Mixing
Apparent contradiction to Poincaré lemma?
$begingroup$
I have learned that, if we have a connected, oriented and compact n-dimensional manifold, the top de Rham cohomology is isomorphic to $mathbb{R}$, i.e
$$ H_{text{dR}}^n(M) cong mathbb{R}.$$
However, Poincaré's lemma states that for any star-shaped subset of $ mathcal{U} subset mathbb{R}^n$, it holds that
$$ {H_{text{dR}}^k(mathcal{U}) = 0} qquad text{for all }k neq 0.$$
Consider now the ball of radius $1$, this is evidently star shaped. Furthermore, it seems to me that it also suffices the properties in the first result. What is wrong in this reasoning?
differential-geometry de-rham-cohomology
asked Jan 24 at 16:14
GregGreg
183112
$endgroup$
|
show 2 more comments
$begingroup$
I have learned that, if we have a connected, oriented and compact n-dimensional manifold, the top de Rham cohomology is isomorphic to $mathbb{R}$, i.e
$$ H_{text{dR}}^n(M) cong mathbb{R}.$$
However, Poincaré's lemma states that for any star-shaped subset of $ mathcal{U} subset mathbb{R}^n$, it holds that
$$ {H_{text{dR}}^k(mathcal{U}) = 0} qquad text{for all }k neq 0.$$
Consider now the ball of radius $1$, this is evidently star shaped. Furthermore, it seems to me that it also suffices the properties in the first result. What is wrong in this reasoning?
differential-geometry de-rham-cohomology
asked Jan 24 at 16:14
GregGreg
183112
$endgroup$
1
$begingroup$
In just the topological setting, the idea is that because $mathcal U$ is star-shaped, it is contractible – i.e. it can be continuously deformed to a point, and because (singular) cohomology is invariant under such continuous deformations, the (singular) cohomology of $mathcal U$ must be that of a point, i.e. trivial for all $k neq 0$. I suspect something similar is true in the de Rham setting?
$endgroup$
– Rylee Lyman
Jan 24 at 16:20
1
$begingroup$
In your first statement you listed the property 'connected' twice. Is that a typo and is there a property missing?
$endgroup$
– quarague
Jan 24 at 16:20
1
$begingroup$
one problem with the open ball is that although it is bounded, it is not compact?
$endgroup$
– Rylee Lyman
Jan 24 at 16:22
$begingroup$
Yes, it had to be compact. Thanks!
$endgroup$
– Greg
Jan 24 at 16:23
8
$begingroup$
There is an additional assumption needed to conclude $H^n(M)cong mathbb{R}$: $M$ must have no boundary. This fails in your ball example, if you take the closed ball.
$endgroup$
– Jason DeVito
Jan 24 at 16:26
|
show 2 more comments
$begingroup$
I have learned that, if we have a connected, oriented and compact n-dimensional manifold, the top de Rham cohomology is isomorphic to $mathbb{R}$, i.e
$$ H_{text{dR}}^n(M) cong mathbb{R}.$$
However, Poincaré's lemma states that for any star-shaped subset of $ mathcal{U} subset mathbb{R}^n$, it holds that
$$ {H_{text{dR}}^k(mathcal{U}) = 0} qquad text{for all }k neq 0.$$
Consider now the ball of radius $1$, this is evidently star shaped. Furthermore, it seems to me that it also suffices the properties in the first result. What is wrong in this reasoning?
differential-geometry de-rham-cohomology
asked Jan 24 at 16:14
GregGreg
183112
$endgroup$
I have learned that, if we have a connected, oriented and compact n-dimensional manifold, the top de Rham cohomology is isomorphic to $mathbb{R}$, i.e
$$ H_{text{dR}}^n(M) cong mathbb{R}.$$
However, Poincaré's lemma states that for any star-shaped subset of $ mathcal{U} subset mathbb{R}^n$, it holds that
$$ {H_{text{dR}}^k(mathcal{U}) = 0} qquad text{for all }k neq 0.$$
Consider now the ball of radius $1$, this is evidently star shaped. Furthermore, it seems to me that it also suffices the properties in the first result. What is wrong in this reasoning?
differential-geometry de-rham-cohomology
differential-geometry de-rham-cohomology
asked Jan 24 at 16:14
GregGreg
183112
asked Jan 24 at 16:14
GregGreg
183112
edited Jan 25 at 16:08
Greg
asked Jan 24 at 16:14
GregGreg
183112
asked Jan 24 at 16:14
GregGreg
183112
asked Jan 24 at 16:14
GregGreg
183112
183112
1
$begingroup$
In just the topological setting, the idea is that because $mathcal U$ is star-shaped, it is contractible – i.e. it can be continuously deformed to a point, and because (singular) cohomology is invariant under such continuous deformations, the (singular) cohomology of $mathcal U$ must be that of a point, i.e. trivial for all $k neq 0$. I suspect something similar is true in the de Rham setting?
$endgroup$
– Rylee Lyman
Jan 24 at 16:20
1
$begingroup$
In your first statement you listed the property 'connected' twice. Is that a typo and is there a property missing?
$endgroup$
– quarague
Jan 24 at 16:20
1
$begingroup$
one problem with the open ball is that although it is bounded, it is not compact?
$endgroup$
– Rylee Lyman
Jan 24 at 16:22
$begingroup$
Yes, it had to be compact. Thanks!
$endgroup$
– Greg
Jan 24 at 16:23
8
$begingroup$
There is an additional assumption needed to conclude $H^n(M)cong mathbb{R}$: $M$ must have no boundary. This fails in your ball example, if you take the closed ball.
$endgroup$
– Jason DeVito
Jan 24 at 16:26
|
show 2 more comments
1
$begingroup$
In just the topological setting, the idea is that because $mathcal U$ is star-shaped, it is contractible – i.e. it can be continuously deformed to a point, and because (singular) cohomology is invariant under such continuous deformations, the (singular) cohomology of $mathcal U$ must be that of a point, i.e. trivial for all $k neq 0$. I suspect something similar is true in the de Rham setting?
$endgroup$
– Rylee Lyman
Jan 24 at 16:20
1
$begingroup$
In your first statement you listed the property 'connected' twice. Is that a typo and is there a property missing?
$endgroup$
– quarague
Jan 24 at 16:20
1
$begingroup$
one problem with the open ball is that although it is bounded, it is not compact?
$endgroup$
– Rylee Lyman
Jan 24 at 16:22
$begingroup$
Yes, it had to be compact. Thanks!
$endgroup$
– Greg
Jan 24 at 16:23
8
$begingroup$
There is an additional assumption needed to conclude $H^n(M)cong mathbb{R}$: $M$ must have no boundary. This fails in your ball example, if you take the closed ball.
$endgroup$
– Jason DeVito
Jan 24 at 16:26
1
1
$begingroup$
In just the topological setting, the idea is that because $mathcal U$ is star-shaped, it is contractible – i.e. it can be continuously deformed to a point, and because (singular) cohomology is invariant under such continuous deformations, the (singular) cohomology of $mathcal U$ must be that of a point, i.e. trivial for all $k neq 0$. I suspect something similar is true in the de Rham setting?
$endgroup$
– Rylee Lyman
Jan 24 at 16:20
$begingroup$
In just the topological setting, the idea is that because $mathcal U$ is star-shaped, it is contractible – i.e. it can be continuously deformed to a point, and because (singular) cohomology is invariant under such continuous deformations, the (singular) cohomology of $mathcal U$ must be that of a point, i.e. trivial for all $k neq 0$. I suspect something similar is true in the de Rham setting?
$endgroup$
– Rylee Lyman
Jan 24 at 16:20
1
1
$begingroup$
In your first statement you listed the property 'connected' twice. Is that a typo and is there a property missing?
$endgroup$
– quarague
Jan 24 at 16:20
$begingroup$
In your first statement you listed the property 'connected' twice. Is that a typo and is there a property missing?
$endgroup$
– quarague
Jan 24 at 16:20
1
1
$begingroup$
one problem with the open ball is that although it is bounded, it is not compact?
$endgroup$
– Rylee Lyman
Jan 24 at 16:22
$begingroup$
one problem with the open ball is that although it is bounded, it is not compact?
$endgroup$
– Rylee Lyman
Jan 24 at 16:22
$begingroup$
Yes, it had to be compact. Thanks!
$endgroup$
– Greg
Jan 24 at 16:23
$begingroup$
Yes, it had to be compact. Thanks!
$endgroup$
– Greg
Jan 24 at 16:23
8
8
$begingroup$
There is an additional assumption needed to conclude $H^n(M)cong mathbb{R}$: $M$ must have no boundary. This fails in your ball example, if you take the closed ball.
$endgroup$
– Jason DeVito
Jan 24 at 16:26
$begingroup$
There is an additional assumption needed to conclude $H^n(M)cong mathbb{R}$: $M$ must have no boundary. This fails in your ball example, if you take the closed ball.
$endgroup$
– Jason DeVito
Jan 24 at 16:26
|
show 2 more comments
1 Answer
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I know the statement in the form: "top de-Rham cohomology of compact, connected, oriented manifold without boundary is isomorphic to $mathbb{R}$".
answered Jan 24 at 16:28
Rybin DmitryRybin Dmitry
1236
$endgroup$
1
$begingroup$
That's where I made my mistake. I swapped books, where in the first one a manifold is defined without boundary, and in the second one it is defined with boundary. Thank you very much!
$endgroup$
– Greg
Jan 25 at 16:09
add a comment |
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$begingroup$
I know the statement in the form: "top de-Rham cohomology of compact, connected, oriented manifold without boundary is isomorphic to $mathbb{R}$".
answered Jan 24 at 16:28
Rybin DmitryRybin Dmitry
1236
$endgroup$
1
$begingroup$
That's where I made my mistake. I swapped books, where in the first one a manifold is defined without boundary, and in the second one it is defined with boundary. Thank you very much!
$endgroup$
– Greg
Jan 25 at 16:09
add a comment |
$begingroup$
I know the statement in the form: "top de-Rham cohomology of compact, connected, oriented manifold without boundary is isomorphic to $mathbb{R}$".
answered Jan 24 at 16:28
Rybin DmitryRybin Dmitry
1236
$endgroup$
1
$begingroup$
That's where I made my mistake. I swapped books, where in the first one a manifold is defined without boundary, and in the second one it is defined with boundary. Thank you very much!
$endgroup$
– Greg
Jan 25 at 16:09
add a comment |
$begingroup$
I know the statement in the form: "top de-Rham cohomology of compact, connected, oriented manifold without boundary is isomorphic to $mathbb{R}$".
answered Jan 24 at 16:28
Rybin DmitryRybin Dmitry
1236
$endgroup$
I know the statement in the form: "top de-Rham cohomology of compact, connected, oriented manifold without boundary is isomorphic to $mathbb{R}$".
answered Jan 24 at 16:28
Rybin DmitryRybin Dmitry
1236
answered Jan 24 at 16:28
Rybin DmitryRybin Dmitry
1236
answered Jan 24 at 16:28
Rybin DmitryRybin Dmitry
1236
answered Jan 24 at 16:28
Rybin DmitryRybin Dmitry
1236
1236
1
$begingroup$
That's where I made my mistake. I swapped books, where in the first one a manifold is defined without boundary, and in the second one it is defined with boundary. Thank you very much!
$endgroup$
– Greg
Jan 25 at 16:09
add a comment |
1
$begingroup$
That's where I made my mistake. I swapped books, where in the first one a manifold is defined without boundary, and in the second one it is defined with boundary. Thank you very much!
$endgroup$
– Greg
Jan 25 at 16:09
1
1
$begingroup$
That's where I made my mistake. I swapped books, where in the first one a manifold is defined without boundary, and in the second one it is defined with boundary. Thank you very much!
$endgroup$
– Greg
Jan 25 at 16:09
$begingroup$
That's where I made my mistake. I swapped books, where in the first one a manifold is defined without boundary, and in the second one it is defined with boundary. Thank you very much!
$endgroup$
– Greg
Jan 25 at 16:09
add a comment |
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$begingroup$
In just the topological setting, the idea is that because $mathcal U$ is star-shaped, it is contractible – i.e. it can be continuously deformed to a point, and because (singular) cohomology is invariant under such continuous deformations, the (singular) cohomology of $mathcal U$ must be that of a point, i.e. trivial for all $k neq 0$. I suspect something similar is true in the de Rham setting?
$endgroup$
– Rylee Lyman
Jan 24 at 16:20
1
$begingroup$
In your first statement you listed the property 'connected' twice. Is that a typo and is there a property missing?
$endgroup$
– quarague
Jan 24 at 16:20
1
$begingroup$
one problem with the open ball is that although it is bounded, it is not compact?
$endgroup$
– Rylee Lyman
Jan 24 at 16:22
$begingroup$
Yes, it had to be compact. Thanks!
$endgroup$
– Greg
Jan 24 at 16:23
8
$begingroup$
There is an additional assumption needed to conclude $H^n(M)cong mathbb{R}$: $M$ must have no boundary. This fails in your ball example, if you take the closed ball.
$endgroup$
– Jason DeVito
Jan 24 at 16:26