Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Just curious as I am studying quadratic forms. Is there a special way of viewing the Clifford algebra $C(q)$, given the diagonal quadratic form $q = \langle a_1, a_2, \ldots, a_n\rangle$, where $a_i \in K^{\times}$, $K$ is a field?

The reason I ask is because I want to show that given a field $K$, where char $K \neq 2$ and $a_i \in K^\times$, then we have

$C_0(\langle a_1, a_2, \ldots, a_n\rangle) \simeq C(\langle b_2, \ldots, b_n\rangle)$, with $b_i = -a_1a_i$.

Note: $C_0$ denotes the even component (sub-algebra) of the Clifford algebra $C$. By $q = \langle a_1, a_2, \ldots, a_n\rangle$, I mean that $q(\sum_{i=1}^n x_ie_i) = \sum_{i=1}^n a_ix_i^2$.

share|improve this question
    
Hi: I made a few tweaks, and I hope you review them to make sure they are in the spirit of what you were asking. –  rschwieb Jan 22 '13 at 14:49
    
Sorry, English is not my first language. But noted! =) –  Eric Jan 22 '13 at 15:14

1 Answer 1

up vote 1 down vote accepted

$\newcommand{\Cl}{\mathrm{C}\ell}$Write $$\tilde{q} = \langle b_2, \dots, b_n \rangle$$ and let $\{e_1, \dots, e_n\}$ be the standard basis for $K^n$. Define a map $$f: K^{n-1} = \operatorname{span} \{e_2, \dots, e_n\} \longrightarrow \Cl^0(q)$$ by $$f(e_i) = e_1 \cdot e_i$$ for $i > 1$ and extending linearly. Given $$x = \sum_{i = 2}^n x_i e_i \in \operatorname{span} \{e_2, \dots, e_n\} = K^{n-1},$$ we have \begin{align} f(x) \cdot f(x) & = \sum_{i,j} x_i x_j e_1 \cdot e_i \cdot e_1 \cdot e_j \\ & = - \sum_{i,j} x_i x_j e_1 \cdot e_1 \cdot e_i \cdot e_j \\ & = \sum_{i,j} x_i x_j (-a_1) e_i \cdot e_j \\ & = \sum_{i = 2}^n (-a_1 a_i) x_i^2 \\ & = \tilde{q}(x) \cdot 1. \end{align} Hence $f$ is a Clifford map and by the universal property of Clifford algebras, it extends uniquely to a map $$\tilde{f}: \Cl(\tilde{q}) \longrightarrow \Cl^0(q).$$ Now it is easy to check that $\tilde{f}$ is an isomorphism, so that $$\Cl^0(\langle a_1, \dots, a_n \rangle) \cong \Cl(\langle b_2, \dots, b_n \rangle).$$

share|improve this answer
    
Thank you Sir! Just a question, is $e_i\cdot e_j = 0$ when $i \neq j$? Is the $\cdot$ here, similar to some kind of inner product? –  Eric Jan 22 '13 at 17:05
    
$e_i \cdot e_j = 0$ when $i \neq j$ since $\{e_k\}$ is the standard basis for $K^n$. I use $\cdot$ to denote Clifford multiplication, i.e. multiplication of vectors in the Clifford algebra. –  Henry T. Horton Jan 22 '13 at 17:07
    
Got it! Thanks! –  Eric Jan 22 '13 at 17:08

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.