Cross Product defined in $\mathbb R^n$

Question

I am in an introductory Linear Algebra course and we were recently introduced to the cross product. I have read online and heard from others that the cross product is defined in $\mathbb R^3$ only. However, when we learnt it my teacher told us that it is defined for any vector space, and mentioned nothing about only in 3 dimensions.The formula that we learnt is this:

If $x_1,x_2,...,x_{n-1}$ are vectors in $\mathbb R^n$, then their cross product $$w=\ast(x_1,x_2,...,x_{n-1})$$ is characterized by $w^Tx=\det(x_1x_2...x_{n-1}x)$, and can be calculated by substituting $x$ as $e_i$ for $1\le i\le {n-1}$.

I was wondering, is this how the actual cross product is defined and most people learn it only for $\mathbb R^3$, or is this my teacher extending the definition to a more general case? I've been trying to find out online but I've had no luck.

Answer 1

The following is why the cross product is defined in $3$ dimensions only.

The issue with defining the cross product is that it's not precisely what you think it is. If you've learned "the right hand rule", you may know that to compute a cross product you:

1. Sweep out a plane with your fingers (For $u\times v$, the plane is just $\text{Span}(u,v)$).
2. Take a specific perpendicular vector (the way your thumb is pointing, not the reverse direction).

This means we can break the "take a cross product" function into a few sub-functions:

1. Given two $u,v$ find the (oriented) plane $\text{Span}(u,v)$
2. Uniquely convert from this oriented plane to a vector.

Step 2 is what fails in general. Specifically, given two vectors, we can always form what's known as a bivector (just the oriented plane mentioned earlier). This is via an operation called the exterior product, which is very similar to the cross product in $\mathbb{R}^3$. Then, we want to convert this bivector into a vector so $f(x,y) = x\times y$ is a function from vectors to vectors (instead of from vectors to "bivectors").

This is done via the Hodge Dual --- given a $k$-vector in $\mathbb{R}^n$ (so an oriented span of $k$ vectors, not just $2$ like with a bivector), this gives a natural way to get a $n-k$ vector. So, for a $2$ vector in $\mathbb{R}^3$, we get a $3-2 =1$-vector, or a normal vector. The Hodge Dual is precisely the (generalized) process of using the right-hand rule, but only in $\mathbb{R}^3$ do we have that $n - 2 = 1$, giving us the conversion back we want.

So, we have an easy way to generalize the cross product to higher dimensions if we're ok with dealing with "bivectors", and in general $k$-vectors. If we aren't willing to do this, it gets much more difficult (There are some oddities like the Seven dimensional cross product, but nothing in general that works and avoids the wedge product).

The wedge product is essentially what your professor has defined for you, so it will be the correct generalization (although the output won't be a "vector" unless you look at the wedge product of $k$ vectors such that $n-k = 1$, so $k = n-1$ vectors).

The field that studies this is known as Geometric Algebra, if you're interested in learning more.

Answer 2

There are two ways to define the cross product (in $\mathbb{R^3})$. Both are equivalent. The easiest way is to define, for $\mathbf a,\mathbf b \in \mathbb{R^3}$ :

$$\mathbf a \times \mathbf b := \text{ the unique vector } \mathbf y \text { having magnitude } \Vert \mathbf y \Vert = \Vert\mathbf a\Vert \Vert \mathbf b \Vert \sin \theta, \text{ where } \theta \text{ is the angle between } \mathbf a \text{ and } \mathbf b \text{ and direction given by the right-hand rule.}$$

This definition effectively defines the cross product between two vectors as the vector having a length equal to the area of the parallelogram formed by them and having a direction that is perpendicular to both of them.

One can then show that in component form, if $ \mathbf a = (a_1,a_2,a_3)$ and $\mathbf b = (b_1,b_2,b_3)$, then

$$ \mathbf a \times \mathbf b = \det \left( \begin{bmatrix} \mathbf i & \mathbf j & \mathbf k \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{bmatrix} \right) $$

Where $\mathbf i = \mathbf e_1$, $\mathbf j = \mathbf e_2$, and $\mathbf k = \mathbf e_3 $.

In theorem-proving classes, specifically analysis, it is normally just defined via the determinant formula. This is because it is generally accepted that one one should not use geometric reasoning in calculus-based proofs. Your proofs and definitions can be motivated by geometric reasoning, but you should not use geometric arguments in your proofs. As such, most theorems involving the cross product make use of that formula alone, and ignore its geometric meaning. In physics, on the other hand, the geometric definitions tends to prevail. Examples of its use include the definition of torque and magnetic flux. Obviously only the second definition can be generalized. This situation is similar to the generalization of the dot product. Geometrically the dot product measures how "close" the vectors are to being perpendicular. Obviously you can't have things be "perpendicular " in the usual sense in settings have more than 3 dimensions. However, in three dimensions, the dot product of the vectors mentioned above turns out to be equal to

$$\mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3$$

This latter notion can be generalized, and it is. The notion of things "being perpendicular" in higher (or even infinite) dimensions tends out to be extremely important, and you will cover this at some point in your class when you study orthogonal vectors.I'm not aware of any generalizations of the cross product. The closest thing to it would be something called the wedge product, which is more in line with what your professor has given you.