# Polarization identity $2(a,b)(c,d)=(ac,bd)+(ad,bc)$

I am interested in following along this Wikipedia article's derivation of properties of composition algebras (in particular, Euclidean Hurwitz algebras). Let $A$ be a unital, not necessarily associative algebra over the reals with a norm $q:A\to\Bbb R$ satisfying the identity $q(ab)=q(a)q(b)$ which leads to a Euclidean inner product $(a,b):=\frac{1}{2}[q(a+b)-q(a)-q(b)]$.

Why does $q$ multiplicative entail the following polarization identity?

$$2(a,b)(c,d)=(ac,bd)+(ad,bc) \tag{\circ}$$

By plugging in $x+y,z$ or $x,y+z$ into $(ab,ab)=(a,a)(b,b)$ I get

$$(xz,yz)=(x,y)(z,z), \quad (xy,xz)=(x,x)(y,z).$$

We can also rewrite $(\circ)$ as

$$[(ac,bd)-(a,b)(c,d)]+[(ad,bc)-(a,b)(c,d)]=0.$$

Either way, I haven't been able to derive $(\circ)$. Ideas?

## 1 Answer

After a bit of fruitless mess, I eventually found the solution is straightforward:

$$\begin{array}{ll} 2(a,b)(c,d) & =[(a+b,a+b)-(a,a)-(b,b)](c,d) \\ & = (a+b,a+b)(c,d)-(a,a)(c,d)-(b,b)(c,d) \\ & = ((a+b)c,(a+b)d)-(ac,ad)-(bc,bd) \\ & =(ac+bc,ad+bd)-(ac,ad)-(bc,bd) \\ & = (ac,ad)+(ac,bd)+(bc,ad)+(bc,bd)-(ac,ad)-(bc,bd) \\ & = (ac,bd)+(bc,ad). \end{array}$$

This uses the $(x,x)(y,z)=(xy,xz)$ property I found in my previous efforts.