.3-rd grade Clifford algebra Cl3,0(R) has division operators defined as the inverse of the product. For example if a,b are Clifford vectors then the division operations / \ of those two produce other Clifford vectors d and s such that
a/b=d <=> a=b*d
a\b=s <=> a=s*b
Those are definitions of the right-hand-side (/) with the right-hand-sode quotient d, and the left-hand-side (\) division operator with the left-hand-side quotient s. Here below we discuss only the right hand side division as an example.
Division operations are restricted to vectors b such that b is not equal 0 and the modulus (length) of b is also non-zero. There is one exception to this restriction: division may be defined for the cases when both a and b are vectors of zero length and are aligned in the same direction (that is have a common zero-length multiplier, for example the (Dirac) spinor base vector (1-f)/2 where f is the unity primary base Clifford vector. Note that any unity vector can be used instead of f , for example (e+f)/sqrt(2) etc. It is interesting to note that the zero length Clifford vectors represent physical objects propagating at the speed of light, such as photons.
An example of a special case when a and b are zero-length and are both aligned to the same zero-base (1-f)/2 is:
a=b=1-f
where f is a unity base vector, one out of the 3 primary Clifford algebra generator vectors:
{e,f,g} (note: ee=ff=gg=1 and efg=i).
Using the definition of the right-hand-side division (/) above, we obtain:
a/b=(1-f)/(1-f) = v + (1-v)(1-f)/2 = 1 + (1-v)(1+f)/2
this can be also written as exp( u*(1+f)/2 )
where v, u are arbitrary vectors or complex numbers
Proof:
(v + (1-v)(1-f)/2)(1-f) = v(1-f) + ( (1-v)(1-f)/2)(1-f) =
v(1-f) + (1-v)(1-f) = (v + 1-v)(1- f) = (1-f)
——
A general formula for the zero by zero quotient aligned with the (1-f)/2 spinor base is as follows:
a/b = q*exp(u(1+f)/2)
(where q is the nonzero vector part of a: a=q(1-f) , b=(1-f) and u is an arbitrary Clifford vector
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