2
$\begingroup$

I'm studying Clifford Algebra $\mathcal{Cl}_2$ and got stuck in an exercise:

Let $\mathbf{a}=e_2+e_{12},\quad \mathbf{b}=(1/2)(1+e_1).$ Compute $\mathbf{ab}$.

The answer is zero, but I can't get to it and I'm having trouble putting $\mathbf{b}$ in the form $\mathbf{b}=b_1e_1+b_2e_2$.

Important information:

  • $(1, e_1, e_2, e_{12})$ form a basis for the Clifford algebra $\mathcal{Cl}_2$
  • The Clifford product of two vectors $\mathbf{a}=a_1e_1+a_2e_2\text{ and }\mathbf{b}=b_1e_1+b_2e_2$ is defined as $\mathbf{ab}=a_1b_1+a_2b_2+(a_1b_2-a_2b_1)e_{12}$
  • And the following multiplication table: $ \begin{array}{cccc} & \mathbf{e_1} & \mathbf{e_2} & \mathbf{e_{12}} \\\\ \mathbf{e_1}& 1 & e_{12} & e_2 \\\\ \mathbf{e_2}& -e_{12} & 1 & -e_1 \\\\ \mathbf{e_{12}}& -e_2 & e_1 & -1 \end{array} $

1 Answers 1

3

$\begin{eqnarray*} 2{\bf a}{\bf b} &=& (e_2+e_{12})(1+e_1) \\ &=& e_2+e_{12}+e_2 e_1+e_{12}e_1 \\ &=& e_2 + e_{12} +(-e_{12}) + (-e_2) \\ &=& 0 \end{eqnarray*}$

  • 0
    @FernandoH.M.Bastos: Glad to help. Cheers!2012-07-26