Do you need a differentiable homotopy for the Poincaré Lemma
up vote
0
down vote
favorite
The Poincaré Lemma (Amann-Escher XI.3.11) states
If $X$ is contractible, then every closed differential form on $X$ is exact,
where contractible means that the identity on $X$ is null-homotic. Amann-Escher however requires such a homotpy to be smooth ($C^infty$).
So my question is whether the homotopy really must be smooth for the lemma to hold. In particular
I guess there are examples of manifolds (or maybe subsets of $mathbb{R}^n$) which are "topologically contractible" but not "smoothly contractible"?
should "topologically contractible" not be strong enough for the theorem to hold, would something like a "$C^2$-contractible" be enough?
differential-geometry differential-topology de-rham-cohomology
asked Jun 1 at 16:02
0x539
414213
add a comment |
up vote
0
down vote
favorite
The Poincaré Lemma (Amann-Escher XI.3.11) states
If $X$ is contractible, then every closed differential form on $X$ is exact,
where contractible means that the identity on $X$ is null-homotic. Amann-Escher however requires such a homotpy to be smooth ($C^infty$).
So my question is whether the homotopy really must be smooth for the lemma to hold. In particular
I guess there are examples of manifolds (or maybe subsets of $mathbb{R}^n$) which are "topologically contractible" but not "smoothly contractible"?
should "topologically contractible" not be strong enough for the theorem to hold, would something like a "$C^2$-contractible" be enough?
differential-geometry differential-topology de-rham-cohomology
asked Jun 1 at 16:02
0x539
414213
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
The Poincaré Lemma (Amann-Escher XI.3.11) states
If $X$ is contractible, then every closed differential form on $X$ is exact,
where contractible means that the identity on $X$ is null-homotic. Amann-Escher however requires such a homotpy to be smooth ($C^infty$).
So my question is whether the homotopy really must be smooth for the lemma to hold. In particular
I guess there are examples of manifolds (or maybe subsets of $mathbb{R}^n$) which are "topologically contractible" but not "smoothly contractible"?
should "topologically contractible" not be strong enough for the theorem to hold, would something like a "$C^2$-contractible" be enough?
differential-geometry differential-topology de-rham-cohomology
asked Jun 1 at 16:02
0x539
414213
The Poincaré Lemma (Amann-Escher XI.3.11) states
If $X$ is contractible, then every closed differential form on $X$ is exact,
where contractible means that the identity on $X$ is null-homotic. Amann-Escher however requires such a homotpy to be smooth ($C^infty$).
So my question is whether the homotopy really must be smooth for the lemma to hold. In particular
I guess there are examples of manifolds (or maybe subsets of $mathbb{R}^n$) which are "topologically contractible" but not "smoothly contractible"?
should "topologically contractible" not be strong enough for the theorem to hold, would something like a "$C^2$-contractible" be enough?
differential-geometry differential-topology de-rham-cohomology
differential-geometry differential-topology de-rham-cohomology
asked Jun 1 at 16:02
0x539
414213
asked Jun 1 at 16:02
0x539
414213
edited Nov 22 at 19:02
asked Jun 1 at 16:02
0x539
414213
asked Jun 1 at 16:02
0x539
414213
asked Jun 1 at 16:02
0x539
414213
414213
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
The answer to this question provides a reference for the fact that homotopic smooth maps are always smooth-homotopic, so in particular $text{Id}_M$ and $constant$ are homotopic by a $C^infty$-map.
A related standard fact you may know is that the singular (“continuous”) cohomology groups and the de Rham cohomology groups are isomorphic for smooth manifolds. This is “de Rham’s theorem”, I believe.
answered Jun 1 at 18:02
mcwiggler
45415
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
The answer to this question provides a reference for the fact that homotopic smooth maps are always smooth-homotopic, so in particular $text{Id}_M$ and $constant$ are homotopic by a $C^infty$-map.
A related standard fact you may know is that the singular (“continuous”) cohomology groups and the de Rham cohomology groups are isomorphic for smooth manifolds. This is “de Rham’s theorem”, I believe.
answered Jun 1 at 18:02
mcwiggler
45415
add a comment |
up vote
1
down vote
accepted
The answer to this question provides a reference for the fact that homotopic smooth maps are always smooth-homotopic, so in particular $text{Id}_M$ and $constant$ are homotopic by a $C^infty$-map.
A related standard fact you may know is that the singular (“continuous”) cohomology groups and the de Rham cohomology groups are isomorphic for smooth manifolds. This is “de Rham’s theorem”, I believe.
answered Jun 1 at 18:02
mcwiggler
45415
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The answer to this question provides a reference for the fact that homotopic smooth maps are always smooth-homotopic, so in particular $text{Id}_M$ and $constant$ are homotopic by a $C^infty$-map.
A related standard fact you may know is that the singular (“continuous”) cohomology groups and the de Rham cohomology groups are isomorphic for smooth manifolds. This is “de Rham’s theorem”, I believe.
answered Jun 1 at 18:02
mcwiggler
45415
The answer to this question provides a reference for the fact that homotopic smooth maps are always smooth-homotopic, so in particular $text{Id}_M$ and $constant$ are homotopic by a $C^infty$-map.
A related standard fact you may know is that the singular (“continuous”) cohomology groups and the de Rham cohomology groups are isomorphic for smooth manifolds. This is “de Rham’s theorem”, I believe.
answered Jun 1 at 18:02
mcwiggler
45415
answered Jun 1 at 18:02
mcwiggler
45415
answered Jun 1 at 18:02
mcwiggler
45415
answered Jun 1 at 18:02
mcwiggler
45415
45415
add a comment |
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.
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%2f2804491%2fdo-you-need-a-differentiable-homotopy-for-the-poincar%25c3%25a9-lemma%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
