4 ms·

Alright, I got nerd sniped halfway through when he said: Scalar triple products only exist in 3D, because they involve cross products which only exist in 3D

Alright, I got nerd sniped halfway through when he said: Scalar triple products only exist in 3D, because they involve cross products which only exist in 3D (please keep the “they exist in 7D too” comments for some other time). There isn't a reference, and me googling "higher dimensional cross product" seems to return results generalising to any number of dimensions. Can anyone explain what he's talking about or provide references? Off the top of my head I can't see why you can't generalise cross products to D > 3. Or if you can't I don't see what's special about dimensions 3 and 7...

mijoharas 3y ago
Ok... I'll leave the comment to stand but apparently I just needed to scroll a tiny bit more and get an answer here[0]. The seven-dimensional cross product is one way of generalizing the cross product to other than three dimensions, and it is the only other bilinear product of two vectors that is vector-valued, orthogonal, and has the same magnitude as in the 3D case.[2] In other dimensions there are vector-valued products of three or more vectors that satisfy these conditions, and binary products with bivector results. [0] https://en.m.wikipedia.org/wiki/Seven-dimensional_cross_product https://en.m.wikipedia.org/wiki/Seven-dimensional_cross_prod...
ninepoints 3y ago
The exterior aka wedge product generalizes, as do determinants. Cross products absolutely do not
- mijoharas 3y ago
  Could you explain some more about why?
  - ninepoints 3y ago
    The simplest counterexample is just to consider vectors in 2D. It is not possible in general to construct a third vector perpendicular to two other vectors unless those other vectors are collinear or anitcollinear. The notion of an exterior product, represented as a signed area subtended by the two vector arguments of the operator sits in 2 space nicely however. In general, signed areas and M dimensional signed volumes are embeddable in N dimensional subspaces for M < N. The notion of a signed volume is directly tied to both the exterior product and the determinant. In higher dimensions, the cross product is equally unhelpful, given that for two vectors, the set of mutually orthogonal vectors abiding by the right hand rule is often infinite.
    - ninepoints 3y ago
      (should be M <= N but HN isn't letting me edit the comment at this point)

mijoharas 3y ago

Ok... I'll leave the comment to stand but apparently I just needed to scroll a tiny bit more and get an answer here[0]. The seven-dimensional cross product is one way of generalizing the cross product to other than three dimensions, and it is the only other bilinear product of two vectors that is vector-valued, orthogonal, and has the same magnitude as in the 3D case.[2] In other dimensions there are vector-valued products of three or more vectors that satisfy these conditions, and binary products with bivector results. [0] https://en.m.wikipedia.org/wiki/Seven-dimensional_cross_product https://en.m.wikipedia.org/wiki/Seven-dimensional_cross_prod...

ninepoints 3y ago

The exterior aka wedge product generalizes, as do determinants. Cross products absolutely do not

mijoharas 3y ago
Could you explain some more about why?
- ninepoints 3y ago
  The simplest counterexample is just to consider vectors in 2D. It is not possible in general to construct a third vector perpendicular to two other vectors unless those other vectors are collinear or anitcollinear. The notion of an exterior product, represented as a signed area subtended by the two vector arguments of the operator sits in 2 space nicely however. In general, signed areas and M dimensional signed volumes are embeddable in N dimensional subspaces for M < N. The notion of a signed volume is directly tied to both the exterior product and the determinant. In higher dimensions, the cross product is equally unhelpful, given that for two vectors, the set of mutually orthogonal vectors abiding by the right hand rule is often infinite.
  - ninepoints 3y ago
    (should be M <= N but HN isn't letting me edit the comment at this point)