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# hw2_sol - 18.966 – Homework 2 – Solutions 1 Equip R 7 =...

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

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