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: $$ \begin{array}{cccc} & \mathbf{e_1} & \mathbf{e_2} & \mathbf{e_{12}} \\\\ \mathbf{e_1}& 1 & e_{12} & e_2 \\\\ \mathbf{e_2}& -e_{12} & 1 & -e_1 \\\\ \mathbf{e_{12}}& -e_2 & e_1 & -1 \end{array} $$

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

  • $\begingroup$ I see what you did, but shouldn't I put b as combination of e1, e2 and then use the Clifford product definition? I mean, can I do that multiplication so directly? p.s.: sorry if it's a silly question, $\endgroup$ – Fernando H. M. Bastos Jul 26 '12 at 0:39
  • $\begingroup$ @FernandoH.M.Bastos: Not a problem. What makes you think you can write ${\bf b}$ in terms of $e_1$ and $e_2$? $\endgroup$ – user26872 Jul 26 '12 at 0:43
  • $\begingroup$ Actually, that was my main issue, cause I couldn't. In a previous execise I could put a vector c=(1+e2) in the form c=c1e1, with c1=(e1+e12), and got to the right answer so thought I should do it everytime. That previous result was just luck or there is any restriciton in this case of b? $\endgroup$ – Fernando H. M. Bastos Jul 26 '12 at 0:52
  • 2
    $\begingroup$ @FernandoH.M.Bastos: The space is spanned by the linearly independent set $\{1,e_1,e_2,e_{12}\}$. By linear independence, you can't write any of the basis elements as a linear combination of the others. Think of it as trying to write $\hat z$ as a linear combination of $\hat x$ and $\hat y$. $\endgroup$ – user26872 Jul 26 '12 at 0:55
  • $\begingroup$ Your last comment made everything clear, thanks a lot! $\endgroup$ – Fernando H. M. Bastos Jul 26 '12 at 1:31

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