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GnomeKat@lemmy.blahaj.zone 7 months agoOh 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.