I am trying to show the three dimensional Clifford algebra over the reals $C_3 \cong \mathbb{H} \oplus \mathbb{H}$.

where $e_1,e_2,e_3$ are generators of $C_3$. I found two idempotents $k_1 = (1 + e_1e_2e_3)/2$ and $k_2 = (1- e_1e_2 e_3)/2$ (orthogonal) so I am trying to use the idea that $C_3 \cong C_3 k_1 \oplus C_3 k_2$.

I am struggling to show $C_3k_1 = \mathbb{H} = C_3k_2$.

  • $\begingroup$ $\mathbb H\times \mathbb H$ is $8$- dimensional. Do you mean that the underlying vector space is $3$-dimensional? Secondly, there are four distinct nondegenerate forms that produce four nonisomorphic clifford algebras. $\mathbb H\times \mathbb H$ is just one of them. So why are you focused on this one? $\endgroup$ – rschwieb Dec 11 '16 at 12:20
  • $\begingroup$ I think you're off the mark trying to write $C_3 \cong C_3 k_1 \oplus C_3 k_2$. You should be considering the even subalgebra $C_3^+$ instead, such that $C_3 \cong C_3^+ k_1 \oplus C_3^+ k_2$. $\endgroup$ – Muphrid Dec 11 '16 at 18:28
  • $\begingroup$ what is the signature of your algebra? $\endgroup$ – user48672 May 19 '17 at 16:26

I assume that you have a signature $\mathcal{C}l(0,3)$ but the method should work with the other signatures as well.

Consider the general element

$A= {{a}_{1}}+{{e}_{1}}\,{{a}_{2}}+{{e}_{2}}\,{{a}_{3}}+{{e}_{3}}\,{{a}_{4}}+{{a}_{5}}\,\left( {{e}_{1}} {{e}_{2}}\right) +{{a}_{6}}\,\left( {{e}_{1}} {{e}_{3}}\right) +{{a}_{7}}\,\left( {{e}_{2}} {{e}_{3}}\right) +{{a}_{8}}\,\left( {{e}_{1}} {{e}_{2}} {{e}_{3}}\right) $

The elements of the event sub-algebra are of the form

$A_2 = {{a}_{1}}+{{a}_{5}}\,\left( {{e}_{1}} {{e}_{2}}\right) +{{a}_{6}}\,\left( {{e}_{1}} {{e}_{3}}\right) +{{a}_{7}}\,\left( {{e}_{2}} {{e}_{3}}\right) $

The dual of an element of the odd sub-algebra is of the form

$-A_1^\star = i A_1 = {{a}_{8}}-{{a}_{4}}\,\left( {{e}_{1}} {{e}_{2}}\right) +{{a}_{3}}\,\left( {{e}_{1}} {{e}_{3}}\right) -{{a}_{2}}\,\left( {{e}_{2}} {{e}_{3}}\right) $,

where $i$ is the pseudo-scalar.

Finally, the bi-vector multiplication table is

$\begin{pmatrix}1 & {{e}_{1}} {{e}_{2}} & {{e}_{2}} {{e}_{3}} & {{e}_{1}} {{e}_{3}}\\ {{e}_{1}} {{e}_{2}} & -1 & -{{e}_{1}} {{e}_{3}} & {{e}_{2}} {{e}_{3}}\\ {{e}_{2}} {{e}_{3}} & {{e}_{1}} {{e}_{3}} & -1 & -{{e}_{1}} {{e}_{2}}\\ {{e}_{1}} {{e}_{3}} & -{{e}_{2}} {{e}_{3}} & {{e}_{1}} {{e}_{2}} & -1\end{pmatrix} $

Which is the quaternion multiplication table.

So we can define a morphism $ H: A \mapsto (A_2, i A_1) $ exhibiting the desired structure.


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