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,

$$aa \equiv g(a,a) = |a|^2, \quad ab = -ba, \quad (uv)w = u(vw)$$

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 $e^t, e^x, e^y, e^z$. Then there is a well-defined volume-form $i$:

$$i \equiv e^t e^x e^y e^z$$

Again, note that I said *orthogonal*. What happens if you take a two-form like $e^x e^y$ and multiply it by $i$?

$$i e^x e^y = e^t e^x e^y e^z e^x e^y =e^t (e^x e^y)(e^x e^y) e^z = - e^t e^z g^{xx} g^{yy} $$

where $g^{xx}$ and $g^{yy}$ 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, $i = e_1 e_2 e_3$, so we get

$$\star e_1 = i e_1 = e_1 e_2 e_3 e_1 = -e_1 e_2 e_1 e_3 = + e_1 e_1 e_2 e_3 = e_2 e_3$$

The other equalities follow similarly.

This post imported from StackExchange Mathematics at 2014-07-21 11:20 (UCT), posted by SE-user Muphrid