Thinking about $\mathbb{C}$:
One can view the multiplicative group of complex numbers to be the matrices in $\mathrm{GL}(2,\mathbb{R})$ such that commutes with the matrix $I=\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$ a quick calculation shows that those are the matrix of the form $\begin{pmatrix} a & b \\ -b & a \end{pmatrix}$.
One can also view the quaternion's multiplicative group as the matrices in $\mathrm{GL}(2,\mathbb{C})$ that satisfies $IX=\bar{X}I$ the same calculation as above shows that those are the matrices of the form $\begin{pmatrix} a & b \\ -\bar{b} & \bar{a} \end{pmatrix}$.
How can i find a matrix that indentifies by a law of the form above the multiplicative group of $\mathbb{H}$ in $\mathrm{GL}(4;\mathbb{R})$?
My attempts:
I've tryed to use the fact that in the matrix representation of complex numbers the conjugation is the transpose operation together with the properties of block matrix under transposition and substituing in the complex matrices the 2×2 blocks that corresponds to the complexes entries, but i didn't succeded.
If someone wants i can give more datails of my attempt. The candidate matrix was a the real-blocks matrix \begin{pmatrix} 0 & -I \\ I & 0 \end{pmatrix} but i am not sure it is enought.
Any hint or suggestion please?
Best Answer
Note that your representation of $\Bbb C$ in $\mathcal M_2(\Bbb R)$ yields a corresponding representation of $GL(\Bbb C,2)$ in $GL(\Bbb R,4)$. In particular, we can take the matrix $$ \pmatrix{a_{11} + b_{11}i & a_{12} + b_{12}i\\ a_{21} + b_{21}i & a_{22} + b_{22}i} $$ and replace each complex number with the corresponding element in $M_{2}(\Bbb R)$ to get the block-matrix $$ \left( \begin{array}{cc|cc} a_{11} & -b_{11} & a_{12} & -b_{12}\\ b_{11} & a_{11} & b_{12} & a_{12}\\ \hline a_{21} & -b_{21} & a_{22} & -b_{22}\\ b_{21} & a_{21} & b_{22} & a_{22} \end{array} \right). $$ Now, we can apply this change to your representation of $\Bbb H$ to go from a subset of $GL(\Bbb C,2)$ to a subset of $GL(\Bbb R,4)$. In other words, we get the map $$ a_{1} + a_{2}i + b_{1}j + b_{2}k \to \pmatrix{a_{1} + b_{1}i & a_{2} + b_{2}i\\ -a_{2} + b_{2}i & a_{1} - b_{1}i} \to\\ \left( \begin{array}{cc|cc} a_{1} & -b_{1} & a_{2} & -b_{2}\\ b_{1} & a_{1} & b_{2} & a_{2}\\ \hline -a_{2} & -b_{2} & a_{1} & b_{1}\\ b_{2} & -a_{2} & -b_{1} & a_{1} \end{array} \right). $$
If we define the matrix $I \in GL(\Bbb R,4)$ to be $$ I = \pmatrix{0&0&1&0\\0&0&0&1\\-1&0&0&0\\0&-1&0&0}. $$ The "quaternions" in $GL(\Bbb R,4)$ are each matrices that satisfy $I X = X^{\Gamma_1} I$, where $X^{\Gamma_1}$ is a partial transpose of $X$. In particular, $$ \pmatrix{A&B\\C&D}^{\Gamma_1} = \pmatrix{A^T & B^T\\C^T & D^T} $$
Writing out the equation in terms of the entries of $X$ yields $$ IX = \pmatrix{x_{31} & x_{32} & x_{33} & x_{34}\\ x_{41} & x_{42} & x_{43} & x_{44}\\ -x_{11} & -x_{12} & -x_{13} & -x_{14}\\ -x_{21} & -x_{22} & -x_{23} & -x_{24}} =\\ \pmatrix{x_{11} & x_{21} & x_{13} & x_{23}\\ x_{12} & x_{22} & x_{14} & x_{24}\\ x_{31} & x_{41} & x_{33} & x_{43}\\ x_{32} & x_{42} & x_{34} & x_{44}} I = \pmatrix{-x_{13} & -x_{23} & x_{11} & x_{21}\\ -x_{14} & - x_{24} & x_{12} & x_{22}\\ -x_{33} & -x_{43} & x_{31} & x_{41}\\ -x_{34} & -x_{44} & x_{32} & x_{42}} = X^\Gamma I. $$ The $a_1$ elements give us the relations $$ x_{13} = -x_{31}, \quad x_{42} = -x_{42}, $$ the $a_2$ elements give $$ x_{11} = x_{33}, \quad x_{22} = x_{44}, $$ the $b_1$ elements give $$ x_{14} = -x_{41}, \quad x_{23} = -x_{32}, $$ and the $b_2$ elements give $$ x_{21} = x_{34}, \quad x_{43} = x_{12}. $$ From there, we could use the additional fact that the quaternions satisfy $JX = XJ^{\Gamma_2}$, where $$ \pmatrix{A & B\\C & D}^{\Gamma_2} = \pmatrix{A & C\\B & D} $$ and $$ J = \pmatrix{0&-1&0&0\\1&0&0&0\\0&0&0&-1\\0&0&1&0}. $$