Do all automorphisms of $\mathbb{H}$ preserve the norm of an element?

Question

Do all automorphisms of $\mathbb{H}$-- the Hamilton quaternions-- preserve the norm of an element? I can't seem to answer this question without using the not-so-elementary fact that all automorphisms of $\mathbb{H}$ are inner-- this tells us that all automorphisms fix $\mathbb{R}$, and so for any given $x=a+bi+cj+dk$, $\phi(x)=a+b\phi(i) + c \phi(j)+d \phi(k)$. Am I making some blatantly incorrect assumption? If so, how would I show this fact about the norm being preserved?

Answer 1

First, note than any ($\mathbb{R}$-algebra) automorphism $\phi$ of $\mathbb{H}$ preserves $1$ and hence (pointwise) $\mathbb{R}$.

Now, any automorphism $\phi$ of $\mathbb{H}$ is given by $\phi(x) := q x q^{-1}$ for some $q \in \mathbb{H}^*$. We may as well pull out the norm of $q$ and absorb it into its inverse, and so assume that $q$ and $q^{-1}$ both have unit length, and in particular that $\phi(x) = q x \bar{q}$.

We show now that any automorphism $\phi$ preserves the squared norm $Q(x) := |x|^2 = x \bar{x}$ of an arbitrary element $x \in \mathbb{H}$:

$Q(\phi(x)) = Q(qx\bar{q}) = qx\bar{q}\overline{qx\bar{q}} = qx\bar{q}q\bar{x}\bar{q}$.

Now, $\bar{q}q = 1$, so this is

$qx\bar{x}\bar{q} = q Q(x) \bar{q} = Q(x) q\bar{q} = Q(x)$,

where the middle equality uses that $Q(x)$ is real.

Remark The fact that any automorphism of $\mathbb{H}$ preserves the norm (or quadratic form $Q$) corresponds to the inclusion $Aut(\mathbb{H}) \cong SU(2) \hookrightarrow SO(4) \cong SO(Q)$. In particular the action of $SU(2)$ preserves the orthocomplement $\mathbb{R}^{\perp} \cong \mathbb{R}^3$ and the restriction of $Q$ to that set and so it defines a homomorphism $SU(2) \to SO(3)$; it turns out to be surjective and have kernel $\{\pm 1\}$, and so it is a double cover.

Remark 2 Note that we can in fact prove that the algebraic structure and norm are compatible in a stronger sense yet, namely that they satisfy $Q(x y) = Q(x) Q(y)$.

Thanks to Jyrki Lahtonen for pointing out an issue with a previous argument.