What is an even Clifford algebra?

Question

I'm reading a paper and he defines $C_0(f)$ to be the "even Clifford algebra over $R$ associated to $f$", where $R$ is a principal ideal domain and $f$ is a non-degenerate ternary quadratic form. What is meant by an 'even' Clifford algebra?

Answer 1

This is from Cassels, Rational Quadratic Forms, chapter 10, especially pages 177-178. We have a basis $e_1, e_2, e_3$ of a vector space $V$ of dimension 3 over a field. These satisfy $$ e_i e_i = f(e_i) $$ and $$ e_i e_j + e_j e_i =0, \; \mbox{when} \; i \neq j, $$ meaning that they are orthogonal. Then a basis of the even Clifford algebra is $$ 1, \; e_2 e_3, \; e_3 e_1, \; e_1 e_2.$$ You can work out things for a PID.

Note that Lemurell uses lower case letter $e$ where Cassels uses upper case, in $$ E_1 = e_2 e_3, \; E_2 = e_3 e_1, \; E_3 = e_1 e_2.$$

Answer 2

Perhaps Wikipedia it...

Clifford Algebras from what I recall are algebraic extensions of the complex plane. ex: The Quaternions.

In C: a Number N = a + bi

In Q (Quaternions): a Number N = a + bi + cj + dk

This is a valid algebra where the following identities:

i^2 = -1, j^2 = -1, and k^2 = -1 ---> ijk = -1

hold.

Even may refer to the number of terms in the algebra (ex: complex have 2 terms, Quaternions have 4, etc...) or it might be more subtle than that. I can't answer that part but it is best you start looking up the terms:

Complex Numbers, Quaternions, Octonion, Split-Complex Numbers

and from there you'll know where to go