I agree quaternions aren't really vectors when used in typical 3D applications but I do usually think of them as vectors when dealing with spacetime: cΔ𝜏=||cΔ𝘵 + 𝘫𝚫𝙭|| and cΔ𝘵=||cΔ𝜏 + 𝘪𝚫𝙭||. Those avoid some silliness with Lorentz factors (𝛾 and 𝛼). I'm using hyperbolic (𝘫²=+1) and complex (𝘪²=-1) notation to simplify things. 𝚫𝙭 can be (Δ𝘹 + Δ𝘺 + Δ𝘻) but often I'm doing 2D spacetime diagram plots so it makes sense to just use vector notation that works just as well for hyperbolic quaternions and regular quaternions. Also, it's faster to multiply them that way than the usual 1ijk notation.
Can you make (a) video(s) about differential forms and/or geometric calculus? These are two approaches to generalize calculus and combine the classical vector calculus theorems into the big Generalized Stokes's Theorem, and a lot of multivariable calculus classes don't cover this.
I think it's kind of cool that Geometric Algebra basically shows that quaternions really were the right choice to generalize the complex numbers into 3D. It just wasn't clear why. It also happens to give an extension into 4+ dimensions, which have some unintuitive properties, like the even subalgebra including the psudoscalar which squares to positive 1 instead of negative 1, and generally not lining up with the usual successor to quaternions, namely the octonions. Also that we've been visualizing quaternions wrong by thinking of them as 4 dimensional vectors.
How do we make sense of the fact that the geometric product of two bivectors in 4 spatial dimensions produces a commutative scalar part, an anti-commutative bivector part and a commutative tetravector part? We can still separate the commutative and anti-commutative parts, but the commutative part will now have a scalar part and a tetravector part. Is this the wedge product of the two bivectors? Is there still any use in considering it if it doesn’t just consist of one grade?
If you replace the dot product with the left contraction product then the geometric product of any two blades can be defined as the sum of the left-contraction, commutator and wedge products. In the case of multiplying two G4 bivectors, the left contraction gives a scalar, the commutator gives a bivector, and the wedge gives a quadvector
R times R(dagger) was equal to one, wasn't it because W and V were actually assumed to be unit vectors in the previous videos, otherwise R times R(dagger) wouldn't have been equal to one. Yes?
I like the perspective of Ie_1 = -i, Ie_2=-j Ie_3=-k where I is the RHR unit pseudo scalar this allows the cross product back into view but also means that the defined algebra elements in q mean what we would want that i relates to a bivector not including e_1 etc.
No. Every subalgebra of a geometric algebra is itself a geometric algebra, and all geometric algebras have a power of 2 number of components. Even subalgebras are just a moderately special case of this.
Yes. In general, unit-magnitude elements of the even subalgebra define transformations. You can debate whether or not certain elements are "rotors" in the sense that they may do 2 perpendicular rotations in 4D, or act as translations in euclidean PGA or hyperbolic rotations in Spacetime Algebra, but they're all still transformations, are applied in the same way, and can be formed through an exponential or product of exponentials. The composition of a translation and a rotation is usually called a "motor" rather than a "rotor" from what I've seen.
On a related note, one good programming trick is to always store G3 bivectors in the form Ayz + Bzx + Cxy, because the bits are identical to a right handed normal vector, so you can push bivectors directly to old fashioned graphics APIs that expect "normal vectors" :)
I love these geometric algebra videos, such a cool perspective
I've been thinking about this stuff over the last couple weeks. This video definitely clears some things up!
I agree quaternions aren't really vectors when used in typical 3D applications but I do usually think of them as vectors when dealing with spacetime: cΔ𝜏=||cΔ𝘵 + 𝘫𝚫𝙭|| and cΔ𝘵=||cΔ𝜏 + 𝘪𝚫𝙭||. Those avoid some silliness with Lorentz factors (𝛾 and 𝛼). I'm using hyperbolic (𝘫²=+1) and complex (𝘪²=-1) notation to simplify things. 𝚫𝙭 can be (Δ𝘹 + Δ𝘺 + Δ𝘻) but often I'm doing 2D spacetime diagram plots so it makes sense to just use vector notation that works just as well for hyperbolic quaternions and regular quaternions. Also, it's faster to multiply them that way than the usual 1ijk notation.
Can you make (a) video(s) about differential forms and/or geometric calculus? These are two approaches to generalize calculus and combine the classical vector calculus theorems into the big Generalized Stokes's Theorem, and a lot of multivariable calculus classes don't cover this.
+Scott Goodson
Yes, the plan is to get there eventually.
Mathoma Awesome! Keep up the good work!
I think it's kind of cool that Geometric Algebra basically shows that quaternions really were the right choice to generalize the complex numbers into 3D. It just wasn't clear why. It also happens to give an extension into 4+ dimensions, which have some unintuitive properties, like the even subalgebra including the psudoscalar which squares to positive 1 instead of negative 1, and generally not lining up with the usual successor to quaternions, namely the octonions.
Also that we've been visualizing quaternions wrong by thinking of them as 4 dimensional vectors.
How do we make sense of the fact that the geometric product of two bivectors in 4 spatial dimensions produces a commutative scalar part, an anti-commutative bivector part and a commutative tetravector part?
We can still separate the commutative and anti-commutative parts, but the commutative part will now have a scalar part and a tetravector part. Is this the wedge product of the two bivectors? Is there still any use in considering it if it doesn’t just consist of one grade?
If you replace the dot product with the left contraction product then the geometric product of any two blades can be defined as the sum of the left-contraction, commutator and wedge products. In the case of multiplying two G4 bivectors, the left contraction gives a scalar, the commutator gives a bivector, and the wedge gives a quadvector
R times R(dagger) was equal to one, wasn't it because W and V were actually assumed to be unit vectors in the previous videos, otherwise R times R(dagger) wouldn't have been equal to one. Yes?
I like the perspective of Ie_1 = -i, Ie_2=-j Ie_3=-k where I is the RHR unit pseudo scalar this allows the cross product back into view but also means that the defined algebra elements in q mean what we would want that i relates to a bivector not including e_1 etc.
Is there a closed 3D subalgebra of G3? I would assume not but I’m curious if there is some way to avoid one of the bases or scalar parts.
No. Every subalgebra of a geometric algebra is itself a geometric algebra, and all geometric algebras have a power of 2 number of components. Even subalgebras are just a moderately special case of this.
I could watch your videos forever, but I have homework to do
Can you please make Clifford algebras course?)
Wooh! Math videos!
So for G(2), the even sub-algebra would just be the complex numbers, with the imaginary unit being the pseudoscalar?
Bingo.
What's the name of the classical music piece? Beautiful!
JS Bach "Concerto for Two Violins in D minor, 1st movement"
rotors make sense because you can multiply by vectors. how do you usually apply a quaternion to a vector?
Two-sided multiplication.
I wonder HOW Hamilton came up with the quaternions. Was he aware, in a different way, of some of the stuff in this video?
check 3blue1brown's quaternion video.
Do rotors generalise to any dimension?
I don't see why they shouldn't, but you never explicitly mention it in your videos, hence my question.
Yes. In general, unit-magnitude elements of the even subalgebra define transformations. You can debate whether or not certain elements are "rotors" in the sense that they may do 2 perpendicular rotations in 4D, or act as translations in euclidean PGA or hyperbolic rotations in Spacetime Algebra, but they're all still transformations, are applied in the same way, and can be formed through an exponential or product of exponentials.
The composition of a translation and a rotation is usually called a "motor" rather than a "rotor" from what I've seen.
If you had set i = e3e1, j = e2e3, and k = e1e2, the weird handedness nonsense wouldn't show up in the algebra.
On a related note, one good programming trick is to always store G3 bivectors in the form Ayz + Bzx + Cxy, because the bits are identical to a right handed normal vector, so you can push bivectors directly to old fashioned graphics APIs that expect "normal vectors" :)