Is Wedge Product the Same as Cross Product?

[hideelo]

I guess the title really says it all. I'm reading "space-time algebra" by David Hestenes, and he explains the properties of the wedge product (area of parallelogram, antisymmetric, etc.) and it sounds just like the cross product to me. He then says clearly however that this is not to be confused eith the cross product since the definition of the a X b depends on the dimensionality of the vector space in which the vectors are embedded whereas a ^ b does not. I don't see how this is the case since he want a ^ b to have an orientation in the space in which it is embedded, if so, does it not also depend on the dimensionality of the surrounding space?

P.S. If you have another way to show the distinction between them aside from his approach, I am open to, and interested in, hearing it.
Thanks.

 

[OldEngr63]

I have published in well known journals using the a^b notation for axb, so for my money, they are the same thing.

 

[Stephen Tashi]

The book "Geometric Algebra for Computer Science" by Dorst, Fontijne and Mann says that
$a \times b = (a \wedge b)^*$, which is the orthogonal complement of $a \wedge b$.

It isn't clear whether the question is "Is the cross product equivalent to the wedge product in 3-D euclidean space?" or "In a space of any given dimensions, is there a cross product equivalent to the wedge product ?".

 

[lavinia]

The wedge product of vectors is not a vector but is an element of a more general space called the Grassmann algebra of a vector space. It's definition does not require a metric so by itself it has no idea of orthogonal complement or area of a parallelogram. The Grassman algebra is defined for any dimension. A typical element is a wedge product of a finite number of vectors.

With a metric one can interpret the wedge product of two vectors as spanning the area of a parallelogram. But these areas are different for different metrics. Note also that this is an oriented area. a^b =-b^a.

Further,without a metric and an orientation there is no idea of orthogonal complement. In 3 space,with the Euclidean metric(actually any positve definite inner product will work) and an orientation, there is a unique vector that is perpendicular to the plane spanned by a and b such that a,b,axb is positively oriented and |axb| is the area of the parallelogram. So in this special case, a^b can be identified as a vector. This unique vector is written as axb mostly in Physics and Engineering books.

But in higher dimensions,there is no such vector naturally corresponding to a^b.. Even with a metric and an orientation,the orthogonal complement will be an element of the Grassman algebra, ie. it will be a wedge product of n-2 vectors where n is the dimension of the space.

The generalization of axb to n dimensions,that works for any n,is to start with the wedge product of n-1 vectors and then there will be a unique vector that is orthogonal to the n-1dimensional hyperplane that they span, is properly oriented and whose length is the volume of the parallelopiped that the vectors span.

 

[mathwonk]

the connection seems to be that in some sense, w is the cross product of v1,...,vn-1, if for all u, w.u = v1^...^vn-1^u, where this last wedge must be interpreted as a scalar, using the given metric.

 

[lavinia]

The general and natural cross product that exists in any dimension above 2 is the one that uses n-1 vectors as described above and Mathwonk has elegantly characterized.

But this still leaves open the question of whether there are any possible ways to define 2 vector cross products in other dimensions. I didn't think of this at first but a web search for generalized cross product came up with this Wikipedia link.

http://en.wikipedia.org/wiki/Seven-dimensional_cross_product

which shows that there is also a 2 vector cross product in 7 dimensions.
Apparently, 7 is the only other dimension possible.

- By cross product what is meant is that the following four conditions should be satisfied:

VxW is bilinear
The vector, VxW, is perpendicular to the plane spanned by V and W
|VxW| equals the area of the parallelogram spanned by V and W
VxW = - WxV

- If I understand the article right, the reason that 3 and 7 dimensions are the only possible for a 2 vector cross product is that whenever there is a 2 vector cross product on an n dimensional vector space there is a multiplication defined in n+1 dimensions by the rule,

(*) (a,V).(b,W) = (ab-V.W, aW + bV + VxW)

where V.W is the inner product of the n-1 dimensional vectors, V and W, and a and b are real numbers
And this defines what is called a normed division algebra.

For the usual cross product in 3 dimensions this 4 dimensional multiplication is the quaternions. In 8 dimensions it is the octonians.

A theorem of Hurwitz then states that the quaternions and the octonians are the only two possible examples of normed division algebras over the real numbers in dimensions greater than 2. Here is a link to the statement of the theorem.

http://en.wikipedia.org/wiki/Composition_algebra

- Equation (*) shows that the cross product can be retrieved from n+1 dimensional vectors whose first coordinate is zero. For quaternions this means that if one interprets two 3 dimensional vectors as pure quaternions, then the cross product is just their product as quaternions. For 7 dimensional vectors, one interprets them as pure octonians.

- According to the first article, the are 480 different cross products in 7 dimensions. Using the formula (*) each of them forms a normed division algebra that is isomorphic to the octonians.

- I am not sure what the orientation properties are of the various 7 dimensional cross products. It would be interesting to look at them.

 

[mathwonk]

so cool!