Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
Have you heard about non-commutative determinant? Or Dieudonnè determinant? – Sigur Jul 24 '13 at 22:18
You can use $(I - M)^{-1} = \sum_0^\infty M^k$ for $\lVert M \rVert < 1$, and then translate the open neighbourhood of $I$ to $A \in GL_n(\mathbb{H})$ by left (or right) translation. – Daniel Fischer Jul 24 '13 at 22:20

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:

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

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

  3. 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.

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

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

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

  7. 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!

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.