Discussing general quaternion algebras and thier Lie algebras

quaternion_algebras_and_their_lie_algebras.txt

% This is the LaTeX file to the video explanation of quaternion algebras here
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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}
Let
$$
i:=
\sqrt{a}\left(
\begin{array}{cc}
 1 & 0 \\
 0 & -1 \\
\end{array}
\right)
$$
$$j:=
\left(
\begin{array}{cc}
 0 & b/c \\
 c  & 0 \\
\end{array}
\right)
$$
The simplest representation is the one with $c=1$. And $c=\sqrt{b}$ gives a symmetric $j$. 
\subsection{Auxiliary definitions}
Let 
$$k:=i\cdot j=
\sqrt{a}\left(
\begin{array}{cc}
 0 &  b/c \\
 -c  & 0 \\
\end{array}
\right)
$$
and 
$$
e := \tfrac{1}{a}\, i\cdot i =
\left(
\begin{array}{cc}
 1 & 0 \\
 0 & 1 \\
\end{array}
\right)
$$
A general element reads
$$v:=t\,\left(
\begin{array}{cc}
 1 & 0 \\
 0 & 1 \\
\end{array}
\right)+x\,\left(
\begin{array}{cc}
 \sqrt{a} & 0 \\
 0 & -\sqrt{a} \\
\end{array}
\right)+y\,\left(
\begin{array}{cc}
 0 & b/c \\
 c  & 0 \\
\end{array}
\right)
+z\,\left(
\begin{array}{cc}
 0 &  \sqrt{a}\,b/c \\
 -\sqrt{a}\,c  & 0 \\
\end{array}
\right)$$

\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.

\subsection{Immediate results/properties}
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.
$$k\cdot k  = (i\cdot j)\cdot (i\cdot j) = -i\cdot (j\cdot j)\cdot i = -(a\, b)\ e.$$
or 
$$i\cdot k = i\cdot (i\cdot j) = i\cdot (- (j\cdot i))  = -(i\cdot j)\cdot i = -k \cdot i $$
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 $\{-1,1\}$. 

{\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.

\section{Explicit product and commutator}
We arrange a bunch of coefficients as
$$
g^i
:= 
\left(
\begin{array}{cccc}
0 & 1 & 0 & 0 \\
1 & 0 & 0 & 0 \\
0 & 0 & 0 & -b \\
0 & 0 & b & 0 %\\
\end{array}
\right), 
\hspace{.5cm}
g^j
:= 
\left(
\begin{array}{cccc}
0 & 0 & 1 & 0 \\
0 & 0 & 0 & a \\
1 & 0 & 0 & 0 \\
0 & -a & 0 & 0 %\\
\end{array}
\right),
\hspace{.5cm}
g^k
:= 
\left(
\begin{array}{cccc}
0 & 0 & 0 & 1 \\
0 & 0 & 1 & 0 \\
0 & -1 & 0 & 0 \\
1 & 0 & 0 & 0 %\\
\end{array}
\right)
$$
Let
$$v:=t\,e+x\,i+y\,j+z\,k, \ \ \ \ V:=T\,e+X\,i+Y\,j+Z\,k$$
E.g., in the 4-dimensional vector space, we find the two-form of $g^j$ given as
$$\sum_{n,m} v^n\, (g^j)_{nm}\, V^m = (T\,y + t\,Y) + a\,(x\,Z - X\,z)$$
Note that, as is clear from the matrix expression, this has a symmetric and an anti-symmetric part w.r.t. $v\leftrightarrow V$:
$$\sum_{n,m} v^n\, ((g^j)-(g^j)^T)_{nm}\, V^m = 2\,a\,(x\,Z - X\,z)$$
The matrix $g^k$ is a permutation with sign, and we go on to define
$$
g^e
:= g^ig^kg^kg^kg^j=
\left(
\begin{array}{cccc}
1 & 0 & 0 & 0 \\
0 & a & 0 & 0 \\
0 & 0 & b & 0 \\
0 & 0 & 0 & -ab %\\
\end{array}
\right),
\hspace{.5cm}
g^kg^eg^k =
\left(
\begin{array}{cccc}
-ab & 0 & 0 & 0 \\
0 & -b & 0 & 0 \\
0 & 0 & -a & 0 \\
0 & 0 & 0 & 1%\\
\end{array}
\right)
$$
After some computation with the 2x2 matrix representations, we find the product evaluates as
$$v\cdot V = \sum_{\alpha\in\{e,i,j,k\}} (\sum_{n,m} g_{nm}^\alpha\, v^n\, V^m)\, \alpha$$
We are interested in the commutator
$$[v,V]:=v\cdot V - V\cdot v,$$
which is now easy to guess.

First, note that the adjoint
$${\mathrm {ad}}_v:=V\mapsto [v,V]$$
is a representation (the map $v\mapsto {\mathrm {ad}}_v$ is a homomorphism, the law turning out to be identical to the Jacobi identity.)\\
Define
$$[v]_\times :=
\left(
\begin{array}{cccc}
0 & 0 & 0 & 0 \\
0 & 0 & -z & y \\
0 & z & 0 & -x \\
0 & -y & x & 0%\\
\end{array}
\right)
$$
which has matrix action
$$[v]_\times V := 0\, e + (y\, Z - z\, Y)\, i + (z\, X - x\, Z)\, j + (x\, Y - y\, X)\, k$$
Observe that this is defined to not involve $t$ and that it maps $T\,e$ to zero.\\
We find
$${\mathrm {ad}}_v = 2\,(g^kg^eg^k)\, [v]_\times,$$
$$[v,V] = 2\,(g^kg^eg^k)\, [v]_\times V$$
so that the general commutator of the quaternion algebra is evaluates in terms of the cross product of the imaginary part.
Also note that for $a=b=-1$, this reduces to just ${\mathrm {ad}}_v = 2\,[v]_\times$. 

We previoulsy saw that the vector spaces spanned by the two quaternion algebra generators $i, j$ and their multiplication is necessarily 4-dimensional. 
The matrices ${\mathrm {gl}}(n)$ make for a Lie-algebra for any $n$, so looking at the linear representations of the quaternion algebra gives us a Lie-algebra.
We now see that we can throw out the commutative dimension/element $e$ and are still left with a nice 3-dimensional Lie-algebra.

Note that $v\mapsto {\mathrm {ad}}_v$ is a Lie-algebra homomorphism (the homomorphism law turns out to be just the Jacobi identity, which is fulfilled by the domain).
As such, we get a 3-dimensional representation of this 3-dimensional Lie-algebra (see isomorphic Lie-algebras $\mathfrak{su}(2)$ and ${\mathfrak{so}}(3,{\mathbb R})$).

\section{Subalgebras of the quaternion algebra}
Assume now that $\sqrt{a}\in\mathbb F$. \\
Let
$$
h:=\tfrac{1}{\sqrt{a}}\, i
= 
\left(
\begin{array}{cc}
 1 & 0 \\
 0 & -1 \\
\end{array}
\right)
$$
$$
l:=\tfrac{1}{\sqrt{a}}\, k
=
\left(
\begin{array}{cc}
 0 &  b/c \\
 -c  & 0 \\
\end{array}
\right)
$$
and

$e_{11} = \tfrac{1}{2}(e + h)$

$e_{12} = \tfrac{1}{2\,b/c}(j + l)$

$e_{22} =\tfrac{1}{2} (e - h)$

$e_{21} =\tfrac{1}{2\,c} (j - l)$

Now
$$[e_{11}, e_{12}] = e_{12},$$
so we find a 2-dimensional Lie-algebra.
This, up to isomorphism, is, in fact, the only 2-dimensional Lie algebra.

We found this as sub-algebra of the quaternion algebras when $\sqrt{a}\in\mathbb F$, 
but this Lie-algebra can also be defined as a standalone Lie algebra.

Consider now
$$[e_{12}, e_{21}] = h$$
$$[h, e_{12}] = +2\,e_{12}$$
$$[h, e_{21}] = -2\, e_{21}$$
We also found this using $e$ and indeed this is, e.g. as Lie-algera over $\mathbb R$, not isomorphic to the
above ``cross product Lie algebra''. 
(This can by shown by showing that the adjoint representation of the cross product algebra is not diagonalizable over $\mathbb R$.)