# Cross Products and Determinants

1. Sep 22, 2006

### Swapnil

Is the fact that
$$\vec{a}\times\vec{b} = \det \begin{bmatrix}\hat{e}_1& \hat{e}_2 & \hat{e}_3 \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{bmatrix}$$

just a coincedence (ie. a mneminic device) or does it have some deep mathematical significance?

edit: Also, is the cross product, like the dot product, defined for higher dimentions?

Last edited: Sep 22, 2006
2. Sep 22, 2006

### Oxymoron

It does indeed has a deeper mathematical significance. Note: Please take the following with a grain of salt because it very brief and I could have skipped stuff or over-simplified some things. Hopefully the gist of it is correct.

Ok, you know that whenever you do determinants you end up calculating a lot of stuff like

$$\det\begin{bmatrix}a & b\\c& d\end{bmatrix} = (ab - bc)$$

you know, when you take the top left entry and multiply by the bottom right entry and then subtract the top right entry multiplied by the bottom left entry - continuing in this fashion more than once for larger matrices and inserting the correct plus or minus sign between calculations. You got that? (you will see later that this is like an anti-symmetric operation) For an arbitrarily large matrix you have the more generalized formula

$$\det M = \sum_{\pi\in S_n} \mbox{sgn}(\pi) M_{1_{\pi_1}}M_{2_{\pi_2}}\dots M_{n_{\pi_n}} = \sum_{\pi \in S_n} \epsilon_{\pi_1\dots\pi_n}M_{1_{\pi_1}}M_{2_{\pi_2}}\dots M_{n_{\pi_n}}$$

which basically says you do

$$\det\begin{bmatrix}a & b \\ c & d \end{bmatrix} = ab - bc$$

or

$$\det\begin{bmatrix}a & b & c \\ d & e & f \\ g & h & i \end{bmatrix} = aei + bfg + cdh - afh - ceg - bdi$$

Now the point is that there is something called an anti-symmetric operator and it just so happens that the act of calculating the determinant of a matrix is an anti-symmetric one. Now in Exterior Algebra we use anti-symmetric operations a lot! Especially when we perform exterior products. Unlike cross products, exterior products are very, very general and work in n-dimensions. They are defined using anti-symmetric operators - which means that they involve formulas that look almost exactly like the formulas involved with determinants.

I dont know exactly what you know but if you were to gain anything from this is that the exterior product is an operator defined on vectors of arbitrary dimension, not just three. The act of performing an exterior product on two vectors involves an anti-symmetric operator (formally, it looks like this:

$$u \wedge v = \mathcal{A}(u \otimes v)$$

where A is the anti-symmetric operator. Whenever you see the antisymmetric operator you should think of calculations like

$$u_i v_j - u_j v_i$$

much like that in the calculation of determinants. Lets look at an example:

$$(u \wedge v)(a,b) = \mathcal{A}(u \otimes v)(a,b)$$
$$=\frac{1}{2}(u(a)v(b) - u(b)v(a))$$

Here you can see that the wedge product of two vectors involves a calculation very similar to that of a determinant.

I guess the moral of this story is that the cross product of two 3-dimensional vectors is really an exterior product that should involve wedge products and anti-symmetrization operations, Hodge stars, etc... all the usual stuff from Exterior algebra. The reason you see it very early instead of in some advanced maths course is because the determinant is kind of like the anti-symmetrization operation, so you can replace all the hard exterior algebra stuff with easy determinants and multiplication, and all the dimensions work out nicely because the Hodge star vanishes. I guess we are just lucky that someone figured out the simplification to allow us to work with the cross product in 1st year instead of 3rd year.

Here is a brief calculation of the relation that exists between the cross product and the exterior product.

Have you heard of the Levi-Civita symbol? If so then you should agree that the cross product may be written in terms of this symbol as such:

$$(\bold{u}\times\bold{v})_i = \epsilon_{ijk}u_j v_k$$

$$=\frac{1}{2}\epsilon_{ijk}(u_j v_k - u_k v_j)$$

$$=\epsilon_{ijk}(u \wedge v)_{jk}$$

$$=\star(u \wedge v)_i$$

This little calculation shows that the cross product of two vectors (in bold because they are in a 3-dimensional manifold) is related to their wedge product and the Hodge star. Note: The last equality is arises due to the fact that the Hodge star of the identity element in 3 dimensions is

$$(\star 1)_{ijk} = \epsilon_{ijk}$$

Last edited: Sep 22, 2006
3. Sep 22, 2006

### Swapnil

Thank you for your elaborate response. I didn't quite understand the things about Hodge Stars and anti-symmetric operators (mainly because I haven't studied Exterior Algebra yet) but I get the overall picture.

4. Jan 30, 2011

### ralphiep

Dear Oxymoron,

Thank you for offering an excellent discussion of the similarities obtaining between cross products and determinants. In my middle sixties I am finally knocking over calculus questions that I could not handle at 20. Go figure! I am, however, a little unsure as to what 'wedge products' are. When you are disposed to could you give a little gloss on that topic. Thanks again for your comprehensive and uncondescending discussion.

Ralphiep