I have read (I think) all of the previous threads on this website (and many others) on this topic & unfortunately have not found an answer to my question. Due to the fact that I am only beginning to study the subjects of tensors, differential forms etc... I appeal to your good nature not to respond with anything too advanced as it just goes over my head. The structure of my question is as follows: first a look at the cross product; then a look at a wedge product calculation & it's similarities (that I think are far more explicit if you interpret it in the way I've explained below) to the cross product & finally 4 questions (in bold) that are motivated by the wedge product calculation. The 4 questions are all I'm hoping for answers to as I barely understand anything beyond what I've written about this subject & just don't understand what a lot of the online fora threads on this topic are saying in the responses to this topic. I think these questions are my way in.

The cross product is a strange animal, it really has very little justification as it is taught in elementary linear algebra books. It took me a long time to learn that the cross product is really no more than the dot product in disguise. It is actually quite easy to derive the result that a cross product gives, through clever algebra, as is done in the cross product pdf's here & here: synechism.org/drupal/cfsv/ (sorry about the text-link!). By doing your own algebra you can justify the anti-symmetric property of the cross product, $\overline{u} \ x \ \overline{v} \ = \ - \ \overline{v} \ x \ \overline{u}$.

So understanding the cross product in this way is quite satisfying to me as we can easily justify why $\ \overline{u} \ x \ \overline{u} \ = \ 0$ without relying on these properties as definitions.

My questions are based on the fact that these properties can be justified in such an elementary way. If you've never seen the cross product explained they way it is in the .pdf's then I urge you to read them & think seriously about it. I'm sure these are justified in more advanced works in other ways but if an explanation can be given at this level I see no reason not to take it.

So lets look at an example & the steps taken that I think have explanations analogous to those of the cross product above:

$\ \overline{v}$ = v₁$\ \hat{e_1}$ + v₂$\ \hat{e_2}$

$\ \overline{w}$ = w₁$\ \hat{e_1}$ + w₂$\ \hat{e_2}$

(where $\ \hat{e_1}$ = (1,0) & $\ \hat{e_2}$ = (0,1)).

v ⋀ w = (v₁$\ \hat{e_1}$ + v₂$\ \hat{e_2}$) ⋀ (w₁$\ \hat{e_1}$ + w₂$\ \hat{e_2}$)

v ⋀ w = v₁w₁$\ \hat{e_1}$⋀$\ \hat{e_1}$ + v₁w₂$\ \hat{e_1}$⋀$\ \hat{e_2}$ + v₂w₁$\ \hat{e_2}$⋀$\ \hat{e_1}$ + v₂w₂$\ \hat{e_2}$⋀$\ \hat{e_2}$

v ⋀ w = v₁w₂$\ \hat{e_1}$⋀$\ \hat{e_2}$ + v₂w₁$\ \hat{e_2}$⋀$\ \hat{e_1}$

v ⋀ w= (v₁w₂ - v₂w₁)$\ \hat{e_1}$⋀$\ \hat{e_2}$

This is interpreted as the area contained in the parallelogram formed by v & w. No doubt you noticed that all of the manipulations with the $\ \hat{e_i}$'s have the exact same form as the cross product. Notice also the fact that this two dimensional calculation comes out with the exact same result as the cross product of

$\ \overline{v'}$ = v₁$\ \hat{e_1}$ +_v₂$\ \hat{e_2}$_+ 0$\ \hat{e_3}$

$\ \overline{w'}$ = w₁$\ \hat{e_1}$ + w₂$\ \hat{e_2}$ + 0$\ \hat{e_3}$

in ℝ³.Also the general

x ⋀ y = (x₁$\ \hat{e_1}$ + x₂$\ \hat{e_2}$ + x₃$\ \hat{e_3}$) ⋀ (y₁$\ \hat{e_1}$ + y₂$\ \hat{e_2}$ + y₃$\ \hat{e_3}$)

comes out with the exact same result as the cross product. The important thing is that the cross product of the two vector results in a vector orthogonal to $\ \overline{v}$ & $\ \overline{w}$ and that the result is the same as the wedge product calculation.

1: Can $\ \hat{e_1}$⋀$\ \hat{e_2}$ be interpreted as $\ \hat{e_3}$ in my above calculation?

What I mean is that can $\ \hat{e_1}$⋀$\ \hat{e_2}$ be interpreted as a (unit) vector orthogonal to the two vectors involved in the calculation that is scaled up by some factor β, i.e. β$\ \hat{e_1}$⋀$\ \hat{e_2}$ where β is the scalar representing the area of the parallelogram.

2: Just as we can algebraically validate why $\overline{u} \ x \ \overline{v} \ = \ - \ \overline{v} \ x \ \overline{u}$ why doesn't the exact same logic validate $\ \hat{e_1}$⋀$\ \hat{e_2}$ = - $\ \hat{e_2}$⋀$\ \hat{e_1}$?

If we think along these lines I think we can justify why $\ \hat{e_1}$⋀$\ \hat{e_1}$ = 0, just as it occurs analogously in the cross product. They seems far too similar for it to be coincidence but I can't find anyone explaining this relationship.

3: In general, if you are taking the wedge product of (n - 1) vectors in n-space will you always end up with a new vector orthogonal to all of the others?

If you are taking the wedge product of (n - 1) vectors then will you end up with λ($\ \hat{e_1}$⋀$\ \hat{e_2}$⋀...⋀$\ \hat{e_n}$) where the term ($\ \hat{e_1}$⋀$\ \hat{e_2}$⋀...⋀$\ \hat{e_n}$) is orthogonal to all the vectors involved in the calculation & the term λ represents the area/volume/hypervolume (etc...) contained in the (n - 1) vectors?

4: I have seen it explained that we can interpret the wedge product of $\ \hat{e_1}$⋀$\ \hat{e_2}$ as in the picture here, as a kind of two-dimensional vector. Still, the result given is no different to that of the 3-D cross product so is it not justifiable to think of $\ \hat{e_1}$⋀$\ \hat{e_2}$ as if it were just an orthogonal vector in the same way you would the cross product if you think along the lines I have been tracing out in this post? When you go on to take the wedge product of (n - 1) vectors in n-space can I not think in the same (higher dimensional) way?

That's it, thanks a lot for taking the time to read this I have tried to be as clear as possible, any contradictions/errors are as a result of my poor knowledge of all of this! :D

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    $\begingroup$ The operation of converting $e_1 \wedge e_2$ into $e_3$ is usually called the "Hodge star" operation. So although the two are not the same objects, there is a standard relationship between them. Yes, in general if you take a wedge of $n-1$ vectors in $\mathbb R^n$, via this construction you've defined a general "cross product" in any dimension. You can also use the determinant formulation of cross product to make this definition. $\endgroup$ – Ryan Budney Mar 7 '11 at 19:43
  • $\begingroup$ See the "three dimensional example" here: en.wikipedia.org/wiki/Hodge_dual $\endgroup$ – Ryan Budney Mar 7 '11 at 19:45
  • $\begingroup$ @RyanBudney Thanks, I was breaking my head on understanding the relationship b/w cross product (in 3D, of course), perpendicular vector in 2D and wedge product in general, now I think I'm starting to understand it. In 2D we do the wedge product with n - 1 i.e. 1 vector which gives the perpendicular vector in 2D space (y, -x) or (-y, x), while in 3D we do the cross product with, again n - 1 i.e. 2 vectors which gives us the vector orthogonal to both. $\endgroup$ – legends2k Feb 19 '13 at 12:41

The best introduction I know of to the exterior product is Sergei Winitzki's free book Linear Algebra via Exterior Products. Chapter $2$ in particular I think addresses all of your questions (it is unclear how much of Chapter $1$ you need to read in order to read Chapter $2$, I guess that depends on how much linear algebra you've had).

  • $\begingroup$ Wow, this book is awesome! $\endgroup$ – Greg Graviton Mar 7 '11 at 22:01
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    $\begingroup$ Here is the correct link if you want the free version. $\endgroup$ – Arnaud Mortier Feb 15 at 8:52

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