I pursue here the Clifford algebra over V, the 4D vector space over the field GF(2). There are just 16 vectors in all, hex v0, ... v9, va, ... vf. Note the hexadecimal modifiers. Pardon me for not using typographical subscripts, but this way is easier to type and read. We will use the basis γ0 = v1, γ1 = v2, γ2 = v4, γ3 = v8.

For a Clifford algebra over this vector space we require a quadratic form on the space. This exhaustive program enumerates just 1024 quadratic forms over the 4D vector space over GF(2). If n>1 then no quadratic form on the n dimensional vector space over GF(2) can be positive definite, for the field is too small. All of the quadratic forms are singular! Nonetheless not all quadratic forms are alike under conjugation. I don’t know what quadratic form to use but I will start with q(v) = parity of count of basis vectors in expansion of v:
q(Σ αiγi) = Σ(αi)2
γiγi = −1 = 1.
γ0γ1 is its own thing and γ0γ1 = γ1γ0, etc.
This seems to imply that this is a commutative algebra. There are 16 basis vectors for the Clifford algebra, one for each subset of the basis for V, as in general Clifford algebras.
The vectors in V are sums of subsets of the basis of V. (Sum of the null subset = 0 = c0000.)
The basis for C is products of the subsets of the basis of V. (Product of the null subset = 1 = c0001.)

Expressing a general member of C as a linear combination of this basis for C requires one coefficient (from GF(2)) for each member of the basis, i.e. 16 bits. c0000 = zero, c0001 = one. Here “0001” in “c0001” is hexadecimal and denotes the basis expansion where only term zero has non-zero coefficient. Basis element zero of C corresponds to the null subset of the basis of V. Thus c0001 is the unique multiplicative identity of C and c0000 is the unique additive identity. c0001 + c0001 = c0000; indeed x+x = c0000. Since xy+yx = xy+xy = 2xy = 0xy = 0, this is a Grassmann algebra and should probably live there.

The four roots of −1 are here: c0002 = γ0, c0004 = γ1, c0010 = γ2, c0100 = γ3.

This is confusing since powerset logic is working at two levels:

• There is just one vector for each subset of the basis for V since the coefficients must be either 0 or 1. There are thus 16 vectors.
• There is just one member of the Clifford algebra for each subset of the basis for C and there is just one member of the basis for C for each subset of the basis for V. There are thus 65535 = 224 elements in the Clifford algebra over the 4D vector space over GF(2). There are 22n elements in the Clifford algebra over the n dimensional vector space over GF(2) if n≥0.
Is this better?:[In (not so) short, the members of the Clifford algebra C, are sums of subsets of the basis for C. The basis for C is products of subsets of the basis of V. V is 4 dimensional and its basis has 4 elements. There are 16 subsets thereof and thus 16 members of the basis of C. There are 216 subsets thereof and thus that many elements of the Clifford algebra.]

c0003*c0002 = (c0002 + c0001)*c0002 = (γ0 + 1)*γ0 = γ0*γ0 + 1*γ0 = 1 + γ0 = c0003.
c0008*c0005 = (γ0γ1)*(γ1 + 1) = γ0 + γ0γ1 = c0002 + c0008 = c000a.

This C code, mulC4, is a brief form of multiply. Clifford addition is merely bitwise exclusive or:

`int addC4(int j, int j){return k^k;}`
This program suggests that the Clifford numbers with inverses are exactly the odd part of C. Indeed if xy=1 then x is odd and x=y.

### Clifford Group

Wikipedia defines the Clifford group, Γ, as those Clifford numbers x with an inverse and such that for all v in V, xvα(x)−1 is in V. By that definition it seems that Γ is the entire odd part of C. In its native setting (Field is reals) the Clifford group provides all of the isometric transformations on V. That is to say that for each linear transformation T that preserves Q, there is a member x of Γ such that for all v in V Tv = xvα(x)−1. When the field characteristic is 2 α(x) = x. This algebra is commutative and thus xvα(x)−1 = v. It is not clear what isometric means here but any of the 24 permutations of the 4 γi’s should be isometric. Γ provides only the identity. Things are not going well for the Clifford Group; but then the theorems promise nothing for our quadratic form is singular.