octanion---
in [Math]
|
From: JEMebius on 2 May 2010 10:27 Aleks Kleyn wrote: > > "JEMebius" <jemebius(a)xs4all.nl> wrote in message > news:4BD8CE6A.5060606(a)xs4all.nl... >> Aleks Kleyn wrote: >>> 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 >> >> >> Different bases of units: this reminds me of quaternions. >> >> In the quaternion algebra one has the well-known units 1, i, j, k with >> the well-known >> multiplication table. >> Now think of the 4D space R^4 spanned by 1, i, j, k as the direct sum >> of the 1D subspace R with basis {1} and the 3D subspace R^3 with basis >> {i, j, k}. >> Also recall the quaternion product formula >> >> p * q = (Sp, Vp) * (Sq, Vq) = (Sp.Sq - (Vp dot Vq), Sp.Vq + Sq.Vp + >> [Vp cross Vq]), >> >> where the quaternions p, q were split into their scalar parts Sp, Sq >> and their vector parts Vp, Vq. >> >> This formula is coordinate-free and rotation-invariant. Therefore, >> whatever set of orthonormal vectors e1, e2, e3 (in this order) one >> takes instead of i, j, k, one will obtain just another copy of the >> quaternion algebra, isomorphic to the original one. >> Any rotation of the basis {i, j, k} will do the job, and no other >> basis transformation will do, so the 3D rotation group SO(3) is the >> automorphism group of the quaternions. >> >> >> By and large the same game is played in R^8 spanned by 1, e1, e2, ... e7. >> We have R^8 = R (+) R^7, where R has basis {1} and R^7 has basis {e1, >> e2, ... e7}. >> To obtain isomorphic copies of the octonion algebra founded on {1, e1 >> ... e7} one has anyhow to have the subspace {1} fixed, like in the >> quaternions. >> >> One could think that one might rotate the basis {e1 ... e7} of R^7 at >> will, but this will not work in general: not the full 7D rotation >> group SO(7), but only a certain subgroup, named G2, is the >> automorphism group. >> >> G2 is the smallest compact exceptional Lie group; it is a >> 14-dimensional subgroup of SO(7), which is itself 21-dimensional. So >> there are 7 relations among the elements of the rotation matrices in >> addition of the 28 relations which exist in general 7D rotation >> matrices. I do not know them by heart, but the existing literature >> will help. >> >> A further complication in understanding what is going on algebraically >> and geometrically, is the circumstance that SO(7) contains an >> infinitude of subgroups G2, any pair of which is conjugate in SO(7). >> This is in stark contrast with the quaternions, where we have only a >> single copy of SO(3). > This note may appear important to me when i finish calculation. > there is some glue in Baez's paper. I understand that for him is more > important symmetry and this the reason why he did not put i j k. and i > will try to follow his paper to understand it better. >> >> Morale of this story: multiplication tables that look completely >> different may actually be essentially identical! >> And one more remark: mind the left-right orientation. A soon as one >> has an odd permutation of the seven elements of the one multiplication >> table onto the seven of the other table, the whole thing reverses >> signs. By the way, this also happens in the quaternions. Observe for >> yourself how things go. >> >> Good luck: Johan E. Mebius > Also a second reply to the original post ---------------------------------------- Two useful Wikipedia lemmas: http://en.wikipedia.org/wiki/Seven-dimensional_cross_product http://en.wikipedia.org/wiki/Cross_product Forgot to add to first reply: Tidied-up versions of the octonion multiplication tables in the original post ----------------------------------------------------------------------------- (Use monospaced font) 1=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 ----------------------------------------------- Good luck: Johan E. Mebius
From: Aleks Kleyn on 2 May 2010 10:42 i yesterday recovered what basis used Baez in his paper. Fano diagram helped me in this. i put attention that in the midle point arrows in Baez paper and in wikipedia are in opposite direction. It means that Baez used e7=l^{-1}=-l. All together it gives e0=1 e1=i e2=j e3=il e4=k e5=kl e6-jl e7=-l i think he used this numeration to show different simmetries in the product. also he did not clearly tell e1=i because there is circle symmetry between i j k. Aleks Kleyn
From: Aleks Kleyn on 5 May 2010 01:19 i yet not finished my calculations. however i discovered one interesting thing. if I assign conjugate for z as z* then i can write linear equation z*=-1/6 ( z + (iz)i + (jz)j + (kz)k + (lz)l + ((il)z)(il) + ((jl)z)(jl) + ((kl)z)(kl)) Aleks Kleyn
|
Pages: 1 Prev: The Klein group (non-cyclic of order 4) Next: Symmetry and Abstraction |