octanion
in [Math]
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From: Aleks Kleyn on 27 Apr 2010 23:05 i want to make some calculation in octanion algebra. however when i looke for table of product i was little bit confused. usually i see table like following e0 i, e1 j, e2 k, e3 l, e4 il, e5 jl, e6 kl, e7 i, e1 -1 k −j il −l −kl jl j, e2 −k -1 i jl kl −l −il k, e3 j −i -1 kl −jl il −l l, e4 −il −jl −kl -1 i j k il, e5 l −kl jl −i -1 −k j jl, e6 kl l −il −j k -1 −i kl, e7 −jl il l −k −j i -1 However in paper arXiv: 0105155 John Baez uses following table of product e1 e2 e3 e4 e5 e6 e7 e1 −1 e4 e7 −e2 e6 −e5 −e3 e2 −e4 −1 e5 e1 −e3 e7 −e6 e3 −e7 −e5 −1 e6 e2 −e4 e1 e4 e2 −e1 −e6 −1 e7 e3 −e5 e5 −e6 e3 −e2 −e7 −1 e1 e4 e6 e5 −e7 e4 −e3 −e1 −1 e2 e7 e3 e6 −e1 e5 −e4 −e2 −1 He does not describes what vectors of basis he uses. Trying i found one possible case e0=1 e1=i e2=j e3=l e4=k e5=kl e6=jl e7=il This choice matches first line. However according to this table e2 * e3= j l = e6 However in corresponding cell in the table i see e5=kl. Did i missed something? Thank you, Aleks Kleyn
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