Clifford's algebras have basis elements squaring either to $1$, $0$ or $-1$. Why not $i$?
Is that already covered by $1$, $0$ or $-1$?
If there is an algebra squaring to $i$, what is his name? I have interest on the geometric interpretation of the geometric product of basis vectors squaring to $i$.
I mean something like $Cl_{p,q,r,s}$ where s is the number of basis vectors $e_j$ such that $e_j*e_j=i$
The situation of real Clifford algebras having bases which "square" to $\pm 1$ or $0$ is just a consequence of Sylvester's law of inertia: every quadratic form on a real vector space can be diagonalized, by a change of basis, with diagonal entries $\pm 1$ or $0$, with a uniquely-determined number of each.
But there is no compulsion to choose such a basis. So we could have a vector space basis whose "squares" were $\pi$, $\sqrt{2}$, and lots of other things.
For real vector spaces, no, we can't get $i$ as a "square", because it's not a real number. But, for complex vector spaces, with quadratic (not hermition) forms, we can certainly have "square" values $i$, or $\sqrt{-19}$, and so on.
Likewise, to make a Clifford algebra with "exotic" values, we just need to have vector spaces over a field that contains those exotic values. :)