[The following exposition covers some familiar territory, but with perhaps
a more perspicuous narrative.]
Factoring/normalizing quaternions
Henry Baker
Posted 10/13/2020 to math-fun
Let Q=A+Bj be an arbitrary quaternion, for arbitrary complex numbers A,B.
I.e., quaternions can be represented as pairs of ordinary complex numbers;
j is a symbol which does NOT commute with the ordinary complex 'i'.
Complex numbers are commutative and always have square roots.
Note that i^2=j^2=-1, and ij=k, and ijk=-1, as is usual for quaternions.
(We won't be needing or using 'k' in this exposition.)
Let A' be the complex conjugate of A, B' be the complex conjugate of B.
Let N(A)=N(A')=AA'=|A|^2. N() is a multiplicative homomorphism.
Now jB = B'j, so we can commute complex numbers with the j quaternion,
so long as we conjugate them when they hop over j in either direction.
We can 'normalize' the A component of Q into a non-negative real number
by pre and post multiplication:
(1/sqrt(A))Q(sqrt(A))' = (1/sqrt(A)) (A+Bj) (sqrt(A))'
= (1/sqrt(A)) (A+Bj) sqrt(A')
= (1/sqrt(A)) A(sqrt(A')) + (1/sqrt(A)) Bj (sqrt(A'))
= sqrt(A) sqrt(A') + (1/sqrt(A)) B (sqrt(A))j
= sqrt(AA') + Bj
= sqrt(|A|^2) + Bj
= |A| + Bj
So Q = A+Bj = sqrt(A) (|A|+Bj) (1/sqrt(A))'
Note that this works even when A is already a negative real number, e.g.,
A=-1, because sqrt(A)=sqrt(-1)=i, and (1/sqrt(A))'=(1/i)'=(-i)'=i, and
A+Bj = -1+Bj = i (1+Bj) i = i^2 + iBji = -1 + i(-i)Bj = -1+Bj.
Similarly, we an 'normalize' the B component of Q into a non-negative
real number by pre and post multiplication:
(1/sqrt(B))Q(sqrt(B)) = (1/sqrt(B)) (A+Bj) sqrt(B)
= (1/sqrt(B)) A (sqrt(B)) + (1/sqrt(B)) Bj (sqrt(B))
= A + sqrt(B) j (sqrt(B))
= A + sqrt(B) sqrt(B') j
= A + sqrt(BB') j
= A + sqrt(|B|^2) j
= A + |B|j
So Q = A+Bj = sqrt(B) (A+|B|j) (1/sqrt(B))
Combining these two operations into one:
(1/sqrt(AB))Q(sqrt(A'B) = (1/sqrt(AB)) (A+Bj) (sqrt(A'B))
= (1/sqrt(AB)) A (sqrt(A'B)) + (1/sqrt(AB)) Bj (sqrt(A'B))
= (1/sqrt(B)) sqrt(A) (sqrt(A'B)) + (1/sqrt(A)) sqrt(B) j (sqrt(A'B))
= sqrt(A) sqrt(A') + (1/sqrt(A)) sqrt(B) sqrt(B'A) j
= sqrt(|A|^2) + sqrt(B) sqrt(B') j
= |A| + sqrt(BB') j
= |A| + |B|j
So,
Q = A+Bj = sqrt(AB) (|A|+|B|j) (1/sqrt(A'B))
We have thus factored our quaternion into two complex numbers
sqrt(AB) and 1/sqrt(A'B), and a quaternion having only real
(rather than complex) components.
If Q is a 'unit' quaternion -- i.e., QQ'=1 -- then
QQ' = N(sqrt(AB)) N(|A|+|B|j) N(1/sqrt(A'B)
= sqrt(N(A)) sqrt(N(B)) N(|A|+|B|j) 1/(sqrt(N(A')) sqrt(N(B)))
= sqrt(N(A)) N(|A|+|B|j) 1/sqrt(N(A'))
= sqrt(N(A)) N(|A|+|B|j) 1/sqrt(N(A)))
= N(|A|+|B|j)
= N(A)+N(B)
= |A|^2+|B|^2
= 1
So in this unit quaternion case, we can set |A|=cos(theta),
|B|=sin(theta), for a suitable real angle theta in the first
quadrant, and
Q = A+Bj = sqrt(AB) (|A|+|B|j) (1/sqrt(A'B))
= sqrt(AB) (cos(theta)+sin(theta)j) (1/sqrt(A'B))
Note for the record that
sqrt(AB)/sqrt(A'B) = sqrt(A)/sqrt(A')
= sqrt(A/A')
= sqrt(AA/AA')
= sqrt(AA)/sqrt(AA')
= A/sqrt(|A|^2)
= A/|A|
= phase(A)
However, our factorization is NOT the same as the more usual
factorization in terms of phases, since |sqrt(AB)| /= 1, and
|1/sqrt(A'B)| /= 1.
We can always recover the more traditional factorization by
taking phases:
Q = A+Bj = phase(sqrt(AB)) (|A|+|B|j) phase(1/sqrt(A'B))
since |sqrt(AB)|/|sqrt(A'B)| = |phase(A)| = 1.
and if QQ'=1, we get the classical factorization
Q = A+Bj = phase(sqrt(AB)) (cos(theta)+sin(theta)j) phase(1/sqrt(A'B))
= exp(i*gamma) (cos(theta)+sin(theta)j) exp(i*delta)
for suitable real angles gamma, delta, theta.
We have one last factorization to make.
If m,n are *real* numbers then m+nj is a quaternion both
of whose complex components are missing their imaginary parts.
We can convert such a quaternion m+nj to the complex number
m+ni by means of following product:
m+ni = -(i+j)(m+nj)(i+j)/2
or
-2m-2ni = (i+j)(m+nj)(i+j)
= m(i+j)^2 + n(i+j)j(i+j)
= m(i^2+j^2) + n(i+j)(ji+jj)
= -2m + n(i+j)(-ij-1)
= -2m - n(i+j)(1+ij)
= -2m - n(i+i^2j+j+jij)
= -2m - n(i-j+j-ij^2)
= -2m - n(i+i)
= -2m - 2ni
But the same transformation also works in reverse!
-2m-2nj = (i+j)(m+ni)(i+j)
= m(i+j)^2 + n(i+j)i(i+j)
= -2m + n(i+j)(i^2+ij)
= -2m + n(i+j)(-1+ij)
= -2m + n(-i -j +i^2j + jij)
= -2m + n(-i -j -j - ij^2)
= -2m + n(-2j -i + i)
= -2m -2nj
This transformation thus allows us to convert
cos(theta)+sin(theta)j into cos(theta)+sin(theta)i,
and then into exp(i*theta).
-2*exp(i*theta) = -2*(cos(theta)+sin(theta)i)
= (i+j) (cos(theta)+sin(theta)j) (i+j)
So our original arbitrary quaternion Q can be factored as
Q = A+Bj = sqrt(AB) (|A|+|B|j) (1/sqrt(A'B))
= -(sqrt(AB) (i+j) (|A|+|B|i) (1/sqrt(A'B)))/2
where |A|+|B|i is now an ordinary complex number in the first quadrant.
We note that |A|,|B| supply only 2 degrees of freedom, so the other two
degrees of freedom are supplied by phase(sqrt(AB)) and phase(1/sqrt(A'B)).
Once again, if Q is a unit quaternion, QQ'=1, then
Q = A+Bj = phase(sqrt(AB)) (cos(theta)+sin(theta)j) phase(1/sqrt(A'B))
= exp(i*gamma) (cos(theta)+sin(theta)j) exp(i*delta)
= -exp(i*gamma) (i+j) (cos(theta)+sin(theta)i) (i+j) exp(i*delta)/2
= -exp(i*gamma) (i+j) exp(i*theta) (i+j) exp(i*delta)/2
for suitably chosen angles gamma, theta, delta.
A unit quaternion has 3 real degrees of freedom, and this
3-parameter factorization provides one convenient representation.
