Unravelling Hodge Duality: A Starter Guide

In summary, the star operator brings about all the Faraday 2-form example by bringing about the equation (*x,*y) = ±1. This property is always true, no matter the order the components are in.
  • #1
Gianni2k
17
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I'm having problems understanding Hodge duality in its most basic form. It relates exterior p forms to exterior n-p forms where n is the dimensionality of the manifold. I can't seem to follow the discussion on the hodge dual operator on this lecture course (page 19):

http://www.damtp.cam.ac.uk/user/gr/about/members/dgnotes3.pdf

How does the star operator bring about all the Faraday 2-form example?

Any help would be appreciated.
Thanks.
 
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  • #2
the * is "perpendicular" to the original

Gianni2k said:
I'm having problems understanding Hodge duality in its most basic form. It relates exterior p forms to exterior n-p forms where n is the dimensionality of the manifold.

How does the star operator bring about all the Faraday 2-form example?

Hi Gianni2k! :smile:

The defining property is that inner product with the * (an ordinary scalar) is ± the same as outer product with the original:

For example, in Minkowski (+,-,-,-) space, the outer product x^y^z^t = 1, so (*x,y^z^t) = ±1, so *x must be ±y^z^t.

(by comparison, x^x^z^t = 0, so (*x,x^z^t) = (y^z^t,x^z^t) = 0)

and similarly outer product x^y^z^t = 1, so (*(x^y),z^t) = ±1, so *(x^y) must be ±z^t.

It's always the "perpendicular" component:

If we think of a p-form as spanning a p-dimensional subspace (yes, I know we shouldn't!), then its * spans the perpendicular, or complementary, subspace.

*x "is" the three dimensions perpendicular to x, and must therefore be ±y^z^t (± according to the metric).

and *(x^y) must be ±z^t.


So in Faraday, Ex is the x^t component, and so its * is ± the y^z component, which is Bx.

It's this equality between inner product with * and outer product with the original that forces the * to "be" the perpendicular element.

Does that help? :smile:
 
  • #3
ok great, so why is the outer product on minkowski x^y^z^t=1? And is (x,x) always =1 ?

Also, so in this formalism is the magnetic field the Hodge dual to the electric field in Minkowski space?

thanks.
 
  • #4
Hi Gianni2k! :smile:
Gianni2k said:
ok great, so why is the outer product on minkowski x^y^z^t=1? And is (x,x) always =1 ?

With the usual (+.-.-.-) metric, (t,t) = 1 and (x,x) = -1.

And the product of all four of x y z and t is ±1, depending on the order they're in (maybe it's x^y^z^t=-1, I haven't checked :redface:)
Also, so in this formalism is the magnetic field the Hodge dual to the electric field in Minkowski space?

Yes (times -1):

*E = -B, *B = E, **E = -E, **B = - B. :smile:
 
  • #5
Gianni, thanks for the link to good lectures.

By the way, these lectures are referred to as Part III. Do you also have links to other parts?
 
  • #6
In three dimensions forms can be visualized like this:
  • A 1-form is a set of equally spaced planes.
  • A 2-form is a set of equally spaced lines.
  • A 3-form is a set of equally spaced points.
The Hodge dual of a 2-form is then a set of planes perpendicular to the 2-form lines. The spacing between the planes is such that the intersection points between the planes and lines are a lattice with one point per unit volume.

In four dimensions it gets slightly more complicated:
  • A 1-form is a set of equally spaced 3-dimensional linear spaces.
  • A 2-form is one or two sets of equally spaced planes.
  • A 3-form is a set of equally spaced lines.
  • A 4-form is a set of equally spaced points.
The Hodge dual of a simple (i.e. it is only one set of planes) 2-form is another set of planes, perpendicular to the first set and spaced such that the intersection points are a lattice with one point per unit 4-volume.

(Ah, and yes, all lines, planes, etc should have an orientation as well...)

Demystifier said:
By the way, these lectures are referred to as Part III. Do you also have links to other parts?
Part II is found a http://www.damtp.cam.ac.uk/user/gr/about/members/gwglectures.html"
 
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  • #7
Demystifier said:
By the way, these lectures are referred to as Part III. Do you also have links to other parts?
Part III is the other name for Cambridge's "http://www.maths.cam.ac.uk/postgrad/casm/": their masters course. Thus, these notes are not the third part of a lecture series, but is a complete course taught during the CASM (or Part III).
 

1. What is Hodge duality?

Hodge duality is a mathematical concept that relates the topological and geometric properties of a space. It is based on the idea that every closed form can be paired with a dual form of complementary dimension.

2. How is Hodge duality used in mathematical and scientific research?

Hodge duality is used in a wide range of fields, including differential geometry, algebraic topology, and theoretical physics. It allows for the translation of complex geometric problems into simpler algebraic ones, making it a powerful tool for solving equations and analyzing data.

3. What is the significance of unraveling Hodge duality?

Unraveling Hodge duality involves understanding the underlying principles and theorems that govern its application. This can lead to new insights and discoveries in various fields, as well as the development of more efficient computational algorithms.

4. Are there any challenges associated with understanding and applying Hodge duality?

Yes, Hodge duality can be a complex and abstract concept, and it requires a strong foundation in mathematics to fully grasp. It also involves understanding the connection between topology and geometry, which can be challenging for some researchers.

5. Can you provide an example of how Hodge duality has been used in a real-world application?

One example is in the field of computer graphics, where Hodge duality is used to create realistic 3D models of objects and environments. It allows for the efficient representation and manipulation of geometric data, making it a crucial tool in the development of visual effects and computer-generated imagery.

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