Riemannian volume form without local coordinates


by Don Aman
Tags: coordinates, form, local, riemannian, volume
Don Aman
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#1
Jun12-06, 02:06 AM
P: 73
The volume form on a Riemannian manifold is usually defined either in terms of a local holonomic frame or orthonormal frame. Since it's defined globally, I would like there to be a global definition, like there is with symplectic manifolds (vol = ωn). Is there one?
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Doodle Bob
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#2
Jun13-06, 04:32 AM
P: 255
I believe it would be: dV=*1, where * is the Hodge star operator and 1 is the 0-form representing the constant function, f(x)=1.
Don Aman
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#3
Jun13-06, 01:01 PM
P: 73
Quote Quote by Doodle Bob
I believe it would be: dV=*1, where * is the Hodge star operator and 1 is the 0-form representing the constant function, f(x)=1.
The Hodge star is usually defined in terms of the volume form, so your proposed definition seems circular. Do you have a different definition of the Hodge star in mind?

Doodle Bob
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#4
Jun15-06, 04:50 AM
P: 255

Riemannian volume form without local coordinates


Quote Quote by Don Aman
The Hodge star is usually defined in terms of the volume form, so your proposed definition seems circular. Do you have a different definition of the Hodge star in mind?
Actually, I believe that the Hodge star operator can be defined locally via an oriented orthonormal basis with no reference to the volume form (you can do it merely on the linear algebra level).

Unless you're restricting yourself to a special manifold, e.g. a space form, I believe that this is the best you can do with regard to a global representation of the volume form.
coalquay404
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#5
Jun15-06, 07:38 PM
P: 218
You can define the Hodge dual in a couple of ways. The simplest way is to assume that you have some basis p-form [tex]\wedge_{i=1}^p\omega^{\mu_i}[/tex]. You then define [tex]\star:\Lambda^p(M)\to\Lambda^{m-p}(M)[/tex] (assuming an m-dimensional manifold), where

[tex]
\star\wedge_{i=1}^{p}\omega^{\mu_{i}}=\frac{\sqrt{|g|}}{(m-p)!}\epsilon_{\phantom{\mu_{1}\ldots \mu_{p}}\nu_{1}\ldots \nu_{m-p}}^{\mu_{1}\ldots \mu_{p}}\wedge_{i=1}^{m-p}\omega^{\nu_{i}}.
[/tex]

where [tex]|g|[/tex] is the absolute value of the determinant of the metric tensor. An alternative way is to note that putting a metric on your manifold automatically defines an inner product on the space of p-forms. Given [tex]\alpha,\beta\in\Lambda^p(M)[/tex] we can define an inner product by

[tex]
(\alpha,\beta)=\frac{1}{p!}\alpha^{\mu_{1}\ldots\mu_{p}}\beta_{\mu_{1}\ ldots\mu_{p}}
[/tex]

Then define the Hodge star as an operator which satisfies

[tex]
(\alpha,\beta)=\frac{1}{p!}\alpha^{\mu_{1}\ldots\mu_{p}}\beta_{\mu_{1}\ ldots\mu_{p}}
[/tex]

for all [tex]\beta\in\Lambda^{m-p}(M)[/tex], and where [tex]\eta[/tex] is the volume form on [tex]M[/tex].
Don Aman
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#6
Jun16-06, 10:55 AM
P: 73
Quote Quote by Doodle Bob
Actually, I believe that the Hodge star operator can be defined locally via an oriented orthonormal basis with no reference to the volume form (you can do it merely on the linear algebra level).

Unless you're restricting yourself to a special manifold, e.g. a space form, I believe that this is the best you can do with regard to a global representation of the volume form.
OK, yes that's right. The Hodge star can be defined in terms of a local orthonormal or holonomic frame, and then the Volume form can be defined in terms of it. But of course this is what I was hoping to avoid. Maybe you're right, this is the best that I can do. What prompted me to ask the question was this post on Usenet:
Quote Quote by Toby Bartels
A metric is a section of the 2nd symmetric power of the cotangent bundle;
that is, it's locally a sum of symmetrised products of 2 1forms.
So if you wedge the metric with itself n times,
then you'll get a section of the 2nd symmetric power of the nform bundle;
I would very much like there to be a construction like this, but I cannot make heads or tails of what Toby is describing. For one thing, he seems to assume that the nth alternating power of the second symmetric power is equal to the second symmetric power of the nth alternating power of a vector space or bundle. I'm pretty sure that in general, alternating powers do not commute with symmetric powers, and I don't think setting the alternating power to n helps this.
Hurkyl
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#7
Jun16-06, 07:55 PM
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I'm pretty sure that in general, alternating powers do not commute with symmetric powers
You sure on that? But in any case, he doesn't need equality, or even a canonical isomorphism; he just needs a canonical embedding of one in the other.
Don Aman
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#8
Jun17-06, 10:33 AM
P: 73
Quote Quote by Hurkyl
You sure on that?
Well, I think so... I mean, I'm pretty sure...

For example, the alternating square of a 2d vector space is 1d, whose symmetric square is also 1d. While the symmetric square of a 2d vector space is 3d, and its alternating square is 3d again. So clearly they cannot be isomorphic.

But why do you ask? You sound like you know something I don't. Please share.
Quote Quote by Hurkyl
But in any case, he doesn't need equality, or even a canonical isomorphism; he just needs a canonical embedding of one in the other.
OK, so is there such an embedding? I'm playing around with it, maybe a map
[tex]\operatorname{Alt}^2(\operatorname{Sym}^2V)\to\operatorname{Sym}^2(\ope ratorname{Alt}^2V)[/tex]
for example I'm thinking given by something like
[tex](a\otimes_\sigma b)\wedge(c\otimes_\sigma d)\mapsto (a\wedge c)\otimes_\sigma(b\wedge d)+(a\wedge d)\otimes_\sigma(b\wedge c)[/tex]
maybe.

But this map is no embedding. It takes
[tex](e_1\otimes_\sigma e_1)\wedge(e_1\otimes_\sigma e_2)[/tex]
to zero.

I am dubious about your suggestion.
Don Aman
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#9
Jun20-06, 05:22 PM
P: 73
Quote Quote by Hurkyl
You sure on that? But in any case, he doesn't need equality, or even a canonical isomorphism; he just needs a canonical embedding of one in the other.
Still there?
Hurkyl
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#10
Jun20-06, 05:34 PM
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I had originally made a mistake in my mental picture of the group of symmetries/anti-symmetries of the arguments. I haven't had anything useful to say since.
Don Aman
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#11
Jun26-06, 03:34 PM
P: 73
Quote Quote by Hurkyl
I haven't had anything useful to say since.
Unfortunately even your original comment was more misleading than it was useful, but I thank you from refraining to make additional useless comments. And thank you to others in this thread who tried to answer. At this point, I guess I have to conclude that perhaps Bartels was mistaken about his description.


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