why there is no four dimensional cross product.

This is great from the point of view of the physics. We define the angular momentum for a particle as the Cross product of the position and the linear momentum, but when you try to do the same for fields, specially when you study classical fields theory, you can't define a four dimensional Cross product in order to define a four dimensional angular momentum. You use instead a tensor that depends on the position and the momentum-energy tensor to construct the angular momentum tensor and the spacial components of that tensor are the ones that gives the angular momentum vector, that corresponds to the old three dimensional definition.

An interesting interpretation of the cross product in R^1: the resulting vector from a cross product must be orthogonal to the input vectors, and two vectors are orthogonal if and only if their dot product is 0. Every non-zero vector in R^1 is parallel to every other non-zero vector, and so their dot product is never 0. When any vector is dotted with the 0 vector, however, you always get 0, in any dimension. This means two neat things: the 0 vector is orthogonal to all vectors (including itself) in any number of dimensions, and, in R^1 specifically, it is the only vector which is orthogonal to any other vector in the space, which means it must be the result of every cross product in R^1. Thanks for the great video, Michael!

Even though there are no cross products in 5D, I still take consolation in the fact that I can nonetheless play "Rock, paper, scissors, lizard, Spock."

Geometric Algebra be like "Am I a joke to you"?

note: i have not seen the video yet i think of the cross product as the dual of the wedge product in G3. if we use this to define an n-dimensional cross product, then the 'reason' there isn't a 4d cross product is that the dual of a wedge product between two vectors in G4 isn't a vector, its another bivector, which doesn't exist in normal linear algebra

Geometric Algebra has this thing called the “wedge product” or “outer product”, which works in any dimension. The cross product is redundant in 3D Geometric Algebra.

The video establishes that the allowed values of n for an n-dimensional cross-product are 1, 3, 7 (very surprised to find that there's a 7-dimensional cross-product!) I note: 1 = the dimension of the "imaginary part" of the complex numbers 3 = the dimension of the "vector part" of the quaternions 7 = the dimension of the "non-real part" of the octonions (is there such a thing as the "non-real part" of the octonions?) And, as shown in an earlier video, the complex nos, quaternions and octonions are the only normed division algebras of dimension > 1 Does the existence of a cross-product of order n imply the existence of a division-algebra of order n+1 (1<=n)? I.e., could we treat the paucity of cross-products as a corollary of Hurwitz's Theorem?

I've always looked at it from the perspective that you input n vectors of n+1 dimensions and it returns a vector orthogonal to all of the inputs. So, in 4-D, you need 3 vectors, and the cross product is rather a function, not an operation.

I think I like Geometric Algebra better. Wedge Product Baby!

I feel like there should have been a disclaimer that one can define a generalisation of the 3D cross product using the hodge star of the wedge product/the Levi-Civita contraction of n-1 vectors in R^n. In a lot of applications this is precisely what one wants.

That was very interesting! When we defined cross product in an abstract way, we said it's the Lie-bracket of a Lie-algebra on R^3, so I wonder how it would change the result if you assumed the Jacobian identity instead of |a×b|^2 = |a|^2|b|^2 - (a "dot" b)^2 . Also, fun fact: the reciprocal of the fine-structure constant is approximatly 137, whose digits are the solution of the problem :)

Surely the cross product is most clearly understood as a determinant x^y = | i j k | | x1 x2 x3 | | y1 y2 y3 | in which i, j and k are the three orthogonal unit vectors for the coordinates. This has an application in the geometry of the projective plane m = x^y where x and y are the homogeneous coordinates of two points, and m is the homogeneous coordinates (known as tangential coordinates) of their line join. It is easily seen that m.x=0, m.y=0 so the points both fall on the line. Here i, j, and k are placeholders for the so-called triangle of reference i.e. the points (1,0,0); (0;1;0); (0,0,1). The dual relationship is x = m^n where x is the point intersection of lines m and n. The generalisation to 4 coordinates is a = | i j k l | | x1 x2 x3 x4 | | y1 y2 y3 y4 | ' | z1 z2 z3 z4 | a is the coordinate set for the plane holding x, y, and z, Again a.x=0, etc. The construction dualises to the point intersection of three planes.

Understanding the relationship between cross product and wedge product is an easier way to explain this. The cross product is the duel of the wedge product in 3D space. More correctly it is the complement of the wedge product in 3D space i.e axb = *a^b where * is the Hodge star operator. The *a^b is only a vector in 3D space, in 4D space it would be a Bi-vector.

Great to see you getting recognized for your great videos by Brilliant. Your work is more than worthy of a sponsorship.

45 min video,michael penn started competing netflix now.

Cross product is awkward anyway. Outer product is the way, and is generalizable.

Would be great to see what you think about Geometric (Clifford) Algebra.

it looks like the axioms don't uniquely define the cross product. for instance in R3, I could have e1xe2 = -e3, e2xe3 = -e1 and e3xe1 = -e2 and all the axioms are satisfied i think.... There is no chirality given by the axioms.

remember about the way to compute the cross product with a determinant? that if you have a x b it would be | i j k | | a1 a2 a3 | | b1 b2 b3 | well I tested it and if you try this same method for a 2D vector, you also get one that is perpendicular | i j | | a1 a2 | if a = (a1, a2) that determinant gives you (a2, -a1), and that vector is always perpendicular to a in 2D if with the same pattern we make this determinant for 4D it would be | i j k w | | a1 a2 a3 a4 | | b1 b2 b3 b4 | | c1 c2 c3 c4 | of course you need a third vector, because in 2D with one vector you can only have 1 direcction that is perpendicular, but in 3D you have infinite directions that are perpendicular to 1 vector, but if you use 2 there is only one direction (positive and negative directions I'm counting as the same) and as might seem logical, in 4D space there must be infinite directions that are perpendicular to 2 vectors, so to get just one you need 3 I don't know how to prove that the vector resultant is always perpendicular to the others, but I tried using a=(1,0,0,0) b=(0,1,0,0) c=(0,0,1,0) and I got the vector (0,0,0,1), so at least there it works also I don't recommend to get stuck in the notation of this kind of stuff, because this being a 3 object operation, you cant really type it out in a simple way like with multiplication or something like that, and to try to derive to a 4dimensional equation using only our 2 dimensional expressions seems kindof weird to me with this I'm reffering to the non comuntative properties of the cross product for example, yes with 2 is simple enough, there are just 2 ways of doing the operation, but with 3 is a whole new thing that should be studied apart from what we know

This video is pretty nice to explain the stuff