Sorry for the delay from last Friday to today, but I was chasing down a good lead.
Anyway, last week I said that I’d talk about a linear map that extends the notion of the correspondence between parallelograms in space and perpendicular vectors.
First of all, we should see why there may be such a correspondence. We’ve identified -dimensional parallelepipeds in an -dimensional vector space with antisymmetric tensors of degree : . Of course, not every such tensor will correspond to a parallelepiped (some will be linear combinations that can’t be written as a single wedge of vectors), but we’ll just keep going and let our methods apply to such more general tensors. Anyhow, we also know how to count the dimension of the space of such tensors:
This formula tells us that and will have the exact same dimension, and so it makes sense that there might be an isomorphism between them. And we’re going to look for one which defines the “perpendicular” -dimensional parallelepiped with the same size.
So what do we mean by “perpendicular”? It’s not just in terms of the “angle” defined by the inner product. Indeed, in that sense the parallelograms and are perpendicular. No, we want any vector in the subspace defined by our parallelepiped to be perpendicular to any vector in the subspace defined by the new one. That is, we want the new parallelepiped to span the orthogonal complement to the subspace we start with.
Our definition will also need to take into account the orientation on . Indeed, considering the parallelogram in three-dimensional space, the perpendicular must be for some nonzero constant , or otherwise it won’t be perpendicular to the whole - plane. And has to be in order to get the right size. But will it be or ? The difference is entirely in the orientation.
Okay, so let’s pick an orientation on , which gives us a particular top-degree tensor so that . Now, given some , we define the Hodge dual to be the unique antisymmetric tensor of degree satisfying
for all . Notice here that if and describe parallelepipeds, and any side of is perpendicular to all the sides of , then the projection of onto the subspace spanned by will have zero volume, and thus . This is what we expect, for then this side of must lie within the perpendicular subspace spanned by , and so the wedge should also be zero.
As a particular example, say we have an orthonormal basis of so that . Then given a multi-index the basic wedge gives us the subspace spanned by the vectors . The orthogonal complement is clearly spanned by the remaining basis vectors , and so , with the sign depending on whether the list is an even or an odd permutation of .
To be even more explicit, let’s work these out for the cases of dimensions three and four. First off, we have a basis . We work out all the duals of basic wedges as follows:
This reconstructs the correspondence we had last week between basic parallelograms and perpendicular basis vectors. In the four-dimensional case, the basis leads to the duals
It’s not a difficult exercise to work out the relation for a degree tensor in an -dimensional space.