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

[camel_jockey]

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, that's 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 :(

 

[zhentil]

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.

 

[Gilmoshar]

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.

 

[Phrak]

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, that's 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?

 

[blue_raver22]

Hi there,

I understand your frustration with trying to find information on creating a Hodge dual without a metric tensor. It can be a tricky concept to grasp, but I will try my best to explain it to you.

Firstly, let's define what a Hodge dual is. A Hodge dual is an operation that takes a k-form and produces an (n-k)-form on an n-dimensional manifold. This operation is typically defined using a metric tensor, but there is a way to define it using only the volume form.

To do this, we need to first understand the concept of a "generalized" Hodge dual. This is an operation that maps a k-vector (a multivector with k components) to an (n-k)-vector. In other words, it maps a k-form to an (n-k)-form.

Now, let's consider the volume form on an n-dimensional manifold. This is a k-form, where k=n. We can define a "generalized" Hodge dual by taking the volume form and raising it to the power of (-1)^k, where k is the dimension of the manifold. This operation will map the volume form to a 0-form, which is just a scalar. This scalar is then the "generalized" Hodge dual of the volume form.

To answer your first question, yes, the double-application of this "generalized" Hodge dual will always reproduce the original multivector. This is because the volume form raised to the power of (-1)^k will always produce a scalar, and the inverse of a scalar is just itself.

To answer your second question, the application of the Hodge dual to "1" will always produce the volume form. This is because raising the volume form to the power of (-1)^k will always produce a scalar, and the inverse of a scalar is just itself.

I hope this explanation helps to clear things up for you. Keep in mind that this "generalized" Hodge dual is not a replacement for the traditional Hodge dual using a metric tensor, but rather a way to define it without a metric tensor. Good luck with your studies!