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?

  • 1
    $\begingroup$ The Hodge dual is a kind of "refined" vector space dual, since it depends on additional information. When you talk about abstract vector space duals you are not allowed to use words like "normal" because there is no inner product, so you have to let go of some of your geometric intuition. $\endgroup$ Commented Jul 28, 2010 at 0:49
  • $\begingroup$ But if I was talking a vector space dual in R^3 with euclidean metric, are the two equivalent? $\endgroup$ Commented Jul 28, 2010 at 0:52
  • $\begingroup$ Vector space duality is part of what goes into a definition of the Hodge dual, but as I said there are other ingredients. See my answer for a precise statement. $\endgroup$ Commented Jul 28, 2010 at 1:22
  • 1
    $\begingroup$ I have never seen the word "blade" before. Is there an area of math where this is a widely used term? I'd call it an elementary wedge product. (Or is it any element of an exterior power of a vector space?) $\endgroup$
    – KCd
    Commented Aug 8, 2010 at 3:15
  • 2
    $\begingroup$ @KCd en.wikipedia.org/wiki/Multivector#Geometric_algebra $\endgroup$ Commented Aug 8, 2010 at 5:28

2 Answers 2


(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.


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

  • $\begingroup$ Thanks, but I'm unclear on the "the latter is one-dimensional statement." I understand how the wedge product will retrieve the pseudoscalar, from the two blades (exterior powers) that are duals, and that it is a bilinear map (a term I am still understanding), but what exactly is one dimensional? $\endgroup$ Commented Jul 28, 2010 at 1:14
  • $\begingroup$ The top exterior power. Given a basis b_1, ... b_n of V, the top exterior power is spanned by b_1 \wedge ... \wedge b_n. $\endgroup$ Commented Jul 28, 2010 at 1:20
  • $\begingroup$ Okay so is it one dimensional, in the sense that there is only one as opposed to some lower dimensional version that has n choose k? $\endgroup$ Commented Jul 28, 2010 at 1:23
  • $\begingroup$ Yep. The fact that it's one-dimensional is the first step in making the definition of the Hodge dual work. $\endgroup$ Commented Jul 28, 2010 at 1:26
  • 4
    $\begingroup$ @QiaochuYuan - the link you gave in the answer is broken. $\endgroup$
    – user297514
    Commented Jan 25, 2017 at 15:47

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


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .