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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?

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    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.2010-07-28
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    But if I was talking a vector space dual in R^3 with euclidean metric, are the two equivalent?2010-07-28
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    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.2010-07-28
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    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?)2010-08-08
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    @KCd http://en.wikipedia.org/wiki/Multivector#Geometric_algebra2010-08-08
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    Jonathan: I saw that wikipedia page before writing my previous comment. What I meant was that before today I had not seen the term blade before. On the wikipedia page http://en.wikipedia.org/wiki/Blade_(geometry) which supposedly is about blades, the concept is defined to be "a generalization..." but that is not a definition in its own right. People in computer graphics may use blade, but I think mathematicians do not. It's like defining a dyad as "an operator represented as a pair of vectors juxtaposed without multiplication." Mathematically that has real problems as a working definition.2010-08-08
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    @KCd I have no idea what is most popular. I am a novice, but if you have any suggestions on the terms I should use feel free to let me know.2010-08-08
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    If you want to talk to mathematicians, far more will understand you if you call a/\b or a/\b/\c "elementary wedge products" (also called simple or decomposable wedge products) rather than 2-blades or 3-blades. But "blade" may be popular outside of mathematics (I mean, you learned it somewhere).2010-08-08
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    @KCd Thanks. I do want to talk to mathematicians and I misuse terminology often so feel free to correct me.2010-08-08

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(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.

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    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?2010-07-28
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    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.2010-07-28
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    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?2010-07-28
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    Yep. The fact that it's one-dimensional is the first step in making the definition of the Hodge dual work.2010-07-28
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    I gotcha there, I'll look over the rest again in bit.2010-07-28
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    Okay, let me see if I can summarize and put it in different words to see if I understand. The anti-symmetric quality of the wedge means that vector spaces have an orientation. This idea is lacking in the dual space of a vector space, but the hodge dual needs this information. However what is confusing me is that would only be a problem when the psuedoscalar (top exterior product) is odd.2010-07-28
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    No; you can define exterior powers and the wedge product without an orientation or a vector space. The problem is twofold: first of all, you need an inner product to identify a vector space with a dual, and second of all, you need an orientation to identify the top exterior power with the base field. In general, you can't identify a vector space with its dual in a canonical way.2010-07-28
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    Ok I'm still not entirely there but I'll mediate on that.2010-07-28
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    @QiaochuYuan - the link you gave in the answer is broken.2017-01-25
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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