Is the vector cross product only defined for 3D?

Question

Wikipedia introduces the vector product for two vectors $\vec a$ and $\vec b$ as $$ \vec a \times\vec b=(\| \vec a\| \|\vec b\|\sin\Theta)\vec n $$ It then mentions that $\vec n$ is the vector normal to the plane made by $\vec a$ and $\vec b$, implying that $\vec a$ and $\vec b$ are 3D vectors. Wikipedia mentions something about a 7D cross product, but I'm not going to pretend I understand that.

My idea, which remains unconfirmed with any source, is that a cross product can be thought of a vector which is orthogonal to all vectors which you are crossing. If, and that's a big IF, this is right over all dimensions, we know that for a set of $n-1$ $n$-dimensional vectors, there exists a vector which is orthogonal to all of them. The magnitude would have something to do with the area/volume/hypervolume/etc. made by the vectors we are crossing.

Am I right to guess that this multidimensional aspect of cross vectors exists or is that last part utter rubbish?

Answer 1

Yes, you are correct. You can generalize the cross product to $n$ dimensions by saying it is an operation which takes in $n-1$ vectors and produces a vector that is perpendicular to each one. This can be easily defined using the exterior algebra and Hodge star operator http://en.wikipedia.org/wiki/Hodge_dual: the cross product of $v_1,\ldots,v_{n-1}$ is then just $*(v_1 \wedge v_2 \cdots \wedge v_{n-1}$).

Then the magnitude of the cross product of n-1 vectors is the volume of the higher-dimensional parallelogram that they determine. Specifying the magnitude and being orthogonal to each of the vectors narrows the possiblity to two choices-- an orientation picks out one of these.

Answer 2

Well, it depends on what you mean by "the vector cross product." There is a generalization to $n$ dimensions which takes $n-1$ vectors as input and returns what can be thought of as a vector orthogonal to all of them. It generalizes to an operation taking $k$ vectors as input where $k \le n$, but then the output is not something like a vector but something more complicated. See wedge product.

There is a more specific generalization to $7$ dimensions coming from multiplication in the octonions in the same way that the cross product can be thought of as coming from multiplication in the quaternions.

Answer 3

The cross product is very closely related to the concept of quaternions. The cross product "identities" that operate on the unit vectors $\rm \mathbf{\hat{i}}, \mathbf{\hat{j}}, \mathbf{\hat{k}}$ are essentially similar to the identities on the quaternion units (basis elements) $i,j,k$.

Notice that quaternions have four units: $1, i, j, k$, and the 3-D cross product works in vector spaces of dimension 4-1 = 3.

The seven-dimensional cross product is analogous to the octonions, and has a similar definition which I do not wish to enumerate here.

These are the only dimensions in which a "cross-product" of such a sort exists. The relation between the vector operation and multiplication of quaternions/octonions is the underlying reason why. Quaternions and Octonions compose what is known as a closed normed division algebra. And real closed division algebras can only have dimensions of 1,2,4,8.

As @QiaochuYuan mentions, the generalization to other dimensions yields something different than a vector -- such an operation exists and can be used in a similar way, but you do not have the niceness of retrieving a vector in the end.

Answer 4

See my answer here to see an example of a generalization of the cross product to 4 dimensions. Notice, this generalization works in $n$-dimensions and always return a vector orthogonal to all $n-1$ vectors you use. And, you calculate it almost exactly the same way you calculate the normal cross product, nothing complicated. To get the cross product of $n-1$ vectors of dimension $n$, you simply make a matrix which has top row with entries $i_1, i_2, \ldots, i_n$, which generalize the normal $i, j, k$ in 3 dimensions. Then, your next $n-1$ rows are the $n-1$ vectors of dimension $n$. Now, take the determinant and you get your $n$-dimensional result.

Reference: I learned about this when I took a 4th semester of calculus in college, where we used Vector Calculus by Susan Jane Colley. It is introduced in the exercises for Section 1.6. One of the exercises is to prove that the new vector is orthogonal to the previous ones.

Answer 5

You can define the ternary cross-product as the determinant $$ \vec v_1 \times \vec v_2 \times \vec v_3 = \left\vert\begin{matrix} \vec i & \vec j & \vec k & \vec l \\ v_{1x} & v_{1y} & v_{1z} & v_{1t} \\ v_{2x} & v_{2y} & v_{2z} & v_{2t} \\ v_{3x} & v_{3y} & v_{3z} & v_{3t} \end{matrix}\right\vert $$ that you have to develop along the first row.

If you have a parametric surface $s(u,v,w) \to \mathbb{R}^4$ then the ternary cross-product $\frac{\partial s}{\partial u}(u,v,w) \times \frac{\partial s}{\partial v}(u,v,w) \times \frac{\partial s}{\partial w}(u,v,w)$ gives the normal at $s(u,v,w)$.