Riemannian volume form without local coordinatesby Don Aman Tags: coordinates, form, local, riemannian, volume 

#1
Jun1206, 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 = ω^{n}). Is there one?




#2
Jun1306, 04:32 AM

P: 255

I believe it would be: dV=*1, where * is the Hodge star operator and 1 is the 0form representing the constant function, f(x)=1.




#3
Jun1306, 01:01 PM

P: 73





#4
Jun1506, 04:50 AM

P: 255

Riemannian volume form without local coordinatesUnless 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. 



#5
Jun1506, 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 pform [tex]\wedge_{i=1}^p\omega^{\mu_i}[/tex]. You then define [tex]\star:\Lambda^p(M)\to\Lambda^{mp}(M)[/tex] (assuming an mdimensional manifold), where
[tex] \star\wedge_{i=1}^{p}\omega^{\mu_{i}}=\frac{\sqrt{g}}{(mp)!}\epsilon_{\phantom{\mu_{1}\ldots \mu_{p}}\nu_{1}\ldots \nu_{mp}}^{\mu_{1}\ldots \mu_{p}}\wedge_{i=1}^{mp}\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 pforms. 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^{mp}(M)[/tex], and where [tex]\eta[/tex] is the volume form on [tex]M[/tex]. 



#6
Jun1606, 10:55 AM

P: 73





#7
Jun1606, 07:55 PM

Emeritus
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PF Gold
P: 16,101





#8
Jun1706, 10:33 AM

P: 73

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. [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. 



#9
Jun2006, 05:22 PM

P: 73





#10
Jun2006, 05:34 PM

Emeritus
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PF Gold
P: 16,101

I had originally made a mistake in my mental picture of the group of symmetries/antisymmetries of the arguments. I haven't had anything useful to say since.




#11
Jun2606, 03:34 PM

P: 73




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