# What is an even Clifford algebra?

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?

• what paper would that be? – Will Jagy Jan 15 '13 at 23:08
• Is the question with "Clifford algebra" or with "even." If it is the latter, then Clifford algebras are naturally $\mathbb{Z}/2$-graded, so the even Clifford algebra is the even part. I.e. the map $r\mapsto -r$ induces an algebra automorphism preserving the form and you get a direct sum decomposition from the positive and negative eigenspaces: $C(R,f)=C_0(f)\oplus C_1(f)$. The even part $C_0(f)$ is a subalgebra, but $C_1(f)$ (the "odd" part) is not. – Matt Jan 15 '13 at 23:34
• @WillJagy 'Quaternion Orders and Ternary Quadratic Forms', by Stefan Lemurell – user53515 Jan 15 '13 at 23:35
• arxiv.org/abs/1103.4922 – Will Jagy Jan 15 '13 at 23:38

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 $$E_1 = e_2 e_3, \; E_2 = e_3 e_1, \; E_3 = e_1 e_2.$$

• Another common term for this is the even subalgebra, to emphasize that it is merely composed of basis elements formed from an even number of products. – Muphrid Jan 15 '13 at 23:42
• Okay, that makes sense, except for $e_i e_i=f(e_i)$; doesn't $f$ take 3 arguments, since it's a ternary quadratic form? – user53515 Jan 15 '13 at 23:49
• No, $f$ takes a single vector argument. Buy Cassels. – Will Jagy Jan 15 '13 at 23:53
• Yes, this is what I said. – Matt Jan 16 '13 at 0:30
• @WillJagy Oops. I realized that probably sounded a little rude. I didn't mean to say that I answered it before you. I just wanted to indicate to the OP that my comment is a different way of saying the same thing since it isn't completely obvious that these are equivalent constructions. – Matt Jan 16 '13 at 2:38

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

• $i^2=j^2=k^2=-1$ does not imply that $ijk=-1$. That is required as another axiom (along with associativity etc.). – mr_e_man Sep 8 '18 at 1:57