# Classification of real algebra with zero divisors (subquotient of Clifford algebra?)

Consider the even-dimensional real vector space $\Bbb R^{2N}$. We construct the algebra as follows:

1. Pick a basis in this space.
2. Partition the $2N$ basis elements into $N$ pairs of zero divisors, called $e_1, \hat e_1, e_2, \hat e_2, ..., e_N, \hat e_N$.
3. For all $n$, $e_n \hat e_n = 0$.
4. For all $n$, $e_n^2 = e_n$, and $\hat e_n^2 = \hat e_n$.
5. For all $m, n$, $e_m e_n = e_n e_m$, $\hat e_m \hat e_n = \hat e_n \hat e_m$, and $e_n \hat e_m = \hat e_m e_n$.

In other words, the basis elements are all idempotent, come in zero-divisor pairs, and commute. The full algebra is the construction then generated by the above relations, given N generators.

This has the structure of a graded algebra, so that for products $e_m e_n$, these exist in a larger space that the original vector space is a subspace of, in a similar to the exterior powers of the exterior algebra. Assuming we throw in the field of scalars as grade-0 vectors, you end up with the full algebra being finite-dimensional.

My question: how do you classify this algebra? I can see this possibly being a subquotient of the Clifford algebra, but it seems messy to look at it that way.

An interesting case is the 3-dimensional real algebra yielded by this construction - you get the field of real numbers, plus two additional elements $i$ and $j$ which have the property that $ij = ji = 0$, $i^2 = i$, $j^2 = j$, and also $(i+j)^2 = i+j$.

• You haven't fully constructed the algebra, as you haven't decomposed products like $e_me_n$ into linear combinations of basis elements. – arctic tern Apr 2 '17 at 4:35
• I tried to note with "I'm envisioning this as a graded algebra, so that for products $e_m e_n$, these exist in a larger space that the original vector space is a subspace of, in a similar to the exterior powers of the exterior algebra." I'm not actually sure how to explicitly formalize that idea, though. I guess the quickest way is as a quotient of the symmetric algebra? – Mike Battaglia Apr 2 '17 at 4:36
• So you want the algebra generated by those elements and relations? – arctic tern Apr 2 '17 at 4:39
• Right - edited to make it clearer. – Mike Battaglia Apr 2 '17 at 4:41
• What is unsatisfactory about the definition by relations? After all, that's a very common pattern for describing Clifford algebras, so nothing would be gained by going through Clifford algebras first. – rschwieb Apr 2 '17 at 12:53

The algebra is a quotient of a polynomial algebra: $$A_N=\Bbb{R}[x_1,y_1,x_2,y_2,…,x_N,y_N]/\langle x^2_i=x_i,y^2_i=y_i,x_iy_i=0\rangle,$$ and $\dim(A_N)=3^N$. In fact, the non zero monomials in that algebra can be reduced to monomials where every variable has at most power one, and if $x_i$ is in that monomial, then $y_i$ is not. So given a monomial, for each $i$ we have three choices: Either $x_i$ is a factor, or $y_i$ is a factor or none of them is a factor. This leaves us with $3^N$ monomials that give the linear basis of the algebra. If for all $i$ the choice is that none of $x_i$, $y_i$ is a factor, then we obtain the scalars with basis element "1". The grading is simply the usual degree of polynomials.
• Thanks, this is helpful. One additional thought: suppose we also want $x_i+y_i=1$. Would it be as simple as adding the ideal ((x_i+y_i-1)) to the kernel? I don't see why not, but I can't tell what might make this turn out to be degenerate for unexpected reasons. – Mike Battaglia Apr 28 '17 at 22:10
• In that case you obtain $B_n=\Bbb{R}[x_1,...,x_N]/\langle x_i^2=x_i\rangle$ – san Apr 29 '17 at 15:34
• Of dimensión $2^N$ and with a surjective morphism from $A_N$ to $B_N$, which sends $x_i$ to $x_i$ and sends $y_i$ to $1-x_i$. – san Apr 29 '17 at 15:45
• But how do you get $(1-x_i)^2 = (1-x_i)$? – Mike Battaglia Apr 29 '17 at 16:25
• $(1-x)^2=1-2x+x^2=1-2x+x=1-x$. – san Apr 29 '17 at 17:56