# Exterior product

1. Nov 8, 2011

### dimension10

I have a question about the exterior product. Is it true that

$$\mathbf{a}\wedge\mathbf{b}=|| \mathbf{a}\times \mathbf{b}||$$

If not, how does one relate the exterior product to the cross product?

2. Nov 9, 2011

### homeomorphic

Presumably, you mean that a and b are vectors in R^3. Wedge product of vectors would be a bivector, so it couldn't possibly be equal to the left hand side.

If you take a wedge b and then the Hodge dual of that with respect to the Euclidean metric, you get the cross product.

Crash course in Hodge duals for this case:

Let e1, e2, e3 be a basis for R^3.

The Hodge dual of e1^ e2 can be written like (e1^e2)* and it will be e3.

And similarly, we have (e2^e3)* = e1 and (e3^e1)* = e2.

Then you can extend by linearity to the vector-space of all bivectors. You can see that this gives you the cross product.

3. Nov 9, 2011

### dimension10

Thanks.

I think I was confused by the picture in Wikipedia which makes it seem like the exterior product is the area of the parallelogram. This time, I noticed that there were 2 arrows around it, making it a bivector. So am I right to say that the magnitude of the exterior product is the same as the magnitude of the cross product whereas the exterior product itself is not the same as the magnitude?

Thanks.

4. Nov 9, 2011

### chiro

Hey dimension10.

Are you familiar with the determinant form of the exterior product?

5. Nov 9, 2011

### dimension10

The alternating exterior product?

6. Nov 9, 2011

### chiro

Just before I give an answer, I just want to be clear: is the wedge product and the exterior product the same thing? (I was under the impression it was).

7. Nov 10, 2011

### dimension10

I think they are.

8. Nov 10, 2011

### chiro

In that case just use the definition of the wedge product (for R3 given two initial vectors) and the cross product and then use the property of norms (i.e. expand out the terms) to show they are equal.