Quaternion division rings in characteristic $0$

Question

Hamilton's original quaternions $\mathcal{Q}$ form a division ring which is $2$-dimensional over $\mathbb{C}$, and which has $\mathbb{R}$ as its center. Define its elements as $a + bi + cj + dk$, where $a, b, c, d \in \mathbb{R}$, $i^2 = j^2 = k^2 = -1$, and $ijk = -1$.

Now suppose $\ell$ is an algebraically closed field in characteristic $0$, so that it contains an involutory automorphism $\sigma$ which is not the identity. The fixed field of $\sigma$ is a subfield $k$ which is real-closed ($[ \ell : k ] = 2$). Now let $i$ be a solution of the equation $X^2 = -1$; this element is not contained in $k$. Introduce symbols $j, k$ as above, and consider the set $\{ a + bi + cj + dk \vert a, b, c, d \in k \}$ with natural addition and multiplication.

My questions:

1. I gather that $\mathcal{Q}(\ell,\sigma)$ defined above is also a "quaternion" division ring which is $2$-dimensional over $\ell$, and whose center is $k$?
2. Are there other quaternion division rings which are $2$-dimensional over $\ell$, and whose center is $k$, or do they all arise in a similar way?

Lastly, a question which is more related to calculations in the context of conjugacy classes. Suppose, in $\mathcal{Q}$, that we consider any element $h$; what are the elements $g \in \mathcal{Q}$ for which $g^{-1}hg \in \mathbb{C}$? Is there an elegant answer?

Answer 1

If $[\overline{k}:k]=2$ then by Artin Scheirer $\overline{k}=k[i]$.

$Q(k) = \{ a+bi+cj+dij, (a,b,c,d)\in k^4, i^2=j^2=-1, iji^{-1}=-j\}$ is a $4$-dimensional $k$-algebra and a division ring.

And any $4$-dimensional $k$-algebra which is a division ring is isomorphic to $Q(k)$.

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Let's prove the last point. Given $D$ such a division ring. Take any $a\in D,a\not \in k$ so $k[a]$ is a subring of $D$, and an integral domain. Let $f\in k[x]$ be the minimal polynomial of $a$. It is irreducible for $k[a]$ to be an integral domain, so automatically $\deg(f)=2$. Whence $k[a]\cong \overline{k}$ contains a root of $x^2+1$ that we'll call $i$, so $k[a]=k[i]$.

Let $T:D\to D, T(b)=i b i^{-1}$.

$T^2 = Id$ and $T\ne Id$ so there is $c\in D-0$ such that $T(c)=-c$.

Again $k[c]\cong \overline{k}$ so $k[c]=k[j]$ where $j$ is a root of $x^2+1$.

$T$ is an automorphism of $k[c]$, it must be the complex conjugaison, so we got that $1,i,j,ij$ is a $k$-basis of $D$ and $i ji^{-1}=-j$ ie. $D\cong Q(k)$.

Answer 2

$ \newcommand\Q{\mathcal Q} \newcommand\Cl{\mathrm{Cl}} \newcommand\k{\mathbb k} $I will write $\k$ for the field and $k$ for the quaternion element so that these are not confused.

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Regarding question 1:

$\mathcal Q(\ell,\sigma)$ is (isomorphic to) a Clifford algebra over $\k^2$, and it is well known that over a vector space of even dimension the center of a Clifford algebra is $\k$.

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Regarding your last question:

I assume by "$\mathbb C$" you mean the set $\{a + bi \;:\; a, b \in \k\}$.

$\mathcal Q = \mathcal Q(\ell, \sigma)$ is the even subalgebra of the Clifford algebra $\Cl_3(\k)$ over $\k^3$ with quadratic form $q(a,b,c) = a^2 + b^2 + c^2$ for $a,b,c \in \k$. Write $e_1, e_2, e_3$ for the standard basis of $\k^3$. Then $\Cl_3(\k)$ is the algebra generated by sums of products of elements $\k^3$ subject to the relations $v^2 = q(v)$ for every $v\in\k^3$; it follows that $e_1^2 = e_2^2 = e_3^2 = 1$ and that $e_ae_b = -e_be_a$ when $a \ne b$. We can then identify $\Q$ as a subalgebra of $\Cl_3(\k)$ by defining $$ i = e_3e_2,\quad j = e_1e_3,\quad k = e_2e_1 $$ with $1$ as the identity element for both $\Q$ and $\Cl_3(\k)$. This allows us to interpret $i, j, k$ as representing the planes spanned by those vectors. The nonzero elements $\Q^\times$ form the even Lipschitz group of $\Cl_3(\k)$, and this group represents the entire special orthogonal group of $q$ via the action $$ v \mapsto g^{-1}vg,\quad g \in \Q^\times,\quad v \in \k^3. $$ This also extends to the interpretation of the set $\Cl^2_3(\k) = \{ai + bj + ck \;:\; a,b,c \in \k\}$ as a space of planes.

If $h = a + bi + cj + dk$, then necessarily $g^{-1}hg$ fixes the scalar part $a$; so we want to map the imaginary part $h_{im} = bi + cj + dk$ to a multiple of $i$; which is to say $$ g^{-1}h_{im}g = \alpha i. $$ Because $g$ represents an orthogonal transformation the norm of $h_{im}$ is preserved and $$ \alpha^2 = b^2 + c^2 + d^2. $$ If this quantity doesn't have a square root in $\k$ then $g$ cannot take $h_{im}$ to a multiple of $i$. Otherwise, what we are looking for is precisely the rotation which which takes the plan $h_{im}$ to the plane $i$; or equivalently which takes the normal vector $h_{vec} = be_1 + ce_2 + de_3$ to the normal vector $\alpha e_1$. We can find such a $g$ by composing two reflections: (1) one which takes $h_{vec}$ to $\alpha e_1$ and (2) one which fixes $\alpha e_1$. In $\Cl_3(\k)$ reflections through a plane are represented by the vectors $r$ orthogonal to that plane: $-r^{-1}vr$ is the reflection of $v \in \k^3$ through the plane orthogonal to $r$. Since $r^{-1} = r/r^2$ it suffices to construct $g$ as $r_1r_2$ where $r_1$ is a vector performing reflection (1) and $r_2$ a vector performing reflection (2). Reflection (1) can be given by the vector $$ r_1 = h_{vec} - \alpha e_1 = (b-\alpha)e_1 + ce_2 + de_3 $$ since $h_{vec}$ and $\alpha e_1$ have the same magnitude under $q$. Reflection $r_2$ can be through any plane containing $e_1$; all of these are of the form $$ (xe_2 + ye_3)e_1 = xe_2e_1 + ye_3e_1 $$ and we take as orthogonal vector $$ r_2 = xe_3 - ye_2. $$ Thus $g$ must be $$\begin{aligned} g &= r_1r_2 = ((b-\alpha)e_1 + ce_2 + de_3)(xe_3 - ye_2) \\ &= x(b-\alpha)e_1e_3 + xce_2e_3 + xc - y(b-\alpha)e_1e_2 - yc - yde_3e_2 \\ &= (x-y)c - (xc + yd)i + x(b-\alpha)j + y(b-\alpha)k \\ &= (x-y)c - (xc + yd)i + \left(b \pm \sqrt{b^2 + c^2 + d^2}\right)(xj + yk) \end{aligned}$$ for any $x, y \in \k$.

To summarize: when $h = a + bi + cj + dk$,

• If $b^2 + c^2 + d^2$ is not a square in $\k$ then there is no $g$ such that $g^{-1}hg \in \mathbb C$.
• If this quantity is a square then all $g$ such that $g^{-1}hg \in \mathbb C$ are given by $$ g = (x-y)c - (xc + yd)i + \left(b \pm \sqrt{b^2 + c^2 + d^2}\right)(xj + yk) $$ for any $x, y \in \k$ which are not both $0$.