# Algebra generators of Clifford Algebra

A Clifford Algebra $$C_k$$ is a Real Algebra of dimension $$2^k$$ with its algebra generators being $$\{e_1,\ldots,e_k\}$$, satisfying the following relations: $$e_i^2 = -1 ~~\& ~~e_je_i = -e_ie_j ~~\text{if} ~~i \neq j.$$ The vector space basis of $$C_k$$ is $$\{e_{i_1}\cdots e_{i_r} : i_1 < i_2 \ldots < i_r, ~0 \le r \le k\}$$

Let $$R^k$$ denote the k-space in $$C_k$$ spanned by $$e_1,\ldots,e_k$$. Now let $$\{u_1,\ldots, u_k\}$$ be some other basis of $$R^k$$. Then my question is following:

Do we still have the following relations?

$$u_i^2 = -1 ~~\& ~~u_ju_i = -u_iu_j ~~\text{if} ~~i \neq j.$$

If no, then under the restriction of $$\{u_1,\ldots,u_k\}$$ being an orthonormal basis, do we have the above-mentioned relations on $$u_i$$s?

What I have tried is given below:

Let $$u_i = \sum_l a_l e_l$$ and $$u_j = \sum_n b_n e_n$$. Then we have $$u_i u_j = \sum_i^k a_ib_i e_i^2 + \sum_{~m,n \\ m \neq n} a_m b_n e_me_n$$ $$\&$$ $$u_j u_i = \sum_i^k b_ia_i e_i^2 + \sum_{~n,m \\ m \neq n} b_n a_m e_ne_m = \sum_i^k a_ib_i e_i^2 + (-1)\sum_{~n,m \\ m \neq n} b_n a_m e_me_n$$ From here I cannot establish the fact $$u_ju_i = - u_iu_j$$

• No, we don't have those relations in general, and yes, we still have them if the $u_i$ are an orthonormal basis. There is a different definition of the Clifford algebra in terms of the inner product that makes this clearer, which you can find on Wikipedia. Commented Aug 29, 2022 at 17:57
• @QiaochuYuan Can you please help me understand why for orthonormal basis we have the relations? Commented Aug 30, 2022 at 9:18

$$\newcommand\Cl{\mathrm{Cl}}$$

Given any $$K$$-vector space $$V$$ with quadratic form $$Q$$, we may define the associated Clifford algebra $$\Cl(V,Q)$$. Canonically identifying $$K$$ and $$V$$ as subsets of $$\Cl(V,Q)$$, its fundamental property is that $$v^2 = Q(v)$$ for all $$v \in V$$. By definition, $$Q$$ has an associated symmetric bilinear form $$B$$, where when the characteristic of $$K$$ is not 2 we have $$B(v,w) = \frac12(Q(v+w) - Q(v) - Q(w)) = \frac12((v+w)^2 - v^2 - w^2) = \frac12(vw + wv)$$ for $$v,w \in V$$. It follows that if $$B(v,w) = 0$$, i.e. $$v$$ and $$w$$ are orthogonal, then $$0 = vw + wv \implies vw = -wv,$$ and it is clear that $$v$$ and $$w$$ anticommute iff they are orthogonal. This is still all true when the characteristic of $$K$$ is 2, but we instead have to define $$B$$ by $$B(v,w) = Q(v+w) - Q(v) - Q(w).$$

For more on defining Clifford algebras, see chapter 14 of Clifford Algebras and Spinors (2001) by Pertti Lounesto.