hw2_sol - 18.966 Homework 2 Solutions. 1. Equip R 7 = Im O...

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Unformatted text preview: 18.966 Homework 2 Solutions. 1. Equip R 7 = Im O = { a + be, Re a = 0 } with the cross-product x y = Im( xy ). By definition of the octonion product, if x, y Im O then Re( xy ) = x, y (the usual Euclidean scalar product on R 7 ). Indeed, Re(( a + be )( a + b e )) = Re( aa b b ) = Re( aa b b ) = a, a b, b . Therefore x y = Im( xy ) xy = x y , with equality iff x y . Let x S 6 R 7 , and let y T x S 6 x R 7 . Then we define J x ( y ) = x y. Note that, since y x , we have J x ( y ) = x y = xy + x, y = xy . We have to prove that J x maps T x S 6 to itself, and that J x 2 = Id. For this purpose, let x = a + be be a unit imaginary octonion ( a = a ) and let y = c + de be any octonion: then x ( xy ) = a ( ac db ) ( ad + c b ) b + ( da + bc ) ae + b ( ca bd ) e = | a | 2 c ( a + db c | b | 2 d | a | 2 c ( a + a ) e | b | 2 de a ) e + b = ( | a | 2 + | b | 2 )( c + de ) = x 2 y = y. If y T x S 6 , i.e. y is imaginary and y x , then x, xy = Re( x ( xy )) = Re( y ) = 0, so J x maps T x S 6 to...
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This note was uploaded on 03/08/2010 for the course MATHEMATIC 570 taught by Professor Denisauroux during the Spring '10 term at Caltech.

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hw2_sol - 18.966 Homework 2 Solutions. 1. Equip R 7 = Im O...

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