In a Clifford Algebra $\mathbb{CL}_{(1,1,1)}$ we have the following relations:
$e_1^2=1$
$e_{-1}^2=-1$
$e_0^2=0$
Question:
Do all the products $e_ie_j=-e_je_i$ anticommute? If so, why?
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We know that for versors $e_1,e_2$ squaring to $1$, the following condition is sufficient to ensure $vv=|v|^2$ when the product of vectors preserves the distributive property over addition:
$e_1e_2=-e_2e_1$
This means that for any vector $v=ae_1+b_2e_2$ the following equation holds:
$vv=a^2e_1^2+b^2e_2^2+ab(e_1e_2+e_2e_1)=a^2+b^2=|v|^2$
Similarly, for $v=ae_{-1}+be_{-2}$ to get $vv=-|v|^2$, is enough to have $e_{-i}e_{-j}=-e_{-j}e_{-i}$
$vv = a^2e_{-1}^2 + b^2e_{-2}^2 + ab(e_{-1}e_{-2} + e_{-2}e_{-1}) = -a^2 - b^2 = -|v|^2$
and for $v=ae_0+be_{0'}$ if $e_{0}e_{0'}=-e_{0'}e_{0}$ then
$vv=a^20+b^20+ab(e_{0}e_{0'}+e_{0'}e_{0})=0$
Can the same line of thought be used to determine whether the products $e_1e_0$, $e_1e_{-1}$ or $e_0e_{-1}$, anticommute? If so, what are the geometric interpretations of these products?
I am aware that different conventions may lead to different applications. However, I am not familiar with any map of conventions $\rightarrow$ applications.
I read Geometric Algebra, which only hands down the case where $vv=|v|^2$.