In a real Clifford algebra $\mathbb{Cl}(2,2)$ over ${\mathbb R}^4$ with the quadratic form defined on the orthogonal basis $e_1,e_2,e_3,e_4$ by $e_1^2=1, e_2^2=1, e_3^2=-1$, and $e_4^2=-1$, find an even number of vectors $a_i$ ($i=1,2,\dots,2k$) such that $a_i^2=1$ for all $i$ and $e_1e_2e_3e_4=a_1a_2\cdots a_{2k}$.
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Can I use the basis vectors, i.e. can it be the case that for all $i$, there exists a $j$ such that $a_i = e_j$? – 11Kilobytes Jan 06 '15 at 12:17
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Some basis vectors square to -1. – Andrey Sokolov Jan 06 '15 at 20:16
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so why is $e_1e_2e_3e_4e_3e_4$ for example not a solution. – 11Kilobytes Jan 07 '15 at 22:36
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Because it's not what I'm asking for. – Andrey Sokolov Jan 08 '15 at 02:14
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Oh, I see, $e_3^2 = -1$, and you're asking for all the $a_i$s to satisfy, $a_i^2 = 1$, silly me. – 11Kilobytes Jan 08 '15 at 14:59
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And it doesn't evaluate to $e_1e_2e_3e_4$. – 11Kilobytes Jan 08 '15 at 15:05
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I suspect there will be infinitely many such vectors because of the any even power will square to 1 or -1. So then a suitable selection of the sign of the expression considering the anticommutativity will yield the result. – user48672 Aug 10 '15 at 23:16
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@user48672 Sure. Can you give a specific example? – Andrey Sokolov Aug 10 '15 at 23:43