(γ0γ1 + γ2)
(γ0γ1 − γ2) = 0 in Cl(3, 0)
(γ3 + γ0γ1γ2)
(γ1 + γ0γ2γ3) = 0 in Cl(4, 0)
(1+γ0γ1γ2γ3)
(1−γ0γ1γ2γ3) = 0 in Cl(4, 0)
(C* (C+ (C* g0 g2) g01234) (-- (C* g0 g2) g01234))
(1 + γ0)(1 − γ0) = 0 in Cl(0, 5) or Cl(1, 3) (where γ02 = 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. γ1γ2 + γ3 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). (γ0γ1 + γ2) and (γ0γ1 − γ2) 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 = γ1 + γ0γ2γ3 is an essential singularity for 1/(e + .000001) is large in the sense of both sp and ip. 1/(εγ3 + γ1γ2 + γ3) 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 = γ0γ1γ2γ3 and U = {a | (magnitude of a) = 1}. Recall that ad is a ZD if d is and that 1+t is a ZD. To find zero divisors near 1+t we take a small ball B around C1 and consider the dimensionality of X = {a(1+t) | a∊B∩U}. X is an ellipsoidal patch on a manifold of ZDs centered at 1+t. B is 2n dimensional and Y = B∩U is 2n−1 dimensional. (n is at least 4.) For each of the 2n−1 basis vectors b of Cl(n,0) excluding 1, bγ0γ1γ2γ3 is another of those basis vectors, or its negative.
We continue these ideas here.
(define a (C+ g12 g3)) ; a = γ1γ2 + γ3 (define b (-- g12 g3)) ; b = γ1γ2 - γ3 (C* a b) ; => C0 ; ab = 0 (define (z p q)(C* (C* b p)(C* q a))) (define s (C+ C1 g0)) ; s = 1+γ0 (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*(γ0γ1γ2γ3 - γ0) (C* B A) ; => 4*(g0123 - g0) not C0 but is a ZD.