Do you need a differentiable homotopy for the Poincaré Lemma
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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
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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
451213
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
451213
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
451213
asked Jun 1 at 16:02
0x539
451213
edited Nov 22 at 19:02
asked Jun 1 at 16:02
0x539
451213
asked Jun 1 at 16:02
0x539
451213
asked Jun 1 at 16:02
0x539
451213
451213
add a comment |
add a comment |
1 Answer
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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
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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 |
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