Wedge product and cross product - any difference?

Question

I'm taking a course in differential geometry, and have here been introduced to the wedge product of two vectors defined (in Differential Geometry of Curves and Surfaces by Manfredo Perdigão do Carmo) by:

Let $\mathbf{u}$, $\mathbf{v}$ be in $\mathbb{R}^3$. $\mathbf{u}\wedge\mathbf{v}$ in $\mathbb{R}^3$ is the unique vector that satisfies:

$(\mathbf{u}\wedge\mathbf{v})\cdot\mathbf{w} = \det\;(\mathbf{u}\;\mathbf{v}\;\mathbf{w})$ for all $\mathbf{w}$ in $\mathbb{R}^3$

And to clarify, $(\mathbf{u}\;\mathbf{v}\;\mathbf{w})$ is the 3×3 matrix with $\mathbf{u}$, $\mathbf{v}$ and $\mathbf{w}$ as its columns, in that order.

My question: is there any difference between this and the regular cross product or vector product of two vectors, as long as we stay in $\mathbb{R}^3$? And if there is no difference, then why introduce the wedge?

Cheers!

Answer 1

No difference at all.

I've been trying to write a little proof, but the software on this page seems to have forgotten how to write maths. :-(

Anyway: I assume that by "regular cross/vector product" you mean the definition with coordinates as in Wikipedia.

Try to compute both sides of your equation $(u\wedge v ) \cdot w = \det (u, v, w)$ with your definition using coordinates for the cross / vector product.

Since this equation defines a unique vector, has just one solution, once you've checked your definition with coordinates verifies it, that means that your definition with coordinates agrees with the present one.

EDIT. I forgot. The adventage of this "new" definition is that it is "coordinate-free". So what? -I can hear someone whispering at the end of the classroom.

So, if you ever wonder if your "former" definition could be affected just in case you change the (positive orthonormal) basis in which you're writing your vectors, the answer, you betcha, is "no".

Answer 2

I would have left this as a comment, but apparently I can't comment yet.

To answer your final question, why introduce the wedge, the point is that the wedge product (as explained in the Wikipedia article) is a notion that generalizes to $\Bbb R^n$ and indeed any vector space---in general the output is what is called a "bivector". Now, it just so happens that in $\Bbb R^3$ there is a natural identification between bivectors and vectors, and the vector that corresponds to the (bivector) wedge product is indeed the cross product, but this does not work in other vector spaces.

So, to sum up, the notion of cross product is specific to $\Bbb R^3$, whereas wedge product makes sense in any vector space.

Answer 3

There is a difference. Both products take two vectors in $\mathbb{R}^3$. The cross product gives a vector in the same $\mathbb{R}^3$ and the wedge product gives a vector in a different $\mathbb{R}^3$. The two output vector spaces are indeed isomorphic and if you choose an isomorphism you can identify the two products. However this isomorphism is a choice, or to put it another way depends on fixing a convention.

In higher dimensions the wedge product gives a vector in a vector space of higher dimension and so no identification is possible.

Answer 4

Yes, this definition is chosen for a reason, as the unique solution to a pedagogical problem.

Do Carmo's definition is awkward and redundant in 3 dimensions, but it is the only one among the usual definitions for the cross product that when generalized to $n$ dimensions (there is a cross-product of $n-1$ vectors in $R^n$) is rigorous, visibly basis independent, and produces a vector in the same space --- all without a long digression on multilinear algebra, duality, and other sophisticated theory. Consider the other definitions:

1. The right hand rule does work in $n$ dimensions, and is coordinate-independent, but it is hard to describe without a formula which hand is the "right" one, and verifying that the result is multilinear can be awkward if one does not have a reliable formalism of $n$-dimensional Euclidean geometry. I doubt most students would consider this approach rigorous even if, in principle, it can be made so.

2. The exterior algebra relies on theory not known by all audiences of Do Carmo's book (e.g, engineers). It does not produce a vector in $R^n$ without even more theory.

3. The definition as a determinant of the vector coordinates can be confusing, with vectors and scalars mixed together as entries of one matrix. It is also geometrically meaningful only when proven independent of the choice of (orthonormal) basis for the coordinates. In 3-dimensions this is known by the equivalence with the right hand rule and by drawing pictures, techniques that require a long development to perform in $n$ dimensions.