Clifford Units

We seek Clifford numbers c such that φ(c) = cvc^–1 leaves all elements v of V fixed, i.e. v = cvc^–1 (↔ cv = vc). The only such numbers I know now are the non-zero reals within Cl(n). We restrict ourselves to values v such that (ip v v) = 1. It suffices to require that b = cbα(c)^–1 for each base element b of V.

An alternate definition of the Clifford group uses the expression cvα(c)^–1 in place of cvc^–1. This defines the same set of Clifford numbers but the orthogonal matrices they lead to have the opposite sign when c is odd.

To find those Clifford numbers c that commute with each member of V we first seek all c such that cγ₀ = γ₀c.
Split c thus c = f + γ₀g where neither f nor g mention γ₀.
cγ₀ = γ₀c ↔ (f + γ₀g)γ₀ = γ₀(f + γ₀g) ↔ fγ₀ + γ₀gγ₀ = γ₀f + γ₀γ₀g ↔
γ₀f = fγ₀ & γ₀gγ₀ = γ₀γ₀g ↔ f and g are both pure even. (Note that γ₀ has an inverse: –γ₀.)
The code corroborates this.

Said another way, a Clifford number commutes with γ₀ iff each of its terms mentions an even number of other base vectors. When we require that a Clifford number commute with each of the base elements we are clearly left with only the reals among the Clifford numbers.