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Let $A$ be a real composition algebra ($A=\mathbb{R}, \mathbb{C}, \mathbb{H}, \mathbb{O}$). I would like to prove that $$ |\lambda|=1 \implies(\lambda u) \overline{(\lambda v)}=u\overline{v}$$


In a composition algebra we have that $\lambda \bar{\lambda}=|\lambda|^2$ but $A$ may be non-associative and non-commutative.

$\bar{x}:=2(x,1)-x$.

Moufang identities hold:

$$(ax)(ya)=a((xy)a) $$ $$a(x(ay))=(a(xa))y$$ $$x(a(ya))=((xa)y)a$$

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This is not true (in the noncommutative case). For instance, in the quaternions, taking $\lambda=u=i$ and $v=j$ we get $$(\lambda u)\overline{(\lambda v)}=i^2\overline{(ij)}=k\neq -k=i\overline{j}=u\overline{v}.$$

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