Using two vectors and the epsilon tensor (Levi-Chivita-symbol), I guess?
Edit: So like this: two circles with one connection each as input, i.e. vectors (a and b), and a tensor with three connections, leaving one, meaning the result will be a vector again.
a o
\
o - (-o c)
/
b o
Multiplying with another vector c gives the parallelepipedial product (a x b) c, a scalar (also = det(a,b,c)).
Yeah, cross product is the determinant with "one strand bent back."
If you take a 3-regular planar graph and put the determinant/Levi-Civita tensor on each vertex, you get plus-or-minus the number of edge-3-colorings of that graph. The four-color theorem is merely to show this tensor contraction is always non-zero :-) (Nobody's managed to prove it this way, yet.)
Here are some notes showing how to contract two Levi-Civita symbols together graphically: https://imgur.com/HmaLSaB.png
This is in Penrose's 1971 paper ("Applications of negative-dimensional tensors"). He went on to notice that you can do a "puff-up-and-contract" construction to completely get rid of all the Levi-Civita tensors: https://imgur.com/eK9ZCkc.png
From here, he recognized that this sort of tensor network on 3-dimensional space was equivalent to an abstract tensor network on "(-2)-dimensional space", where there is an identity that let him remove all the places the strings cross https://imgur.com/65z0zov.png (This element placed at each edge in (-2)-dimensional space is known as the "second Jones-Wenzl projector for the Temperley-Lieb algebra.")
From a more recent point of view, what he did was find a graphical way to go from an SO(3) tensor network to an SL(2) tensor network (which should be possible since both Lie groups are isomorphic).
> From a more recent point of view, what he did was find a graphical way to go from an SO(3) tensor network to an SL(2) tensor network (which should be possible since both Lie groups are isomorphic).
Locally isomorphic, but not isomorphic: SL(2) is the simply connected double cover Spin(3) of SO(3). As you probably know, the non-simply-connectedness of SO(3) is what's witnessed by Dirac's belt trick (https://en.wikipedia.org/wiki/Plate_trick).
Yeah, and this works for any "algebra" (ie. vector space with a way of multiplying vectors.) The three legged thing is what they call the "structure constants" for the algebra.
Using two vectors and the epsilon tensor (Levi-Chivita-symbol), I guess?
Edit: So like this: two circles with one connection each as input, i.e. vectors (a and b), and a tensor with three connections, leaving one, meaning the result will be a vector again.
Multiplying with another vector c gives the parallelepipedial product (a x b) c, a scalar (also = det(a,b,c)).Cool :)