denote z^t the conjugate of z in H, (i.e. if z=a+b*I+c*J+d*K then z^t = a-b*I-c*J-d*K); then (z1*z2)^t=z2^t*z1^t, as explained in video and for 3 variables z1,z2,z3 we have (z1*z2*z3)^t = z3^t*z2^t*z1^t; g is in S3 so you have g*g^t=1, g^t=g^-1; also v^t=-v because v is pure; if w=g*v*g^-1 then in order to show that w is pure you have to show w^t = -w; we have w^t = (g*v*g^-1)^t = (g^-1)^t*v^t*g^t = (g^t)^t * (-v) * g^-1 = g*(-v)*g^(-1) = -w /q.e.d.
Alternatively: qp = q × p - q · p if q and p are pure imaginary. (pv)p¯¹ = (p × v - p · v)p¯¹ = (p × v)p¯¹ - (p · v)p¯¹ = -((v × p)p¯¹ + (v · p)p¯¹) (v · p)p¯¹ might look familiar, especially if the inverse is expanded like -(v·p/p·p)p = -Proj_p(v) (v × p)p¯¹ looks similar, but not as familiar. Well: q × p = qp + q · p, so (v × p)p¯¹ = (vp + v · p)p¯¹ = vpp¯¹ + (v · p)p¯¹ = v - Proj_p(v) = Rej_p(v). The projection of v onto p + the part of v orthogonal to p together add back to v. Now we can return to pvp¯¹ = -((v × p)p¯¹ + (v · p)p¯¹) = -(Rej_p(v) - Proj_p(v)) = Proj_p(v) - Rej_p(v) = A reflection of v across p. g in the example isn't a pure imaginary quaternion, but any arbitrary quaternion can be formed as the product of two pure quaternions: g = qp gvg¯¹ = (qp)v(qp)¯¹ = qpvp¯¹q¯¹ = q(pvp¯¹)q¯¹, which is just two reflections, both of which keep v as a pure imaginary. It's also a known theorem that two reflections make a rotation, which applies completely in this case.
Geometric algebra aka Clifford algebras are a much larger subject than just that of quaternions, as I am sure you are aware. They can be used to construct all the Spin(n) groups and are extremely useful. They also come with their own thorny, complicated issues that are often swept under the rug by people introducing the subject, but I embrace the thorny issues because they are learning opportunities, similar to how the fact that there is debate over whether the traditional choice of + and - in EM is unfortunate or not can lead to a lot of interesting discussions and learning. The term "Geometric Algebra" is indeed just a buzz word and you can say "Clifford algebra" if you prefer.
It's good to understand quaternions first partly as a motivation for exploring Clifford algebras, and to have the simplest non-commutative example of a Clifford algebra as a toy to play with while studying the subject.
Exceptionally good math explainer, and a hero for the sharing of knowledge.
At 9:16, why is gvg^-1 a pure "imaginary" quaternion?
denote z^t the conjugate of z in H, (i.e. if z=a+b*I+c*J+d*K then z^t = a-b*I-c*J-d*K); then (z1*z2)^t=z2^t*z1^t, as explained in video and for 3 variables z1,z2,z3 we have (z1*z2*z3)^t = z3^t*z2^t*z1^t; g is in S3 so you have g*g^t=1, g^t=g^-1; also v^t=-v because v is pure; if w=g*v*g^-1 then in order to show that w is pure you have to show w^t = -w; we have w^t = (g*v*g^-1)^t = (g^-1)^t*v^t*g^t = (g^t)^t * (-v) * g^-1 = g*(-v)*g^(-1) = -w /q.e.d.
Alternatively: qp = q × p - q · p if q and p are pure imaginary.
(pv)p¯¹ = (p × v - p · v)p¯¹ = (p × v)p¯¹ - (p · v)p¯¹ = -((v × p)p¯¹ + (v · p)p¯¹)
(v · p)p¯¹ might look familiar, especially if the inverse is expanded like -(v·p/p·p)p = -Proj_p(v)
(v × p)p¯¹ looks similar, but not as familiar. Well: q × p = qp + q · p, so (v × p)p¯¹ = (vp + v · p)p¯¹ = vpp¯¹ + (v · p)p¯¹ = v - Proj_p(v) = Rej_p(v). The projection of v onto p + the part of v orthogonal to p together add back to v.
Now we can return to pvp¯¹ = -((v × p)p¯¹ + (v · p)p¯¹) = -(Rej_p(v) - Proj_p(v)) = Proj_p(v) - Rej_p(v) = A reflection of v across p.
g in the example isn't a pure imaginary quaternion, but any arbitrary quaternion can be formed as the product of two pure quaternions: g = qp
gvg¯¹ = (qp)v(qp)¯¹ = qpvp¯¹q¯¹ = q(pvp¯¹)q¯¹, which is just two reflections, both of which keep v as a pure imaginary. It's also a known theorem that two reflections make a rotation, which applies completely in this case.
Geometric algebra aka Clifford algebras are a much larger subject than just that of quaternions, as I am sure you are aware. They can be used to construct all the Spin(n) groups and are extremely useful. They also come with their own thorny, complicated issues that are often swept under the rug by people introducing the subject, but I embrace the thorny issues because they are learning opportunities, similar to how the fact that there is debate over whether the traditional choice of + and - in EM is unfortunate or not can lead to a lot of interesting discussions and learning. The term "Geometric Algebra" is indeed just a buzz word and you can say "Clifford algebra" if you prefer.
Wow
Down with quaternions! Up with geometric algebra!
It's good to understand quaternions first partly as a motivation for exploring Clifford algebras, and to have the simplest non-commutative example of a Clifford algebra as a toy to play with while studying the subject.