4 ms·
How would one write a cross-product with that? Using two vectors and the epsilon tensor (Levi-Chivita-symbol), I guess? Edit: So like this: two circles with o
by _Microft 7y ago
How would one write a cross-product with that?
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)).
Cool :)
- mathgenius 7y agoYeah, 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.
- _Microft 7y agoWould it make sense/work to do something like this ? (a x b )(c x d) => ??? o o o o \ / \ / o - o => o / \ / \ o o o o Would the circle in the middle of the cross be a tensor of order 4 then? Edit: It's been a while since I did that but wouldn't that be the following in sum notation? eps_ijk a_j b_k eps_ilm c_l d_m = (new tensor)_jklm a_j b_k c_l d_m
- mathgenius 7y agoYes it really is just like lego, you can make all kinds of bizarro shapes.
- kgwgk 7y ago“Chivita“ is written Civita. Chivita would be pronounced “Kivita”. :-)
- kmill 7y agoYeah, 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 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 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 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).
- _Microft 7y agoNegative-dimensional tensors... now that's wild. Thanks for the sketches, I'll have a look!
- JadeNB 7y ago> 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 https://en.wikipedia.org/wiki/Plate_trick).
- kmill 7y agoThanks, that's right. (I had the Lie algebras in mind with that statement. The Lie brackets are placed at the vertices in the construction.)