2
$\begingroup$

To me it seems the way to motivate the Cayley-Dickson construction is to prove Hurwitz's theorem, which is done over at Wikipedia. The theorem states the only real division algebras equipped with a norm associated with a Euclidean inner product are $\Bbb R$, $\Bbb C$, $\Bbb H$, and $\Bbb O$ (reals, complex numbers, quaternions and octonions). To begin with, I was eventually able to follow along the proof of all the properties of such algebras listed in the Definition section (although it took me a bit to get the polarization identity. Then I started reading the proof.

We begin with a Euclidean Hurwitz algebra $A$, and consider a unital subalgebra $B$. Because it's unital, it contains $\Bbb R$, so if $b\in B$ then ${\rm Im}(b)=b-{\rm Re}(b)\in B$ and $b^*={\rm Re}(b)-{\rm Im}(b)\in B$ so we conclude $B^*=B$, i.e. $B$ is closed under conjugation and taking real/imaginary parts. If we pick an element $\ell\in A\setminus B$ (Wikipedia uses $j$, I want to use $\ell$) then we may shear and normalize, so wlog $\|\ell\|^2=1$ and $\ell\perp B$ (i.e. $(\ell,b)=0$ for all $b\in B$). We prove $B\oplus B\ell$ is an orthogonal direct sum easily enough, but what I don't understand is:

Why is $B\oplus B\ell$ closed under multiplication?

Evidently $BB\subseteq B$. Using $\ell b=b^*\ell$ and $(xy)^*=y^*x^*$ I can see that $(B\ell)B\subseteq B\oplus B\ell$ and $B(B\ell)\subseteq B\oplus B\ell$ are equivalent, so we need only prove the latter. Wikipedia's first and third bullet points in the Classification section cover $B(B\ell)$ and $(B\ell)(B\ell)$. The technique used there is to invoke the fact that $(a,x)=(b,x)\forall x\in A$ implies $a=b$. However I am unable to follow the equalities in those bullet points. How does $(b(c\ell),x)=(b(\ell x),\ell(c\ell))$? Since the LHS has one $\ell$ and the RHS has three, I imagine the facts $(u,v)(w,w)=(uw,vw)$, $(w,w)(u,v)=(wu,wv)$ were used with $(\ell,\ell)=1$, but I don't see how. It's also unclear to me how $-x^*$ may be replaced with $x$ in the last equality, given that $x\in A$ is arbitrary. I was eventually able to see the first equality in the third bullet point, but it took me five lines!

So I suppose my subquestion is:

What are the missing steps in those two bullet points?


I found this helpful note. Based on $2(a,b)(c,d)=(ac,bd)+(ad,bc)$ from my previous question, if $(u,v)=0$ we can conclude $(au,bv)=-(av,bu)$. Therefore

$$\begin{array}{ll} (a(b\ell),x) & = (b\ell,a^*x) \\ & -(b\ell,x^*a) \\ & =(ba,x^*\ell) \\ & = (ba,(-x+2(x,1)1)\ell) \\ & = -(ba,x\ell)+2(x,1)(ba,\ell) \\ & = -(ba,x\ell) \\ & =((ba)\ell,x) \end{array} $$

for all $x\in A$, which proves $a(b\ell)=(ba)\ell$ when $a,b\in B$.

It proves $(a\ell)(b\ell)=-b^*a$ using the Moufang identities though. Can we finish off without?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.