Non-trivial squares of bases in Clifford algebras?

Question

I've been learning about Clifford algebras recently, and I know that bases can square to +1, -1, or 0 (but are not real numbers themselves).

Is there any value in considering bases that square to other numbers? In particular, are there possibly bases that square to $\pi$, exp(1), and/or the prime numbers (but are not real numbers themselves). Perhaps such bases wouldn't correspond to well-behaved algebras like Clifford algebras. Is there a type of algebra that these bases would fit into, if not Clifford?

Answer 1

It's still a Clifford algebra even when the bases square to other numbers. The reason we tend to work with bases that square to +1, -1, or 0 is that we can do so without loss of generality.

For instance, in simple cases (where the quadratic form of the Clifford algebra is straightforward), if you have some basis element that squares to exp(1), you can just scale it down by a factor of sqrt(exp(1)) and have a basis that squares to 1.

Concrete example: consider $\mathbb R$ as a vector space over the reals, and consider the quadratic form $Q := x \mapsto x^2$. Then the Clifford algebra $\mathrm{Cl}(\mathbb R, Q)$ is 1-dimensional, and any non-zero real number $a$ gives rise to a basis, with the basis vector (represented by) $a$ squaring to $a^2$, which is not in general equal to +1, -1, or 0.

Answer 2

Clifford algebras are associated to quadratic forms. Any real non degenerate quadratic form has a basis $(e_1...e_n)$ in which it can be written as $$ q(x_1e_1+...+ x_ne_n) = x_1^2+...+x_r^2 -x_{r+1}^2-...-x_{n}^2$$ for some r between 0 and n. This is the reason only these quadratic forms are considered in the theory of real Clifford algebras.

Answer 3

The square of a vector $x$ in a Clifford algebra $Cl(V,Q)$ is equal to $Q(x)$, where $Q$ is the quadratic form. It so happens that if $V$ is a real vector space, we can find a basis such that $Q(e_i)\in\{-1,0,1\}$, but this needn't be true over other fields. Over the field of complex numbers $\mathbb C$, you can always find a basis such that $Q(e_i)\in\{0,1\}$. Over the field of rational numbers $\mathbb Q$, bases behave in a much more complicated way.