What is the relationship between the exterior and cross products?

[Swapnil]

How are the exterior products and the cross products related?

Wikipedia says: "The cross product can be interpreted as the wedge product in three dimensions after using Hodge duality to identify 2-vectors with vectors."

 

[Swapnil]

Anyone? ......

 

[pkleinod]

The cross product can only be defined in three dimensions. It is *defined* to
be the the dual of the exterior product:
a x b := -I a^b, where I=e1e2e3 is the pseudoscalar and e1, e2, and e3 are
orthogonal unit vectors that span the space. See the following link for more:
http://www.mrao.cam.ac.uk/~clifford/publications/abstracts/imag_numbs.html

 

[mathwonk]

the cross product of 2 vectors in R^3 is another vector in R^3. The exterior product of two vectors in R^n is a bivector in a space of dimension "n choose 2".

Thus we get an object in a 3 dimensional space from the exterior product of 2 vectors in R^3, which by choosing a basis, of merely volume form, we can view as a vector.

In R^n we could similarly view a product of n-1 vectors as a vector, so we could take the cross product of more than 2 vectors in higher dimensions.

the geometry is that if we have n-1 vectors they usually span an n-1 dimensional block. so they act on vectors as follows: given another vector, all together we get an n dimensional block and we can take its volume.

thus n-1 vectors assign a number to another vector, the volume of that block.

moreover if the last vector chosen is in the spane of the first n-1, the number assigned is zero. so we could represent this action by dotting with some vector perpendicular to the span of the first n-1 vectors. this last named vector would be called the cross product of the first n-1.

i.e. the cross product of n-1 vectors is a vector perpendicualr to them, whose length equals the volume of the n-1 block they span, and whose orientation with them gives an oriented n block.

the exterior product of n-1 vectors is a gadget representing the n-1 block they span, including its span and its volume.

thus one could also form the cross product of k vectors in n space, getting an (n-k) multivector. you just need enough to fill out an n block.