It’s interesting to look at what happens when we apply the Hodge star twice. We just used the fact that in our special case of we always get back exactly what we started with. That is, in this case, — the identity operation.
It’s easiest to work this out in coordinates. If is some -fold wedge then is times the wedge of all the indices that don’t show up in . But then is times the wedge of all the indices that don’t show up in , which is exactly all the indices that were in in the first place! That is, , where the sign is still a bit of a mystery.
But some examples will quickly shed some light on this. We can even extend to the pseudo-Riemannian case and pick a coordinate system so that , where . That is, any two are orthogonal, and each either has “squared-length” or . The determinant is simply the product of all these signs. In the Riemannian case this is simply .
Now, let’s start with an easy example: let be the wedge of the first indices:
Then is (basically) the wedge of the other indices:
The sign is positive since already has the right order. But now we flip this around the other way:
but this should obey the same rule as ever:
where we pick up a factor of each time we pull one of the last -forms leftwards past the first . We conclude that the actual sign must be so that this result is exactly . Similar juggling for other selections of will give the same result.
In our special Riemannian case with , then no matter what is we find the sign is always positive, as we expected. The same holds true in any odd dimension for Riemannian manifolds. In even dimensions, when is odd then so is , and so — the negative of the identity transformation. And the whole situation gets more complicated in the pseudo-Riemannian version depending on the number of s in the diagonalized metric tensor.