It may sometimes be easier to compute Hodge duals using an orthogonal basis and clifford algebra.
Clifford algebra allows you to write a lot of complicated tensor expressions as simple products of vectors, using the "geometric product". If a,b are orthogonal vectors with respect to some metric, and u,v,w are arbitrary vectors, then under the geometric product,
aa≡g(a,a)=|a|2,ab=−ba,(uv)w=u(vw)
So this captures the inner product, encoded in the metric; the need for antisymmetric tensors in ab=−ba, and associativity allows us to drop unneeded parentheses and talk about long products of vectors as entities in their own right.
Now then, suppose you have orthogonal coordinate one-forms et,ex,ey,ez. Then there is a well-defined volume-form i:
i≡etexeyez
Again, note that I said orthogonal. What happens if you take a two-form like exey and multiply it by i?
iexey=etexeyezexey=et(exey)(exey)ez=−etezgxxgyy
where gxx and gyy arose from the inner products of forms. The sign might be "wrong" for a given convention, but this is doing exactly the same sort of thing the Hodge dual does.
I'll illustrate that, in the case of 3d with a Cartesian metric, you get the same results this way as you would using the index definition. In 3d, i=e1e2e3, so we get
⋆e1=ie1=e1e2e3e1=−e1e2e1e3=+e1e1e2e3=e2e3
The other equalities follow similarly.
This post imported from StackExchange Mathematics at 2014-07-21 11:20 (UCT), posted by SE-user Muphrid