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How to make a hodge dual with no metric, only volume form

 
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Feb10-11, 06:58 AM   #1
 

How to make a hodge dual with no metric, only volume form


Hey guys!

I am going crazy... most books don't cover this and instead assume that the manifold is Riemannian or pseudo-Riemannian and has a metric tensor defined on it. I want a "generalized" hodge star.

I have an orientable smooth manifold, thats IT. I have heard that there is a way to formally create a Hodge star/dual between multivectors/forms using only the volume form (a volume form always exists on an orientable manifold). I have really been struggling to do this.

Two things I have asked myself (though do not lead to solutions)

Firstly, does double-application of this kind of "generalized" hodge dual always reproduce the original multivector? I think that it should.

Secondly, does an application of the Hodge dual to "1" (the zero-vector/function = 1 everywhere on the manifold) always need to produce the volume form?

I am really going crazy... please assist :(
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Feb10-11, 02:06 PM   #2
 
How do you have a volume form without a metric? Are you just taking a non-vanishing n-form and declaring it to be your volume form?

How would you define a Hodge star without reference to the norm of the differential form?

The way to produce duality is obvious - wedge two forms and integrate. But the way to identify a finite-dimensional vector space with its dual requires (you guessed it) an inner product.

The higher-level look makes it obvious that your approach is doomed to fail: if you change the metric, the hodge dual changes. So to make your fake Hodge dual well-defined, you're going to have to make a lot of choices that (in the end) amount to specifying a metric.
Feb10-11, 04:16 PM   #3
 
The only 'generalized hodge dual' that I'm familiar with doesn't take forms to forms, but rather [using math's conventions] alternating contravariant [tex]k[/tex]-tensors to alternating covariant [tex]n-k[/tex]-tensors and vice versa. Let [tex]V[/tex] be your volume form, take the simple tensor [tex]v_1 \otimes \dots \otimes v_k[/tex] to the function on (T_pM)^{n-k} that takes [tex](w_1,\dots,w_{n-k}) \mapsto V(v_1,\dots,v_k,w_1,...,w_{n-k})[/tex]; you can show that this is well defined and alternating, thus corresponds to an n-k-form. This defines your generalized hodge star.

Of course without a metric there's no natural way to switch between contravariant and covariant tensors, so perhaps this won't help you.
Feb11-11, 04:29 AM   #4
 

How to make a hodge dual with no metric, only volume form


Quote by camel_jockey View Post
Hey guys!

I am going crazy... most books don't cover this and instead assume that the manifold is Riemannian or pseudo-Riemannian and has a metric tensor defined on it. I want a "generalized" hodge star.

I have an orientable smooth manifold, thats IT. I have heard that there is a way to formally create a Hodge star/dual between multivectors/forms using only the volume form (a volume form always exists on an orientable manifold). I have really been struggling to do this.
What I have been looking for is an oriented manifold defined by a Levi-Civita tensor at each point, rather than a metric. The Levi-Civita tensor is not to be confused with the Levi-Civita symbol which consists of alternating ones and negative ones and zeroes except in orthonormal coordinates.

If this would work-out, the metric should be definable in terms of the Levi-Civita tensor rather than the other way around.

Is this close to what you have in mind?
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