# Cross product of 4D Vectors

I am trying to understand how to do the cross product of two four-dimensional vectors. From what I understood it's not possible, unless the vectors are of the form $\mathbb R^3 \times \{0\}$ or $\{0\} \times \mathbb R^3$.

So for example, if I have $u = (1, \frac12, 0, 0)$, $v = (0, 2, 1, 0)$, what is the cross product $w = u \times v$?

• It's not usually defined. If you're using a source that does define some cross product in $\Bbb R^4$ (or $\Bbb R^3\times\{0\}$ or whatever), then you'll have to look at how they define it to say for sure (though if I had to guess, I'd say they probably just define it as $(u_2v_3-v_3u_2, u_3v_1-u_1v_3, u_1v_2-u_2v_1, 0)$).
– user137731
Jun 10, 2017 at 18:32

Four-dimensional Euclidean space does not have a binary cross product. (If it did, you could use it to define a five-dimensional division algebra, which isn't possible.) However, if you choose a three-dimensional subspace of $$\mathbb{R}^4$$, then that subspace inherits the inner product, which uniquely determines a cross product (up to choice of orientation).

You seem to be talking about $$\mathbb{R}^3\times\{0\}$$ as a 3D subspace of $$\mathbb{R}^4$$, in which case to calculate the cross product of two vectors (in this 3D subspace) you simply ignore the fourth coordinate (which is $$0$$) and do the calculation with the first three coordinates.

There is a ternary cross product on $$\mathbb{R}^4$$ in which you can compute a vector perpendicular to three given ones, with size and orientation based on the parallelotope generated by the three vectors (instead of a parallelogram as with two vectors). This can be calculated with differential forms if one was so inclined.

• Correct, cross products are only defined in $\mathbb{R}^3$ and $\mathbb{R}^7$, because of the relationship of the cross-product to quaternions and octonions respectively. en.wikipedia.org/wiki/Seven-dimensional_cross_product Aug 26, 2020 at 18:59
• How does one compute the ternary cross product? Jan 5, 2022 at 18:15

Cross products are only defined in $$\mathbb{R}^3$$ and $$\mathbb{R}^7$$, because of the direct relationship of the cross-product to quaternions and octonions respectively.

$$\mathbb{R}$$ (the reals), $$\mathbb{C}$$ (the complex numbers), quaternions and octonions are the only examples of division algebras that exist. Moving from $$\mathbb{C}$$ to the quaternions, you lose commutativity; moving from the quaternions to the octonions, you lose associativity.

If the reason why you are trying to take the cross product of two four-dimensional vectors is because you want to find another vector that is orthogonal to both vectors, then you could consider this suggestion, although you need to use three vectors in $$\mathbb{R}^4$$ to find a fourth vector that is orthogonal to all three.

Being a vector, the vector product has a magnitude and direction.

For the magnitude, you can take the usual definition, i.e., $$||w||=||u||\cdot||v||\cdot \sin(u,v),$$ with $\cos(u,v) = u\cdot v / (||u||\cdot ||v||)$ and $\sin^2 + \cos^2 = 1$.

For its direction, you could impose the usual $w \perp u$ and $w \perp v$, i.e., $w \cdot u = 0$ and $w \cdot v = 0$.

Presumably, an extra condition is required to define $w$ uniquely in the fourth dimension.