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  Duality in arbitrary finite dimension using the Levi-Civita tensor

+ 2 like - 0 dislike

In 4-D flat metric E&M context, given a rank $p$ tensor, one can construct dual of $4-p$ rank tensor by Levi-Civita tensor. Here dual is not in the same sense of mathematical dual. I do not know where it comes from. The followings are my questions:

  1. Is it necessarily true for arbitrary dimension that $n$, given a rank $p$ tensor, one always can have a rank $n-p$ tensor constructed by Levi-Civita tensor dual to the rank $p$ tensor?

  2. If it is true, in any dimensional space+1 dimension time, any tensor and its dual contain the same information. Shouldn't the bigger the space, the more information be thrown in? Why am I expect this duality is true up to any dimension?

  3. It seems somehow there is some sort of one to one correspondence between dual and double dual itself where dual is still not the familiar mathematical dual. I knew in math double dual space isomorphic to dual space which is different from what we are talking here. What is the correspondence in the E&M dual context?

This post imported from StackExchange Physics at 2014-06-29 09:35 (UCT), posted by SE-user user45765
asked Jun 27, 2014 in Theoretical Physics by user45765 (20 points) [ no revision ]
retagged Jun 29, 2014
There's probably a good question in here somewhere, but it's hard to parse. Perhaps you could put in some paragraph breaks, and make it clear what the one key conceptual issue you want to know is? In particular, it would help a lot to have a title that actually asks what you want to ask, rather than "two dumb questions on duality".

This post imported from StackExchange Physics at 2014-06-29 09:35 (UCT), posted by SE-user David Z
The relation between the double dual of a $p$-form and the $p$-form in dimension $n$ is given by $\star\star \alpha = (-1)^{s+ p(n-p)} \alpha$, where $s$ is the number of minus signs in the signature of the metric and $\star$ denotes the Hodge dual. This is easy to prove by doing the permutations over the indices. Note that the Hodge dual acts only on forms, namely totally anti-symmetric covariant tensors.

This post imported from StackExchange Physics at 2014-06-29 09:35 (UCT), posted by SE-user auxsvr

The key constraint is antisymmetry--- it's only the totally antisymmetric tensors, aka one-forms, which are dual to each other (it's called Poincare duality).

1 Answer

+ 5 like - 0 dislike

Your question is one about true mathematical duality, you just do not know it. What you are looking for is Hodge duality, which holds in the exterior algebra of any vector space, and the differential forms one looks at in EM, GR and elsewhere are just elements of the exterior algebra of the tangent space (or, equivalently, of deRham cohomology).

It is a basic result (its combinatorial - if you can count, you can prove it!) that the $p$-th degree of the exterior algebra over a space of dimension $n$ has dimension $n \choose p$, and since $\binom{n}{p} = \binom{n}{n-p}$, the spaces of the $p$-th and $n-p$-th degree have indeed the same dimension, and sending their basis vectors to each other via the epsilon tensor defines an isomorphism.

This post imported from StackExchange Physics at 2014-06-29 09:35 (UCT), posted by SE-user ACuriousMind
answered Jun 27, 2014 by ACuriousMind (910 points) [ no revision ]
Thanks a lot. That is what I need.

This post imported from StackExchange Physics at 2014-06-29 09:35 (UCT), posted by SE-user user45765

I realise this is nit-picking, but you need more structure than that of a vector space in order to define the Hodge dual.  You need a vector space with an inner product and a choice of orientation.  The choice of orientation is the "epsilon symbol", but the epsilon symbol alone does not suffice to define Hodge duality: since it maps $\Lambda^p V$ to $\Lambda^{n-p}V^*$.  You then need to identify $V$ with $V^*$ and this is where the inner product comes in.

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