Is every top dimensional form a volume form?
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Assuming that I have an orientable Riemannian manifold, can every top dimensional form be shown to be a volume form? If not, can all top dimensional forms be decomposed into a volume form and an exact form?
differential-geometry riemannian-geometry
asked Jan 14 at 21:01
AdityaAditya
1275
$endgroup$
|
show 1 more comment
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Assuming that I have an orientable Riemannian manifold, can every top dimensional form be shown to be a volume form? If not, can all top dimensional forms be decomposed into a volume form and an exact form?
differential-geometry riemannian-geometry
asked Jan 14 at 21:01
AdityaAditya
1275
$endgroup$
2
$begingroup$
What's a "volume form"? And is this a way of asking if the top de Rham cohomology group is $Bbb R$?
$endgroup$
– Lord Shark the Unknown
Jan 14 at 21:05
$begingroup$
By a volume form, I am referring to a differential form of top degree which allows for integration on Riemannian manifolds. (See here for more information en.wikipedia.org/wiki/Volume_form)
$endgroup$
– Aditya
Jan 14 at 21:09
$begingroup$
You can integrate any top-dimensional form (at least on a compact orientable manifold). Your Wikipedia reference gives "volume form" as a synonym for top-dimensional form.
$endgroup$
– Lord Shark the Unknown
Jan 14 at 21:53
1
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Personally, I think Wiki is wrong. A volume form should be nonzero at each point, which occurs if and only if the manifold in question is orientable.
$endgroup$
– Ted Shifrin
Jan 15 at 19:32
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To elaborate on my previous comment: Without that requirement, we end up with volume $0$ if we start with an exact $n$-form on a compact (orientable) manifold without boundary. That is not what should be happening, IMHO.
$endgroup$
– Ted Shifrin
Jan 15 at 19:46
|
show 1 more comment
$begingroup$
Assuming that I have an orientable Riemannian manifold, can every top dimensional form be shown to be a volume form? If not, can all top dimensional forms be decomposed into a volume form and an exact form?
differential-geometry riemannian-geometry
asked Jan 14 at 21:01
AdityaAditya
1275
$endgroup$
Assuming that I have an orientable Riemannian manifold, can every top dimensional form be shown to be a volume form? If not, can all top dimensional forms be decomposed into a volume form and an exact form?
differential-geometry riemannian-geometry
differential-geometry riemannian-geometry
asked Jan 14 at 21:01
AdityaAditya
1275
asked Jan 14 at 21:01
AdityaAditya
1275
edited Jan 14 at 21:06
Aditya
asked Jan 14 at 21:01
AdityaAditya
1275
asked Jan 14 at 21:01
AdityaAditya
1275
asked Jan 14 at 21:01
AdityaAditya
1275
1275
2
$begingroup$
What's a "volume form"? And is this a way of asking if the top de Rham cohomology group is $Bbb R$?
$endgroup$
– Lord Shark the Unknown
Jan 14 at 21:05
$begingroup$
By a volume form, I am referring to a differential form of top degree which allows for integration on Riemannian manifolds. (See here for more information en.wikipedia.org/wiki/Volume_form)
$endgroup$
– Aditya
Jan 14 at 21:09
$begingroup$
You can integrate any top-dimensional form (at least on a compact orientable manifold). Your Wikipedia reference gives "volume form" as a synonym for top-dimensional form.
$endgroup$
– Lord Shark the Unknown
Jan 14 at 21:53
1
$begingroup$
Personally, I think Wiki is wrong. A volume form should be nonzero at each point, which occurs if and only if the manifold in question is orientable.
$endgroup$
– Ted Shifrin
Jan 15 at 19:32
$begingroup$
To elaborate on my previous comment: Without that requirement, we end up with volume $0$ if we start with an exact $n$-form on a compact (orientable) manifold without boundary. That is not what should be happening, IMHO.
$endgroup$
– Ted Shifrin
Jan 15 at 19:46
|
show 1 more comment
2
$begingroup$
What's a "volume form"? And is this a way of asking if the top de Rham cohomology group is $Bbb R$?
$endgroup$
– Lord Shark the Unknown
Jan 14 at 21:05
$begingroup$
By a volume form, I am referring to a differential form of top degree which allows for integration on Riemannian manifolds. (See here for more information en.wikipedia.org/wiki/Volume_form)
$endgroup$
– Aditya
Jan 14 at 21:09
$begingroup$
You can integrate any top-dimensional form (at least on a compact orientable manifold). Your Wikipedia reference gives "volume form" as a synonym for top-dimensional form.
$endgroup$
– Lord Shark the Unknown
Jan 14 at 21:53
1
$begingroup$
Personally, I think Wiki is wrong. A volume form should be nonzero at each point, which occurs if and only if the manifold in question is orientable.
$endgroup$
– Ted Shifrin
Jan 15 at 19:32
$begingroup$
To elaborate on my previous comment: Without that requirement, we end up with volume $0$ if we start with an exact $n$-form on a compact (orientable) manifold without boundary. That is not what should be happening, IMHO.
$endgroup$
– Ted Shifrin
Jan 15 at 19:46
2
2
$begingroup$
What's a "volume form"? And is this a way of asking if the top de Rham cohomology group is $Bbb R$?
$endgroup$
– Lord Shark the Unknown
Jan 14 at 21:05
$begingroup$
What's a "volume form"? And is this a way of asking if the top de Rham cohomology group is $Bbb R$?
$endgroup$
– Lord Shark the Unknown
Jan 14 at 21:05
$begingroup$
By a volume form, I am referring to a differential form of top degree which allows for integration on Riemannian manifolds. (See here for more information en.wikipedia.org/wiki/Volume_form)
$endgroup$
– Aditya
Jan 14 at 21:09
$begingroup$
By a volume form, I am referring to a differential form of top degree which allows for integration on Riemannian manifolds. (See here for more information en.wikipedia.org/wiki/Volume_form)
$endgroup$
– Aditya
Jan 14 at 21:09
$begingroup$
You can integrate any top-dimensional form (at least on a compact orientable manifold). Your Wikipedia reference gives "volume form" as a synonym for top-dimensional form.
$endgroup$
– Lord Shark the Unknown
Jan 14 at 21:53
$begingroup$
You can integrate any top-dimensional form (at least on a compact orientable manifold). Your Wikipedia reference gives "volume form" as a synonym for top-dimensional form.
$endgroup$
– Lord Shark the Unknown
Jan 14 at 21:53
1
1
$begingroup$
Personally, I think Wiki is wrong. A volume form should be nonzero at each point, which occurs if and only if the manifold in question is orientable.
$endgroup$
– Ted Shifrin
Jan 15 at 19:32
$begingroup$
Personally, I think Wiki is wrong. A volume form should be nonzero at each point, which occurs if and only if the manifold in question is orientable.
$endgroup$
– Ted Shifrin
Jan 15 at 19:32
$begingroup$
To elaborate on my previous comment: Without that requirement, we end up with volume $0$ if we start with an exact $n$-form on a compact (orientable) manifold without boundary. That is not what should be happening, IMHO.
$endgroup$
– Ted Shifrin
Jan 15 at 19:46
$begingroup$
To elaborate on my previous comment: Without that requirement, we end up with volume $0$ if we start with an exact $n$-form on a compact (orientable) manifold without boundary. That is not what should be happening, IMHO.
$endgroup$
– Ted Shifrin
Jan 15 at 19:46
|
show 1 more comment
1 Answer
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active
oldest
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The answer is no. To be a volume form, there is an extra condition. The condition is that the volume form must have length one everywhere. An explicit counter example; If a top dimensional form $omega$ may be written as $omega = fepsilon$ where $epsilon$ is a volume form, then $omega$ is not a volume form unless the scalar function $f=1$.
answered Jan 15 at 0:01
AdityaAditya
1275
$endgroup$
1
$begingroup$
What do you mean by length here? Do you not use the volume form (or even more structure, a Riemannian metric) to define that? Shouldn't the condition instead be that the form be nowhere-zero?
$endgroup$
– Ted Shifrin
Jan 15 at 19:33
add a comment |
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$begingroup$
The answer is no. To be a volume form, there is an extra condition. The condition is that the volume form must have length one everywhere. An explicit counter example; If a top dimensional form $omega$ may be written as $omega = fepsilon$ where $epsilon$ is a volume form, then $omega$ is not a volume form unless the scalar function $f=1$.
answered Jan 15 at 0:01
AdityaAditya
1275
$endgroup$
1
$begingroup$
What do you mean by length here? Do you not use the volume form (or even more structure, a Riemannian metric) to define that? Shouldn't the condition instead be that the form be nowhere-zero?
$endgroup$
– Ted Shifrin
Jan 15 at 19:33
add a comment |
$begingroup$
The answer is no. To be a volume form, there is an extra condition. The condition is that the volume form must have length one everywhere. An explicit counter example; If a top dimensional form $omega$ may be written as $omega = fepsilon$ where $epsilon$ is a volume form, then $omega$ is not a volume form unless the scalar function $f=1$.
answered Jan 15 at 0:01
AdityaAditya
1275
$endgroup$
1
$begingroup$
What do you mean by length here? Do you not use the volume form (or even more structure, a Riemannian metric) to define that? Shouldn't the condition instead be that the form be nowhere-zero?
$endgroup$
– Ted Shifrin
Jan 15 at 19:33
add a comment |
$begingroup$
The answer is no. To be a volume form, there is an extra condition. The condition is that the volume form must have length one everywhere. An explicit counter example; If a top dimensional form $omega$ may be written as $omega = fepsilon$ where $epsilon$ is a volume form, then $omega$ is not a volume form unless the scalar function $f=1$.
answered Jan 15 at 0:01
AdityaAditya
1275
$endgroup$
The answer is no. To be a volume form, there is an extra condition. The condition is that the volume form must have length one everywhere. An explicit counter example; If a top dimensional form $omega$ may be written as $omega = fepsilon$ where $epsilon$ is a volume form, then $omega$ is not a volume form unless the scalar function $f=1$.
answered Jan 15 at 0:01
AdityaAditya
1275
answered Jan 15 at 0:01
AdityaAditya
1275
answered Jan 15 at 0:01
AdityaAditya
1275
answered Jan 15 at 0:01
AdityaAditya
1275
1275
1
$begingroup$
What do you mean by length here? Do you not use the volume form (or even more structure, a Riemannian metric) to define that? Shouldn't the condition instead be that the form be nowhere-zero?
$endgroup$
– Ted Shifrin
Jan 15 at 19:33
add a comment |
1
$begingroup$
What do you mean by length here? Do you not use the volume form (or even more structure, a Riemannian metric) to define that? Shouldn't the condition instead be that the form be nowhere-zero?
$endgroup$
– Ted Shifrin
Jan 15 at 19:33
1
1
$begingroup$
What do you mean by length here? Do you not use the volume form (or even more structure, a Riemannian metric) to define that? Shouldn't the condition instead be that the form be nowhere-zero?
$endgroup$
– Ted Shifrin
Jan 15 at 19:33
$begingroup$
What do you mean by length here? Do you not use the volume form (or even more structure, a Riemannian metric) to define that? Shouldn't the condition instead be that the form be nowhere-zero?
$endgroup$
– Ted Shifrin
Jan 15 at 19:33
add a comment |
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$begingroup$
What's a "volume form"? And is this a way of asking if the top de Rham cohomology group is $Bbb R$?
$endgroup$
– Lord Shark the Unknown
Jan 14 at 21:05
$begingroup$
By a volume form, I am referring to a differential form of top degree which allows for integration on Riemannian manifolds. (See here for more information en.wikipedia.org/wiki/Volume_form)
$endgroup$
– Aditya
Jan 14 at 21:09
$begingroup$
You can integrate any top-dimensional form (at least on a compact orientable manifold). Your Wikipedia reference gives "volume form" as a synonym for top-dimensional form.
$endgroup$
– Lord Shark the Unknown
Jan 14 at 21:53
1
$begingroup$
Personally, I think Wiki is wrong. A volume form should be nonzero at each point, which occurs if and only if the manifold in question is orientable.
$endgroup$
– Ted Shifrin
Jan 15 at 19:32
$begingroup$
To elaborate on my previous comment: Without that requirement, we end up with volume $0$ if we start with an exact $n$-form on a compact (orientable) manifold without boundary. That is not what should be happening, IMHO.
$endgroup$
– Ted Shifrin
Jan 15 at 19:46