  Octonions & Sedenions
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 Representation:     Since octonions do not form an associative algebra they can not be represented by matrices. However this does not preclude us from constructing matrices with octonion entrices or representing adjoint algebra of octonions in terms of matrices.   The following describes two methods of representing octonions. (1) Quaternionic representation: An octonion, A, is written as an ordered pair of two 4-dimensional quaternions, Q1 and Q2 as         A := ( Q1; Q2 )  . For the algebra of octonions, O , to satisfy the basic properties outlined in Introduction one needs to define the multiplication rules of two octonions,  A and B.  Writing B in terms of two quaternions, Q3 and Q4, we define the multiplication of A and B, via Cayley-Dickson process by         AB := ( Q1; Q2 )( Q3; Q4 )                 := ( Q1Q3 + bQ4*Q2; Q2Q3* + Q4Q1 ) , where Q* is a conjugate of Q, and  b is a field parameter. (2) Coordinate representation:     An octonion, A, may be written componentwise as         A := m=0S7  am im  =  a0 i0  +  j=1S7  aj ij , where im are the basal elements of O. The basal element, i0, is chosen as  the basis of reals, R.  For the algebra of octonions to satisfy the basic properties outlined in Introduction one needs to define the multiplication rules for the basis. There are many choices for this task but we will describe just two of them below.      (I) via congruence modulo 7:             Define     ij ij = -1 ;    ij ij+1 :=  ij+3 :=  ij-4 ,        (mod 7)                with         i2k i2m = i2n,    whenever ik im = in . Then the multiplication rule for the seven basal elements, ij  ( j = 1,...7) is summed up in the following seven cycles:         (1,2,4) ,  (1,3,7) ,  (1,5,6) ,  (2,3,5) ,  (2,6,7) ,  (3,4,6) ,  (4,5,7) . The cycle (j,k,m) is to mean         ijik= im ,   ik im = ij ,   ik ij = - im  .     (II) via Cayley-Dickson process from the quaternion basis: Write octonion basal element as ordered pair of quaternion basal elements     im  := ( im ; 0 ) ,      and       im+4  := ( 0 ; im ),     for  m = 0,1,2,3.     e.g., i3 = ( i3 ; 0 ) ,  and i6 = ( 0 ; i2 ) .  Then by Cayley-Dickson process the multiplication rule can be found, for example,   i3 i6= ( 0 ; i2 i3) = ( 0 ; i1 ) = i5, and summarized in the following seven cycles: (1,2,3) , (1,4,5) , (1,7,6) , (2,4,6) , (2,5,7) , (3,4,7) , (3,6,5) .            ..... details
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