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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 2x2 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?

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    $\begingroup$ Rather than thinking of "the set of matrices that commute with I", it is easier to think in terms of a ring homomorphism from $\phi:\Bbb C \to M_{2 \times 2}(\Bbb R)$. In particular, it must holds that $\phi(\operatorname{id}_{\Bbb C}) = \operatorname{id}_{M_{2\times 2}}$, and if $\phi(I) = I$, then it follows that $$ \phi(a + bi) = \phi(a \cdot 1) + \phi(b \cdot i) = a \cdot \phi(1) + b \cdot \phi(i) = a \operatorname{id}_{M_{2\times 2}} + b I. $$ $\endgroup$ Dec 21 '20 at 23:09
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    $\begingroup$ Because $\phi$ is injective, it defines an isomorphism between $\Bbb C$ and its image. It follows that the multiplicative group of units of $\Bbb C$ is isomorphic to the multiplicative group of units of $\phi(\Bbb C)$. The quaternions can be attained as a subring of the $2 \times 2$ complex matrices in a similar way after noting that the multiples of the Pauli matrices $i\sigma_x,i\sigma_y,i\sigma_z$ satisfy the relations among $i,j,k$ of the quaternions. $\endgroup$ Dec 21 '20 at 23:12
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    $\begingroup$ Note that looking for the matrices that commute with $I$ is problematic if we are looking for a representation of the non-zero complex numbers in a larger matrix space such as $GL(4,\Bbb C)$. $\endgroup$ Dec 21 '20 at 23:18
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    $\begingroup$ I think i understand what you said about Pauli matrices, but i don't see if it helps, can you add more details please? I was looking for a matrix in $Gl(4;\mathbb{R})$. $\endgroup$
    – yo yo
    Dec 21 '20 at 23:48
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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 $$ J = \pmatrix{0&-1&0&0\\1&0&0&0\\0&0&0&-1\\0&0&1&0}. $$

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    $\begingroup$ It is exactly what i have done, it is in the description of the answer, but i want to find a matrix that do it naturally, so that the "commutation" condition force every blocks to be of that form. $\endgroup$
    – yo yo
    Dec 22 '20 at 0:13
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    $\begingroup$ @yoyo It wasn't clear what you meant by "i didn't succeed" $\endgroup$ Dec 22 '20 at 0:18
  • $\begingroup$ @yoyo I've added something, maybe that's what you're looking for $\endgroup$ Dec 22 '20 at 0:28
  • $\begingroup$ Yes, it is I will check, i remember i tryed that matrix also but i was unable to show that if $X$ satisfies the condition it is formed by blocks of "complex"( from the system of equation that comes out) . If for you a straightforward calculation works i will try with more decision. $\endgroup$
    – yo yo
    Dec 22 '20 at 0:42
  • $\begingroup$ @yoyo I was wrong. The characterization that you used for the $GL(2,\Bbb C)$ version does not work in $GL(4,\Bbb R)$. See my edit. $\endgroup$ Dec 22 '20 at 1:07

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