(γ_{0}γ_{1} + γ_{2})
(γ_{0}γ_{1} − γ_{2}) = 0 in Cl(3, 0)

(γ_{3} + γ_{0}γ_{1}γ_{2})
(γ_{1} + γ_{0}γ_{2}γ_{3}) = 0 in Cl(4, 0)

(1+γ_{0}γ_{1}γ_{2}γ_{3})
(1−γ_{0}γ_{1}γ_{2}γ_{3}) = 0 in Cl(4, 0)

(C* (C+ (C* g0 g2) g01234) (-- (C* g0 g2) g01234))

(1 + γ_{0})(1 − γ_{0}) = 0 in Cl(0, 5) or Cl(1, 3) (where γ_{0}^{2} = 1) but not in Cl(n, 0)

This feels like stamp collecting. These samples suggest that if ab = 0 then ba =0 but see “does not imply” below for counterexample. None of these are near the Clifford group. I have no reason to think that zero divisors are important except the general feeling that they are obstacles to be avoided which means knowing where they are. See this for an unlikely conjecture.

The product of a ZD (zero divisor) and any Clifford number is either 0 or a ZD. Indeed if ab = 0 then for any x and y: (xa)(by) = x(ab)y = x0y = 0 and thus both xa and by are zero divisors.

If φ is an automorphism on Cl(n) and z is a zero divisor then φ(z) is too.
γ_{1}γ_{2} + γ_{3} is a zero divisor for any three pairwise orthogonal members of V.
This provides a manifold of zero divisors shaped like O(n).
I think that all automorphisms of C stem from orthogonal transformations on V, and thus from members of the Clifford group.
Here we examine the interaction of these two modes of deriving ZDs from ZDs.

If x is a zero divisor of Cl(n) then <x, 0> is a zero divisor of Cl(n+1).
(γ_{0}γ_{1} + γ_{2}) and
(γ_{0}γ_{1} − γ_{2})
are zero divisors of Cl(n) for n≥3 and the expressions make no sense otherwise so we need not qualify the order of the algebra.
Cl(2) is isomorphic to the quaternions which have no zero divisors.
Thus Cl(n) has zero divisors iff n>2.

e = γ_{1} + γ_{0}γ_{2}γ_{3} is an essential singularity for 1/(e + .000001) is large in the sense of both sp and ip.
1/(εγ_{3} + γ_{1}γ_{2} + γ_{3}) seems unbounded for small ε.

Here is what I have discovered, but read on here to see how I found it.

Use these definitions for the following:

(define g12 (C* g1 g2)) (define g012 (C* g0 g12)) (define g023 (C* g0 (C* g2 g3))) (define g0123 (C* g012 g3)) ; the next 6 lines each demonstrate a zero divisor. (C* (C+ g12 g3) (-- g12 g3)) (C* (C+ g3 g012) (C+ g1 g023)) (C* (C+ C1 g0123) (-- C1 g0123)) (C* (C* (C+ C1 (sm .01 (Cr))) (C+ C1 g0123)) (C* (-- C1 g0123) (C+ C1 (sm .01 (Cr))))) (let ((c (Cr))) (C* (C* (C+ C1 g0123) (C/ c)) (C* c (-- C1 g0123)))) (C* (nc om (C+ g3 g012)) (nc om (C+ g1 g023)))Now we let t = γ

We continue these ideas here.

(define a (C+ g12 g3)) ; a = γ_{1}γ_{2}+ γ_{3}(define b (-- g12 g3)) ; b = γ_{1}γ_{2}- γ_{3}(C* a b) ; => C0 ; ab = 0 (define (z p q)(C* (C* b p)(C* q a))) (define s (C+ C1 g0)) ; s = 1+γ_{0}(define A (C* s a)) ; A = sa (define B (C* b s)) ; B = bs (C* A B) ; => C0 ; ab=0 does not imply ba=0: ; AB = 0 but BA = 4*(γ_{0}γ_{1}γ_{2}γ_{3}- γ_{0}) (C* B A) ; => 4*(g0123 - g0) not C0 but is a ZD.

See this for comments on and references to notes on zero divisors in the Cayley-Dickson algebras.

Exploring zds