3
$\begingroup$

I have seen the wedge-product as being defined in differential geometry in the definition of a differential form or p-form. Now in the course we have proven the basic properties of this product and how to take the differential.

Now when we apply this to the differentiation of a function in on a curved manifold, we make the following change in the integral $\int dx^n\rightarrow\int dx^0\wedge dx^1\wedge...\wedge dx^{n-1}$.

I don't see how the wedge product is related to the volume element, I was hoping that you guys might be able to clear that up for me ?

Second, I already asked around about this, the wedge product is said to be the generalisation of the cross vector product and the 3D volume element to higher dimensions. I was woundering of anyone could explain that ?

$\endgroup$

1 Answer 1

5
$\begingroup$

The right place to start is with determinants. The determinant of a 2 by 2 matrix can be thought of geometrically as the (signed) area of the parallelogram formed by the column vectors. The exterior product is a way of formalizing this intuition. The expression $dx^0\wedge dx^1\wedge...\wedge dx^{n-1}$ is by definition the volume element. The wedge product is a "generalisation" of cross product only in the sense that both involve the area of the parallelogram in one way or the other. Otherwise there are significant differences between them; thus, the exterior product is associative but the cross product is not. There is a good introduction to these ideas here.

$\endgroup$
1
  • $\begingroup$ Thanks a lot, the link was extremely helpful ! $\endgroup$
    – Nick
    Commented Jun 23, 2013 at 21:31

You must log in to answer this question.

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