Differential forms and divergence

[Damidami]

Hello everyone, I'm new to this forum.
I have a doubt about differential forms, related to the divergence.
On a website I read this:
"In general, it is true that in R^3 the operation of d on a differential 0-form gives the gradient of that differential 0-form, that on a differential 1-form give the curl of that differential 1-form, and that on a differential 2-form gives its divergence."

My question is: In R^2 how can I obtain de 2-d version of the divergence by differentiating a form? Because if I differentiate a 0-form it gives me the 2-d version of the gradient, and if I differentiate a 1-form it gives me the 2-d version of the curl.

 

[HallsofIvy]

You don't. Any 1-form is of the form f(x,y)dx+ g(x,y)dy and any 2-form is h(x,y)dxdy. Strictly speaking the differental $d(fdx+ gdy)= f_x dxdx+ f_y dydx+ g_x dxdy+ g_ydydy$ but because the "product" is skew symmetric, dxdx and dydy are both 0 so [itex]d(fdx+ gdy)= (g_x- f_y)dxdy. There cannot be a product of three of "dx" and "dy" so the differential of any 2-form is 0.

(In R³, the differential of a 2-form is [itex]d(fdxdy+ gdydz+ hdxdz)= f_zdzdxdy+ g_xdxdydz+ h_ydydxdz= (f_z+ g_x- h_y)dxdydz so I am not at all sure what is meant by "its divergence" there.)

 

[Hurkyl]

The website is correct, under a suitable interpretation of the word "gives".

For example, strictly speaking, divergence is an operator that takes a vector field and returns a scalar field. So, you can't even apply it to a differential 1-form!

But once you've chosen a metric, there are two duality operations you can apply: there is the transpose (a.k.a. "raising" and "lowering" indices) and there is Hodge duality.

So, to produce divergence in 3-space, you have to do the following:

(1) Apply a transpose to convert your vector field into a differential 1-form
(2) Apply Hodge duality to produce a differential 2-form
(3) Apply d to produce a differential 3-form
(4) Apply Hodge duality to produce a scalar field. (a.k.a. differential 0-form)

If (x, y, z) are orthonormal coordinates, then in these coordinates we start with

$$f \frac{\partial}{\partial x} + g \frac{\partial}{\partial y} + h \frac{\partial}{\partial z}$$

transposing gives

$$f \, dx + g \, dy + h \, dz$$

Hodge duality gives

$$f \, dy dz + g \, dz dx + h \, dx dy$$

differentiation gives

$$(f_x + g_y + h_z) \, dx dy dz$$

and hodge duality gives

$$f_x + g_y + h_z$$

So

$$\mathop{\mathrm{div}} v = \mathord{*} d \mathord{*} v^T$$

 

[timur]

Cool. Is the transpose notation common?

 

[Hurkyl]

I don't think so, but I don't know for sure; this isn't my specialty. I know I've heard the operation called the "metric transpose" often enough, but I'm not sure if I've actually seen that notation used. Some sort of index notation (abstract or concrete) is usually used.

 

[Sophus]

for correct transformation vector field to 2 form you need to use the notion of so-called "volume element" ($$\Omega$$=*1=dxdydz)

 

[Chris Hillman]

Often called volume form and written $\omega = dx \wedge dy \wedge dz$. More generally, whenever you have a coframe field in a Riemannian or Lorenztian n-manifold, $\omega = \sigma^1 \wedge \sigma^2 \wedge \dots \sigma^n$ The transpose is also known by the overused word dual. Did anyone mention the Hodge star operator? The book by Flanders is a good source of information for all these topics.

 

[timur]

In functional analysis one would say that the transpose here is actually the Riesz map (or the inverse?) acted pointwise.

 

[dextercioby]

That's mainly because in functional analysis we're more interested in topological duals of a TVS, rather than its algebraic dual. The Riesz map establishes the homeomorphism between a Hilbert space and its topological dual.

 

[Sophus]

In general you must be oriented not only for notation, which can be different in various branch of science, but also on some mathematical package which you are going to use. As for me I prefer Reduce with "excalc" and "eds" packages.