I think you can make arbitrarily complicated roots if you move over to Gn which includes the R and C roots…
For example the grade 4 blade (3e1e2e3e4)^2 = 9 in G4
Complex roots are covered because the grade 2 blade (e1e2)^2 = -1 making it identical to i so Gn(n>=2) includes C.
Gn also includes all the scalars (grade 0 blades) so all the real roots are included.
Gn also includes all the vectors (grade 1 blades) so any vector with length 3 will square to 9 because u^2 = u dotu= |u|^2 where u is a vector.
All blades will square to a scalar but blades are not the only thing in Gn so things get weird with the multivectors(sums of different grades). Any blade with grade n%4 < 2 will square to a positive scalar and the other grades will square to a negative, with the abs of the scalar equal to the norm2 of the blade. Can pretty much just make as many roots as you want if you are willing to move into higher dimensional spaces and use a way cooler product.
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.
Oh no, you were right on the money. In G2 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 G2.
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 G3 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 G2 the bivectors + scalars form C you get the quaternions in G3. 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.
I think you can make arbitrarily complicated roots if you move over to Gn which includes the R and C roots…
For example the grade 4 blade
(3e1e2e3e4)^2 = 9in G4Complex roots are covered because the grade 2 blade
(e1e2)^2 = -1making it identical toiso Gn (n>=2) includes C.Gn also includes all the scalars (grade 0 blades) so all the real roots are included.
Gn also includes all the vectors (grade 1 blades) so any vector with length 3 will square to 9 because
u^2 = u dot u = |u|^2whereuis a vector.All blades will square to a scalar but blades are not the only thing in Gn so things get weird with the multivectors(sums of different grades). Any blade with grade
n%4 < 2will square to a positive scalar and the other grades will square to a negative, with the abs of the scalar equal to the norm2 of the blade. Can pretty much just make as many roots as you want if you are willing to move into higher dimensional spaces and use a way cooler product.You lost me at “arbitrarily complicated,” sorry.
Lost me at “I think”. I don’t, apparently.
I don’t think therefore I never was
I wish it was that easy! I stopped thinking and I’m still stuck here… Stupid Descartes
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.
Oh no, you were right on the money. In G2 you have two basis vectors
e1ande2. The geometric product of vectors specifically is equivalent touv = 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, thedotgoes to 0 and you are just left withu wedge v. Soe1e2returns a bivector with norm 1, its the only basis bivector for G2.e1e2^2 = (e1e2)*(e1e2) = e1e2e1e2A 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 soe2e1 = -e1e2. Meaning you can rewrite the above ase1*(-e1e2)*e2 = -(e1e1)*(e2e2) = -(e1 dot e1)*(e2 dot e2) = -(1)*(1) = -1… Thuse1e2squares to -1 and is the same asi. And now you can think of the geometric product of two vectors asuv = u dot v + u wedge v = a + biwhich is just a complex number.In G3 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 bivectorse1e2, e2e3, e3e1. You can flip them to bee2e1, e3e2, e1e3without any issues its just convention and then its the same as quaternions. They all square to -1 ande2e1*e3e2*e1e3 = -e2e1e2e3e1e3 = e2e1e2e1e3e3 = e2e1e2e1 = -1which is the same as i,j,k of quaternions. So just like in G2 the bivectors + scalars form C you get the quaternions in G3. 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.Then you can extend to arbitrary algebra