Clifford algebras are proposed as a tool to explore rotations in n-dimensions. For many years I thought that n by n orthogonal matrices were the last word in rotations. These are known as O(n) and form a group. Rotations of 2π are unlike rotation of 0π or 4π. The Clifford algebra Cl(n) captures this while O(n) does not. Clifford algebras are the only access to this fact that I know of when n>3. This seems to be important in physics.

The most fruitful web sources that I have found are the Wikipedia article and John Baez’s Clifford stuff. I shall repeat little that is available there. You will find much there that is not here.

O(n), Cl(n) and quaternions are all redundant in how they represent rotations. The group of rotations in n-space has n(n−1)/2 dimensions. Cl(n) has 2ⁿ dimensions and is the most redundant. There is a subset of Cl(n) called the “Clifford group” which identifies rotations via a map φ from the group to O(n). (The function Om in the tools is just this map.) I follow wikipedia here and define the group as those elements x that have an inverse and for each member v of V, xvα(x)⁻¹ is in V. There are at least two inverse images for any member of O(n) and these two distinct members of the group maintain the strap twist distinction. I do not know what other inverse images there are of a member of O(n). For a positive real x, and a member g of the group φ(g) = φ(xg). φ(x) = φ(−x) but x and −x represent distinct ‘strap rotations’.

Since all rotations can be had as the product of some set of n or fewer reflections, products of up to n members of V will generate enough members of the group to cover O(n). (Recall that φ(cd) = φ(c)φ(d).) We can further restrict ourselves to members of V of length 1.

Perhaps there are Clifford numbers c, d for which φ(c)=φ(d) but for which c/d is not real, but I have found none. (Division in Clifford algebras is possible except for a set of zero divisors of measure 0.) If e = c/d for such a collision then φ(e) = the identity matrix. I may write a program to look for such e’s numerically. An exercise described here discovered two products of orthogonal matrices near the identity. One was surprisingly near the other merely because many products were examined. The product of the corresponding Clifford values produced the Clifford identity which gives very weak evidence that the product of the vectors yield all of the Clifford group.

Zero Divisors

When xy = 0 and neither x nor y is 0 then x is a left zero divisor and y is a right zero divisor. For instance:
(1+γ₀γ₁γ₂γ₃) (1−γ₀γ₁γ₂γ₃) = 0 and
(1+γ₀γ₁γ₂) (1−γ₀γ₁γ₂) = 0.

All other left zero divisors that I have found are generated from the above by these observations: If z is a left zero divisor, c is a Clifford number and φ is an automorphism then cz and φ(z) are left zero divisors. The values of cz form an 8D subspace for a given ZD z, at least for the ZDs that I have found. For a given z the values φ(z) generate a 6D curved manifold, shaped like O(4). Perhaps together they generate a 14D manifold. Perhaps not. x is a left zero divisor iff x^† is a right zero divisor.

The multiplication operation for a Clifford algebra C is a map from C×C → C. We are looking for the pre-image of 0 in C×C. The shape is cone-like in that it is composed of (8D) linear subspaces of Cl(4).

1+γ₀γ₁γ₂ is in the ZD subspace spanned by these 8 ZDs (note optional γ₃ at end of each):
(1+γ₀γ₁γ₂)[γ₃], (γ₀−γ₁γ₂)[γ₃], (γ₁+γ₀γ₂)[γ₃], (γ₀γ₁−γ₂)[γ₃]