Automorphism of quaternions

Question

Let $a, b \in \mathbb{H}$, where $\mathbb{H}$ denotes the algebra of the quaternions, with $a^2=b^2=-1$ and $ab = -ba$.

Show that $\phi: \mathbb{H} \longrightarrow \mathbb{H}, x + yi + zj + wk \mapsto x + ya +zb + wab$ with $x,y,z,w \in \mathbb{R}$, is an algebra automorphism. I was already able to show that this is an algebra homomorphism. I am stuck on the bijection part.

Answer 1

The algebra endomorphism is nontrivial so its kernel is a proper two-sided ideal. Since $\mathbb{H}$ is a division algebra, the only nonzero two-sided ideal is improper, so it is an automorphism. This is the quick way to do this, but I feel it's instructive to talk about other ways there are to see this.

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You can show $\{1,a,b,ab\}$ is a basis to see the range is all of $\mathbb{H}$. First I want to establish some background which is useful not just for this problems but most problems about quaternions.

For vectors (i.e. pure imaginary quaternions) $\mathbf{a},\mathbf{b}$ we have the formula

$$ \mathbf{ab} = \underbrace{-\mathbf{a}\cdot\mathbf{b}}_{~\mathrm{scalar}}+\underbrace{\mathbf{a}\times\mathbf{b}}_{\mathrm{vector}}. $$

The idea for this comes from simply looking at scalar and vector components of the multiplication table for the basis elements $\{\mathbf{i},\mathbf{j},\mathbf{k}\}$. Both sides are linear in $\mathbf{a}$ and $\mathbf{b}$ so it suffices to check the equation is true for basis elements and then it's true for all vectors. A couple of consequences:

• $\mathbf{ab}$ is real $\iff \mathbf{a}\,\|\,\mathbf{b}$ are parallel $\iff \mathbf{a},\mathbf{b}$ commute,
• $\mathbf{ab}$ is a vector $\iff \mathbf{a}\perp\mathbf{b}$ are perpendicular $\iff \mathbf{a},\mathbf{b}$ anticommute.

We can generalize a little bit to arbitrary quaternions $a$ and $b$:

• $a,b$ commute $\iff$ their vector parts are parallel,
• $a,b$ anticommute $\iff$ they are perpendicular vectors.

Use the distributive property to show this! Continuing on, we also have:

Exercise. Every nonzero quaternion has a polar form $r\exp(\theta\mathbf{u})$ for some unit vector $\mathbf{u}$, magnitude $r>0$, and convex angle $\theta$. It is unique for all but negative reals (then $\mathbf{u}$ is any unit vector).

This also has a couple of consequences:

• The square roots of $+1$ are precisely $\pm1$,
• The square roots of $-1$ are precisely the unit vectors.

Now we're given the hypotheses $a^2=b^2=-1$ and $ab=-ba$. The latter implies $a$ and $b$ are perpendicular unit vectors, which implies $ab$ is also a vector (and a unit vector, since the norm is multiplicative). We can check $ab$ anticommutes with $a$ and $b$ so it is also perpendicular to them (or, we can check directly using the real inner product $\langle a,b\rangle=\mathrm{Re}(\bar{a}b)$; note $\overline{a}=-a$ for vectors $a$).

In conclusion, $\{1,a,b,ab\}$ is an orthonormal basis.