Quaternion algebras via their 2x2Mat(F) representations

Quaternion_algebras.tex

% This is the LaTeX file to the video explanation of quaternion algebras here
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% https://youtu.be/tkS_6xY132g

Let $a,b,c$ be non-zero elements in a field $\mathbb F$ with characteristic not $2$. \\
In this video we're discussing an unital algebras in $\mathrm{Mat}_{2\times 2}$ for some field.\\
More concretely, they are representations of (abstract) quaternion algebras are denoted by $(a,b\, |\, F)$.

\subsection{Generators}
$$
i:=
\sqrt{a}\left(
\begin{array}{cc}
 1 & 0 \\
 0 & -1 \\
\end{array}
\right)
$$
$$j:=
\left(
\begin{array}{cc}
 0 & b/c \\
 c  & 0 \\
\end{array}
\right)
$$
The simplest representation is $c=1$ and $c=\sqrt{b}$ gives a symmetric $j$. 
In any case, this $c$ factors out of the determinant and is thus largely invisible. \\

The above generate the algebras, as we will now see.

\subsection{Auxiliary definitions}

Let 
$$
e :=
\left(
\begin{array}{cc}
 1 & 0 \\
 0 & 1 \\
\end{array}
\right)
$$
and 
$$k:=i\cdot j=
\sqrt{a}\left(
\begin{array}{cc}
 0 &  b/c \\
 -c  & 0 \\
\end{array}
\right)
$$

\subsection{Defining properties}

The elements $i\cdot i$ as well as $j\cdot j$ lies in the center,
$$
i\cdot i = a \  e
$$
$$
j\cdot j = b \  e
$$
Finally $i$ and $j$ are anti-symmetric:
$$
j \cdot i = -i \cdot j
$$
The above three relations are the defining property of an abstract quaternion algebra.

It follows that the algebra is always $4$-dimensional and all products are determined by the $3$ equations above. 
E.g. 
$$i\cdot k = i\cdot (i\cdot j) = (i\cdot i)\cdot j = a\ e\cdot j  = a\ j $$
In fact, centrality and anti-commutativity follows for three generators, see e.g.
$$i\cdot k = i\cdot (i\cdot j) = i\cdot (- (j\cdot i))  = -(i\cdot j)\cdot i = -k \cdot i $$
or 
$$k\cdot k  = (i\cdot j)\cdot (i\cdot j) = -i\cdot (j\cdot j)\cdot i = -(a\, b)\ e.$$

Looking now at the last equation, note that all squares are central/of degree $2$. 
As vector spaces, we can always work with a rescaled basis and so one must effectively only think of $a, b$ in $\{-1,1\}$. Powers alternating in sign will lead to more compact objects (also technically speaking). 

{\bf{Special cases}}: In light of our calculation for $k\cdot k$, the case $a=b=-1$ has $-(a\, b)=-1$ and is thus the one which makes the the non-central elements behave the same.

\subsection{Span}
With $t,x,y,z\in{\mathbb F}$, let
$$V := x\ i + y\ j + z\ k$$
and consider a general element
$$v := t\ e + V = 
\left(
\begin{array}{cc}
 t+\sqrt{a}\, x & (y + \sqrt{a}\, z)\, b/c \\
 (y +(-\sqrt{a})\, z)\, c  & t+(-\sqrt{a})\, x \\
\end{array}
\right)
 = (t\ e + x\ i)\cdot e + (y\ e + z\ i)\cdot j
$$
The map from $(t,x,y,z)\mapsto v$ canonically and injectively embeds into $\mathrm{Mat}_{2\times 2}(\mathbb{F}(\sqrt{a}))$. 

This map even spans $\mathrm{Mat}_{2\times 2}(\mathbb{F})$ whenever $\sqrt{a}\in \mathbb{F}$, because then e.g. $(y,z)\mapsto y +(-\sqrt{a})\, z$ has $\mathbb{F}$ in its kernel and we can surject on all four elements.
And this is always true if (but not only if) $\mathbb{F}$ is algebraically closed.

The top rows are two elements of $\mathbb{F}(\sqrt{a})$, as such it's a two dimensional module space over the algebraic structure defined by $t\ e + x\ i$. 

\subsection{Functions}
As the quaternion algebra is so low dimensional, most of the important functions are largely spared of mixing between coordinates and the end up being essentially functions of the determinant. 

The determinant (a.k.a., here, reduced norm, quadratic form) for general $v$ itself is particularly simple and, as noted, $c$ even disappears. It is also almost symmetrical in $a$ and $b$. 
$$
{\det}(v) = t^2 + (-a)\, x^2 + (-b)\, y^2 + (-a)\, (-b)\, z^2 = (t,x,y,z)\cdot \mathrm{diag}(1,-a, -b,a\,b)\cdot (t,x,y,z)^T
$$
{\bf{Special cases}}: The formula tells us that the algebra has the best division properties for $a,b$ negative.

For $v=t\, e + V$, we may define
$${\bar v} := t\ e - V$$
and find
$$v^{-1} = \tfrac{1}{{\det}(v)} \, {\bar v}.$$
and also leads to a nice central product
$${\bar v}\cdot v = t^2\, e - V^2 = \det(v)\ e.$$
The normal square, however, ends up having mixing terms of $t$ with each of the other coordinates. One may write (see also Cayley-Hamilton theorem) 
$$
v\cdot v = - {\det}(V)\ e + t\, (t \, e + 2 V)
$$
This still makes for easy computations of power series. With the series

$\cosh(X) := \sum_{n\in{\mathbb N}} \frac{1}{(2n)!}X^{2n},\ \ \  \sinh=\cosh',\ \ \   \exp:=\cosh+\sinh$

and $r_V := \sqrt{-{\det}(V)}$ and $n_V := \tfrac{1}{r_V}V$ we get (after some calculation)
$$
{\mathrm e}^{-t} \exp(v) = \exp(V) = \cosh(r_V) \, e - \sinh(r_V)\, n_V
$$
and which has
$$\det(\exp(V)) = 1,$$
where you may want to use $\cosh(X)^2-\sinh(X)^2=1$.\\

So that $\exp$ of any $v$ with $t=0$ is actually in ${\mathrm{SL}}_{2\times 2}({\mathbb F(\sqrt{a})})$.

$1\longrightarrow {\mathbb F}^\times\longrightarrow (a,b\, |\, F)^\times \longrightarrow \mathrm{SO}(Q) \longrightarrow 1$

where $\mathrm{SO}(Q)$ is the group of oriented isometries of the quadratic form given with the algebra.

\subsubsection{Product preserving actions}
Let $g, h$ two matrices with $h$ invertible and consider
$$v \mapsto v'=g \cdot v \cdot h^{-1}$$
which is linear and so has a representation in $T_{g,h}\in\mathrm{Mat}_{4\times 4}(\mathbb{F}(\sqrt{a}))$, and which has
$$\det(v')=\frac{\det(g)}{\det(h)}\cdot \det(v).$$ 
Whenever $g$ and $h$ have the same determinant, this map leaves the determinant of the argument $v$ invariant.
In our context this can be read as an inner product being invariant.

Note that a $\frac{\det(g)}{\det(h)}$ has a nontrivial kernel. In particular, since we cal always consider an overall sign flip, we see that so this map is not injective for any such pair.

\section{Special cases}
For $a=b=-1$, we get

Then $\mathrm{diag}(1,-a, -b,a\,b)=\mathrm{diag}(1,1,1,1)$ 

and the determinant thus represents an Euclidean norm.

\subsection{The Hamiltonian quaternions $\mathbb H$}
Set ${\mathbb F}={\mathbb R}$, the elements

$v := t\ e + V = (t\ e + x\ i)\cdot e + (y\ e + z\ i)\cdot j$

can be viewed as complex numbers with another complex coordinate $y+z\sqrt{-1}$ attached to each imaginary coordinate on $\sqrt{-1}x$. 

None of the directions is particularly special: For any point $(x,y,z)$ that lies on $S^2$, we have $V^2=-e$ and thus functions as imaginary unit spanning an imaginary dimension. I.e. $\mathbb H$ is a space of one real dimension (the center) and a sphere $S^2$ full of imaginary units which don't cummute with each other. The imaginary dimensions (the elements obtained by moving radially along any of the complex units) naturally form the ``$\mathbb R^3$ imaginary dimension''. The general matrix product of elements in that space is given as 

$v\cdot w = \langle v\vert w\rangle \  e + v\times w$

where $\langle v\vert w\rangle$ is the inner product and $v\times w$ is the matrix corresponding to the imaginary elements with components computed from the $\mathbb R^3$ cross product of the vectors $v$ and $w$.

\subsubsection{Product preserving actions}
Here $\exp$ of any $v$ with $t=0$ is actually in the standard representation of $SU(2)$.

Let $n\neq 2$, then

$1\longrightarrow {\mathbb Z}/2{\mathbb Z}\longrightarrow \mathrm{Spin}(n) \longrightarrow \mathrm{SO}(n) \longrightarrow 1$

where, $\mathrm{Spin}(n)$ is the dimension $\tfrac{n(n-1)}{2}$ double cover of the rotation group $\mathrm{SO}(n)$.

In particular $\mathrm{Spin}(3)=\mathrm{SU}(2)$ and $\mathrm{Spin}(4)=\mathrm{SU}(2)\times \mathrm{SU}(2)$ and those can both be realized with the action defined further above.