Quaternionic general linear group is open Is there an elegant proof of the following fact: "The quaternionic general linear group $GL(n, \mathbb{H})$ is open in $M_n(\mathbb{H})$", where $M_n(\mathbb{H})$ is the set of all $n \times n$ square matrices with coefficients in the quaternions and the general linear group is the subset of invertible matrices.
Since the usual determinant does not exist for quaternionic matrices, the usual proof that works for the reals or the complex numbers does not work here.
Thanks in advance
 A: I suppose that by $\mathbb{H}$ you mean real quaternions, i.e. quaternions $q=x_{0}+x_{1}\mathbf{i}+x_{2}\mathbf{j}+x_{3}\mathbf{k}$ with coefficients $x_{i}$ real.
You can make the idea of the usal proof work, i.e. you want to show that $GL(n,\mathbb{H})=\phi^{-1}(\mathbb{C}-\{0\})$ for a suitable continuous map $\phi:M_n(\mathbb{H})\rightarrow \mathbb{C}$.
The steps below will lead you to $\phi$, which will turn out to be related to the determinant of a square matrix with complex entries:


*

*Show that every quaternion $q$ can be uniquely expressed as $q=a_1 + a_2 \mathbf{j}$, where $a_1, a_2\in \mathbb{C}$.

*Show that a matrix $A\in M_n(\mathbb{H})$ can be uniquely expressed as $A=A_{1}+A_{2}\mathbf{j}$, where $A_{1},A_{2}\in M_{n}(\mathbb{C})$.

*If $A\in M_{n}(\mathbb{H})$ let $\overline{A}$ denote its conjugate (recall that the conjugate of a real quaternion $q=q_{0}+q_{1}\mathbf{i}+q_{2}\mathbf{j}+q_{3}\mathbf{k}$ is $\bar{q}=q_{0}-q_{1}\mathbf{i}-q_{2}\mathbf{j}-q_{3}\mathbf{k}$).
Show that the map $\varphi:M_n(\mathbb{H})\rightarrow M_{2n}(\mathbb{C})$ defined by 
$$A=A_{1}+A_{2}\mathbf{j}\longmapsto\chi_{A}:=\begin{pmatrix}
  A_{1} & A_{2}\\
  -\overline{A_{2}} & \overline{A_{1}}
 \end{pmatrix}.
$$
is continuous.

*Show that if $A,B\in M_{n}(\mathbb{H})$ and $AB=I$, then $BA=I$.

*Show that if $A,B\in M_{n}(\mathbb{H})$ then $\chi_{AB}=\chi_{A}\chi_{B}$. Thus $\chi_{A^{-1}}=\chi_{A}^{-1}$.

*Show that $A\in GL(n,\mathbb{H})$ iff $\chi_{A}\in GL(2n,\mathbb{C})$.

*Let $\phi=\varphi\circ \det$. Show that $A\in GL(n,\mathbb{H})$ iff $\phi\neq 0$, and that $GL(n,\mathbb{H})=\phi^{-1}(\mathbb{C}-\{0\})$.
Have fun!
