Automorphism group of quaternions

Question

Let $\mathbb{H}$ be the algebra of quaternions. How to prove that the automorphism group $Aut(\mathbb{H})$ is isomorphic to the group $SO_3$?

I found that all derivations on algebra $\mathbb{H}$ are inner. Moreover, they are isomorphic to the Lie algebra $so_3$. It remains to understand whether it is simply connected or not. But I don’t know how to show that the group $Aut(\mathbb{H})$ is not simply connected.

Are you assuming the automorphisms are $\Bbb R$-linear, or that they are continuous? — anon, Nov 06 '20 at 22:40
@JasonDeVito: This post claims that $Aut(\Bbb H=Q_8)=S_4\simeq \Bbb Z_2\ltimes \Bbb Z_2^$. So is it true that $\mathrm{SO}(3)\simeq \Bbb Z_2\ltimes \Bbb Z_2^$? — C.F.G, Nov 08 '20 at 13:37
@C.F.G.: The group $Q_8$ refers to the finite group ${\pm 1, \pm i,\pm j, \pm k}$, not to the full set of quaternions. — Jason DeVito - on hiatus, Nov 08 '20 at 14:08

anon · Accepted Answer · 2020-11-09T14:57:58.870

4

An automorphism $\alpha$ must preserve $1$, so by $\mathbb{R}$-linearity it preserves the real line.

Also $\alpha$ must preserve the set of square roots of $-1$, which is the unit sphere $S^2$, and all of their real multiples, which forms the pure imaginary subspace $\mathbb{R}^3$ of 3D vectors.

If $\mathbf{u}$ and $\mathbf{v}$ are vectors, the real part of $\mathbf{uv}$ is minus the dot product, i.e. $-\mathbf{u}\cdot\mathbf{v}$. Since $\alpha$ preserves real parts, this means it preserves dot products, so it acts as a linear isometry of $\mathbb{R}^3$. So $\alpha\in\mathrm{O}(3)$.

Similarly the imaginary part of $\mathbf{uv}$ is the cross product $\mathbf{u}\times\mathbf{v}$. See if you can show $\alpha$ preserves cross products, and this will further mean $\alpha\in\mathrm{SO}(3)$.

Conversely, any element of $\mathrm{SO}(3)$ preserves dot products and cross products, so preserves the quaternion multiplication (which is built out of them). Because of $S^3\to\mathrm{SO}(3)$, this means all automorphisms of $\mathbb{H}$ are inner.

edited Nov 09 '20 at 14:57

answered Nov 06 '20 at 22:48

anon

151,657

Why couldn't $\alpha \in O(3)$? – Jason DeVito - on hiatus Nov 08 '20 at 13:08
@JasonDeVito Good point. – anon Nov 08 '20 at 19:01
Why is it the whole SO(3) group? – Invincible Nov 09 '20 at 14:52
@Vladislav Also good point. – anon Nov 09 '20 at 14:58
What do you mean by $S^3 \to SO(3)$? – Invincible Nov 09 '20 at 15:17
I mean the homomorphism $S^3\to\mathrm{SO}(3)$ that is used to represent 3D rotations by quaternions (namely, conjugating by $e^{\theta{\bf u}}$ is rotation around $\bf u$ by $2\theta$). – anon Nov 09 '20 at 15:26

Automorphism group of quaternions

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