What is the relationship between the Hodge dual of p-vectors and the dual space of an ordinary vector space?

I understand what the Hodge dual is, but I can't quite wrap my head around the dual space of vector space. They seem very similar, almost the same, but perhaps they are unrelated.

For instance, in $\mathbb{R}^3$, the blade $a \wedge b$ gives you a subspace that's like a plane, and the dual is roughly the normal to the plane.

Is there a similarly simple example for the dual space of a vector space, or is there a way to describe the vector space dual in terms of the Hodge dual?

Solution 1:

(Edit: I have edited this answer several times because my understanding of the situation has been improving.)

It is always profitable to understand these kind of constructions by understanding exactly what information they depend on. The Hodge dual depends on a surprising amount of information: you need a vector space $V$ which is equipped with both an inner product and an orientation, which is essentially a choice of which bases of $V$ are "right-handed." So let's see what we can say ignoring all this information first.

Any abstract vector space $V$ of finite dimension $n$ has exterior powers $\Lambda^2 V, \Lambda^3 V, ... \Lambda^n V$, the last of which is one-dimensional. The vector spaces $\Lambda^k V$ and $\Lambda^{n-k} V$ always have the same dimension, so we would like to be able to define some sort of "canonical" map between them. What can we say? Well, they are always dual: the wedge product defines a natural bilinear map $\Lambda^k V \times \Lambda^{n-k} V \to \Lambda^n V$, and since the latter is one-dimensional this means (once you've proven nondegeneracy) that the two vector spaces are in fact dual.

But duality does not give you a map between them. When two vector spaces $V, W$ are dual, meaning there is a nondegenerate bilinear map $V \times W \to F$ (where $F$ is the ground field), all you get is an isomorphism $V \simeq W^{\ast}$. Here you get an isomorphism $\Lambda^k V \simeq \Lambda^{n-k} V^{\ast}$, once you have specified an isomorphism $\Lambda^n V \simeq F$. This is equivalent to picking out a distinguished vector in $\Lambda^n V$, which there is no way to do in general.

So the answer is to introduce extra data. To identify $\Lambda^{n-k} V^{\ast}$ with $\Lambda^{n-k} V$, we need an inner product. An inner product gives you two distinguished vectors in $\Lambda^n V$, as follows: take any orthonormal basis $b_1, ... b_n$. Then wedging together the $b_i$ in any order gets you one of two elements of $\Lambda^n V$, depending on whether the corresponding permutation is even or odd. But without any extra data, there is no way to identify one of these elements with $1$ and one of these elements with $-1$.

The extra data that does this is an orientation on $V$, which tells you which bases are "right-handed" and which are "left-handed." So an oriented orthonormal basis gives you a distinguished element of $\Lambda^n V$, which gives you a distinguished isomorphism $\Lambda^k V \simeq \Lambda^{n-k} V^{\ast}$, which composed with the isomorphism $\Lambda^{n-k} V^{\ast} \simeq \Lambda^{n-k} V$ is the Hodge dual.

Phew.

This is explained in these notes I just found on Google.

Solution 2:

The basic concept of the Hodge dual has been around since Grassmann -- it is essentially Grassmann's complement. The treatments of the Hodge dual you can find on the web are usually sketchy and unmotivated. My notes on vector spaces give a more thorough treatment than most of what you can find on the web; just google "Vector Spaces, Vector Algebras, and Vector Geometries". --Richard