20:58
$begingroup$
See that's what I guessed, but Tu never properly defines "nowhere vanishing" so I thought I was mistaken. Thank you!
$endgroup$
– Emilio Minichiello
Jan 3 at 21:01
$begingroup$
See that's what I guessed, but Tu never properly defines "nowhere vanishing" so I thought I was mistaken. Thank you!
$endgroup$
– Emilio Minichiello
Jan 3 at 21:01
add a comment |
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$begingroup$
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!
$endgroup$
– Ted Shifrin
Jan 3 at 19:28
$begingroup$
But I multiplied the vector field by a smooth function that vanishes at say a single point.
$endgroup$
– Emilio Minichiello
Jan 3 at 20:07
1
$begingroup$
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.
$endgroup$
– Amitai Yuval
Jan 3 at 20:58
$begingroup$
See that's what I guessed, but Tu never properly defines "nowhere vanishing" so I thought I was mistaken. Thank you!
$endgroup$
– Emilio Minichiello
Jan 3 at 21:01