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 $q(\sum_{i=1}^n x_ie_i) = \sum_{i=1}^n a_ix_i^2$.

  • $\begingroup$ 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. $\endgroup$ – rschwieb Jan 22 '13 at 14:49
  • $\begingroup$ Sorry, English is not my first language. But noted! =) $\endgroup$ – Eric Jan 22 '13 at 15:14

$\newcommand{\Cl}{\mathrm{C}\ell}$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 $$\Cl^0(\langle a_1, \dots, a_n \rangle) \cong \Cl(\langle b_2, \dots, b_n \rangle).$$

  • $\begingroup$ 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? $\endgroup$ – Eric Jan 22 '13 at 17:05
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    $\begingroup$ $e_i \cdot e_j = 0$ 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. $\endgroup$ – Henry T. Horton Jan 22 '13 at 17:07
  • $\begingroup$ Got it! Thanks! $\endgroup$ – Eric Jan 22 '13 at 17:08

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