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I thought this would be related to quaternions, octonions etc. but no, it’s multivectors and wedge products. Very neat, I didn’t know you could use them like that.

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GnomeKat@lemmy.blahaj.zone ⁨1⁩ ⁨year⁩ ago
Oh no, you were right on the money. In G^2^ you have two basis vectors e1 and e2. The geometric product of vectors specifically is equivalent to uv = u dot v + u wedge v… the dot returns a scalar, the wedge returns a bivector. When you have two vectors be orthonormal like the basis vectors, the dot goes to 0 and you are just left with u wedge v. So e1e2 returns a bivector with norm 1, its the only basis bivector for G^2^.

e1e2^2 = (e1e2)*(e1e2) = e1e2e1e2

A nice thing about the geometric product is its associative so you can rewrite as e1*(e2e1)*e2… again that middle product is still just a wedge but the wedge product is anti commutative so e2e1 = e1e2. Meaning you can rewrite the above as e1*(-e1e2)*e2 = -(e1e1)*(e2e2) = -(e1 dot e1)*(e2 dot e2) = -(1)*(1) = -1… Thus e1e2 squares to -1 and is the same as i. And now you can think of the geometric product of two vectors as uv = u dot v + u wedge v = a + bi` which is just a complex number.

In G^3^ you can do the same but now you have 3 basis vectors to work with, e1, e2, e3. Meaning you can construct 3 new basis bivectors e1e2, e2e3, e3e1. You can flip them to be e2e1, e3e2, e1e3 without any issues its just convention and then its the same as quaternions. They all square to -1 and e2e1*e3e2*e1e3 = -e2e1e2e3e1e3 = e2e1e2e1e3e3 = e2e1e2e1 = -1 which is the same as i,j,k of quaternions. So just like in G^2^ the bivectors + scalars form C you get the quaternions in G^3^. Both of them are just bivectors and they work the same way. Octonions and beyond can be made in higher dimensions. Geometric algebra is truly some cool shit.

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